This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
p holds. If a < p, then p iv. If a = p, then p = v. If a > p, then p 2 v. Thus at least one of the three relations must hold. Corollary 3. For an arbitrary infinite cardinal number p, ap
= p.
Proof. Take a set M with /MI = p. By use of the theorem we well-order M denoting its ordinal number with T. Namely,
If for an ordinal number a, { p 1 p < a} has no last number, then we call a a limit number. Then we can easily verify that M is decomposed as
M
=U
{Ma[a is a limit number with 0 s a < T},
where
M,
= {x,, Xa+l,x ,,
. . .>.
20
INTRODUCTION
[I.4
Since Mun Mu. = O for different ordinal numbers a, a’, A4 is decomposed into disjoint countable subsets. Therefore p = av, where v is a cardinal number, and hence u p = u2v = a v = p. (See Exercise I.) Let B be a given set; then we consider a property P on subsets of B. If P is satisfied by a subset B‘ of B if and only if it is satisfied by every finite subset of B‘, then we call P a finite property on subsets of B. Now, we suppose a is an element of a partially ordered set A. If a satisfies a = a’ for every element a’ of A such that a S a’ ( a 3 a’), then we call a a maximal (minimal) element of A. In other words, a maximal element is an element a such that there is no element a‘ with a ’ > a. As the reader may notice, a partially ordered set can have more than one maximal or minimal elements.
Example 1.11. Let A be a partially ordered set; then ‘totally-ordered’ is a finite property on subsets of A, because a subset A’ of A is totallyordered if and only if every two point subset of A’ is totally-ordered. Let us consider the partially ordered set 2M defined in Example 1.8. We say that a subset A of 2M satisfies P if and only if every finite number of elements M,, . . . ,M~ of A satisfies nt,M , # O. Then P is a finite property. 0 is a minimal and also the smallest element of ZM. If we consider the partially ordered set 2M - {O}, then each one point set of M is a minimal element. In the forthcoming discussions we shall often use the foIlowing convenient lemmas which are equivalent with Zermelo’s axiom of choice. Zorn’s lemma. Let A be a non-empty partially ordered set. If every totally-ordered subset of A has a supremum, then A has a maximal element.
Hausdorff’s lemma. Let P be a finite property on subsets of a set A. Then there exists a maximal set among the subsets of A which satisfy P. Theorem 1.4. Zermelo’s axiom of choice, Zom’s lemma and Hausdofls lemma are equivalent. Proof. Zermelo’s axiom j Zom’s lemma. To begin with, we define a mapping f of A, a given partially ordered set satisfying the condition of Zorn’s lemma, into itself as follows: To each element a of A we assign a subset A, defined by
1.41
ZERMELO’S THEOREM AND ZORN’S LEMMA
A,
=
{ {{”b I b
E A,
21
if a is a maximal element of A , b > a } if a is n o maximal element of A .
Then by use of Zermelo’s axiom of choice, we can define a mapping f such that f(a)EA,,
aEA.
It is obvious that f ( a ) is a mapping of A into itself such that (1) a sf@>, (2) a = f ( a ) if and only if a is a maximal element of A. Since A is non-empty, we choose a fixed element a, of A. Then we can construct a totally-ordered subset A, of A such that (9 a, E A,, (ii) f(Ao)C A,, and (iii) sup A, E A,. To construct A,, we consider subsets A’ of A satisfying (i), (ii) and (iii)’ for every non-empty totally-ordered subset B of A’, sup B E A ’ . For example, A itself satisfies those conditions. We denote by A, the intersection of all subsets A’ satisfying (i), (ii) and (iii)’. Then, observe the following fact that will be needed later. Since the set { a 1 a E A, a z= a,} obviously satisfies (i), (ii) and (iii)’,
a 3 a, for every a E A , .
(3)
It is also clear that A, satisfies (i), (ii) and (iii)’. Therefore all we have to prove is that A, is totally-ordered. For this purpose, we put Ah = {a’ I a‘ E A,; f(u) s a’ or a ’ s u for every a E A,} .
(4)
Now, let us show that A ; = A,. To do so, it suffices to show that A; satisfies (i), (ii) and (iii)’, because A, is the smallest set satisfying the three conditions while A ; C A,. We shall first show that for a E A, and a’ E Ah (5) either a 6 a’ or !(a’) s a. Denote by A: the set of all elements a E A, satisfying (5) for every a ’ € Ah. Then we can assert that A: satisfies (i), (ii) and (iii)’. It follows from (3) that a, < a’ for every a’ E A;, i.e. a, E A:, proving (i) for A:. To
22
INTRODUCTION
tI.4
deal with (ii), we suppose a E As and a’ E Ah. Then by the definition of A:, either a < a’ or a = a’ or f(a‘)S a. If a < a‘, then by (4) f(a) S a’. If a = a’, then f(a)=f(a’). If f ( a ’ ) S a, then from (1) we get f ( a ’ ) s f ( a ) . Therefore f(a) satisfies the condition of a in (9,and hence f ( a )E A:, i.e. (ii) is satisfied by A:. As for (iii)’, let B be a non-empty totally-ordered subset of A: and a’ an element of Ah. Note that by t h e hypothesis of Zorn’s lemma s u p B exists and belongs to A, (recall that A, satisfies (iii)’). If a s a’ for every element a of B, then sup B S a’. If f(a’) a for some a E B, then f(a’)S sup B (see (5)). Thus sup B satisfies the condition of a in (5) for every a’€ Ah, and hence sup B € A:, i.e. (iii)’ is also satisfied by As. Since A: C A, and A, is the smallest set satisfying (i), (ii) and (iii)’, we have A: = A,. Thus (5) is established for every a E A,. Now, let us turn to the proof that Ah satisfies (i), (ii) and (iii)’. It clearly follows from (3) and (4) that (i) is satisfied by Ah. To see (ii), let a’ E Ah, a E A,. Then by (5) either a < a’ or a = a’ or f(a’) S a. In the first case we obtain from (4) that f(a)S a’, which combined with (1) implies that f(a)s f ( a ’ ) .In the second case we obtain f(a)= f(a’).Thus f(a’)satisfies either f(a) S f ( a ’ )or f(a’)S a for each a E A,, and hence we conclude, in view of (4),f(a’)E Ah, proving (ii) for Ah. Finally, referring to (iii)’, we suppose B is a totally-ordered subset of Ah and a is a given element of A,. If a 3 a’ for every a ’ € B, then a 2 sup B. If a # a’ for some a’ E B, then it follows from (4) that f(a)S a’ S sup B. Therefore in view of (4), we obtain s u p B E A 6 . Thus Ah is a subset of A, satisfying (i), (ii) and (iii)’, and hence Ah = A,. Therefore from the definition (4)of Ah as well as (l), we can conclude that any two elements of A, are comparable, i.e. A, is totally-ordered. Hence A, is a non-empty totally-ordered set satisfying (ii), (iii)’ and accordingly (iii), as well. Now we put a, = sup A,; then by (iii), a1E A,, and hence by (ii), f(a,)E A,. Since a, is the supremum of A, and a, sf(al) (by (l)),we obtain a, = f(al). Thus, in view of (2), we have reached the conclusion that a, is a maximal element of A,, which proves Zorn’s lemma. Zorn’s l e m m a 3 Hausdofls lemma. We denote by d the collection of the subsets of A with the finite property P. We define order between two elements of d by the ordinary inclusion relation, regarding d as a partially ordered set. Let d‘be a totally-ordered subcollection of d ; then we easily see that A ’ = U { BI B E d’}
has P, because P is a finite property. Therefore A’ = sup d’in d.Thus we
IS]
TOPOLOGY OF EUCLIDEAN PLANE
23
can apply Zorn’s lemma on d to get a maximal element A,, which is the maximal set having the property P. Hausdorffs lemma j Zermelo’s axiom. Let d be a collection of non-empty sets. We denote by 2 the set of the pairs (A, a ) where A is an element of d and a is an element of A. A subset 2’of 2 is said to have property P if and only if every two elements of 2’ have distinct first coordinates, i.e., for every (A, a ) E Y and (A’, a ’ )E Y , A = A’ implies a = a’. Then P is clearly a finite property. Therefore by use of Hausdorff’s lemma we can find a maximal subset Lf,, of Lf having the property P. Then every element A of d appears in one and only one element of z0. For, if we assume A E d does not appear in any element of .=Yo,then we take a E A and construct
2 Soand has P, contradicting the fact that z0is a maximal set Then z1 having the property. Therefore for every A E d,an element a E A is determined by the condition (A,a ) E 2Z0. Now, we can define a mapping f over d by
Since a E A, Zermelo’s axiom is established.
5. Topology of Euclidean plane In the present section we shall deal with point sets in the Euclidean plane E 2 to help the reader to understand the concept of topological space. As a matter of fact, the theory of point sets in Euclidean spaces gives the simplest example of general topology, and historically the investigation of the former theory by G. Cantor in the late 19th century led to the establishment of the concept of topological space by F. Hausdorff, M. FrCchet, C. Kuratowski and the other mathematicians in the early 20th century, while the foundation of further development of general topology was established by A. Tychonoff, P. Urysohn, P. Alexandroff’ and others. ‘The paper [3] of P. Nexandroff is highly recommended to the reader as a historical survey of the modern development in general topology. See his paper [ S ] too, for recent developments.
24
INTRODUCTION
[IS
As is well known, the concept of convergence of a point sequence is very significant in E’, especially in analysis of E 2 . We may say it is fundamental in the study of E 2 . This concept is closely related with such other concepts as neighborhood, closure, open set, etc. as seen in the following. We denote by p ( p , 4)the distance between two points p, q of E 2 and by S , ( p ) for a positive E the domain in the circle with center p and radius E , i.e.
S , ( p ) is often called the &-neighborhood of p. Moreover, for a point p and a point set A of E 2 (i.e. a subset of E’), we define the distance between p and A by
Let p be a point of E 2 and U a point set of E’, which contains p. If U contains an &-neighborhood of p for some E > O , then we call U a neighborhood (or nbd’ for brevity) of p. Example 1.12. We consider an x - y coordinate system in E 2 with the origin po = (0,O). Then for every E > 0 and 6 > 0, the following sets are all examples of nbds of po: S,(po), U = {(x, y ) 1 x’/E’ + y 2 / S 2< l}, V = {(x, y ) 11x1 < E and I y I < S}, W = {(x, y ) I ( x - & / 2 ) ’ + y 2 S E’}. On the other hand P = {(x, y ) I ( x - E ) ’ + ( y - 8 ) ’ s E ’ + S’} and Q= {(x, 0 ) I 1x1 < E } are not nbds of po although they contain po. We often adopt notations like U ( p ) , V ( p ) , etc. to denote nbds of p. We can easily see that nbds have the following properties. A) (i) Let U ( p ) and V ( p ) be nbds of a pointp; then U ( p )n V ( p ) is also a nbd of p, (ii) if a set U contains a nbd of p, then U itself is also a nbd of p, (iii) every nbd U ( p ) of a point p contains a subset V containing p such that V is a nbd of every point belonging to V.
‘We pronounce it ‘neighborhood’.
IS]
TOPOLOGY OF EUCLIDEAN PLANE
25
Proof. Since the proofs of (i) and (ii) are easy, they are left to the reader. Referring to (iii), U ( p ) contains an E-nbd, S,(p), which can be easily seen to satisfy the required condition for V. The convergence' of a point sequence is characterized by use of the concept of nbds, and vice versa, as seen in the following assertions whose easy proofs will be left to the reader.
B) Let {p, 1 n = 1,2, . . .} be a point sequence of E 2 . Then it converges to a pointp of E 2 if and only if for every nbd U ( p )of p and for some number n,
pi E U ( p ) for every i 2 n . C ) A set U is a nbd of p if and only if for every point sequence {pn I n = 1,2, . . .} converging to p, we can choose a number n such that
pi E U for every i a n . Let A be a point set of E 2 and p a point of A. If A is a nbd of p, then p is called an inner point of A. If q is an inner point of E 2 - A, then it is called an exterior point of A. If a point p of E 2 is neither an inner point nor an exterior point of A, then it is called a boundary point of A. We call the set of all boundary points of A the boundary of A. We can define open sets and closed sets in terms of the concept of nbd as follows. A point set U of E 2is called an open set if every point p of U is an inner point of U,or we may express this definition as follows: U is an open set if for every point p E U, we can choose a positive E such that S,(p) C U. A point set F of E 2 is called a closed set if and only if E 2 - F is an open set. We obtain the following easily shown but important properties of open sets whose proofs will be left to the reader.
D ) (i) E 2 and 0 are open sets, (ii) the intersection of finitely many open sets is open, (iii) the union of (not necessarily finitely many) open sets is open.
'
As is well known, a point sequence {pn I n = 1,2, . . .} is said to converge to a point p if for every E > 0 there is n such that p ( p , pi) 4 E whenever i 3 n.
26
INTRODUCTION
[IS
By use of de Morgan’s rule, we can deduce from D) the following properties of closed sets.
E) (i) E and 0 are closed sets, (ii) the union of finitely many closed sets is closed, (iii) the intersection of (not necessarily finitely many) closed sets is closed. In view of the condition (iii) of A), we can characterize nbds of a point by use of open sets as follows:
F) A subset U of E is a nbd of p if and only if there is an open subset V such that p E V C U.
Example 1.13. Let us consider the sets cited in Example 1.12. Every point of S,(p,) is easily seen to be an inner point of it, and accordingly S,(p,) is an open set. In the same way, U, V are also open sets, but W, P, Q are not open. The inner points of W are the points which satisfy ( x - s/2)’+ y 2 < s2,while Q has no inner point. The boundary of S,(p,) is the set of the points on the circle, i.e. {(x, y ) I x 2 + y 2 = s2},and the boundaries of U, V, W, P are the closed curves which surround the respective domains, while all points of Q plus (s, 0), (-s, 0) form the boundary of Q. W and P are examples of closed sets. Although S,(p,) itself is not closed, the union of S,(p,) and its boundary forms a closed set. E 2 and 0 are the only subsets of E 2 which are open and closed at the same time. Roughly speaking, closed sets are sets fringed with the boundary, and open sets are sets stripped of the boundary. Note that ‘non-closed’ does not necessarily mean ‘open’ in topology, as in an ordinary conversation; for example, the set Q in Example 1.12 is neither closed nor open. Also note that condition (ii) of D) (E)) fails to be true for infinitely many open (closed) sets. The intersection of the l/n-nbds Sl,,,(p), n = 1, 2 , . . . is the one point set { p } which is closed but not open. In the above we have learned that the concept of nbds is equivalent with that of convergence, in the sense that the former is characterized by the latter, and vice versa. Likewise, the concept of open sets, and also that of closed sets, is equivalent to that of nbds and therefore to convergence, too. Now, let us turn to another important concept, closure of set, which is also closely related to the concept of convergence; we may
1.51
27
TOPOLOGY OF EUCLIDEAN PLANE
even say that they are equivalent. Let A be a subset of E 2 .We denote by A the set of the points p with p ( p , A ) = 0 and call it the closure of A. Example 1.14. Consider the open set S,(po) in Example 1.13; then
where B(S,(po))denotes the boundary of S,(po). As a matter of fact, we can assert that A = A U B(A) for every subset A of E 2 .Thus we may say that the closure of a set is the original set with its boundary added. We should note that the concept of boundary in its exact definition may not always agree with our intuition. For example, consider the set R of all rational points in E 2 , i.e. the points whose x - y coordinates are both rational. Then the boundary of R coincides with the whole of E 2 and accordingly R = E Generally such a set D that satisfies D = E is said to. be dense in E 2 . The set S of all irrational points is also dense in E 2 .
’.
We obtain the following properties of closure:
G ) (i) (ii) (iii) (iv)
= 0,
A 3 A, AUB = A u B, A= = A.
Proof. (i) and (ii) are obvious. In view of (ii), are also obvious. Therefore we shall show A U B C A U B and
z A and A 3
UB 3 A U B
ZCA.
To show the first relation, suppose p is a given point of A U B . Then p ( p , A U B) = 0 which means that there exist points p,, E A U B, n = 1, 2, . . . , such that p ( p , p,,)< l / n . Since p,, E A U B, either A or B contains an infinite subsequence p,,, p n 2 ,. . . of {p,, I n = 1, 2,. . .}. For example, suppose A does so; then we can easily see that p ( p , A ) = 0 which means p E A.Thus we have verified
AUBCAUB. -
verify 2 c A, we suppose p is a given point of A. Then p ( p , A ) = o which means that there exist points p, E A, n = 1, 2 , . . . , such that TO
28
[IS
INTRODUCTIlON
p ( p , p,,) < l/n. Since p,, E A, there exist points q,, E A, n that p(p,, q,,) < l / n . Thus
=
1, 2, . . . , such
for q,, E A, n = 1, 2, . . . . Therefore p(p, A ) = 0, which means p E A,and thus the assertion is proved. The relationship between closure and convergence is described in the following assertions.
H) A = { p I there exists a point sequence {p,, 1 n = 1, 2, . . .} converging to p such that p,, E A, n = 1, 2, . . .}.
I) A point sequence {p,, 1 n = 1,2, . . .} converges to a point p if and only if for every subsequence {p,,,I i = 1, 2, . . .} of {p,, I n = 1, 2 , . . .}, p E {pfli} holds. Proof. We shall prove only I). If {p,,}converges to p, then its subsequence { p f l j }also converges to p . Therefore for every E > O and for some i, p ( p , p,,) < E , which means p ( p , A ) = 0, where A = {P,,~}.Hence p E A. Conversely, if {p,,} does not converge to p, then for some E > 0, there exists an arbitrarily large number n such that p ( p , p,,) 3 E . Therefore we can choose a sequence n1 < n, < * * such that p ( p , p , ) a ~ , i = 1, 2 , . ...
Thus p ( p , A ) a E for A = { p n i } ,and hence p fZ A. We can also establish a direct relationship between closure and nbds as the reader may have already realized. But now, we do not like to be involved in such detailed discussions because we shall handle them in more generality in the following chapters.
Example 1.15. Concepts such as limit point, continuous functions, etc. which are defined in terms of convergence, can be also defined by use of one of the equivalent terminologies, nbd, open set, closed set or closure. As is well known, a point p is called a limit point of a set A if for every E > 0, there are infinitely many points q of A for which p ( p , q ) < E . To
IS]
TOPOLOGY OF EUCLIDEAN PLANE
29
define this terminology by use of closure, for example, we can say that p is a limit point of A if and only if
Now, we define convergence of set sequence which will help the reader to understand the concept of filter in the following chapter. We mean by a set sequence a sequence A,, A,, . . . of non-empty subsets of E 2 such that
The set sequence {A,, 1 n = 1, 2 , . . .} is said to converge to a point p if for every E > 0 there exists an n for which A, C S , ( p ) . We take a point p,, E A,, from each member, A,,, of the set sequence. Then we obtain a point sequence {p,,} which is called a point sequence derived from {A,}. We can easily show that the set sequence {A,,}converges to p if and only if every point sequence derived from {A,} converges to p. Conversely, we consider a given point sequence {p,,}. Putting A, = {p,, p , + , , . . .} we obtain a set sequence {A,,}. Then we call {A,} the set sequence derived from {p,}. A point sequence {p,} converges to p if and only if the set sequence derived from {p,,}converges to p. Thus we can regard the two concepts, convergence of point sequence and convergence of set sequence, as equivalent.
Example 1.16. The concept of set sequence often appears in an introductory part of calculus where, for example, the following proposition plays a significant role: Every set sequence {A,} converges to a point p if limn+* diameter of A,, = 0; if moreover each A, is a closed set, then n;=,A,, = { P I . We should take note that all the discussions (except some examples) in this section are also valid for three-dimensional Euclidean space E 3 and more generally for every metric space X, where we call a set X a metric space if to every two elements p, q, of X , a non-negative real number p ( p , q ) called the distance between p and q is assigned, such that (i) p ( p , q ) = 0 if and only if p = q, (ii) P(P9 4 1= P(4, P 1, (iii) p ( p , q ) p ( p , r ) + p(r, q), for every r E X .
30
INTRODUCTION
Exercise I’ 1. Prove
where A, A,, etc. denote given sets. 2. Prove that the following sets are countable sets: The set of the solutions of all algebraic equations with rational coefficients. The set of disjoint open intervals on the real line (-m,m). The set of the points at which a given monotone increasing (decreasing) real-valued function of one real variable is not continuous.
3. Let A,
p, v
be given cardinal numbers. Then prove:
4. Prove: A totally-ordered set A is well-ordered if anc.. only if A does not contain a sequence {aiI i = 1, 2, . . .} of elements such that a, > u2> a3
> . . ..
5. We consider a decomposition of a given totally-ordered set A such
that A = B U C, B n C = (3, and b < c whenever b E B and c E C. Prove that A is well-ordered if and only if C has the first element for any such decomposition. 6. Prove: For a given partially ordered set A, there is a totally-ordered subset A’ of A such that every upper bound of A ‘ belongs to A’. As a rule the exercises in this book do not contain new topics which result from rather extensive studies or which could make additional sections of the text. Such supplementary topics will rather be found in footnotes, in examples, or in descriptions without proof. The exercises do not contain many special types of questions which require artificial techniques of proof, but consist mostly of questions that are rather easy to prove and are often related to handy small theorems or examples. They are primarily aimed to aid understanding, while at the same time serving as part of the main discussion. Proofs of the propositions which were left to the reader (e.g. 5.B))may be also regarded as exercise problems though usually they are not restated in the exercise section.
I1
EXERCISE
31
7. Let A be a subset of E 2 . Then prove that
A = A U B ( A )= A U A * , where B ( A ) and A* denote the boundary and the set of the limit points of A, respectively. Furthermore, prove that A is a closed set if and only if
A=A.
CHAPTER I1
BASIC CONCEPTS IN TOPOLOGICAL SPACES
At the end of the last section of Chapter I, we noted that the whole discussion there, essentially based on the concept of convergence in E 2 , could be extended t o metric spaces. However, even the existence of distance is no prerequisite for the concept of convergence. In fact, we can discuss convergence in a topological space, which is even more general than a metric space. We may say a topological space is a set endowed with the concept of convergence. From a practical point of view, however, lefining a topological space by means of convergence itself is not the best yay. Rather, we prefer to define a topological space first with the concept If open sets, nbds or closure, which are essentially equivalent with :onvergence. Then we define convergence of the space. We shall find )pen sets, nbds and closure are often more convenient than convergence lot only for defining a topological space but also for making a study If the space. In this chapter we shall begin with the concept of open sets and derive the other concepts from it.
1. Topological space
Definition II.1. Let X be a set and 0 be a collection of subsets of X which satisfies: (i) 0 E 0, X E 0, k (ii) if q. E 0, i = 1 , . . . , k, then n,,, V , E 0, (iii) if U y € O ,y € r , t h e n U { U , ( y E T } E O , where the index set r is not necessarily finite. Then we call every set belonging t o 0 an open set and X a topological space or a T-space.The collection 8 is called the topology of X.'
'
To be precise, we should call the pair (X, 0)a topological space. But for brevity we usually say 'X is a topological space'. The same applies to metric spaces and other spaces with various structures that will appear later in this book.
11.11
TOPOLOGICAL SPACE
33
Definition II.2. Consider two topologies 0 and 0' for a given set X. If 0 C O',then the topology 0 is called weaker than 0' (0' is stronger than 0).
Example II.1. The Euclidean plane E 2 is one of the most popular examples of a topological space. In fact, we have seen in I.5.D) that the collection of the open sets of E 2 satisfies (i), (ii) and (iii) of Definition 11.1.
Generally every metric space X is a topological space with the topology 0 = {U1 U C X, and for every p E U and for some E > 0, S , ( p ) C U}, where S , ( p ) = {q I q E X,p ( p , q ) < E } (0 is called the metric topology of X). Among examples of metric spaces are n-dimensional Euclidean space E n = {(q,. . . , x , ) I xi, i = 1, . . . , n, are real numbers} with distance
and Hilberf space H = {(q, x2, . . .) 1 xi, i = 1,2, . . . , are real numbers, xf < + m} with distance
xyZl
To give an example of a non-metric topological space, we consider the set R, of all ordinal numbers of the countable well-ordered sets. Namely, R , = {a 1 0 s a < W l } , where w1 denotes the smallest non-countable ordinal number. We define a collection 0 of subsets of R, as follows:
0 = { U l U C R , , for every a € U with a > O and for some P < a,(P, a]C U }U (0), where
(P, a ] ={ Y IP < Y
O is a nbd of p, while in R , every set containing (p, a ] for some fl < a is a nbd of a if a # 0, and every set containing 0 is a nbd of 0. A) With respect to Definition 11.3, we denote by % ( p ) the collection of all nbds of p. Then it satisfies: (9 E %(P), (ii) if U E % ( p ) , then p E U, (iii) if U E % ( p ) , V > U, then V E % ( p ) , (iv) if U, V E % ( p ) , then U n V E % ( p ) , (v) if U E % ( p ) , then there exists a set V s u c h thatp E V C U and such that V E 92 ( 4 ) for every point q E V.
x
Proof. The conditions (i), (ii) and (iii) are direct consequences of Definition 11.3. The condition (iv) is derived from the definition combined with the condition (ii) of open sets in Definition 11.1. The condition (v) is directly derived from the assertion that if U E % ( p ) , then there is an open set V such that p E V C U, and this assertion is implied by Definition 11.3 itself.
B) A subset U of a topological space X is an open set if and only if U is a nbd of each point p of U.
Proof. Suppose U is an open set of X and p E U. Then it follows directly from Definition 11.3 that U is a nbd of p.
11.11
TOPOLOGICAL SPACE
35
Conversely, suppose U is a subset of X satisfying the condition of B). Then, by Definition 11.3, for each point p of U we can select an open set V ( p )such that p E V ( p )C U. Therefore
which means that U is the union of open sets, and hence from (iii) of Definition 11.1 it follows that U is an open set. Though we have defined a topological space initially by means of open sets in Definition 11.1, we can start with nbds to define a topological space if we want to do so. C ) Let X be a set in which to each element p a collection 42 ( p) of subsets of X is assigned. Suppose % ( p ) satisfies (i)-(v) of A). Then we call each element of 42 ( p ) a nbd of p. Now, we define that a subset U of X is an open set if and only if it satisfies the condition B). Then thus defined open sets satisfy the conditions (i)-(iii) of Definition 11.1, i.e. X is a topological space.
Proof. The conditions (i), (ii) and (iii) of Definition 11.1 are direct consequences of (i), (iv) and (iii) of A), respectively. Proposition C) shows that to make a topological space from a set we may first define the totality of nbds instead of defining open sets. But, now, there arises a question. Let X be a topological space defined by use of Definition 11.1. Using Definition 11.3, we can define the nbds in X. They, of course, satisfy (i)-(v) of A), and hence we can again define open sets by use of the nbds as shown in C). Do those newly defined open sets coincide with the original open sets? If not, we might get confused about the concept of open sets in a topological space. Fortunately, there is no chance for such confusion to occur. Namely we can assert the following:
D) We suppose X is a topological space defined by Definition 11.1, denoting its topology by 0,i.e. the collection of the open sets of X . We define the nbds of X by use of 0 and Definition 11.3. By use of those nbds and C ) we again define open sets and denote by 0’ the collection of the newly defined open sets. Then 0 = 6”.
Proof. It is directly derived from B) which implies that 6 has the same relation to the nbds as 0’ does.
36
BASIC CONCEPTS IN TOPOLOGICAL SPACES
[II.l
The relationship between the two concepts, open sets and nbds, are reciprocal. The following proposition shows that we shall likewise have no trouble even if we begin with nbds to define a topological space.
E ) Let X be a topological space defined with nbds satisfying (i)-(v) of A). We denote by % ( p )the collection of the nbds of each point p of X . B y use of those nbds and C ) we define open sets of X . Then by use of the open sets and Definition 11.3 we define nbds for each point p of X denoting by % ' ( p ) the collection of those newly defined nbds of p. Then % ( p )= OU ' ( p )for each point p of X . Proof. Let U E % ( p ) ;then by (v) of A) and C) there is an open set V such that p E V C U. Hence by Definition 11.3, U E % ' ( p ) . Conversely, let U E % ' ( p ) ;then by Definition 11.3, there is an open set V such that p E V C U. Since V is an open set containing p, it follows from C) that V E % ( p ) .Therefore by (iii) of A) U E % ( p ) .Thus % ( p )= % ' ( p )holds for each point p of X.
In view of propositions A), C), D) and E) we can conclude that there is no difference between the results whether we adopt open sets or nbds to define a topological space. Now, let us proceed to extend the other terminologies on point sets of E * to topological spaces. Definition 11.4. Let X be a topological space. A subset F of X is called a closed set if X - F is an open set.'
F) The collection %' of the closed sets of X satisfies the following conditions: (i) 0E%,X E % , k (ii) $4 E %, i = 1 , . . . , k, then U,,,F, E %', (iii) if F, E %, y E r, then n { F , 1 y E r }E V, where the index set r is not necessarily finite.
Proof. We shall prove only the condition (iii). By use of de Morgan's rule,
X-~{F,~~E~}=U{X-F,~~E~}. Since F, is closed, X - F , is open, and hence it follows from (iii) of Definition 11.1 that U{R - F, 1 y E r }is open. Therefore its complement, n{F, I y E is a closed set.
r},
'
Generally, we may denote by Fc the complement X - F of a subset F of X if there is n o fear of confusion.
11.11
TOPOLOGICAL SPACE
37
Since the definitions of open set and closed set are reciprocal, we can define a topological space by means of closed sets as well, but the detailed discussion will be left to the reader. A set A of a topological space X is called a G,-set (an F,-set) if it is the intersection of countably many open sets (the union of countably many closed sets, respectively). Let A be a set of a topological space X and p a point of X. If A is a nbd of p, or in other words if there is a nbd U ( p ) of p such that U ( p ) C A , then p is called an inner point of A. We denote by A" (or Int A ) the set of the inner points of A and call it the interior or innerpart of A . If p is an inner point of X - A, then it is called an exterior point of A. If p is neither an inner nor an exterior point of A, i.e. if every nbd U of p intersects both X - A and A, then p is called a boundary point of A. We denote by B ( A ) (or Bdr A ) the set of t h e boundary points of A and call it the boundary of A.
G ) A" is an open set. B ( A ) is a closed set.
Proof. The easy proof of this proposition will be left to the reader. Example II.3. Let R , be the topological space of ordinal numbers given in Examples 11.1 and 11.2. Then F = [w, 2 w ] = {a I w =sa S 2w) is a closed set for which F" = U and B ( F ) = { w } , where U = (w, 2w]. U is an open set for which U "= U, B ( U )= 0. Note that U is a closed set at the same time. As a matter of fact, R, contains, besides R, and 0, infinitely many sets which are open and closed at the same time. (Such a set is called a clopen set.) Example II.4. Let 93 be a collection of subsets of a topological space X. $23 is called a Borel field if it satisfies: (i) A~E 3, i = 1 , 2 , . . . , imply n;=,A~E 8, (ii) A E 93 implies X - A E 93. For a given collection d of subsets of X, we denote by % ( d )the intersection of all Borel fields containing d as a subcollection. Then a(&)is the smallest Borel field containing d. Let us denote by %' the collection of all closed sets of X. Then every set B E $23(%') is called a Borel set. F,-sets and G,-sets are Borel sets, but the Borel sets are not exhausted by those two types of sets.'
' For the theory of Borel sets, see C. Kuratowski [3].
38
BASIC CONCEPTS IN TOPOLOGICAL SPACES
[II.2
2. Open basis and neighborhood basis Reviewing Section 5 of Chapter I, we notice that in the definition of nbds of E 2 ,the &-nbd,S,(p), plays an important and specific role. In fact, we often define a topological space by first defining some specific nbds and then, in terms of these, all nbds, rather than defining all the nbds at once. The same circumstances occur when defining a topological space by means of open sets.
Definition II.5. Let X be a topological space, 0% a collection of open sets of X. If every open set of X can be expressed as the union of sets belonging to 0%, then 0% is called an open basis of X (or base for X). Let Y be a collection of open sets in X. If the collection AY of all finite intersections of sets belonging to "Ir is an open basis of X , then "Ir is called an open subbasis of X (or subbase for X). Definition II.6. Let 9 be a collection of closed sets of X.If every closed set of X can be expressed as the intersection of sets belonging to 9, then 9 is called a closed basis (closed base) of X. A collection X of closed subsets of X is called a closed subbasis (closed subbase) if the collection of all finite sums of elements of X is a closed basis of X. A) A collection 0% of open sets of a topological space X is an open basis of X if and only if for each open set V of X and each point p of V, there is some
UE%suchthatpEUCV. Proof. The easy proof is left to the reader. B) Let 0% be an open basis of a topological space X ; then it satisfies (i) B E 0%, (ii) if U,, U, E 0% and p E U, n U2, then there is U3E 0% such that p E u3c U , n u,, (iii) U{U( UE%}=X. Conversely, let X be a set and 0% a collection of subsets of X which satisfies (i)-(iii). Then, if we denote by 0 the collection of the unions of sets belonging to %, then 0 satisfies (i)-(iii) of Definition 11.1. Thus X is a topological space with 0% as an open basis.
Proof. The first half of the assertion is almost obvious in view of the property of open sets and A). Therefore we shall prove only the latter half, i.e. that 6 derived from 0% satisfies (i)-(iii) of Definition 11.1.
11.21
39
OPEN BASIS AND NEIGHBORHOOD BASIS
The condition (i) of Definition 11.1 follows from (i) and (iii) of B). The condition (iii) of Definition 11.1is obvious from the definition of 0. Thus it suffices to verify (ii) of Definition 11.1. To do so, we note that by repeated use of (ii) of B) the following assertion can be verified: (1) If U,, . . . , u k E %, p E U, n * * * n uk,then there is some U E % such that p E u u,. Now, assume that V,, . . . , V, E 0 and p € f-If=l V,; then p E V,, i = 1,. . . , k. Since each V, is the union of sets belonging to %, for each i there is some E $2 such that
cnt,
u.
pEV,.CV,. The statement (1) assures us that there is some U E % such that k
p~
ucn V , i=l
Therefore
p
k
~
i=l
~
~
n
~
,
where U E %. This means that r?fZl V, can be expressed as the union of sets belonging to % and hence n,,, V, E 0 proving (iii) of Definition 11.1. Finally it is obvious that % is an open basis of the topological space X with the topology 0. C ) Let Ce be a closed basis of a topological space X ; then it satisfies: (i) X E Ce, (ii) if G,, G, E Ce and p E GIU G,, then there is some G3E Ce such that P 6z G3 z l Gl u G2, (iii) n { F 1 F E Ce} = 0. Conversely, let X be a set and Ce a collection of subsets of X which satisfies (i)-(iii). Put 011 = { X - G I G E Ce}, then 021 satisfies (i)-(iii) of B). Thus X is a topological space with Ce as a closed basis. Proof. The easy proof is left to the reader.
Definition II.7. Let p be a point of a topological space X. A collection Y ( p )of nbds of p is called a nbd basis of p if for every nbd U of p, there exists V E Y ( p ) such that p € V C U.
40
BASIC CONCEPTS IN TOPOLOGICAL SPACES
[II.2
D) Let V ( p )be a nbd basis of a point p of X . Then it satisfies the following conditions: 0) V ( P )# 0, (ii) if U E V ( p ) ,then p E V, (iii) if U E V ( p ) and V E V ( p ) , then there exists W E Y ( p ) such that
wc u n v,
(iv) if U E V ( p ) ,then there exists a set V such that p E V C U and such that for every point q E V, there is some W E V ( q )satisfying W C V. Proof. The easy proof is left to the reader. To define a topological space, we may give only a nbd basis of each point instead of the totality of nbds as seen in the following.
E ) Let X be a set in which to each element p of X a collection V ( p ) of subsets of X is assigned. Suppose V ( p )satisfies (i)-(iv) of D). If we define a collection % ( p) for each p E X by % ( p ) = { U I U > Vforsome V E V ( p ) } , then % ( p )satisfies the conditions (i)-(v) of l.A). 7hus X is a topological space with % ( p )as the collection of the nbds of pointp while V ( p )forms a nbd basis of p. Proof. The easy proof is left to the reader.
Example II.5. In view of Definition 11.3, we know that in every topological space X the collection of the open nbds of a point p form a nbd basis of p. In fact, we can get along with only open nbds in most discussions related to nbds. In E 2 , or more generally in every metric space X , the collection V ( p )of the E-nbds, S,(p), for E > O , forms a nbd basis of p. The subcollection V ' ( p )= {Sl,n(p)1 n = 1,2, . . .} of V ( p ) also forms a nbd basis of p. It is also easily seen that U { V ( p ) l p E X } and U { V ( p )1 p E X } are open bases of X . Generally two nbd bases (open bases) which generate the same topology are called equivalent. Thus V ( p ) and s"(p) are equivalent. As for R, in Example 11.1, the collection % ( a )= {(p, a ] I p < a} for a f 0 and % ( a ) ={{0}} for a = 0 forms a nbd basis of a, and {{0}, (p, a]1 0 6 p < a < q}is an open basis. Let us give two more examples of topological spaces defined by means of nbd bases.
11.31
41
CLOSURE
Let R, be the set of all real numbers. For each point p of R,, we define a collection % ( p ) of subsets by % ( p ) = { ( p- E , p
+ E ) I > 0)'
for p f 0 ,
Then we can easily show that %(p)satisfies the conditions (i)-(iv) of D). Thus R, turns out to be a topological space with a different topology from the usual metric topology. Let
R,= {(x, y ) Jy ==o,- W < X
. Proof. It is left to the reader. A decomposition 9of a topological space X is called upper semi-continuous if for each D E 9and each open set U containing 0,there is an open set V such that D C V C U, and V is a union of members of 9.
I ) The natural mapping cp of X onto X(9) is closed if and only decomposition 9is upper semi-continuous.
if the
Proof. Let 9 be upper semi-continuous. Suppose F is a closed set of X . Put
11.71
SUBSPACE, PRODUCT SPACE, QUOTIENT SPACE
65
Since
we shall prove that p(U) is open to show that p(F) is closed. Assume q E p(U). Then pO-'(q) is a member of 9 contained in X - F. Since 9 is upper semi-continuous, there is an open set V which is the union of members of 9 satisfying p-'(q) C V C X - F. Thus p( V ) is an open set of X(9) satisfying
Therefore p(U) is a nbd of q, which means that p(U) is an open set. Thus p is a closed mapping. Conversely, suppose p is a closed mapping. Assume U is an open set of X containing D E 9.Then, since X - U is closed, p(X- U ) is a closed set of X(9). Therefore W = X ( 9 )- p(X - U ) is an open nbd of p(D). Hence V = p-'(W) is an open set and is the union of sets belonging to 9 satisfying
DCVCU This proves that 9is upper semi-continuous. Finally, let us define the concept of inverse limit space.
Definition II.18. Let A be a directed set and {X, I a E A } a collection of topological spaces. We suppose that to every pair a,p of elements of A with a > p, there is assigned a continuous mapping n; (called a bonding mapping) of X, into X, such that
IT; 1 a, p E A, a > p } an inverse system Then we call the system {Xe, (inverse spectrum) of topological spaces. If A is the set of all natural numbers with the natural order, then we call the inverse system an inverse sequence. Now, for a given inverse system {X,, T;}, we define a subspace X of the product space {X, 1 a E A } as follows: X = { p 1 p = {p, 1 a E A } and
n
66
BASIC CONCEITS IN TOPOLOGICAL SPACES
[II.7
r ; ( p u )= ps for every a, p E A with a > p}. Then we call X the limit space of the inverse system and denote it by
X = lim{Xu, r ; 1 cy, p E A, cy > p } t
Let us define the concept of direct limit space (a less used twin of the above defined concept of inverse limit space). Generally, let {XuI a E A} be a collection of disjoint topological spaces; then we introduce a topology into the sum X = U{Xu I a E A} by defining that a subset U of X is an open set if and only if U f l Xu is an open set of Xu for every cy E A. Then X with this topology is called the discrete sum or topological sum of Xu, a E A. Now, assume that {Xu1 a E A} is a given collection of topological spaces and A a directed set. Furthermore, we assume that for every pair a, p of elements of A with a < p, there is assigned a continuous mapping p; from Xu into Xs such that
Then we call the system {X-, p i I a, p E A, a < p } a directed (inductive) system of topological spaces. We may use the term 'spectrum' instead of 'system' as we did for an inverse system. Let X denote the discrete sum of Xu, a E A. Two points pa and ps of X are called equivalent if pu E Xu, ps E Xs and pp(pu)= p : ( p s ) for some y. Now the quotient space X * of X with respect to this equivalence is called the limit space of the directed system {Xu,p;} and denoted by
X * = lim{Xu, p i 1 a, j3 E A, a < p } . + Example 11.17. Let I" = { ( x l , . . . , x,,) I 0 S xi S 1, i = 1, . . . , n}, n = 1, 2,. . . , be n-dimensional cubes. For natural numbers, n, m with n > m, of I" onto I" by we define a mapping
Then we can easily see that the limit space
I " = lim{I", t
r k 1 n, m = 1 , 2 , . . . ; n > m }
11.71
SUBSPACE, PRODUCT SPACE, QUOTIENT SPACE
67
of the inverse sequence { I ” ,T:} is homeomorphic to Hilbert cube I” by the mapping
Example II.18. Let A’ be a cofinal subset of a directed set A . Then
X
I
= lim{X,, T; a,p E t
A, a > p }
and
X‘ = lim{X,., mz: 1 a’, p’ E A’,a’> p } t
are homeomorphic with each other. To see it, we define a mapping f of X onto X ’ by
Then f is one-to-one. For, if {pa 1 a E A } # { q , 1 a E A } , then p, f q, for some a E A. Since A’ is cofinal in A, there is a’E A’ with a’> a. It follows from
It is clear that f is continuous. To see that f-’ is continuous, we consider a nbd U ( p , ) of pa0 in X., with a,tiZ A’. Then
is a nbd of p = {pa I a E A } . Choose a1E A’ with al > a,;then there is a nbd V(p,,) of pa, in X,, such that
because
IT:;
is a continuous mapping with m:;(p,J = p%. Now, putting
68
BASIC CONCEPTS IN TOPOLOGICAL, SPACES
[I13
we obtain a nbd V of p ’ = {pusI a’€ A’} such that f -’( V )C U. Of course, U is a special type of nbd of p, but in view of the above argument, we can easily see that f -’( V )C U holds for every nbd U of p and for some nbd V of p’. Thus f is a homeomorphism of X onto X‘. If all X, are the same topological space X, and each T; is the identity mapping, then lim{X,, r; 1 a, /3 E A , a > /3} is homeomorphic to X. If A t contains a cofinal subset A’ such that X,. = X for all a ’ € A’ and r ; : ( p )= p, a‘, P’E A’, we obtain the same conclusion.
8. Connectedness
Definition 11.19. A topological space X is called connected if it cannot be decomposed into the sum of two non-empty, disjoint closed sets. A subset X ’of X is called connected if the subspace X ‘ is a connected space. A topological space X is called locally connected if every point p of X has a nbd basis consisting of connected sets. A) Let A,, y € r , be connected subsets of a topological space X . If n{A,1 y E # 0, then U { A , I y E r }is a connected set.
r}
Proof. Assume the contrary; then
U A , =F U G,
F n G=0
for some non-empty closed sets F, G of the subspace U A , . Choose a point p E n A , ; then either F or G contains p . Suppose, for example, p € E Since G # 0, G n A, # 0 for some y. Now putting
FnA,=F‘,
GnA,=G’,
we get non-empty closed sets F’ and G’ of the subspace A,. Moreover A, = F’ U G’ and F‘ n G‘ = 0 are obvious, and hence A, is not connected, contradicting the initial hypothesis. Thus U A, must be connected.
B) Let A be a connected subset of a topological space X. If A then B is connected. Thus A is also connected.
CB C
A,
11.81
69
CONNECTEDNESS
Proof. Assume the contrary, i.e. let
where F and G are non-empty closed sets of the subspace B. Note that F and G are also open sets of the subspace B. Choose a point p E F. Then, since p E F, p E A,and F is a nbd of p in the subspace B, we obtain
In the same way we obtain G
F’=FnA,
n A # 0. Therefore putting
G’=GnA,
we get non-empty closed sets F’ and G‘ of the subspace A . Moreover A = F‘ U G’ and F ’ n G ’ = 0 are obvious, and hence A is not connected, contradicting the initial hypothesis. Thus B must be connected.
C ) Let p be a point of a topological space X . If we denote by K ( P ) the sum of the connected subsets of X which contain p , then K ( P ) is a connected closed set. -
Proof. By A), K ( P ) is a connected set. Therefore by B), K ( P ) is also a connected setwhich contains K(P). Hence it follows from the definition of K ( P ) that K ( P )C K ( P ) . This means that K ( P ) is a closed set. Definition 11.20. The connected closed set K ( P ) given in C) is called a component of X . If every component of a topological space X contains only one point, then X is called a totally disconnected (or hereditarily disconnected) space. It is easily seen that if two components intersect, then they coincide. Hence we can assert the following:
D) Every topological space X is decomposed into the sum of disjoint components. E) Let Xu, a E A, be connected topological spaces; then their product space X = {Xu 1 a E A} is also connected.
n
70
BASIC CONCEPTS IN TOPOLOGICAL SPACES
[11.8
Proof. Assume that X is decomposed into the sum of two disjoint, closed sets F and G. Let F # 0; then we can select a point p = { p a I a E A} E F. Now, we can assert that every point q = {q, 1 a E A} which has at most finitely many coordinates different from those of p , belongs to F. To prove this, it suffices to show that if q = {q, 1 a E A} differs from p by only one coordinate, i.e. q., # pa,, qa = p a for (Y f ao, then q E F. The above is sufficient because the general case can be proved by repeated use of this assertion. Put
P = { { p : ] a E A } \ p L = p , for a
f
ao};
then we can easily see that P is a subset of X which is homeomorphic
to
X,,. Therefore P is connected. On the other hand, P is decomposed into two disjoint closed sets P n F and P f l G. Since p E P fl F f 0, it must be that P n G = 0, i.e. P f l F = P. Thus we have proved that
Now, we put Q = ((4,
I a € A} I q, = pa except for at most finitely many a}.
Then, as shown in the above, Q C F, and hence Q C F. On the other hand, by the definition of product space, Q is obviously a dense subset of X Therefore 0 = X which implies F = X . Since F is closed, we obtain F = X , G = 0.Thus X is a connected space. Example 11.19. Among the most popular examples of connected spaces is En.To see the connectedness of E', we suppose E' = F U G for disjoint closed sets F, G. If both F and G are non-empty, then there is an interval I = [a, b ] such that I F # 0. I n G # 0. Assume, for example, b E I r l G. Let c = sup ( I f l F ) . Since I f l F is closed, c E I nF. Hence c < b and J = ( c , b ] C l n G . Since I n G is closed, c € J C I f l G which is a contradiction. Thus E' is connected. Therefore by E), E" is connected for every n. On the other hand, the subspace consisting of the rational points of E n is a totally disconnected space. Let X be a topological space and 9the collection of the components of X . Then the decomposition space X(9) consists of a point if X is connected. Generally we can assert that X(9) is a totally disconnected space. For, let K be a component of X(9); then the inverse image rp-'(K) of K by the natural mapping rp is a closed set of X and contains at least
11.81
CONNECTEDNESS
71
one component of X . To show that p-'(K) is connected, assume the contrary. Then p-'(K) = F U G for non-empty disjoint, closed sets F and G of p-'(K). Every component contained in p-'(K) is contained either in F or in G, for the component cannot meet both of F and G because of its connectedness. Now, let p(F) = L, p(G) = M ; then by the above remark, L and M are non-empty, disjoint sets whose union is K. On the other hand it is clear that p-'(L) = F, p-'(M) = G, and hence L and M are closed sets of K by the condition of decomposition space. This contradicts the fact that K is connected. Therefore p-'(K) is connected. Consequently it coincides with a component P of X. This implies that K = p(P) contains only one point proving that X(9) is totally disconnected. Example II.20. Suppose every two points p, q of a topological space X can be joined by a curve, i.e. there is a continuous mapping f of the unit segment [0, 11 into X such that f(0) = p, f(1) = q. Then we call X arcwise connected or path -connected. Every arcwise connected space is connected, but the converse is not true. For example, the subspace {(x, y ) 1 y = sin l/x, x # 0) U ((0, y ) 1 -1 S y s 1) of E2 is connected but non-arcwise connected. Definition 11.21. Let p be a point of a topological space X. Then we define a quasi-component Q ( p ) by
Q ( p ) = n{A 1 A is a clopen set of X such that p E A}
F) Generally K ( p )C Q ( p ) . Proof. The easy proof is left to the reader. G ) Every topological space X is decomposed into the sum of mutually disjoint quasi-components.
Proof. Observe that for every p, q E X , either K ( p ) C Q ( q ) or K ( p ) n Q ( q )= 0 holds. Because otherwise there is a clopen set C such that
contradicting the connectedness of K ( p ) . Thus the present proposition follows from D).
72
BASIC CONCEPTS IN TOPOLOGICAL SPACES
Example II.21. O ( p )and K ( p ) can differ. Let
F, = { ( x , y ) E E 2 1 y= l / n } , n
=
1 , 2 , .. .
Then consider the subspace m
Y=
u Fn U { ( O , O ) , (1,O))
of E 2
n=l
In Y,
Exercise II 1. Let X be a given set. Then 0 = { U I U C X , X - U is finite or U = 0)is a topology of X. (Such a topological space is called a cofinite space.) 2. For a set X and a collection % ( p ) ,p E X, of subsets of X , the conditions (i)-(v) of 2.D) are equivalent with (i)-(iv) plus (v)' if U E % ( p ) , then there exists a set V ' E % ( p ) such that for every point q E V' there is W E % ( q ) with W C U. (To derive (v) from (i)-(iv) plus (v)', put V = { q I W C U for some W E % ( q )}.) 3. A" is the largest open set contained in A. Consequently A is an open set if and only if A" = A. 4. A is a closed set if and only if A 3 B ( A ) .
5. An open set U is called a regular open set if 0" = U. Then A" is a regular open set for every subset A of a topological space X.(A closed set F is called a regular closed set if F"= F.) ' A s for further discussions on connectedness, see, e.g., books like C. Kuratowski [4], C. T. Whyburn [l], R. Engelking [Z], where one can find rather extensive information on this aspect; the last gives detailed arguments on disconnectedness and 0-dimensionality. As for articles on various topics, see, e.g., W. Sierpinski [l], [2] for interesting classical theorems, B. Banaschewski [l], G. T. Whybum [2], J. de Groot-R. H. McDowell[2], M. Henriksen-J. Isbell [l]for local connectedness, A. P. Kombarov [l], E. Pol-R. Pol [l] for totally disconnectedness, E. Michael [lo], J. T. Goodykoontz, Jr. [l] for connectedness of spaces of subsets and of functions and also T. Tanaka [l], B. Knaster-A. Lelek [l], J. Mycielski [l], A. Lelek [l], R. Duda [ l ] for various interesting results.
111
73
EXERCISE
6. Give the relationship between the two concepts, nbd basis and open
basis.
A
7. Let 9 be a closed basis of a topological space X . Then = n{GI A C G E %} for every subset A of X . A mapping f from Z into X is continuous if f-'(G) is closed in 2 for every G E 3. 8. A = X - ( X - A ) " . 9. Prove A n B C A n B. Give an example to show the equality does not necessarily hold. 10. A is a closed set if and only if every cluster point of A belongs to A . 11. For a closed set F, F = F" U B ( F ) , F" = F"".
12. A subset A of a topological space X is nowhere dense if and only if
for every non-empty open set U of X , there is a non-empty open set V such that V C U and V n A = 0. 13. A filter 9 converges to p if and only if p is a cluster point of every filter which contains 9 as a subcollection. 14. A filter 9of a topological space X is maximal if and only if for every subset A of X, either A or X - A belongs to 9. 15. A net derived from a maximal filter is maximal. 16. The filter derived from a maximal net is maximal.
17. Discuss the relationship between cluster points of nets and those of filters in view of 4.D), E). (Revise Definition 11.12 of derived net slightly to allow A, and A,, coincide for different S and S'.) 18. A subset U of a topological space X is a nbd of p E X if and only if every filter converging to p contains U as a member. 19. Let A be a subset of a topological space X and put B = { P I there is a filter 9 such that A E S + p } . If a filter 9' satisfies B E P+q, then
q E B. 20. Let X be a set. Suppose convergence of filters in X is defined so that:
(i) if for a filter 9 and a point p of X , { p } E 9, then 9 + p , (ii) if 9 + p and % is a filter with 9 3 9,then % + p , (iii) t h e condition of Exercise 19 is satisfied. Now, we define that a subset W of X is a nbd of a point p of X if and only if U and p satisfy the condition of Exercise 18. Then t h e thus defined nbds satisfy (i)-(v) of l.A), and therefore X is a topological space.
74
BASIC CONCEPTS IN TOPOLOGICAL SPACES
[I1
21. A subset U of a topological space X is a nbd of p if and only if for every net p(A I >) converging to p, U f l p(A) # 0. 22. Let p(A I >) be a net; then prove: (i) if p(A 1 > ) + p , then for every cofinal subset d ' of A, p(A' I > ) + p , (ii) if p(A 1 > ) + p , pS(TS I >)+ p(6), then there is a net $(rI > ) + p such that C U{pps(I's) 1 6 EA}.
$(r)
23. Let X be a set. If convergence of nets of X is defined so that they satisfy (i), (ii) of Exercise 22 and (iii) if p(6) = p, 6 E A, then p(A 1 > ) + p , then defining nbd by the condition of Exercise 21 we see (i)-(v) of l.A) is satisfied, i.e. X is a topological space. 24. If a covering % of a topological space is locally finite, then locally finite.
4 is also
25. If % is an open covering of a topological space X , then for each subset A of X, S"(A,%) = U:=, S"(A,%) is a clopen set of X. 26. I f f is a mapping of a topological space X onto a topological space Y and 9 is a filter of X, then = { f ( F ) F E 9} is a filter of S. If f is simply an into mapping, then {G 1 G C Y, f-'(G)E 9) is a filter of Y.
f(a
I
27. Let f be a mapping of a topological space X into a topological space Y. Then f is continuous if and only if for every net p(6 1 >) of X with p(A I > ) + p , f.p((d I > ) + f ( p ) holds in Y. 28. A continuous mapping of a topological space X into a topological space Y is a continuous mapping of X onto the subspace f ( X ) of Y. 29. Let f be a mapping from a topological space X onto a topological space Y. Then f is closed and continuous iff f(A) = f o for every A C X.
30. Let X 3 X' 3 A, where X is a topological space. We denote by A"' and B'(A) the interior and boundary of A in the subspace X'. Then prove A"n X ' C A"',
B ( A )f l X ' 3 B'(A).
Give examples to show that the equality does not necessarily hold in either of the cases. 31. Let A be a dense subset of a topological space X and f, g continuous maps from X into a topological space Y whose every distinct points have mutually disjoint nbds. I f f = g on A, then f = g o n X.
111
75
EXERCISE
32. Let f be a continuous mapping of a topological space X onto a topological space Y. Suppose that for any topological space 2 and any mapping g of Y into 2, the continuity of g o f implies the continuity of g. Then Y is homeomorphic to the decomposition space X(9), where 9 = { f - ' ( q )I q E S } , i.e. f is a quotient mapping.
r}
33. (i) Let {X, 1 y E be a collection of topological spaces. Assume U, c X,, y E then { U, y E r }= { 0, I y E in the product space IIX,. (ii) Let A C X , B C Y for topological spaces X and Y. Then B ( A x B) = ( B ( A )x B)U ( A x B ( B ) ) in X x Y , where boundary and closure of A and B are those in X and Y, respectively.
r;
n I
n
r}
34. Let X,,y E r, be topological spaces. For each point p = { p , 1 y E r } of the Cartesian product X = n { X , 1 y E r},we define a collection W ( p ) of subsets of X as follows: W ( p )= { n { U ,I y E r }1 U, is a nbd of p, in X , for each y E r }where all U, may differ from X,. Then W ( p )satisfies (i)-(iv) of 2.D), i.e. X turns out to be a topological space called the product space with the strong topology or box topology. The product space defined in Definition 11.16. is sometimes said to have the weak topology. 35. Let f be a continuous mapping of a topological space X onto a topological space Y.If X is connected, then Y is also connected. 36. Let F and G be closed sets of a topological space X such that F f l G and F U G are connected. Then F and G are connected sets.
37. Give an example of a topological space which is connected but not locally connected and of a topological space which is locally connected but not connected. 38. A topological space X is arcwise connected if and only if X is connected and each point of X has an arcwise connected nbd. 39. Let f be a one-to-one continuous mapping from the real line onto itself. Then f is a topological mapping.'
40. If X is locally connected, then each component of X is an open set. The converse is not true. 41. Construct a closed subset C of [O,1] as follows: Let J , = [0, 11, J1= [0, 1/31 U [2/3, 11, J, = [0, 1/91 U [2/9,3/9] U [6/9,7/9] U [8/9, 11, . . .. Generally
'
It is an interesting problem to find out more general conditions of spaces under which one-to-one continuous mappings are topological mappings. See E. Duda [l].
76
BASIC CONCEPTS IN TOPOLOGICAL SPACES
PI
J, is obtained from J,-, by subtracting the middle one third open interval from each closed interval constituting J,-l. Then C = n;=J,is called Cantor discontinuum. (i) Cis the set of all real numbers of the form a,/Y for a, = 0 or 2, (ii) C i s homeomorphic to the product space of countably many copies of the two point discrete space (0, l}, (iii) C is a totally disconnected space with no isolated point.
cz=,
CHAPTER I11
VARIOUS TOPOLOGICAL SPACES
1. T I , T,, regular and completely regular spaces
Definition III.1. A topological space X is called a Tl-space if for every two distinct points p, q of X, there are nbds U of p and V of q such that q $Z U and p $Z V. X is called a T2-space or a Hausdofl space if for every two distinct points p, q of X , there are nbds U of p and V of q such that
unv=~. A Tl-space is characterized as a topological space in which every point forms a closed set. The following proposition characterizes T2-spaces. A) A topological space X is a T2-space if and only to a t most one point.
if every filter converges
Proof. Let X be a T2-space and 9a filter of X. Suppose 9 + p and p Z q. By Definition 111.1, there are disjoint nbds U of p and V of q. Since 9+p, U E 9. If V E 9, then U n V = 0 for two members of the filter, contradicting the definition of filter. Therefore V e 9, and hence 9 does not converge to q. Conversely, let X be a non T2-space. Then there are two distinct points p, q, of X such that every pair of nbds U of p and V of q intersect. Thus 9= { M I M 3 U n V for some nbds U of p and V of q } is a filter. Since we get 9 + p , every nbd of p as well as every nbd of q belongs to 9, 9+q. Thus the assertion is proved. Example III.1. Every indiscrete space is not a TI-space if it contains at least two points. The topological space Xl in Example 11.7 is a Tl-space but not T,.
78
VARIOUS TOPOLOGICAL SPACES
[III.l
Definition III.2. A Tl-space X is called a regular space' if for every point p of X and every closed set F of X satisfying p & F, there are open sets U and Vsuch that p E U, F C V and U n V = 0 . A Tl-space X is called a completely regular space' (or Tychonoff space) if for every point p of X and every closed set F of X satisfying p & F, there is a continuous function f over X such that f(p)=O,
f ( F ) = l and O < f < l .
B) A TI-space X is regular if and only if for every open nbd W of each point p of X, there is an open set U such that pEUCUCW. Proof. Let X be a Tl-space satisfying the condition. Given a point p and a closed set F such that p F, then W = X - F is an o p m nbd of p . Therefore we can choose an open set U such that
pE
u c uc w .
Then U and V = X are open sets satisfying the condition of Definition 111.2, and hence X is regular. Conversely, let X be regular and W an open nbd of a point p of X. Then there are open sets U and V such that PEU,
x - w c v ,
unv=0.
Hence U C X - V,which implies
ucx- v=x-vc w since X - V is closed. Then U satisfies the desired condition. Example III.2. The space R, in Example 11.5 is T, but not regular. To construct a regular space which is not completely regular,' we consider a subset E = {(x, y ) I y 2 0} of EZ and add one point p to E to define R ; = E U { p } . We define a special topology for R ; as follows. 'Some people mean by a regular (completely regular) space merely a T-space satisfying this condition. *We owe this simple example to A. Mysior [Z].
111.11
TI, T2.REGULAR AND COMPLETELY REGULAR SPACES
79
Every point (x, y) E E with y > 0 is defined as an isolated point of R;. To each point (x, 0)E E we assign two segments I, and J,, defined by
I 0 G y' S 2}, J, = {(x', y ' ) E E I x' = y' + x, 0 =Sy' s 2).
I,
= {(x, y ' ) E E
Then {I, U J, - P I P is a finite set which does not contain ( x , 0)) is defined as a nbd base of (x, 0) in R;. On the other hand {H,, 1 n = 1,2,. . .}, where
is defined as a nbd base of p in R;. Then it is easy to see that R j is a regular space. To prove that R ; is not completely regular, let
F,, = {(x,O)I n - 1c x s n}, n = 1 , 2 , . . .. Then Fl is a closed set such that p Z ! F,. Assume that f is a real-valued continuous function on R ; satisfying
f(Fl)=1,
OsfGl.
We can prove that f(p) = 1 (and accordingly f ( p ) # 0) as follows. For that purpose, we shall show that f-'(l) r l F,, is infinite for every n, by use of induction on n. The assertion is obviously true for n = 1. Assume that f-'(1) n F,, is infinite. Select a countable subset A of f-'(l)r l F,,. For each (x, 0) E A, m
J, - f-'(l)= U G, , fl=l
where Gn= {Z E J, I O S f ( z ) G 1- l/n}.
c,,,
Observe that each G,, is a closed set of R;. Since (x, 0) E G,,, (x, 0)E and hence G,, is a finite set. Thus J, -f-'(l) is at most countable, and so is A' = U {J, -f-'(l) I (x, 0) E A } ,
Now, denote by
7~
the projection of E onto the x-axis, i.e. ~ ( ( xy)) , = x.
80
VARIOUS TOPOLOGICAL SPACES
[III.l
Then .rr(A’)f l F,,, is at most countable. Hence F,,, - T ( A ‘ is ) infinite. Let ( x ’ , 0) E F,,, - T(A’); then
I,,nJ,nf-’(i) # 0 f o r a l l x E A Thus, for every nbd U of ( x ’ , 0),
unJ,nf-’(i)#O
forsome x E A
This implies that
i.e. f-’(I)n F,,, is infinite, proving our assertion. Since H, 3 F,+,, it follows that
H, nf -l(1)Z 0 for every n Therefore f ( p ) = 1, i.e., R ; is not completely regular.
C ) Every subspace of a Tl (T2,regular, completely regular) space is TI (T2, regular, completely regular). The product space of Tl (T,, regular, completely regular) spaces is Tl (T2, regular, completely regular). Proof. The easy proof is left to the reader. Completely regular spaces are characterized by the following important theorem which is very interesting, since it gives a concrete expression to the (abstract) completely regular spaces. Theorem III.1 (Tychonoff’s imbedding theorem). A topological space X is completely regular if and only if it is homeomorphic with a subspace of the product space of copies of the unit segment [0,1].
Proof. The ‘if’ part is a direct consequence of C), because [0,1] is obviously completely regular. To prove the ‘only if‘ part, we suppose X is a completely regular space.
111.11
TI, T2. REGULAR AND COMPLETELY REGULAR SPACES
81
Let {f, 1 a E A} denote the totality of continuous functions over X such that 0 s f , S 1. For the index set A, we construct the product space
of the unit segments I, point
= [0,1], a E
A. To every point p of X , we assign a
of P. Thus we get a mapping f of X into P. Now, all we have to do is to show that f is a topological mapping. If p, q are distinct points of X , then since X is completely regular, and therefore TI, {q} is a closed set. Hence by Definition 111.2, there is a continuous function fa such that
f,(q)
f , ( p ) = 0,
=
1 and 0 < f a s 1 .
Hence f ( p ) and f ( q ) are different in their a-coordinates, i.e. f ( p ) # f ( q ) . Thus f is a one-to-one mapping. To see the continuity of f, we consider a given point p E X and a given nbd V of f ( p ) in P. By the definition of product space, there are nbds Vmiof f J p ) in IOi, i = 1, . . . , k, such that k
V' =
n VUi n {I, 1 x
a # ai, i = 1, . . . , k } C V
i=l
Since f&), such that
i = 1, . . . , k, are continuous functions, there is a nbd U of p
for every point q E U. Hence q E U implies
i.e. f ( U )C V. Thus by Definition 11.13, f is a continuous mapping. Finally, to prove the continuity of the inverse mapping f-' of f, we suppose U is a given nbd of a given point p of X . Since X is completely regular, there is a continuous function f, such that
82
VARIOUS TOPOLOGICAL SPACES
f,(p)=O,
[III.2
f u ( X - U ) = l and O s f , s l .
Though X - U is not necessarily closed, we can construct such a function f , by considering an open nbd U’ of p which is contained in U. Now
for Uu= [0, l), is a nbd of f ( p )in P. If p’ Sr U, then fu(p’)= 1, which implies f(p’)!iZ V. Therefore f ( p ’ ) E V implies p’E U. In other words, q ’ E V nf ( X ) implies f - l ( q ’ )E U. Thus f is also a continuous mapping, and hence f is a topological mapping of X onto a subspace f ( X ) of P.
-’
Generally a topological mapping f from X into Y is called a topological imbedding.
D ) Let f be a mapping from a topological space X into the product space {Xu1 a E A} of topological spaces X,, a E A. Then f is continuous if and only if nu.f is continuous for every a E A.
n
Proof. It is practically proved and used in the proof of Theorem 111.1.
2. Normal space and fully normal space Definition III.3. A Tl-space X is called a normal space if for every pair of disjoint closed sets F, G of X, there are open sets U, V such that U 3 F, V>Gand UnV=O. A Tl-space X is called a fully normal space if for every open covering 92 of X, there is an open covering 7f such that “Y.* < 92.’ A) A TI-space X is normal if and only if for every closed set F and every open set W with F C W, there is an open set U such that
F c U cucW . Some people mean by a normal (fully normal) space a T-space satisfying this condition. The six conditions given in Definitions 111.1, 2, 3, are called separation axioms. There are some other separation axioms which are generally not so important. We shall deal with another separation axiom, TO,in Chapter VII.
111.21
NORMAL SPACE A N D FULLY NORMAL SPACE
83
Proof. The proof of this assertion is quite analogous to that of l . B ) and is left to the reader. B) A TI-spaceX is fully normal if and only if for every open covering % of X , there is an open covering "y. such that 7 f A< %. Proof. This is a direct consequence of II.5.B).
Example III.3. The topological space R, in Example 11.5 is completely regular, but not normal. It is clear that R, is T,. To see that R, is completely regular, let p E R, and F a closed set of R, such that p fZ F. If p is not on the x-axis, then it is clear that there exists a continuous function f satisfying the condition of Definition 111.2. Therefore we assume that p is a point on the x-axis, i.e. p = (x, 0). Then we can choose a nbd U,(p)= S : ( p ) U {p}such that V , ( p )n F = 0, where we denote by SL(p)the open set bounded by the circle with radius E touching the x-axis at p, i.e. S : ( p ) = S,((x, E ) ) . For every point q E S:(p), we consider the circle passing through q and touching the x-axis at p and denote by 77 (q) its radius. Then we define a continuous function g over R, by
Putting f = ( l / & ) g , we obtain a continuous function f over R, satisfying f(p)=O,
f ( F ) = l and O s f S l
Therefore R, is completely regular. To show that R, is not normal, we consider two sets G = {(x, 0) 1 x is rational},
H = {(x, 0 ) I x is irrational}, on the x-axis. Then G and H are obviously disjoint closed sets of R,. Now, we can assert that if U and V are open sets such that U 3 G and V 3 H, then U n V # 0. For, if we assume the contrary, then U fl V = 0. Put
84
VARIOUS TOPOLOGICAL SPACES
[III.2
Every point p = (x,0 ) E G has a nbd U , ( p ) = S:(p) U { p } with E < 2/n which is contained in U and accordingly does not meet V. Hence it is easily seen that the open interval (x - E , x + E ) on the x-axis does not meet H,. From now on, let us regard the x-axis as E' with the usual topology. Then it follows from the above argument that x @ H,, for every x E G. This implies
G c E'
- H,
in E'. Hence
and hence H, is a nowhere dense set. Therefore H is a set of the first category, but this is a contradiction since H is of the second category, as shown in Example 11.8. Thus we obtain U fl V f 0,i.e. R , is not normal. R, is often called Niemytzki space. The topological space R , in Example 11.1 is normal, but not fully normal. It is clear that R , is T I .To see that R , is normal, let F and G be disjoint closed sets of R,. For every point a E F U G, we choose, by transfinite induction on the ordinal number a, an open nbd U ( a )of the form ( p , a ] such that U ( a )n U ( p )= 0 if a E F, p E G, U(0)= (0) if OEFUG. Assume that we have defined U ( p ) for every p E F U G with p < a and that a E F U G. For example, let a E F ; then there is y < a such that ( y , a ] n G = 0 because G is a closed set which does not contain a (see Example 11.5). Putting U ( a )= ( y , a ] ,we get a nbd of a which does not meet U ( p )for every p E G with p < a. Thus we can define desired open nbds U ( a ) for every a E F U G. Then U = U { U ( a )I a E F } and V = U { U ( p )I p E G } are disjoint open sets containing F and G respectively. Thus R , is normal. Now observe that every increasing point sequence a1< a2< ag< . . . of R , converges to its supremum sup{ai I i
=
1,2, . . .} = min{p I p E R,, p
2
ai for i = 1,2, . . .}
111.21
85
NORMAL SPACE A N D FULLY NORMAL SPACE
(Note that there is a countable ordinal number p such that p z a ; , i = 1,2,. . .. Because, if not, then R , = { p 1 p is an ordinal number such that p < aifor some i}, which is impossible because R, is uncountable while the set on the right of the above equality is countable.) Assume that R, is fully normal. Then for the open covering 42 = {[O, a ] 1 a E R,} we can select an open covering Y such that 'VA< 42. Select an arbitrary point a, of R,. Then S ( a l , 'V) C [0, PI] for some p1E R, Select a, > pl. Then
S(a,, 'V) C [0, p,]
for some p2E R, .
Repeating the same process we get a point sequence a; 6 p1< a 2 s p 2< a 3 sp3 < *
* '
such that S ( a i , 'V) c [O,
pi],
i = 1,2, . . . .
Then, as observed in the above, {a;}converges to a a > p i , i = 1 , 2 , . . . , and hence
= sup{a;}.
Then
m
a 5z
u S(a;,Y ) . i= 1
On the other hand, since S ( a , 'V) is a nbd of a, a; E S(a, 'V) for some i .
This implies a E S ( a j , 'V), a contradiction. Hence R, is not fully normal. A property which distinguishes normal and fully normal spaces from spaces in the former section is that a subspace of a normal space is not necessarily normal. In fact, as shown later in Example IV.3, even a subspace of a fully normal space may fail to be normal. Furthermore, the product space of two fully normal spaces is not necessarily normal, as shown in Example 111.5. We shall show later in Example V.2 that even the product space of a fully normal space and a metric space need not be normal.
86
VARIOUS TOPOLOGICAL SPACES
p1.2
Theorem III.2 (Urysohn's lemma). A T,-space X is normal if and only if for every disjoint closed sets F, G of X , there is a continuous function f defined over X such that
f(F)=O,
f ( G ) = l and
Osfsl.
Proof. Sufficiency. Let F and G be disjoint closed sets and f a function satisfying the condition. Then putting
u = {P I f ( P )
v = {P I f ( P ) >
9
we obtain open sets U, V which satisfy
UZM,
VIG,
unv=0.
Therefore X is a normal space. Necessity. Let F and G be disjoint closed sets of a normal space X. Then by A) we can construct an open set U such that
F C U C U C X - c. We put U = U(1/2).By use of a similar process, we can construct open sets U(1/22),U(3/22)such that
By repeating this process, we define open sets U(m/2"),m = 1,2,. . . , 2" - 1, for every positive integer n. To give a precise definition by induction on the number n, we assume that we have defined U(m/2"-'), m = 1 , . . ., 2"-'- 1, such that
2"-'
-
c u (2"-' " ) c u(5 " ) c * .cu(+)cx-G. Since U(2mI2") (= U(m/2"-')), m
=
1, . , . ,2"-l- 1, are already defined,
111.21
NORMAL SPACE AND FULLY NORMAL SPACE
we construct, by use of A), open sets U((2rn such that
Thus we can construct open sets U(rn/2"),rn positive integer n such that
87
-
1)/2"), rn = 1 , . . . ,2"-1,
=
1 , . . . ,2" - 1, for every
m-1
2" - 1
c- . c U ( F )
c X - G.
Adding to those U(rn/2") we define
U(1)= X
U(O)=0 .
and
It follows from (1) that
for any r, r' with r < r', and r, r' E (rn/2" 1 rn = 0, . . . ,2" ; n = 1,2, . . .}. For, if r < r', then we can find a common denominator 2" to express r and r' as r = m / Y , r' = rn'/2" with m < in'. Now, we define a real-valued function f over X by
where we note that r is a rational number of the form 4 2 " with OGrn s 2 " . This function f obviously satisfies 0 s f Sl. Since F C U(1/2"), n = 1,2, . . . ,f ( F ) = 0 is clear. On the other hand, since each point of G belongs to U ( r )only for r = 1, f ( G )= 1 is also clear. Thus all we have to show is the continuity o f f . Let p be a given point of X with 0 0, we select a positive integer n with 1/2"-' < E . Furthermore, we take the positive integer rn defined by rn-1 2"
-< f ( P )
rn
s 2"
88
VARIOUS TOPOLOGICAL SPACES
[III.2
Then by the definition of f ( p ) , we obtain
This is true because, if we assume p E V((m - 1)/2"), then by (l), p E U(r) for every r > (m - 1)/2" which implies f ( p ) S ( m - 1)/2" contradicting (2). On the other hand, we also obtain m+l
1
This follows since, by the definition of f ( p ) , p 62 V((m + 1)/2") implies f ( p ) 2 ( m + 1)/2", contradicting (2). Thus we can conclude that ~~
m+l
p E u(T)
-
u("-') 2"
=
v,
i.e., V is an open nbd of p. Suppose q is a given point of V ; then it is obvious that
This combined with (2) implies that
We have assumed 0 < f ( p ) < 1, but in the special case, f ( p ) = 0 or f ( p ) = 1, we can find, in a similar way, a nbd V of p satisfying the same condition. Therefore f is continuous at every point p and hence it is continuous over X.This proves the theorem.
'
We can assume rn as we like.
+1
2" without loss of generality, because we may choose as large n
111.21
NORMAL SPACE AND FULLY NORMAL SPACE
89
Theorem III.3 (Tietze’s extension theorem). Let X be a T,-space. Then X is normal if and only if for every closed set F of X and every (real-valued) continuous function f over F, there exists a continuous extension off over X .
Proof. Sufficiency. Let F and G be disjoint closed sets. Then we define a continuous function f over the closed set F U G by f ( F ) = 0,
f ( G )= 1 .
Since F and G are disjoint, f is certainly a continuous function over the subspace F U G. Hence f can be extended to a continuous function cp over X. Put
Then by the continuity of cp, U and V are open sets satisfying
Therefore X is a normal space. Necessity. Let f be a real-valued continuous function defined over a closed set F of a normal space X. First we assume that f is bounded, i.e.,
we obtain disjoint closed sets G, H. Since X is normal, by Urysohn’s lemma, there is a continuous function g over X such that
we get a continuous function
cpl
over X satisfying
90
VARIOUS TOPOLOGICAL SPACES
IIII.2
By applying an argument analogous with that for f, but this time for the function
we obtain a continuous function p2 satisfying
By repeating this process we obtain continuous functions p f l over X and +, over F such that
It follows from $1
=
f
- (P1
2
+2=
+1-(P2j
+,=
+2-(P3,
+,, = $ 4 - 1
- (Pn 7
that n
+n
=f
-
c
(Pi?
i=l
and hence n
111.21
NORMAL SPACE AND FULLY NORMAL SPACE
91
x:=l
It follows from (1) that rpi(p) uniformly converges over X.' Therefor q ( p ) = Xy=, cpi(p) is a continuous function over X. In view of (2) we know that
i.e., rp is a continuous extension off over X. We should note that by (1) rp satisfies the same condition, lrpl S a,over X as f does over F. Now, let us deal with a non-bounded continuous function f over F. Then tan-'f is a continuous function over F satisfying
Therefore, using the preceding result, we can construct a continuous extension @ of tan-lf over X such that S r / 2 . Put
)@I
G = { p I @(p) = 7r/2 or @(p) = -7r/2} ;
then G is a closed set of X which does not intersect F. By use of Urysohn's lemma we define a continuous function g over X such that g(F)=l,
g(G)=O and O G g S l
Putting Gr= g@,we obtain a continuous function
arover X such that
I@q < d 2 ' and @ ' ( p )= tan-'f(p),
Thus rp
=
p EF.
tan @' is a continuous extension off over X .
Corollary. We denote by S" the n-dimensional sphere, i.e. the subspace
A sequence {f,I n = 1,2, . . .} of functions over a topological space X is said to uniformly converge to a function f if for every E > 0, there is no such that l f , ( p ) - f ( p ) l < E for every n 2 no and p E X. As well known in elementary analysis, {fn 1 n = 1,2,. . .} uniformly converges if and only if for every E > 0, there is an no such that Ifn(p)-fm(p)I< E for every n, rn 3 no and p E X. In this case, iff,,, n = 1,2, . . . , are continuous functions, then their limit function f is also continuous.
92
VARIOUS TOPOLOGICAL SPACES
p11.2
n+l
{
S" = (xl,. . . , xn+J
of Let f be a continuous mapping of a closed set F of a normal space X into S". Then we can continuously extend f over an open set of X which contains F.
Proof. Let
Then f l , . . . ,fn+, are continuous functions over F satisfying n+l
cf X P )
=1
.
i=l
Therefore by the theorem we can extend f i to a continuous function pi over X . Putting
we get an open set U of X containing F. Now ( p E U )
is a continuous mapping of U into S" satisfying
Example III.4. The condition that F is closed is crucial in Tietze's extension theorem. If we drop it, then the theorem does not hold even for El. Let
f ( x ) = sin l/x, x E F
=
(0, w)
,
Then f is continuous over F, but we cannot continuously extend it over El, because lim,,,, f ( x ) is indeterminate.
111.21
NORMAL SPACE AND FULLY NORMAL SPACE
93
Example IlI.5. In Example 111.3 we showed that the Niemytzki space R, is not normal. Here we shall prove the same by use of Tietze's theorem. Generally we denote by C ( X )the set of all real-valued continuous functions defined on a topological space X.Now, assume that R, is normal and put
Then F is a closed set of R, and it is a discrete space when regarded as a subspace of R,. Hence every real-valued function defined on F is continuous and accordingly can be continuously extended over R, by use of Tietze's theorem. Thus we obtain
where c denotes the cardinal number of the set of all real numbers. On the other hand, A = {(x, y ) E R , I x and y are rationals}
is dense in R,. Observe that every continuous function on R, is decided by the values on A. Namely, iff, g E C(R,) satisfy f(p) = g ( p ) for all p E A, then f = g. (See Exercise 11.31.) Thus
IC(R,)( = IC(A)I c" = c , because A is countable. But this contradicts (1).Hence R, is not normal. Next, we shall show, by an example, that the product of two normal spaces is not necessarily normal. Let S be the set of all real numbers. We define the topology of S by use of the open base {(x - E,
XI
I x E s, E > 01 .
Then S turns out to be a topological space called Sorgenfrey line. S is normal, because if F and G are disjoint closed sets of S, then for each x E F and y E G find E ( X ) > 0 and 6(y) > 0 such that
then it is easy t o see that U n V = 0, which proves the normality of S.
94
VARIOUS TOPOLOGICAL SPACES
[III.2
To see that S x S is not normal, assume the contrary. Then note that
D
= {(x, y ) E
s x s I x + y = 0)
is a closed set of S and a discrete subspace. Also observe that A
= {(x,
y ) E S x S I x and y are rationals}
is a countable dense set of S x S. The rest of the proof goes on like the above proof on R,. Namely, by use of Tietze's theorem,
On the other hand,
IC(S x S)l = IC(A)I c"
=c
,
which is a contradiction. Thus S x S is not normal. We shall prove later in Example 111.8 that S is fully normal. Also observe that S is totally disconnected. Theorem III.4. In the following sequence of classes of topological spaces, each class is included by its precursor: T I , T,, regular, completely regular, normal, fully normal. This relation is non-reversible.
Proof. By use of Urysohn's lemma, we can see that every normal space is completely regular. To show that every fully normal space is normal, we consider two given disjoint closed sets F, G of a fully normal space X . Since % = {X- F, X - G} is an open covering of X, there is an open covering 'V of X such that 'VA< %. Putting
U = S(F, 'V) and
V = S(G, 'V) ,
we obtain open sets U, V containing F and G respectively. If we assume that U n V # 0, then there are V,, V, E 'V such that
V,nF#0,
V,nG#0
Hence, for a point p E V, fl V, ,
and
V , n V,#0.
111.21
95
NORMAL SPACE A N D FULLY NORMAL SPACE
which implies
i.e., S(p, V ) is contained in no member of %. This contradicts V A< %, and hence
Therefore X is normal. The rest of the proof is easy and so it is left to the reader. Remember that counter examples showing that this relation is non-reversible were given in Examples 111.1, 2, 3. The following assertion will be used later.
C ) Let % = {U, 1 LY E A} be a point-finite open covering of a normal space X. Then there is an open covering V = { V, 1 a E A} such that
v,
C
U,, a € A .
In this case we say we can shrink 42 to
v.
Proof. By use of Corollary 1 to Theorem 1.3 we well-order the elements of % so that
where T denotes a definite ordinal number. Now, we shall define an open set V, for every a such that
For that purpose we shall use transfinite induction on the ordinal number a. First, let us define V,. Since % is an open covering, X - U { U, y > 0} is a closed set contained in U,. Therefore by A) we can construct an open set V, such that
I
96
VARIOUS TOPOLOGICAL SPACES
[III.3
Assume that we have defined V, for every p < a. Then note that
forms a covering of X . For, if p is a given point of X which is not contained in U, for every y 2 a, then, since OU is point-finite, there is the last ordinal number p such that p < a and p E Us. If p Pr Va. for every p' < p, then
which is contained in Vs by the induction hypothesis (l),. Hence p E Vp, which proves that OU' is a covering. Thus X - [U { Vs 1 p < a} U U { U, 1 y > a}] is a closed set contained in U,, and hence by use of A) we can construct an open set V, such that
x-[U {Vp 1 p < a } u u {U, I y > a } ]c v, c v, c u, . Thus we can construct V, satisfying (l), for every a with a < T. Since each V, is an open set satisfying C U,, we must only prove that {V, I a < T} is a covering of X , but the proof is quite analogous to the proof that 42' is a covering, so it is left to the reader.
v,
3. Compact space and paracompact space In the preceding sections we have learned a sequence of conditions which belong to the category of separation axioms. Now, here is another important group of conditions for a topological space, i.e. compactness and related conditions, which are rather different from the separation axioms in their nature. Definition III.4. A topological space X is called a compact space (or bicompact space) if every open covering 92 of X contains a finite subcovering "Ir, i.e., a finite open covering "Ir such that 7fC %. A topological space X is called a paracompact space if for every open covering % of X , there is a locally finite open covering "Ir such that 7 f < %.'
'
A subset of a topological space is called a compact (paracompact) set if it is compact (paracompact)as a subspace. A paracompact Tz-space is often called a paracompact space, but in this book we do not assume T2 when speaking of a general paracompact space.
111.31
COMPACT SPACE A N D PARACOMPACT SPACE
97
The following is a direct consequence of Definition 111.4. A) Every compact space is paracompact.
B) Let f be a continuous mapping of a compact space X onto a topological space Y. Then Y is also compact.
Proof. Let "Ir be an open covering of Y. Then f-l("Ir)={f-l(V)/ V E "Ir} is an open covering of X. Since X is compact, there is a finite subcovering {fl( V,), . . . ,f-'( V,)} off -'("Ir). Then { V,,. . . , V,} is a finite subcovering of "Ir, proving the compactness of Y.
C ) Every closed set of a compact space is compact. Every compact set of a T2-spaceis closed.
Proof. Let us prove only the latter half. Let A be a non-closed set of a T2-spaceX. Then there is a point p E A - A. For each point q E A, select an open nbd U ( q )of 4 and a nbd V,(p) of p such that U ( q )n V,(p) = 0. Then
is an open covering of A. But for every k,
which implies
Thus "u contains no finite subcovering, proving that A is non-compact. Later, in Theorem IV.l, we shall see that the product space of compact spaces is also compact. D) Let X be a compact space and Y a T2-space. Then every continuous one-to-one mapping of X onto Y is a topological mapping.
98
[III.3
VARIOUS TOPOLOGICAL SPACES
Proof. All we have to prove is that f is closed. Suppose F is a closed set of X ; then it is compact by C). Hence by B) f ( F ) is compact, and hence by C) it is closed in Y , proving that f is a closed and, consequently, a topological mapping. Theorem III.5. A topological space X is compact if and only if it satisfies one of the following conditions : (i) every closed collection with finite intersection property has a non-empty intersection, (ii) every filter of X has a cluster point, (iii) every maximal filter of X converges.
7
Proof. Compactness3 (i). Let X be a compact space and Ce closed collection with f.i.p. Suppose (B has empty intersection. Then {X- G I G E $2) is an open covering of X. Hence there is a finite subcovering { X G,, . . . , X - G,}. This implies Gj = 0 contradicting the f.i.p. of 97. Therefore Ce has non-empty intersection. (i)+ (ii). Suppose 9 is a filter of a topological space X satisfying (i). Then
n
is a closed collection with f.i.p. Therefore by (i), its intersection contains at least one point p. This means that p E F for every F E 9, i.e. p is a cluster point of 9. (ii) j (iii). If X is a topoIogica1 space satisfying (ii), then every maximal filter of X has a cluster point which is, by 11.4.C), simultaneously a convergence point of the filter. (iii) j compactness. Finally, we suppose % is a given open covering of a topological space X satisfying (iii). Assume that % has no finite subcovering. Then
has f.i.p. Therefore, by 11.4.A), we can construct a maximal filter 9which contains 9 'as a subcollection. By (iii) we have 9 + p for some point p of X. Since by II.4.C) p is a cluster point of 9, p E F for every member F of 9 and especially pEX-U=X-U
111.31
COMPACT SPACE AND PARACOMPACT SPACE
99
for every U E %. But this cdntradicts the fact that % is a covering of X. Thus % must have a finite subcovering, and hence X is compact.
Corollary. A topological space X is compact if and only if it satisfies one of the following conditions: (i) every net of X has a cluster point, (ii) every maximal net of x converges.' Proof. Compactness3 (i). Let cp(A I >) be a net of compact space X. Then by (ii) of the theorem, the filter derived from cp has a cluster point which is, as easily seen, a cluster point of rp, too. (i) 3 (ii). If X satisfies (i), then every maximal net cp of X has a cluster point p . Then for each nbd U of p , cp is residual either in U or in X - U. But, since cp is cofinal in U, it cannot be residual in X - U. Thus cp is residual in U, proving rp(A I > ) + p . (ii)Jcompactness. Suppose 9 is a maximal filter of a topological space X satisfying (ii) and cp(A I >) is a net derived from 9. Then rp is maximal (Exercise 11.15), and hence cp (A I >)+p for some p E X. Now p is easily seen to be a cluster point of 9, but since 9 is maximal, by 11.4.C), 9 + p . Thus by the theorem X is compact.
E) Every paracompact T,-space X is fully normal. Proof. First let us prove that X is regular. Suppose F is a closed set of X and p is a point of X which is not in F. Since X is T,, for every point q of F, there is an open nbd V ( q )of q and a nbd U , , ( p )of p such that
-
Hence p @ V(4).We assign to each point q of F such an open nbd V ( q ) and let
V = { X - F , V(q)IqEF}. Then V is an open covering of X , and hence by the paracompactness of X there is a locally finite open covering W of X such that W < V. Putting
'
G. Helmberg [l] gave another interesting necessary and sufficient condition for compactness with respect to nets of continuous functions.
100
VARIOUS TOPOLOGICAL SPACES
[111.3
W = S(F, W ),
we obtain an open set W which contains F. Then we can assert that For, if
p
w.
thensince W < "1', W'C V ( q ) for some q E F. This combined with p V ( q )implies that p !?k I%". Therefore, by use of 11.5.C),we obtain
=U{W'I w n F f 0 , W ' E % q =
w.
This means that p has an open nbd U satisfying W n U = 0, proving that X is regular. Now, to prove the normality of X , we suppose F and G are given disjoint closed sets of X . To every point p of F, we assign an open nbd V ( p )such that
this is possible by virtue of the regularity of X (see 1.B)). Putting
we get an open covering "1' of X . Since X is paracompact, there is a locally finite open covering Q such that % i,. Therefore W' is locally finite. This proves that X is paracompact. By use of 3.E), we also conclude that X is fully normal. Example III.7. The real line E' satisfies the second axiom of countability. For % = {S,,"(p)1 p is a rational number, n = 1,2, . . .} is easily seen to be a countable open basis of E l . Therefore, by A), every Euclidean space E" and the Hilbert cube I", too, satisfy the second axiom of countability. Furthermore, every subset of I" also satisfies the same axiom. It is obvious that every compact space is Lindelof. For example, R, given in Example 111.6 is compact and hence Lindelof, too. On the other hand, since the supremum of a countable set of R,- {w,} belongs to R,- {q}, we can easily show that w 1 has no countable nbd base. Therefore R,does not satisfy the first axiom of countability. To the contrary, R,in Example 11.1 satisfies the first axiom of countability, but is not Lindelof. For, if we choose for every a E R, different from 0 an ordinal number /?(a) with /?(a)< a,then
is an open covering of R, which has no countable subcovering. It is also easy to see that neither R, nor R, is separable. R,in Example 11.7 is separable, because the rational numbers on E' form a dense subset of R,, but p has no countable nbd basis, i.e., R,does not satisfy the first axiom of countability. We can also easily see that R, in Example 11.5 is separable, but not Lindelof. R, also shows us that a subspace of a separable space does not need to be separable, because the x-axis is a nonseparable subspace of R,.
111.41
AXIOMS OF COUNTABILITY
107
Example III.8. Let us prove that the Sorgenfrey line S defined in Example 111.5 is Lindelof. Let % be a given open covering of s. It sufficies to prove that every interval [ a , b] in S is covered by countably many elements of %, because S is a countable sum of such intervals. Let
L = {x E [ a , b ] I [x,b ] can be covered by countably many elements of %} . Since b E L, L # 0. Put c = inf L It is obvious that c < b. Assume a < c. Choose xl,x2,. . . E [a , b] such that c < * * * < x2 < x1< b. Then obviously xi E L, i = 1,2, . . . , and hence c E L, because
Now, for some E > 0 such that a < c - E < c, (c - E, c] is contained in some element of %. Thus c - 4 2 E L, which contradicts the definition of c. This proves that a = c, and therefore a E L , i.e., [a , b] is covered by countably many elements of %. Hence S is Lindelof. By C) S is paracompact and fully normal, too. On the other hand, as we saw in Example 111.5, S X S is not normal. Hence it is neither paracompact nor Lindelof. This example shows that the product space of two paracompact (Lindelof, fully normal) spaces is not necessarily paracompact (Lindelof, fully normal). Also observe that S is separable and 1st countable but not 2nd countable. In Section 2 we have characterized all completely regular spaces as subspaces of the product space of unit segments. Now, we can give a concreter materialization to regular spaces satisfying the second axiom of countability . Theorem m.6 (Urysohn's imbedding theorem). A regular space satisfies the second axiom of countability if and only if it is homeomorphic with a subset of the Hilbert cube I". Proof. Since the 'if' part is a direct consequence of A) (see Example III.7), we shall concern ourselves only with the 'only if' part. Let OU be a
108
VARIOUS TOPOLOGICAL SPACES
[III.4
countable open basis of a regular space X satisfying the second axiom of countability. We consider a pair P = (U, V) of members of % satisfying fl C V and denote by 9 = {Pi I i = 1,2, . . .} the totality of such pairs. Note that 9 is a countable collection since % is countable. Let P, = (Ui, V,) be where C V,. Then, since X is normal, by B) and C), we a member of 9, can construct, by use of Urysohn's lemma, a continuous function fi over X such that
fi(U,)=O,
f i ( X - V,)= 1 and 0 S f . S l .
Now we define by
a continuous mapping of X into I". Let W be a nbd of a point p of X; then, since % is an open basis of the regular space X, we can choose U, V E % such that
pE
u c u c v c w.
Thus (U, V) is a member of 9; hence we may suppose, say P, = (U, V), i.e. U = Ui, V = V,. Therefore
which implies that
It is easily seen that the last statement implies that f is one-to-one, and f-' is continuous. Thus f is a homeomorphism of X onto a subset of I". The following corollary, which is a modification of Tychonoffs imbedding theorem, follows by use of the technique of the proof of Theorem 111.6. Corollary. A topological space X is a Tychonoff space with weight S p if and only if it is homeomorphic to a subspace of 1', where p denotes an infinite cardinal number and I' the product of p-copies of the segment [O, 11.
111.51
METRIC SPACE
109
5. Metric space
Definition III.6. Let X be a set. We call X a metric space if for every pair p, q of elements of X,the distance (or metric) p(p, q) between them is defined satisfying the following conditions: (i) p(p, q) is a non-negative real number, (ii) p ( p , q) = 0 if and only if p = q, (iii) P(P, s) = P(4, P ) , (iv) P(P, q ) p ( p , r ) + p(r, q ) for every r E X . A) Let X be a metric space. If we define by % ( p )= { S I l n ( p1)n = 1,2, . . .} the nbd basis of each point p of X , then X is a fully normal topological space satisfying the first axiom of countability.’ Proof. It is clear that X is a topological space satisfying the first axiom of countability. To prove that X is fully normal, we consider a given open covering % of X.To every point p of X , we assign a spherical nbd SE(,)(p) such that & ( p ) s1 and S,,,,(p)C U for some U E 42. Then Y = {SE,,)(p)1 p E X} is an open covering of X with Y < %. Now, we can show that the open covering Y = {S,,,),(p) I p E X} is a delta-refinement of 9’. For, let p be a given point of X; then we put
It follows from the definition of 7 that p E Sc(&/4(po)for some po with
.(Po)’577.
(1)
Let S,(q)/4(q)be a given member of Y’ which contains p. Then for each we obtain, in view of (l), point r E S,(q)/4(q),
and hence r E S,,,&po), i.e.,
Remember that S,(p) = {q I p ( p , q ) < E } . S , ( p ) is called the E-nbd or a spherical nbd of
P.
110
VARIOUS TOPOLOGICAL SPACES
[IIIS
This means that
Therefore Y'is a delta-refinement of Y a n d accordingly of %, too. Thus X is fully normal.
B) For metric spaces, the three conditions, 2nd axiom of countability, Lindelof and separability are equivalent. Proof. In 4.B), we have proved that the second axiom of countability implies the other two conditions. To prove the converse, suppose X is a separable metric space; then there is a countable dense set D = { p * ,p2, . . .} of x.Put
then Y is a countable open basis of X . To see it, let p be a point of an open set U of X . Then S,(p) C U for some E > 0. Choose a positive integer n with 2Jn < E ; then, since D is dense in X , there is pi E D with p ( p i ,p ) < l/n. Now
and hence Y is an open basis of X.In a similar way we can also prove that Lindelof implies the 2nd axiom of countability. Example III.9. Euclidean space E" and Hilbert space H are the most popular examples of separable metric spaces. In H the subset A = { ( x 1 7x2, . . .) I for some n, x l , . . . , x, are rational, and xi = 0 for all j 3 n + 1> forms a dense countable set. Let B be a set of real-valued bounded functions over a topological space X . Defining the distance between f,g E B by
we obtain a metric space called a function space. Function spaces are not generally separable, but we can construct a simpler example of a nonseparable metric space by considkring the discrete sum of uncountably
111.51
111
METRIC SPACE
many non-empty metric spaces. Let I,, a E A , be copies of t h e unit segment [0,1]. In their union U {I,I a E A } we identify all zeros to get a star-shaped set S ( A ) . Then we define the distance between two points of S ( A ) by if P? q E I, if p E I , , q E I , , where a # p 9
p+q
Now, we can easily verify that S ( A ) is a metric space, which is called a (metric) star-space with the index set A. A star space is non-separable if its index set A is uncountable. S ( A ) may be called sometimes a (metric) hedgehog. Here is another interesting example of non-separable metric space. Let A be a set and
be the Cartesian product of the copies E, of the real line E’
Then for each pair p = { p , I a E A} and q H ( A ) we define the distance by
=
= (-m,
00).
Put
{q, 1 a E A } of elements of
Then we can easily show that H ( A ) is a metric space. This metric space is called a generalized Hilbert space (with the index set A ) . If A is a countable set, then H ( A ) is the separable Hilbert space. If A is uncountable, then H ( A ) is non-separable. Baire’s zero-dimensional space is also an interesting metric space. Suppose A is a given set. We denote by N ( A ) the set of all sequences (a1,cy2,. . .) of elements ai of A. The distance between two points a = (a1, a 2 , .. .) and /3 = (p,,p2,.. .) of N ( A ) is defined by p ( a , p ) = l/min{k I ak#
pk}
if a # p,
p ( a , a )= 0
Then N ( A ) turns out to be a metric space called Baire’s zero-dimensional
112
VARIOUS TOPOLOGICAL SPACES
[IIIS
space (or generalized Baire’s zero-dimensional space) with respect to A. If A is the set of all natural numbers, then by mapping a = (a1,a2,. . .) E N ( A ) to the continued fraction 1 ff1+
a,+-
1
a , + ...
we know that N ( A ) is homeomorphic with the subset P = {x E E’ I x is irrational and >1} of El. Finally, let us give a method to introduce a metric into the collection of the closed subsets of a metric space. Let X be a metric space with a bounded metric p and 2x the collection of all non-empty closed sets of X . For E > 0 and C E 2x, let S,(C) = { p I p ( p , C) < E } . ~For C, K E 2x, we define p’(C, K ) = inf{&1 C C S,(K) and K C S,(C)}.
Then p’ satisfies the conditions of metric and is called the Hausdorff metric of Example III.10. Every subspace of a metric space X is also a metric space with the same distance as in X. Let X and Y be metric spaces with metrics p1 and p2 respectively; then their Cartesian product X x Y turns out to be a metric space with the metric function
which induces the topology of the product space. ‘Generally for a subset A of a metric space, we define its diameter by S(A)= sup{p(p,q)Ip,qEA}, and the distance between two sets A and B by p ( A , B ) = i n f i d x , Y > I x E A, Y E BI. ’There is another, interesting method to introduce a metric into the space of compact subsets of a metric space. See K. Borsuk [ 3 ] ,J. de Groot [ 2 ] .See also J. Flachsmeyer [ 3 ] .Let X be a regular space and ZX the collection of all non-empty closed subsets of X. For any . . . , Un}of open sets of X, we define a subcollection of 2x by finite collection {V,,
Let all such collections (Uj, . . . , U,,) form an open basis of 2x. Then 2x turns out to be a Tz-space. This topology is called the finite topology of 2X. An interesting result on this topology is due to J. Keesling [ l ] , who proved that continuum hypothesis implies that 2X is normal if and only if X is compact. Because 2x is compact for compact X, this result means that 2x is normal if and only if it is compact.
III.~]
113
METRIC SPACE
Now, let X i , i = 1 , 2 , . . . , be metric spaces with metric functions pi, i = 1,2, . . . respectively. Then
is a metric function such that pi S l/i. We denote by Xi the metric space with metric p:. Xiand Xi are easily seen to have the same topology. Now m
p’(P,4)=
(c p:(Pi, qil2)
1/2 9
p = (PI, P27 * . .> 4 = (41, 42,. . .) 7
7
i=l
rIy=l
gives a metric of the Cartesian product X i . The metric p’ induces the topology of the product space n;=lK. In other words, the metric space rI;=,X i with metric p’ is homeomorphic with the product space rI;=l Xiby the identity mapping.
Definition III.7. A mapping of a metric space X into a metric space Y is called a uniformly continuous mapping, if for every E > 0, there is 6 > 0 such that p ( p , p’) < 6 for p, p‘ E X implies p ’ ( f ( p ) ,f ( p ’ ) )< E in Y, where p and p’ denote the metrics in X and Y , respectively. This definition is a generalization of the well-known definition of a uniformly continuous function on a Euclidean space. The following proposition is also a generalization of the theorem in calculus: A continuous function defined over a bounded closed set A of En is uniformly continuous. C ) Every continuous mapping f of a compact metric space X into a metric space Y is uniformly continuous.’
Proof. Suppose E is a given positive number. Then for every point p of X , there is 6 ( p )> 0 such that
Since X is compact, we can cover X with finitely many of S8(p),2(p), say, s8(pi)/2(~i), i = 1, . . . , n. Put
J. de Groot [l] proved another interesting property for such a mapping f; namely f is topological on a Ga-set X ’ of X such that f ( X ’ )is dense GS in f ( X ) .
114
VARIOUS TOPOLOGICAL SPACES
[IIIS
s = min{;S(pi) 1 i = I , . . . , n )
we obtain
Therefore f is a uniformly continuous mapping. Example III.11. Proposition C) does not hold if we drop the compactness of X. For example, f ( x ) = tan x, 1x1 < 7~/2is a continuous mapping of I IT/^, ~ / 2 into ) E l , but it is not uniformly continuous.
Definition III.8. Let 9 be a filter of a metric space X. If for every there exists a point p ( ~E)X and a member F ( E )of 9 such that
E
>O,
then we call 9 a Cauchy filter. Let cp(A 1 >) be a net of a metric space X. If for every E > 0, there exists a point p ( ~E)X such that cp(A I >) is residual in S,(P(E)),then we call cp a Cauchy net. If every Cauchy filter of X converges, then we call X a complete metric space.
D ) If a Cauchy filter of a metric space has a cluster point, then it converges to the cluster point. Proof. The easy proof is left to the reader. E ) A metric space X is complete converges.
if and only if every Cauchy net of X
Proof. Note that every net derived from a Cauchy filter is Cauchy and the filter derived from a Cauchy net is Cauchy. The proof in detail is left to the reader.
111.51
115
METRIC SPACE
We can describe the condition of completeness in terms of ordinary point sequences as follows:
F) A metric space X is complete if and only { p i I i = 1,2, . . .} of points converges.
if every Cauchy sequence
Proof. In view of E), the necessity of the condition is clear because a Cauchy point sequence is a Cauchy net on the directed set N of the positive integers. To see the sufficiency, we suppose 9 is a given Cauchy filter of X. For n = 1,2, . . . , we choose points p ( l / n ) E X and F ( l / n )E 9 such that
Then { S , / , , ( p ( l / n )1)n = 1 , 2 , . . .} has f.i.p. because 9 does so. This implies that { p ( l / n )1 n = 1 , 2 , . . .} is a Cauchy sequence of points. For, let E be a given positive number; then choose no such that 2/n0< E. Then it is easy to show p ( l / n )E S E ( p ( l / n o )for ) every n 3 no by use of the fact that
This means that { p ( l / n )I n = 1 , 2 , . . .} is Cauchy. Therefore by the hypothesis, { p ( l / n ) }converges to a point p of X. Now we can easily see that 9 also converges to p proving that X is complete. The following theorem is often used to prove topological theorems and shows the significance of complete metric spaces. Theorem III.7 (Baire’s theorem). Let V,, n = 1 , 2 , . . . , be open dense subsets of a complete metric space X . Then nz=, U, is dense in X . Proof. Let q be a given point of X and V ( q )a given open nbd of q. Since U, is dense, there is a point p l E V ( q ) nU,. Since U , is open, we can choose > 0 such that E , < 1 and
Since U, is open dense, we can choose a point p2 E S E l ( p lf ) U, land such that e2 < and
E~
>0
116
VARIOUS TOPOLOGICAL SPACES
[IIIS
Repeating this process, we can choose points p l , p2, . . . of X and positive numbers E ~ E, ~ ., . . such that
Since (2) implies
( p , 1 n = 1 , 2 , . . .} is a Cauchy sequence of points. Hence by F) it converges to a point p of X. It follows from (3) that PE
fiqa
n=l
(4)
This combined with (2) implies that p E nx=, U,,. On the other hand, it follows from (4) and (1) that p E V(q).Therefore
holds for every nbd V ( q ) of a given point q of X. This means that
n:=,U,, is dense in X . Corollary. Let Un,n T,-space X . Then
=
1 , 2 , . . . , be open dense subsets of a locally compact
n:=,U,, is dense in X.'
Proof. The proof is quite similar to that of the theorem, so it is left to the reader. Let us turn to another important condition for a metric space. We shall later generalize Theorem 111.7 and the corollary to eech complete spaces. J. de Groot [3] established a new class of spaces called subcompact spaces including complete metric spaces and locally compact Tz-spaces as special cases and proved Baire's theorem for subcompact spaces. Generally, a topological space is called a Baire space if the intersection of countably many open dense sets is always dense.
111.51
117
METRIC SPACE
Definition III.9. Let X be a metric space. If for every E > 0 the open covering { S , ( p ) I p E X } contains a finite subcovering, then we call X a totally bounded metric space. G ) Every totally bounded metric space satisfies the 2nd axiom of countability. Proof. Using the condition of totally boundedness, for every positive integer n, we can choose Sl/n(py),i = 1 , 2 , . . . , k ( n ) , such that
Now, it is easily seen that {Slln(p;)1 i = 1, . . . , k ( n ) ; n = 1,2, . . .} is a countable open basis of X , and hence X satisfies the 2nd axiom of countability.
H ) A metric space X is totally bounded of X is Cauchy.
if and
only if every maximal filter
Proof. Let 9 be a maximal filter of a totally bounded metric space X. Suppose E is a given positive number. Then k
for a finite number of points p l , . . . ,pk. Now we can assert that F C & ( p i )for some F E 9 and i. For, if we assume the contrary, then for each i
( X - ,$,(pi))r l F f
0
for all F E 9.
Therefore
X
- S,(pi) E g,
i = 1, . . . , k
since 9 i s maximal (11.4.B)). But we also have
118
VARIOUS TOPOLOGICAL SPACES
[IIIS
which contradicts f.i.p. of 9. Thus F C &(pi) for some F E 9and i, which means that 9 is a Cauchy filter. Conversely, if X is a non-totally bounded metric space, then there is E > 0 such that any finite number of &-nbdsd o not cover X . Therefore
has f.i.p., and hence we can construct a maximal filter 9which contains Y a s a subcollection (11.4.A)). Now, we can show that 9 i s not Cauchy. Because, if F C S,(p) for some F E 9 and p E X , then
contradicting the fact that
Hence F f S , ( p ) for every F E 9 and p E X , proving that 9 is not Cauchy. Now, combining D) with the definition of a complete metric space and H), we obtain the following important theorem for metric space which shows that compactness of a metric space is a combination of completeness and totally boundedness.
Theorem III.8. A metric space X is compact if and only and totally bounded.
if it is complete
Example I11.12. All bounded subsets of a Euclidean space E n are totally bounded, and they are the only totally bounded subspaces of En.On the other hand the closed sets and only those are complete subspaces of En. Generally, we can prove that every generalized Hilbert space H ( A ) is a complete metric space (but non-compact). To show it we suppose
p(i)= {p!) I a E A}, i = 1,2, . . . , form a Cauchy point sequence of H ( A ) . Since for each 0 , {p!'I i = 1,2, . . .} is a Cauchy point sequence of E l , it converges to a point pa of E l , i.e.,
IIIS]
Let
119
METRIC SPACE
E
> 0 be given; then there is i, such that for every i, i’ 3 i,,
Hence for every finite subset A’ of A, we obtain
Letting i’+
03
in this inequality, we obtain
( c (p!’-
c&.
aEA‘
This implies that
Since X a E A(pj1”)’< +m, we get
X a E Ap i < +m,
i.e.,
On the other hand, (1) implies
Therefore { P ( ~I )i = 1,2, . . .} converges to p, proving that H ( A ) is complete. Let C * ( X )be the collection of all bounded continuous functions over a topological space X . We introduce into C * ( X ) the metric given in Example 111.9. Then we can easily see that C * ( X )is a complete metric
120
VARIOUS TOPOLOGICAL SPACES
[111
space. If X is the unit segment I = [0,1], then it follows from Weierstrass' theorem' that C * ( I ) (= C ( I ) )is also separable and therefore it satisfies the 2nd axiom of countability. However, it is not compact.
Exercise I11 1. Test separation axioms, compactness and axioms of countability for a cofinite space defined in Exercise 11.1. 2. In a T,-space X, every nbd of a cluster point of a given set A contains infinitely many points of A. The set of cluster points of A is closed. 3. A subset A of a topological space X is called a retract if there is a continuous mapping f from X onto A such that f(x) = x for x E A. Every
retract of a T2-space is a closed set. 4. A topological space X is T2if and only if every net of X converges to at most one point. 5. Let f be a continuous mapping of a topological space X into a T2-space Y . Then {(x, y ) I x E X , y E Y,f(x) = y } is a closed set of X x Y .
6. A topological space X is T, if and only if A = {(x, x ) 1 x E X } is closed in X x
x.
7. Let f be a continuous mapping of a closed set F of a normal space X into the n-dimensional cube I" = {(x,, . . . ,x,) I 0 C xi C 1, i = 1, . . . , n}. Then f can be continuously extended over X. 8. A TI-space X is normal if and only if for every finite open covering
{ U , ,. . . , U,} there is an open covering { V , ,. . . , V , } such that c U,, i = 1, . . . , k (or a closed covering or a covering by cozero sets satisfying the same condition). ( V is called a cozero set of X if there is a real-valued continuous function f on X such that V = { x E X 1 f ( X )# O } . ) 9. A T,-space X is normal if and only if for every finite open covering %, there is an open covering 'Ir with 7 f A< %. (This proposition shows that full normality is a natural generalization of normality. The proof is analogous to the last part of the proof of 3.E).)
'
For every f E C(1)and E > 0, there is a polynomial g such that p(f, g) < E . Therefore the set of all polynomials with rational coefficients forms a countable dense set of C(1).
1111
EXERCISE
121
10. Let a topological space X be t h e union of normal closed subspaces F,, . . . , Fk ; then X is normal. 11. Every F,-set of a normal space is a normal subspace (Yu. Smirnov's theorem). 12. Give examples of a space which is locally compact but not paracompact and a space which is paracompact but not locally compact. 13. Let S be a locally compact subspace of a T,-space. Then S is open in 3. Thus a subspace S of a locally compact T,-space X is locally compact if and only if S is t h e intersection of a closed set and an open set of X . 14. A real-valued continuous function defined over a compact space X is bounded and takes on its largest and smallest values at some points of X . 15. Let X be a compact space satisfying the 1st axiom of countability. Then
every point sequence of X has a convergent subsequence. 14. Let X be a topological space and
9a decomposition of X into closed
sets. If X is T , , then the decomposition space X ( 9 ) is T,. If 9 is upper semi-continuous and X is normal, then X ( 9 ) is normal. If 9is an upper semi-continuous decomposition into compact sets, then X ( 9 ) is T2, is regular or satisfies the 2nd axiom of countability provided X has the respective property. 17. A topological space X is compact if and only if it satisfies either one of
the following conditions: (i) every filter basis has a cluster point, (ii) every closed filter has a cluster point, (iii) every maximal closed filter converges. 18. A topological space X is compact if one of the following conditions is satisfied. (i) There is an open base % of X such that if %' is a covering of X satisfying %' C %, then %' has a finite subcovering. (ii) There is a closed base 59 of X such that if Ce' is a subcollection of Ce with f.i.p., then fl Ce' # 0. 19. Let X be a compact T2-space.Then it is totally disconnected if and only
if every point has a nbd base consisting of clopen sets. (To prove 'only if', show that Q(x) = K ( x ) = {x} for every x E X.) 20. We add a new point po to R,in Example 11.5 and define a nbd basis "u(p,,)
of Po by
122
VARIOUS TOPOLOGICAL SPACES
[III
Then the thus obtained space R:, which contains R, as a subspace, is separable but satisfies neither the 1st axiom of countability nor the Lindelof property. 21. Let X be a Lindelof (separable) space andf a continuous mapping of X onto a topological space Y ; then Y is Lindelof (separable). 22. Each F, set of a Lindelof space is also Lindelof. 23. Each open subspace of a separable space is separable. 24. Every separable paracompact space is Lindelof.
25. Let X be a topological space satisfying the 2nd axiom of countability and P a property for closed sets of X such that if F, 3 F,3 * and each F, satisfies P, then F, also satisfies P. Then there is a minimal closed set which satisfies P. (Brouwer’s reduction theorem.)
n;=l
26. S ( A ) and N ( A ) in Example 111.9 are both complete. 27. A normal space X is called perfectly normal if every closed set of X is an F,-set. A normal space X is perfectly normal if and only if every closed set F of X is a zero set, i.e., there is a real-valued continuous function f on X such that F = {x E X I f ( x ) = 0). Every metric space is perfectly normal. Every perfectly normal space is hereditarily normal. 28. C ( I )in Example 111.12 is not locally compact. 29. Assume that every totally bounded subset of a metric space X is compact. T h e n X is complete.
30. Let f be a uniformly continuous mapping from a totally bounded metric space X onto a metric space Y. Then Y is totally bounded. Does the same hold for completeness? 31. Every non-empty open subset of a Baire space is of second category. 32. Let f be a mapping of a complete metric space X into itself, satisfying p ( f ( p ) ,f ( 9 ) ) S A p ( p , 9) for a definite positive number A with A < 1 and any
1111
EXERCISE
123
pair p, q of points of X. Then there is one and only one point po of X such that f ( p o ) = p o . (S. Banach’s theorem. Consider a point sequence { p ,f ( p ) ,f ( f ( p ) ) ,. . .}, noticing that f is continuous.) 33. A TI-space X is hereditarily normal if and only if for any two sets A and B satisfying A f l B = 0,A f l B = 0,there are open sets U and V such that U 3 A, V 3 B,U f l V = 0.A TI-space X is hereditarily normal if and only if each open set in X is normal.
34. Let f be a closed continuous mapping from a normal space X onto a topological space Y. Then Y is normal. 35. A topological space X is called extremely disconnected if the closure of every open set of X is open. Every extremely disconnected T,-space is totally disconnected, but the converse is not true. 36. Let F and G be disjoint zero sets of a topological space X. Then there is a real-valued continuous function on X such that f ( F )= 0, f ( G )= 1, 0cfsl.
CHAPTER IV
COMPACT SPACES AND RELATED TOPICS
In Section 4 of the last chapter, we learned some elementary properties of compact spaces. The purpose of the present chapter is to give an account of more advanced theories on compact spaces.
1. Product of compact spaces
r,
be compact Theorem IV.l (Tychonoff's product theorem). Let X,, y E spaces. Then their product space X = { X , 1 y € r } is also a compact space.
n
Proof. We denote by r, the projection of X ontoX,. Namely,
Suppose 9 is a given maximal filter of X. Then we can assert that for each y E
r
9,= { r , ( A )I A E 9) is a maximal filter of X,. First it is clear that 0sf sy. Secondly, if B 3 .rr,(A) for some A E 9 in X,, then r i 1 ( B 3 ) A in X , which implies r ; ' ( B ) E 9. Hence
B
=
Thirdly, if B
B
fl
r Y ( r y l ( B )E) 9,. =
r,(A), B'= ry(A') for some A, A ' € 9, then
B'> r,(A nA').
Since A fl A ' € 9, and hence r , ( A n A') E 9,. this implies (as has just
IV. I]
PRODUCT OF COMPACT SPACES
125
been proved)
Therefore 9,is a filter of X,. To show that 9,is maximal, we suppose B is a subset of X, intersecting every member of 9,. Then n-;'(B)is a subset of X intersecting every member of 9. Hence by 11.4.B), n-il(B)E 9. Therefore
B
=
n-y(n-;l(B)) E 9, .
Applying again II.4.B) to 9,we , conclude that 9,is a maximal filter. Now, since X , is compact, by Theorem 111.5, 9,converges to a point 9, E X,. Then we can assert that
To show it, suppose U is a given nbd of q. Then we can find a nbd U ' of q with U ' C U and of the form k
U'=
n u,,
x
nix, 1 y z 'yi, i = I , . . . , k ) ,
i= 1
where each Uyiis a nbd of q,, in Xyi.Note that U ' can be expressed as
On the other hand, it follows from 9,+ ,qy, that U,, E 9,; Hence U,, = n-,,,(B,)for some B, E 9.Since n-;,'(U,,)3 B,, we obtain n-;,'(U,,>E 9. Hence, in view of (l),we conclude that U' E 9. Therefore 9converges to 9. Thus, by Theorem 111.5, we can conclude that X is a compact space. Corollary. A topological space X is compact T2 if and only if it is homeomorphic to a closed set of the product of copies of the unit segment [O, 11. Proof. Combine Theorems 111.1 and IV.l.
126
pv.1
COMPACT SPACES AND RELATED TOPICS
Example IV.l. The Hilbert cube I"' is a compact metric space, because it is the product of countably many closed segments which are compact. We can prove Theorem IV.l using nets instead of filters. Suppose n-(A I >) is a given maximal net of X . Then for each y E cp,, ( A 1 >) = (n-,op)(A 1 >) is easily proved to be a maximal net of X,. Therefore cp,+qy for some point q, of X,. Now, it is also easy to prove that cp ( A 1 >)+ q = { q,, I y E r }in X . This means that X is compact by virtue of the corollary to Theorem 111.5.
r,
Theorem IV.2. For every completely regular space X , there exists a compact T,-space p ( X ) such that (i) X is a dense subset of p ( X ) , (ii) every bounded real-valued continuous function over X can be extended to a continuous function over p ( X ) . Moreover, such a space p ( X ) is uniquely determined by X in the sense that i f p ( X ) and p ' ( X ) are compact T,-spaces satisfying (i), (ii), then p ( X ) and p ' ( X ) are homeomorphic by a topological mapping which leaves invariant every point of X . Proof. By Theorem 111.1, X is homeomorphic with a subspace X ' of the product space P = n{(I,I a E A } of the copies of the unit segment [0, 11, where {f, I a E A } is the totality of the continuous functions over X such that 0 zs f , zs 1. Since X and X ' are homeomorphic, we may identify them. Namely we identify each point p of X with { f , ( p ) I a E A } of X ' . Thus we can regard X as a subspace of P. (See the proof of Theorem 111.1.) By Theorem IV.l, P is a compact space and hence is also a compact space since it is a closed subset of the compact space P. Putting = p ( X ) we obtain the desired space. (Note that p ( X ) is also T2 and thus it is a compact, fully normal space by III.3.F).) First of all, (i) is obviously satisfied by X and p ( X ) . To see (ii) we assume f is a given bounded continuous function defined over X . Then for appropriate positive numbers, E and k , 0 s ~ (+fk ) s 1 holds, i.e.
x
for some a E A . At every point p = {pa 1 a E A } of X , f, takes the value f , ( p ) = p,. Hence putting g,(p) = p, for every point p = { p , 1 a E A } of = p ( X ) , we obtain a continuous extension g, of f, over p ( X ) . Then (l/&)g,- k is a continuous extension of f over p ( X ) , which proves (ii).
x
IV. I ]
PRODUCT OF COMPACT SPACES
127
Finally to show the uniqueness of p ( X ) , we assume that P ’ ( X ) is a given compact T,-space which satisfies (i), (ii). Let f , be a given continuous function over X such that 0 s f a s 1. Then by (ii) there is a continuous extension cp, of f , over p ’ ( X ) . Since X = p ’ ( X ) by (i), it is clear that 0 s cp, zz 1. Therefore
is a mapping of p ’ ( X ) into P which leaves each point of X fixed. It follows from the continuity of cp, that cp is a continuous mapping. Hence
On the other hand, since p ’ ( X ) is compact, ( p ( p ‘ ( X ) is ) also compact by III.3.B), and hence by III.3.C) it must be a closed subset of p ( X ) containing X . Thus from
it follows in p ( X ) that
Since X
=
P ( x )we , get
To prove that cp is one-to-one, we note that by (i) the continuous extension mentioned in (ii) is uniquely determined (Exercise 11.31). Suppose that p’ and q’ are distinct points of p ‘ ( X ) . Because P ’ ( X ) is completely regular, there is a continuous function (1 over p ’ ( X )such that (1(p’)=O,
( 1 ( q ’ ) =1 and
OS(1Sl.
We denote by f,, t h e restriction of (1 to X . Then as noted in the above, the continuous extension p‘, of f,, over p ’ ( X ) is unique and hence = (1. Therefore qC ,l
This implies that cp(p’)# cp(q’), and hence cp is one-to-one. Thus it
128
COMPACT SPACES AND RELATED TOPICS
pv.1
follows from I11.3.D) that cp is a homeomorphism of p ' ( X ) onto p ( X ) , proving the theorem. Corollary 1. Let f be a continuous mapping of a completely regular space X into a compact T,-space Y. Then f can be extended to a continuous mapping of p ( X ) into Y, and the extension is unique.
Proof. Since Y is completely regular, by Theorem 111.1, we can imbed it in the product space P of unit segments I,, a E A. Therefore we may represent the mapping f by
where each fa is a real-valued function satisfying 0 S f a S 1. Since f is continuous, so is f,. Therefore, by (ii) of Theorem IV.2, we can extend f a to a continuous function cp, over p ( X ) . It follows from p ( X ) = r7 that 0 s cp, S 1 and hence
is a continuous mapping of p ( X ) into P. On the other hand, since cp is continuous, we obtain in P
cp(P(X))=& ? ) C c p ( X ) = f ( X ) C
Y =Y
(Note that Y is compact and therefore closed in P.) Thus cp is the desired continuous extension off over p ( X ) .We can also easily prove the uniqueness of the extension using the fact that p ( X ) = X.
Corollary 2. Let B ( X ) be a compact T,-space in which X is a dense subset. Then there is a continuous mapping g of p ( X ) onto B ( X ) which keeps every point of Xfixed and maps p ( X )- X onto B ( X )- X . Proof. Let us denote by f the identity mapping which maps X C p ( X ) onto X C B ( X ) , namely f ( p ) = p for every p E X . By use of Corollary 1, we can extend f to a continuous mapping g of p ( X ) into B ( X ) . Since g is continuous, g ( p ( X ) ) is compact and therefore closed in B ( X ) . This combined with r7 = B ( X ) implies g ( p ( X ) )= B ( X ) . To prove g ( P ( X ) X ) C B ( X )- X , we suppose that p is a given point of p ( X )- X . Put
9 ={ U n X I U is a nbd of p in p ( X ) } ;
IV. 11
PRODUCT OF C O M P A m SPACES
129
then 9 is a filter basis in p ( X ) converging to p . Since g is continuous, by 11.6.B),
Observe that g ( 9 ) = 9 because each element of 9 is a subset of X on which g = f. Thus 9+g ( p ) in B ( X ) . Since 9 converges to n o point of X , we conclude that
which proves the assertion. Definition IV.l. The compact T,-space p ( X ) obtained in Theorem IV.2 is called the cech-(M.H.) Stone’s compactification of X . A) Assume that X is a non-compact Tychonoff space. Then every pointy of
p ( X )- X has no countable nbd base. Proof. Assume to the contrary that y has a countable nbd base U,, U,, . . .. Then we may assume
U i + l ~ ~i =,l , 2 , . . . . Observe that U, f l X is infinite for every i, because y E .%. Thus we can select two point sequences
such that P
n Q = 0 and
Then we can find open nbds C (Uni -
V, of pi such that
Uni+, U Q) n X .
Since X is Tychonoff, there are continuous functions f i from X into [0,1] such that
130
COMPACT SPACES AND RELATED TOPICS
f,(p,)=l,
[IV.1
f,(X-V,)=O.
Then put m
f=Ch. i= 1
Observe that {V, I i = 1,2, . . .} is mutually disjoint and locally finite in X. Thus it is easy to prove that f is a continuous function from X into [0,1]. (The detailed proof is left to the reader.) It is also obvious that
Note that p i + y and 9 , + y in p ( X ) . Thus f cannot be extended over p ( X ) , contradicting the property of p ( X ) . Thus y has no countable nbd base.
Example IV.2. As indicated by A), p ( X ) is not 1st countable even if X is a metric (non-compact) space. Generally speaking, the structure of p (X) is not simple even if X is such a simple space like the real line. A rather exceptional simple example is p(R,) = R,. (See Example 111.6.) It is obvious that R, is a compact T,-space containing R , as a dense subspace. On the other hand, it is easy to show that for every real-valued, bounded, continuous function f defined on R,, there is a E R, such that f ( x ) = f ( a ) for all x E R , with x
2 a.
Thus we can extend f to the continuous function cp over R, by defining that
Therefore R, is the tech-Stone’s compactification of R,. On the other hand, (0, 1) is a dense subset of [0,1], but by A) the latter is not the tech-Stone’s compactification of the former, because [0,1] is 1st countable. Similarly, the 2-dimensional sphere S 2 is not p ( E 2 ) ,though S 2 is a compact T2-space in which E 2 is dense. (See Example 11.12.)
Example IV.3. In Example 111.3 we showed that R, is completely regular but not normal. On the other hand, p ( R 4 )is compact T2and therefore fully normal. Thus a subspace of a fully normal space does not need even to be normal.
IV. 13
PRODUCT OF C O M P A e SPACES
131
We can prove the uniqueness of P ( X ) in a slightly different way from the proof of Theorem IV.2 as follows. Assume that X and X’ are homeomorphic and that P ( X ) and P’(X’)are T,-compactifications of X and X‘, respectively, satisfying (i) and (ii) of Theorem IV.2. Let f be a homeomorphism from X onto X’ and i and i’ the identity mapping of X and X’, respectively. By Corollary 1 we extend f and f-’ to cp and cp’, respectively, where cp is a mapping from p ( X ) into p’(X’) and cp‘ from p’(X’) into p ( X ) . Then cp’ocp is a continuous extension of i, and hence cp’o cp = j, where j is the identity mapping of P ( X ) .In a similar way cp 0 cp’ = j ’ , the identity mapping of P‘(X’). Thus it is easy to see that cp is a homeomorphism from p ( X ) onto p ’ ( X ’ ) such that cp = f on X .
Example IV.4. We gave in Example 111.2 a regular space R; which is not completely regular. The following classical example due to A. Tychonoff [ I ] is much more complicated but interesting in its own right. We consider R, in Example 111.5 and define
RA = (0, 1 , 2 , . . . , w , } , where wo denotes the first countable ordinal number, and topology of RA is the order topology. Let
1 0 s a < ao,0 =S P 0, Po> 0, where
= (a,, Po)E
P
It is easily seen that P is a compact T,-space because both of R, and RA are compact T2. Therefore P is regular. Put
P’ = P - ( w , , w,) ; then P’ is also a regular space. To proceed with the construction of the desired space, we need the following properties of P‘.
132
[IV.l
COMPACT SPACES AND RELATED TOPICS
(a) For each a with a < w , and n with n < wo, we put
If U is an open set of P' containing some XL, then U contains some Y ; . Proof. Since U is open, for each x such that
Hence for each
5 with
a
=
(5,wo)E XL, there is a nbd
UC,(x)
< 5 <w1
Since each nz is an integer, there is some n such that nf = n for uncountably many 5, i.e. { 5 I nf = n } is cofinal in { 6 I 0 S 5 < w , } . Then it is easy to see that YAC U. (b) If U is an open set of P' containing some YL, then X A.
u contains some
Proof. Since U is open, for each x = (al, 7 ) E YL, there is a nbd UoIq7.(x) C U. Hence
Then it is easily seen that XAC countable which implies a < w , .
u, because
YA and hence {a,} is
Now, we denote by R ' the union of countably many copies PIk, k = 1, 2, . . . , of P'. We consider { P r kI k = 1,2,. . .} as forming a discrete open covering of R'. Then R' is a regular space. We denote (5,v ) , XA, YL in PIk by (6, T ) ~Xa', , Yt', respectively. We consider a decomposition 9 of R' consisting of
IV. 11
PRODUCT OF COMPAm SPACES
133
and the other points of R'. Consider the decomposition space
R = R'(9). Since 9 is easily verified to be upper semi-continuous, by Exercise 111.16, R is a regular space. Let us denote by f the natural mapping of R' onto R and put, for brevity,
Now, we construct a space R , by adding a point y o to R. For each positive integer 1, we put
Then we define that {U,(yo)1 I = 1, 2, . . .} is a nbd basis of yo in R,. We can easily verify that R3 is a regular space. But, now, we can assert that R, is not completely regular. For, if R , is completely regular, then we can construct cech-Stone's compactification P(R3) of R , and assert that P(RJ 2 R,, because R , is easily seen to be non-compact. (Generally, for a given completely regular space X , X = P ( X ) holds if and only if X is compact.) Hence we can select
Since P(R,) is regular, so is
and hence there are nbds U N ( y o )of yo and V(zo)of zo in Ro such that
134
COMPACT SPACES AND RELATED TOPICS
[IV.1
We may suppose without loss of generality that N is odd. Let us continue our discussion in R,. We choose a sequence { V,(z,) I m = 1, 2, . . .} of nbds of zo such that
then it is clear that
and each W,,, is an open set of R, containing yo. From (l), (2) it is also clear that
which implies
w ,3 X : + l , because
PN+I3 X:+l . Putting a, = 0, we express this relation as
Starting here, we proceed with an inductive argument although the detailed proof is left to the reader. By (a) this implies that
W ,3 Y:'
for some n 1 .
Combining this with (6) and (4), we get
Since N is odd,
IV. 11
PRODUCT OF COMPAm SPACES
135
Ygtl = Y:,
follows from the definition of the decomposition 9. Therefore
By (b) this implies that
W21xr2
for some a2
Hence
w33x
p Y;,
follows from (4). Since by the definition of 9,
Xt2= x
y,
we obtain
w,3 x;-1 . Now, we can apply the same discussion to W, and X:-' as the one applied t o W , and X F ' ; then we get ng such that
w'Jx;-:-'u yZ-1.
(9)
Repeating this process we get aN and nN+,for which
In view of (4), (7), (8), (9), (lo), we know that for every k with 1 S k s N,
W N + 2X3& , U Y:(k) for some a ( k ) and n ( k ) . (3), ( 5 ) it follows that On the other hand, from (2), (T),
(Note that (2), ( T ) ,(3), (5) imply
(11)
136
COMPACT SPACES AND RELATED TOPICS
[IV.2
Let k be a natural number with 1 < k S N . Recall that W,+, is an open set of R, containing yo. Then ( R ,- W,+,) f l P k is homeomorphic with P ' - W, where by (11) W is an open set of P' such that
On the other hand, PI- W = P - [ W U { ( W , , W , ) } ] . Modifying the proof of (b), we can derive from (13) that W U { ( w ' , w,)} is w,,)}] and accordingly an open set of P. Since P is compact, P - [ W U {(q, P ' - W are compact. Thus ( R ,- W,,,) f l Pk is compact. Therefore from (12) it follows that
is a compact subset of R, containing VN+,(z,,) fl R,. Thus
zo E VN+2(z,) fl R, C Z?
=
K
in R,. On the other hand,
which is a contradiction. Thus R, is not completely regular.'
2. Compactification It is an interesting problem to find a compact space X * for a given topological space X such that X is homeomorphic with a dense subset X ' of X * . Generally, such a space X * is called a cornpactification of X . The concept of compactification appears now and then in fields other than ' A s a matter of fact, there is a regular space in which every continuous function is constant! See J. Novak [l], E. Hewitt [l] and A. Mysior [2].
IV.21
COMPACTIFICATION
137
general topology. In fact, we are familiar with a compactification in the theory of functions, where the complex number plane with the point at infinity is a compactification of the plane. We have already studied a compactification p ( X ) in the preceding section. The purpose of the present section is to develop in further detail investigations on compactifications. Now, let us suppose X is a given topological space; then we consider a space X * = X U { p } adding a new point p to X . If we define that p has the only nbd X * , then X * turns out to be a compact T-space containing X as a dense subset. Thus we can easily obtain a compactification for every topological space, but such a compactification is not even T I .What we are concerned with is compactifications with more interesting properties like the tech-Stone’s compactification. Definition IV.2. Let X be a given topological space. Add a point of infinity to X and put a ( X )= X U {w}. We define a topology of a ( X ) as follows. Each point x E X has the family of all nbds in X as a nbd base in a ( X ) . On the other hand, {U 100 E U, X - U is a compact closed set of X } is defined as a nbd base of in a ( X ) . Then it is easy to see that a ( X ) is a compactification of X . We call a ( X ) Alexandrofs one-point compactification of X . A) a ( X ) is T2 if and only if X is locally compact and T2. Proof. The ‘only if’ part is easy and is left to the reader. Let X be locally compact and T2.Suppose x E X . Then there is a compact nbd G of x in X . Then a ( X ) - G is a nbd of which is disjoint from G. Hence a ( X ) is T2. Next, we are going to discuss a general method of H. Wallman and N.A. Shanin to construct compactifications. For that purpose we generalize the concept of filter and its convergence which we originally defined in Definitions 11.9 and 11.10. Definition IV.3. Let 93 be a collection of subsets of a topological space X such that B, C E 93 implies B fl C E 93. A subcollection 9of 93 is called a B-filter if (i) 0 @9, (ii) ifB E 9 and B C C E 9, then C E 9, (iii) if B, C E 9, then B fl C E 9.
138
COMPACT SPACES AND RELATED TOPICS
pv.2
A 9-filter 9 is called maximal if for every 93-filter Ce satisfying 92 3 9, it holds that Ce = 9.
B ) Every 93-filter 9is a base of a filter. Thus 9 converges to a point x of X if and only if for every nbd U of x there is B E 9such that B C U. A point y of X is a cluster point of 9 if and only if y E F for all F E 9. Proof. 9 is obviously a base of the filter 9’= { Y 1 Y is a subset of X containing some member of 9).Thus the rest of t h e proof follows from the definition of convergence and cluster point of a filter base given after II.4.C). Example IV.5. If 93 is the collection of all closed sets, then a 9-filter is called a closed filter as already defined in IV.4. If 93 is the collection of all zero sets of X , then we call a 9-filter a zero-filter. We can generalize some basic propositions previously proved for filters as follows. (The proofs are left to the reader.)
C ) A %-filter 9 is maximal if and only if it satisfies the following then B E 9. condition: If B E 93 and B n F # 0 for all F E 9, D) Let 9’be a subcollection with f.i.p. of 9.Then there is a maximal 93-filter Ce such that Ce 3 9’. Definition IV.4. Let 93 be a closed base of a TI-space X . Then 9 is called a TI-base if it satisfies (i) 0E 93, (ii) B,, B, E 9 implies B, n B, E 9 and B , U B, E 9, (iii) if x fZ B E 9,then there is B ’ E 9 such that x E B’, B’nB = 0. If, moreover, 9 satisfies the following condition, then we call it a normal base: (iv) if B,, B, E 9,and B , n B, = 0,then there are B ; , B; E B such that B, n B ; = 0,~ , B;= n 0, B ;u B;= x. Theorem IV.3. Let X be a TI-space and 9 a TI-base of X . Then let a ( X , 93) = the set of all maximal .%-filtersof X . For each B E 93 we define
IV.21
COMPACTIFICATION
139
Introduce a topology into u ( X , 93) by taking 6 as a closed base. Then u ( X , 93) is a TI-compactificationof X satisfying (i) {BI B E B} is a closed base of u ( X , a), (ii) if B,, B,E 93, then B,n B,= B,n B,, where, both in (i) and (ii), B denotes the closure in u ( X , 93). Furthermore, u ( X , 93) is T, if and only i f 9 3 is a normal base. Proof. We shall prove the theorem step by step as follows. (1) B, U 6, = (B, U B2)-and B, f l B, = (B, n B2)-hold for every B,, B,E 93. Let us prove only t h e first equality. B, U B2C (B, U B2)-is obvious. Assume 9sf B, U B,; then B,fZ 9 and B, sf 9. Thus, by C), there are F,, F, E 9 such that
F, f l B,= 0,
F 2 n B,= 0.
Hence
Fl f l F, E 9 and (F, n f’,)n (B, u B,)= 0 Therefore again by use of C), we obtain B,U B,Sf 9, i.e. 9Sf (B, U B2)-. This proves B, U B23 (B, U B2)-and eventually (1). (2) 8 satisfies the condition for a closed base given in II.2.C). This is obvious; in fact it satisfies a stronger condition,
u(X,93)=2€6, 0 E 6 , B, u B, E for every B,, B2E 6. Thus u ( X , B ) turns out to be a topological space with the topology defined by the closed base 6. (3) u ( x , 93) is a TI-space. Let 9,Ce E u(X,a), 9Z Ce. Then there are F E 9, G E Ce such that F n G = 0. Now, P E 6 satisfies 9E fi and Ce sf which implies that a ( X , 93) is TI. Define a mapping cp from X into u ( X , 93) by (4) cp(x)= PX= {BI x € B E B } . It follows from (iii) of Definition IV.4 that PXis a maximal ,%-filter. We can show that cp is a topological imbedding. Let x and y be two distinct points of X. Then, since X is TI,there is B E 93 such that x 5Z B 3 y. Now
140
COMPACT SPACES AND RELATED TOPICS
pv.2
which proves p(x) # p(y), i.e. cp is an injection. Let B E 4. Then
Hence cp-'(cp(X)f l B) = B. Thus by Exercise 11.7 (p is continuous. Similarly we can show that cp is closed. Thus cp is a topological imbedding. From now on we identify p(X) with X to regard X as a subspace of u(X 9).Note that then B f l X = B for every B E 93. (5) B = B for every B E 9,where B denotes t h e closure in u ( X , 93). By Exercise 11.7,
This proves that u ( x , 93) satisfies the required conditions (i) and (ii). Thus we obtain = X = u ( X , 93). (6) Hence X is a dense subset of u ( X , 93). (7) u ( X , 9)is compact. Let B f C 9 and suppose that I B E %'} has f.i.p. Then it follows from (1) that a f has f.i.p. By D) there is a maximal 9-filter 9 such that 9 ' c 9. Then 9 E ndf holds in u ( X , 3).Thus by Exercise 111.18 u ( X , 93) is compact. Namely, u ( X , 93) is a TI-compactification of X satisfying (i), (ii). Assume, furthermore, 9 is a normal base. Then for any two distinct points 9, % of u(X, 9)we find B , E 9, B, E % such that B , f l B, = 0. By (iv) of Definition IV.4, there are B ; , B ; E % such that
a r ={a
B , n B ; = 0, Then B ; , B;E
B , n B;= 0,
B ; u B;= x.
4 satisfy
i.e. a ( X , 93) (denoted by u in the above) is T,. On the other hand, assume that u(X, 9) is normal, and B,, B,E 93, B , n B, = 0. Then B,IIB,= 0. Choose open sets U,, U, of u ( X , 9) such that
B,cu,,
B,cu,,
There are B ; , B ; E 93 such that
u,nu,=0.
IV.21
141
COMPACTIFICATION
B,n B; = 0,
B; 2 ( ~ (B) x ,- U , ,
B2n B;= 0 ,
E ; 3 a ( X , 9)-U,. Then it is obvious that condition (iv) of Definition IV.4 is satisfied by B,,
B,,B;,B;. E) u ( X , 93) is unique. Namely, ifa’(X,93) is another TI-compactificationof Xsatisfying (i), (ii) of Theorem IV.3, then there is a homeomorphism between cr(X, 93) and d ( X , 93) which leaves every point of Xfixed. Proof. Let x E a ( X , 93); then x can be regarded as a maximal 9-filter and accordingly as a filter base in d ( X , 93). Hence it has a cluster point y in a’(X,9), because cr’(X,93) is compact. Thus (1) y E fl {F 1 F E x } , where denotes the closure in a(X,9). By condition (i) of Theorem IV.3 we have (2) iY } = n{BI E B,B E Consider any B E 9 such that y E B. Then for any F E x we obtain F n B = F n B # 0,because of (ii) of the theorem. Thus F n B f 0,which implies that B E x. Hence, from (1) and (2) it follows that
a}.
{ y } = n{F1 F E x }
.
Now, let U be a given nbd of y in a‘(X,93).Then there is B E 93 such that y 6Z B 3 o’(X, 93)- U. Hence there is F E x such that F fl B = 0, because otherwise F n B # 0 for all F E x , and accordingly B E x follows. This implies y E B,a contradiction. Thus F C U. Namely x, as regarded as a filter base, converges to y in d ( X , 93). Now, we define a mapping cp from a ( X , 93) into d ( X , 93) by
It is obvious that p leaves each point of X fixed. Assume that x, x’ are distinct points of a(X,93). Then there are F E x , F ’ E x ’ such that F n F’ = 0. Hence F n F’ = 0, where closure is taken in o’(X, 9).Since cp(x) E F and cp(x’) E F’, we obtain cp(x) # cp(x’), i.e. cp is one-to-one. Next, cp is onto. Because, let y E d ( X , 93); then by (i) of the theorem
where B“’denotes the closure in a’(X 93). Let 9 ={ B E 93 1 y E B”’}.
142
COMPACT SPACES AND RELATED TOPICS
[IV.2
Then by (ii) of the theorem 9is a 9-filter. If B E 9 satisfies B n F # 0 for all F E 9, then y E B"', i.e. B E 9. (Because otherwise from the compactness of a ' ( X , 93) it follows that c"'nB"' = 0 for some C E 93 with yE i.e. C E 9. Thus C n B = 0, a contradiction.) Hence 9 is a maximal %-filter by C). Namely, 9 can be regarded as a point, say x, of u ( X , 9).Now, it is obvious that ~ ( x=)y , which proves that cp is onto. It is also easy to see that cp is closed and continuous. Because for each B E 9 it holds that
c"',
which proves that cp is closed and continuous (Exercise 11.7). To prove (3), let x E B" = B (the last symbol from the theorem). Then B E x, and hence cp(x) E B"'. Thus cp-'(B"')3 B". Conversely, if x 6! B" = B,then F n B = 0 for some F E x. Hence p(x) E F"' c a ' ( X , 9) - B"' ,
proving
cp-'(B"')C B". Thus cp
is the desired homeomorphism.
Definition IV.5. a ( X , 93) is called Shanin's (or Wallman-Shunin's) compactification of X with respect to 93. If 93 is the collection of all closed sets of X , then a ( X , 9)is denoted by w ( X ) and called Wallman's compactijication of x.'
Example IV.6. Let X be a locally compact T,-space and 3 = {B I (i) B is a compact set of X , or (ii) B is a closed set of X such that B U B' = X for some compact set B'}. Then it is easy to see that 9 is a normal base of X . Let B be a closed set of a ( X ) such that me B. Then B satisfies (i) in the above, and thus B E 93 and B = B. Let C be a closed set of a ( X ) such that CQ E C. Suppose y E a ( X )- C. Then there is an open nbd U of y in X such that the closure 0 in X is compact. Then X - U satisfies (ii), and thus X - U E 3. It is easy to see that y f Z X - U 3 C in a ( X ) . Thus {BI B E 9}is a closed base of a ( X ) . Generally B E 3 satisfies B = B in a ( X ) if B satisfies (i), and B = B U {m} in a(X)if B satisfies (ii). Hence B,n B, = B, n B2 holds in a ( X ) whenever B,, B, E 93. Thus it follows from E) that a ( X )= u ( X , 93). I H. Wallman [l] defined w ( X ) , and N. A. Shanin [l], [2], [3] generalized Wallman's idea to define u ( X , a ) ) .In fact Shanin discussed his compactification under a more general
condition than we do here.
IV.21
COMPAmIFICATION
143
This relation does not hold in general. Suppose X , is a cofinite space of infinitely many points. Then 93,= the collection of all finite subsets of X , is the only Tl-base of X,. Now, for every F E % ~0 I,, F in a(X,). Thus 4, is no closed base of a(Xo).Namely, a(X,) is not Shanin's compactification of x,.
F) Let X be a Tychonoffspace and B the collection of all zero sets of X . Then p ( X )= u(X,%).
Proof. It suffices to prove that p ( X ) satisfies (i) and (ii) of Theorem IV.3. Let x E p ( X ) - F for a closed set F of p ( X ) . Then there is a zero set Z of p ( X ) such that F C Z " C Z Z 4 x in p ( X ) . Then Z f l X E %. and F
C
~
~
C inZp ( x ~).X
Thus condition (i) holds for p ( X ) and 2. To prove (ii), assume Z , , Z , E 3,Z , f l Z, = 0. Then there is a realvalued continuous function f on X such that
f(Z,)=O,
f(Z,)=l,
OGfGl
We extend f to a continuous function g on p ( X ) and put
z;= { x E P ( X ) 1 g ( x ) = O }
1
z;= { x E p ( x ) 1 g ( x ) = 1 ) .
Then z;nz;=0,
Z,cz;,
Z,CZ;,
nz,
where the closure is taken in p ( X ) .Thus 2, = 0. (Namely 2,f l 2, = Z , n Z , holds.) We shall use this observation to prove (ii) in general. Assume that Z , and Z , are elements of B which are not necessarily disjoint. Suppose x E 2,f l Z,, where here and in the following the closure is taken in p ( X ) . Let U be a zero-set nbd of x in p ( X ) . Then
Hence by use of the above observation we obtain
144
COMPACT SPACES AND RELATED TOPICS
[IV.2
Hence x E Z, f l Z,, which proves (ii). Therefore p ( X ) = a ( X ,2 ) follows from E).
G ) p ( X ) = w ( X ) holds if and only if X is a normal space.
Proof. If p ( X ) = w ( X ) , then p ( X ) = w ( X ) = a ( X , %), where % is the collection of all closed sets of X. Since p ( X ) is T,, by Theorem IV.3 % is a normal base of X, i.e. it satisfies condition (iv) of Definition IV.4. Hence X is a normal space. The converse can be proved in a similar way as the proof of F).' Example IV.7. Let us study some other aspects of w ( X ) . Generally we define (covering) dimension, dim X of a topological space X as follows: d i m 0 = -1 ; for a non-negative integer n, dim X zs n means that for every finite open covering % of X there is a finite open covering "1' of X such that "1' < %, and ord "1' = max {il there are i distinct elements of "y. whose intersection is non-empty} S n + 1. Then dim X = y1 means that dim X G n holds, but dim X s n - 1 does not. (dim X = if dim X s n does not hold for any integer n.) Then it is easy to see that dim w ( X )= dim X for every TI-space X. Thus dim p ( X )= dim X holds if X is a normal space. A lattice L is called distributive if it satisfies
a v (b A c ) = ( a v b)A ( a vc),
a
A
(b v c ) = ( a A b ) v ( a A c ) .
H. Wallman [l] constructed a compact TI-space A ( L ) from a given distributive lattice L with the smallest element 0 and the largest element 1 as follows. We call a subset P of L a maximaZ dual idal if (i) OtZ P, (ii) b > a E P implies b E P, (iii) a, b E P implies a A b E P. Then, let A ( L )
'
It is possible to discuss more general conditions in order that u(X, 93) = u(X, 93'). See N. A. Shanin [3].
IV.31
MORE OF COMPACTIFICATIONS
145
be the set of all maximal dual idals of L. The topology of A ( L ) is defined by use of the closed base ’3 = {ii 1 a E L}, where = { P E A ( L ) 1 a E P}. As a matter of fact, if L is the lattice of all closed sets of a TI-space X , then A ( L )= w ( X ) .Wallman proved that L is isomorphic to the lattice of a closed base of A ( L ) if and only if L satisfies the disjunction property: For every pair of distinct elements a, b of L, there is c E L such that one of a A c and b A c is 0 and the other is not 0. Thus every distributive lattice with 0, 1 and the disjunction property can be represented as a closed base of a compact T,-space. The same idea was originated by M. H. Stone [I], who proved that every Boolean algebra can be represented as a closed base of a totally disconnected compact T,-space, and conversely, the set of all clopen sets of such a space is a Boolean algebra, where a Boolean algebra is a distributive lattice with 0 and 1 satisfying the following condition: For every element a there is an element a’ for which a r \ a ’ = O , a v a ’ = l .
3. More of compactifications As we saw in the previous section, tech-Stone’s compactification p ( X ) as well as Alexandroff’s compactification a ( X ) (in case that a ( X ) is T,)
are special cases of Shanin’s compactification, which indicates that the method used to construct c+(X,3) is quite general. Thus the following question occurs: Is it possible to construct every T,-compactification 8 of every Tychonoff space X by use of that method? In other words, is it true that 8 = m(X, 3) for some normal base 3 of X?’ The purpose of this section is to discuss this problem, especially the partial positive answers due to J. M. Aarts [l], A. and E. Steiners [I] and C. Bandt [l], and the eventual negative answer due to V. M. Ul’janov [2]. All spaces in this section are at least T,. A) Let
2 be
a T,-compactification of a Tychonoff space X . Then
8=
u(X,3 )for some normal base 3 of X if and only if X has a closed base Ce such that (i) C,, C, E V implies C, U C,E V and C, f C, lE Ce, (ii) for every F E %, F = F f l X , where the closure is taken in 2.
’
0. Frink [I J constructed a compactification (probably without knowing Shanin’s work) by a method which is essentially the same as Shanin’s, and he asked the above question. (He called such a compactification ‘of Wallman type’.) So this question is sometimes called Frink’sproblem.
146
COMPACT SPACES AND RELATED TOPICS
[IV.3
Proof. Assume that 2 =a(X, 93) for a normal base B of X. Then % = {BJ B E B} satisfies the conditions. Condition (i) is obvious. To prove (ii), let F = B E V, where B E B. Then
FnX=BnX=B n X = B = F . Conversely, assume that % is a closed base of Then put
2
satisfying (i) and (ii).
It is obvious that B is a closed base of X satisfying the conditions (i), (ii) of Definition IV.4. To see (iii), let x E X - F n X, where F E V. Then, since 2 is compact and V is a closed base of 2,there are F,,. . . , FkE % such that
Then x € F, n . fl Fkfl X E 93, satisfying (iii). In a similar way we can prove (iv) of Definition IV.4, too, by use of normality and compactness of 2.Thus 3 is a normal base of X. Let B € B ; t h e n B = F n X f o r s o m e F E V . H e n c e
B =FnX=F,
a}
which means that {B 1 B E = %. Thus condition (i) of Theorem IV.3 is satisfied. Finally,' suppose B,, B, E 9; then
B , = F, f Xl , Hence
B, = F, f l X
for F,, F2E V .
B,n B2= F, n x n F2n x = F, n F~= F, n F2n x = B , n B,, ~~
proving (ii) of Theorem IV.3. Thus from E) it follows that
8 = a(X,93). Definition IV.6. Let F and G be subsets of a space X. Then
- -
d(F, G ) = F - G f l G - F .
IV.31
MORE OF COMPACTIRCATIONS
147
A closed base B of X is called a A-base if
A(EG1-0
foreveryF, G E W .
A compact T,-space X is called a A -space if it has a A -base. Example IV.8. Let I be a closed segment, say I = [0,d?]. Let B be the collection of all finite unions of disjoint closed segments in I with a rational left end and an irrational right end. Then it is easy to see that 99 is a A -base of I. Thus I is a A -space.
B) If X is a A -space and Y C X , then Y has a A -base. Proof. The easy proof is left to the reader.
C) If X is a A-space, then it has a A-base W satisfying (i) of A) and consisting of regular closed sets.
Proof. Let %? be a A-base of X . Put
W o= {BE % I B
=
F }.
Then W,is a closed base of X . Because, if x E X - F for a closed set F of X , then there is an open nbd U of X such that r? n F = 0. Pick B E W o such that X - U C B S x. Then obviously F C E S x and B"E 98,. Thus W, is a closed base. Since B0C %, W o is a A-base consisting of regular closed sets. Let W be the minimal closed collection satisfying (i) of A) and W 3 W,,. Then W is what we want. Obviously it suffices to prove the following: Let W ,be a A-base consisting of regular closed sets, then W 2= {A f l B, A U B I A, B E B l } is also a A-base consisting of regular closed sets. Now, assume A , B E W,. Then ( A U B ) " I I A " U B " =PUB"= A U B
Thus A U B is a regular closed set. Let X E F ~Then, F . since A ( A , B ) = 0 , either x e A - B or x bZ B - A holds. Assume, e.g., x E A - B. Consider any open nbd U of x such that U n (A - B) = 0. Then U fl A"# 0. Pick y E U fl A". To prove y E B", assume the contrary; then, because U n A" is a nbd of y , we have
148
[IV.3
COMPACT SPACES AND RELATED TOPICS
U r l A"f l ( X - B ) # 0. This implies that U n (A - B ) # 0, a contradiction. Hence y E A ' n B " , and thus U n A " f l B " #0. This proves x E A"f l B". Thus AnB
=
(A n B ) " ,
A"nB"CA"n=
proving that A n B is a regular closed set. Suppose A, B, C, D E B,. For brevity we denote X A', B', etc. Then A(A U B, C U D ) = A U B
-
-
A, X
-
B, etc. by
C U D fl CUD -A UB
=(AuB)nenocn(cuo>nAcnBc
=A
(A, C ) U A (A, D ) U A (B,C ) U A (B,D ) = 0.
In a similar way we can prove A(A n B, C f l 0 )= 0,
A(A U B, C n D ) = 0 .
Hence B2is a A-base consisting of regular closed sets.
Theorem IV.4. Let X be a A -space and Y a dense subset of X . Then X is a Shanin's compactification of Y. Proof. By C) X has a A-base Ce satisfying (i) of A) and consisting of regular closed sets. Let F E Ce and x E F. Then, since F is a regular closed then U n F" # 0,and hence set, x E F . Let U be a given open nbd of x;~U n F" f l Y # 0, because = Y. Thus x E F" n Y c F n Y . This proves that F C F fl Y ,i.e. F = F f l Y .Hence condition (ii) of A) is satisfied by C). Therefore, by A), X is a Shanin's compactification of Y. Corollary. If X is a compact T2-space with a closed base satisfying (i) of A) and consisting of regular closed sets, then X is a Shanin's compactification of its every dense subset.
IV.31
149
MORE OF COMPACTIFICATIONS
n
D ) Let I‘ = {I, 1 t E T } denote the product space of T copies of the closed segment I = [O, 11, where T is a cardinal number, and IT I = T . Zf U is an open set of 1‘, then there is a countable subset T’ of T such that for an open set w of I*‘ = Il {I,I t E T I } , v = w x I I (4 I t E T - T I }satisfies v c u and where closures are taken in IT.
v = u,
Proof. We assume U # 0, # X . Let us denote by 93, a countable base of I, such that I, E 93,. Let Bj E g I i ,i = 1 , . . . , k. Then we call the open set B = B,X X BkX n{I, I t f t,, . . . ,t k } of I’ a basic rectangle and denote it by B = (B,t,B,t,. . . Bktk).Assume B, # I,i for i = 1,. . . , h, and Bj = I , for i = h + 1 , . . . ,k ; then we define that
---
r
Denote by B t h e collection of all basic rectangles of I‘ and by the collection of all non-empty finite subsets of T. Let B E B and S C T ; then we denote by P(B, S ) the projection of B in I s . Further, define that
93’= { B E B 1 B c u,B
#
0).
Then U = U{B 1 B E B‘}. Let 9 ( S ) = {P(B,S ) 1 B E B’} for each S E
r.
Now, we choose a collection 2 of finite alternating sequences {S,P,S,P, . . . SkPk}of Sj E and P, E ?P(Si)such that S, r l Sj = 0 for i # j . First, observe that there is S , E r such that S , f l KB# 0 for all B E 3’ unless 0 = X . Fix an S , satisfying this condition. Then, let
r
9, = { P ( B ,S , ) 1 B E B’} . Observe that PI is at most countable. For each P , E 9,we proceed as follows. Put
%(PI) = { B E B’ 1 P(B, S , ) = PI}. (i) If there is S, E r such that S , n S, = 0, and S, n KB# 0 for all B E %(P,),then we fix such an S, and proceed in the same way as in the above (to complete the incomplete sequence {SIP&). (ii) Otherwise, there is a (at most) countable subset B,, of g ( P , ) such
150
[IV.3
COMPACT SPACES AND RELATED TOPICS
that for every B E B(P,) there is B, E B0satisfying
In this case we define that {SIP,}E 2 and put 9,= B(S,P,). This process must end with case (ii) after a finite number of steps, say k . Then we define that
and assign to it a (at most) countable subset 9((s,Pl.. . S,P,) of 9 ( P l .. . P,) = {BE 9'1 P(B, Si) = Pi, i = 1, . . . , k } such that for every B E 9(PI. . . P,) there is B,E 9(S,P, . . . S,P,) satisfying
(A detailed argument is left to the reader.)
Now, it is obvious that 2 is at most countable, and so is
then T' is countable. Put
W = U { P ( B ,T ' )I B E B f f }; then W is an open set of I T ' such that V = W X n{I, 1 t E T - T ' } = U ( B 1 B E Bff}C U .
v = u.
u.
Further, it is easy to see that To see it, let x E Then for each basic rectangle B containing x, B n B ' f 0 for some B'E 9'. Thus B f l B' = B " E 9'. We can select (S,P, . . . SkPk)E 2 such that P(B",S,) = P,, i = 1, . . . , k . Thus there is B,E B(S,P, . . . S,P,) C 9Iffsuch that
Thus B" f l B, # 0, which implies B n V #
u.
0. Hence x E
v,proving v
=
IV.31
MORE OF COMPACTIFKATIONS
151
E ) Every regular closed set of I' is a zero set.
Proof. It suffices to show that every regular open set U of I' is a cozero set. By D) there is an open set W of IT',where T' is at most countable, such that V = W X {I, t E T - T'} satisfies = 0. Since U is regular open, = U. Hence U = To. Since 7"is the product of an open set of the metric space IT' and 1 t E T - T'}, it is a cozero set. Hence U is a cozero set.
u"
n I n{I,
v
n{I,
Generally, let A be a subset of I' = 1 t E T } . If there are {x, I t E T }E A and {xi I t E T }sl A such that x, = xi, then we say that A depends o n the t-coordinate. The open set V in E) depends only on (at most) countably many coordinates, and so does U. Thus we obtain the following proposition. F ) Every regular closed set of I' depends only on (at most) countably many coordinates.' G ) Let B and C be non-empty closed sets of I' = I , x 12, where I;= [0, 11, i = 1,2, such that B = F x I, # I*,C # I,, A (B, C) = 0.Then C depends on the first coordinate.
Proof. Let xo = inf F. We assume xo> 0 and (xo, 0) jZ C, leaving check-up of the other cases to the reader. Put
(Observe that if there is no y for which (xo, y ) E C, then the proof is over.) Then yo > 0, and
Since (xo,y ) E B - C for every y < yo, (xo, yo) E B - C. Since A (B, C) = 0, this implies (xo, yo)6Z C - B. There is E > 0 satisfying
'
In fact, as we shall see in VII.1, a closed set o f I' depends only o n (at most) countably many coordinates if and only if it is a zero set.
152
COMPACI‘ SPACES AND RELATED TOPICS
Since (xo- E , xo)x { y o }n B
= 0,
[IV.3
(2) implies
Hence (xo- 4 2 , y o ) E C. Thus from (1) it follows that C depends on the first coordinate. Now, we can prove C. Bandt’s [l] first theorem.
Theorem IV.5. I‘ is no A -space whenever r 3 K,. Proof. Assume that I‘ is a A-space. Then by C) X has a A-space 93 consisting of regular closed sets. Let IT= {I,1 t E T}. Choose a subset T‘ of T such that IT’J= K,. Denote by 0 the point of I‘ whose all coordinates are zero and by H, the closed set of all points of I‘ whose t-coordinate is 1. Then for each t E T there is B, E 93 such that H, C B, 3 0. By F) every element of 93 depends only on countably many coordinates. Thus {B,I t E T’} depends only on TI‘ C T with JT”I<XI.’ Select t ’ E T - T“. Then for each t E T’, B, does not depend on the t’-coordinate. Consider a closed subset 1’ = { X , , I, t” E T } of IT,where X, = I,, X,. = I,,,X,,,= (0) for t“ # t, t‘. Then B = B, fl1’ and C = B,. f l I’ satisfy the condition of G). Hence C and accordingly B,. depends on the t-coordinate. Thus B,. depends on the uncountably many coordinates T ’ , which is a contradiction. Hence I‘ is no A-space.
n
H) Let X be a compact T,-space and {Bi I i = 1,2, . . .} a sequence of closed sets of X such that A (B;,B,) = 0 whenever i # j . Then for any disjoint closed sets D and E of X there is a closed set C such that D C C, C n E = O , A ( C , B i ) = O f o r i = 1, 2, . . . . Proof. Let h be a real-valued continuous function o n X such that
h(D)=O,
h(E)=l,
Ochcl.
Let gi be a real-valued function on X defined by g;(x) =
’
[Fi
if x E Bi, otherwise
Precisely this means that if x, y E I‘ have the same t-coordinate for every t E T” and x E €3,. for some r‘ E T‘, then y E €3,. .
IV.31
Put g
MORE OF COMPACIIFICATIONS
=
153
x;=, gi, and c={ x E X I g ( x ) 2 h ( x ) } .
(1)
Then C is the desired set. First, to see that C is a closed set, let x @ C. Then
follows. Select a natural number rn such that 3-" < e. Put
U = h-'(S,,*(h(x))- U {Bi 1 i G m , and x Sr B i } Then obviously U is a nbd of x. We claim that U n C = 0. To prove it, we suppose y E U. Then we have h ( y ) > h(x)- &/2.
(3)
Also note that x @ B, and i G rn imply y @ B,, and thus it follows that m
m
Hence from (4), (2), ( 3 ) it follows that m
m
g ( y ) S cgi(y)+3-"/2G i= I
gj(x)+3-"/2Sg(x)+3-"/2 i= I
This means that y @ C. (See (l).) Hence U f l C = 0, proving that C is a closed set. It is obvious that D C C and C n E = 0. Finally we can prove that A (C,B i )= 0 for each i. Assume the contrary, x E A ( C , B m ) # O forsome m .
(5)
We claim that there is a nbd U of x satisfying: (6) If y E U n ( C - B m ) , z E U n (B, - C), then {i E N I i < rn, Y E B i }C {i E N I i < m , z E B,}, where N denotes t h e set of natural numbers. Assume t h e contrary and define F ( U ) for each nbd U of x by F ( U )=
154
[IV.3
COMPACT SPACES AND RELATED TOPICS
{i E N I i < m , there are y E U n (C-B,) and z E U n (B, - C) such that y E Bi and z fZ Bi}. Then F ( U ) # 0 for every U follows from the k
assumption. Since F(fl,,lU,) cnt,F ( U , ) for any nbds U,, i = 1,. . . , k , of x , {F(U)I U is a nbd of accordingly a non-empty intersection. Hence there is i < m such that for every nbd U of x there are y E z E U n (B, - C) such that y E Bi and z fZ Bi. Thus
finite number of x } has f.i.p. and
a natural number U f l (C-B,) and
and thus
Namely x E A (Bi,B,) # 0, which is a contradiction. Hence there is a nbd U of x satisfying (1). We may assume without loss of generality that
Now, since x satisfies ( 5 ) , we can choose y E U n (C-B,) and z E U f l (B, - C). Then from (1) and (6) it follows that
Since z E B,, g,(z) = 3-", and hence g(z) - g,(z)
+ 3-"/2 = g(Z)
-
3-"/2 < h ( ~-)3-"/2,
because z fZ C. (See (l).)Thus we obtain
On the other hand, since
IV.31
MORE OF COMPACT'IFICATIONS
155
which contradicts (7). Hence we conclude that
proving the proposition. Now, we are in a position to prove the second theorem of Bandt.
Theorem IV.6. I' is a A -space for every T S K,. Thus every compact T,-space X with w(X)s K , is a A-space and accordingly a Shanin's compactifcation of each dense subset.
Proof. Let 93 be a closed base of I' such that 1931 = K,. Well-order the collection
to put B ' = {(BY, B;) 1 0 s a <w , } .
E,
Apply H) to get a closed set C, for the disjoint closed sets B, and where ( B l ,B2) is the first element of 93'. Namely, C,, is a closed set such that
Assume that 0 < p < w , , and for all a < p closed sets C, have been defined to satisfy (1) A (C,, C,,) = 0 for every a, a ' < p, (2) BYc c,, c, n (B;Y = 0 . Then by use of H) we can construct a closed set C, such that
Now we obtain a collection V = {C, 10 S p < T} of closed sets such that A (C,, C p )= 0 whenever a # p and such that Bf C C,, C, f l = 0 for every element (Bf,B!) of 93'. We claim that V is a closed base of 1'. Assume x !Z F for a closed set F and a point x of 1'. Since 93 is a closed base, there is B, E 93 such that F C B,9 x. There is an open nbd U of x
156
[IV.3
COMPACT SPACES AND RELATED TOPICS
u
f l B , = 0. There is B, E 93 such that X - U C B, 3 x. Then such that B , fl = 0. Thus (B,,B2)= (B:, B;) for some a < w l . Hence F C BY C C,, C, f l (B;y = 0, x E (B;)’C (B”;. Therefore F C C, i 4 x, proving that (6’ is a A -base of IT.Hence I‘ is a A -space.
The following important Corollary 1 was first proved by J. M. Aarts [l] and A. K. Steiner-E. F. Steiner [l] independently; the latter also proved Corollary 2. Corollary 1. Every compact metric space is a Shanin’s compactification of each of its dense subsets. Corollary 2. Every product of compact metric spaces is a Shanin’s compactification of each of its dense subsets.
Proof. Use the corollary of Theorem IV.4. (It is easy to see that if each of Xu, a E A, has a closed base satisfying (i) of A) and consisting of regular closed sets, then so does n,X,.) Example IV.9. As indicated by Corollary 2, the converse of Theorem IV.4 is not true. In fact I‘ (for every cardinal number T) is a Shanin’s compactification of each of its dense subsets. However, V. M. Ul’janov [2] proved the following extremely important theorem: Let 2‘ 3 K,. Then there is a Tychonoff space X such that = T and a is not a Shanin’s T2-compactification2 of X such that w ( X )= T,and compactification of X . In his proof Theorem IV.5 plays an essential role. From his result it follows that the continuum hypothesis CH is equivalent with that every T,-compactification of every separable Tychonoff space is a Shanin’s compactification. To see it, assume CH and let 2 be a T,-compactification of a separable Tychonoff space X . Then is also separable, so it has a countable dense subset A. Note that the collection 93 of all regular open sets of 2 is a base of 2 and that every regular open set U is decided by U fl A . Thus 1931 s 2n = K,. Hence by Theorem IV.6, X is a Shanin’s compactification. Conversely, assume that CH does not hold. Then 2” 3 K,. Thus by the above-mentioned Ul’janov’s theorem there is a Tychonoff space X such that = a and a non-Shanin’s T,-compactification 2 of X such that w(X)= 2“ . I
1x1
1x1
’ See J. v. Mill-H.
Vermeer [ I ] for further discussions on this aspect
IV.41
COMPACT
SPACE AND THE LATI-ICE OF CONTINUOUS FUNCTIONS
157
4. Compact space and the lattice of continuous functions Let C ( X )denote the set of all real-valued continuous functions over a topological space X. Then it can be regarded as a lattice with respect to the order: f S g if and only if f ( p ) S g ( p )at every point p of X. The supremum and infimum of two elements f , g of C ( X ) are given by
respectively. We may also regard C ( X )as a ring with respect to the usual sum and product operations. The algebraic structure of C ( X ) as a lattice or as a ring has a deep connection with the topological structure of X itself. Especially, if X is a compact T,-space, then X is topologically characterized by the algebraic structure of t h e lattice C ( X ) .Furthermore this implies that X is topologically characterized by the ring C ( X ) ,too. A theory of this kind was first established by I. Gelfand and A. Kolmogoroff [ 11 regarding C ( X ) as a ring, and later it was extended to the lattice case by I. Kaplansky [l].' In the present section we shall give an account of their theory beginning with the lattice case. We mean by a prime ideal of C ( X ) the inverse image of 0 under some lattice homomorphism of C ( X ) onto the two-element lattice (0, 1). A prime ideal P of C ( X ) is said to be associated with a point p of X if it satisfies the following condition: Iff E P and g ( p ) 6 if
and only if f - If1 E J if and only if
M ) C ( X ) l J is totally ordered. Proof. It is easy to prove that C ( X ) / J is partially ordered by the above defined order, so it is left to the reader. Now, let f € C ( X ) ;then
because % ( J ) is maximal. Namely,
Hence by K)
proving that C ( X ) / J is totally ordered.
IV .6]
REALCOMPACT SPACE
181
N) The mapping O(r)= f, r E R, is an order-preserving, one-to-one homomorphism from the field R of all real numbers into the ring C ( X ) / J .
Proof. Assume 8 ( r ) = F = 0 in C ( X ) / J .Then r E J. Since J C ( X ) ,r = 0 follows. Namely, 8 is one-to-one. It is obvious that O is order-preserving.
Definition IV.13. Generally, let A be a totally ordered field such that the set N of all natural numbers is contained in A . If there is r E A such that r 2 n for all n E N , then A is called non-Archimedean. Otherwise A is called Archimedean. We need the following facts in algebra, whose easy proofs are left to the reader.
Remarks. (1) Let A be a commutative ring with 1 and J its maximal ideal; then the quotient ring A/J is a field. (Thus C ( X ) / J is a field.) (2) Every Archimedean field is isomorphic to a subfield of the real field R. (3) Let cp be a one-to-one homomorphism from the real field R into itself. Then cp is the identity mapping of R. Definition IV.14. A maximal ideal J of C ( X ) is called a real ideal if C ( X ) / J is isomorphic to the real field R. An ideal J of C ( X ) is called fixed to a point x of X if
Example IV.16. Let x be a point of a Tychonoff space X. Then
is obviously a real ideal of C ( X )and fixed to x. On the other hand, consider the spaces R, and R, in Example IV.2. Put
where f denotes the extension o f f over R,. Then it is easy to see that J is a real ideal, because for each f E C(R,)there is a real number r ( f ) such that f ( [ a ,q)) = r ( f ) for some a < w 1 ,
182
[IV.6
COMPACT SPACES AND RELATED TOPICS
7
and the mapping which maps to r ( f ) is an isomorphism between C(R,)/J and R. But J is fixed to no point of R,.
0) A maximal ideal J of C ( X ) is real if and only dean.
if C ( X ) l J is Archime-
Proof. The necessity of the condition is obvious. Assume that C ( X ) / J is Archimedean. By Remark (2) there is an isomorphism cp from C ( X ) / J to a subfield of R . Let t,$ be the natural homomorphism from R into C ( X ) / J . Then p o t , $ is a one-to-one homomorphism from R into R . Thus by Remark (3) p o t , $ is onto, and so is cp, proving that J is a real ideal. P) A maximal ideal J of C ( X ) is real if and only if T ( J )is real.
Proof. Assume that T ( J )is not real. Then there are f, E J, n satisfying
=
1,2,. ..,
Define
Then g 2 0, g E C ( X ) , and hence g On the other hand.
and hence
2
fi in C ( X ) / J .
g > fi. For each natural number
g ( x ) 6 2-"
II,
we obtain
on Z(fJ n . . . n Z(f,,)E % ( J )
Thus by L) C (2)-", i.e. s 2", n = 1,2, . . . , in C ( X ) / J .Hence C ( X ) / J is non-Archimedean. Conversely, assume that J is not real, i.e. C ( X ) / J is non-Archimedean because of 0).Thus there is f E C ( X ) such that
IV.61
REALCOMPACT SPACE
183
fsri, n = 1 , 2 , . . . . Then
n P ( f - n )= 0 , n=l
while P ( f - n ) E .%(J), n = 1, 2, . . . , because of L). Thus .% is not( real. .I)
Q) Let y E P ( X ) . Then, since p ( X ) = u(X,.%), y may be regarded as a maximal tero-filter of X , which we denote by TY.Then y E y(x) if and only ifSYis real. Proof. Suppose ZY is not real. Then there are Zj E TY,i = 1 , 2 , . . . , such that fI;=,Zj = 0. Let Z, = Z ( f , ) ,where f, E C ( X ) . Then construct g E C ( X ) as in the proof of P). Then l/g E C ( X ) , and for every natural number m l / g 3 2"'
on Z , n . . . n Z,,,
Since y E Z , fl-. . n Z, holds in P ( X ) , we obtain
proving that y !2 y ( X ) . Conversely, suppose that
Then there is f E C ( X ) such that f * ( y ) = m. Put
Then PnE .2FY is obvious, and so is nr=, Pn = 0. Thus .Yyis not real. The following theorem is due to E. Hewitt [3] and T. Shirota [l].'
Theorem IV.ll. For a Tychonoff space X the following conditions are equivalent : (i) X is realcompact, (ii) every real filter .% of X converges, (iii) every real ideal J of C ( X ) is fixed to a point of X .
' To be precise, Hewitt proved ( i i ) e (iii), and Shirota ( i ) O (ii).
184
COMPACT SPACES AND RELATED TOPICS
[IV.6
Proof. X is realcompact if and only if y ( X )= X because of Corollary 1 to Theorem IV.10. Now, assume the condition (ii); then from Q) it follows that y(X)= X,i.e. (i) follows. Conversely, if y(X)= X,and if 9’ is a real filter of X , then 9’= TYfor some y E p ( X ) . By Q), y E y ( X )= x . Thus 2’- y in X . Therefore (i) and (ii) are equivalent. From J) and P) it follows that (ii) and (iii) are also equivalent. Corollary 1. Two realcompact spaces X and Y are homeomorphic only if C ( X ) and C ( Y )are ring isomorphic.
if and
Proof. It follows from (i)W(iii) in the above theorem, by use of the method given in Example IV.ll. Corollary 2. Every regular Lindelof space is realcompact.
Proof. It follows from (i)@ (ii) in the above theorem. Example IV.17. Let S be the Sorgenfrey line. Then, as proved before, S is Lindelof while S X S is not Lindelof. Thus by D) S x S is realcompact.
M. Katetov [2] proved that a paracompact T,-space X is realcompact if and only if every closed discrete subspace of X is realcompact.’ A discrete space Y is realcompact if and only if IYI is non-measurable. A cardinal number Y is called non-measurable if every countably additive (at most) two-valued measure m on 2’ satisfying m ( { y } )= 0 for every y E Y is the trivial measure 0. It is easy to see that a and c, e.g. are non-measurable. It is also known that the statement ‘Every cardinal number is non-measurable’ is consistent with the ordinary axioms of set theory. Thus the discrete space R, of all real numbers is realcompact, while there is a continuous one-to-one mapping from R, onto R,, which is not realcompact .*
’
A proof will be given later in VIII.4. ’It was proved by V. Ponomarev [7] and Z. Frolik [4] that the image of a realcompact space by a perfect mapping is not necessarily realcompact. Thus realcompactness is quite different in this respect from compactness, which is preserved by a continuous mapping. For further results o n realcornpact spaces, see e.g. I. Glicksberg [l], M. Henriksen-L. Gillman [I], G. Aquaro [I], T. Ishiwata [l], W. Comfort [3], M. HuSek [l], W. McArthur
IVI
EXERCISE
185
Exercise IV 1. A topological space X is compact if and only if every maximal closed filter of X converges.
2. Let X be a closed subbasis of a topological space X . Then X is compact if and only if every subcollection of X with f.i.p. has a nonempty intersection (J. Alexander’s theorem). Use this theorem to prove Theorem IV. 1 .
3. The product space P = n { D , I (Y E A} of the two point discrete spaces 0, is compact, but it is not sequentially compact unless the cardinal number of A is less than c. 4. (i) If X is dense and C*-embedded in a Tychonoff space Y, then
Y cP ( X ) . (ii) Let F be a closed subset of a normal space X . Then P ( F ) = F, where F denotes the closure of F in P ( X ) . 5. P(R, x R5)= R , x R , (cf. Example IV.2). 6. a ( X ) is metrizable (i.e., it is homeomorphic to a metric space) if and only if X is a locally compact T,-space satisfying the second axiom of countability. 7. Let X be a Tychonoff space. Then there is a normal base 9 3 of X such , = w(X). that w ( c ~ ( X3))
8. P ( X ) is connected if and only if X is connected. 9. P ( X ) is extremely disconnected if and only if X is extremely disconnected. 10. A completely regular space X is locally compact if and only if it is open in every T,-compactification if and only if it is open in some T,-compactification. (footnote continued from p. 184) [I], H . Ohta [l], [2]. A. Mysior [ I ] . R. Engelking-S. Mrowka [ l ] generalized the concept of realcompact space as follows. Let E be a space; then a space is called an E-compact space if it is homeomorphic t o a closed subset of the product of copies of E. See S. Mrowka [4], [ 5 ] o n this aspect. If E = I, E l , {O, 1) with the discrete topology, then E-compact means compact Tz, realcompact, totally disconnected and compact Tz, respectively. O n e of the other interesting cases is N-compact space, where N is the discrete space of all natural numbers. See e.g. P. Nyikos [l]. To the reader who is interested in more extensive study of realcompact spaces, we also recommend L. Gillman-M. Jerison’s [ 11 book and J. v. d. Slot’s [ 1 ) survey article.
186
COMPACT SPACES AND RELATED TOPICS
[IV
11. Let f be a bounded continuous function defined over a TI-space X . Then f can be continuously extended over the Wallman’s compactification w ( X ) of X . 12. Let X be a TI-space, A a subset of w ( X ) and p a point of w ( X ) . Then p E A in w ( X ) if and only if n { p r1 p ’ E A} C p . In t h e last inequality we regard p and p f as closed collections in X (i.e. maximal closed filters).
13. Let X be a TI-space and p a point of w ( X ) . If F is a closed set of X satisfying F kZ p , then
is an open nbd of p and { U ( F )1 F is a closed set of X satisfying F S f p } forms a nbd basis of p in w ( X ) . 14. Reprove 2.F) by use of Theorem IV.2.
15. Let X be a cofinite space with infinitely many points. Then X is a compactification of each of its infinite subsets Y but n o Shanin’s compactification unless X = Y. 16. Let X be a compact T,-space and C ( X ) the ring of all real-valued continuous functions over X . Then every maximal ideal I of C ( X )has the form
for some point p of X.(See Example IV.ll. Hint: If the assertion is false, then for every point p E X , there is f , E I such that f , ( p ) # 0. Thus we can find an open nbd U ( p )of p in which f i > 0.) 17. Discuss a method to define p ( X ) by use of the ring C * ( X )of all real-valued, bounded, continuous functions on the Tychonoff space X . 18. Let C ‘ ( X )denote the ring of all complex-valued continuous functions on a Tychonoff space X . Then two compact T,-spaces X and Y are homeomorphic if and only if C f ( X )and C r ( Y )are ring isomorphic.
19. The product space of countably many sequentially compact spaces is sequentially compact. 20. X is pseudo-compact if and only if every countable cover of X by
cozero sets has a finite subcover.
IVI
EXERCISE
187
21. A Tychonoff space X is pseudo-compact if and only if every maximal ideal of C ( X ) is real.
22. Let X and Y be T,-spaces. If X is compact or sequentially compact, and Y is countably compact, then X x Y is countably compact. 23. Let X be compact or sequentially compact and Y pseudo-compact. Then X x Y is pseudo-compact.
24. A sequential TI-space is sequentially compact if and only if it is countably compact. 25. A mapping of a k-space X into a topological space Y is continuous if and only if it is continuous over every compact set of X .
26. The product space of a locally compact T,-space and a k-T,-space is a k -space. 27. Let X be the quotient space obtained from the real line by identifying all integers. Then X is sequential but not first countable (due to S. Franklin).
28. Every sequential T,-space is a k-space. 29. Let f be a perfect map from a Tychonoff space X onto Y. If Y is realcompact, then so is X . (It is known that the perfect image of a realcompact space is not necessarily realcompact.)
30. Let X and Y be Tychonoff spaces in which every singleton is a G,-space. Then X and Y are homeomorphic if and only if C ( X ) and C ( Y )are ring isomorphic.
CHAPTER V
PARACOMPACT SPACES AND RELATED TOPICS
The purpose of this chapter is to give a rather detailed account of the theory of paracompact spaces. Since the concept of paracompactness was invented by J. DieudonnC [l] and the fundamental theorem by A. H. Stone [l] was established soon after, there has been a great development in the theory of this class of spaces, and now it is a major condition, frequently used in topology. A significance of this newer category of spaces is in the fact that it contains two important classical categories, compact spaces and metric spaces, as special cases and still it is concrete enough to allow fruitful theories in itself.
1. Fundamental theorem
In the present chapter we need some new terminologies for coverings and collections which recently have proved to be powerful tools for study not only of paracompact spaces but also of metric spaces. Definition V.l. Let % be a collection in a topological space X . If every point p of X has a nbd which intersects at most one member of %, then % is called discrete. If every member U of % intersects at most finitely many (countably many) members of %, then % is called star-finite (sturcountable). If for every subcollection %' of %,
U{Vl U E % ' } = U { U I U E W } , then % is called closure-preserving. Let % and 2' be collections; then we say that V is cushioned in % if we can assign to each V E 2' a U ( V) E % such that for every subcollection 2" of 2',
U{Vl V E V ' } C U { U ( V ) J VE V'}
v.11
FUNDAMENTAL THEOREM
189
We often use the above terminologies for coverings, i.e., we use terminologies like star-finite covering, closure-preserving covering, etc. In particular, we call a covering % a cushioned refinement of a covering "1' if % is cushioned in 'V. Let % be a collection (covering) such that m
%=
u %i,i = 1 , 2,..., i=l
where each %i is a locally finite collection. Then % is called a o-locally finite collection (covering). In a similar way we can define u-discrete, o-closure-preserving, u-star-finite, etc. Example V.l. It is clear that discrete is the strongest among the conditions in Definition V.l, and that relationships between them are given by the following scheme (see 11.5.C)): discrete 3 star-finite
+ locally finite j closure preserving 3 cushioned.
In this scheme the implication 'star-finite jlocally finite' is valid only for open coverings, and 'closure-preserving j cushioned' should be understood to mean the following: If % is a closure-preserving collection, then % is cushioned in 4. W e can easily show that those conditions are essentially different from each other. For example, in E' {(-n, n ) 1 n = 1,2, . . .} is a closure-preserving, but non-locally finite open covering.
To prove A. H. Stone's fundamental theorem in an efficient manner, we need the following Theorem V.l, due to E. Michael [l], which is interesting in its own right. Let us begin with Proposition A) which is a lemma for Theorem V.l. A) A regular space X is paracompact if and only if every open covering of X has a locally finite (not necessarily open) refinement.
Proof. It suffices to prove only the sufficiency of the condition. Let X be a regular space satisfying the condition, and % a given open covering of X. Then by use of the hypothesis, we can find a locally finite (not necessarily open) covering d such that d < %. Since d is locally finite, there is an open covering 9 ' each of whose members intersects only finitely many members of d. Since X is regular, we can construct an open covering 9 with 5 < 9. Again by use of the hypothesis, we can construct a
190
PARACOMPACT SPACES AND RELATED TOPICS
[V.1
locally finite covering 93 with 93 < 9.Then B is easily seen to be a locally finite covering satisfying n ( q ) , then
p
by the definition (1) of n ( p ) , and hence p ff W ( q )follows from (3). Finally, we put
for each U E %. Then A = { M ( U )I U E %!}
is a cushioned refinement of %. Because, if
for a subcollection %' of %, then for every U E %!' and for each point p E M ( U ) ,we obtain
which implies
Therefore it follows from (4)that p @ W(q).Hence
for every U E
%!I.
This means that
v.31
205
COUNTABLY PARACOMPACT SPACE
proving that .dl is cushioned in %. Since A is obviously a covering of X , it is a cushioned refinement of %. Thus it follows from Theorem V.3 that X is paracompact. Theorem V.4 practically implies Theorem V.l and the following corollary. Corollary. A regular space X is paracompact if and only covering of X has a a-closure preserving open refinement.’
if every open
3. Countably paracompact space and collectionwise normal space
In the present section we shall define countably paracompact space and collectionwise normal space and study their basic properties. These conditions are somewhat weaker than paracompactness and full normality, respectively, and considerably important in general topology. We shall discuss more of their properties later in this book. Definition V.2. A topological space X is called countably paracompact if every countable open covering of X has a locally finite open refinement. Example V.3. It is clear that every paracompact space is countably paracompact, but the converse is not true. The space R, in Example 11.1 is as shown in Example IV.3, normal but not fully normal, and therefore non-paracompact. On the other hand, R, is countably compact and therefore countably paracompact.
First we shall give some conditions (due to C. H. Dowker [3], M. Katgtov [3]) for a normal space to be countably paracompact. Theorem V.5. The following properties of a normal space X are equivalent: (i) X is countably paracompact,
’See H. Tamano [3], [4], Y. Katuta [l], [3], B. H. McCandless [l] and J. Mack [2] for other interesting characterizations of paracompact space. See also H. Tamano-J. Vaughan [l].
206
[V.3
PARACOMPACT SPACES AND RELATED TOPICS
(ii) every countable open covering of X has a point-finite open refinement, (iii) every countable open covering {U, 1 i = 1,2, . . .} has an open refinement { V, 1 i = 1,2, . . .} with V; C U,, (iv) for every countable open covering {U,I i = 1,2, . . .} with U, C U,+l, there is a closed covering {F, 1 i = 1,2, . . .} such that F, C U,. (v) for every sequence {F; 1 i = 1 , 2 , . . .} of closed sets with F, 3 F,+l, f7:=lF, = 0, there is a sequence {U, 1 i = 1,2, . . .} of open sets such that
U, 3 F,, n;=lU, = 0.
Proof. (i) j (ii) is clear because every locally finite covering is point-finite. ( i i ) j (iii). Let W be a point-finite open refinement of {V, i = 1,2, . . .}. Then putting
I
w . = U { W I W E W , W C U , , W C V , f o r j = 1 , ..., i - I } , we get a point-finite open covering { I i = 1,2, . . .} with Wi C U,. Hence by III.2.C) we can construct an open covering {V, I i = 1,2,. . .} with V; c This implies V; c (iii) .$ (iv) is clear. ( i v ) j (i). Let iU,I i = 1,2, . . .} be a given countable open covering of X with U, C U,,,. Then by (iv) there is a closed covering {F, I i = 1 , 2 , . . .} such that F, C U,.. Since X is normal, for each i there is an open set V, such that
w.
u,.
F,cV,cV;cU,
w
i-1
-
I
Let = U, - U j = l V,. Then {Wii = 1,2,. . .} is a locally finite open refinement of {U,}. Now, suppose { U ;1 i = 1,2,. . .} is a given countable . we open covering of X which does not necessarily satisfy U ;C U ; + lThen Put i
U,=u u;,
i = l , 2) . . . )
j= 1
to obtain an increasing open covering {U, I i = 1,2, . . .}. Construct a locally finite open refinement {Wi I i = 1,2, . . .} of {U, 1 i 1, 2, . . .} using the process mentioned in the above and put
W;= U (j=l
y nu;.
=
v.31
207
COUNTABLY PARACOMPACT SPACE
Then { W l [i = 1,2, . . .I is a locally finite open refinement of {UlI i = 1,2,.. .), because its local finiteness follows from that of (Wi). On the other hand, each point p of X belongs to W: for the first number i satisfying p E U:.Thus X is countably paracompact. It is obvious that (iv) and (v) are equivalent. Corollary. Every perfectly normal space X is countably paracompact.
Proof. Let {U, 1 i = 1,2, . . .} be an increasing open covering of X. Since X is perfectly normal, for each i there are closed sets F,,, s = 1,2, . . . , such that m
U, = U F,, and
F,,CF,,+l
s=l
Put i
Gi= U F,i. j= 1
Then GiC U, and {Gi 1 i = 1,2, . . .} is a closed covering of X . Hence by (iv) of the theorem, X is countably paracompact. Without assuming normality, we can characterize countably paracompact spaces as follows.' Theorem V.6. A topological space X is countably paracompact if and only if for any decreasing sequence {F, I i = 1 , 2 , . . .} of closed sets with fl;=, F, = 0, there is a decreasing sequence { U, 1 i = 1,2, . . .} of open sets satisfying U, 3 F, and n;=, = 0.
q.
Proof. Assume the condition is satisfied by X. Let 'V = {V, I i be a given countable open covering of X. Then put
F, = X - V , U . . * U V,, i
=
=
1,2, . . .}
1,2,. . .,
to obtain a decreasing sequence {F, I i = 1,2, . . .} of closed sets such that fly=, F, = 0. Let {U, I i = 1,2,. . .} be a decreasing sequence of open sets satisfying the condition. Then put
'Due to F. Ishikawa [l].
208
PARACOMPACT SPACES AND RELATED TOPICS
[V.3
w i = x - c ;
then
{y. 1 i = 1,2, . . .} is an open covering of X such that
then {Pi I i = 1,2, . . .} is a locally finite open refinement of 7f, and hence X is countably paracompact. Conversely, assume that X is countably paracompact. Let be a decreasing sequence of closed sets with fI;=l 4 = 0. Then there is a locally finite open covering W of X such that W < { X - F; I i = 1,2, . . .}. Now put
{e}
U , = U { W € U r I W @ X - F , f o r j = I ,..., i } ,
i = l , 2,.... (I)
Then F, C U,. is obvious. It is also obvious that 3 U,.+],because F, 3 F,+]. Finally, to prove f-I;=l = 0, let x EX. Then there is a nbd V of x which intersects only finitely many members of W, say W,, . . . , W,. Suppose Wi C X - F.dJ). ' j = i, . . . , k ,
and io = max{i(l),
. . . ,i ( k ) } .
Then W 3 W It X - Fb implies W # W,, . . . , W, V n W = 0. Thus from (1) it follows that
Namely x $Z r",, proving that fIy=l
=
and accordingly
0.
Definition V.3. A TI-space X is called collectionwise normal if for every discrete closed collection {F, I a E A} in X there is a disjoint open collection {U, 1 a E A} such that F, C U, for all a E A. A) Every fully normal space is collectionwise normal.
Proof. Let {Fa 1 a E A} be a discrete closed collection in a fully normal
v.31
209
COUNTABLY PARACOMPACT SPACE
space X . Then there is an open covering "Ir of X each of whose members intersects at most one member of {F,}. Then let W be an open covering of X such that W A< W: Put
U, = S(F,, W ) , a E A . Then {U, 1 a E A} satisfies the required condition.
Example V.4. R , in Example 11.1 is collectionwise normal but not fully normal.' B ) Let X be a collectionwise normal space and {F, 1 a E A } a discrete closed collection in X. Then there is a discrete open collection { V, 1 a € A } such that F, C U, for all a E A.
Proof. Select a disjoint open collection {U, I ct E A } such that F, C U,. Since X is normal, there is an open set W such that
u{F, I
E A}C
w c W C U {u,I
E A}.
Now put
v, = w n v,,
(Y
EA.
Then { V, I a E A} satisfies the desired condition.
C ) A TI-spaceX is collectionwise normal if and only if for every closed set F of X and every locally finite open covering 021 = { U, I a E A } of F, there is a locally finite open cooering "Ir = { V, I a E A } of X such that F n C U, for all a E A.
v,
Proof. Assume that X satisfies the said condition and that {F, 1 ct E A} is a discrete closed collection in fi. Then observe that F = U {F, i LY t ,-ii is a closed set of X and that {F, I ct E A} forms a discrete open covering of F. Thus there is a locally finite open covering "Ir = { V, 1 a E A } of X such that
Fn
v, c Fa
for all
(Y
EA.
Thus it follows that See R. H. Bing 111 for an example of a normal but non-collectionwise normal space.
210
PARACOMPACT SPACES AND RELATED TOPICS
F
[V.3
n v, = F n V, = F, .
Put
u,=V , - U { V , I P # a } Then { U, I a E A} is a disjoint open collection such that FaC U ,. Hence X is collectionwise normal. Conversely, suppose that 4‘2 = {U, 1 a E A} is a locally finite open covering of a closed set F of a collectionwise normal space X. Then by Corollary 4 to Theorem V.2 there is a cr-discrete closed covering 9 = Uy=lq.of F such that 9 a we have
G’C Uy- U GB, P‘Y
because
and hence
G’ fl (:y
GB) Z 0 .
Thus a’ > a cannot happen, i.e. a’ = a. Hence
G ’ C U , - U GBCU,. P 0, --m < x < +4,where V,(x) = {(u, v ) E X I u - v = x and 0 G v < l / n , or u + v 0 s v < l / n }.
=x
and
Then it is easy to see that X is metacompact but not subparacompact. It is not so difficult to prove that the product S X S of the Sorgenfrey line with itself is not metacompact. (Actually it is subparacompact.) Thus metacompactness is not preserved by a finite product. On the other hand, J. M. Worrell [2] proved that a T,-space is metacompact if it is the image of a metacornpact space by a closed continuous map. K. Alster-R. Engelking [l] gave an example of a paracompact T,-space X such that X X X is not subparacompact. T. Przymusiliski [l] gave a Lindelof regular space X and a separable metric space Y such that X X Y is not 0-refinable.
E) Every collectionwise normal &refinable space X is paracompact. Proof. Let % be a given open cover of X. Then there are open covers {V, n = 1,2, . . .} satisfying the condition of Definition V.5. Suppose V,, = { V, 1 a E A,}. For a fixed n and for an arbitrary choice a l ,. . . , (Yk of finitely many distinct elements of A,, we put
Then {F,(a)1 a E A,,} is a discrete closed collection. Hence there is a discrete open collection {U,,(a)I a E A,,} such that U,,(a)3 F,(a). We can select U,,(a)contained in some element of % because V,,< %. Put
a 2 ) - U i 1 a l ,a 2 E A,, a1# a2}is a discrete closed collection. Now {Fn(a1, Because, if x E X - U!,, then ord,"lr, 2 2. Hence there are distinct a l , a,€ A,, such that x E Valf l V.,. Now Va, n V., intersects F,(a;,a;) only if { a l ,aJ r {a;, a;}. Hence the concerned collection is discrete. I t is also obvious that each F,,(a,,a 2 ) - U!, is closed. Thus there is a a2) al,a2E A,, a1# a,} such that discrete open collection {U,(al,
I
222
PARACOMPACT SPACES AND RELATED TOPICS
[V.4
Un(a1,a2)3 F,(al, a2)- U i and such that Un(al, a2)is contained in some element of V,. Then put
Continue the same process to define a discrete open collection
such that
u:.,=,
Now %& is obviously a c+-discrete open refinement of %. Hence by Michael's theorem, X is paracompact.
In a d-refinable space compactness and countable compactness coincide as it was the case for a paracompact space. F) Every countably compact &refinable space X is compact. Proof. First observe that every discrete collection of non-empty closed sets of X is finite because X is countably compact. Let % be a given open cover of X and {V,,I n = 1,2,. . .} open refinements of % satisfying the condition of Definition V.5. We use the same symbol F,(al,. . . , a k )as in the proof of D). Then {F,(a) I a E A} is finite because of the above observation. Hence it is covered by a finite subcollection V,, of V,. Let V,, = U Vnl. Then {F,(a,, a3 - V,, I al,a, E A,, a , # aJ is a discrete closed collection and hence finite. Thus it is covered by a finite subcover Vn2of V,,. Continue the same process; then we obtain a sequence {V,; I i = 1,2, . . .} of finite subcollections of V, such that U {Vfli1 i = 1 , 2 , . . .} covers U {F,(a,, . . . (Yk) I a,,. . . , ffk E A, ; k = 1, 2,. . .}. Hence U{V,; 1 i = 1,2,. . . ; n = 1,2,. . .} is a countable subcover of Uz=,V,. Hence % has a countable subcover and accordingly a finite subcover. Thus X is compact.'
'See P. Zenor [l], Y . Yasui [3] for another modification of paracompactness called %-property.
v.51
CHARACTERIZATION BY PRODUCT SPACES
223
5. Characterization by product spaces As seen in Example V.2, the product space of two normal spaces may fail to be normal even if one of them is metric and the other is paracompact. This fact leads us to the following problem: Let P be a class of normal spaces; then what is the necessary and sufficient condition for a normal space X in order that the product space of X with every space belonging to P be normal? We shall discuss this problem in cases that P is compact metric (due to C. H. Dowker [3]), compact (due to H. Tamano [l]) and metric (due to K. Morita [3]). The former two cases will be the topic of the present section while the last will be left to Section 7 of the next chapter. It is interesting that in the former cases we obtain a new characterization of countably paracompact normal space and of paracompact T,-spaces, respectively. On the other hand, the last case (that P is metric) leads us to a new class of spaces. In the present section we use K. Morita’s [4]method.
A) Let X be a paracompact T,-space and Y a compact T,-space. Then X x Y is paracompact T,. Proof. It suffices t o prove that X X Y is paracompact. Suppose W is a given open covering of X X Y. Then for each x E X we can select open nbds q ( x ) , i = 1,. . . , k ( x ) , of x in X and an open covering {V, I i = 1, . . . , k ( x ) } of Y such that each V , ( x )x V, is contained in some member of W, because Y is compact. Put
Then, since X is paracompact, there is a locally finite open covering 9 of X such that
9 < { V ( x )I x EX} . To each P E 9 we assign x ( P ) E X such that
P Then put
c U ( x ( P ) ).
224
[V.5
PARACOMPACT SPACES AND RELATED TOPICS
W ‘ = { V ( x ( P ) )x U,(x(P))I i = 1,. . . , k ( x ( P ) ) ;P E Y}. Now, it is easy to see that wr’ is a locally finite open covering of X X Y such that W’ < W. Thus X X Y is paracompact.
Definition V.6. Let p be a real-valued continuous function on X XX, where X is a topological space. Then p is called a pseudo-metric of X if it satisfies p(x, Y ) 3 0, P ( X ,x) = 0, P ( X , Y ) = P(Y, x) and P ( X , Y ) + P(Y, z >a p(x, z) for every x, y, z E X . (Namely, a pseudo-metric is not required to satisfy p(x, y ) > 0 for x f y like a metric.) Let p be an infinite cardinal number and X a topological space. If every open covering 021 of X with 10211 < p has a locally finite open refinement, then X is called p paracompact. (Thus X is paracompact if and only if it is p-paracompact for every p.) We are going to modify some corollaries of Theorem V.2 to use them in the following arguments.
-
B ) Let p be a pseudo-metric of a topological space X . Define that x y for x, y E X if and only if p(x, y ) = 0. Then denote by Y the set of the equivalence classes and by rp the natural mapping from X onto Y,i.e. for each x E X rp(x) is the class which contains x. Define a real-valued function p’ on Y x Y by
Then p’ is a metric of Y,i.e. Y is a metric space, and mapping.
rp
is a continuous
Proof. It is easy to verify that p’ is well-defined and satisfies the conditions for a metric. So the detailed check-up is left to the reader. Now, let y E Y and E > 0. Then S , ( y ) = { y ‘ E Y I p’(y, y’) < E } is a basic nbd of y in the metric space Y.Suppose x E rp-’(y). It is obvious that
where B , ( x ) = {x’ E X I p(x, x ’ ) < E } . Since B , ( x ) is a nbd of x because of the continuity of p, rp is continuous.
C ) Let 021 be a normal covering of a topological space X . Then % has a locally finite open refinement which consists of cozero sets.
v.51
CHARACTERIZATION BY PRODUCT SPACES
225
Proof. T h e r e is a normal sequence {%; I i = 1 , 2 , . . .} such that
W e define a real-valued function p o n X x X as follows. For every rational number of the form k/2", k = 1 , 2 , . . . , 2" - 1; n = 1, 2 , . . . , we define an open covering "tr(k/2")as follows:
Generally, assume we have defined V ( k / 2 " ) , k define "tr(k/2"") by
if k = 2 k ' + 1, where 1s k' Moreover, we put
6 2" -
1.
"tr(1)= {X}.
We can derive from t h e definition that
=
1, . . . , 2 " - 1. Then we
226
[VS
PARACOMPACT SPACES AND RELATED TOPICS
V ( v )< V ( v ' ) if v < v' .
(3)
The validity of (1) and (2) is clear. To see (3), we shall show, by induction on the number n, that
S( V, %!") C V' for every V E V(k/2") and for some V' E
k+l
V(l"),
1 s k s 2" - 1 .
(4)"
It is easily seen that (4),,, n = 1,2, . . . , imply (3), because v and v' can be expressed as fractions with a common denominator. First, V ( i )= %!l and 'V(g) = {X}. Hence (4), is obviously true. Assuming the validity of (4),,, we can show that (4)"+' is also true. To do so, we note that
and
k'+ 1
We divide the proof into the two cases, k V E V(2k'/2"+'),then by (2)
= 2k'
and k
= 2k'+
1. If
V' = S( v,%,,+1)E V f F ) , and hence the assertion (4)n+1 is true in the case k V((2k'+ 1)/2"+')and k' 1, then by (2)
V = S(V,,, %n+l)
for some V,E
V
(3 -
.
Therefore by the induction hypothesis, we obtain
').
S( V,, %!J C V' for some V'E 'I(k; f -
= 2k'.
If V E
v.51
CHARACTERIZATION BY PRODUCT SPACES
227
This combined with 021 Xtl < 021, implies
which proves (4)n+lin the case k = 2 k ' + 1 . Finally, in case k = 1, (4)fl+1is a direct consequence of ( 1 ) and 021~+1wl>w;>.-. Now
Vi= f-l(Wi), i = I, 2 , . . . , are open coverings of X satisfying
Hence 021 is a normal covering. E) Let 021 be an open covering of a topological space X . If there is a normal covering 'V of Xsuch that for each V E Y the restriction { V f l U I U E %} of 021 to V is a normal covering of V, then 021 is a normal covering of X . Proof. Since 'V is normal, by D) there is a locally finite open refinement ' V I of v consisting o f cozero sets. We may assume V 1< Y. Let 'V'= {V, I a E A } , V, = { x E X 1 f , ( x ) > O},
where f a is a continuous function over X with 0 S f a S 1. To each V, we assign a member V (V,) of 'V containing Since { V (V,) f l U 1 U E %} is a normal covering of V(V,), we can construct a locally finite cozero covering V ' , of the subspace V ( V , ) such that
v,.
v, OU; > Q,> %< > .. ..
246
METRIZABLE SPACES AND RELATED TOPICS
[VI.1
Since X is T,, by the condition (ii) we can choose an n for which
Hence rp(p, 4 ) s 1/2", and hence p ( p , 4 ) 3 1/2", i.e. p ( p , 9) f 0. Thus p is a metric of X. Now, let us turn to the proof of that p is compatible with the topology of X. Since p is continuous on X x X and p ( p , p ) = 0, for each natural number i, there is a nbd V ( p )of p such that
Conversely, let V ( p )be a given nbd of p. Then by (ii) there is an n such that
Take a natural number i with l / i < 1/2" and suppose p ( p , q ) < l / i for a point 9 of X.Then
Therefore by the definition of cp
This proves that {Sl,i(p)1 i = 1,2, . . .} forms a nbd base at p. Thus p is a metric compatible with the topology of X , and hence by A) X is a metrizable space.
Corollary. A T,-space X is metrizable if and only if it is paracompact and developable, where a topological space is called developable if it has a sequence {%,, %, . . .} of open covers such that { S ( p ,%,) 1 n = 1,2, , . .} is a nbd base of each point p of X , and {%" I n = 1,2, . . .} is called a development of X.
VI. 11
METRIZABILITY
247
Proof. The easy proof is left to the reader. (Use full-normality of a paracompact T,-space.)
Theorem VI.2. A TI-spaceX is metrizable if and only if for each point p of X, there exist two sequences {U,(p) 1 n = 1,2, . . .} and {V,,(p)1 n = 1 , 2 , . . .} of nbds of p such that (i) { U,(p) I n = 1,2, . . .} is a nbd basis of p, (ii) q ~ , , ( pimplies v,,(q) n v , ( p ) = 0,’ ) (iii) q E V , ( p ) implies V , ( q ) C U,(p). Proof.’ The necessity is almost clear if we put
Therefore we shall only prove the sufficiency. First of all we can prove that X is paracompact if it satisfies the conditions of the theorem. To show this we consider a given open covering % = {U, I a E A}.Put
Then by (i) w^ = Ui=,‘V,, is an open covering of X such that w^ < %. We can prove, moreover, that each ‘V,, is cushioned in %. To see it, we suppose that A’ is a given subset of A. Take a point q !Z U { U , I a E A’} and a point p satisfying U , ( p ) C U, for some a E A’. Then q !Z U,,(p), which combined with (ii) implies that
follows from (l),and hence q !Z U {V, I a E A’} ‘This condition implies that V , , ( p ) C U , , ( p )for each p . We owe the idea of deriving Theorem VI.2 from Theorem V.4 to E. Michael [ S ] .
248
METRIZABLE SPACES AND RELATED TOPICS
[VI.l
Thus we have verified that
U{V,
1 a E A ' )C U{U, I a E A ' ) ,
proving that "Ir, is cushioned in %. Hence "Ir is a u-cushioned open refinement of %. Thus by Theorem V.4, we can conclude that X is paracompact. Since X is TI, the conditions (i) and (ii) obviously imply that X is T2. To complete the proof, let us show that if we put
then {S(p, W m I)m = 1,2,. . .} is a nbd basis of each point p of X . To do so we first note that we can assume m 2 n implies U m ( p ) C U , ( p ) and V m ( p ) CV , ( p ) . Because if not, we replace U , ( p ) and V , ( p ) with I-Iy=l U , ( p ) and f-Iy=l V , ( p ) , respectively. Then they satisfy the desired conditions besides the conditions (i)-(iii). Let U ( p ) be an arbitrary nbd of p E X , and choose n for which U , ( p )C U ( p ) .Then by use of (i) we can find m a n such that U m ( p ) C V , ( p ) . If p E (Vm(q))" for some point 4, then
which combined with (ii) implies
Hence it follows from (iii) that
This of course implies
Therefore we have
This means that { S ( p , W m1)m = 1,2,. . .} is a nbd basis of p . Note that each Wmis an open covering of X .
VI. 1J
METRIZABILITY
249
Thus by the Corollary of Theorem VI.1, X is metrizable.
Corollary. A T,-space X is metrizable if and only if it satisfies one of the following conditions : (i) There exists a nbd basis { W,( p ) 1 n = 1,2, . . .} for each point p of X such that for every n and p E X , there exists an m = rn (n,p ) for which wrn(p)nW r n ( q ) # 0 implies Wrn(q)C W ( P ) . (ii) There exists a sequence {F, I n = 1,2, . . .} of closure-preserving closed coverings of X such that for every nbd U ( p ) of each point p of X , there is an n for which S ( p , g,,) C U(p). (iii) There exists a sequence {%,, 1 n = 1,2, . . .} of open coverings of X such that { S 2 ( p ,%,,) I n = 1,2, . . .} is a nbd basis of each point p of X.’ Proof. The necessity of the conditions is almost obvious. We shall verify only the sufficiency. The sufficiency of (i). We first note that we may assume without loss of generality that m 2 n implies W r n ( pC) W n ( p )and that m (n, p ) n. Putting
wn = { W n ( P ) we define nbds U , ( p ) , V n ( p )of each point p of X by
Then we can assert that
To see it, suppose that V n ( p )n V n ( q )f 0; then
’
We originally owe (i) t o A. H. Frink [ 1J and (ii) and (iii) to K. Morita 13). We can derive several other conditions for metrizability from Theorem VI.2. For example, a TI-space X is metrizable if and only if it satisfies either of the following conditions: (i)’ There is a non-negative valued function q ( p , q ) over X x X such that (a) q(p, q ) = q ( q , p ) , (b) for each fixed closed set F, d(p, F) = inf{q(p, q ) I q E F } is a continuous function of p, (c) {SA(p)I n = 1.2,. . .} where S : ( p ) = {q 1 q(p, q ) < l/n} is a nbd basis of each point p of X. (ii)’ There is a collection {fa 1 (I E A } of real-valued functions over X such that (a) for every subset B of A, sup {fa 1 p E B} and inf {fs 1 p E B} are continuous, (b) for every nbd U of each point p of X , there is a E A and a real number F for which f , ( p ) < E , f a ( X - U )2 F . See J. Nagata [3]. In (i)’ d(g, F) may be regarded as a map from the closed sets of X into C ( X ) . P. Zenor 121 gave a metrizability condition in terms of such maps.
250
METRIZABLE SPACES AND RELATED TOPICS
[VIS (2)
In the case that m ( n , p ) G m(n, q ) we have
because, as implied by (2),
Therefore
In the case that m(n,p ) 3 m ( n , q ) we have
by an argument quite similar to that of the previous case. Thus p E W,,(q) which, combined with the definition of U , ( p ) , implies W , ( q ) C U,(p). Therefore
(note that m (n,q ) 3 n). Thus the validity of (1) is proved. It is clear that (1) implies (ii) and (iii) of Theorem VI.2. To prove (i) of Theorem VI.2, we suppose U ( p )is a given nbd of a point p. Then W,,(p)CV ( p )for some p. Now, from the hypothesis of the theorem we obtain
proving (i). Thus by Theorem VI.2 we can conclude the metrizability of X. The sufficiency of (ii). Since .Fflis closure-preserving,
=
x - u { F I p @ F E Pfl}
VI.11
METRIZABILIW
251
is an open nbd of each point p of X. For each pair (n, m ) of natural numbers, we define nbds U n m ( p )and V n m ( p )of p as follows: If S(p, Sm) C N n ( p ) ,then
Since Smcovers X, we obtain
N m ( p )C S ( p , Sm)for each p E X (see (3)).
(4)
Suppose q $Z U n m ( p )then ; U , , ( p ) # X , i.e., it is the case that
which implies
Therefore it follows from (3) and (4) that
On the other hand, it follows from (3)-(5) that q E V n m ( pimplies )
By the hypothesis on {S,, I n = 1 , 2 , . . .}, { U n m ( p1 )n, m = 1, 2 , . . .} is easily seen to be a nbd basis of p. Thus { U n m ( p )and } {Vnm(p)} satisfy all the conditions required in Theorem VI.2. Hence by the same theorem, X is metrizable. The sufficiency of (iii). Put U n ( P ) = SZ(P9en) and V , ( p )= S(P, Then all the conditions required in Theorem VI.2 are clearly satisfied, and hence X is metrizable.
252
METRIZABLE SPACES AND RELATED TOPICS
[VI.l
Theorem VI.3 (Nagata-Smirnov’s metrization theorem). A regular space X is metrizable if and only if it has a cr-locally finite open basis. Proof. Necessity. Suppose X is a metric space and put
Then, since X is paracompact by Corollary 1 to A. H. Stone’s coinIt is cidence theorem, there is a locally finite open refinement V,, of 9,. clear that U,=] V,, is a cr-locally finite open basis of X . Sufficiency. Let V = UL, V,, be a cr-locally finite open basis of a regular space X , where each V, is a locally finite open collection. For every pair (n, m ) of natural numbers and each point p of X , we define nbds U , , ( p ) and V,,(p) as follows. Put m
Then V , ( p ) is an open nbd of p because of the local finiteness of V,. Now, if
p
E
u c V c v,(p)
for some
u E vrn,
then we put
otherwise we put
Since Vmis locally finite, V,,(p) is an open nbd of p . Suppose W ( p )is a given nbd of a point p of X ; then since V is an open basis and X is regular, we can choose n, m and U E Yrnsuch that p E
u c uc V,(p)C W ( p ) .
Hence by the definition of U , , ( p ) ,
VI. 11
253
METRIZABILITY
proving that { U , , ( p ) 1 n, m = 1,2, . . .} is a nbd basis of p. On the other hand, let 4 6Z U , , ( p ) ; then this implies that U , , ( p ) # X , and therefore that there is some U E Y ' , such that p EUC Hence q tif
u C V , ( p )= U , , ( p )
and
V,,(p) C U .
u,which combined with the definition of V , , ( q ) implies
V,,(q)CX-
0.
Since V , , ( p ) c U, we get
Finally, suppose q E V,,(p). Since V , , ( p ) C V , ( p ) follows from the definition of V , , ( p ) ,
Since V , , ( q ) C V , ( q ) ,we obtain
in the case that U , , ( p ) = V , ( p ) . If U , , ( p ) = X , then V , , ( q ) C U , , ( p ) is clear. Thus { U , , ( p ) I n, rn = 1, 2 , . . .} and { V , , ( p ) I n, m = 1, 2,. . .} satisfy the conditions (i), (ii) and (iii) of Theorem VI.2, and hence X is rnetrizable.' As a corollary of this theorem we obtain the following which gives a complete answer to the metrization problem in the separable case and is a variation of Theorem 111.6.
Corollary 1 (Urysohn's metrization theorem). A topological space satisfying the 2nd axiom of countability is metrizable if and only if it is regular.
'
As for other proofs of this theorem, see J. Nagata [2] or Yu. M. Smirnov [2]. A sketch of the former proof will be found in Example VI.1.
254
METRIZABLE SPACES AND RELATED TOPICS
[VI.1
Corollary 2. A compact T,-space X is metrizable if and only if in the product space X x X the diagonal set A = {(p,p ) 1 p E X } is G8.' Proof. The necessity of the condition is clear, because X X X is metrizable. To see the sufficiency, we use the following proposition B) to get a S ( p , %,,) = { p } for sequence %,, %*, . . . of open covers such that nz=, each p E X . Since X is compact and regular, there are finite open covers V,, n = 1,2, . . . , such that < %,. Then nz=, S(p, V n )= { p } for every p E X . L e t B = { X - V , U * . . U V , I V , , . . . , V , E U z = , V n ; k = l , 2,...}. Then 93 is a countable open collection. Now we claim that 93 is a base for X . To prove it, let p be a point of X and U an open nbd of p. Then for each x E X - U, there is V(x)EUz=,V, such that P E W . Cover X - U by finitely many of V(x)'s, say V(x,), . . . , V(x,). Then X V(x,) U . . U V(x,) E 93, and
v,,
-
p E X - V ( X , ) U * * * V(X,)CU. U Thus 93 is a countable base for X , and hence by Corollary 1 X is metrizable. B ) Let X be a topological space. Then A = {(p,p ) I p E X } is G8 in X x X if and only if X has a sequence %,, % 2 , . . . of open covers such that n ~ ( pa,) , = { p }for every p E X .
;=,
Proof. Assume that A is G,. Then, let A = nz=, W,, where each W, is an open set of X x X . To each x E X we can assign an open nbd U,(x) such that Un(x) x U,(x)C W,. Put %" = {U,(x>I x E X } .
Then it is easy to see that %,, % 2 , . . . satisfy the above condition. Conversely, if a,,%, . . . is a sequence satisfying the condition, then put W, = U { U x U I U E a,}. It is easy to see that W,, n = 1,2, . . . , are open sets of X x X such that m
nW,=A. ,=I
' For brevity such a space X
is said to have a Gs-diagonal.
VI. 11
255
METRIZABILITY
Example VI.2. We can derive Theorem VI.3 directly from AlexandroffUrysohn's metrization theorem as follows. Suppose U;=, Ynis a c+-locally finite open basis of X, where we assume each Yn is a locally finite open covering of X. First, X is normal. For, let F and G be disjoint closed sets of Put
x.
Un=U{V]VEY,,,vVnG=O} and
v,,=u(vlv~~,,,Vn~=0}. Then Unand Vn are open sets satisfying
u n ~ G = O , VnnF=O, m
m
U U n >F and
U Vn> G
n=l
n=I
Therefore
U=
6(Un-6q) i= 1
n=l
and
n=l
i= 1
are disjoint open sets containing F and G respectively. Secondly, X is perfectly normal. For using the notation in the above, we obtain m
F = n(X-v,), n=l
which proves that F is Gp Thirdly, it follows from the normality of X that for every locally finite open covering % of X , there is a locally finite open covering X such that X A< %. The method of proof is quite analogous to the last part of the proof of III.3.E) and therefore it is left to the reader. Now, let V ', = { V, I ct E An};then using the perfect normality of X, we express X - V, as m
X-v,= n v:, k=l
where V ; are open sets. For each finite subset A' of A, we put
256
METRIZABLE SPACES AND RELATED TOPICS
W ( k ,A') = [fl{V, I (Y E A'}]f[fl{V: l
((Y
IVI.1
E A, - A ' } ] .
Then W n k= { W ( k ,A')I A' is a finite subset of A,} is a locally finite open covering of X. On the other hand, for every nbd W ( p )of each point p of X , there are n and k for which
Thus using the third assertion in the above and an argument analogous with that in the last part of the proof of Theorem VI.2 we can construct a sequence of open coverings satisfying the condition of Theorem VI.l. Therefore X is metrizable.
Theorem VI.4 (Bing's metrization theorem). A regular space X is metrizable if and only if it has a u-discrete open basis.
Proof. The sufficiency of the condition is a direct consequence of Theorem VI.3 because every discrete collection is locally finite. As for the necessity, by Corollary 2 of Theorem V.2, for each n the open covering 9,= {S,,,(p) I p E X } has a cr-discrete open refinement %., Thus Ui=,%, is easily seen to be a u-discrete open basis of X.'
Corollary. A topological space X is metrizable if and only if it satisfies one of the following conditions: (i) X is regular, and there exists a sequence {%, I n = 1 , 2 , . . .} of closure-preserving open collections of X such that { S ( p ,%,) I n = 1,2,. . . , S(p, %,) # 0) is a nbd basis of each point p of X . (ii) X is collectionwise normal and developable. Proof. Since the necessity of either condition is clear, we shall only prove the sufficiency. The sufficiency of (i). Using Zermelo's theorem, we regard each %, as a well-ordered collection,
where
7 ,
denotes a fixed ordinal number. For every triplet (1, m, n ) of
See R.H. Bing [l] for a direct proof.
VI.l]
METRIZABILITY
257
natural numbers, we define an open collection W,,, as follows:
Note that { W (U,,) I 0 S a < 7,) is closure-preserving, because %, is so. Then W,,, is discrete. For if p 6Z W(U,,) for every a, then X U{W ( U , , ) 10 s a < 7,) is an open nbd of p intersecting no member of W,,,. Otherwise, we denote by LY the first ordinal number for which
Then [X - U{W(U,,) 1 p < a}]n V(U,,) is an open nbd of p which intersects no member of W,,, except W a n .To see that U{Wl,, 11, m, n = 1, 2 , . . .} is an open basis of X,we suppose N ( p ) is a given nbd of a point p of X. Then there is an n for which
Let a be the first ordinal number for which p E U,, . Now, since X is regular, there are m, I, U , E %, and U, E %, such that p E U2C
v*cU , C v,c U,, .
Therefore p E W(U,,,)for the triplet (1, m, n). Since by t h e definition of a,
p
E U,,
3 V(U,,)
for every p with P < a ,
it follows from the definition (1) of W,,, that
This proves that {W,,, 11, m,n = 1,2,. . .} is an open basis. Thus by Theorem VI.4, X is metrizable. The sufficiency of (ii). In view of the corollary of Theorem VI.1, all we
258
METRIZABLE SPACES AND RELATED TOPICS
[VI.1
have to show is that X is paracompact. Let W = {W, 10 S a < T} be a given open covering of X. Then for each natural number n and ordinal number a with 0 zs a < T, we put
Then { F a , 10 S a < T} is easily seen t o be a discrete closed collection. Hence by virtue of V.3.A), there is a discrete open collection { U , , 1 0 zs a < T} for which U , , > F,,. We can choose U , , such that U , , C W,, because F,, C W,. Now % = { U , , I 0 ~ Clr,>v;>-*. . Then f -'("Ir1), f-'("Ir2), . . . is a sequence of open coverings of
'
x satisfying
A topological space is called Dieudonne' complete when it admits a complete uniformity compatible with the topology. * Spaces for which every continuous function becomes uniformly continuous were studied by J. Nagata [ l ] and A. Atsuji [ l ] and others. Such a space is especially interesting in the metric case, when various equivalent conditions are given by J. Nagata [l], M. Atsuji [l], J. Rainwater 111, S. Willard [l]. A further development in this direction can be seen in A. Hohti [2]. See also S. Ginsburg-J. Isbell [ l ] and A. Hohti [l].
VI.51
293
UNIFORM SPACE
f-'(olr)>f-'(olr])*
>f-'(olrl)>f-1(olr2)*
>..* .
Hence f-'(olr) is a normal open covering, i.e., it is a uniform covering of the uniform space X since X has the a-uniformity. This proves that f is uniformly continuous. Combining this theorem with Theorem VI.17, we get:
Corollary. Let X be a compact uniform space. Then every continuous mapping of X into a uniform space Y is uniformly continuous. Let (P) be a property of topological spaces or of uniform spaces. A uniform space X is called uniformly locally (P) if it has a uniform covering 011 each of whose elements has (P). In this connection, the following proposition may be of some interest.
H) Let X be a connected, uniformly locally compact, uniform space. Then it is the union of countably many compact spaces, i.e., it is a-compact. Proof. Since X is uniformly locally compact, there is a uniform covering OUu consisting of compact sets. Take a uniform covering OUp with 3 ; < OU, and a non-empty member U of OUp. Then we put
F, =
u
F,,,
= S(F,, O U B ) ,
and
n = 1,2, . . . .
Then Fl is compact because
for some member U' of Q U which is compact. W e shall prove by induction o n n that the F,, IZ = 1,2,. . . , are compact. Assume that F, is compact. Then
F, CU{V, I i
=
1,. . . , k }
for some V,.E OUp, i = 1, . . . , k. Since OU; < OUu, we can choose Ui E OUu, i = 1,. . . , k, for which
294
[VIS
METRIZABLE SPACES AND RELATED TOPICS
For these U:,we can prove that
F , + , C U { U ~i(= 1,.. . , k } . Because, if U E implies that
QS
and U fl F, # 0, then U rl V, #
0 for
some i, which
u;cu{u;
u c S(U,, Q S ) C 1 i = 1 , . .. ,k ) Therefore S ( F , , Q S ) C U { U : l i = l , . . . , k} . Since each U : is compact, it is closed and hence U { U iI i = 1,. . . , k} is a closed compact set. Therefore
F,+l C U { U ;1 i = 1, . . . , k } (see (1)). This means that F,,, is compact since it is a closed subset of a compact set. Now, we can prove that U:=,F, = X follows from the connectedness of X. For if p E U:=, F,, then p E F, for some n. Hence by ( 1 )
Thus U:=, F, is open. On the other hand, if p @ UFa, then
For, if we assume the contrary, then
S ( p , Q B )f lF, # 0 for some n . This implies that p E F, (see ( l ) )which , is a contradiction. Thus U % ,F, is open and closed. Therefore it follows from the connectedness of X that m
UF, =X , n=l
and hence X is the countable sum of the compact sets F,, n = 1,2, . . . ,
VI.51
295
UNIFORM SPACE
A complete uniform space 2 is called a completion of a uniform space
X if X is unimorphic with a dense subspace of the uniform space x. The
position which is occupied by completion in the theory of uniform spaces is somewhat like that of compactification in the theory of topological spaces. Now, let us consider the problem of constructing a completion of a given uniform space X. If X is a metric space, then by Example VI.6 it is isometrically imbedded in a complete metric space C * ( X ) . Thus the closure of X in C * ( X )is a completion of X (see F)). If X is a general uniform space, then by use of the following proposition we can uniformly imbed X in the product of metric spaces. Therefore X is unimorphic to a subspace of a complete uniform space because the product space of complete uniform spaces is complete (see G)). Thus in the product space is a completion of X.
I) Every uniform space X is unimorphic to a subset of the product space of metric spaces.
Proof. Let us denote by {p: 1 a E A} the collection of all uniformly continuous functions p: over X X X which satisfy
For each a E A, the relation pA(p, q ) = 0 is an equivalence relation between two points p, 4 of X . Therefore classifying all points of X by use of this relation we obtain a decomposition ga of X. Define a function pa@, D’) on pairs of members of aaby
where p E D, p‘ E D’. W e can easily verify that pa defined in this manner satisfies the conditions for a metric, and hence gabecomes a metric space which we denote by X,. Moreover we denote by cp, the natural mapping of X onto Xu, i.e.,
cp,(p) = D E ga for p E D .
I
Now define a mapping cp of X into the product space P = n { X , a E A}
296
METRIZABLE SPACES AND RELATED TOPICS
[VIS
of the uniform spaces X, as follows:
Since we can easily see that p is a one-to-one uniformly continuous mapping, we shall prove only the uniform continuity of q-’. Let % be a given uniform covering of X ; then we choose a sequence {en1 n = 1 , 2 , . . .} of uniform coverings such that
For convenience of description, we put
Q0
= {X}. For
p, 4 E X , we define
Then we can show that p’ is a member of { p : I a E A}. It is easy to prove P Y P , 4 ) = P’(4, P) 3 0, P ’ b , P) = 0 and PYP, 4 1+ P’(4, r ) 2 P ’ h r ) from the definition of p’. To see its uniform continuity, assume
proving that p’(p, 4 ) is uniformly continuous. Thus we assume
Now, let us prove that
for any choice of po, p l , . . . ,pk E X .
VI.51
UNIFORM SPACE
297
We use induction on the number k. For k = 1, the inequality (2) is obviously true. Assume it is true for all k < 1. Put
and denote by m the largest number such that
which implies u ( p m + lpm+J , +*
. . + d P / - l ?P,) =z
Therefore it follows from the induction hypothesis that
On the other hand, u(pmtpm+JS s is obvious. Denote by n the smallest number such that 2-" s s; then, in view of the definition of a, we can assert
Because, e.g. if a(po,p,) > 2-", then a(po,p,) = 2-' and t < n. Thus 2-' s s, which contradicts the definition of n. Hence from the definition of u it follows that
Since %
A. Thus for a given C C X , either BSC or C B A holds. In the latter case there is a point p E A - C, i.e., { p } S A and p E X - C . Hence it follows from B) that ASX- C . Thus we have proved that for every C C X , either BSC or A S X - C holds. Hence using (iv), we get BSA. Therefore all the conditions for closure are satisfied, i.e., X is a topological space. It follows from (v) that { p } = ( p } for each point p of X, and hence X is a TI-space.
F) Let X be a uniform space with uniformity {%,, and B of X, we define ASB if and only i f S(A, %,,)
f lB #
I y E r}.For two sets A
0 for every
y E
r
Then X is a proximity space.
Proof. The conditions (i), (ii) and (v) of Definition VI.6 are clearly satisfied. It is also clear that ASB or ASC implies ASB U C. Now, let ASB
and ASC
at the same time. Then S(A, %a) fl B
=
0
and S ( A , % p )
for some a,/3 E f.Then S(A, %,,)fl (BU C ) = 0
f l
C=0
308 for %,
METRIZABLE SPACES AND RELATED TOPICS
=
[VI.6
Qc A Q,. This means that
A ~ u B c, proving (iii). Finally, suppose ASB; then
for some y E
r. Take p E r for which %21T,
< Q y . Then put
Now, it is easy to see that
Therefore
A8C and
B8X-C.
proving (iv) of Definition VI.6.
u.
Let % = { I i = 1, . . . , k } be a finite covering of a proximity space X . If there is a covering 'V = { V, I i = 1, . . . , k } of X such that
V,8X- U,, i = 1, . . . , k , then % is called a p-covering. Let f be a mapping of a proximity space X into a proximity space Y. If A6B in X implies f(A)Gf(B)in Y, then f is called a p-mapping. A one-to-one mapping of a proximity space X onto a proximity space Y is called p-homeomorphic if f and f -' are p-mappings. Let X be a topological space and 6 a proximity relation in X . If p E A is equivalent with pSA, then the proximity relation is called compatible with the topology of X . Let X be a uniform space with uniformity (92, 1 y E and 6 a proximity relation in X . If A6B if and only if S(A, %, ) r l B # 0 for every y E I', then the uniformity is called compatible with the proximity relation. Now, we shall study when we can introduce a compatible proximity relation into a given topological space and when we can introduce a compatible uniformity into a given proximity space.
r}
VI .6]
309
PROXIMITY SPACE
G) Let X be a given proximity space, then the collection of all p-coverings of X is a basis of a totally bounded uniformity which is compatible with the proximity of X.
r}
Proof. We denote by { % y 1 y E the collection of all p-coverings of X. Given two members 021, and %B of this collection, we suppose that = {U,
1 i = 1 , . . . , m}
and
021@
=
{V, l j = 1,. . . , n }
Then there exist coverings 021; = {Ui 1 i = 1 , . . . , m} and
%; = {Vl I j = 1 , . , . , n}
such that
u ~ ~ x - and u , v;~x-v,. It follows from B) that
U :n V;SX
-
U, and
U ;n V;&X - V,
which combined with (iii) of Definition VI.6 implies that
u;n v ; S ( x - u,)u(x- v,), i.e., u;nv;S(x-u,nv,),
i = l , . . . , m ; j = l , ..., n .
Since { V ;n V; I i = 1 , . . . , m ; j = 1,. . . , n} is a covering, 92, A %@ is a p-covering. To prove that {4!lYI y E r }satisfies (iii) of Definition VI.3, we suppose that 021, = {U, 1 i = 1,. . . , m} is a p-covering of X. Then there is a covering W = {%.I i = 1 , . . . , m} such that %.&(X - v,) .
By (iv) of Definition VI.6, there is an Miwith W,g(X - Mi)and X - V.6Mi. For different integers i,, . . . , ik chosen from (1,. . . , m}, we define
310
[VI.6
METRIZABLE SPACES AND RELATED TOPICS
~ ( i , .,. . , i k ) =
[niv,,l j = 1 , . . . , k ) ] n [ n { xy- I i f
i,, . . . , ik}]
(2) and
~ ( i ,. ,. . , ik) = [ n{q., Ij
=
1, . . . , k } ] n [nix- MiI i
f
i,, . . . ik}l .
(3) Then % = {N(il,.. . , i k ) 11 s i,,
. . . , ik S
m;
k
=
1 , . . . ,m }
is a covering of X , because %a is a covering, and W,C V , as seen from W,s<X- U , ) combined with C). For this covering % we can assert that %A < %a. For, let p be a given point of X . Then there is an i for which p E y..Now, suppose that p E N(Z,, . . . , i k ) ; then by the definition of N ( i l , .. . , i k ) it must be true that i j = i for some j . Therefore N(i,,. . . , i k ) CV,..This means that S(p, %) c
v,E % a .
Thus %' < %&. Furthermore, we can prove that % is a p-covering. To accomplish this, it suffices to show that
M(i l ,. . . , ik )EX
- ~ ( i ,. ,. . , i k ),
because Ju
=
{ M ( i , , .. . , ik) 1 1s i,, . . . , ik s m , k
=
1,. . . , m }
is obviously a covering of X . Since it follows from (1) that
in view of D) and (3) we obtain
Thus by (2) w i 1 , . . , ik)SX- N ( i , , . . . , i k ), f
proving our assertion that % is a p-covering. Suppose p, q are different points of X . Then by (v) of Definition VI.6, p6q. Hence by (iv), there is a set C for which
VI.61
311
PROXIMITY SPACE
p8C and q 8 X - C . Since {C, X - C } is a covering of X, % ' = { X - p, x - q 1
is a p-covering of X satisfying p $Z S ( p , %'). Thus the collection {ayI y E of all p-coverings satisfies the conditions for a basis of uniformity, i.e. (i) and (iii) of Definition VI.3. Since each covering %v is finite, the uniformity is totally bounded. Finally, we shall prove that this uniformity is compatible with the proximity of X. Suppose that A S . . Then by (iv) of Definition VI.6, there is a set C with
r}
A8X-C
and B8C.
Hence the covering %={X-A,X-B}
is a p-covering, i.e., a uniform covering satisfying S(A, %) f l B
=0
.
Conversely, let
for sets A , B and a uniform covering %. Take a p-covering {V, 1 i = 1,. . . , m} with
Since
%y
%y
=
is a p-covering, there is a covering 'V = { V, 1 i = 1, . . . , m } with
We denote by by D)
yl,.. . , Vjkall the members of
'V which intersect A. Then
312
METRIZABLE SPACES AND RELATED TOPICS
u{y;1 / = 1, . . . , k } S n { X - V , 1 j
=
1, . . . , k } .
IVI.6 (5)
Since
A C U { V , l j = 1, . . . , k } and
X - S ( A , % 7 ) C n { X - U, l j = 1,.. ., k } , we obtain from B) and (5) that
A6X - S(A, 4!17).
In the same way, we obtain BSX - S(B, % y ) . Since
S(A,
)c
x
-
S(B, % 7 )
follows from (4),by (7) and B) we obtain
Therefore, in view of (6),by (iv) of Definition VI.6 we can conclude that
ASB . This completes the proof of the proposition.
Theorem VI.21. Let X be a proximity space. Then there is one and only one totally bounded uniformity compatible with the proximity of X . Proof. The existence of a totally bounded uniformity compatible with the proximity of X was proved in G). Let us consider a given totally bounded uniformity compatible with the proximity of X . Then, since it is totally bounded, as seen easily, the uniformity has a basis consisting of finite uniform coverings. Therefore we denote by {Y7I y E r }the basis of the uniformity consisting of all finite uniform coverings. Then for each y E I', there is p E such that ?"; < Y 7 .Thus we may assume that
r
V7= { V, I i
=
1, . . . , n } ,
Yo = { Vi 1 i = 1, . . . , n }
and S ( V i , Y P ) Cy
,
i = 1,. . . , n ,
VI .6]
PROXIMITY SPACE
3 13
where some V : might be empty, and V, and V, may coincide for different i and j . Since the uniformity is compatible with the proximity,
Therefore V,, is a p-covering. Conversely, let V = { V, 1 i = 1, . . . , n} be a p-covering of X . Then there is a covering V'= { Vi 1 i = 1, . . . , n} such that
v$(x- y ) , Hence there are yi E r, i
Choose y E
i = I , . . . ,n . =
1, . . . , n, for which
r for which
Then V,,< "Y. For, let V E "Y,, ; then Vfl V:# 0 for some member V:of 7' since V' is a covering. This combined with (1) and (2) implies that
proving our assertion. Thus V is a uniform covering belonging to {Y,,1 y E Therefore the only totally bounded uniformity compatible with the proximity of X is the one induced by the totality of p-coverings of
r}.
x.'
H) Let X and Y be proximity spaces and f a mapping of X into Y. I n view of G) we can consider X and Y to be totally bounded uniform spaces. Then f is a p-mapping if and only if it is a uniformly continuous mapping of the uniform space X into the uniform space Y .
Proof. It is left to the reader.
'
Note that this is the weakest uniform topology compatible with the proximity. To the contrary, there is a proximity space which has no strongest compatible uniformity. See E. M. AIfsen and 0. Njistad [l].
314
METRIZABLE SPACES AND RELATED TOPICS
[VI.6
I) Let X be a proximity space and A and B subsets of X . Then ASB if and only if there is a p-mapping of X into [0, 11 such that f ( A )= 0 and f ( B )= 1. Proof. Generally for subsets A , B of a uniform space X , we can easily show, in a way similar to the proof of Urysohn's lemma, that there is a uniformly continuous mapping f of X into [0,1] such that f ( A )= 0 and f ( B )= 1 if and only if S(A, %) fl B = 0 for some uniform covering "21 of X . The present proposition follows directly from the above observation combined with H).
Theorem VI.22. Let X be a topological space. Then we can introduce a proximity compatible with the topology of X if and only if X is completely regular. Proof. If X is a proximity space, then by G) there is a uniformity compatible with its proximity. This uniformity is easily seen to be also compatible with the topology of X . Therefore X is completely regular by Theorem VI.15. Conversely, if X is completely regular, then by Theorem VI.15, there is a uniformity (92, I y E r }compatible with the topology of X . If we define that ASB if and only if S(A, %,) f l B f 0 for every y € r, then this proximity is easily seen to be compatible with the topology of X . Corollary. Let X be a compact T,-space. Then there is one and only one proximity compatible with the topology of X . Proof. Since the existence of a proximity is a direct consequence of Theorem VI.22, we shall prove only the uniqueness. Let 6 be a proximity compatible with the topology of X . Then by Theorem VI.21, there is a totally bounded uniformity compatible with S. Since the uniformity is also compatible with the topology of X , and X is compact, the collection of all finite open coverings of X forms a basis of the unique uniformity of X . Suppose A and B are two sets of X such that A fl B = 0.Then the finite open covering % = { X - A, X - B } satisfies S(A, %) f l B = 0. Therefore
ASB. Conversely, if A f Bl# 0, then we take a point p every finite open covering % of X, S(A, %)IA 3 p ,
€ A nB. Now,
for
VI.61
315
PROXIMITY SPACE
i.e. S(A, %) is a nbd of p . Hence from p E B it follows that S(A, % ) n Bf 0 .
Therefore A6B. Thus A6B if and only if A n B # 0. Namely the proximity 6 is uniquely defined by the topology of X . Let X be a proximity space and Y a subset of X. If for every pair of subsets A and B of Y we define that A6B in Y if and only if ASB in X , then Y also turns out to be a proximity space called a subspace of the proximity space X. We know that not every uniform space can be imbedded as a dense subspace of a compact uniform space. In fact, totally bounded uniform spaces, and only those, can be dense subsets of compact uniform spaces. But for proximity space the circumstance is different as seen in the following.
Theorem VI.23. Every proximity space X is a dense subspace of a compact T2proximity space 2 which is uniquely defined by X . Every p-mapping of X into a compact proximity space Y can be extended to a p-mapping of X into Y . Proof. We regard X as a totally bounded uniform space with the uniformity given in G). We construct the completion r?. of the uniform space X.Then, since X is totally bounded, X is a compact T,-space. We may consider 2 to be a proximity space with the proximity defined by its uniformity. Thus X is easily seen t o be the desired proximity space. Conversely, if X * is a given compact proximity space which contains X as a dense subspace, then by the corollary to Theorem VI.22, we can regard X * as a totally bounded uniform space with the uniformity given in G). We can assert that this uniform space coincides with the uniform To prove this, it suffices to show that X * is a completion of the space uniform space X. Since X * is a complete uniform space and contains X as a dense subset, all we have to prove is that X is a subspace of the uniform space X * . Let {4!Ly I y E r }be the uniformity of X * ; then we can prove that {%;I y E r},where %; = { U fl X I U E QY}, is a uniformity compatible with the proximity of X . Suppose that A and B are subsets of X ; then ASB in X if and only if A6B in X * which is true if and only if S(A, Q Y ) f l B = 0 for some y E I'. The last statement is true if and only if S(A, %;) f l B = 0. Thus the uniformity {%; y E r } is compatible with
I
316
LVI.7
METRIZABLE SPACES AND RELATED TOPICS
the proximity of X . Since {'4!ly I y E I'} is a totally bounded uniformity, so is {"u: 1 y E Therefore by Theorem VI.21, it is the uniformity of X considered at the beginning of this proof. Therefore X is a subspace of the uniform space X * , i.e., X * is a completion of X . Thus by the uniqueness of completion (4.R)), 2 and X * are unimorphic by a unimorphism which keeps X fixed. Therefore by H), 2 and X * are p homeomorphic by a mapping which keeps X fixed. Finally, we can easily verify the last part of Theorem VI.23 by use of Theorem VI.20.'
r}.
7. P-space
In the present section we shall give an account of the theory of K. Morita [4], [5] which is aimed at characterizing spaces whose products with metric spaces are normal and leads us, as a result, to a new category of spaces which contains metric spaces and compact spaces as special cases. Definition VI.7. A topological space X is called a P-space' if for every open collection {U(a,,. . . , a , ) I a l , . . . , a, E A ; i = 1,2, . . .} in X satisfying the condition
u(al,. . . , ai)C U ( a l ,. . . , a,,
a l , . . . , a,+,E A ; i
=
1,2, . . . ,
there exists a closed collection { F ( a,,..., a i ) I a l, . . . , a , E A ; i = l , 2 , . . .} in X satisfying: ' W e can define terminologies like totally bounded proximity space and complete proximity space. See Yu. M. Smirnov 141, IS], S. Leader [l], E. M. Alfsen and 0.Njistad [l]. V. A. EfrernoviT: and A. S. &arc [ l ] considered a necessary and sufficient condition for a proximity space to be metrizable. Some mathematicians are studying theories unifying those of topological space, uniform space and proximity space. See A. Csiszir [l], D. DoiCinov [l]. The concept of category is also useful for giving a systematic description of a formal part of the theory of topological spaces, uniform spaces and proximity spaces though the usefulness is restricted. See, e.g., J. R. Isbell 111, H. Herrlich 111, H. Herrlich-G. E. Strecker [l], 0. Wyler [l]. 'This terminology is sometimes used in a completely different sense. So, to be precise, we should call a space satisfying this condition a Moritu's P-space.
VI.71
P-SPACE
317
(i) F(a,,. . . , a i ) c U ( a , ,. . . , a;), (ii) if U;=l U ( a , ,. . . ,a i )= X for a sequence {ai I i = 1,2, . . .}, then F(a,, . . . ,a;)=
u;=,
x.
A) Every countably compact space is a P-space.
Proof. Let {U(a,,. . . , a;)I a,,. . . , ai E A ; i = 1 , 2 , . . .} be a given open collection satisfying the condition of Definition VI.7. Then we define F(a1, . . ' ai) by 7
. . . , ai) if U ( a , ,. . . , a;)= X ,
F(a,, . . .
otherwise.
Then {F(a,, . . . , a i )I al,. . . ,ai E A ; i = 1,2, . . .} satisfies (i) and (ii) of the definition, because U;=, U ( a , ,. . . , a;)= X implies that U ( a , ,. . . , a ; )= X for some i, since X is countably compact. B) Every fully normal, Cech complete space is a P-space.
Proof. Since the validity of this proposition is implied by A) if X is compact, we may assume that X is not compact. Then by 2.D), there is a sequence si of collections of zero sets with f.i.p. satisfying (i) and (ii) of the same proposition. We note here that (ii) holds for every nonbecause X is normal. Putting convergent maximal closed filter 9, %; = { X - F 1 F E q },
we obtain open coverings %i of X. Since X is fully normal, we can choose open coverings Vi, i = 1 , 2 , . . . , of X for which
We may assume that
y.+,< T;.Now, suppose that
{U(a,,. . . , a;)I a , , . . . , a; E A; i = 1 , 2 , . . .) is a given open collection satisfying the condition of Definition VI.7. Then we define closed sets F ( a I , .. . , a;)by
F(a,, . . . ,a;)= x - S(X - U ( a , ,. . . , a;),T ; ).
(2)
318
METRIZABLE SPACES AND RELATED TOPICS
[VI.7
It is clear that F ( a l ,. . . ,a ; )satisfies (i) of Definition VI.7. To see (ii), we suppose that m
u U(a,,. . . ,
ff;) =
x.
i= t
(3)
If we assume that U;=,F(a,,. . . , a ; )# X , then there is some m
n ( X - F ( ~ ,. ., . , i=l
Then
Y = {S(p,, Vi) I i = 1,2, . . .} u { X - U(a,,. . . ,a ; )I i = 1,2, . . .}
has f.i.p., because if i, 2 i,
3
-
*
. 3 i, then
since po E X - F ( a l , . . . , ail)(see (2)). Hence there is a maximal closed filter 9 which contains Y as a subcollection. It follows from (3) that 9is non-convergent. On the other hand, for each i,
which combined with (1) implies that X - F E 9 for some F E Si, i.e. F $? 9; hence 9' !Z 9. But this contradicts the condition (ii) of 2.D) (in a modified form as noted before). Thus we have proved m
UF(ffl,. . . , ai)= x,
1=l
i.e. X is a P-space.
C ) Every normal P-space is countably paracompact. Proof. Let {V, I i = 1,2, . . .} be an open covering of a normal P-space X such that U, C Ui+].Then, since X is a P-space, there is a sequence {F, ] i = 1,2, . . .} of closed sets such that
VI.71
319
P-SPACE m
F , C q . and
UF,=X. i=l
Therefore by Theorem V.5, X is countably paracompact.
D) A covering % of a topological space X is normal if and only if there is a u-locally finite open refinement "Ir of % such that each member V of Yf is a cozero set, i.e.,
for some real-valued continuous function f over X with 0 S f
S
1.
Proof. Let % = {U, I a < T } be a normal covering. Then there exists a sequence a1,Q 2 , . . . of open coverings such that
Using the argument in the proof of Theorem V.2, we can define open sets W:, a < 7,n = 1 , 2 , . . . , such that
Then by a process similar to that in the proof of Urysohn's lemma, we can define a continuous function f : such that
then N : is a cozero set satisfying
It easily follows from
%i$3 < %n+l and (1) that
S ( N 1 , %,,+J n N ; = 0 if a # p .
320 Thus
METRIZABLE SPACES AND RELATED TOPICS
7f = {NZ I
CY
[VI.7
< 7,n = 1,2, . . .}
is the desired c+-locally finite open refinement of %. Conversely, we consider a covering % of X and a g-locally finite open covering m
7f=
u 'v, O at every p E X . Further, we put
Then g,,,
CY
E A,, n = 1,2, . . . , are continuous functions satisfying
Let us consider a metric space Y whose points are the points {x,, I CY E A,, n = 1 , 2 , . . .} of the Cartesian product of the copies I,", CY E A,,
VI.71
P - SPACE
321
n = 1 , 2 , . . . , of the unit segment [0, 11, satisfying
and whose metric is defined by
for x = {xu,} and y = {yu,}. To every point p of X , we assign a point
of Y (see (2)). Then g is seen to be a continuous mapping of X into Y. For, suppose p o is a given point of X , and E is a positive number. Then we can choose (a1, nl), . . . , (ak, nk)and a nbd U ( p o )of po such that
Then, in view of the fact that
322
METRIZABLE SPACES AND RELATED TOPICS
[VI.7
for every p E U(p,,),which proves the continuity of g. Put
is an open covering of g ( X ) satisfying
Since g ( X ) is a metric space, we can construct open coverings W,, W 2 , . . of g ( X ) such that
are open coverings of X satisfying
Hence % is a normal covering.
E) Let 4!l be an open covering of a topological space X . If there is a normal covering 7f of X such that for each V E 'V, the restriction { V r l U I U E %} of % to V is a normal covering of V,then 3, too, is a normal covering of X. Proof. Since "1' is normal, by D) there is a a-locally finite open refinement Uy=, of 'If where , each is a locally finite collection of cozero sets. We may assume that
where fmi is a continuous function over X with 0 zs fai S 1. To each Vaiwe assign a member V(Vai)of 'V containing Voi. Since { V(Vai)f l U I U E %}
VI.7)
P-SPACE
323
is a normal covering of V(V,;),we can construct a d o c a l l y finite open covering U;=l 7fi; of the subspace V(V,;)such that
) consists of cozero sets of where each T i ; is locally finite in V ( V n j and V(V,;).Now put
Then it easily follows from the local finiteness of 'Ti that
is a locally finite open collection of X . Thus m
w = u wj i, j =l
is a a-locally finite open refinement of %. On the other hand, for each V E 7 f ; ; there is a continuous function g, with 0 S g S 1, over V (V,;)such that
Then h is a continuous function over X satisfying
Thus Wr consists of cozero sets of X . Hence by D), % is a normal covering.
F) Every a-locallyfinite open covering % of a countably paracompact, normal space X is normal. Proof. Let % = U;=l %; be a u-locally finite open covering of X , where each 0.li is locally finite. Putting
324
METRIZABLE SPACES AND RELATED TOPICS
[VI.7
V , = u { u I u E a;}, we obtain an open covering {U, 1 i = 1,2, . . .} of X . Since X is countably paracompact, we can construct a locally finite open refinement {V, I i = 1,2,. . .} of {V, I i = 1,2,. . .} with V, C V,. Put yj =
{v, n u 1 u E
1;
then m
7'=UYi i=l
is a locally finite open refinement of 3. Let
7' = {V, I a E A} ; then, since X is normal, there is a closed covering {F, I a E A } of X such that F, C V,. For each a E A, we consider a continuous function f , with f,(F,)=l,
f,(X-V,)=O
and O S f , S l ,
and put
w,={PIf,(P)>ol. Then W={W,IaEA}
is a locally finite open refinement of ?'f consisting of cozero sets. Thus by V.5.D), % is normal. Corollary. Every u-locally finite open cover % of a countably paracompact normal space X has a locally finite open refinement.
Proof. The cover 7' in the proof of F) is the desired one. G ) Let H be an F,-open set of a countably paracompact normal space X . Then H is also countably paracompact and normal. Proof. Since X is normal, by Urysohn's lemma we can construct cozero sets V,., i = 1,2, . . . , such that
qCU,+l,
m
Uq=H. i=l
P-SPACE
VI.71
325
Let Y'" = { V,, V,, . . .} be a given countable open covering of H. It suffices to prove that Y'" is normal. For, if it is so, then there is an open covering w^, of H with Y'"T .
For given a , , . . . , a; E A and a positive integer k, we define an open set L(a, . . . ai; k ) of X by L ( a l . .. a i ;k ) = U { LI L x V ( a , .. . a i ) C M k L , is an open set o f X } .
Note that some of the L(a, . . . ai; k ) may be empty. Then L(aI. . . a;; k ) satisfies
328
[VI.7
METRIZABLE SPACES AND RELATED TOPICS
L(a, . . . a; ; k ) x V(al. . . a i )c Mk
and
{L(al. . . a; ; k ) x V ( a l .. . a ; )1 al, . . . , ai E A ; i = 1,2, . . .;
k forms an open covering of X X Y, because
=
1,2, . . .}
{ V ( a l . . a ; )I a,, . . . , a; E A ; i = 1,2,. . .}
is an open basis of Y. Note that for j
G
i,
L(a, . . . ai; k ) x V ( a ,. . . aj. . . aj)c
c L ( a , . . . a; ; k )
x V ( a ,. . . a ; )c Mk
.
Hence putting i
u L(al . . . aj ; k ) ,
U ( a l. . . ai ; k ) =
j=l
we get open sets of X satisfying U ( a l .. . a; ; k ) x V ( a l .. . a i )c Mk
and U ( a l .. . a, ; k ) c U(al. . . ";ai+l; k).
Furthermore, we put m
U(a1.. . a i )=
u U ( a l .. . a ; ;k ) ; k=l
then the open sets U ( a ,. . . a i )clearly satisfy
U ( a l. . . a i )c U(al. . . a;ai+l). We can also prove that m
U U ( a l .. . a ; )= X if ( a l ,az,. . .) E Y . i=l
(4)
For, given a point p of X , then ( p , ( a l ,az,. . .)) is a point of X X Y. Since we have chosen L ( a l .. . a; ; k ) so that {L(a,. . . ai ; k ) x V ( a l .. . a ; ) } forms a covering of X x Y,
VI.71
P-SPACE
329
( p , (a1,a2, . . .)) E L(a, . . . a; ; k ) x V(a1.. . a ; ) for some i and k. Therefore it follows from (1) and (3) that p € L(a, . . . a1 ; k ) c U ( a l . . . ai ; k ) c U ( a l. . . a ; ) , proving (4). Since X is a normal P-space, by J) there is a collection
{ H ( a , . .. a ; )1 a,, . . . , a; E A; i
=
1 , 2 , . . .}
of open F,-sets of X such that
H ( a l . .. ai)cU ( a , . .. a ; ) ,
(5)
m
U H ( a , . . . a;)= X
if (al,az, . . .) E Y .
i=l
Note that (6) follows from (4).Since X is countably paracompact and normal by C), it follows from G) that H ( a , . . . a j ) is countably paracompact and normal. Hence, by F), the open covering { U ( a l . . . a; ; k ) f l H ( a l . . . a i )I k = 1 , 2 , . . .} of H ( a l . . . a ; )is normal. (See (3) and (5) for the reason why it is a covering of H ( a l . . . a;).)Hence by I),
% ( a l . .. a ; )= { [ u ( .~ . a,; .;k ) n H ( ( Y , .. . a;)] x V ( a l .. . a ; )I k
=
1 , 2 , . . .}
(7)
is a normal covering of the subspace H ( a , . . . a ; )x V ( a l. . . a ; )of X X Y. Since H ( a l . . . a ; ) and V(a,. . . a ; ) are open F,-sets of normal spaces X and Y respectively, by H) they are cozero sets. Hence
for some continuous functions f and g over X and Y, respectively, with OSfSl andOSgS1.Thus
330
[VI.7
METRIZABLE SPACES AND RELATED TOPICS
is also a cozero set of X x Y. Since { V ( a l . .. a ; )1 a,,.. . ,aiE A ; i = 1,2, . . .} is a cr-locally finite (actually c+-discrete) open collection of Y, X
= { H ( a l . .. ai) x
V ( a ,. . . a;)I a,,. . . , aiE A ; i = 1,2, . . .}
is a c+-locally finite open collection in X X Y. As a matter of fact, by virtue of (6), X is easily seen to be a c+-locally finite open covering of X x Y whose members are cozero sets. Hence it follows from V.5.D) that X is normal. Now, we consider the restriction A ( ' Y ,. .. a i )=
n [ H ( c Y.~. ai) . x v ( ~ . . .,C X ~ ) I] k
=
1 , 2 , . . .}
of the original open covering A
= {Mk
I k = 1,2, . . .}
to each member H ( a , . . . ai)X V ( a ,. . . a;)of X. Then it follows from (2) that
which implies that %(al. . . ai) < A ( a ,. . . ai) (see (7)).
Since by (7) %(al. . . a i )is a normal covering of H ( a l . . . a i )x V ( a ,. . . aj), so is A ( a l . .. a i ) . Thus by virtue of V.5.E), the given open covering A is a normal covering of X because X is normal. Therefore X X Y is normal and actually countably paracompact, too.'
M) Let X be a topological space such that the product space X x Y of X with every subset Y of N ( A ) for every A is normal. Then X is a normal P-space.
'
Actually K. Morita proved that if the product X X Y of a normal space X and a metric space Y is countably paracompact, then it is normal.
P-SPACE
VI.71
33 1
Proof. First we can prove that X X Y is countably paracompact for every subset Y of N ( A ) . Let us denote by D the discrete space of two points 0, 1.' Then we define a mapping f of Y X N ( D ) into N(A U D ) by
We can easily see that f is a topological mapping. Therefore Y x N ( D ) is homeomorphic with a subset of N(A U D). Thus X X ( Y X N ( D ) ) is normal by the hypothesis of this proposition. Since X X ( Y X N ( D ) ) is homeomorphic with (X x Y) x N ( D ) ,the latter is also normal. Thus from Theorem V.8 it follows that X x Y is countably paracompact and normal. We suppose that { U ( a ,. . . ai) 1 a,, . . . , aiE A; i = 1,2,. . .} is a given open collection of X such that
U(a,,. . . Vj)c U(al. . . (Yi(Yi+,). Putting m
az,. . .) I (a,, a2,. . .) E N ( A ) , U U ( a ,. . . ai)= X
( ~ 1 ,
i=l
we get a subset Y of the Baire's zero-dimensional space N ( A ) . For each (a1,. . . , ai) with aj E A, j = 1,. . . , i, we define an open covering V(a,. . . ai)of Y by
W = { U ( a , ,. . . ai)x V(al. . . ai)I a,,.. . ,ai E A ; i
=
1,2,. . .}
is an open covering of X x Y . Furthermore, for each i { V ( a ,. . . a i )I a l ,. . . , ai E A} is locally finite in Y, and hence W is a d o c a l l y finite open covering of X X Y . Since X x Y is countably paracompact and normal, by the corollary of F), we can 'Suppose A f l D = 0. Note that N ( A ) is homeomorphic with AKo,the product of countably many copies of the discrete space A.
332
METRIZABLE SPACES AND RELATED TOPICS
[VI.7
construct a locally finite open covering
of X x Y such that
9<w .
(2)
Putting L(a1.. . a , ; A)=U{U I U XV ( a , .. . a i ) cLA, U is an open set of X}, we can construct an open covering
9= { L ( a , .. . a;;A)x V ( a , .. . a i )I a,... ai€ A , A € A , i = 1 , 2 , . . .} of X X Y such that
L ( a l .. . ai; A ) x V(a,. . . a ; )C L A ,
(3)
where we note that some of the L(a, . . . ai ; A ) may be empty. On the other hand, it follows from (2) that for each A E A, there is a finite sequence ( P I , .. . , pi) of elements of A for which
For every a,,.. . , ai€ A, A E A , we put M(a,. ..ai; A ) = U { L ( a , .. . a,;A ) ) j s i , L(a, . . . aj; A ) C U ( a ,. . . a j .. . a i ) }. (5)
Now we can prove that A = { M ( a , . .. a; ; A ) x V ( a ,. . . a;)I a,,.. . , ai E A,
A € A , i = 1 , 2 , ...} is an open covering of X X Y. To see it, suppose that ( p , a ) is a given point of X x Y.Then since T'is a covering of X x Y , ( p , a)E L ( a l . .. ai ; A ) x V ( a ,. . . a;)
for some al,. . . , ai E A and A E A
VI.71
P-SPACE
Then it follows from (3) and (4) that ( p , a)E L(a,. . . ai;A ) x V ( a ,. . . a i )
.
V ( a ,. . . a ; )C LA
C L(a, . . a;; A ) X
for some pl, . . . , pj E A. If j
G
i, then
p l = a l , . . . ) p.= I 9, and hence
u ( p , . . . pi) = U ( a , . .. a i ) c U ( a , .. . ai... a ; ) , which combined with (6) implies
L(a,. . . a i ;A)C U ( a , .. .ai) Therefore it follows from (5) that
L ( a l . .. a i ;A ) C M ( a , . . . a i ;A) Since p E L ( a , . . . a;; A)
(see (6)),
we obtain ( p , a)E M ( a , . . . a;; A )
X
V ( a ,. . . a ; ) .
If j > i, then
pl=al, ...) pi=q (note that a = (a,. . . , ai . . .) E V ( p ,. . . pi)). Thus by (6) we obtain
L(a, . . . a;; A )
C
U ( a ,. . . aiPi+,. . . pi).
Therefore by (5) L(a, . . . a;; A ) C M ( a , . . . aipi+,. . . pi ; A ) ,
333
334
METRIZABLE SPACES AND RELATED TOPICS
[VI.7
which combined with (6) implies (p, a)E M ( a , . . . aipj+,. . . pi; A )
X
V ( a ,. . . aiPi+l. . . pi).
In any case ( p , a ) belongs to a member of .A, and hence .A is an open covering of X x Y.Furthermore, we note that M ( a I .. . a;; A ) x V ( a ,. . . a;)C LA
(7)
follows from (3), (5) and that
M ( a , . . . ai ; A )
C
U ( a ,. . . a;)
follows from (5). Finally, we put F ( a , ...a;)= U { M (a l . . . a jA; ) ~ A E A } Then, since Y = {LAI A E A } is locally finite, it follows from (7) that { M ( a , . . . a;; A ) 1 A E A } is locally finite, and hence F ( a , . . . a;)is a closed set of X . On the other hand, (8) implies
F ( a , . ..a;)c U ( a , .. . a i ) . Since dl = { M ( a ,. . . ai ; A ) X V ( a ,. . . a i )1 a,,. . . , a;E A , A E A , i = 1, 2, . . .} is a covering of X x Y,{ F ( a ,. . . a;)x V ( a I. . . a;)I a l , . . . , a;E A ; i = 1, 2, . . .} is also a covering of X X Y.Therefore, if m
u U ( a , .. . a;)= x, i=l
then by (1) a = (a,, a*,. . .)E
Y,
and hence for each point p of X , we obtain ( p , a)E F ( a , . . . a;)x V ( a ,. . . a ; ) for some i.
This implies p E F ( a I .. . a ; ) ,and hence
VI.71
P-SPACE
335
m
U F ( a , .. . a i ) = X . i=l
Thus we have proved that X is a P-space. Since X is normal as a closed set of the normal space X X Y, the proof of this proposition is complete. Now, we can prove the following theorem.
Theorem VI.24. Let X be a topological space. Then the product space X x Y of X with every metric space Y is normal if and only if X is a normal P-space. Proof. The necessity of the condition follows directly from M). To prove the sufficiency, let X be a given normal P-space and Y a metric space. Then by Corollary 2 to Theorem VI.14, there is a subspace Z of a Baire's zero-dimensional space N ( A ) and a perfect mapping f of Z onto Y.Put
Then we can assert that g is a closed continuous mapping of X X Z onto X X Y. Since g is clearly continuous and onto, we shall prove only that g is closed. Given a closed set F of X x 2, we consider an arbitrary point ( p , q ) of X x Y with
Since f - ' ( 4 ) is compact, there is an open set V of Z containing f - l ( q ) and an open nbd U of p such that
By the definition of g,
which is easily seen to be closed because f ( Z - V ) is a closed set of Y. On the other hand, it is clear that
336
METRIZABLE SPACES AND RELATED TOPICS
[VIA
and hence
w = x x Y-g(XXZ-ux V) is an open nbd of ( p , q ) in X
X
Y. It is also clear that (1) implies
w n g ( F )= 0. Therefore g ( F ) is a closed set of X X Y proving that g is a closed mapping. Thus there is a closed continuous mapping of X x Z onto X X Y. On the other hand, it follows from L) that X X Z is a normal space, and hence X x Y is normal (Exercise 111.33).' Corollary 1. If a normal space X is either countably compact, fully normal and cech complete or perfectly normal, then the product space X x Y of X with every metric space Y is normal.
Proof. This is a direct consequence of Theorem VI.24 combined with A), B) and K). Corollary 2. Let X be a topological space. Then the product space X X Y of X with every separable metric space Y is normal if and only if X is a normal space satisfying the condition of Definition VI.7 for an arbitrary countable set A .
Proof. We can apply the proof of Theorem VI.24 to this case, but considering that A denotes a countable set. The details are left to the reader.
8. Various generalized metric spaces
We have already learned quite a few generalizations of a metric space,
' It is known that if X
is not only normal P but also paracompact, then the product space Y is paracompact. See K. Morita [5]. M. E. Rudin-M. Starbird [l] proved the following important results: (i) Suppose X is metric and Y is p-paracompact. If X X Y is normal, then X X Y is p-paracompact. (ii) Suppose X is metric and X x Y is normal. If there is a closed continuous map from Y onto Z, then X x Z is normal.
X
X
VI.81
VARIOUS GENERALIZED METRIC SPACES
337
e.g. developable space, uniform space, proximity space and P-space. In the present section we are going to discuss more generalizations which are becoming increasingly important in recent research of general topology. A direct method to obtain a generalization of a metric space is to relax the conditions for the metric function. We defined ‘pseudo-metric’ in such a way. Another example is ‘semi-metric’ defined in the following.
Definition VI.8. Let X be a topological space and p a real-valued function defined on X X X . If p satisfies the following conditions, then it is called a semi-metric: 0) P ( X , Y ) a 0, (ii) p ( x , y ) = 0 if and only if x = y , (iii) P ( X , Y ) = P ( Y , X I . (Namely, a semi-metric is not required to satisfy the triangle axiom among the conditions for a metric.) If {S,(x)I E >O}, where S,(x) = { y E X I p ( x , y ) < E } , is a nbd base at each point x EX, then X with p is called a semi-metric space. If there is such a semi-metric of X, then X is called semi- m etrizable. Every semi-metric space is obviously a TI-space.’ Another method of generalization is to consider a metric function which takes on values in an abstracter range, abstracter than the non-negative real numbers, e.g. in a totally ordered commutative semi-group. In fact various interesting results are being obtained in this way?
’
A similar but somewhat weaker condition is ‘symmetrizability’. A topological space X is called symrnetrizable if X has a semi-metric p such that a subset F of X is closed if and only if p(x, F) > 0 for every x E X - F. See A. Arhangelskii [6] for results surrounding this concept. * See G . Kurepa [l], Z. MamuziC [Z], M. Antonovskii [1], M. Antonovskii-V. Boltyanskii-T. Sarymsakov [l], F. W. Stevenson-W. J. Thron [l], H.-C. Reichel [l], [2], [3], H.-C. Reichel-W. Ruppert [I], Y. Yasui [2] for generalizations in this direction. ‘Statistical metric space’ is an interesting generalized metric space initiated by K. Menger [I], where the metric function takes o n values in a set of functions. Let X be a set and 9 a mapping defined on X x X such that for every x, y EX, 9 ( x , y ) = Fly is a real-valued left-continuous (i.e. for each r E R, FxyI(--,,l is continuous at r ) non-decreasing function defined on the real line R taking values between 0 and 1 satisfying: (i) Fxy(0)= 0, (ii) FXy= Fyx for every x, y E X , (iii) F,(r) = 1 for all r 3 0 if and only if x = y , (iv) F,(r) = 1 and Fy2(s)= 1 imply F,(r + s) = 1. Then (X, 9) is called a sratisrical metric space. For intuitive understanding we may regard F,(r) as the probability for the distance between x and y t o be smaller than r. Namely, the distance between two points is given only in a probability. Now, let X be a given statistical
338
METRIZABLE SPACES AND RELATED TOPICS
[VI.8
Another group of generalized metric spaces was obtained by modifying various metrizability conditions. In the following are those obtained by modifying Nagata-Smirnov's metrization theorem and also related spaces. Definition VI.9. A regular space X is called an M,-space if it has a a-closure-preserving base; a regular space X is called an M2-space if it has a u-closure-preserving quasi-base %, i.e., for each x E X and every nbd W of x, there is U E % satisfying x E U"C U C W ; a TI-space X is called an M,-space (or a stratifiable space) if it has a a-cushioned pair base, where a collection 9 of ordered pairs (P,, P2) of an open set P , and a subset P2 of X is called a pair base if for each x E X and every nbd U of x there is (P,, P2)E 9 such that x E PI C P2C U, and a pair base 9 is called a-cushioned if it is t h e sum of countably many subcollections Pi, i = 1,2,. . . , such that for each i {P, I (Pl,P 2 ) E P i } is cushioned in {P2 I (PI9 PJ E pi}* Let X be a topological space and 0 its topology (the collection of all open sets) and (e the collection of all closed sets of X . Furthermore, we denote by N the set of all natural numbers in the rest of this section. Then X is called semi-stratifiable if there is a map G from B X N into % such that: (i) U = Uz=,G(U,n) for each U E 0, (ii) G(U,n) C G( V, n), n = 1,2, . . . , whenever U C V. The map G is called a semi-stratification of X . Let % be a collection of subsets of a topological space X . If each open set of X is a sum of members of 3,then % is called a network of X.(Thus every base is a network. On the other hand, members of a network are not required to be open.) A topological space X is called a a-space if it has a u-locally finite network.' (Footnote continued from p. 337)
metric space, x E X and
E,
N*(E,6) = { Y E X
6 > 0; then we put
1 F r y ( & ) > 1 - 81
Then {Nx(&,6) I E , S > 0 ) does not necessarily satisfy the conditions for a nbd basis at x. But if it does for every x E X and accordingly induces a topology T, then (X, 7 ) is known t o be semi-metrizable. Conversely every semi-metric space is homeomorphic to a statistical metric space with the topology induced by INI(&,6)}. In this sense topological statistical metric spaces are equivalent with semi-metrizable spaces. See e.g. B. Schweizer-A. Sklar [I], B. Schweizer-A. Sklar-E. Thorp 111, E. Thorp [I], J. Brown [l] and B. Morrel-J. Nagata [l] for results in this aspect. We owe the definitions of 4 - s p a c e s for i = 1, 2, 3 t o J. Ceder [I], semi-stratifiable space to G. Creede [l], network to A. V. Arhangelskii and c-space to A. Okuyama [3].
'
VI.81
VARIOUS GENERALIZED METRIC SPACES
339
A ) A TI-spaceX i s stratijiable if and only if there is a map G from 6 X N into V such that: (i) U = U l , G(U,n ) = Ui=,(G(U,n))ofor each U E 0, (ii) G(U,n ) C G( V,n), n = 1,2, . . . , whenever U C V, where 6 and % denote the collection of all open sets and of closed sets, respectively. Such a map G is called a stratification of X .
Proof. Let X be a stratifiable space with a a-cushioned pair base 9 = Ub, 9,.Then for each U E 6 and n E N , we define
G(U,n ) = U{P, 1 (PI,P2)E 9,, P2C U } . Then it is obvious that G satisfies the said conditions. Conversely, let G be a stratification of X . Then it is obvious that X is regular. For each n E N we consider
9,= {(G(U,n)O, U )1 U E 6). Then it is obvious that Ur=, 9,is a cr-cushioned pair base of X . Thus X is stratifiable.
Theorem VI.25. Let X be a TI-space and suppose that a sequence {U(n,x) 1 n = 1,2, . . .} of open nbds of x is defined at each point x of X . Then we consider the following conditions for the sequence: (i) {U(n,x) 1 n = 1 , 2 , . . .} is a nbd base of x, (ii) if y E U ( n ,x), then U ( n ,y ) C U ( n ,x), (iii) if x kZ F for a closed set F, then x !2 U { U(n,y ) 1 y E F } for some n, (iv) if x $Z F for a closed set F, then x !2 U{ U(n,y ) 1 y E F } for some n, (v) if {x, x,}C U ( n ,y,), n = 1,2,. . . , then x is a clusterpoint of {x,}. Now, we can characterize various generalized metric spaces in terms of {U(n,x)} as follows: Semi-stratifiable = (iii), c+ = (ii) and (iii), stratijiable = (iv), M2 = (ii) and (iv), semi-metrizable = (i) and (iii), developable = (v), where, e.g., the first equality means that X is a semistratifiable space if and only if there is a sequence { U ( n ,x) 1 n = 1,2, . . .} of open nbds of each x E X satisfying the condition (iii).' Proof. The methods of proofs are somewhat similar in all cases. So we shall prove only the first case and the last two cases to leave the rest to 'These characterizations are due to G. Creede [I], R. Heath-R. Hodel [I], R. Heath [2], J. Nagata [7], respectively, and the last two cases to R. Heath [l].
340
METRIZABLE SPACES AND RELATED TOPICS
[VI.8
the reader. Assume that X is a semi-stratifiable space with a semistratification G. Then put
U(n,X) = X
- G ( X - {x}, n ) .
To see that (iii) is satisfied by the sequence { U(n,x) I n = 1,2, . . .}, let x $2F for a closed set F. Then for some n we have xEG(X-F,n)CX-F. Then for each y E F it holds that G ( X - F, n ) C G ( X - {y }, n), because X - F C X - { y }. This implies that
proving (iii). Conversely, assume that {U(n,x) I n = 1,2, . . .} is given to satisfy the condition (iii). Then define a map G from 0 x N into % by
G ( U , n ) = X - U{U(n,X) I x E X - U }. It is obvious that G is a semi-stratification of X . Suppose ( X ,p ) is a given semi-metric space. Then put
Now it is obvious that {U(n,x) I n = 1,2, . . .} satisfies (i) and (iii). Conversely, assume that {U(n,x)} satisfies (i) and (iii). Note that we may assume U(n,x) 3 U ( n + 1, x), n = 1 , 2 , . . . , at each x E X . Then define a function p on X x X by p ( x , y ) = inf { l / n 1 x E U(n,y ) or y E U(n,x ) }
It is obvious that p(x, x) = 0 and p(x, y ) = p ( y , x ) . Now, to prove that {Sl/,(x) I n = 1,2,. . .} is a nbd base at each x EX, let U be a nbd of x. Then there is n for which U(n,x) C U and x !Z U { U ( n ,y ) I y E X - U } hold at the same time. Then for each y E X - U, p ( x , y ) 2 l/n, i.e. S,,,(x) C U. Thus X is a semi-metrizable space. Let X be a developable space and {%,, 1 n = 1,2, . . .} a development of X . Then for each n E N and x E X we fix an element U(n,x) of %, such
VI.81
VARICWS GENERALIZED METRIC SPACES
341
that x E U ( n ,x ) . Then the condition (v) is obviously satisfied by {U(n,X N . Conversely, assume that { U ( n ,x ) } satisfies (v). Then put %, = { U ( n ,x ) 1 x E X } .
To prove that {%,,} is a development, suppose U is a given nbd of x E X. If S(x, "11,,)CU for n = 1 , 2 , . . . , then select x,, E S(x, "11,)- U, n = 1 , 2 , . . .. Then there are y,,, n = 1 , 2 , . . . , such that {x, x , } C V ( n ,y,,). Since x,, IZV,n = 1,2, . . . , x cannot be a cluster point of {x,,} contradicting (v). Thus S(x, %,)C U for some n, proving that {%,,} is a development of X. Hence X is a developable space.
Corollary. A T,-space X is a semi-metrizable space countable and semi-stratifiable.
if and only if it is first
Proof. The 'only if' part is obvious because of the above theorem. Assume that X is first countable and semi-stratifiable. Then it has a sequence { U ( n ,x ) } satisfying the condition (i) of the theorem and a sequence { U f ( n ,x ) } satisfying the condition (iii). Put U"(n, x ) = U ( n ,x ) n U'(n,x ) Then { U f f ( nx,) } satisfies (i) and (iii) at the same time, and hence X is semi-metrizable.
B) Let X be a T,-space. Then relations between various generalized metric spaces are given in the following diagram :
+ M3 (stratifiable)j u j semi-stratifiable
metrizable j M , j M2
% developablej + semi-metrizable 4 where, e.g., Ml j M2 means that every M,-space is M2.
+
+
Proof. The implications metrizable MIj M2 M3 are obvious. The implications u j semi-metrizable and semi-metrizable j semi-stratifiable follow from Theorem VI.25. Let X be a developable space with a development {%,,}. Then define p by
342
METRIZABLE SPACES AND RELATED TOPICS
tVI.8
Now it is obvious that p is a semi-metric of X such that {S,(x) 1 e > 0) is a nbd base at each x E X. Hence X is semi-metrizable. Suppose
Then each Ceflm is a discrete closed collection. To prove that Ce = %nm is a network of X , suppose that U is a nbd of x E X. Then S(x, %,,) C U for some n. Suppose x E U, E %, and x 6 Up for all U, E %, with P < a.Select m for which S(x, em) C U,. Then it is obvious that
U:,,=,
x E GzmC U, C U .
Thus Ce is a network of X, proving that X is a u-space. The most difficult part is the proof of the implication M,J u,which was first proved by R. W. Heath [3] as follows. Let X be an M,-space; then by Theorem VI.25 there is a sequence {U(n,x) I n = 1,2,. . .} of open nbds of each x E X satisfying the condition (iv). We may assume that U(n,x) 3 U ( n + 1, x) for every n and x. Well-order the points of X to put
Now we define
Then it is obvious that D(a, i, n ) C U(i,x,). Let
9(i, n ) = {D(a,i, n> 1 O =Sa < T } ; then each 9(i, n ) is obviously a discrete closed collection. Further define
VI.81
VARIOUS GENERALIZED METRIC SPACES
343
I
E(a, i, n, m ) = { z E D(a, i, n ) x, E U ( m , z ) } , O c a < T , (i, n, m ) E N x N x N ,
(2)
%(i, n, m ) = { E ( a ,i, n, m ) 10 a < T} , 8 = U { %(i,n, m ) 1 (i, n, m ) E N
X
N x N }.
Since 9 ( i , n ) is discrete, so is 8(i, n, m ) . Thus 8 is a c+-discrete collection. We claim that 8 is a network of X. To prove it, let W be an open nbd of x E X . For each i E N , let ai be the least ordinal number such that x E U(i,xS). Then it follows from the condition (iv) of Theorem VI.25 that xmi+= x. By the same condition there is m E N such that
x @ U { U ( m y, ) l y E X - W ) .
(3)
Thus there is i 2 m such that
Since x E U ( i ,X J , there is n E N satisfying
Now, we can prove that x E E(ai,i, n, m ) c W .
First x E E(ai,i, n, m ) follows from (5), x E D(ai,i, n ) and (2). (Note that x E D(ai,i, n ) follows from (6), the definition of ai and (l).) E(a,, i, n, m ) c W follows from the fact that y 6Z W implies U ( m , y ) 3 X,, (because of (4)), i.e. y fZ E(ai,i, n, m ) (because of (2)). Hence 8 is a c+-discrete network of X , and hence X is a u-space.
C ) Every M,-space is hereditarily paracompact. Proof. It directly follows from Theorem V.4.
D ) Every semi-stratifiable space X is subparacompact. Thus compactness
344
METRIZABLE SPACES AND RELATED TOPICS
[VI.8
and countable compactness coincide for semi-stratifiable spaces, and every collectionwise normal semi-stratifiable space is paracompact. Proof. Let '42 = {U, 10 s a < T} be an open cover of X and G a semistratification of X. Put
Then 9, =, {FunI 0 s a < T) is discrete, and U:=, 9"is a u-discrete closed refinement of '42. Hence X is subparacompact. See V.4 for the rest of the claim. Theorem VI.26. For a regular space X the following conditions are equivalent : (i) X has a u-closure-preserving network, (ii) X is a a-space, (iii) X has a a-discrete network.'
Proof. The implication ( i i i ) j ( i i ) j (i) is obviously true. To prove ( i ) j (iii), assume that X has a u-closure-preserving network 9 = U l , %, where we may assume that %, is a discrete closed collection because X is regular. Let (n, m ) be a fixed element of N x N. Then for each G E %, we define
is a closed set. Further we define a closed collection % ,, Note that Hm(G) by
Then we claim that each X,,, is discrete. To prove the claim, it suffices to show that X,,,,, is disjoint and closure-preserving. Since each X m ( G )is disjoint, xm,, is obviously disjoint. Assume that
' Due to F. Siwiec-J. Nagata [l].
VI.81
VARIOUS GENERALIZED METRIC SPACES
345
r}
where {H, 1 y E C Xmn. Then from the definition of Xm,,it follows that U = X - U { G I xe G E Y, U Y m }is an open nbd of x such that U f l H = 0. This proves that H is closed. Hence Xmnis closure-preserving and accordingly discrete. To show that X = U:,,=, Xmnis a network of X , let W be a nbd of x E X . Then there is n E N and GoE Y, such that x E G,C W. Select m E N such that x E GI C X - U { G 1 x 6Z G E Y,,}
for some G IE Ym.
(1)
Define H E Xmnby
Then x E H follows from (1) and (2), because x $Z G E %, implies that x E GI C H,(G). On the other hand, H C GoC W follows from (2). Thus X is a discrete network of X , proving (iii).
Corollary. Let f be a closed continuous map from a regular c+-spaceX onto a topological space Y ; then Y is a u-space. Proof. Consider a u-closure-preserving closed network of X. Then its image by f is a u-closure-preserving network of Y. Hence by the theorem Y is a u-space. Definition VI.10. Generally, let X be a class of topological spaces. Then we consider various conditions to be satisfied by X,as follows: (a) If X ’ C X E X,then X ’ E X. (b) If Xi E X,i = 1,2, . . . , then rI;=]X, E X. (c) If X E X,and there is a closed continuous map f from X onto a topological space Y,then Y E X. (d) If X = U;=,X, for closed subsets X, E X7 i = 1 , 2 , . . . , of X , then
x E X.
I
(e) If X is dominated by a closed cover {X, a E A } such that X , E 3% for all a, then X E X, where the closed cover {X, I a E A } is said to dominate X if the following holds: A subset F of X is closed if there is a subcollection { X , I a E A’}of the cover such that U {X , I a E A ’ }3 F and such that X, n F is closed for every a E A’.’
’
It is obvious that every locally finite closed cover of X dominates X , and every dominating cover is closure-preserving.
346
METRIZABLE SPACES AND RELATED TOPICS
[VI.8
(a’) If X ’ is a closed subset of X E X,then X ’ E X. (c’) If X E X,and there is a perfect map f from X onto Y, then Y E X. Recent developments of the theory of generalized metric spaces are motivated by several factors; one of them is that some generalized metric spaces behave, as a class, better than the metric spaces. The class A of metrizable spaces satisfies only (a), (a’), (b) and (c’). The non-metrizable hedgehog S ’ ( N ) gives an example to show that A satisfies no other condition in Definition VI.10. On the other hand, e.g. the class of semistratifiable spaces satisfies all conditions there. Thus S’(N) is semistratifiable. There are many other such spaces which are non-metrizable but ‘near-metrizable’ in the sense that they are constructed from metric spaces through a simpler process. Thus it is desirable to systematically study those non-metrizable spaces.
E) M,-spaces satisfy (b). M,-spaces satisfy all conditions in Definition VI.10 except (d). u-spaces satisfy all conditions if X is regular in (c) and (e). Semi-stratifiable spaces satisfy all conditions. Semi-metrizable spaces satisfy (a), (a‘) and (b), and so do developable spaces. Proof. We have already proved that (c) is satisfied by cr-spaces. We shall prove here only some of the non-trivial claims and leave the rest to the reader. To prove (e) for the class of semi-stratifiable spaces, let { X , I a E A } be a closed cover dominating X , where each X, is semi-stratifiable. Then we denote by G, a semi-stratification of X . Well-order all members of the cover t o put
{X, I a E A }= {X, I 0
a
< T}
Suppose U is an open set of X and n E N. Then we define G(U,n ) by
Then it is easy to see that G is a semi-stratification of X . To see that G(U,n ) is a closed set of X,let x E X - G(U,n ) . Suppose x E X,, and x for all?!, < a . Then V = X-U{X, J P< a } is an open nbd of X , which is disjoint from U {Fp 1 /3 < a}.(Note that { X , } is closure-preserving.) On the other hand, x has an open nbd W, in X, such that
exa
VIA]
VARIOUS GENERALIZED METRIC SPACES
347
W, f l F, = 0. Now, for each y with a S y < 7,we can construct an open set W, in Y, = U {X,, 1 a s a ' s y } such that W, fl Y,. = W,. whenever CY S y' < y < 7,and such that W, f l F, = 0.Then put
w = U{ w,I a ==y < 7). Since {X,1 CY E A} dominates X , W is an open nbd of x in X , because W C U{X,1 a s y < T}, and W fl X, is open in X,.Also observe that W fl F, = 0 for all y 2 a. Thus V fl W is a nbd of x which is disjoint from G (U, n). This proves that G( U, n ) is closed. Hence X is semi-stratifiable. To prove (e) for a-spaces, let {X,I a E A} be a closed cover dominating X, where each X, is a regular a-space. Let 9, = Ub, %,ibe a a-closure-preserving closed network of X,, where each %,iis closurepreserving. Then it is easy to prove that %;' = U{%,; 1 CY E A} is closureis a a-closure-preserving network of preserving for each i. Thus U b , X,proving that X is a a-space. Finally, let us prove that M,-spaces satisfy (c). Let X be an M,-space with a stratification G. Then to every pair (F, U ) of a closed set F and an open set U such that F C U, we can assign a closed set H(F, U ) satisfying
x
-
U c (H(F,U)>oc H(F, U )c x - F ,
H(F, U ) > H ( F ' , U ' ) whenever F
C F'
(1)
and U C U'.'
To do so, we put
then it is easy to prove that the desired condition is satisfied. (The detail is left to the reader.) Now, assume that f is a closed continuous map from X onto Y.Then we can define a stratification G' of Y as follows. Let V be an open set of Y. Then F(V, n ) = f(G(f-'(V), n ) ) is a closed set of Y satisfying m
u F(V, n ) = v n=l
'
Generally a TI-space X is called monotonically normal if for every pair (F, U ) of a closed set F and an open set U 3 F, there is a closed set H(F, U ) satisfying this condition.
348
METRIZABLE SPACES AND RELATED TOPICS
[VI.8
Then f -‘(F( V, n ) ) is a closed set of X contained in f-’(V ) .Put
Then H’(V, n ) is an open set of Y contained in V such that
F( V,n ) C H’( V,n ) C H‘( V,n ) C V ,
(3)
which follows from (1). Further we put
G’( V,n ) = H’( V,n ) . Thus it follows from (2) and (3) that m
m
m
n=l
n=l
n=l
u H ’ ( V , n ) = U ( G ’ ( V n ) ) ” =U G ’ ( V , n ) = V .
It is obvious that G’(V,n ) C G’(W,n), whenever V C W for open sets V and W. Thus G’ is a stratification of Y, proving that Y is M3. The condition (e) for M3-spaces will be proved later in VII.5. In the following is a generalization of a proposition on metrizable spaces. F ) Separability and Lindelof property coincide for every collectionwise normal semi-stratifiable space and accordingly for every M3-space as well.
Proof. Let X be Lindelof and {U(x,n ) I n = 1,2,. . .} a sequence of open nbds of x satisfying the condition (iii) of Theorem VI.25. Then for each n there is a countable subcover of { U ( n ,x) I x E X } which we denote by { U ( n ,x l ) 1 i = 1,2,. . .}. Now, we can show that { x l I i, n = 1 , 2 , . . .} is dense in X.To this end, let x E X and consider a given nbd U of x. Then there is n for which
e
Thus x? E U for some i, because otherwise x U;=l U ( n ,x)), which is impossible. This proves that {x)} is dense, and hence X is separable. (Actually we used only semi-stratifiability of X to prove this part of the proposition.)
VI.81
VARIOUS GENERALIZED METRIC SPACES
349
Conversely, let X be separable and {xi I i = 1,2, . . .} a countable dense subset of X. Suppose % is a given open cover of X. Then, since X is paracompact by D), there is a locally finite open refinement V of %. It suffices to show that 7 f is countable. To each xi we assign Ti= { V € V 1 xi E V}. Then Vi is a finite subcollection of V. Since V = U;=l Viis obvious, 7 f is countable. This proves that X is Lindelof. Example VI.13. It follows from E) that the non-metrizable hedgehog S’(N)is stratifiable. In fact one can prove that S’(N)is an MI-space. (The easy proof is left to the reader.) The non-metrizable space X given in Example VI.3 is a first countable Ml-space. It is known that M2 and M3 coincide, but it is an open question if MI and M2coincide or not; we shall discuss this later. Niemytzki space is developable and accordingly a a-space but not stratifiable, because it is not paracompact. Also observe that it is separable but not Lindelof. See E. S. Berney [ l ] for an example of a Tychonoff semi-metric space which is not a a-space (and accordingly not developable). The above example S ’ ( N ) shows that developable spaces do not satisfy (c). On the other hand, J. M. Worrell [l] proved that developable spaces satisfy (c’). The condition (d) is not satisfied by M3-spaces. In fact R. Heath [4] gave an example of a countable space which is not M3.We can show that (c’) is not satisfied by semi-stratifiable spaces. Consider X = {(x, y ) < x, y < +a} with the following special topology: Let p = (x, y ) E X and n E N; then put
I
--cc)
* ( y ’ - y + - ( nxl ’ - x )
)
) has a cluster point q Ef(C). Then we can easily show that q is also a cluster point of $(A I >). Hence q E Y - U = Y - U, which is a contradiction. Thus (1) is proved. Now, suppose that g is a member of F(X, Y ) such that g E S ( f , %(p)) (see Definition VII.3),
where p E A is chosen to satisfy 3 ; < %., Then
for every point p, especially for every point of C. Therefore from (1) it follows that
i.e. g E U ( C ;U ) . Thus we have proved that
This means that U ( C ;U ) is an open set for the topology of uniform convergence. Hence every open set of F ( X , Y ) for the compact open topology is also open for the topology of uniform convergence, which means that the latter topology is stronger than the former. Now, let us suppose that X is compact, and f is a member of F ( X , Y ) . To complete our proof it suffices to show that for every a E A, Scf, %(a)) is also a nbd of f for the compact open topology. Since f ( X ) is compact, we can cover it with a finite number of members of say U,, . . . , Uk. Since f ( X ) is normal, we can construct a closed covering {GI,.. . , G k }of f ( X ) such that G, C U,, i = 1 , . . . , k . Put f - ' ( G , )= C , ; then C,, i = 1 , . . . , k , are closed, and therefore are compact sets of X satisfying k
f ( C , ) c V,
and
U C,.= X . i= I
VII. 11
371
MAPPING SPACE
Therefore U ( C ,. . . ck;U , . . . Uk) is an open nbd off. To prove that this nbd is contained in S ( f , %(a)),we take a given point g E U ( C , . .. c k ; V , .. . U k ) .Suppose that p is an arbitrary point of X; then p E for some i. Hence g ( p )E U, ; thus it follows from f ( p )E Uj E %a that
c.
proving that S ( f , %((a)) proposition.'
is a nbd of f . This ends the proof of the
Example VII.3. We can modify the definition of the topology of uniform convergence for F ( X , Y) as follows: Let V be a collection of subsets of X. Suppose C E V and f E F ( X , Y ) . Then for each uniform covering %a of Y, we put
I
Then {%(C,a ) C E V,a E A } defines for F(X, Y) the uniformity (and the topology) of uniform convergence with respect to V. J) If V is the collection of all compact sets of X and F ( X , Y ) is a collection of continuous mappings, then the topology of uniform convergence with respect to V coincides with the compact open topology.
Let F(X, Y) be a collection of continuous mappings of a topological space X into a uniform space Y with uniformity {%a 1 (a E A}. If for each cr E A and for each p E X , there is a nbd N ( p ) of p such that 'Generally, a topology of F(X, Y ) is called joint continuous if the mapping P of F ( X , Y )x X into Y defined by
P(f*P) = f(P) is continuous. The topology of uniform convergence is an example of joint continuous topology. R. H. Fox [l] studied relations between joint continuity and the compact open topology. See J. L. Kelley [l], too.
372
TOPICS RELATED TO MAPPINGS
[VII.l
for every f E F ( X , Y), then F ( X , Y) is called equicontinuous. K ) Let F ( X , Y)be a collection of continuous mappings of a compact space X into a uniform space Y. If F ( X , Y ) is equicontinuous, then the weak topology coincides with the topology of uniform convergence.
Proof. Since by E) and I) the topology of uniform convergence is stronger than the weak topology, we need only prove that the converse is also true. Suppose {%, I a E A} is the uniformity of Y. Let f E F ( X , Y) and V be a given nbd off with respect to the topology of uniform convergence. Then we can assert that V is also a nbd of f with respect t o the weak topology. First we take a E A for which
Take p E A such that %; < %., Since F ( X , Y )is equicontinuous, we can assign an open nbd U ( p )t o every p E X such that
Since X is compact, we can cover X with a finite number of the U ( p ) , say U ( p , ) ,. . . , U ( p k ) .W e consider an open nbd
of f with respect t o the weak topology (see Definition VII.l). Suppose g E U,(f); then for each p E X , there is i for which p E U ( p , ) .It follows from the definition (1) of U ( p , )that
O n the other hand, the definition (2) of V,(f) implies
Combining (3), (4) and (5) we get
VII.11
MAPPING SPACE
373
This implies
because %; < %a. Since this is valid for all p E X, we conclude
which means that
Therefore V is also a nbd off with respect t o the weak topology. Thus the weak topology is stronger than the topology of uniform convergence, and consequently they coincide.
Theorem VII.1. Let C(X, Y ) be the space of all continuous mappings of a compact space X into a compact uniform space Y. We assume that C(X, Y ) has the topology of uniform convergence. Then a subset c' of C(X, Y) is compact if and only if C' is equicontinuous and closed in C(X Y).
Proof. Assume that C' is equicontinuous and closed in C ( X , Y) with the topology of uniform convergence. First we note that by K) the topology of C' coincides with the weak topology. Let us show that with respect to the weak topology, C' is closed in the space Y x of all mappings of X into Y. Suppose f E Y x- C'. If f E C(X, Y), then by the hypothesis f sl in C(X, Y) for the topology of uniform convergence. To prove that f 6 6' is true for the weak topology of C(X, Y), We assume the contrary: f E C?' in C(X, Y).Then for a given a E A, using the equicontinuity of c' we take p E A with %; < %a and assign an open nbd U ( p )to each p E X such that
c'
We cover the compact space X with a finite number of the U ( p ) , say U ( p , ) ,. . . , V ( p , ) . Putting
374
[vn.1
TOPICS RELATED TO MAPPINGS
we obtain a nbd o f f with respect to the weak topology. It follows from our assumption that
Take g E U o ( f )n C ' ; then for each p E X we choose i with p E U(pi). From the definitions of U,,(f)and U ( p i )we obtain
Therefore
follows from 021;
O and every compact set K of X , there is a
378
[VII.l
TOPICS RELATED TO MAPPINGS
polynomial P ( y ) such that IP(y)- u ( y ) l < where k
= max{hi(x)
I
E
whenever 0 c y s k ,
E K}. Therefore
Since P ( h i ) E D by J) we obtain u, E fi. Now, we note that fi is also a subring of C ( X ) because, as easily seen, the operations f +g, fg, af are continuous. Putting u i = 1- u,, we obtain an element uh of D satisfying u&)
=
1,
u;(F) = 0 and 0 s u i s 1
Furthermore, we put V ( q ) = { x E X Iu i ( x ) > k } . Then V ( q )is an open nbd of q. Since
and G is compact, we can cover G with a finite number of V ( q ) ,say V(q,),.. . Put f
m
u’(x) =
2u p ) ,
xE
x.
i= 1
Then u’ED,
u’(F)=O,
u’(G)>k,
OSu’Grn.
We again use the real-valued continuous function u in the above argument and put p(x) = u ( u ’ ( x ) ) , x E X .
Then p(F)=O,
cp(G)=l
and O G c p s 1 .
Using Weierstrass’ approximation theorem, for every polynomial P‘ such that IP’(u’(x))- p(x)l
0 we can find a
VII. 11
MAPPING SPACE
379
Since P’(u’)E D, by J) this implies that rp E 5 = 0. Thus the proposition is proved. From proposition L) we can derive the following useful theorem which is an extension of Weierstrass’ approximation theorem in analysis. Theorem VII.2 (M. H. Stone-Weierstrass’ approximation theorem).’ Let X be a T2-space and D a ring of continuous functions with real scalar multiplication satisfying the conditions in L). Then D is dense in the mapping space C ( X ) of all continuous functions over X with the compact open topology.
Proof. By virtue of J) it suffices to prove that for a given f E C ( X )and for every compact set of X and E > 0 there is cp E fi satisfying Icp(x)- f ( x ) l < E for all x E K. Since f is bounded on K , we assume If(x)l S a for all x E K. Then we choose al,. . . , a, such that -a
=
a 1< a2< *
*
. < a,
=
a
and a ; - a , - , < ~ , i = 2, . . . , n .
Put
F, = { x I x
E K and
f ( x )2 a;}
and G, = {x I x E K and f ( x )G a i - J . Using L) we construct continuous functions piE D such that
cp;(F,)= 1, Put
rpi(Gj)= 0 and 0 S piS 1
n
p ( x )=
2 (a;- a;-l)rpi(x)+ a], x E x. i=2
Then cp is the desired function. Because, if p E K, and aj-1
Sf(p)
aj,
then pEF,,
i = 2 , . . . ,j - 1 ,
P E G , , i = j + l , ..., n .
’
The present version of the proof is essentially due to H. Nakano’s proof of the following corollary, which was first proved by M. H. Stone [l]. There are various other generalizations of Weierstrass’ theorem; see for example M. Krein and S. Krein [l] and E. Hewitt [2].
380
[VII.l
TOPICS RELATED TO MAPPINGS
Therefore
I
n
Since p E D is obvious, p satisfies our requirement. Thus the theorem is proved.
Corollary. Let X be a compact T2-space and D a ring of continuous functions with real scalar multiplication satisfying the conditions in L). Then for every continuous function f over X and for every E > 0 , there is a member of D for which
+
Proof. Combine the theorem with G) and I).
n
Example VII.4. Let F be a zero set of X = I T = {I, 1 a E T } , the product of closed segments I,, a E T. For each finite subset A of T we put I A= {I, I a E A}. Denote by C A ( X )the subset of C ( X )consisting of functions which depend only on A (namely, functions which can be expressed as p r TAfor the projection n-l from I T onto I Aand a function p defined on I " ) . Now put
n
0
D
=U
{C,(X) I A is a finite subset of T }.
Then D satisfies the conditions in L), and hence by the above corollary D is dense in C ( X ) with the topology of uniform convergence. Suppose F = {x E X I f ( x ) = 0}, where f E C ( X ) . For each n E N there is f n E D such that
VII.21
METTUC SPACE, PARACOMPACT SPACE AND CONTINUOUS MAPPING
381
Assume f , E CAn(X), n = 1,2,. . .. Then it is obvious that f depends only on A ’ = U i = l A , . Hence F depends on at most countably many coordinates for cy E A’.
2. Metric space, paracompact space and continuous mapping The purpose of the present section is to learn to what extent a continuous mapping satisfying certain conditions transfers properties of the domain space to the range space (or properties of the range space to the domain space). From a little different point of view we may ask, following P. S. Alexandroff (a) which spaces can be represented as images of ‘nice’ spaces under ‘nice’ continuous mappings? and (b) which spaces can be mapped onto ‘nice’ spaces under ‘nice’ continuous mappings? We have already dealt with such problems in the previous chapters and applied a result of t h e study to prove normality of a product space. In fact, in the final section of Chapter VI, we used Corollary 2 to Theorem VI.14, which asserts that every metric space is the image, by a perfect map, of a subspace of Baire’s zero-dimensional metric space. To accomplish a more detailed study of this subject, we shall adopt, in this section, chiefly metric spaces and paracompact spaces as our ‘nice’ spaces. To begin with, we give here some additional terminologies for mappings. Let f be a mapping of a topological space X into a topological space Y. Suppose that % is an open covering of X . If there is an open covering Y of Y such that f - ’ ( Y )= {f-’(V ) V E Y }is a refinement of % in X , then f is called a %-mapping. We often assume certain property of the inverse image f-’(q) of each point q of Y.Thus if f-’(q), for each point q of Y,is compact, countably compact, etc., then we call the mapping f compact, countably compact, etc., respectively. As we defined before, a compact, closed and continuous map is called a perfect map. We call a countably compact, closed and continuous map a quasi-perfect map.
I
‘P. S. Alexandroff [4]. This paper also contains various results in this aspect of investigation. We should also point out that there are many conditions of maps and related results which we could not discuss in this book. A. V. Arhangelskii’s extensive survey article 161 is recommended to readers who are interested in more details of this aspect. Also see E. Michael 191, F. Siwiec 111, and J. Nagata [7]. Some mathematicians studied multivalued mappings. For example, see V. Ponomarev [3], A. Okuyama [l], C. J. R. Borges 121, J. Nagata [8].
382
TOPICS RELATED TO MAPPINGS
[VII.2
First, we shall deal with metric spaces and their images and inverse images by continuous mappings. We showed in Section 4 of Chapter VI that the image of a metric space by a certain type of closed continuous mapping is also metric. As for open continuous mappings, we obtain the following. A) A TI-space X is the image of a metric space by an open continuous mapping if and only if X satisfies the 1st axiom of countability.'
'
This proposition is due to V. Ponomarev [l] and S. Hanai [l]. Ponomarev also proved that a topological space is the image of a metric space X by an open continuous mapping such that each f - ' ( q ) satisfies the 2nd axiom of countability, if and only if X has a point-countable open basis. S. Hanai (loc. cit) and A. Arhangelskii [2] proved that if a collection-wise normal space Y is the image of a metric space X by a compact open continuous mapping, then Y is metrizable. Actually, the latter proved that the TI-spaces with a uniform open basis, and only they, are open compact images of metric spaces. H. Wicke [2] characterized the regular spaces which are open continuous images of complete metric spaces. E. Michael [8], A. Arhangelskii [S]and S. Franklin [ l ] characterized the images of metric spaces by (continuous) bi-quotient, (continuous) pseudo-open, and quotient mappings, respectively. A mapping from X onto Y is bi-quotient if for each y E Y and each open collection % in X which coversf-'(y), there is a finite subcollection Q' of Q such that U { f ( U )I U E a'} is a nbd of y. Such mappings (invented by 0. Hijek and extensively studied by E. Michael [8]) have some interesting properties; e.g., the product of bi-quotient mappings is bi-quotient while the same is not true for quotient mappings (where for a collection {f, 1 a E A} of mappings from X, into Y,,the mapping f ( x ) = {f.(x,) 1 a E A}, x = {x,} from X, into Y, is the product of the mappings), and a continuous mapping f from a space X onto a T2-space Y is bi-quotient if and only if the product f x iz is a quotient mapping for every space 2, where iz is the identity mapping from 2 onto Z. A mapping f from X onto Y is pseudo-open (due to A. Arhangelskii and Yu. Smirnov) if for each y E Y and each open set U > f - ' ( y ) , f ( U ) is a nbd of y. Let f be a quotient mapping from X onto Y ; then it is known that the restriction off to f-'(Y'),where Y' is a subset of Y, is not necessarily a quotient mapping. Namely, 'quotient' is not hereditary. But pseudo-open continuous mappings are known to be precisely hereditarily quotient mappings. N. s. LaSnev [2] characterized the images of metric spaces by closed continuous mappings. F. G. Slaughter [l] proved that such images are MI-spaces. Interrelations between various (continuous) mappings are shown in the following diagram:
n o E ~ nu,,
open jbi-quotient jpseudo-open jquotient
.h. perfect
fi
3
closed
A collection 9 ' of subsets of a topological space X is called a pseudo-base of X if whenever C C U with C compact and U open in X , then C C P C U for some P E B. A regular space X is called an KO-space if it has a countable pseudo-base. E. Michael [7] defined KO-space and studied its properties. All separable metric spaces are KO. Every KO-space is Lindelof separable and M I . Michael characterized KO-spaces as the images of separable metric spaces by compact-covering mappings. A continuous mapping f from X into Y is called compact-covering if every compact set of Y is the image of a compact set of X. The same author also studied properties of the continuous images of separable metric spaces.
VII.21
METRIC SPACE, P A R A C O M P A ~SPACE AND CONTINUOUS MAPPING
383
Proof. Since the necessity is clear, we shall prove only the sufficiency. Let {%, I a E A} be the collection of all open sets of a topological space X satisfying the 1st axiom of countability. Then using its index set A, we construct Baire’s zero-dimensional space N ( A ) . We define a subset S of N ( A ) by S
= { ( a la , z ,. .
.)I
U u IU,,, , . . .form a nbd basis of a point p of X}.
Now we put f ( a )= p
if a
= (a,, a2,.
. .) and {Uul,Uu,,. . .} is a nbd basis of p .
Since X is T,, it is clear that this uniquely defines a mapping of S onto X. T o prove the continuity of f, we suppose that U is a given nbd of f ( a )= p, where a = (a,, a*,. . .). Then p E Uui CU
for some i .
If p ( a , a’)< l / i for a and a point a‘= (a;, a;, . . .) of S, then a ; = a I ,. . . , a : = ai.Therefore f(a‘)€
A UUkc u,
k=l
which proves that f is continuous. Finally, to prove that f is open, it suffices to show that f ( S , , , ( a ) ) is open for every natural number k and a E S, because { & / k ( a ) 1 k = 1,2, . . . ; a E S} is an open basis of S. To this end we suppose a = (a,,a 2 , . .); then we can show that
is clear, to verify t h e inverse relation, we take a given point p E ni = l Uui. We choo:e a sequence U P k t U l , P k + 2. ., .of open sets such that CUB,l j = k + 1, k + 2,. . .} is a nbd basis of p. Then a‘= (a,,. . . , f f k , & + I , . . .) is a point of S for which f(a’)= p. Since p ( a , a’)< l / k by the definition of the k
384
TOPICS RELATED TO MAPPINGS
[VII.2
metric of N ( A ) ,p E f ( S , , , ( a ) ) .Thus f ( S , , , ( a ) )= f7:=lU,, is proved. Since k is an open mapping.
n,,, Umiis open, we have proved that f
In the following is a result of the problem (b).
B) A T,-space X is paracompact if and only if for every open covering % of X , X can be mapped onto a metric space X by a continuous %-mapping.' Proof. Let f be a continuous %-mapping of a topological space X onto a metric space Y. Then by the definition of %-mapping there is an open covering 'V of Y such that
Since Y is metric and therefore paracompact, there is a locally finite open refinement Y f ' of 'V. Then f - ' ( V ) is obviously a locally finite open refinement of %. Therefore X is paracompact. To prove the converse, let % be a given open covering of a paracompact T,-space X. Then by III.2.C) there are locally finite open coverings 7'" and W of X such that
For the index set A, we consider the generalized Hilbert space H ( A ) . For each a E A, we define a continuous function f , over X such that
Then it is clear that f is a continuous mapping of X onto a subspace Y of H ( A ) , because 'V is locally finite. To prove that f is a %-mapping, we suppose that p is a given point of X ; then f ( p ) = { f , ( p ) 1 a E A } . Now, p E W,, for some member W,, of the covering W ; then for this index a. we can prove that
'
Essentially proved by C. H. Dowker [ 2 ] .V. Ponomarev [4] obtained interesting results in this aspect.
VII.21
METRIC SPACE, PARACOMPACT SPACE AND CONTINUOUS MAPPING
385
For if p‘ E f - ‘ ( S l ( f ( p ) ) )then ,
Since f,&p) = 1 by the definition (1) of f,,, f,,(p’) > 0 follows from the above inequality. Therefore p’ E Va,by (l), i.e. (2) is proved. Since Vaois contained in some member of % (remember that “Ir < %), we have proved that f is a %-mapping. We can extend Corollary 2 of Theorem VI.14 to paracompact spaces as follows. C ) Let X be a T2-space. Then there is a zero-dimensional paracompact T,-space S and a perfect mapping f of S onto X if and only if X is paracompact.’
Proof. Since the necessity of the condition is a direct consequence of Corollary 2 of Theorem V.3, we shall prove only the sufficiency. We denote by {9,1 A E A } the totality of the locally finite closed coverings of t h e paracompact T2-space X. Suppose
then we construct the product space P = n{A, 1 A € A } of the discrete spaces A,. We define a subset S of P by
It is clear that if a E S, then point a = {a,1 A E A } of S,
n{FUA1 A E A } is a point of X. Put, for every
Then f is a mapping of S onto X. Let a = (a,1 A E A} be an arbitrary ‘ D u e to K. Nagami 111. V. Ponomarev [2] proved a similar theorem for normal spaces.
386
TOPICS RELATED TO MAPPINGS
[V11.2
point of S and V a nbd of f ( a ) = p . Then there is a binary (and accordingly locally finite) closed covering SA0 such that
s(p,
SAo)
= FaAo
u(aA,J= { P I
P
%Ao
={PA
and
FaAo
v'
I A E A > E s, P A o = aA,J
is a nbd of a such that f(U(a,,)) C V. Hence f is continuous. Let p be a given point of X. For each A E A,
B, = {a' I a' E A,, p E Fa.} is finite, i.e., B, is a compact subspace of A,. Since
f - ' ( p ) is compact. To see that f is closed, we consider a closed set G of S and a point p of X such that
f - ' ( p ) n G = 0 in S We can assert that there are A,, . . . , A , E A such that for every a = {aAJ A E A } E f-'(p) and P = {PA1 A E A } E G, (a,,l,. . . , a h k ) f (pA1,.. . ,p,,). For, if not, then we denote by A the set of all finite subsets of A. We consider A as a directed set with respect to the usual inclusion relation. For each 6 = (A,, . . . , A,) E A we choose p(6) Ef-'(p) and $(a)€ G such that the coordinates for A,, . . . , A , of ~ ( 6 ) coincide with those of $(6). Since f - ' ( p ) is compact, the net p(A I >) has a cluster point a E f - ' ( p ) . It is easily seen that a is also a cluster point of $(A 1 >). Thus a E G, contradicting the fact that G is closed. Now we choose A,, . . . , A, E A satisfying the above condition. For each point p = (0, I A € A } of G, we consider F n - . .r l FpAk;then this 4I closed set obviously contains f ( p ) but not p. For if
VII.21
METRIC SPACE, PARACOMPACT SPACE AND CONTINUOUS MAPPING
387
then there is a
={a,(A€A)€f-'(p)
such that aAi= PA,, i = 1,. . . , k ,
which is impossible. Thus
9 =A { 9 , , 1 i = l ,..., k } is a locally finite closed covering of
X satisfying
f(G) C U { F I p !Z F E 9). Hence
v = x -u{F I p !Z F € 9)
is an open nbd of p which does not intersect f(G). Thus f(G) is a closed set of X,i.e. f is a closed mapping. It follows from V.2.D) that S is paracompact T2.Finally we must prove that S is a zero-dimensional space. Let 92 be a given finite open covering of S. Note that for each point a = {a,1 A E A } of S, {U(a,,)fl. * * n U ( a A tI )A , , . . . , A, E A , k
=
1,2, . . .}
forms a nbd basis consisting of open closed sets of S, where we put
Therefore there is an open closed covering "Ir with "Ir < OU. Since for each p E X , f-l(p)is compact,
for a finite number of elements Vp,, . . . , Vpmcp, of 7'. Putting
we get an open nbd W ( p ) of p since f is a closed mapping. Since
388
TOPICS RELATED TO MAPPINGS
[VII.2
{ W ( p )1 p E X } is an open covering of the paracompact T,-space X , there is an index A, E A for which
E AAo} is an open covering of S (see (l),(2)) such Then %' = { U((y,,)1 aAo that
We well-order all the points of X and put
where p is regarded as a variable ordinal number. Then we put
U p = U { U IU E % ' , U C V , , U ! Z V ,f o r e v e r y q > p } . Then {UpI p E X } is an open covering of X with order c1 satisfying UpC Vp for all p . Further we put
t
Up;= Upn V,
-
U
j=l
, i
=
1 , . . . , m ( p ) (see (3))
Then, since each Vpi is open and closed, Up; are open and mutually disjoint satisfying
up; c vp; fl up 7
m(P)
upc u up;. r=l
Thus %" = {Up;I i = 1 , .
. . , m ( p ) ;p
EX }
is an open refinement of 011 with order s l , because V, E w^ < 011. This proves that S is zero-dimensional.
VII.21
METRIC SPACE, P A R A C O M P A ~SPACE AND CONTINUOUS MAPPING
389
In the following is a classical theorem of P. S. Alexandroff, who pioneered to represent spaces as images of zero-dimensional spaces.’ The proof is similar to the previous proof. D) A T,-space X with weight p is compact if and only if there is a closed subset C of D’ (see Theorem V.8), and a continuous mapping f of C onto X . (Note that such a closed set C is a zero-dimensional compact T2-space.)
Proof. The sufficiency of the condition is clear. To prove the necessity, we assume that X is a compact T2-space with weight p and denote by {V,1 A E A } an open basis of X with IA I = p. We repeat the process in the proof of C), replacing the collection of all locally finite closed coverings there with the collection {9’,1 A E A } of binary closed coverings, where 9,= { , X - U, }. Put 9’’ = {Fa1 a E A, }. Then we can prove in a similar way that there is a continuous mapping f of a subset C of D” = {A, I A E A} onto X , where
u,
n
Now we have to prove that C is closed. To do so, suppose y
=
I
{ y , A E A} fZ C where y, E A,, A E A. Then by the definition of C,
Since X is compact,
“{FyAiI i = 1 , . . . , k } = 0 for some A,, . . . ,A, E A. Therefore, U ( y , , )fl.. . f l U(y,,) is a nbd of y in D’ which does not intersect C, where
I
u(yA) = { y ’ 1 7’ = (7; A
1 D’,
y; =
YA} ‘
Hence C is closed.
’
There is an interesting result concerning inverse image of a compact space. A. Gleason [ 11 proved that for any compact T2-space X, there is an extremely disconnected, compact, T2-space S and an irreducible continuous mapping 4 from S onto X and moreover that such (S, 4) is unique in the sense that if there is another ( S , 4’) satisfying the same condition, then there is a topological mapping (I, from S’ onto S satisfying 4‘ = 40 (I,. A continuous mapping 4 from S onto X is irreducible if for any closed set C 2 S, 4(C) f X. This uniquely determined space S is called the absolute of X. Theory of absolute wasextended by V. I. Ponomarev [5],S. Iliadis [l], J. Flachsmeyer [2] and others to more general spaces. See also P. Alexandroff-V. Ponomarev [2], V. I. Ponomarev [6].
390
TOPICS RELATED TO MAPPINGS
[VII.2
In the following is an answer by K. Morita [5] and A. V. Arhangelskii [4] to Alexandroff's problem in a very important case.
Theorem W.3. A TI-space X is a n M-space if and only if there is a quasi-perfect m a p f from X onto a metric space Y. hoof. Let f be a quasi-perfect map from X onto a metric space Y with a sequence {ViI i = 1,2, . . .} of open covers satisfying the conditions of Axexandroff-Urysohn's metrization theorem. Then %; = f -'(Vi), i = 1,2,. . . , form a normal sequence of open covers of X . It also satisfies the wA-condition in Definition VI.11. To see it, assume xi E S(x, %;), i = 1,2, . . . , and { x i } has n o cluster point. If xi E f-'Cf(x)) for infinitely many i's, then { x i } has a cluster point, because f is countably compact. Otherwise xi f -'( f ( x ) )for i 2 i,. Then F = {xi 1 i 3 i,} is closed in X while f ( F ) is not, because f ( x i ) + f ( x ) in Y and f ( x ) e f ( F ) .This is a contradiction. Hence { x i } must have a cluster point, proving that X is an M-space. Conversely, assume that X is an M-space with a normal sequence {%i I i = 1,2, . . .} of open covers satisfying wA-condition. Define equivalence in X by
e
-
x
-y
m
if and only if y E f l S(x, %;). i= 1
Then we denote by Y the quotient space XI- and by f the quotient map from X onto Y. It is obvious that Y is a Tl-space and f is a countably compact continuous map. To see that f is closed, suppose F is a closed set of X and x E f - ' ( f ( F ) ) . Put f ( x ) = y ; then x E f - ' ( y ) C X - F . Thus for some n S(x, %,) f l F = 0. Because otherwise there are x,, E S(x, %,,) n F, n = 1,2,. . . , and thus a cluster point x' of { x , } , which satisfies x ' E F f l f - ' ( y ) , a contradiction. Hence
Therefore x $Zf-'Cf(F)). This proves that f-'Cf(F)) is a closed set of X. Hence f ( F ) is a closed set of Y proving our claim. Hence f is a quasi-perfect map. Put
I
V , = { Y - f ( X - U ) U E %,} . Then, as easily seen, {V,,I n
=
1,2, . . .} is a sequence of open covers of Y
VII.21
METRIC SPACE, PARACOMPACT SPACE AND CONTINUOUS MAPPING
391
satisfying the conditions of Alexandroff-Urysohn's metrization theorem. Thus Y is a metrizable space. Corollary 1. A TI-spaceX is paracompact and M perfect map from X onto a metric space.
if and only if there is a
Proof. If there is such a perfect map, then X is M by the theorem and also paracompact by V.2.D). Conversely, if X is paracompact and M, then there is a quasi-perfect map f from X onto a metric space Y. Since f - ' ( y ) is paracompact and countably compact for each y E Y, it is compact. Namely, f is a perfect map. Corollary 2. Let X i , i = 1,2, . . . , be paracompact T I and M ; then so is the product space =; X,.
n
Proof. Note that the product of perfect maps is a perfect map; then this corollary follows from the previous one. Corollary 3. A T,-space X is paracompact and M if and only if it is homeomorphic to a closed set of the product of a compact T2-space and a metric space.'
Proof. The sufficiency of the condition is obvious. Assume that X is paracompact T2 and M. Then by Corollary 1 there is a perfect map f from X onto a metric space Y. Then we define a map cp from X into p ( X )x Y by
where i denotes the imbedding of X into p ( X ) .Then it is obvious that cp is a topological imbedding of X i n t o p ( X ) X Y. To prove that cp(X) is closed in p ( X )x Y, let (u, v ) E p ( X ) x Y - cp(X). Then u I f f - ' ( u ) holds in p ( X ) . Since f - ' ( v ) is compact, it is a closed set of p ( X ) . Hence there are open sets U and V of p ( X ) such that
'
Due to J. Nagata [ 111. A. Kato [ 1) and D. Burke-E. v. Douwen [ 11 gave an example of an M-space which is not homeomorphic to a closed set of any product of a countably compact space and a metric space. See also A. Kato [2].
392
TOPICS RELATED TO MAPPINGS
[VII.2
Put W = Y - f ( X - V). Then W is an open nbd of u in Y. Thus U x V is a nbd of (u, v ) in p ( X )x Y satisfying U x V n p ( X ) = 0. This proves that p ( X ) is closed in p ( X ) x Y. Corollary 4. Every paracompact T2-M-spaceX is a P-space. Proof. By Corollary 2 the product of X with every metric space is normal. Hence this corollary follows from Theorem VI.24. Corollary 5. A Tychonoff space X is paracompact and Cech complete if and only if there is a perfect map f from X onto a complete metric space Y.' Proof. Assume that such f and Y exist. Then, by use of the method of the proof of Corollary 3 to Theorem VII.3, we can show that X is homeomorphic to a closed set of the product space p ( X ) X Y. Since the last space is t e c h complete and paracompact, so is X . Conversely, assume that X is paracompact and t e c h complete. Then by Theorem VII.3 there is a perfect map f from X onto a metric space Y. By Exercise VI.30 Y is t e c h complete. Hence by tech's theorem Y is completely metrizable. Example W . 5 . Sorgenfrey line S is a P-space. For, let {U(a,,. . . , ai) I a , , . . . , a; E A ; i = 1 , 2 , . . .} be an open collection of S satisfying U(a,,. . . , a;)C U(a,,. . . , a;+,). For each i we put F(a,, . . . , ai)= the closure of the set {x 1 x E S, (x - l/i, x + l / i )C U(a,,. . . , ai)or (x - E, x + E ) f2' U ( a , ,. . . , a;)for every E > 0 and [x, x + l / i )C U(a,,. . . , a;)}. Then it is easy to see that F(a,, . . . , ai)satisfies (i), (ii) of Definition VI.7. But S is not an M-space. Because if it were, then S X S would be paracompact T2 and M, which is not true. Example VII.6. Let X be a Tychonoff space. Then we denote by p ( X ) the completion of X with respect to its strongest uniformity. Suppose T is an arbitrary metric space and f a continuous map from X into T. Then, since p ( T )= T, f can be extended to a continuous map p ( f ) : p ( X ) + T. Since X is dense and C-embedded in p ( X ) , p ( X ) C y ( X ) C p ( X ) follows. Also observe that p ( X ) is the smallest DieudonnC complete space in which X is dense and C-embedded (C*-embedded). Because if X is dense and C*-embedded in a DieudonnC complete space Y, then X C
' Due to Z . Frolik [3].
VII.21
METRIC SPACE, PARACOMPACT SPACE AND CONTINUOUS MAPPING
393
Y C P ( Y )= P ( X ) . Let cp be the topological imbedding of X into Y. Then p ( p ) maps p ( X ) into p ( Y ) = Y. Since p ( 9 ) is easily seen to be a topological imbedding, p ( X )C Y follows. Thus p ( X )can be expressed as p ( X ) = {x E p ( X ) every continuous map from X into any metric space T can be continuously extended to x } = fl {P(f)-'(T) is a continuous map from X into a metric space T },
If
where P ( f ) : P ( X ) + P ( T ) denotes the continuous extension off. Now assume that X is an M-space. Then it is easy to see that p ( X )= fl { P ( f ) - ' ( T1)f is a quasi-perfect map from X onto a metric space T }.
Because, let f be a continuous map from X into a metric space T. Then there is a metric space T' and a quasi-perfect map g from X onto T' and a continuous map h from T ' into T such that f = hog. Then P(g)-'(T')C P ( f ) - ' ( T )follows, and so does the above expression of p ( X ) . Suppose f : X + T and g : X + T ' are onto quasi-perfect maps, where T and T ' are metric spaces. Iff = h o g for a continuous map h from T' into T, then h is a perfect map. Hence it is easy to see that P(h)-'(T)=T', and accordingly
Thus p ( X )= P(f)-'(T)for any metric space T and a quasi-perfect map f from X onto T.This implies that the restriction p v ) of P ( f ) to p ( X ) is a perfect map. Thus X is a paracompact M-space.'
E) Every normal M-space X is countably paracompact and collectionwise normal. Proof. By the theorem there is a quasi-perfect map f from X onto a metric space Y. Suppose {Fa I a E A} is a given locally finite closed collection in X . For each y E Yf-'(y) is countably compact, and hence it We owe the concept and properties of p ( X ) in the above to K. Morita [6].See the same paper for further results on p ( X ) .p ( X )may be called the paracornpactificarion of X when it is paracompact.
394
TOPICS RELATED TO MAPPINGS
[vII.2
hits at most finitely many of F,, a E A . Thus {f(F,) I a E A } is pointfinite. Since f is closed, {f(F,) 1 a € A } is a closure-preserving and accordingly locally finite closed collection in Y. Hence there is a locally finite open collection {V, a E A } in Y such that f(F,)C V,. Now { f - ' ( V , ) I a E A } is a locally finite open collection in X satisfying F, C f-'( V,). Thus the proposition follows from V.3.D).
I
F) Let f be a quasi-perfect map from a normal M-space X onto Y. Then Y is an M-space.' Proof. Let {%i / 1 , 2 , . . .} be a normal sequence of open covers of X satisfying wA-condition. By virtue of V.5.C) and VI.7.D) we may assume that each %; is locally finite. Then by III.2.C) we can shrink % l i to a locally finite closed cover 9',. Since f-'(y) is countably compact for each y E Y, it intersects at most finitely many members of 9?Thus Yi =f(R.) is a point-finite closed cover of Y. Since 3; is closure-preserving because of the closedness off, it is locally finite. Put
Note that Y is collectionwise normal and countably paracompact because of E) and Exercises V.5 and V.14. Thus by V.3.D) each G, can be expanded to an open set P, such that Pa C W, and such that W,= { W, I a E A,} is locally finite. Hence W, is a normal open cover of Y such that W, < 3 : . We can show that {W,} satisfies wA-condition. Suppose y, E S ( y , W,),i = 1,2,. . . , where y is a fixed point of Y. Then there are G',, G;, G; E 3, and y',, y; E X such that
Thus there are Fi, F;, F j E
R
such that
Due t o T. Ishii [3] and K. Morita [ 8 ] . V. Filippov [2] proved it in case that X is paracompact.
VII.21
METRIC SPACE, PARACOMPACT SPACE AND CONTINUOUS MAPPING
395
Pick x, E f-'(yi), i = 1,2,. . .. Then we can prove, by use of quasiperfectness of f and wd-condition and normality of {%;}, that ( x i } has a cluster point. Thus it follows that { y , }has a cluster point. The detail is left to the reader. Select a normal sequence {Wi} of open covers in Y such that < W,.Then this sequence also satisfies wd -condition. Hence Y is an M-space.' The following interesting theorem was first proved by N. S. Lasnev [l] for metric spaces and extended by V. Fillipov [l] to paracompact M spaces and the present form is due to T. Ishii [4].2
Theorem VII.4. Let f be a closed continuous map from a normal M-space X onto Y. Then Y can be decomposed as Y = U Y,, where Y, is a discrete (closed) subset of Y for each n 3 1, and f - ' ( y ) is countably compact for every y E Y o .
z=o
Proof. Let {a,} be a normal sequence of open covers of X satisfying wA-condition. Defining equivalence in X by
-
x
-y
m
if and only if y E
n S(X,a,,), n=l
we decompose X into the sum of disjoint equivalence classes, which are countably compact closed sets of X and denoted by Fa, a E A. Put Van= S(Fa,a,,), n = 1,2, . . . . Then {Uan1 n = 1,2, . . .} is an outer base of Fa. In the rest of the proof we mean by a sequence a point sequence, say { y , } , such that y , # yj whenever i # j . For each n E N we put
Y, = { y E Y [ for each sequence { y , } in Y, there is a E A for which F, nf-'(y) f 0 and Vann (U{ f l ( y i ) I i E N ' } ) = 0 for some infinite subset N' of N } . Also put
'
The image of a general M-space by a perfect map is not necessarily an M-space. The images of M-spaces by perfect maps and by closed continuous maps were characterized by J. Nagata [S] and K. Morita-T. Rishel [l],respectively. H. Wicke [l]characterized T2-spaces which are open continuous images of paracompact T2-M-spaces. See K. Morita [7] for further results on M-spaces. 'Ishii proved this theorem under a more general condition. A. Okuyama [ S ] and R. A. Stoltenberg [l] proved the same theorem for a normal cr-space and for a normal semistratifiable space, respectively.
396
TOPICS RELATED TO MAPPINGS
[VII.2
m
Y o =Y -
u Y,. ,=I
Let us prove that Y, is discrete for each n 2 1. Assume the contrary; then c f - ' ( y ) 1 y E Yn} has a cluster point x EX, i.e., each nbd of x intersects f - ' ( y ) for infinitely many y ' s , because f is closed. Suppose x E F,. Then there is a sequence { y , } in Y, such that
U,,#0,
i = l , 2, . . . .
i = 1,2,.. .} has no discrete subsequence.
(1)
(2)
Let N' be an arbitrary infinite subset of N. If f - ' ( y i ) f l F, # 0 for infinitely many i E N ' , then, since F, is countably compact, { f - ' ( y l ) I i E N ' } is not discrete. Assume f - ' ( y , ) f l F, = 0 for all i E N ' with i 3 i,. Then F = U { f - ' ( y , ) i E N ' , i 2 i,} is disjoint from F,. Since F fl U,, # 0 for all i follows from (l), F is not closed. Hence { f - ' ( y , ) 1 i E N ' } is not discrete, proving our claim (2). E A such that Since y , E Y,,there is (~(1)
I
and a subsequence N2 of N , = N such that
Let i(1) = 1 and choose i(2)E N2 such that i(2) > i(1). Since Y , ( ~E) Y,, there is a (2) E A such that
and a subsequence N3 of N2 such that
Choose i(3) E N3 such that i(3) > i(2). Repeating the same process we obtain a subsequence i(l)< i ( 2 ) c e . a of N, a sequence { a ( k ) I k = 1,2, . . .} of elements of A and a sequence N , 3 N23 N33 * of infinite
vII.21
METRIC SPACE, PARACOMPACT SPACE AND CONTINUOUS MAPPING
397
subsets of N such that
Then the sequence { x , } does not cluster. Because, if x is a cluster point of i X k } , then
and also by (3) m
u
Ua(k).
k=l
This is a contradiction. because
s(ku- I
%n)
=
u
k=l
Ua(k)n
.
Hence { x k } is a discrete sequence. Hence { f ' ( ~ ~ 1 k (=~1,2, ) ) . . .} is also discrete, which contradicts (2). Thus Ynis discrete Now, let y be a fixed point of Yo. Then we claim that f - ' ( y ) is countably compact. Suppose
If Fa(;)coincide for infinitely many distinct i's, then { x , } has a cluster f point, because Fa is countably compact. So we assume that whenever i f j . Now we assume that {Fa(i)I i = 1,2, . . .} is discrete. Then, since X is normal, there is a discrete open collection {V, I i = 1,2, . . .} such that V,3 Fa(i). Select a subsequence { I ( i ) 1 i = 1,2, . . .} of N such C Ui. For each n 5 1 there is a sequence { y n I i = 1,2, . . .} in that Urn(;),(;) Y such that for every a with Fa n f - ' ( y ) # 0 and for every subsequence N' of N,
398
TOPICS RELATED TO MAPPINGS
[VII.2
because y @ Y,. Then there is j ( n ) E N such that
We can select { j ( n ) }such that j(1) <j(2) K O for some n. For distinct elements U,, . . . , U, of % we
Now, fix a point x, E X. Then for some U, E % with x, E U, we have lZ(U,)l > KO, because Z,, = U { Z ( U )I x E U E Q}. Hence V , # X and hence we can select x2 E X - U,. There is U2E % with x2 E U, and
Continue the same process until getting 2 ( U , ,. . . , U,,) such that
However, 2 ( U l ,. . . , U,,)consists of the only element q0= {U,, . . . , U,,}. This contradiction proves the proposition.
Theorem VII.6. Let % be a point-countable p-base of a countably compact T,-space X . Then X is metrizable.'
'
Due to A. MiSEenko [l]. In this proposition X may be just a set, because no topology of X is used in the following proof. * Due to A. MiSEenko [l].An open collection 43 in X is called a p-base if for each x E X
n{uI
E LIE
431 =
VII.31
METRIZATION OF M-SPACES
405
Proof. First we claim that every cover V of X by elements of % has a countable subcover and accordingly a finite subcover, too. To prove it, assume t h e contrary. Pick xIE X ; then S(x,, 7'") f X , because "Ir is countable at xl. Select x* E
x
-
S(X], V )
Then S(X], V )u S(X,, V ) #
x.
Select X3
E
x - S(X,, V )u S(X,,
V ).
Continue the same process t o get a sequence { x i } such that n-1
x, E
x - u S(X;, V ) . i= 1
Then { x i } has no cluster point, which is impossible. Thus Zr must have a countable subcover. By B) there are only countably many minimum covers by members of %, which we denote by V , ,V2,. . . . Let x and y be distinct points of X . Then we can show that y @ S(x, V n )for some n. Since 42 is a p-base, there is U E % such that
For each z E X - U, choose U ( z )E % such that
zE
U(Z),
x
e U(z)
Then, as observed before, { U, U ( z )1 z E X - U } contains a minimum subcover, say Vn.It is obvious that S(x, V n )= U 3 y. Thus the sequence {V, n = 1,2, . . .} satisfies
I
m
n S(x, V n )= {x}
for each x E X .
n=l
Namely, X has a G,-diagonal. Hence by Theorem VlI.5 X is metrizable.
406
[VII.3
TOPICS RELATED TO MAPPINGS
Theorem VII.7. A T,-space X is metrizable if and only if it is M and has a point-countable p - base.' Proof. We shall prove only the 'if' part. It follows from Theorem VII.3 and Theorem VII.6 that X is the inverse image of a metric space by a perfect map, and hence it is paracompact. Hence we can select a sequence { % i } of locally finite open covers of X satisfying wd-condition. We put %, = {V, I a E A,} and denote by 7"a point-countable p-base of X . Then for a fixed a there are at most countably many (finite) minimum covers of U, by members of 7"(B)), which we denote by Yarn,m = 1,2, . . . . We can express that m
u
Yam= {VahJ h
=
1,2,. . .},
m=l
because U",=, Yamis a countable collection. For each (n,1 ) E N define
X
N, we
W,,,={U(a,hl ,...,h , ) I a E A , ; l G h l ,..., h k S l ; k = l , . . . , f},
(1)
where
U ( a ,hl, . . . , hk) = U, - V,,, U . . . U V,hk.
(2)
Then it is obvious that Wnlis locally finite. Now we claim that U:,,=, Wn, is a network of X . Suppose P is a given open nbd of a given point x of X . Put m
c = n qX,a,,). n=l
Then C and accordingly C- P, too, are compact. Thus there are V,,. . . , V, E "Ir such that C - P C V , U . . * U v,,
V, n (C - P ) f 0, i = 1, . . . , k , xsf
'
(3)
v , u . . . u v,.
This theorem was first proved by V. Filippov [ l ] in the form that a paracompact Tz-space is metrizable if and only if it is M and has a point-countable base, and then improved by several people including J. Nagata [12], T. Shiraki [l], F. G. Slaughter and others.
METRIZATION OF M-SPACES
VII.31
407
For each i with 1 s i S k, we can construct a minimum cover V; of C by members of V such that V, E V;. Then
where V:,=U{Vl V E
V;}.
Now, we can find n E N and a E A, such that
Then we have U,-PCV,U*..U
v,,
because x'E U, - P implies
Note that V,, . . . , V, € U : = , V,,, because by (4) and (5) V; is a minimum cover of U,, i.e. V, E V; = Vmmi for some m ; . Now, assume that V,=Vuhi, i = l , ..., k , and h
= max{h,,
. . . , h,}. Then from (l),(2), (3) and (6) it follows that
.x E U ( a ,h,, . . . , h,) C P
and
U ( a ,h,, . . . , h k )E Wnh.
This proves that uz,h=, W,, is a network of X . Hence X is a u-space and accordingly semi-stratifiable. Thus X is metrizable by Corollary 2 of Theorem VII.5. There are many other remarkable results obtained on metrizability of generalized metric spaces. For example, A. V. Arhangelskii [12] proved that every regular space which is hereditarily Lindelof M is secondcountable. A T,-space which is hereditarily M is called an Fpp-space.Z . Balogh and A. V. Arhangelskii proved that every Fpp-space contains a dense metrizable subset.' The latter also gave a characterization of ' S e e Z . Balogh [l], [2] and A. V. Arhangelskii [ll].
408
TOPICS RELATED TO MAPPINGS
[VII.4
non-metrizable Fpp-space and showed that every perfectly normal F,space is metrizable. Recently, subconditions of metrizability like developability, p and m are further factored into weaker conditions. R. Hodel and others obtained many interesting results in this aspect.' Some examples of such factorization are: A regular space is developable if and only if it is semi-stratifiable and wA.* A T,-space is a a-space if and only if it is a Z-space with a G,-diag~nal,~ where a topological space X is called a 2-space if it has a sequence PI,.F2,.. . of locally finite closed covers such that x, E f7 {F 1 x E F E 9,,}, n = 1,2, . . . , for a fixed point x implies that {x,} has a cluster point. Every M-space and every regular u-space are 2-space, and Theorem VI.27 can be extended to 2-spaces. In the following is an interesting metrization theorem of 2-space due to E. Michael, T. Shiraki and F. Slaughter: A topological space is metrizable if and only if it is a collectionwise normal 2-space with a point-countable base4; another interesting theorem' in this respect is that a semi-metric space with a point-countable base is developable.
4. Theory of inverse limit space
In preceding sections we have learned to characterize certain types of spaces as the continuous images or inverse images of concreter spaces. In the present section we shall try to characterize spaces in terms of the concept of inverse limit space, which was defined in Definition 11.18. Namely, some types of compact spaces will be characterized as the limit spaces of concreter spaces. Let E be a finite set of elements which are called vertices and K a collection of subsets of E such that every subset of a set belonging to K also belongs to K ; then we call K an (abstract) complex. We call a set of n t 1 vertices a,, . . . , a, which belongs to K an (abstract) n-simplex and denote it by la,. . . a , / . An m-simplex ( a 6 .. . a i m (whose vertices are chosen from {a,,,. . . , a,} is called an m-face of ( a , .. . a,(. 'See R. Hodel [l], [2], [ 5 ] . 'Due to C. Creede [l]. H. Wicke-J. Worrell [l] and H. Brandenburg [l], [2] also gave interesting characterizations of developability . K. Nagami [2] invented the concept of P-space and proved this theorem for paracompact spaces, which was then improved by several other people. See E. Michael [ll] and T. Shiraki [l]. Due to R. Heath [ S ] .
VII.41
409
THEORY OF INVERSE LIMIT SPACE
Now, let us consider r + 1 points
ai = (uy),. . . , a:)), i = 0 , . . . , r , in an n-dimensional Euclidean space E". If the rank of the matrix (by)), where bj" = a?) for i = 0, . . . , r, j = 1 , . . . , n, and bill = 1 for i = 0, . . . , r is equal to r 1 , then the points { a , , . . . , a,} are called linearly independent. For linearly independent points a,, . . . , a,, we define a subset
+
[ a , . . . a,] = {(xl, . . . , x,) 1 xi = ,+,a$)+ . . . + A,ay), i = 1, . . . , n , A,+.
. .+A,
=
1 , O S A,, . . . , A,
S
1)
of E n and call it the geometrical r-simplex spanned by a,, . . . , a,. The latter are called the vertices of the simplex. The point (xl, . . . , x,) of the simplex with A, = A , = * * = A, = l / ( r + 1) is called the burycenter of the simplex. Each geometrical s-simplex spanned by s + 1 points chosen from a,,, . . . , a, is called an s-face of [ a , . . . a , ] . Geometrical 1-, 2- and 3simplices are a segment, triangle and tetrahedron, respectively. Suppose K is a finite collection of geometrical simplices in E". If it satisfies: (i) every face of a simplex belonging to K also belongs to K , (ii) the intersection of two simplices belonging to K is either empty or a face of each of those simplices, then K is called a geometrical complex. The union of t h e simplices (regarded as a subset of E n ) belonging to a geometrical complex forms a closed subset of E n and is called a polyhedron in E".
Example VII.8. Suppose K is a geometrical complex. We consider all vertices of the simplices belonging to K and consider the vertices a,, . . . , a, to form an abstract simplex if and only if they are the vertices of a simplex belonging to K. In this way every geometrical complex determines an abstract complex. Another significant example of abstract complex is a nerve. Suppose X is a topological space and % = {U, I i = 1, . . . , n } is a covering of X . Define a subset {qk I k = 0, . . . , r } of OU to be a simplex if and only if
k=O
Then we obtain an abstract complex having {U, 1 i
=
1,.
. . , n } as its
410
TOPICS RELATED TO MAPPINGS
[VII.4
vertices. This complex is called the nerve of the covering %. The concept of nerve is a key which connects general topology with algebraic topology. Let K be a geometrical complex. Then we define the barycentric subdivision K , of K as follows: For each n-simplex X " of K,we take the barycenter a ( x " ) of x". Then K , is the collection of all the simplices [ a ( x ; ) a ( x : ; ) . . . a(x:;)],where
and XZ is a face of x?;;. As easily seen, K , is also a complex. It is called the barycentric subdivision of K. We often consider successive barycentric subdivisions K,, K,, K3,. . .. Then it is easy to see that mesh K,, + 0 as n+w. First we shall characterize every compact T,-space as the limit space of polyhedra. A) Let { X u ,n$ I a, p E A, a > p } be an inverse system of compact T2spaces. Then the inverse limit space
X
= lim {X,, t
T;}
is a compact T,-space.
Proof. Since X is a subspace of the product space P of the compact T,-spaces X,, a E A, all we have to prove is that X is closed in P. Let
Then there are a,p E A such that p < a and 7r;(pa)f p p in Xp. Since Xp is T,, we can choose disjoint nbds U and V of n-;(pa) and p p , respectively. Since n-; is continuous, there is a nbd W of p , in X , such that T;(W )C U. Then the nbd
of p in P does not intersect X , because r;(q,) # qa holds for every q E N ( p ) . Therefore X is a closed set of P, i.e. X is a compact T,-space.
VII.41
411
THEORY OF INNVERSE LIMIT SPACE
B) Let F be a closed set of a finite-dimensional cube I" and U an open set of I " such that F C U. Then there exists a polyhedron P such that F C Int P C P C U . Proof. Since I" is a polyhedron, we can construct its successive barycentric subdivisions K,, K 2 , .. .. We denote by KA the complex which consists of the simplices of K , intersecting F and of their faces. Let P, be the polyhedron determined by KA ; then since mesh K ; + 0, m
m
Int P, = n=l
n P, = F . n=l
Hence by the compactness of I",P, C U for some n. (Note that P, 3 Pn+l.) Thus P, = P is the desired polyhedron.
Theorem VII.8. A topological space X is compact T2 if and only if it is homeomorphic with the inverse limit space of an inverse system of polyhedra. Proof. The sufficiency is implied by A). To prove the necessity we assume that X is a given compact T2-space. Then by Tychonoff's imbedding theorem it is homeomorphic with a closed set X' of the product space I A = {I, 1 a E A } of closed unit segments I, = [0, 11, a E A. Let B be a subset of A ; then we denote by I B the product space I B = {I, a E B}. Suppose B and C are subsets of A with B 3 C ; then we denote by f: the projection of onto I", i.e.,
n
for each point {x, I a E B } of I B . Therefore I B onto itself. Furthermore we denote by
I
ft is the identity mapping of
the collection of all polyhedra P, contained in I B for some finite subset B of A satisfying
' Namely I" is the product space of n copies of the unit segment [0,1].
412
[VII.4
TOPICS RELATED TO MAPPINGS
f G ( X ’ )c Int P,
in I B
Incidentally, we denote by B ( y ) the smallest finite subset of A such that P, is in IB(,).We define an order between two members y and 6 of r as follows: y
< 6 if and only if B ( y ) C B ( 6 ) and fi::;(P8)C P, .
To prove that
r is a directed set, we suppose y, 6 E r. Put
is a polyhedron in I c satisfying f;((x’)
c Int P, ,
Thus P E E P ’ ,and E is a member of Therefore is a directed set. Now, we consider an inverse system
r
where
TT;
is the restriction of
X, = lim {P,, T;} c
fi$; to P,.
r
such that
E >
y and
&>a.
Let
;
then we can prove that X ’ and X , are homeomorphic. To this end we consider a mapping f of X ‘ into X , defined by
From the property of projection, it is clear that f ( x ) is a continuous mapping of X ’ into X,. To prove that f is onto, take a point y = {y, I y E of X,. For each a E A , I, E 9; hence we put I, = PY(,), where
r}
VII.41 y ( a )E
THEORY OF INVERSE LIMIT SPACE
r. Now x = {x, 1
seen that f ( x ) = y if
413
E A } for x, = y,(,) is a point of I A . It is easily x E X ’ . Assume x E I A - X ’ ; then since X ’ is closed, a!
for some finite subset B of A. By B) there is a polyhedron P, in I B such that
f;(x’) c Int P, c P,
3f i ( x ) .
(1)
We may assume without loss of generality that B ( y ) = B. On the other hand, from the definition of x it follows that
where we put I B = P,.. However, since
P,
B ( y ) = B ( y ‘ ) = B,
C P,’
,
we have y > y‘. Hence
This combined with (1) and (2) implies y , $2P,, which contradicts that y = { y , I y E r }is a point of X,. Therefore x E X ’ , i.e. f is a continuous mapping of X ’ onto X,. Finally, to prove that f is one-to-one, we assume that x and x‘ are different points of X ’ . Then’
f t ( x ) # f t ( x ’ ) for some Therefore, assuming that P,
=
a!
EA .
I,, we obtain
’ Actually the subindex a means the one-element set {a}.
414
TOPICS RELATED TO MAPPINGS
[VII.4
Applying III.3.D) we can conclude that f is a homeomorphic mapping of X ' onto X,. Thus X and X , are homeomorphic. Revising the proof of the preceding theorem slightly we get the following theorem originally due to H. Freudenthal [l]. Corollary.' A topological space X is a compact metric space if and only if it is homeomorphic with the limit space of an inverse sequence n-i 1 i, j = 1,2, . . . ; i >j } of polyhedra Pi.
{c.,
Proof. To adjust the preceding proof to the present case, we imbed X into the product I" of countably many unit segments and consider a sequence of finite dimensional cubes I", n = 1,2, . . . , and a sequence of polyhedra P,, in I", n = 1 , 2 , . . . , such that
n-;(X')C Int P,,
c Sl,n(n-z(X'))
and n-:+'(~,,+~) c P,
in I " ,
where n-f: denotes the projection of I k onto I". Then we can show that X ' is homeomorphic with the limit space lim{Pi, T;}. The detailed proof is t left to the reader. Next, let us turn to the theory, which was originated by P. S. Alexandroff, of representing a given compact space as the limit space of finite spaces. We shall describe the theory along the line of J. Flachsmeyer [l] who used decomposition spaces while the original method of Alexandroff used the concept of nerve. Definition VII.5. A topological space X is called a To-space if for each pair p, q points of X , there is either a nbd U of p with q sf U or a nbd V of q with p Sr V. Tois a weaker condition than Tl.
Let us temporarily denote by p < q the relation between two points p, q of a T,-space X that p E (4). Suppose { X , n-; 1 a, p E A, a > p } is an inverse system of T,-spaces X, ; then we define order between two points
'
This type of theorem is especially interesting in connection with dimension. As a matter of fact, Freudenthal proved that every compact metric space of dimension G n is homeomorphic with the limit space of an inverse sequence of polyhedra of dimension G n. S. MardeSit [l] proved that every compact Tz-space of dim =zn is an inverse limit of compact metric spaces of dim S n.
VII.41
415
THEORY OF INVERSE LIMIT SPACE
I
p = { p , 1 a € A } and q = {q,l a € A } of the product space n { X , a € A } of X , as follows: p c q if and only if p ,
q, for all a E A
Now in view of the fact that X , is To,n X , , and consequently the limit space X , of the inverse system, turns out to be a partially ordered set with respect to this order. Then we mean by a minimal point of X , a minimal element of this partially ordered set.
C) Let {X,, T; I a, p E A, a > p } be an inverse system of finite To-spaces, X,. Then the subspace X of the minimal points of the limit space X , is a compact T,-space. Proof. Suppose p , q are different points of X.Then since p , q are minimal points,
p g q
and
qgp,
i.e.,
Pa@=
and
qp@(p,)
for some a, p E A. This implies that p @ (9) and q @ (p>, i.e., X satisfies the condition for T I . To prove the compactness of X,we suppose that cp(A I >) is a maximal net of X. We denote by P the set of the convergence points of cp in X , , i.e., P = { P I P P ( A I> ) + P E X m ) .
To prove P f 0, denoting by ( ~ ~ the ( 6 )a-coordinate of (p(6),we note that for each a E A, cpo,(A1 >) is a maximal net of X,. Since X, is finite, cp,(A 1 >) is residual in some point p , E X , . We take such a point p, for X,. Then clearly each a E A to construct a point p = { p , I a E A} of cp(A I >)-.p. Let a > p , a, p E A; then
n
which implies
because ~ ( 6E )X , . Thus p E X , , i.e. p E P, which implies P = 0.
416
[VII.4
TOPICS RELATED TO MAPPINGS
Suppose P' is a totally ordered subset of P. Then for each a E A, we Put
this is possible because X , is finite. Then p' = {pi I a E A} satisfies p' s p for every point p E P'. Let a,p E A, a > 0; then there is p EP' such that ps = pi. Then .rr;(p,) = ps since p E X,. Since pi Spa, pi E {pa}. Therefore from the continuity of .rrg it follows that .rrg(pi) E i.e. .rrg(pL) < pi. Thus by the definition of p i we obtain .rr;(pL)= pi. This means that p' E X , . It is also easily seen that p(A I > ) - + P I . Thus p' is a greatest lower bound of P' in P. Hence by Zorn's lemma there is a minimal element q of P. To prove that q € 2,we assume
m,
If r # q, then from the definition of q it follows that r @ P , i.e. p(A I > ) + r . Hence there is an open nbd U ( r ) of r in X , such that p(A I >) is residual in X - U ( r ) .Therefore from p(A 1 >)+ q, we obtain that q $Z U(r),i.e., q, @ U,(ra)for some a E A and some nbd U,(ra)of r, in X,. Therefore r, $ qa, which contradicts the assumption r S q. Thus we obtain r = 9, i.e., q is a minimal point of X , and hence q E 2.Since q E P, p(A >)+ q in 2. This proves that 2 is compact.
I
Suppose X is a TI-space and {%, 1 a E A} the collection of all finite open coverings of X . Then for each a, we define an equivalence relation between two points p, q of X as follows: p -,q means that for each U E %, p E U if and only if q E U. Then we denote by 9,the decomposition of X with respect to the relation -,. From now on, for brevity, we shall denote by X , the decomposition space X ( 9 , ) of X with respect to 9,. The following proposition is easy to prove, so its proof is left to the reader.
D) Every X , is a finite To-space,where X , is the space defined in the above. We define order between two elements a, p of A by
p < a if and only if 9,< g P . Then A turns out to be a directed set. For, let a, p E A ; then
%7
=
VII.41
THEORY OF INVERSE LIMIT SPACE
417
%, A %p is a finite open covering and it is easily seen that p --yq implies p - , q and p -pq, i.e. y > a, p. Suppose a > p ; then to each D E $3, we assign D ' E 91p such that D ' 3 D. It is easily seen that this defines a continuous mapping r; of X, onto Xa. Namely {X,, n-; 1 a, p E A, a > p } forms an inverse system. It follows from C) that C X , = lim{X,, r;} t is a compact TI-space. Let us denote by fa the natural mapping of X onto the decomposition space X,, i.e., f , ( p ) is the member of 9,which contains p. Now, we define a mapping f of X into X , by
Since r ; ( f , ( p ) ) = f p ( p ) obviously holds for every a, p E A with a > p, this definition determines a mapping of X into X,. Since each natural mapping f a is continuous, f is a continuous mapping. We can say more about f.
E) f is a topological mapping of X into X,. Proof. Suppose U ( p ) is a given open nbd of a point p of X. Let
then '?!la is a finite open covering of X, i.e., it is a member of {'?!la I a E A}. Then 91a, and accordingly X,, consists of three elements { p } , U ( p ) - { p } and X - U ( p ) .We denote them, regarded as the points of X,, by q,, q: and q: respectively. Then V, = {q,, q:} is a nbd of q, in X,, and therefore
is a nbd of f ( p ) such that
Thus f maps every open set of X to an open set of f ( X ) . On the other hand, if p, p' are different points of X, then, since X is TI,X - { p ' } is an open nbd of p . Hence there is a nbd V of f ( p ) in X , such that
418
TOPICS RELATED TO MAPPINGS
EVII.4
Therefore f ( p ’ ) # f ( p ) , i.e. f is one-to-one. Thus f is a topological mapping of X into X,.
F) f ( X )C X.
Proof. Suppose p is a given point of X and q a point of X , such that q # f(p). Then for some LY E A, q, # f , ( p ) , where f, denotes the natural mapping of X onto X,. Hence q,, considered as a member of 9,, is a set D E 9,with p @ D. Put
because every point p’ different from p is contained in a member of 021, which does not contain p , and hence p’ is contained in a member of 9,, which is contained in X - { p } . On the other hand,
easily follows from the fact that q, # f , ( p ) and accordingly q, f f , ( p ) . Therefore f , ( X - { p } ) is an open nbd of qy in X , which does not contain f , ( p ) . This implies that f ( p ) is a closed set in X,. Thus f ( p ) is a minimal point of X,, i.e. f(p) E 2. This proves our assertion. Identifying X with f ( X ) , we regard X as a subspace of X.Generally, for a subset F of X we put
F* = { p ( p = { p , } € X p , c F ’ for some a € A } . G) If F is a closed set of X , then its closure F in 2 coincides with F *. Proof. If F = 0 or X , then the proposition is obviously true. So we assume F # $3, F # X . Let
’ Remember that p.
is a member of the decomposition 5j&, at the same time.
VII.41
THEORY OF INVERSE LMIT SPACE
419
(Namely %P = { X ,X - F}.) Then we regard F as a point of X p and denote it by x,. Then it is obvious that F* = p , ' ( X , ) ,
6
where denotes the projection of r?- into XP.Since {x,} is a closed set of X P ,F * is closed in 2.Observe that F C F * ; then we obtain F C F * . Conversely, let p = { p , } be a given point of F * . Then pm0C F in X for some a, E A. Let PI,. . . , p,, E A be arbitrary. Choose y E A such that y > ( Y ~ ,PI, . . . , &. Then, since p E X,,
where p y , pa0,etc. are regarded as subsets of X . Pick up a point x from p y . Then x can be expressed in 2 as x = {x, 1 (Y E A } satisfying xa, = poi, i = 1 , . . . , k . Namely, the &-coordinates of x are equal to the &-coordinates of p for i = 1, . . . , k . Since x E F, this implies that every nbd of p in X intersects F. Thus p E F, i.e. F * C k? Therefore our assertion is proved. Theorem W.9.
X is the Wallman's compactification of X .
Proof. Identifying X with f ( X ) ,we regard X as a subspace of 2.Then by virtue of IV.2.B), all we have t o prove is that the conditions of Theorem IV.3 are satisfied by X , 2 and the collection 3 of all closed sets of X . It is clear that R is a TI-compactification of X . To prove that {F I F is a closed set of X } forms a closed basis of X,we suppose G is a given closed set of and p = { p , } is a point of X with p @ G. Then we can find PI, . . . , Pk E A and open nbds V,(p,) of psi in Xsi, i = 1,. . . , k , such that
where we may assume that
for an open set V,. of X , which contains poi E go,as a subset and is a sum of elements of gFi. Then from (1) it follows that
420
TOPICS RELATED TO MAPPINGS
[VII.4
k
f7v.CX-G
inX.
i= I
Put k
u=nq; 1=1
then F = X - U is a closed set of X satisfying G C F*. For, if 4 (4,)Sf F * in 2, then
=
and especially qai n U, f
0, i = 1,. . . , k .
Therefore
U,., i = 1,. . . , k (see (2)),
4a, C
proving k
4E
nWP,) n x
i= I
{X, I a E A, a
f
Pi)
9
i.e. 4 fZ G (see (1)). This proves G C F * . On the other hand, we can easily see that U * is a nbd of p in X such that
(Note that U is a union of members of 9,.for each y > Pi, i = 1, . . . , k . ) Thus by G), F is a closed set of X, satisfying
proving our assertion. Finally, to prove
k=l
k=l
in 2 for closed sets Gk,k = 1,. . . , n, of X, we take an arbitrary point p = { p a } of n Gk. Then by G),
VII.41
THEORY OF INVERSE LIMIT SPACE
421
which implies pa, C Gk, k = 1, . . . , n, for some akE A . Take
(Y
such that
(Y
> a k ,k =
1 , . . . , n ; then
p , c p m P , k = 1 , . . . , n, in X , pa C
A Gk.
k=I
Therefore
proving
Conversely,
holds in every topological space. Therefore n;=,Gk = n;=,6, is concluded. Thus by IV.2.E) is the Wallman's compactification of X . Corollary. A topological space X is a compact TI-space if and only if it is homeomorphic with the set of all minimal points of the limit space of an inverse system of finite To-spaces.
Proof. This proposition is a direct consequence of Theorem VII.4 combined with C).'
'
J. Flachsmeyer [l] also constructed bch-Stone compactification by use of inverse limit space. As for non-compact spaces, B. Pasynkov [l] obtained interesting results with respect to inverse systems. See also B. Pasynkov (31 and V. KljuSin [l].
422
[VII.4
TOPICS RELATED TO MAPPINGS
Theorem W.10. A T,-space X is paracompact and M if and only if it is the inverse limit space of an inverse system of metric spaces with perfect bonding maps.' Proof. Suppose X = @{X,, n-i I a,p E A, a > p}, where each X , is metric and each z-; perfect. Fix p E A and denote by f p the projection from X into X p . We shall prove that f p is a perfect map. Let y E X p be a given point and x E f i ' ( y ) . Then for-1 any a E A choose y = y ( a ) E A such that y a, p. Then f , ( x ) EfL(fi ( y ) ) . Thus f;'(y) is a closed subset of n { f : ( f F ( y ) )I a E A} and hence it is compact. Let F be a closed set of X and assume y i E f p ( F ) , i = 1 , 2 , . . . , satisfying y ' + y in X,. Suppose
x i E F,
I
x i = { x : a E A},
f p ( x i )= x i
= y',
i = 1,2, . . . .
We claim that for each a E A { x ' } has a cluster point in X,. Because if -1 not, then select y E A such that y > a, p. Then (x'J I i = 1 , 2 , . . .} is discrete in X,. Thus { x i I i = 1 , 2 , . . .} and accordingly { f ; ( x i ) I i = 1 , 2 , . . .} = { x ; i = 1,2,. . .}, too, is discrete, which is not true. Now, fix a cluster point x , of { x b } for each a f p. Then we claim that {x, I a E A} is a cluster point of { X I I i = I, 2 , . . .} in II, x,. TO see it, let a,,. . . , akE A and let U, be a nbd of x, in X,, j = 1,. . . , k. Select a E A such that I I I a > a,,j = 1, . . . , k , and a nbd Uu of x, in X , such that fEl(U,) C U.,, j = 1, . . . , k . For any i, E N there is i 3 i, such that x : E U,. Then
{fz
I
f z j ( x L ) = x b , E U,., j
=
1,. . . ,k .
n,
Thus { x , I a E A} is a cluster point of { x i } in Xu. As implied by the proof of A), X is closed in Xu. Hence {xu} E X , i.e. {x,}F ~ in X. Hence { x , } E F follows because F is closed. Hence y = xp E f p ( F ) , proving that f p ( F ) is closed in Xp. Thus f p is a perfect map from X onto a metric space. Namely, X is paracompact and M by Corollary 1 of Theorem VII.3. Conversely, let X be paracompact T2 and M. Denote by A the set of all normal covers of X satisfying wA-condition. Let p = {%, I i = 1,2, . . .} E A. Then define equivalence relation - p in X by
n,
m
x
-fi
y if and only if y E fl S ( x , ai). i=l
' The 'only if' part is due to V. Kljusin 111 and the 'if' part to K. Morita [6].
VII.51
423
THEORY OF SELECTION
We define direction between two elements p, v of A by p
> v if and only if x
-,y implies
x -,y
.
Then (A, >) is a directed set. As we saw in the proof of Theorem VII.3, the quotient space X, = X/-, is a metrizable space, and the quotient map f, from X onto X, is a perfect map. For each p, v E A with p > v, we define a map f from X, into X, by
fr
Then it is easy to see that is a perfect map from X, onto X,. (The detail is left to the reader.) It is also obvious that f r o f k =f^,whenever A>p>v.Let
We define a map cp from X to X as follows. Consider a given point 2 = {x, I L,L E A} of Then {f;'(x,) I p E A} is a closed filter base of X consisting of compact sets, and hence its intersection is non-empty. If x and y are two distinct points of X, then there is {%;} = p E A such that y fZ S(x, % i ) for some i. Hence n{f;'(x,) I p E A} is a singleton {x}. We define that cp(2) = x. It is easy to see that q is a homeomorphism from onto X. Thus the theorem is proved.
x.
5. Theory of selection In the present section we are going to generalize Tietze's extension theorem by use of a new point of view due to E. Michael [3]. Tietze-type extension theory is treated as a special case of Michael's selection theory, which gives new characterizations of normal spaces, collectionwise normal spaces and fully normal spaces.
Definition W.6. Let X and Y be topological spaces and 2' the collection of all non-empty subsets of Y.In this section we call a mapping cp of X into 2' a currier. Suppose F is a subset of X ;then we shall denote by qF the restriction of q to F. cpF is then a carrier of F into 2'. A carrier cp of X into 2y is called lower semi-continuous (l.s.c.) if for each open set V of Y,
424
TOPICS RELATED TO MAPPINGS
[VIIS
is an open set of X . Suppose cp is a carrier of X into 2'. If a continuous mapping f of into Y satisfies
then f is called a selection for
x
cp.
Example W.9. A mapping cp of X into Y may be considered a special carrier of X into 2' such that each ~ ( pis)a one-point subset of Y. Then this carrier is I.s.c. if and only if the mapping cp is continuous, because cp-'{ V}= cp-'( V )for every subset V of Y. Let F be a closed set of X and f a continuous mapping defined over F. Then the carrier cp defined by
is also I.s.c. If there is a selection g for cp, then it is clearly a continuous extension off over X. A) A carrier cp of X into 2' is 1.s.c. if and only if for every p E X, q E q ( p ) and nbd V of q, there is a nbd U of p such that for every p r E U, p ( p ' ) n v f 0.
Proof. Suppose cp is a I.s.c. carrier and p E X , q E q ( p ) and V is a nbd of q. We may assume that V is an open set of Y. Hence by the definition of I.s.c., U = cp-'{ V} is an open nbd of p . If p' E U, then cp(p') f l V f 0,and hence the condition is necessary. Conversely, assume the condition is satisfied by a carrier cp of X into 2'. Suppose V is a given open set of X , and p E p-l(V}. Then
Take q E ~ ( pfl)V ;then V is a nbd of q, and hence by the hypothesis, there is a nbd U of p such that p' E U implies
VII.51
425
THEORY OF SELECTION
which means that cp-’{V}is an open set of X , i.e. cp is a I.S.C.carrier. B ) Let X and Y be topological spaces and assume that to each point q of Y, an open nbd V ( q )is assigned so that if q E V(q’),then there are nbds W of q and W’ of q‘ such that q” E W’ implies W C V(q”).Suppose cp and CC, are 1.s.c. carriers of x into 2’ satisfying
Proof. Let q E 8 ( p ) and M a given nbd of q in Y.Since q E V ( + ( p ) ) , there is q’E $ ( p ) such that q E V(q’). We take a nbd W of q and a nbd W’ of q‘ satisfying the condition mentioned in the proposition. Then, since CC, is I.s.c., there is a nbd U of p such that p‘ E U implies
On the other hand, since q E cp(p) and cp is I.S.C., there is a nbd U ’ of p such that p’ E U ’ implies rp ( p ’ )f l W fl M #
0.
Thus U“= U fl U ’ is a nbd of p such that p’ E U ” implies
8 ( p r )n M = c p ( p rn) v($ ( p r ) )n M 3 cp(p’)n w n M
f
0
(Note that $(p’) f l W‘ # 0 and thus we can select q” E $(p’) fl W’ for which V ( q ” ) >W and hence V ( $ ( p ’ ) ) > W.) Therefore 8 ( p ‘ ) n M # 0. Hence b y A), 8 is I.s.c. C) Let X and Y be topological spaces and d a subcollection of 2’ containing all one-point subsets as members. Then the following two conditions are equivalent: ( i ) If cp is a 1.s.c. carrier of X into d,then for every closed set F of X , each selection for 40, can be extended to a selection for rp. (ii) Every 1.s.c. carrier rp of X into d has a selection.
426
TOPICS RELATED TO MAPPINGS
[VIIS
Proof. Since (i) clearly implies (ii), we shall prove only that (ii) implies (i). We suppose g is a selection for (pF, i.e., g is a continuous mapping of F into Y satisfying
We define a carrier
We can assert that
+ of X
into d by
+ is 1.s.c. For, let
V be an open set of Y. Then
and hence in view of the fact that (D is 1.s.c. we can verify that t+-'{V} is open. This means that CC, is I.s.c. Thus it follows from (ii) that there is a selection h for Now it is obvious that h is an extension of g over X , and a selection for cp.
+.
Let us recall Definition IV.2, to observe some simple properties of normed linear spaces.
D ) Let Y be a normed linear space. We consider mappings f (x, y ) = x + y and g(a, x) = a x of Y X Y into Y and of E' X Y into Y. Then f and g are both continuous mappings.'
E) Let a, A and V be a real number, a set and an open set of a normed linear space Y, respectively. Then A + V = {x + v 1 x E A, v E V } and a V = {av I v E V } are open sets of Y.
F ) Let Y be a normed linear space. Then for every nbd U of 0 there is a convex symmetric open nbd V of 0 such that V C U.
Proof. Take a spherical nbd S,(O) C U. Then it is easy to see that V = S,(O) is a desired nbd of 0.
'
If a linear (not necessarily normed) space Y is a T2-space at the same time, and f and g are continuous, then Y is called a linear topological space.
VII.S]
THEORY OF SELECTION
427
G ) Let X be a paracompact T,-space. Then for every open covering % of X , there is a collection { f , I a E A } of non-negative, real-valued continuous functions over X such that: (i) if, I a E A} = 1, (ii) every point p of X has a nbd in which all but finitely many of the f , vanish, (iii) for every a, there is U E 91 for which f,(X - U )= 0. Conversely, if X is a TI-space,and for every open covering % there is a collection {f , I a E A } of non-negative, real-valued continuous functions over X satisfying (i) and (iii), then X is paracompact T,.'
Proof. Suppose X is paracompact T,. We take a locally finite open refinement "Ir = { V, 1 a E A } of the given open covering %. Since X is normal, there is an open covering W = { W, I a E A } with W,C V,. Using Urysohn's lemma, for each a we construct a continuous function g, over X such that g,(Wu)=l,
g,(R-V,)=O
and O S g , G l .
Putting
we get continuous functions f,, a E A, satisfying the desired conditions. Conversely, suppose X satisfies the condition. Let p and F be a point and a closed set of X respectively, such that p F. Since { X - { p } ,X - F } is an open covering of X , we can construct a collection {f , I a E A} of continuous functions satisfying (i) and (iii). By virtue of (i) there is an a for which f,(p) > 0. Then f,(F) = 0 follows from (iii). Hence
is an open nbd of p, satisfying
which implies that X is regular. Now, let % be a given open covering and { f , I a E A } a collection of continuous functions satisfying (i) and (iii). Then for each pair a E A and Such a collection
{fa
I a E A } is called a partition of unity subordinated to %.
428
TOPICS RELATED TO MAPPINGS
[VII.5
a natural number n, we put
It is easy to see that {U,, 1 a E A } is locally finite for each n. For, let p be a given point of X , then by (i) f , ( p ) = 1, and hence there are a;, i = 1 , . . . , k , such that k
1
Cfai(P)> I---. i=l n Putting
i=l
n
we get an open nbd V of p. Then it follows from (i) that V does not intersect U,, if a # a;,i = 1, . . . , k . Thus {U,, 1 LY E A } is locally finite. Hence by (iii) {U,, I a E A, n = 1,2, . . .} is a cr-locally finite open refinement of %. Thus by Theorem V.l, X is paracompact. In the rest of this section we denote by X ( Y ) , F ( Y ) and W(Y)the collections of all non-empty convex sets, all non-empty convex closed sets and all non-empty convex compact sets plus Y, respectively, of a linear space Y.
+
H ) Let X be a paracompact T,-space, Y a normed linear space, and a 1.s.c. carrier of X into X ( Y ) . Suppose V is a convex open nbd of 0 E Y. Then there is a continuous mapping f of X into Y such that
Proof. For each point 9 of Y, we put
I
+
where q - V = {q - u v E V}. Since q - V is an open set of Y and is l.s.c., U, is an open set of X . Let p be a given point of X . Then for an arbitrary point q E i,b(p),
VII.51
429
THEORY OF SELECTION
and hence p E U,. Therefore % = {U, 1 q E Y}is an open covering of X. Since X is paracompact T,, we can construct functions f,, a E A, which satisfy the conditions of G). For each a E A, we choose q ( a ) such that
we get the desired mapping. (Note that all f , ( p ) vanish except at most finitely many.) The continuity off follows from (ii) of G). On the other hand, by (2), f , ( p ) # 0 implies p E U,(,),and hence
by the definition (1) of U&). Since f u ( p )# 0 holds only for finitely many a,it follows from (i) of G) that
because both of $ ( p ) and V are convex sets of Y and therefore $ ( p ) + V is convex.
Theorem W.11. A TI-spaceX is paracompact T2 if and only if for every Banach space Y and every 1.s.c. carrier cp of X into 9 ( Y ) , there is a selection. Proof. Necessity. By F) there is a nbd basis {V, 1 i = 1,2, . . .} of 0 in X consisting of symmetric, convex, open sets V,. In view of E) we may assume that V, satisfies
Now, we can construct continuous mappings fi, i such that
=
1 , 2 , . . . , of X into Y
430
[VIIS
TOPICS RELATED TO MAPPINGS
For i = 1, the existence of such a mapping f l is a direct consequence of H). To define fi by induction on i, we assume that f,,. . . ,fi-l have been defined. Since
and
yFlis symmetric,
Since we can easily show that {q + V,-l I q E Y } satisfies the condition of { V(q)I q E Y } in B), by the same proposition we can conclude that
defines a I.s.c. carrier of X into X(Y ) .(we regardfi-,(p) as a I.s.c. carrier.) It follows from H) that there is a continuous mappingf;. of X into Y such that
f i ( p ) E cpi(p)+ V, for every p E X This combined with (2) implies
(Note that
(1) combined
v, c ;V,-l c v,-,.)
with
the
convexity
of
Y-l
implies
Thus f i ( p ) satisfies the desired condition. Hence it follows from (1) that
As shown in E), we define that 2K-1 = {2u 1 u E K-I}, which is, by the convexity of equal to V-l + K - l = { u + u' I u, u' E K-3.
K-l,
VII.51
431
THEORY OF SELECTION
which implies
Therefore { f ; ( p ) } is a Cauchy point sequence of Y , and hence { f ; } uniformly converges to a continuous mapping f of X into (see 1.H)). If we assume f ( p )JZq ( p ) for some p E X , then
q ( p ) n ( f ( p )+ V,)= 0 for some i , because q ( p ) is closed, and { f ( p )+ V, I i = 1,2, . . .} is a nbd basis of f ( p ) . Hence
follows from (1). Since f ; ( p ) + f ( p ) , there is some k
2
i + 1 such that
On the other hand, from (3) it follows that
which contradicts (4).Therefore f ( p ) E cg(p), i.e. f is the desired selection. Sufficiency. Let UU be a given open covering of X. We define a normed linear space X = ll(UU) as follows. The points of fl(UU) are the real-valued functions y defined over '% such that
and the norm of y is defined by
We can easily verify that Y = 1,(%) is a Banach space. Put
C = { y 1 Y E Y , y ( U ) a O for all U E UU, and llyll= 1).
(5)
Then it is almost obvious that C E 9 ( Y ) .It is also clear that, for each PGx,
432
[VIIS
TOPICS RELATED TO MAPPINGS
Theref ore, C n C ’ ( p )E 9( Y ) for every p E X. Put cp(p)=
cn CYP),
PEX ;
(7)
then we can assert that cp is a I.s.c. carrier of X into 9 ( Y ) . To prove that cp is I.s.c., we show that given y E C and there are y ’ E C and U,, . . . , U, E 0% such that
y(U,.)>O,
i = 1 , . .. ,k ,
and
y ’ ( U ) = O if U f U,, i = 1, . . . , k . For, since it follows from (5) that
we can choose U,, . . . , U, E % such that y ( U,)+ . . . + y ( u,) = 6 > 1 -
;
&
and
y(U,)>O,
i = 1, ..., k .
Now, we define a function y’ over 0% by
y ’ ( U ) = O if U Z U , . , i = l , . . . , k , y’(U,)=Y(U,)+ 1 - 6 and
y ’ ( U , ) = y ( U , ) , i = 2,. . . , k
Then it is clear that y‘ E C (see (5)). Furthermore,
E
> 0, then
VII.51
THEORY OF SELECTION
= 1- S
+ 2 { y ( U )I U f V,, i = 1,. . . , k }
= 2(1-
S)
O. Then we choose U,, . . . , U, E % and y’ E C which satisfy (8)-(10). Since by (9)
it follows from the definition (6) of C ( p ) that
p E U i , i=1, ...,k .
-
Therefore V,= U , f l * * fl U, is a nbd of p. Suppose p ‘ is an arbitrary point of U,; then it follows from (10) that y‘( V )= 0 for every U E % with p’Z ! U.
Hence y ’ E C ’ ( p ’ )follows from (6). This combined with y ’ E C implies that y ‘ E cp(p‘) (see (7)). Therefore by (8)
i.e., ~ ( p ‘fl) S , ( y ) # 0 for each point p‘ of the nbd U, of p. Thus by A), cp is I.s.c. Now, using the hypothesis we can choose a selection f for p. For each U E 031, we define a real-valued function fu over X by
Since f is continuous, fu is also continuous by the definition of the topology of l,(Q). Furthermore, we note that it follows from f(p)E q ( p )C C and (5) that f , ( p ) 0 and { f u ( p ) 1 U E a}= 1, p E X. On the other hand, if p 6Z U, then it follows from f(p) E cp(p)C C ’ ( p )and (6) that
c
434
CVII.5
TOPICS RELATED TO MAPPINGS
Therefore { f u I U E %} satisfies the conditions (i) and (iii) of G), and hence X is paracompact T2by virtue of the last part of G). We need the following properties of normal spaces and collectionwise normal spaces to extend selection theory to those spaces.
I) Every point-finite, countable, open covering of a normal space has a locally finite open refinement. Proof. The proof is quite analogous to the proof ( i v ) j (i) of Theorem V.5. so it is left to the reader. J) Let % be a point-finite open covering of a collectionwise normal space X . Then there is a locally finite open refinement of %. Proof. Suppose % = { U, I a E A } ; then since X is normal, there is an open covering 7"= { V, I a E A } such that C U,. For each finite subset { a l , . . , ak} of A we define
v,
I
U(al . . . ak)= { p p E Uai,i = 1, . . . , k ; p for (Y # airi = 1,. . . , k } .
I!..[*
Now, we shall define open sets V ( a , .. . ak)and V ; which satisfy the following conditions (l),(2) and (3):
U,,
n . ' n U,, *
3
v(
a1
k-1
. . . (Yk ) 3 U((Y1 . . . (Yk ) - u
vi
9
(1)
i=l
where
q.= u { V ( a ,. . . ai) 1 a,,. . . , aiE A } , { V ( a l.. . , (Yk) I a l , . . . , akE A } is discrete for each k .
(3)
Note that (1) and (2) imply k
U V:>U{U( a I . . ai)l . a1, . . . , a i E A , i = 1, . . . , k } . i= 1
(4)
We shall define such open sets V ( a l. . . (Yk) and V ; by induction o n the number k .
VII .5]
THEORY OF SELECTION
435
For k = 1, { U ( a l )1 a1E A} is a discrete closed collection, and hence by the collectionwise normality of X there is a discrete open collection { V ’ ( a l I)al E A } satisfying U ( a l )C V’(a,) (see V.3.B)). Since U ( a l )C U,,, putting
we get V ( a , ) ,a1E A, which satisfy (1)and (3). Assume we have constructed V ( a , .. . a;) and accordingly V : (by (2)) for i = 1, . . . , k - 1. Now we can assert that { U ( a ,. . . ak)k-1 U V :1 a l ,. . . , akE A } is a discrete closed collection of X. To prove k-1 that U ( a ,. . . ak) - U i = l Vi is closed, take a point k-1
u v;.
pbZU(a, . . . a k ) -
i=l
If k-1
pEU
v:,
i=l
then U ;:;
u;:; v;.If
V : is an open nbd of p which does not intersect U ( a ,. . . a k ) k-1
pbZ
u v:,
i= 1
then it follows from (4) that p is contained in at least k members of %. Since p E U ( a l .. . a k )holds by (5), p E U, for some a E A with a # a;, i = 1,. . . , k . Then U, is an open nbd of p which does not intersect k-1 U ( a ,. . . ak)- U i = l V : .Thus in any case k-1
p @ U ( @ [* . @ k ) f
u vi
9
i=l
which proves our assertion. Observe that for each x E X either U: V : or W = n{U, 1 x E U, E %} is a nbd of x which intersects at most one member of the concerned collection. Thus the collection is a discrete closed collection. Therefore by the collectionwise normality of X, there is a discrete open collection { V ’ ( a ,. . . ak)1 a I ,. . . , (Yk E A} such that k-1
U ( a ,. . . a k ) -
u v;cV ‘ ( a ,. . . a k ) .
i= 1
436
TOPICS RELATED TO MAPPINGS
[VIIS
Since k-l
W a , . . . ffk) - U V ;c u., n . . . n uok i= 1
is obvious, putting
we get open sets V ( a , .. . ak)which satisfy (1) and (3). Finally we put
This completes our induction process to define V ( a ,. . . ak)and V ; . Now we note that { V ; I k = 1,2, . . .} is a point-finite open covering of X . Because if p E V ; ,then p E V(a,. . . f f k ) for some a,, . . . , ffk E A, and hence by (1) p is contained in at least k members of %. Since 0% is point-finite, this means that { V ; I k = 1,2, . . .} is point-finite. Thus by I), we can construct a locally finite open refinement {W, 1 k = 1,2, . . .} of { V ;I k = 1 , 2 , . . .} such that Wk C v;.Put
then { W ( a ,. . . f f k ) I a,, . . . , akE A, k = 1,2, . . .} is the desired locally finite open refinement of % (see (lt(3)).
K) Let X be a collectionwise normal space, Y a normed linear space and rC, a 1.s.c. carrier of X into %( Y ) . Then for a given convex open nbd V of 0, there is a continuous mapping f of X into Y such that
Proof. Put
u = { p E x I 0 E $ ( p ) + V }. Then we can easily see that U is an open set of X , because is a 1.s.c. carrier. Since Y is paracompact, there is a locally finite open covering W = { WpI p E B} satisfying
VII.51
THEORY OF SELECTION
437
Now it is obvious that {$-'{ W,} 1 p E B} is an open covering of X . W e can say moreover that the covering is point-finite on X - U. For, if p E X - U, then
which implies that $ ( p ) # Y. Therefore $ ( p ) is compact. Hence $ ( p ) intersects at most finitely many of W, because W is locally finite. This implies that p is contained in at most finitely many of $-'{W,}. Since X - U is a closed set of a collectionwise normal space, it is also collectionwise normal. Therefore, by J), there is a locally finite open covering { VL I P E B } of X - U such that
v;c I,-'{ w,}n (x- U ). Thus by V.3.C) there is a locally finite open covering {V, I p E B} of X such that
then 'V is a locally finite open collection of X and covers X - U. Thus 'V' = 'V U { U } is a locally finite open covering of X . Therefore by G), there is a collection { f , 1 a E A } of non-negative, real-valued, continuous functions over X satisfying (i)-(iii) of G). Especially in view of the condition (iii), t o each a E A we can assign @ ( a E ) B such that
unless f,(X - U )= 0. To each a E A we assign q ( a )E Y as follows. If f,(X - U )= 0, then q ( a )= 0. Otherwise q(a) is a point of Y satisfying
By (ii) of G), f is a continuous mapping of X into Y.
438
TOPICS RELATED TO MAPPINGS
[VIM
+
To prove f ( p ) E $ ( p ) V, let p E U. (In the case that p Sr U the proof proceeds in a similar manner.) Then for each CY such that f a ( p )# 0, f,(X - U )f 0, we obtain
and hence
which implies
For each CY such that f a ( p )# 0, f,(X - U )= 0, we obtain p E U and accordingly
Since $ ( p ) and V are both convex, $ ( p ) + V is also convex. Therefore, from (i) of G), we get
proving the assertion.
Theorem VII.12. A TI-spaceX is collectionwise normal if and only if for every Banach space Y and for every 1.s.c. carrier 9 of X into %( Y ) ,there is a selection. Proof. The 'only if' part is directly derived from K) in a quite similar way as Theorem VII.ll was derived from H). The 'if' part is a direct consequence of the following proposition.
Corollary. A T,-space X is collectionwise normal if and only if every continuous mapping of every closed set F of X into a Banach space can be continuously extended over X.'
' Essentially proved first by C . H. Dowker [4].
VII.51
THEORY OF SELECTION
439
Proof. The 'only if' part is a direct consequence of the 'only if' part of the theorem. To prove the 'if' part, we denote by {F, I a E A } a discrete closed collection of X. We denote by H ( A ) the generalized Hilbert space with t h e index set A. Define a mapping f of the closed set U {F, 1 a E A } of X into H ( A ) by
where p a = {xu,1 a' E A}, x, = 1, x,, = 0 for a' # a. Then f is easily seen to be continuous over U {F, 1 a E A ) , and hence by the hypothesis there is a continuous extension g off over X . We consider a spherical nbd Sll,(pU) for each p'. Then
S,/,(pa)n ~ , , , ( p " = ' ) 0 if
(Y
# a'.
Therefore the sets
form a disjoint open collection in X satisfying U, 3 E,. This proves that X is collectionwise normal.
L) Let {F, 1 i = 1,2, . . .} be a countable, discrete, closed collection in a normal space X , then there is a discrete, open collection { V, 1 i = 1,2, . . .} such that V, 3 F,. Proof. Since Fl and U=,; F; are disjoint closed sets, there is an open set U , which satisfies m
F,culculcx-LIF,. i=2
Since F, and 5,U (U;=,F,) are disjoint closed sets, there is an open set U, which satisfies
(i,U F ,1
F,cu,cIs,cx-u,u
Repeating this process, we obtain disjoint open sets U,, U,, . . . such that V , 3 F,. Since U I;I and X - U U, are disjoint closed sets, there is an open V which satisfies
;=,
=;,
440
TOPICS RELATED TO MAPPINGS m
x - ui=
m
U,c
vc V C X - U F , i=l
1
Now V, = U, -
[VIIS
V, i = 1,2,. . . , are the desired
open sets.
M ) Let F be a closed set of a normal space X and % = {U,. I i = 1,2, . . .} a locally finite, countable, open covering of F. Then there is a locally finite open covering 'V = {V, i = 1 , 2 , . . .} of X s u c h that V, f l F C U,.
I
Proof. Using L) we can easily prove this proposition (see V.3.C) for a similar proof).
N) Let X be a normal space, Y a separable normed linear space, cp a 1.s.c. carrier of X into %( Y ) .Then, for a given convex open nbd V of 0, there is a continuous mapping f of X into Y such that
Proof. The proof is quite analogous to that of K). This time we must take a countable locally finite open covering W such that
For the rest of the proof, we can proceed in a quite similar way as in the proof of K), but considering only countable coverings and using I) and M). The proof in detail is left to the reader.
Theorem W.13. A T,-space X is normal if and only if one of the following conditions is satisfied: (i) for every 1.s.c. carrier cp of X into %(El),there is a selection, (ii) for every separable Banach space Y and every 1.s.c. carrier cp of X into %( Y ) ,there is a selection. Proof. Assuming that X is normal, we can derive the condition (ii) from N) as we derived the necessity of Theorem VII.12 from K). It is obvious that (ii) implies (i) and that (i) implies the normality of X (Tietze's extension theorem). Example W.10. Let us denote by C" and S"-' the n-dimensional open ball with radius 1 and its boundary in E n ,i.e.,
v11.q
441
THEORY OF SELECTION
C" = { ( X I , . . . , x,) 1 x:+ . * .
-
C" = { ( x , ,. . . , x,)
+xi
x and for some m .
Thus {%i 1 i = 1,2,. . .} is a development of X , proving that X is a developable space. Example VIII.2. Sorgenfrey line S has a G,-diagonal but is non-metrizable. Thus F) cannot be extended to GO-spaces.
Definition VIII.3. Let X be a GO-space. Suppose ( A , B ) is an ordered pair of disjoint open sets of X such that: (i) X = A U B, (ii) a < b whenever a E A and b E B.
' Due to D. Lutzer [l].
VIII. 11
LINEARLY ORDERED SPACE
461
Then (A, B)is called a gap if it satisfies (i), (ii) and (iii) A has no maximal point, and B has n o minimal point. If furthermore A = 0 or B = 0,then (A, B) is called an end gap. (A, B ) is called a pseudo-gap if it satisfies (i), (ii), (iv) A f 0, B f 0, and (iv), or (iv), stated in the following: (iv), A has no maximal point, and B has a minimal point, (iv)r A has a maximal point, and B has no minimal point. Generally, let A be a subset of a linearly ordered set X. A subset A’ of A is called cofinal (coinitial) in A if for each a E A there is a ’ € A’ such that a ’ s a ( a ’ sa ) . Suppose ( A , B ) is a (pseudo-) gap of a GO-space X. If there are discrete subsets A’ of A which is,cofinal in A and a discrete subset B’of B which is coinitial in B,then (A, B) is called a Q- (pseudo-) gap.
Theorem VIII.2. A GO-space X is compact if and only if X has no gaps and no pseudo-gaps. Proof. Necessity of the condition is obvious. Suppose that X satisfies the condition. Then X has the minimal point a and the maximal point b. Suppose % is a given open cover of X by non-empty convex sets. Put Y = { x E X I there are finitely many elements U,, . . . , U, of % such that a E U,, x E U,, q. U q.+l is convex for i = 1,.. . , k - 1). If Y f X, then (Y, X - Y) is a gap or pseudo-gap of X. Because, Y and X - Y are obviously open sets. It is also easy to see that either Y has no maximal point, or X - Y has no minimal point. This contradicts that X has neither gap nor pseudo-gap. Hence Y = X follows. Thus b € Y, which implies that X = [a, b ] is covered by finitely many elements of %. Hence X is compact. Example VIII.3. Let X be a LOTS. Then X has no pseudo-gap. We denote by X + the set of all gaps of X plus X itself. We can define a linear order for X + in a natural way. Namely, if (A, B)= c is a gap of X , then we define that
a < c for all a E A
and
c < b for all b E B .
Introduce the order topology into X’; then X + turns out to be a compactification of X. More generally, let X be a GO-space. Then (X * )’ is a compact space which contains X as a subspace, where X * denotes
462
OTHER ASPECTS
[VIII.1
the LOTS given in B). The closure of X in (X*)' is a compactification of X , which we call Dedekind compactification. Let X be a LOTS and U a subset of X . A gap (A,B ) of X is said to be covered by U if there is a convex set V such that V C U, V f l A # 0 and V n B # 0. A cover 021 of X is said to cover the gap (A,B ) if % has an element which covers (A,B). G) A n open cover % of a LOTS X has a finite subcover if every gap of X is covered by %. Proof. We consider Dedekind compactification X' of X . For each U E % we define an open set U' of X + by
U' = U U {all gaps of X which are covered by U } Then %' = {U' 1 U E %} is an open cover of X . Thus %' has a finite cover V ' . Then 7" = { V' f l X I V E V ' }is a finite subcover of %. Theorem VIII.3.' Every GO-space X is countably paracompact. Proof. By virtue of B) we may assume that X is a LOTS. Let % = {Ul, U,, . . .} be a given countable open cover of X such that U, C q.+l. It suffices to show that % has a locally finite open refinement. Decompose each element Ui of % into its convex components V,,, a E Ai, to denote m
7"; = {V, I a E A ; } ,
V = UVi. i= 1
For each V E V i ,we define W ( V ) = V1U v * u . . . , where V,=V,
V;.+lEVi+j and
V;.CV;+, f o r j = 1 , 2, . . . .
Then X is decomposed into the sum of the disjoint clopen sets W (V)'s. If { V, 1 i = 1,2, . . .} has a locally finite open refinement in W (V) for each V, then 7.f and accordingly %, too, has a locally finite open refinement in X . Thus we may assume without loss of generality that 021 consists of convex sets.
' Due to L. Gillman-M. Henriksen 111.
VIII. 11
LINEARLY ORDERED SPACE
463
Pick a fixed point x,E X . Assume that [x,, w)C q., i = 1,2, . . . . Then select xiE [x,, w) - q., i = 1,2, . . . , such that xo< x1< x2 < . . .. Then {xi I i = 1,2, . . .} is cofinal in X . Since { x i } is discrete, there is a discrete open collection {P, I i = 0,1,. . .} in X such that x, E P, and such that P, is contained in some element of %. Now, {(xi, xi+J, P, I i = 0, 1 , . . .} is a locally finite open collection in X which covers [x,, to) and refines %. In a similar way (-w, xO]can be covered by a locally finite open collection which refines %. Thus % has a locally finite open refinement proving that X is countably paracompact. Theorem Vm.4.' A GO-space X is paracompact if and only of X is a Q-gap, and every pseudo-gap is a Q-pseudo-gap.
if every gap
Proof. Suppose that X is paracompact and (A,B)a gap of X with A # 0. Then % = {(-w, a ) 1 a E A } is an open cover of A. Let "Ir be a locally finite open refinement of 92. Decompose each element of "Ir into its convex components and denote by W the open cover of A consisting of the convex components. Obviously W is point-finite. Let 9 be a locally finite open refinement of W. Pick x ( P ) E P for each P E 9.Then { x ( P ) JP E 9}is discrete. Let a € A . Select x , E A such that x , > a . Suppose x1E PI E 9,P , C W,E W. If PI f l (-w, a ] # 0, then a E W,. Suppose W,C (-00, a , ] ;then select x2 E A such that x2 > a , . Let x2 E P2 E 9,P2 C W, E W. If P 2 n (-w, a ] # 0, then a E W,. We cannot continue this process indefinitely because W is point-finite. Hence
P,
n (-w, a ] = 0 for some n ,
and thus x(P,) > a. Therefore { x ( P )1 P E 9}is cofinal in A. In a similar way we can select a discrete set which is coinitial in B.Hence (A,B) is a Q-gap. Similarly, every pseudo-gap is a Q-pseudo-gap. Conversely, assume that X satisfies the said condition. Then it is easy to see that every gap of X * (defined in B)) is a Q-gap. Thus by virtue of B) we may assume without loss of generality that X is a LOTS whose every gap is a Q-gap. Let % be an open cover of X . Denote by F the set of all gaps of X which are not covered by %. We regard F as a subset of X'. Then F is obviously a closed set. Now we decompose X + - F into its convex components G,, y E to put
r,
Hy=G y n X
'
Due to L. Gillman-M. Henriksen [l] in case that X is a LOTS. This general form is due to M. J. Faber [l].
464
OTHER ASPECTS
[VIII.l
r}
is a disjoint open cover of X by Then it is obvious that {H, I y E convex sets. Regard H, as a LOTS covered by the open cover %. Then % covers every gap of H, possibly except its end gaps. Select an inner point a of H,. If H, has the maximal point, then by G) H i = { x E H, I x 2 a } is covered by finitely many elements of %. If (I$,,,@)is an end gap of H,, then it determines a Q-gap of X. Thus there is a discrete set cofinal in H,. Thus by use of G) and an argument similar to the one in the proof of Theorem VIII.3, we can prove that H i is covered by a locally finite open collection which refines %. We apply the same argument on the left half of H, to find a locally finite open cover of H, which refines %. Thus X is covered by a locally finite open cover which refines %. Namely X is paracompact. Corollary. Let X be a GO-space with a G,-diagonal. Then X is hereditarily paracompact.
Proof. It suffices to show that X is paracompact because every subset of X has a G,-diagonal. Suppose not; then by Theorem VIII.4 X has a non-Q-gap or a non-Q-pseudo-gap. Assume, e.g., that ( A , B ) is a gap such that A has n o discrete cofinal subset. Let {%, 1 n = 1 , 2 , . . .} be a sequence of open covers of X by convex sets such that %, > %n+l and such that m
n S(x, %,)
= {x}
for each x E X .
n=l
We claim that for each n there is x, E A such that S(xn,42,) r lA is cofinal in A. Because, otherwise, by use of a transfinite induction we can select an increasing (transfinite) sequence {x" I a < T} such that x p @ S(x", %,) whenever a < p < T, and {xa I a < T} is cofinal in A. Then { x " } is obviously discrete contradicting our assumption. Thus our claim is verified. It follows from our assumption that {x,} is not cofinal in A. Suppose a E A satisfies a a x , , n = 1 , 2 , . . .. Then S(a, %,) is cofinal in A for every n. Since S(a, %,) is convex, S(a,%,)>{x€AIx2a}, which contradicts that
n = l , 2,... ,
n:=,S(a, a,,) = { x } . Thus X
is paracompact.
VIII. 11
LINEARLY ORDERED SPACE
465
H) Let X be a GO-space with a G,-diagonal. If H ( X )= {x E X I [x, to) E 0 - O( f ( x ; ) + 1 , i = 1,2, . . . . Put 2,= f - ' ( f ( x , ) ) ; then (2;I i = 1,2, . . .} is a collection of zero sets in X . Obviously there is a discrete collection { q.I i = 1,2, . . .} of cozero sets satisfying 2,C U,. Hence 2; = U;=kZi is a zero set for each k. Thus Po= (2;I k = 1,2, . . .} is a decreasing sequence of non-empty zero sets
' Due to A. D. Alexandroff [ l ] and I. Glicksberg 111
VIII .4]
MEASURE AND TOPOLOGICAL SPACE
487
with an empty intersection. Denote by 9 a maximal filter such that 9 3 So. Define p E Y(X) by use of D) and 9.Then p ( Z ; ) = 1 for k = 1,2, . . . , while flL=l Z ; = 0, which means that p is not u-additive. Hence Y(X) # YJX).
Theorem VIII.13.' The following conditions are equivalent for a space X : (i) X is compact, (ii) A ( X ) = A , ( X ) , (iii) Y(X) = Yr(X), (iv) Y(X) = 9 ( X ) . Proof. (i) j (ii) .$ (iii) .$ (iv) is obvious. To prove ( i v ) j (i), suppose that 9 is a zero filter of X , and X satisfies (iv). Define p E Y(X) by use of D) and 9.Then p E 9(X)follows from (iv). Thus there is x E X such that x E 2 for every 2 E % ( X )with p ( Z ) = 1. Hence x E { Z I 2 E 8, which means that ,F+ x. Therefore X is compact.
Theorem VIII.14: X is realcompact
if and only if Y u ( X )= 9(X).
Proof. Let X be realcompact and p E Y u ( X ) . Then 9 = { Z E S ( X )I p ( Z ) = 1) is a maximal zero filter with c.i.p., because p is regular and a-additive. Hence 9 converges to x E X . Then p = 8, E 9(X) is easy t o see, as follows. Suppose x E B E d ( X ) .If p ( B )= 0, then there is U E 9 ( X ) with U 3 B, p ( U )= 0. Thus X - U E % ( X ) satisfies p ( X - U )= 1 , while ,F+ x E U implies X - U 6Z 9, which is a contradiction. Hence p ( B ) = 1. Suppose x Z ! B E d ( X ) ; then x E X - B E d ( X ) , and hence p ( X - B ) = 1. This implies p ( B )= 0. Namely p = 6, E 9(X). Therefore YJX) = 9 ( X ) is proved. Conversely, assume F u ( X )= 9 ( X ) and that 9 is a maximal zero filter with c.i.p. Define p E F(X) by use of D) and ,F. Then p E Fc(X) can be proved as follows. Suppose {Zi 1 i = 1 , 2 , . . .} is a decreasing sequence of zero sets with n;=,Z,=0. If p ( Z i ) = l , i = l , 2, . . . , then Z i E 9 , i = l , 2, . . . , which contradicts c.i.p. of 9. Thus p ( Z i )= 0 for some i, and hence by A) p is c+-additive. Thus p E 9(X). Suppose p = 6,. Then 9 converges to x. Therefore X is realcompact.
Corollary 1. Every Baire set B of a realcompact space X is realcompact. 'Due to A. D. Alexandroff [l]. 'Due to E. Hewitt [4].
488
[VIII.4
OTHER ASPECTS
Proof. It suffices to show Y v ( B ) C9 ( B ) . Suppose p E YJB). Then we define a map fi : d ( X ) + ( 0 , l ) by fi(B’)= p ( B ’ n B) for B‘ E d ( X ). It is obvious that fi E Yn(X). Since X is realcompact, fi E 9 ( X ) follows from the theorem. Suppose fi = a,, and x EX. It suffices to show that every Baire set of X is an intersection of members of d ( X ) . Because, assume the claim is true, and x 65 B. Then B C B’3 x for some B’ E Sa(X). Thus f i ( B ‘ )= p ( B ) = 1, which contradicts that fi = 6,. Hence x E B, i.e. p E 9 ( B ) .Hence the corollary is proved. Now, to prove our claim, we define a transfinite sequence {B,10 S a < w , } of subsets of X by induction o n a as follows. 3,= d ( X ) . For a with 0 < a < w l ,
Then it is obvious that % , ( X )= U{%, 10 a < w,}. Now we prove the following assertion by use of induction on a. Let x 65 B,E 93, or x E X - B,E Li3, ; then there is B‘ E d ( X ) such that B,C B’ax. The assertion is obviously true if a = 0. Assume that the assertion is true for all ,O
