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, and
The the
quantifier 3. Suppose that in the middle of a computation we know that P V Q is true.
Then we should be able to continue our computation in one of two
ways depending upon whether P is true or Q is true (we assume that if P and Q are both true, then we are content to proceed in either of the two possible ways).
But this implies that if we assert P V Q, we must be
prepared to assert P or to assert Q. We
may
consider
the
statement
3xP(x)
as
disjunction, asserting that P(x^) or P(%2> or ... as xt universe of discourse. produce an element a
a
multiple
runs over the
If we assert 3xP(x), then we must be able to such that P{a).
The assertion P^ V P2 may be
interpreted as 3i P t , where the universe of discourse for i is {1,2}.
3. Informal intuitionistic logic
11
The statement P ^ Q means that Q holds under the assumption that P holds: we show P =* Q by deriving Q from the hypothesis P. The statement P means that P * Q where Q is contradictory (for example, Q is (0 = 1)). Note that the constructive interpretation of P * Q is weaker than its classical interpretation, ~P V Q, because from >P V Q and P we can derive Q; but it is stronger than the intepretation  Q and P A Q. The classical principle  W * P is rejected in constructive mathematics because if P = 3xQ(x), then  W merely says that it would be absurd to deny that there is x such that Q(%), but gives no clue about how to construct x. On the other hand it is readily shown that ^,p , p. For, under the assumption ...P, we can prove P by deriving a contradiction from P: indeed, if P, then >P is absurd; whence  W , which contradicts IW. The principle >W => 
(x = 0), then x ^ 0. (iii) For each real number x, if  has always been controversial, and this remains the case in the constructive context.
The
intuitionists
interpret P => Q prooftheoretically, to mean that we have a finite routine for converting any proof of P to a proof of Q.
For a discussion of
another interpretation, due to Gbdel, see Bishop's paper Mathematics as a numerical
language
(in Intuitiontsm
and
Proof Theory,
NorthHolland,
1971); for a deeper analysis of the intuitionistic interpretations of the logical connectives and quantifiers, see Dumnett. Brouwer
referred
to
the
principle
P * P
as
the
simple
principle of the reciprocity of absurdity (Brouwer, page 11), because it says, essentially, that if Q is equivalent to the absurdity of P, then P is equivalent to the absurdity of Q.
Brouwer disliked logical shorthand
and expressed the theorem riiP => iP by absurdity of absurdity of absurdity is equivalent to absurdity (Brouwer, page 12). Although we refer to 'choice axioms', the practitioners of constructive mathematics do not
see
themselves
as exercising
choice;
rather, they see the choice function as inherent in the meaning of the Statement Vn3mP(n ,m). In Axiom of choice and complementation (Proc. Amer. Math. Soc. 51 (1975), 176178), Diaconescu shows that in a topostheoretic setting, the axiom of choice implies the law of excluded middle.
The settheoretic
version that we use is found in Choice implies excluded middle, by Goodman and Myhill (Z. Math. Logik Grundlagen Math. 23 (1978), 461).
Chapter 2.
Constructive Analysis
In u>hich various examples are given to Illustrate techniques and results in constructive analysis. Particular attentio?! ts paid to results which are classically trtutal, or haue little or no classical content, but uihose constructiue proof requires considerable ingenuity. The final section drams together uarious ideas from earlier parts of the chapter, and deals with a result which is interesting both in itself and for the questions raised by its proof. 1. Complete metric spaces The appropriate setting for constructive analysis is in the context of a metric space: there is, as yet, no useful constructive notion corresponding to a general topological space in the classical sense. We assume that the reader is familiar with the elementary classical
theory
of
metric
spaces
and
normed
linear
spaces.
Most
definitions carry over to the constructive setting, although we may have to be careful which of several classically equivalent definitions we use. For example, some classical authors define a closed set to be one whose complement
is open, whereas others define a closed
set S
to be one
containing all limits of sequences in S; it turns out that the latter is the more useful notion, and hence the one adopted as definitive, in the constructive setting. We shall use B(a,r) (respectively, B{a,r)) to denote the open (respectively, closed) ball with centre a and radius r in a metric space. The prototype complete space is the real line R. (1.1) Theorem Proof.
R is a complete metric space.
Consider any Cauchy +
sequence
integer k, choose IV^ in IN such that
(xn) in K.
For each positive
1. Complete metric spaces
19
Write
Let x^ be the 2kt*1 term of the regular sequence xvi^y n
in
m
v
(TO
) '
' v (m)
i> (n) '
< (2m)"1 + (2m)"1 + (^T 1 Hence x = ("^nl ^ s a  C 0
x
I
.
Then we have ' v (n)
n'
+ (2n)~X = m"1 + n"1.
rea
^ number. Moreover, if n > v(k), then
I
(O
 x n  < x
0 0   C 0

 x n  + x n  xv{n)\

I
+ xu(n) 
xj
< n'1 + (2T1)"1 + < (3k)'1 + (Wi)" 1 + (2k)" 1  h~X. Thus (xn) converges to the real number x". For
those concerned with
n
countable
instances of that axiom in the above proof.
choice, there
are
two
The first occurs when we
construct the sequence (%); the second when we construct the sequence (x^) by treating
the sequence
(xn) of real numbers as if it were a
sequence of regular sequences of rational numbers. A real number b is called the supremum, or least upper bound, of a subset S of R if it is an upper bound of S, and if, for each e > 0, there exists x in S such that x > b  e.
It is readily shown that the
supremum of S, if it exists, is unique; we denote it by sup S.
The
definition of infimum, or greatest lower bound, of S (written inf S ) is left to the reader. It
is trivial
to show that LPO is a consequence
of
the
classical leastupperbound principle: a nonvoid set of real numbers that is bounded above has a supremum.
As an application of the completeness of
IR we have the following constructive leastupperbound principle. (1.2) Theorem above.
Let S be a nonvoid set of real numbers that is bounded
Tlien. sup S exists if and only if for all a,P in IR with a < P,
either ft is an upper bound of S or there exists x in S with x > a. Proof.
If M s sup S exists and a < p, then either M < p or H > a; in the
latter case, we can find x in S with x + (M  a) > H and therefore x > a. Assume, conversely, that the stated condition holds. an upper bound u Q of S, and e > 0, so that [uoe,ug] n S is nonvoid.
Choose With
20
Chapter 2. Constructive analysis
en = e(3/4) n , construct a sequence (un) of upper bounds of S inductively, such that for each n, (i) ["n^n^l 0 S is nonvoid; (ii) either u n + 1  u R  £ f /4 or u n + 1 = utl. To do so, assume that U Q , ,un have been constructed with the appropriate properties. Then either un  tn/4 is an upper bound of S, or else there exists f in S with f > un  e n + 1 . In the first case, set u n + 1 = un  e n /4; in the second, set u n + ^ = un. This completes the inductive construction. Using (ii) we easily see that
whenever m > n. Then (un) is a Cauchy sequence, so its limit um exists by (1.1). Clearly un  en < um < un for each n. Since each u R is an upper bound of S, so is u^. Therefore, by (i), [i^e^u,,,] n S is nonvoid for each n; so um is the supremum of S. a This last proof is an illustration of dependent choice: the element u n+ i is not unambiguously specified, as the two conditions in question may both hold and the possible choices for u n + ^ depend on u n . However we can modify the proof so that only countable choice is involved (Problem 1 ) . A fundamental property of complete metric spaces is given by Baire's theorem: (1.3) Theorem The intersection of a sequence of dense open sets in a complete metric space X is dense in X. The standard classical proof of this theorem carries over into the constructive setting. We shall return to that proof, which involves dependent choice, in Section 2 below. Let X be a set with an inequality *. If for each sequence (xn) of elements of X there exists x in X such that x * x n for each n, then X is said to be uncountable. Using Baire's theorem, we can prove Cantor's theorem. (1.4) Theorem. K is uncountable; in fact, if (xJt) is any sequence of real numbers, then {x K: x ji xn for all n) is dense in R.
1. Complete metric spaces Proof. define
21
Consider any sequence (xn) of real numbers, and for each n in IN Vn = {x 1R: x ? xn}.
Then B(x,x  x n ) c Un for all x in Un, so that Vn is open. To show that Un is dense in K, consider any real number x. For each e > 0, either xn > x, in which case x Vn; or xn < x + e, so that x + e Un and x  (x + e )   e. Thus there are elements of Un arbitrarily close to x, and so Vn is dense in R. Applying Baire's theorem, we see that {x R : x * xn for all n) = (%ml Vn is dense in R. o
2. Baire's theorem revisited We often find that a single classical theorem gives rise to several constructively inequivalent results. A particularly common example of this occurs when a classical theorem p * q is written as q' =» p', where p',q' are classically, but not constructively, equivalent to "p,q respectively. To illustrate this, we prove versions of Baire's theorem that are classically equivalent to the contrapositive of (1.3), and therefore to (1.3) itself, but whose constructive content is altogether different from that of (1.3). Before doing so, however, we must introduce several notions, the first of which has no classical significance but is often essential in constructive analysis. A subset S of a metric space (X,p) is located in X if the distance p(x,S) = inf{p(x,s) : s S} from x to S exists for each x in X. Note that if S is located, then so is the closure S of S in X; if also T is dense in S, then T is located in X, and p{x,T) = p(x,S) for each x in X. There are Brouwerian examples of subsets of R that are not located (Problem 3). The metric complement of a set S in a metric space X is the set X  S of all points x of X that are bounded away from S, in the sense that p(x,s) > a for some a > 0 and all s in S. Note that X  S is open in X; and that if S is located in X, then X  S  {x e X : p(x,S) > 0}. By a construction similar to that extending Q to R, an
22
Chapter 2. Constructive analysis
arbitrary metric space X can be embedded isometrically as a dense subset of a complete metric space 1, called the completion of X; then X is a complete metric space if and only if X = X. The essence of the standard proof of Baire's theorem is the following lemma. (2.1) Lemma
Let Wn)
be a sequence of dense open sets in a metric space
X, V = f\r=i Un, xg a point of X, and T Q > 0.
Then there extsts a point xm
of the completion of X such that (i) P(X O T ,XQ) < r Q , and (ii) if xm X, then xro U. Proof.
Since U^ is dense and open in X, there exist x^ in X and a real
number r^ such that 0 < r^ < 2~* and B(x^,r^) C V^ D B(xQ,rQ).
Similarly,
there exist X2 in X and a real number r^
< 2
B(x2,i~2) C
O2 0 Btxj^r^).
Carrying
on
such that 0 < ^ in
this
way,
we
and
construct
inductively a sequence (xn)"mQ in X, and a sequence (rn)"_o of positive numbers converging to 0, such that if m > n > 1, then x m B{xn,rn) C Un n B U ^ j y  ^ ) .
Let xro be the limit of (xn) in the completion X of X.
Then xro belongs to the closure of B(xg,rg) in X, so that P(XB,,XQ) Also, if Xj,, e X, then for each n > 1, xm B(xn,rn)
< TQ.
belongs to the closed subset
of X, and therefore to Un; whence xm V.
a
We can now give the Proof of Baire's theorem (1.3).
Given a
sequence
{Vn)
of dense
open
subsets of a complete metric space X, let X Q be any point of X, and T Q any positive number. Lemma
(2.1).
Construct the point xm
Since X
is complete, xm
U D B(xg,rQ), where U = nJ^jU^. that V is dense in X.
of the completion of X as in belongs to X and therefore to
As X Q and rg are arbitrary, it follows
o
A metric space X is said to be incomplete if there exists a Cauchy sequence (xn) in X that is eventually bounded away from any given point of X, in the sense that for each x in X there exist N in IN and a > 0 such that p(x,xR) > a for all n > N.
Note that X is incomplete if and
only if there exists a point f in the completion of X such that p(f,x) > 0 for each x in X. Since Markov's principle is true in CLASS and RUSS, the next theorem can be viewed as another constructive version of Baire's theorem.
2. Baire's theorem revisited
(2.2) Theorem
23
Markov's principle is equivalent to the following state
ment : (2.2.1)
If (Un) is a sequence of dense open sets in a metric space X, such that the metric complement of f\i=jVn is nonvoid, then X is incomplete.
Proof.
Assume that Markov's principle holds, and let (Un) be a sequence
of dense open subsets of a metric space X such that the metric complement of U = (\x=]Pn is nonvoid.
Choose X Q in XU and rg > 0 such that p(x,xg) >
2rg for all x in II, and construct the point xm in the completion of X as in (2.1).
If x e X and xro = x, then by (2.1), ^ £ 1 ) and P U ^ X Q ) < r Q ,
which contradicts our choice of I Q .
Thus for all x in X, (x^,  x ) , and
therefore p{xm,x) > 0 by Markov's principle.
Hence X is incomplete.
Conversely, assume (2.2.1), and consider an increasing binary sequence (an) such that Nn(an = 0 ) .
Let 0 = rQ,r^,... be an enumeration
of the rational points of [0,1], and consider the metric space X = {0} U {x Q : 3n(a n = 1 and 0 < x < 1/n)}, with the metric inherited from K. U
n =
X
 Km
For each n, define : m
^"'
a
m
= 1
}
Then Un is dense and open in X: for if an  0, then U?l  X; while if an 1, then X = xn] is located and open in X for each n, that 1 e V = n £ = 1 U n , and that if 0 e U, then Vnfx^ = 0 ) .
Hence p(0,x) = 1 for all x in 17, and so 0
X  U.
By (2.4.1), there exists n such that X  Un
xn = 1.
Thus Markov's principle holds, The proof of Theorem
is nonvoid; whence
a
(2.4) illustrates a technique widely
employed in the presence of completeness.
Given a complete metric space X
and properties P(n), Q(n) of a positive integer n, such that for each n either P(n) or Q(n) holds, we want to show that Q(rc) holds for some n. construct an increasing binary sequence ^
= 0, and Q(it) holds if X,x = 1.
We
(Xrl) such that P(n) holds if
Using the information provided by the
properties P(n) and Q(n), we then construct a Cauchy sequence (x n ) in X such that xn = x n _^ if X^ = 1, and such that the behaviour of the limit of {xn) in X will enable us to compute a value of n for which X^  1 and therefore Q(n) holds. We shall return to this technique in Section 3.
In the
meantime, we have two more definitions. A subset S of a metric space X is colocated in X if its metric complement X  S is located; and S is bilocated i n X if it is both located and colocated in X. Lemma (2.3) also leads to a version of Baire's theorem that does not involve Markov's principle. (2.5) Theorem
Let (Cn) be a sequence of closed, bilocated subsets of a
complete metric space X, such
that X = U^ = ]C n .
Then C^ has nonvoid
interior for some n. Proof.
The proof is left as an exercise (Problem 5 ) .
n
26
Chapter 2. Constructive analysis
We invite the reader to examine the standard classical proof of the open mapping theorem for bounded linear maps, in order to confirm that, in the light of Theorem
(2.5), we can establish
the
following
constructive version of the open mapping theorem. (2.6) Theorem
Let u be a bounded linear mapping of a Banach space X onto
a Banach space Y, such that u(B(0,l)) is bilocated in Y.
Then u is an
open mapping  that is, u maps open subsets of X onto open subsets of Y. D
3. Located subsets As Theorems (2.4) and (2.5) suggest, locatedness is a significant property of a subset of a metric space, and one which merits further investigation. Even if a subset is located, we may not be able to prove that its metric complement is located (Problem 8 ) .
In certain cases, however,
we can do so. (3.1) Proposition.
If S is a located, convex subset of IRn such that
is notivoid, then R^S is located. Proof.
As the reader may verify, it is enough to prove that T £ DfS is
located relative to the metric p^ on IR", where Pl<x,y) = sup{xf  y t  : 1 < i < n} for all x = (x 1( ...,x n ) and y = (yp ... ,ytl) in tRn.
Accordingly, consider
any x in K n , and for each r > 0 let B^{x,r) be the closed ball with centre x
and
radius
r
relative
to
the
metric
p^.
Then
B^(x,r)
is
an
ndimensional cube with centre x and sides of length 2r. Consider any two real numbers a,p such that a < p. B^(x,^(a + p)).
Let t>(l),
,i>(2") be the vertices of
It is left as an exercise for the reader to show that for
all sufficiently small 6 > 0, and all points u>(l), p^uU ),u>(i)) < {iu(l),
,iu(2n)}
6
for
all
i
in
contains B^(x,a).
{1,...,2"}, Fix
such
a
the
,u)(2") such that convex
number
Either p^(u(l),S) < 6 for all i, or p^(u(t),S) > 0 for some i. first
case,
for
each
i
in
{1,...,2"}
choose
u>(l)
in
p^(u(i ),u>(i )) < 6; then S contains the convex hull of {ui(l), therefore contains B,(x,a).
hull
6 < jd 3
S
of a).
In the so
that
ri
,iu(2 )J and
Thus pi(x,y) > a for all y in T.
On the
other hand, if, for some i, we have p^(u(t),S) > 0, then u(i) T and
3. Located subsets
27
p±(x,v(i)) < ^(a + p) < p. principle (1.2).
Thus p^XjS) exists, by the leastupperbound
a
Two subsets S and T of Iff1 are said to interlace if there exist s,s' in S, X > 1, and fi in (0,1) such that both As + (1  A)s' and us + (1  u)s' belong to T.
If S and T are disjoint, and either S or T is
convex, then S and T do not interlace.
Thus Proposition
(3.1) is a
special case of the following general result, which is proved in the paper Locating metric complements in Krl, in Springer 873. (3.2) Theorem
If S
is a
located
subset of Iff mlth nonvoid metric
complement T, such that S and T do not interlace, then T is located.
D
The completeness technique used in the proof of Theorem (2.5) can be applied to produce a very useful result with almost no classical content. (3.3) Lemma
Let S be a complete, located subset of a metric space X, and
x a point of X.
Then there exists y in S such that p(x,y) > 0 entails
p(x,S) > 0. Proof.
Construct a binary sequence (A^) such that A^ = 0 => p(x,S) < n" 1 ,
and A n = 1 =» p(x,S) > (n+ir 1 . Note that (A^) is increasing.
Let z be any point of S.
x^ = z; if A7l » 0, choose xn in S with p(x,xrl) < n set xn = x n _^.
If A^ = 1, set
, if n > 1 and A^ = 1,
Then {xn) is a Cauchy sequence in X  in fact, p(x m ,x n ) < 2n~
(m > n ) .
As X is complete, (xn) converges to a limit y in X satisfying p(y,x n ) < 2H" 1
for all n.
If p(x,y) > 0, choose N in IN+ so that p(x,y) > 3N"1.
Then piXfXff) > p(x,y)  p(y,x N ) > N~ , so that A N ? 0, and therefore A N = 1.
Hence p(x,S) > (N+l)" 1 > 0.
a
To prove (3.3) classically, we choose y to be x if p(x,S) » 0; otherwise, we choose y to be any point in S.
28
Chapter 2. Constructive analysis
(3.4) Corollary and
x
is a
p(x,S) > 0.
If S Is a complete, located subset of a metric space X,
point
of X
such
that
p(x,y) > 0 for all
y
in S,
then
Corollary (3.4) provides conditions under which a uniformly
continuous, positivevalued function  namely, the function taking y to p(x,y) on S  has positive infimum.
As we shall see in Chapter 6, an
arbitrary uniformly continuous mapping of a compact metric space into K + need not have positive infimum.
4. Totally Bounded Spaces Many of
the most
important
results of
classical
analysis
depend on an application of the property of compactness of a metric space X.
This property is usually defined in one of two classically equivalent
ways.
The first  every open cover of X contains a finite subcover  is
of no use in constructive analysis as, for reasons that will be clear after Chapters 3 and 5, we cannot prove that the interval [0,1] is compact in that sense. sequence
of
The second notion, that of sequential compactness (every points
of
X
contains
a
convergent
subsequence),
is
inapplicable because even the twoelement set {0,1} cannot be shown to be sequentially compact. There is, however, a third property which, for a metric space, is classically equivalent to compactness in either of the above senses, and which is widely applicable in constructive mathematics.
It is this
third notion that we shall dignify by the title of 'compactness'. Let (X,p) be a metric space. subset Y of X p(x,y) < e.
such that for each x
An eapproximation to X is a
in X
there exists y
in Y with
The space X is said to be totally bounded if for each e > 0
there exists a finite eapproximation to X.
A complete, totally bounded
metric space is said to be compact. A
totally bounded
space X
is bounded,
p(x,y) < c for some c in K and all x,y in X.
in the
sense
that
A dense subset of a totally
bounded space is totally bounded. A bounded, closed interval is compact relative to the standard metric on K. bounded.
More generally, closed balls in 0?" and C11 are totally
4. Totally bounded spaces
29
Recall that a set X is finite, under the denial inequality, if there is a oneone mapping
e. Proof.
Let {xj^,
,xn} be an eapproximation to X, and let 77 > e.
proceed by induction on n.
We
Either p ( x t , x ) > 0 whenever i * j , or
p(xt,xj) < TJ  e for some pair of distinct indices i,j.
In the former
case, {xp.. .,xn} is a metrically finite Tjapproximation.
In the latter
case, we can delete x from {x^,...,xn} and get a finitely enumerable e'approximation with e' = e + p(xt ,x .) < r\. n A metric space is separable if it contains a countable dense set. (4.2) Lemma Proof.
A totally bounded space is separable.
If, for each positive integer n, Fn is a finite n~ approximation
to the metric space X, then of X.
F H L ^=i F rl
is
a
countable dense subset
D Many applications of total boundedness depend on the next
result. (4.3) Proposition
If X is a totally bounded space, and f a uniformly
continuous map of X into a metric space X', then f(X) = {f(x) : x X} is totally bounded. Proof.
Consider any e > 0, and choose 6 > 0 such that p(f(x),f(y)) < e
whenever x,y X and p(x,y) < 6. to X.
Construct a 6approximation { x ^ , , x n )
Then for each x in X there exists i such that p(x,x^) < 6 and
therefore
p(f(x),f(xt)) < e.
Thus
enumerable eapproximation to f(X).
If(x1),...,f(xn)}
is
a
finitely
Reference to (4.1) completes the
30 proof.
Chapter 2. Constructive analysis D To apply Proposition (4.3), we need a lemma.
(4.4) Lemma
If S is a nonvoid totally bounded subset of IR, then inf S
and sup S exist . Proof.
Let a,p be real numbers with a < p, and set e = {p  a)/4.
Let
{x^,...,xn} be an eapproximation to S, and choose N in {l,...,n} such that x n > sup{xj_,...,x(l}  e.
Either a < x N or x N < a + 2e.
In the
latter case, if s is any point of S, and i is chosen so that s  x t  < e, we have s
< xt+e
< Xjy + 2e
so that p is an upper bound of S.
< a + 4e
= P,
Thus sup S (and similarly inf S)
exists, by the constructive leastupperbound principle (1.2). (4.5) Theorem
If f is a uniformly continuous map of a totally bounded
space X into K, then inf f = inf f(X) and sup f = sup f(X) exist. Proof.
Apply (4.3) and (4.4).
n
In particular, a uniformly continuous map f of a totally bounded space X into IR is bounded, in the sense that f(x) i c for some c in IR and all x in X. (4.6) Corollary
A totally bounded subset S of a metric space X is
Iocated . Proof.
Let X Q be any point of X.
Then the map taking x to P ( X , X Q ) is
uniformly continuous, so that p(xo,S) = exists, by (4.5).
inf{p(x,x0) : x S}
o
The next theorem ensures that a totally bounded space has a plentiful supply of totally bounded subsets. (4.7) Theorem
Let X be a totally bounded space, X Q a point of X, and r a
posittue number.
Then there exists a closed, totally bounded subset K of
X such that B(x o ,r) C K C B(x Q ,8r). Proof.
With F^ = {xg}, we construct inductively a sequence
finitely enumerable subsets of X such that
(F n ) of
4. Totally bounded spaces
31
(i) p(x,Fn) < 2"+1r for each x in B{xQ,r), (ii) p{x,Fn) < 2"n+3r for each x in To
this
end,
assume
that
appropriate properties.
F]_,...,Fn have
F^.
been
constructed
with
Let [x^,... ,x^} be a 2~"rapproximation
the to X.
Partition {1,...,N} into subsets A and B such that p(xt,Fn) < 2~n+3r for i in A, p(x i ,F n ) > 2~n+2r for t in B. Then, clearly, Fn+^ = {xt : i e A} satisfies the appropriate instance of Let x be any point of B(xg,r).
By the induction hypothesis, there
exists y in Fn with p(x,y) < 2~ n+1 r.
(ii).
Choosing I in {1,...,N} such that
n
p(x,x t ) < 2~ r, we have p(xt,Fn)
n and y Fm, then by (ii) we can
p(y,Fn)
0, construct an e/32approximation
{x^,...,xn} to X.
By (4.7), for each i in {l,...,n} there exists a closed, totally bounded
32
Chapter 2. Constructive analysis
set Kt such that B(xt,e/32) C K { C B(x t ,e/4).
Clearly, X  L/J»iKt.
p(x,y) < e/2 for all x,y in Kt , so that diam K^ < e/2 < e. The standard notion of
Also,
a
(pointwise) continuity of functions
between metric spaces is not sufficiently powerful for most constructive purposes.
Uniform continuity, however, is a concept of
considerable
constructive power (witness, for example, Proposition (4.5)).
As will be
made clear in Chapter 6, we have no hope of proving the classical uniform continuity theorem (a pointwise continuous function on a compact space is uniformly
continuous);
nor
are
we
likely
to
find
a
constructive
counterexample to that theorem. The
reader
familiar
with
the
classical
theory of
uniform
spaces will appreciate that the following result is a constructive version of the theorem that the unique uniform structure U compatible with the given
topology
on
a
compact
Hausdorff
space
X
is
induced
by
the
Uuniformly continuous mappings of X into IR; the latter theorem is the key to the standard classical proof of the uniform continuity theorem in the general context of uniform spaces (4.9) Proposition space
X,
such
Let h be a mapping of a metric space into a compact
that
f°h
continuous map f :X » IR. Proof.
Given
any
is
uniformly continuous
positive
approximation to X.
for
each uniformly
Then h is uniformly continuous. number
e,
let
{x^,...,xR}
be
an
p{xt,x) for i = l,...,n.
It suffices to show that if If^x)
< e/3 for each t, then p(x,x') < e. < e/3, and therefore ft(x') < 2e/3.
ft(x')\
Choose i such that f t (x) = p(xt,x) Then
p(x,x' ) < p(xt,x) + p(xt,x' ) = ft(x) + ft(x' ) < e/3 + 2e/3 = e. It
is
e/3
Define uniformly continuous maps ft :X » IR by ft(x) =
instructive
to
compare
classical approach to Proposition (4.9). mind goes as follows.
this
proof
with
a
a natural
The classical proof we have in
Suppose that h is not uniformly continuous on X.
Then there exist (sic) a > 0 and sequences (x n ), (yn) in X, such that for each n, p(xn,yn) < n exists
and p{h(xn),h(yn)) > a.
(sic) a subsequence
(xn(kj )£ = 1 of
converges to a point z of X'.
As X' is compact, there
(xn) such that
(Mx,,^)) >fe=i
The uniform continuity of the map taking x
to p(h(x),z) on X now ensures that (Myn(fe)) )£_i also converges to z; whence p(h(xn,hs)»Mun(fc)))
< a for all sufficiently large k.
Since this
4. Totally bounded spaces
33
contradicts our choice of (xn) and (u,,)» w
conclude that h is uniformly
continuous on X after all. A nonvoid metric space X is locally totally bounded if each bounded subset of X is contained in a totally bounded subset; if, in addition, X is complete, then it is said to be locally compact.
For
example, the real line R, the complex plane C, and the euclidean spaces IR and C are locally compact. The following lemma prepares the way for our next theorem, which deals with
certain
fundamental
properties
of a locally
totally
bounded space. (4.10) Lemma
Let L be a located subset of a metric space X, and T a
totally bounded subset of X.
Then there exists a totally bounded set S
such that r n L C S C t . Proof. to T.
For each positive integer n, let T n be a finite 1/happroximation Write T n as a union of finite sets An and Bn such that p(t,L) < 3/h
for all t in A(1, and p(t,L) > 3/2n for all t in B[X. choose s in L such that p(t,sn) < 3/h. S be the closure of U^_jSIX in L.
For each t in A R ,
Write S R = {sn : t An], and let
T O prove S totally bounded, consider
positive integers m,n with 6m < n, and any s in S n .
As p(s,T) < 3/h, we
have p(s,Tm)
0 so that if Given such y, we have y =
u(x), where x s 2?«i\ x t and llxll < 1. Thus B(0,e) C u(B(O,l)). (5.6) Lemma
D
Let B be a totally bounded, balanced neighbourhood of 0 tn a
normed linear space Y containing a nonzero uector. Then inf{t : y tB} exists for each y in Y. Proof.
Let R = sup{llxll : x e B}, which is positive as B is a neigh
bourhood of 0 and Y is nontrivial. y e tB} = Ilyll/R. (5.7) Lemma space X, {ei
Then for each y in Y, inf{t :
D
Let S be a finitedimensional subspace of a normed linear e
i } a metric basts of S, and e a uector in X such that
p(eR,S) > 0. Then the set {e^,...,eR} is a metric basis of its span. Proof.
It suffices to show that {{*!>...,\) : 112^X^11 < 1}
i s bounded. As { e ^ , . . . ,efl_^} i s a metric basis of S, i t suffices to find bounds on 1^1 and H2*^~Jxtet II when ll2r^=1Xtet II < 1. To this end, suppose that ll2n=1Aletll < 1. If lAj > l/p(e n ,S), then
which is a contradiction. So 1^1 < l/p{en,S) and
(5.8) Lemma
Let X be a normed linear space.
The set {ei,...,e } is a
metric basts of X if and only if (i) it spans X, and (ii) if X^,...,^ are scalars such that 2^_^ Xt  > 0, then H5:"_^A1elII > 0. Proof.
If {e1,...,en} is a metric basis, then conditions (i) and (ii)
clearly hold.
Conversely, suppose those conditions obtain.
If n » 1,
then {ejj is easily seen to be a metric basis of X. If n > 1, assume that {e^t ...,en_i) is a metric basis of its span, S. In view of (5.7), it will suffice to prove that p(en,S) > 0.
As S is finitedimensional, it is
complete and located; whence, by (3.3), there exists s e S such that if llea  sll > 0, then p(en,S) > 0.
But lle^  sll > 0, by (ii) and the
assumption that {e^,...,en_^} spans S; so p(en,S) > 0.
D
38
Chapter 2. Constructive analysis
We now return to the Proof of Theorem (5.4).
Let u be a bounded linear map of X onto a
finitedimensional normed linear space Y, and write B = u(B(0,l)).
If u
is compact, then it follows from (5.5) and (5.6) that in£{t : y tB} exists for each y in Y; whence, by (5.2), Ker u is located.
Conversely,
if Ker u is located, then llu(x)H'
=
p(x,Ker u)
defines a norm, with a corresponding equality, on T. that the open unit ball of < r, II II') is u(B(0,D).
It is easily seen Let {u(x x ),... ,u(xn)}
be a metric basis for (Y,ll II).
By (5.8), if A^,...,^ are scalars with
2 r t l =1 \ t  > 0, then I M S ?  ! ^ ^ ^ t ) "
> 0; as u is continuous, it follows that
"2t=l \"(*i)ll' = p{^"=i\xitKer Hence
{ u U ^ , . . . ,u(xn)}
is
a
metric
basis
u) > 0. of
(Y,ll II'), by
(5.8).
Therefore the identity maps from (Y,H II) to (r,ll II'), and from (r,ll II') to (Y,ll II), are
uniformly
continuous.
By
(4.12),
u(B(0,l))
is
totally
bounded with respect to II II'; whence, by (4.3), it is totally bounded with respect to II II. Thus u is compact.
n
In classical mathematics, every bounded linear mapping of a normed
linear
compact.
space
onto a
finitedimensional
normed
linear
space
is
The reader is encouraged to construct a Brouwerian example of a
bounded linear map u with range R
such that u(B(0,1)) is not totally
bounded (Problem 9 ) . The theorem.
remainder
of
this
section deals with
the
HahnBanach
We will prove this theorem for real vector spaces, and then
extend to the complex case.
The first lemma shows how to extend a linear
functional from a finitedimensional subspace to a subspace with one more dimension. (5.9) Lemma space
Y
over
p(yQ.X) > 0.
Let X be a finitedimensional subspace of a normed linear IR, and
suppose
that
Y
is
spanned
by
X U (yn), utfiere
Let u be a nonzero normable linear functional on X.
Then
for each e > 0, there exists a normable linear functional v on Y such that v = u on X, and Hull < Hull + e. Proof.
We may assume that Hull = 1.
For all x,x' X we have
u ( x ) + u ( x ' ) = u ( x + x ' ) < llx + x ' l l < llx  ygll + l l x ' + J / Q I I ;
5. Bounded linear maps
39
whence u(x)  llx  u 0 " ^ " * ' + yo" ~
u ( x
' >'
and, a f o r t i o r i , f(x)
= u(x)  (ltt)llx  yoll < (l+t)llx' + yoll  u ( x ' ) .
Choose r > 0 so t h a t f(x)
< (1+e)IIL/QII whenever llxll > r .
Then
s = sup{ 2~rl~1, then p(x,s) > 2"114"1 for all s in S. Prove that S is totally bounded, (cf. Lemma (3.3).) 14. Let f :X » IR be a continuous function on a compact metric space X,
46
Chapter 2. Constructive analysis
and for each a in 1R define X a E {x X : f(x) < a } . Prove that there is a sequence (a=n) of real numbers greater than m = inf f such that Xa is compact whenever a > m and a ji an each n.
for
(For each positive integer k write X as a finite union
U ^ k ) X i fe °f compact sets, each of diameter Define
(an)
to
be
an
enumeration
less than k
of
the
.
numbers
inf{f(x) : x e X kJ for all relevant values of j and k.) 15. Let h be a mapping of a compact space X into a metric space X', such
that
f°h
is
uniformly
continuous map f:X' » 1R.
continuous
for
each
uniformly
Prove that h is pointwise continuous,
that h(X) is bounded, and that h is uniformly continuous if and only if h(X) is totally bounded.
Prove that if X' is locally
compact, then h is uniformly continuous,
(cf . Proposition (4.9))
16. Let E be a balanced subset of C containing a nonzero number. Prove
that 1
inf {111""
£
is
locally
totally
bounded
if
and
only
if
: t e E, t / 0} exists.
17. Prove Proposition (5.1). 18. Construct a Brouwerian example of a linear subset V of IR and a bounded
linear
normable,
and
functional u (ii) u
does
on V, not
such
extend
that to
a
(i) u bounded
is
not
linear
functional on IR. 19. Show that the positive number e is needed in the HahnBanach theorem, by constructing a Brouwerian example as follows. 2
be R , with norm given by ll(x,y)ll = sup{ x, y  } , small
real number, and let p =
(l4a,l).
Let V
let a be a
Define u:Rp > K by
u(rp) = rllpll, and suppose u extends to a linear functional v on V of norm 1. vie^) > 0
Let e^ = (lta,0) and e 2 = (0,1). or
v(e 2 ) > 0;
and
that
v(e 2 ) = 0
Show that either if a > 0, while
v(e 1 ) = 0 if a < 0. 20. Show that if X is a Banach space, and Y is a subspace of X, then X/Y is a Banach space. 21. Prove Lemma (6.3). 22. A
normed
linear
space X
is infinitedimensional
if X  V
is
Problems
47
nonvoid for each finitedimensional subspace V of X. if V
Prove that
is a locally compact subset of an
infinitedimensional
Banach space X, then X  Y is dense in X.
(Use (6.7), Baire's
Theorem (1.3), and Problem 11.)
NOTES One might
argue
that
a
sequence
of
real
numbers
is, by
definition, a sequence of regular Cauchy sequences of rational numbers; thus there would be no second instance of countable choice in the proof of (1.1).
Rephrased, this says that any function from IN to R is a function
from IN to the set of regular Cauchy sequences of rational numbers.
No one
would make this last assertion with IN replaced by R, as it would imply LPO; so there appears to be something special about IN, which is exactly what countable choice says. Classical proofs of Baire's
theorem
and
the
open
mapping
theorem are found in Chapter 5 of Rudin, Real and Complex Analysis (McGraw Hill, 1970). mapping
For more information on constructive versions of the open
theorem,
see
the
forthcoming
paper
Open
and
unopen
mapping
theorems, by Bridges, Julian, and Mines; and Stolzenberg's unpublished paper A critical analysts of Banach's open mapping theorem
(Northeastern
University, 1971). An
elementary
proof
of Cantor's
theorem
(1.4) appears
in
Chapter 2 of BishopBridges. The
completeness
technique
described
after
the
proof
of
Theorem (2.4) first appeared in Bishop (Chapter 6, Lemma 7, p. 177). The technique has many interesting applications: for example, it is used at a key step in the proof of a constructive version of the fundamental theorem of approximation theory (BishopBridges, Chapter 7, (2.8)). The
proof
of
Theorem
(4.7)
is
taken
from
Aspects
of
Constructiutsm, an unpublished version of Errett Bishop's lectures at the Mathematics Symposium held at New Mexico State University in December, 1972. The approach to Theorem (4.11), via Lemma (4.10), is neater and more natural than the standard one (found, for example, in BishopBridges, Chap. 4, (6.2) and (6.3)).
48
Chapter 2. Constructive analysis
Bishop defines a continuous mapping from a compact space X to a metric space Y to be one that is uniformly continuous; he defines a continuous mapping on a locally compact space X to be one that is either continuous
on
each
compact
subset of X,
or, equivalently,
uniformly
continuous on each bounded subset of X. The standard classical proof of the uniform continuity theorem in the context of uniform spaces is found in Volume
I of Bourbaki's
General Topology (AddisonWesley, 1966, Chap. II, §4.1). The locatedness of the range of a selfadjoint operator on a Hilbert space is discussed in Operator ranges, tntegrable sets, and the functional calculus (Houston J. Math. 11 (1985), 3144). Proposition (1.10)) without
(5.3) is proved
the help of Proposition
in BishopBridges (5.2).
Theorem
(Chapter
7,
(5.4) first
appears, with a slightly different proof, in Bounded linear mappings of finite rank (J. Functional Anal. 43 (1981), 143148). The HahnBanach theorem is proved as a consequence of the separation theorem in Bishop and BishopBridges. The material of Section 6 is drawn from Compactly generated Banach spaces, Archiv der Math. 36 (1981), 239243.
Chapter 3.
Russian Constructive Mathematics
In which the fundamentals of RUSS are presented loithtn a framework constructed from BISH and an axiom deriued from Church's thesis. The notion of a programming system Is introduced In Section 1, and Is used In discussions of omniscience principles, continuity, and the Intermediate ualue theorem. Section 3 deals with Specter's remarkable construction of a bounded Increasing sequence of rational numbers that is euentually bounded au>au from any giuen (recursiue) real number. The next section contains a strong counterexample to the HelneBorel theorem, and a discussion of the Lebesgue measure of the unit Interual and Its compact subsets. The last tioo sections of the chapter deal with Ceitln's theorem on the continuity of realvalued functions. 1. Programming Systems and Omniscience Principles The ChurchMarkovTuring hailed
as
a milestone
notion of computability
in our understanding
However, its use as a general method
of
from a
Recursivefunction
formulations of theorems appear unnatural to the practising
mathematician, technical
justly
effective procedures.
for viewing mathematics
constructive point of view has serious disadvantages. theoretic
is
and
their
terminology
proofs
are
often
fraught
that obscures the essential
with
ideas.
forbidding Indeed, an
analyst might be forgiven for believing that the advantages to be gained by an understanding of the theorems of recursive analysis were not worth the effort necessary to achieve this understanding.
This is unfortunate,
because the essence of the basic theorems in recursive function theory can be obtained without a complicated technical formulation.
We shall arrive
at these theorems within the informal framework of BISH by appending an axiom that is a simple consequence of Church's thesis. One of the insights of recursive function theory is that the proper study of computation involves partial functions rather than just
50
Chapter 3. Russian constructive mathematics
functions. Recall that a partial function from a set X to a set Y is a function f from a subset of X to Y.
If x X, we say that f (x) is defined
if x e dom f, and that fix) is undefined if x e dom f is impossible.
A
partial function f is a restriction of a partial function g if dom f C dom g and f (x) = g(x) for each x in dom f. equal if each is a restriction of the other. is total.
Thus a total
Two partial functions are If dom f » X we say that f
(partial) function from X to y is simply a
function from X to Y. Partial functions from IN to IN arise from computer programs operating on inputs in IN.
Imagine that each step of a program is executed
in a fixed amount of time, and that the program signals when it stops executing, if ever, at which time the output can be viewed. associate with
We may
such a process a mapping A from IN x IN to IN U {1}
by
setting A(x,n) equal to 1 if the program executes n+1 steps without stopping, and equal to the output if the program stops before completing n+1 steps. IN U {1}
With this in mind, we say that a mapping A from IN x IN to
is a partial function algorithm if A(x,n+1) = A(x,n) whenever
A(x,n) ??  1 . The partial function f associated with A has domain dom f
=
{x e IN : A(x,n) t 1 for some n in IN};
if x e dom f, we define fix) = A(x,n) for any n such that A(x,n) #  1 . Such a partial function can be characterized by the property that its domain is countable.
Before we prove this, note that a subset of IN is
countable if and only if it is the union of a sequence of finite sets. (1.1) Lemma
A partial function from IN to IN has a countable domain if and
only if it is equal to the partial function associated loith some partial function algorithm. Proof.
Let f be a partial function from IN to IN.
If f is associated with
the partial function algorithm A, let D{k) = {x < k : A(x,k) jt  1 ) .
Then
D(k) is a finite subset of IN for each k, and dom f is the union of the sets D(k).
Thus dom f is countable. Conversely, suppose that dom f is the union of a sequence
(D{h)) of finite subsets of IN.
Define A(x,n) s f(x) if x D(k) for some
k < n, and A(x,n) = 1 otherwise.
Then A is a partial function algorithm,
and f is the associated partial function.
1. Programming systems and omniscience principles
51
The reader should prove that the composition of two partial functions IN to IN with countable domain is also a partial function with countable domain. If L is a sufficiently powerful programming language, then we can write a program in L that enumerates all programs of L. Thus the set of partial functions arising from such programs is countable. If we believe that all functions worthy of the name are computable by the language L, then we are naturally led to accept the following axiom. CPF
There is an enumeration