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0, we define recursivel' ~ I o.p.en I::~ and II~ of subsets of X as follows: Y amIhes
t
I::~ = T, II~ = {X \ A : A E lJo} I::~ = {U M : M E [U/3 k. Hence x b E cl(g(brk~O)) and XC E cl(g(brk~1)). By (iii), x b -:j:. xc. Now consider the function
' < '" for all A < "'.
14.
TREES
31
From now on we will call NI-Kurepa trees simply Kurepa trees. The following statement is known as the K urepa Hypothesis. (KH) There exists a K urepa tree. Theorem 22.8 shows that KH is relatively consistent with ZFC. On the other hand, if the existence of an inaccessible cardinal is consistent with ZFC, then so is the negation of KH. In many applications of trees it is crucial to know whether a given tree has at least one cofinal branch. The tree in Example 14.6 has no cofinal branches, but rather large levels (of size h'1). On the other hand, the slender trees of Examples 14.2 and 14.4 have the property that every branch is cofinal. It is natural to conjecture that trees with relatively small levels will always contain cofinal branches. Let us explore this hunch. LEMMA 14.2 (Konig). Let (T, :ST) be a tree of height w such that T(n) is finite for all nEw. Then T contains a cofinal branch. PROOF. We construct recursively a sequence (tn)n<w of elements of T so that for all n < w:
(i)
tn 0 is a limit ordinal of countable cofinality, and (~n)nEw is an increasing sequence of ordinals with limnEw ~n = 8. For simplicity, suppose that e~n ~ e~n+l for all nEw. EXERCISE 14.28 (G). Convince yourself that if UnEw rng(e~J = there is no eo EEl such that e~n ~* eo for all nEw.
WI,
then
How can we prevent the situation described in Exercise 14.28 from ever arising? Evidently, the requirement that IWI \rng(e~)1 ~ NI is not sufficient. We can make the requirement more stringent by fixing a proper ideal I of subsets of WI and considering the following subset of E I :
E2 = {e E <W2 WI
:
e is an injection and rng( e) E I}.
14.
TREES
39
What properties should I have so that our construction will work? First of all, we will want I to contain some uncountable sets. Let us require that I has the following property: (i) VA E I3B E [W1]Nl (A n B = 0" A U BE I). Second, in order to get our construction past limit stages of countable cofinality, we will want I to be countably complete; i.e., we want I to satisfy the following:
(ii)
If A E [I] No , then
UA E I.
Note that if I satisfies (ii), 8 < W2 is a limit ordinal of countable cofinality, (~n)nEw is an increasing sequence of ordinals cofinal in 8, and for each nEw we have constructed een E E2 such that een ~ een+l for all nEw, then we can simply define eli = UnEw een' It remains to' consider limit stages 8 of cofinality W1. Let us assume that (~Q)Q<Wl is an increasing sequence of ordinals cofinal in 8, and the ee", 's have already been chosen from the set E 2 . For the sake of simplicity, let us assume that ee", ~ ee13 for all a < f3 < W1· If there exists eli E E2 such that eea :::;* eli, there must exist B E I with Irng(eeJ\BI < ~1 for all a < W1' In other words, we need to require the following property of I: (iii) If A E [I]Nl, then there exists B E I such that IA\BI < ~1 for all A E A. It turns out that this property also suffices. Given B as above, use (i) to find C E [wd N1 that is disjoint from B and still a member of the ideal I. Now define recursively eea as follows: If ee", (1]) E B, then let e~J 1]) = eea (1]). If eea (1]) tJ. B, then let eea (1]) be the smallest ordinal in C that has not yet been used in the construction. Finally, let eli = UQEWl e~a' EXERCISE 14.29 (G). Convince yourself that this construction yields a sequence (ee)e<Wl of elements of E2 that satisfies conditions (a) and (b). But does there exist an ideal I that satisfies conditions (i)-(iii)? In Chapter 21 (Lemma 21.11) we will prove the following: LEMMA 14.10. Let", be a regular uncountable cardinal. Then there exists an ideal N SI< on '" with the following properties: (i) VA E NSI 1. Let r, s be two different elements of 8(ao). Pick ~ < WI such that r ~ such that s (3 such that there is no t E Rbs) with s 1 for each (a, b) E T(o:). Then we let T(o: + 1) = U{I(a,b): (a, b) E T(o:)}. At limit stages 0 > 0, if T(o:) has already been constructed for all 0: < 0, let T(o) be a family I of pairwise disjoint elements of I(L), maximal with respect to the property that if (c,d) E I, then Vo: < 03(a,b) E T(o:) ((c,d) ~ (a,b)). Finally, let T = U"'<Wl T(o:). EXERCISE 14.52 (G). (a) Convince yourself that (T,~) is a tree whose o:-th level is equal to T( 0:) for each 0: < WI' (b) Show that if (a, b), (c, d) E T are incomparable with respect to ~, then (a,b) n (c,d) = 0.
It follows immediately from the assumption on (L, '5.d and point (b) of the last exercise that (T,~) does not contain uncountable antichains. Moreover, it follows from the remark made in the construction of successor stages that T is splitting. Hence, by Exercise 14.34, T does not contain uncountable chains either. It is somewhat less obvious that T has height WI' SO let 0:0 is the smallest ordinal 0: such that T(o:) = 0. Clearly, 0:0 can be neither a successor ordinal nor O. Thus, 0:0 must be a limit ordinal :2: w. Consider the set E = {c E L: 3(3 < 0:03(a,b) E T((3) (c = a V c = b)}. EXERCISE
14.53 (G). Show that E is dense in (L, '5.L), and conclude that
~=~.
0 Mathographical Remarks
If you are looking for a source to learn more about trees and their applications, we recommend the article Trees and linearly ordered sets, by Stevo Todorcevic in the Handbook of Set-Theoretic Topology, K. Kunen and J. E. Vaughan, eds., North-Holland, 1984,235-293. More information about equivalences of Konig's Lemma and weak versions of the Axiom of Choice can be found in the article by J. Truss, Some cases of Konig's Lemma, in: Set Theory and Hierarchy Theory. A Memorial Tribute to Andrzej Mostowski, Bierutowice, Poland, 1975, edited by W. Marek, M. Srebrny, and A. Zarach, Lecture Notes in Mathematics 537, Springer Verlag, 1976.
CHAPTER 15
A Little Ramsey Theory Here is a bit of common sense: If more than n pigeons occupy n pigeonholes, then at least one pigeonhole must be occupied by at least two pigeons. Lemma 14.3 is a mathematical counterpart of this folk wisdom. Let us reformulate it here. THEOREM 15.1 (Pigeonhole Principle). Assume,.. is an infinite cardinal, A is a set of cardinality,.., and A = Ui" are infinite, (J 2': 2, >.. is regular, and I\, f+ (>..)~. Then 21< f+ (>..)~+1. PROOF. We prove how to step up from superscript 2 to superscript 3 in the case when (J 2': w. For the complete proof of Lemma 15.16, as well as some variations on it, we refer the reader to Lemma 24.1 of [EHMRJ (see Mathographical Remarks). So let F : [I\,J2 --+ (J be a coloring without a homogeneous subset of size >... For {J, g} E [1. < K. We show that K is not weakly compact by constructing a function f : [KF ~ K without homogeneous subset of size K. Let (Ki)i. be an increasing sequence of cardinals with SUp{Ki : i < >.} = K. Define f by: f( {a,,B}) = 0 iff Vi < >. (a < Ki +-+,B < Ki). Now suppose X ~ K is homogeneous for f. If f({a,,B}) = 1 for all {a,,B} E [XF, then IXI ~ >. < K. If f( {a,,B}) = 0 for all {a,,B} E [X)2, then X ~ Ki for some i < K, and again we have IXI < K. 0 COROLLARY 15.20. Every weakly compact cardinal is strongly inaccessible. Thus weakly compact cardinals are another example of large cardinals: Their definition generalizes a property of w, their existence implies the consistency of ZFC,7 and therefore their existence is not provable in ZFC. EXERCISE 15.22 (PG). One can show that the smallest strongly inaccessible cardinal is not weakly compact. Use this information to show that the existence of a weakly compact cardinal implies consistency of the theory ZFC + "there exists a strongly inaccessible cardinal." Hint: Use the same technique as in the proof of Corollary 12.23. Their dubious existence notwithstanding, weakly compact cardinals have some neat properties that make them worth studying. THEOREM 15.21. If K is weakly compact, then K has the tree property. PROOF. Assume K is weakly compact. Let (T, ~T) be a K-tree, i.e., a tree of height K such that every level contains fewer than K elements. Fix a wellorder ~ of the nodes of T of order type K, and let ~l be the lexicographical ordering on T 7See Chapter 12 for a discussion of the same phenomenon for inaccessible cardinals.
15. A LITTLE RAMSEY THEORY
59
induced by::;, as defined in Chapter 14. Now define a partition P = {Po, Pd of [TF by putting the pair {s, t} into Po if and only if the relations ::;t and ::; agree on {s,t}, i.e., if s < t implies s 1, I'\, is as in the assumptions, A < 1'\" and let f : [I'\,jn -+ A. We have the following analogue of Lemma 15.14 LEMMA 15.23. There exist a I'\,-tree (T, ::;T) and an injection h : T -+ I'\, such that for all s, t E T with S sup Ui3 2; in fact, one can show that w· k ~ (w· k , 3) whenever 3 ::; k < w. For k = w, the situation changes again; by a result of Chang, w. w -+ (w' w , m) for all mEw. We will prove Specker's Theorem by showing that, in a sense, the functions fi constructed in Exercise 15.31 are the only counterexamples witnessing that w· 2 ~ (w· 2 , w). More precisely, we will prove the following: LEMMA 15.34. Suppose g : [w x wj2 -+ 2 is a coloring without O-homogeneous set of order type w· 2 (in the lexicographical order on w x w) and without infinite I-homogeneous set. Then there exist A E [wt o and i < 6 such that gf[A6.]2 = fir[A~12.
15. A LITTLE RAMSEY THEORY
64
PROOF. Let 9 be as in the assumptions. The idea of the proof is to whittle down the domain of 9 so that what remains will resemble more and more closely one of the fi's. As a first step, consider the coloring hI : [wP --+ 2 defined by:
h 1 ({a,b,e}) = g((a,b), (a, e)) for a < b < e < w. Let C E [wt o be homogeneous for hI' It follows from the definition of hI that gnC.6.]2 is homogeneous on all vertical sections, and since 9 has no infinite 1homogeneous sets, all these vertical sections must be O-homogeneous. Now consider colorings h2 : [C]4 --+ 2 and h3 : [CP --+ 2 defined by
h 2({a,b,e,d}) = g((a,b), (e, d)) for a < b < e < d < w, h3({a,b,e}) = g((a,b), (b,e)) for a < b < e < w. Applying Ramsey's Theorem twice, we find B E [cto that is a homogeneous set for both h2 and h3. EXERCISE
15.32 (G). Show that B must be homogeneous of color 0 for both
h2 and h 3· We need to consider three more colorings h4' h 5, h6, where h4 : [BP defined by h4({a,b,e}) = g((a,e), (b,e)) for a < b < e < w, and h 5, h6 : [B]4 --+ 2 are defined as follows:
--+
2 is
h 5({a,b,e,d}) = g((a, d), (b, e)) for a < b < e < d < w, h6({a,b,e,d}) = g((a,e), (b,d)) for a < b < e < d < w. Applying Ramsey's theorem three more times, we find a set A E [BP~o such that A is homogeneous for h4, h5, and h6 simultaneously. EXERCISE 15.33 (PG). (a) Show that A is I-homogeneous for at least one of the functions h5, h6. (b) Show that A is I-homogeneous for all three functions h4' h5, h6 if and only if gf[A.6.j2 = hf[A.6.]2. (c) Formulate and prove analogues of (b) for the functions fi where i < 5. 0
The example we constructed to show that 2~o f+ (Wl)~ was a rather unnatural concoction. One might ask whether it is possible to show that every "nice" partition of [lR]2 into two subsets has an uncountable homogeneous set. To make mathematical sense of this question, we have to specify what "nice" means. One possible interpretation would be "topologically uncomplicated." Suppose X is a topological space, and P = {Po, Pd is a partition of [xj2. One can identify Po with the set p;ym = {(x,y) : {x,y} E Po}. We say that P is an open partition if p;ym is an open subset of X x X with the product topology.1O The following statement is known as the Open Coloring Axiom (abbreviated OCA): If X is a separable metrizable space and (OCA) P = {Po, Pd is an open coloring oftxj2, then • either there exists an uncountable O-homogeneous subset of X, • or X is a union of countably many I-homogeneous sets. 10 Note
that P is open if and only if p:ym is a closed subspace of X x X\{(x,x): x EX}.
MATHOGRAPHICAL REMARKS
65
Note that, in particular, OCA implies that no open partition of a set of pairs of reals witnesses the negative partition relation 2No f+ (wd~. The Open Coloring Axiom is not a theorem of ZFC. However, Stevo Todorcevic has shown that if ZFC is a consistent theory, then so is the theory ZFC + OCA. Thus, every statement provable in ZFC + OCA is relatively consistent with ZFC. Let us conclude this chapter with an interesting consequence of OCA. THEOREM 15.35. Suppose OCA holds and f is a function from an uncountable subset of the reals into the reals. Then there exists an uncountable subset D ~ dom(f) such that f D is weakly monotonic. In fact, if there is no uncountable D ~ dom(f) such that f D is strictly increasing, then there exists a sequence (fn)nEw of nonincreasing functions such that f = UnEw fn.
r
r
PROOF. It suffices to prove the second statement, since the first one follows from it. Let f be as in the assumptions, and let X = {(x, f(x)) : x E dom(f)} be the graph of f. Then X is a subspace of R2; hence X is separable metrizable. Let Po
= {{ (xo, f(xo)),
(Xl,
f(xd)} E [XF : Xo
0 be a positive natural number, and consider the following statement: (Ll n )
Every uncountable B ~
[Wlt
contains an uncountable Ll-system.
CLAIM 16.2. (Ll n ) holds for every natural number n > O. EXERCISE 16.1 (G). Derive Theorem 16.1 from Claim 16.2. PROOF OF CLAIM 16.2. By induction over n > O. To get started, note that every B as in (Ll l ) is a Ll-system with root 0. Now suppose (Ll n ) holds, and let B = {be : ~ < Wl} ~ [Wl]nH. In order to show that B contains an uncountable Ll-system, we distinguish two cases. Case 1: For each a E Wl, the set {b E B : a E b} is countable. In this case, for each countable C C Wl the set {~ < Wl : be n U'1 EC b'1 =f. 0} is countable. We define recursively a function h : Wl - Wl as follows: h(O) = OJ h(~) = min{l1 : b'1
n U bh «()
=
0}.
«e
The family A = {bh(e) : ~ < wd is a Ll-system with kernel 0. Case 2: There is an a E Wi such that I{b E B : a E b}1 = Ni . Fix such a. Let C = {b E B : a E b} and C' = {b \ {a} : b E C}. By (Ll n ), there is an uncountable Ll-system A' ~ C'. Let r denote the root of A', and let A = {a U {a}: a E A'}. This is a Ll-system with root r U {a}. 00 EXERCISE 16.2 (R). Let B be an uncountable family of finite subsets of Wi. Show that there exists a countable subset N C Wi such that every b E B which is not a subset of N is a member of some uncountable Ll-system A ~ B with kernel r ~ N. l Theorem 16) can be generalized as follows: THEOREM 16.3 (Ll-System Lemma). Let K, and A be infinite cardinals such that A is regular and the inequality 11 n for all nEw. For k E w, define recursively a function fk E Ww as follows: fO(n) = n, fk+l(n) = f(fk(n)) for all nEw. By the assumption about f, the sequence (fk(O))kEW is strictly increasing. Thus we can associate with f a set b(n n}. EXERCISE 17.28 (G). Let a E [w]No, and assume that f E Ww is such that f(n) > n for all nEw and f(a) l{I. THEOREM 17.15. Assume s > l{I. Then for every partition of W x WI into disjoint sets Hand K there exist a E [w]No and B E [WI]Nl such that a x B ' = 2No , so cf(2)') = A, which contradicts Konig's Theorem (Corollary 11.24). 0 Next we are going to discuss several equivalent formulations of Martin's Axiom. The first of these is topological. Consider the following statement. (MAtop)
No compact Hausdorff space with the C.C.c. is a union of less than 2No nowhere dense subsets.
LEMMA 19.6. MA
---t
MA top .
PROOF. Let X be a compact Hausdorff space, and let JP> be the family of all closed subsets F ~ X such that F has nonempty interior. If X is a c.c.c. topological space, then (JP>,~) also satisfies the c.c.c. Let K. < 2No , and let {NO!: 0: < K.} be a family of nowhere dense subsets of X. For 0: < K., let DO! = {F E JP>: F n NO! = 0}. Then DO! is dense in (JP>, ~). To see this, consider FE JP>. Since NO! is nowhere dense in X, we can pick a nonempty open U ~ F such that Un NO! = 0. Let x E U. By regularity of X, there is a closed neighborhood V of x such that V ~ U. Clearly, V E DO!, and V ~ F. Now assume MA holds, and let G be a filter in (JP>, ~) such that G n DO! =f: 0 for each 0: < K.. Then G is a family of closed subsets of X with the finite intersection property (it is even true that H has nonempty interior for every H E [G] <W). By 0 compactness of X, there exists some x EnG. Clearly, x ~ UO! = {[a, b] : ~ a < b ~ 1, a, bE Q}, and mimic the proof of Lemma 19.6.
°
Now let us consider two versions of Martin's Axiom for Boolean algebras. (MA Ba ) If B is a Boolean algebra that satisfies the c.c.c., and if V is a family of dense subsets of B with IVI < 2No , then there exists a V-generic filter in B. (MA cBa )
If B is a complete Boolean algebra that satisfies the c.c.c., and if V is a family of dense subsets of B with IVI < 2No , then there exists a V-generic filter in B.
Recall from Section 13.3 that if B is a Boolean algebra, then its Stone space st B is the topological space (Ult B, r), where Ult B is the set of all ultrafilters in Band r is the topology generated by the base U = {Ua : a E B}, where Ua = {F E Ult B': a E F}. We have shown in Section 13.3 that Stone spaces of Boolean algebras are compact Hausdorff spaces. EXERCISE 19.5 (PG). Let B be a Boolean algebra. Show that a set D ~ B+ is dense in B if and only if the set D* = {F E Ult B ; F n D ;f 0} is dense and open in st B.
19.
98
MARTIN'S AXIOM
Now suppose MAtop holds, let B be a Boolean algebra that satisfies the c.c.c., and let V be a family of fewer than 2No dense subsets of B. By Exercise 19.5, for each D E V the set UltB\D* is a nowhere dense subset of stB. Since stB is a compact Hausdorff space that satisfies the c.c.c., there exists F E Ult B such that F E D* for all D E V. Then F n D =I- 0 for all D E V, i.e., F is V-generic. We have proved the following: LEMMA 19.7. MA top
--+
MA Ba .
Clearly, MABa implies MAcBa. The latter statement in turn implies the original version of Martin's Axiom. LEMMA 19.8. MAcBa
--+
MA.
PROOF. Assume MAcBa and let", < 2No. We want to show that MA(",) holds. Consider the following statement: Let (lP,~) be a c.c.c. p.o. of cardinality ~ "', and let V be a family of dense subsets of lP with IVI ~ "'. Then there exists a V-generic filter in lP.
MA -("') appears to be weaker than MA(",) , but in Chapter 21 (Theorem 21.26) we will show that MA( "') and MA - ("') are equivalent statements. Thus, in order to prove MA(",), we can restrict our attention to p.o.'s of cardinality ~ "'. So assume that (lP, ~) is a c.c.c. p.o. such that IlPl ~ "'; let A ~ "', and consider a family V = {Da : Q < A} of dense subsets of lP. By Theorem 13.34, there exist a complete Boolean algebra B and a function i : lP --+ B+ such that (i)
rng(i) is dense in B+;
(ii) Vp,q E lP(p ~ q --+ i(P) ~B i(q)); (iii) Vp, q E lP (p..l q +-d(p) ..lB i(q)). Fix such i, B, and for
Q
< A, let i[Dal be denoted by Ea·
EXERCISE 19.6 (G). Convince yourself that B satisfies the c.c.c. and each Ea is dense in B. By MAcBa , there exists an ultrafilter IF in B such that IF n Ea =I- 0 for each < A. Define G = i-I [lFl. It appears that we have found a V-generic ultrafilter in lP: Clearly, G n Da =I- 0 for each Q < A. If pEG and p ~ q, then i(P) E IF
Q
and by (ii), i(p) ~B i(q). Since IF is a filter, i(q) ElF, and hence q E G. Similarly, (iii) implies that the elements of G are pairwise compatible in lP. But this does not quite suffice: The definition of a filter requires that if p, q E G, then there exists a lower bound for p and q in G, not just in lP. EXERCISE 19.7 (G). Assume that i satisfies the following strengthening of (ii): (ii)+
Vp, q E lP (p ~ q
+-t
i(p) ~B i(q)).
Show that in this case G is a V-generic ultrafilter in lP. Unfortunately, in general i satisfies just condition (ii), not (ii)+. But by using a trick we can make sure that G is as required. For p E lP, let Dp = {r E lP : r ~ p V r..l pl.
19.1. MA ESSENTIALS
99
EXERCISE 19.8 (G). (a) Show that for every p E lP' the set Dp is dense and open in lP'. (Recall that the latter means that Vq, r E lP' (q E Dp /\ r :::; q ---t r E Dp)). (b) Show that for all p, q E lP' the intersection Dp n Dq is dense in lP'. Here comes the trick: Since 1lP'1 :::; /'i., we may assume without loss of generality that for all p, q E lP', the set Dp n Dq is an element of V! Now let G be as above, let p, q E G, and let rEG n Dp n D q. Since G consists of pairwise compatible elements, we must have r :::; p, q. Thus, p and q have a lower bound in G, and we are done. 0 We have proved the following: THEOREM 19.9. Each of the statements MA top , MA Ba , and MAcBa is equivalent to Martin's Axiom. EXERCISE 19.9 (R). Show that MAcountable is equivalent to cov(M) = 2No. Hint: Use Corollary 25.4. Now let us strengthen Theorem 19.1 by showing that MA(N 1) implies that every Aronszajn tree is not only not Suslin, but even special. THEOREM 19.10. If MA(N 1) holds, then every Aronszajn tree is special. PROOF. Assume MA(NI), and let (T, :::;T) be an Aronszajn tree. We want to construct a function f : T -+ w such that f(t) = f(s) implies that s and tare incomparable with respect to :::;T. In order to use MA(N 1), we need to work with a suitable C.c.c. partial order (lP', :::;). The elements of lP' will be approximations to our function f. More precisely, let lP' denote the set of all functions p such that (i) (ii) (iii)
dom(p) E [T] a such that q{3 E IF n A. Hence IF n A is an uncountable centered subset of A, as required. 0 Since precaliber Nl implies Property K, Lemma 18.24 and Theorem 19.13 yield the following: COROLLARY 19.14 (MA(Nd). The simple product of any two C.C.c. p.o. 's satisfies the c. c. c. In particular, the product of any two topological spaces with the C.C.c. satisfies the C.C.c.
102
19. MARTIN'S AXIOM
Corollary 19.14 in turn gives together with Theorem 16.4: COROLLARY 19.15 (MA(l'~l)). The Tychonoff product of any nonempty family of topological spaces with the c.c.c. satisfies the c.c.c. Corollary 19.14 can also be used to prove the following: THEOREM 19.16 (MA(l'~l)). Every c.c.c. partial order of cardinality Nl is centered.
(j-
PROOF. Assume MA(N 1 ) and let (IP',~) be a c.c.c. partial order such that IIP'I = N1 . Let Q = <wIP', and define a p.o. relation ~ on Q as follows: ql ~ qo if and only if dorn(qo) ~ dom(ql) 1\ Vi E dom(qo) (ql(i) ~ qo(i)). EXERCISE 19.12 (G). Show that that the p.o. (Q,~) satisfies the c.c.c. Hint: Note that for every uncountable A ~ Q there must be nEw such that A n nIP' is uncountable. Then use Corollary 19.14 and induction over nEw. Now define for p E IP':
Dp = {q E Q: 3i E dorn(q) (q(i)
~ pH.
It is not hard to see that each of the sets Dp is dense in (Q, ~). Since IIP'I = Nl and Q satisfies the c.c.c., MA(Nl) implies that there exists an ultrafilter IF in Q such that IF meets each of the sets Dp. Fix such IF and define for i E w: IP'i = {q(i) : q ElF}.
o
Then each of the sets IP'i is centered and IP' = UiEW IP'i.
19.2. MA and cardinal invariants of the continuum In this section we shall examine what Martin's Axiom implies about certain cardinal invariants of the continuum. One result in this direction was already proved in Chapter 18. Recall that a
= min{IAI : A
is an infinite maximal almost disjoint family of subsets of w}.
In this new terminology, Theorem 18.2 asserts that a > translates into the inequality a > ~l.
~o,
and Statement 18.3
EXERCISE 19.13 (G). Convince yourself that Theorem 18.8 remains true if "MACK)" is replaced by "MAa-centered(K)." If a, b ~ w, then we write a ~* b as shorthand for la\bl < ~o. If a ~* b, then we say that a is almost contained in b. We say that a and b are almost equal, and write a =* b, if a ~. b and b ~* a. We write a c· b if a ~. b and a #* b. A subset b of w is called a pseudo-intersection of a family C ~ P(w) if b is infinite and b ~. c for all c E C. We say that a family C ~ P(w) has the strong finite intersection property (sfip) if I nHI = ~o for all H E [C] I(n)). Show that (lP,:::5) is a a-centered p.o., and find a family V of fewer than 2No dense subsets of lP such that for every V-generic filter G in (lP,:::5) the set gIG defined by gIG = U{ s : 3F ((s, F) E G)} is a function from w into w with I max{mo, m{3}, there are qo, q{3 E Pn such that qo ::; Po and q{3 ::; P{3· Since Pn is centered, qo and q{3 are compatible, and so are Po,P{3' It follows that L is linked. Are there sets C and L that satisfy (A) and (B)? Not necessarily. It may happen that the families Am defined in Exercise 19.27(b) only have the finite intersection property, but not the strong finite intersection property. If this is the case, then we cannot expect to find an infinite C as above. Fortunately, in such a situation Theorem 19.25 has a rather simple proof. CLAIM 19.26. Suppose that there exists mEw such that the family Am does not have the sfip. Then there exists nEw such that Pn intersects all Do's.
n
PROOF. Let mEw be as in the assumptions. It suffices to show that Am =f. 0. Suppose this were not the case, and let {Po, ... ,Pk-d s:;; Pm and {ao, ... ,ak-d C It be such that ni JTJ for all ~ < 'r/ < WI. If c E n~<wl J~, then a~ c* c c" b~ for all ~ < WI, which is ruled out by the assumption that ((a~ : ~ < WI), (b~ : ~ < WI)) is a gap. 0 Claim 20.3 is akin to Lemma 17.10, but the latter uses the extra assumption b = t{I, while the former is a theorem of ZFC. Theorem 20.2, Claim 20.3, and Lemma 17.9 together show that uncountable, perfectly meager subsets of P{w) can be constructed using only the axioms of ZFC. More precisely, we have the following:
set
COROLLARY 20.4. For any Hausdorff gap ((a~ : ~ < WI), (b~ : ~ < WI)), the : ~ < wd U {b~ : ~ < wd is a perfectly meager subset ofP{w).
{a~
Corollary 20.4 has a measure-theoretic dual: {a~
LEMMA 20.5. For any Hausdorff gap ((a~ : ~ < WI)' (b~ : ~ < WI))' the set ~ < wd u {b~ : ~ < wd has universal measure zero.
:
PROOF. Let ((a~ : ~ < WI), (b~ : ~ < WI)) be a Hausdorff gap, and let J.L be a Borel measure on P{w) that vanishes on the singletons. For ~ < WI, let J~ be defined as in Claim 20.3. Since each J~ is J.L-measurable and contains all but countably many elements of {a~ : ~ < wd u {b~ : ~ < wd, it suffices to show that J.L(J~) = 0 for sufficiently large ~. Suppose towards a contradiction that this is not the case. Since the J~ 's form a decreasing transfinite sequence, there must be ~o < WI and co > 0 such that J.L(J~) = co for all ~ > ~o. Define: c = {n E w: J.L{{d E J~o : nEd}) ~ co/2}, and let
6 > ~o be such that c fj.
J~I
.
EXERCISE 20.2 (G). Let c, 6 be as above. Show that in the definition of c, the subscript ~o can be replaced by 6, i.e., show that c = {n E w : J.L( {d E J~I : nEd}) ~ co/2}. As in the proof of Claim 20.3, for nEw, let K~I ,n
= {d ~ w : a~1 \n ~ d\n ~
b~1 \n}.
The K~I,n's form an increasing sequence of J.L-measurable sets with union J~I' Hence, there must exist no E w such that J.L(K~I,no) > co/2. Fix such no. Since c ¢ K~I,no, there must exist nI > no such that either ni E c\b6 or nI E a~1 \c. In the former case, by Exercise 20.2, the set L = {d E J6 : ni E d} has measure lSome authors reserve the term Hausdorff gap only for (Wl,wi}-gaps with special properties (like Property (iv) in the proof of Theorem 20.6).
20.
HAUSDORFF GAPS
119
at least eo/2 and is disjoint from K~l,no' which is impossible, since J.L(K~l U L) ~ eo· In the latter case, the set M = {d E J~l : ni ¢ d} has measure at least eo/2 and is disjoint from K~l ,no, which again leads to a contradiction. 0
J.L(J~1) =
We shall prove a version of Theorem 20.2 that differs from the one originally stated. Let A, B ~ [W]N o • We say that A and B are almost disjoint, and write A .1 B, if la n bl < ~o for all a E A and b E B. We say that A and B can be separated if there exists d ~ W such that a ~* d and Ib n dl < ~o for every a E A and b E B. Of course, if A and B can be separated, then A and B are almost disjoint. However, the converse is not true. EXERCISE 20.3 (G). Let ((a~ : ~ < K.), (bry : TJ < >.)) be a (K., >.*)-pregap, let < >., and let A = {a~ : ~ < K.}, B = {Cry: TJ < >.}. (a) Convince yourself that A .1 B. (b) Show that A and B can be separated if and only if ((a~ : ~ < K.), (b.,., : TJ < >.)) is not a (K.,>.*)-gap.
Cry
= w\b.,., for all TJ
The above exercise implies that Theorem 20.2 is equivalent to the following statement: THEOREM 20.6. There exist transfinite sequences (a~ : ~ < WI) and (b~ : ~ < of infinite subsets of W such that a~ c* ae: and b~ c· be: for all ~ < ( < WI; and the families {a~ : ~ < wt} and {b~ : ~ < wt} are almost disjoint, but cannot be separated.
WI)
PROOF. We shall construct sequences (a~ : ~ < WI) and (b~ : ~ < WI) simultaneously by recursion over WI in such a way that for all ~ < ( < WI: (i) Iw\(a~ U b~)1 = No; (ii) a~ c* ae: and b~ c* be:; (iii) a~ n b~ = 0.
Condition (i) is a technical requirement that allows us to keep going. Conditions (ii) and (iii) together not only ensure that our sequences are increasing in the sense of c*, but also that the families {a~ : ~ < wt} and {b~ : ~ < wt} will be almost disjoint. In order to ensure that these families cannot be separated, Hausdorff devised an ingeneous trick. Let us say that b ~ W is close to A ~ P(w) if for every nEw the set {a E A : a n b ~ n} is finite. EXERCISE 20.4 (G). (a) Suppose that b is close to a set A and b ~* d ~ w. Convince yourself that d is also close to A. (b) Suppose that b is close to a set A and B ~ A. Convince yourself that b is also close to B. We are going to require that for every ( < WI: (iv) ~
be: is close to {a~ : ~ < (}.
If condition (iv) holds for all ~ < ( < WI, then the sets {a~ : ~ < WI} and {b~ : < WI} cannot be separated. This is a consequence of the following observation.
LEMMA 20.7. Let {a~ : ~ < wt} and {b~ : ~ < WI} be separated families of infinite subsets of WI such that a~ n b~ = 0 for all ~ < WI. Then there exists an EX. uncountable set X ~ WI such that (a~ n be:) U (ae: n b~) = 0 for all
e, (
120
20. HAUSDORFF GAPS
PROOF. Let d ~ W be such that a~ ~* d and Ib{ndl < ~o for all ~ < WI. By the Pigeonhole Principle, there exists nEw such that the set Xn = {~< WI : (a~\d) U (b~nd) ~ n} is uncountable. Moreover, using the Pigeonhole Principle again we can see that there are s, t ~ n such that the set X = {~E Xn : a~\d = s /\b~ nd = t} is uncountable. It is straightforward to verify that the above set X is as required. D EXERCISE 20.5 (G). Use Lemma 20.7 to show that if condition (iv) holds for all ~ < ( < WI, then the sets {a~ : ~ < WI} and {b~ : ~ < WI} cannot be separated. Hint: For X as in the lemma, consider the (w + l)st element ( of X and derive a contradiction. Having all ideas in place, let us now describe the construction itself. To get started, let {Co, Cb C2} be a partition of W into three disjoint infinite subsets. Define: ao=co, bO=Cl. Now suppose 0 < ( < Wb and for all ~ < ( the sets a~, b~ have been defined in such a way that conditions (i)-(iv) hold. If ( = 1/ + 1 for some 1/, then we choose a partition {do, d 1, d2 } of w\(ary U bry) into pairwise disjoint infinite sets and define:
a( = ary U do,
b( = bry U d 1 .
A straightforward argument shows that in this case, conditions (i)-(iv) continue to hold at stage (. Now let us consider the case where ( is a limit ordinal. Let us begin by choosing sets x ~ wand d ~ x such that Iw\xl = ~o, a~ ~* d, and b~ ~* x\d for all ~ < WI' This is possible by Theorem 20.1. Think of (x\d, d) as a first approximation to (a(, be:). This approximation satisfies conditions (i)-(iii), but not necessarily (iv). To see to what extent (iv) fails, fix a strictly increasing sequence of ordinals (1/n)nEw such that 1/0 = 0 and ( = SUP{1/n : nEw}, and define for each nEw:
Bn = {1/ : 1/n :::; 1/ < 1/n+1 /\ d n ary ~ n}. Since (iv) holds for every ( = 1/n+1 and since bryn+l ~* d, each of the sets Bn is finite. Thus, if we define B = U nEw Bn, then B either is finite, or has order type w. Let us consider these two cases separately. Case 1: B is finite. Then for all but finitely many n the set = {~ < ( : d n a~ ~ n} is contained in 1/n. Since bryn ~* d and, by the inductive assumption, bryn is close to {a~ : ~ < 1/n}, so is d. Therefore, the set {~ < ( : d n a~ ~ n} is finite. Since this argument applies to all but finitely many nEw, d is close to {a~ : ~ < (l. Case 2: B has order type w. In this case, d may fail to be close to {a~ : ~ < (}. However, d misses the mark of being close just barely.
en
EXERCISE 20.6 (G). Show that d is close to the set {a~ : ~ E (\B}. Let (~n)nEw be an enumeration of B in increasing order. For every nEw, pick Pn E x n a~n \ Uihical Remarks If you would like to learn more about gaps, we recommend the survey article Gaps in w w, by Marion Scheepers, in Israel Mathematical Conference Proceedings 6 (1993), AMS, 439-561.
CHAPTER 21
Closed Unbounded Sets and Stationary Sets 21.1. Closed unbounded and stationary sets of ordinals Let, be a limit ordinal. A set C S;;; , is bounded in , if there exists {3 < , such that C S;;; {3. Otherwise C is unbounded in ,. C is closed in , if Q E C for all limit ordinals Q such that 0 < Q < , and C n Q is unbounded in Q. C is club i in, if C is closed and unbounded in ,. EXERCISE 21.1 (G). Show that C is closed in , iff C is closed in the sense of the order topology on ,. EXAMPLE 21.1. If {3 < , then the set ({3,,) = {Q : {3 < Q < ,} is closed and unbounded in " but, if, is a limit ordinal, then this set is not closed in any ordinal Q > ,. If cfb) = w, then every cofinal subset of order type w is club in ,. The set, n LIM of limit ordinals smaller than, is closed in any ordinal " and, if cfb) > w, then this set is also unbounded in ,. EXERCISE 21. 2 (G). Find limit ordinals " 0 of countable cofinality that are not cardinals so that LIM n, is unbounded in, and LIM n 0 is bounded in o.
Let, E ON. We define: CLUBb)
= {A S;;;,: :3C S;;; A (C is club in ,)}.
EXAMPLE 21.2. If A = w U {{3 < WI : w < {3 and {3 is a limit ordinal}, then A E CLU B(wI), but A is not club since w S;;; A but sup(w) = w ¢ A. EXAMPLE 21.3. Let K be a cardinal of uncountable cofinality, and suppose A is an unbounded subset of K. We let
A'
= {Q < K
:
Q = sup( A n Q)}.
We call A' the derived set of A. Note that if we consider K with the order topology, then this notion coincides with the concept of derived set used by topologists. EXERCISE
in
21.3 (G). (a) Show that A' is club in
K
whenever A is unbounded
K.
(b) Show that if A is club, then A' S;;; A. EXERCISE 21.4 (PG). Let K be an infinite cardinal such that Cf(K) > ~o. Let ). denote Cf(K), and let {co: Q < ).} and {do: Q < ).} be club subsets of K with Co < C{3 and do < d{3 for Q < {3 < ).. Show that the set C = {co: Q < ). 1\ Co = do} is club. Hint: To show that C is unbounded, let {3 < K and define recursively a sequence (Qn)nEw by letting: Qo = min{Q : Co ~ {3}, Q2n+l = min{Q : Q ~ Q2n 1\ do ~ C02n } ; Q2n+2 = min{Q: Q ~ Q2n+ll\co ~ d o2n +J· lSome authors write "cub." Remark 21.29.
The reason why the word club is italic here will appear in
123
124
21. CLOSED UNBOUNDED SETS AND STATIONARY SETS
EXAMPLE 21.4. Let", be a cardinal of uncountable cofinality, and let "'. We say that I is normal if it satisfies the following conditions: (i) If a < 13 < "', then I(a) :S l(f3)j (ii) If I' < '" is a limit ordinal, then Ih) = sup{J(a) : a < I'}j (iii) I(a) ~ a for all a < "'.
I: '" ~
Note that (i) says that I is nondecreasing, whereas (ii) means that I is continuous in the order topology. Note also that if I is strictly increasing, then (iii) is automatically satisfied. EXERCISE 21.5 (G). Suppose F ~ It", is a set. of normal functions such that for all a < "', sup{J(a) : IE F} < '" exists and is smaller than "'. Show that the function 9 : '" ~ '" defined by g(a) = sup{J(a): IE F} is also normal. LEMMA 21.1. II", is a cardinal 01 uncountable cofinality and I E It '" is normal, then the set Fix(f) = {a < "': I(a) = a} is club in "'. PROOF. In order to see that Fix(f) is closed, either recall that the set of fixed points of any continuous function of a Hausdorff space into itself is closed (see Theorem 26.1), or note that if the set {a : I(a) = a} is cofinal in 13, then sup{J(a) : a < f3} = sup{a : a < f3} = 13· We show that Fix(f) is unbounded: Let 0 < "'. We want to show that there exists some a E Fix(f) with 0 < a. To this end, we construct recursively a sequence (f3n)nEw as follows: 130 = o. Given f3n, let f3n+l > l(f3n). This construction yields an increasing sequence with
0=130 :S 1(130) < 131 :S 1(131) < ... < f3n :S l(f3n) < f3n+l < .... It follows that if a w} = a.
= sup {f3n:
nEw}, then a
> 0 and I(a) = sup {J(f3n):
n E
0
EXERCISE 21.6 (PG). Let D = [0, l)nQ, let order on D x WI defined by:
~a
denote the antilexicographical
(d,a) :Sa (e,f3) iff a < 13 V (a = 13/\ d:SQ e) (for the natural orders :SQ on Q and :S on WI respectively), and let A, B ~ WI \ {a}. Prove that ((D x wd\( {a} x A), ~a) is order-isomorphic to ((D x WI)\( {a} x B), ~a) if and only if there exists a club C ~ WI such that An C = B n C. EXAMPLE 21.5. Let", be a regular uncountable cardinal, let nEw \ {O}, and let 9 be any function from ",n into "'. Then the set
Mg
=
{a < "': rng(gfa n )
~
a}
is club in "'. To see this, consider the function Ig : '" ~ '" defined by Ig (a) = sup ( {g(jJ) + 1 : jJ E an} U {a}). Note that Ig is a normal function. Moreover, Ig has been defined so that Fix(f) = Mg. We shall refer to Mg as the set of ordinals a that are closed under g. Note that this use of the word "closed" has nothin~ to do with the order topology. In a certain sense, Examples 21.4 and 21.5 are the only examples of club sets. THEOREM 21.2. II I' is an ordinal and C is club in 1', then there exists a ' normallunction I : I' ~ I' such that C = Fix (f) .
21.1. CLOSED UNBOUNDED AND STATIONARY SETS OF ORDINALS
125
PROOF. Let C be club in 'Y. Define a function f : 'Y --+ 'Y by f(a) = min {C\a}. A straightforward verification shows that f is as required. 0 EXERCISE 21.7 (PG). (a) Show that for every club C ~ function g : WI --+ WI such that C = Mg. (b) Let C = [WI,W2). Then C is club in W2. Show that if g: nEw \ {O}, then C =I- Mg.
WI
there exists a
w2
--+ W2
for some
THEOREM 21.3. Let'Y be an ordinal of uncountable cofinality A. If K < A and {Co: : a < K} is a family of club subsets of 'Y, then the intersection C
=
nCo:
is club in 'Y.
PROOF. For each a < K, let fo: : 'Y --+ 'Y be a normal function such that Fix(fo:). For (3 < 'Y, let f((3) = sup{Jo:((3) : a < K}. Since K < cf("(), f((3) < 'Y for each (3. By Exercise 21.5, f is a normal function from 'Y into 'Y. Now the theorem follows from the next exercise. Co:
=
EXERCISE 21.8 (G). Convince yourself that Fix(f)
= no: w. Then CLUB(,,() is a
PROOF. Let K < A, and suppose {Ao: : a < choose a club set Co: ~ Ao:. Then
CLUB(,,(). For each a
. = cfb)· Then the set 8 = {{3 < 'Y : cf({J) = K.} is stationary in 'Y.
w
S
o
77. By Theorem 21.24, if A E [X], the set {a E E : A n a = Aa} is stationary in 1>,. Of course, O(E) is the same as OWl (E). Instead of 0,,,(1),) one usually writes
0",. THEOREM 22.13. Assume V = L. Let I>, be an uncountable regular cardinal, E a stationary subset of 1>,. Then O",(E) holds.
Mathographical Remarks
The O-principle is due to R. B. Jensen. The "principle is also called the Ostaszewski Principle. In his paper On countably compact, perfectly normal spaces, Journal of the London Mathematical Society 14 (1967), 505-516, A. Ostaszewski used" + CH to construct a first countable, perfectly normal, hereditarily separable, locally countable, locally compact, count ably compact Hausdorff space that is not Lindelof (and hence not compact). K. Devlin later noticed that" + CH already implies O. For variations on the T-principle, see S. Broverman, J. Ginsburg, K. Kunen, and F. D. Tall, Topologies determined by G"-ideals on WI, Canadian Journal of Mathematics 30(6) (1978), 1306-1312. The construction of a model for ",CH appeared in S. Shelah, Whitehead groups may not be free even assuming CH, II, Israel J. Math. 35 (1980), 257-285. The proof of Theorem 22.13 can be found in K. Devlin's book Constructibility, Springer Verlag, 1984. The latter reference also contains information on several other useful combinatorial principles that are consequences of V = L.
,,+
CHAPTER 23
Measurable Cardinals DEFINITION 23.1. Let -+ [0,1] such that (1) 1"(0) = OJ
be a cardinal. A probability measure on K is a function
K
I" : P(,..)
(2) (3)
J.L(K) = Ij J.L(UnEw An) = LnEw J.L(An) for every countable family {An: nEw} of pairwise disjoint subsets of
K.
Property (3) has several names in the literature: a-additivity, countable additivity, or Nl -additivity of 1". If I" is such that 1"( {O) = for every ~ E K, then we say that I" vanishes on the singletons. The family II-' = {N ~ K: J.L(N) = O} is called the ideal of null sets of 1".
°
EXERCISE 23.1 (G). Show that II-' is indeed an ideal on EXAMPLE 23.1. Let
~
E
K,
K.
and define:
J.L(A) = Then I" is a probability measure on the singletons.
{01"
K.
if ~ if ~
E Aj If A.
Of course, this measure does not vanish on
EXAMPLE 23.2. Let U be a count ably complete ultrafilter on
J.Lu (
A) = {I, 0,
K.
Define:
if A E Uj if A If U.
Then J.Lu is a measure on K. The measure of Example 23.1 is a special case of J.Lu for the principal ultrafilter U generated by {O. Note that J.Lu vanishes on the singletons if and only if U is a nonprincipal ultrafilter. EXERCISE 23.2 (G). Show that if I" is any measure on a cardinal K, then the family UI-' = {A ~ K: J.L(A) = I} is a count ably complete filter on K. Moreover, show that UI-' is an ultrafilter on K if and only if rng(J.L) = {O, I}. (The latter property will be referred to by saying that I" is a two-valued measure.) How is Definition 23.1 related to more familiar probability measures, like Lebesgue measure on [0, I]? Of course, the latter is not defined on sets of ordinals. But this is only a superficial difference, since any one-to-one correspondence between the cardinal 2No and the unit interval [0,1] allows us to treat a measure defined on P(2 No) as one defined on P([O, 1]) and vice versa. More relevant for the present purpose is the fact that Lebesgue measure is defined only on a certain a-algebra of subsets of [0,1]' whereas the measures considered here are defined for all subsets of a cardinal K. We shall say that Lebesgue measure can be extended to P([O, 1]) if
there exists a probability measure J.L : P([O, 1]) 147
-t
[0,1] such that J.L([a, b]) = b - a
148
23.
MEASURABLE CARDINALS
for all a, b with 0 ~ a ~ b ~ 1. It is clear that each such measure vanishes on the singletons (consider the degenerate intervals [a, a] = {a}). The following question provided the initial motivation for the notions studied in this chapter. QUESTION 23.2. Does there exist a probability measure on P([O, 1]) that extends Lebesgue measure? Note that this is not the same as asking whether there exists a subset of [0,1] that is not Lebesgue measurable. The latter question has been answered in the negative by Theorem 9.23 of Volume 1. Incidentally, the Vitali set constructed in the proof of that theorem also demonstrates that Lebesgue measure cannot be extended to a translation-invariant measure defined on all subsets of JR. THEOREM 23.3 (Ulam). Let K be an infinite cardinal, and let A = K+. Then there exists an indexed family {Aa,e : 0: < K, ~ < A} of subsets of A such that: (1) Aa,e n Ap,e = 0 for all 0: < f3 < K and ~ < Ai (2) Aa,e n A a,7] = 0 for all 0: < K and ~ < "l < Ai (3) Ua O. Now the Pigeonhole Principle implies that for some Q < W the set
Xa is uncountable. Fix such an
Q.
n(~) =
= {~:
For
~
Q(~)
= Q}
E X a , let
min{n E W : f..L(Aa,~)
> Tn}.
Using the Pigeonhole Principle once more, we find an uncountable subset Y £; Xa and nEw such that n(~) = n for all ~ E Y. This leads to a contradiction: Let {~i : 0::::; i ::::; 2n} be the first 2n + 1 elements of Y. Since Aa'~i nAa,~j = 0 whenever i ¥ j, we have 2n
f..L(Aa,~o U··· U A a,6 n ) =
L f..L(Aa,~.) ~ (2n + 1) . Tn > 1, i=O
contradicting the assumption that f..L is a probability measure.
o
COROLLARY 23.6. There does not exist a nonprincipal, countably complete ultrafilter on WI . PROOF. This follows from Example 23.2 by an argument as in the proof of Corollary 23.5. 0 COROLLARY 23.7. Suppose E is a stationary subset of WI such that 0, and let (A€ : ~ < 8) be a sequence of subsets of B such that A€ ~ A'1 for all ~ < "l < 8, and each A€ is the universe of an elementary submodel of~ (i.e., !2l€ -< ~, where!2l€ = (A€, (A;o(i) nRi)iEl, {A?(j)+1 nFj)jEJ, (Ck)kEK)). Then U€'2n+l s:;; H,,'}. Let Aw = sUPnEw An. Then Aw ~ A. By (24.27) and (24.28), Aw E E. We have D shown that E is unbounded in ON. Now suppose we are given a formula cp E FOTm~~1 and sets ao, ... ,an-I, and we want to prove, with the help of elementary submodels, that cp( ao, . .. ,an-I) holds. Since every closed unbounded class of cardinals is in particular a closed unbounded class of ordinals, and since the intersection of two closed unbounded classes of ordinals is a closed unbounded class of ordinals,S Theorem 24.14 and Lemma 24.15 imply that there exists a cardinal A such that H>. = V>., and H>. F cp[ao, ... ,an-I] if and only if V F cp[ao, ... ,an-I]' In most cases, we may as well start our argument with picking A as above, and then restrict the proof to showing that H>. F cp[ao, ... ,an _I].9 Why did we write "in most cases"? Don't Theorem 24.14 and Lemma 24.15 imply that there is always a suitable A around? True enough, but you have to be careful about what you can and what you cannot assume about A. Note that Theorem 24.14 guarantees that Ccp contains arbitrary large cardinals, even cardinals of arbitrary large cofinality, but Theorem 24.14 does not guarantee that Ccp contains any regular cardinals. As long as you only need to assume that A is sufficiently large, or even that Cf(A) > w, you may safely use Theorem 24.14 and Lemma 24.15. But if for some reason your argument uses the assumption that A is a regular uncountable cardinal, you are back to square one and you must either verify absoluteness of CPH".[ao, ... , an-d directly, or modify your argument. Let us summarize what we have said so far by giving an "official" definition of the phrase "sufficiently large A" that will appear at the beginning of several arguments later in this chapter: If ao, ... , an-l are given and if we want to prove a formula cp(ao, ... , an-I) with the help of elementary submodels, then A is "sufficiently large" if an, ... , an-l E H>., the formula cp is absolute for H>., and H>. satisfies all the (finitely many) axioms of ZFC that will be used (implicitly or explicitly) in the proof. lO Theorem 24.14 and Lemma 24.15 imply that such A exists. However, note that our approach does not guarantee that if A is suitable and II: > A, 8This was shown in Exercise 21.19. 90f course, we might as well show that V",
F c,o[ao, ...
,an-I], but since the structure of H",
is easier to describe, set theorists usually prefer working with H).. IOThis is what we mean when we speak of a "sufficiently large fragment of ZFC."
24.
170
ELEMENTARY SUBMODELS .
then K is also suitable. We shall always put the phrase "sufficiently large" in scarequotes to alert you that it should not be taken too literally. While we are at it, let us mention another important application of Theorem 24.14. Recall from Chapter 12 the definitions of the La-hierarchy and of the class L. EXERCISE 24.15 (PG). Show that the class D = {a E ON : La = L n Va} is closed and unbounded in ON. Hint: Modify the argument of the proof of Lemma 24.15. The following result appeared in Volume 1 as Lemma 12.35. COROLLARY 24.16. Let a E ON and x,ao, ... ,ak-I E La, and let cP E Form£;I. Then there exists j3 > a such that for every z Ex:
(L{3, E)
F cp[z, ao, ...
,ak-I]
iff L
F cp[z, ao, ...
,ak-I].
PROOF. Let a E ON, and let cP, x, ao, ... ,ak-I be as in the assumptions. Let '¢ be a representation of L, i.e., let '¢ be a formula with one free variable such that L = {y : '¢(Yn. By Theorem 24.14 and Exercise 21.19, there exists j3 > a such that for all z E V{3:
V
(24.29)
F ,¢[z]
iff V{3
F ,¢[z];
and (24.30)
V
F cp/dz,ao, ...
,an-d iff V{3
F cp/dz,ao, ...
,an-I].
The left hand side of (24.30) is equivalent to (24.31)
L
F cp[z,ao, ...
,an-I];
and by (24.29), the right hand side of (24.30) is equivalent to (24.32)
V{3 n L
F cp[z, ao, ... , an-I].
By Exercise 24.15, we can choose j3 as above in such a way that V{3 n L = L{3, which allows us to rewrite (24.32) as: (24.33)
(L{3, E)
F cp[z, ao, ... ,an-I].
Since (24.31) is equivalent to (24.33) for all z E L{3, and since x ~ L{3 by transitivity of L{3, Corollary 24.16 follows. 0 Now let us prove Theorem 24.14. PROOF OF THEOREM 24.14. Let cp be as in the assumption. Let CPo, ... ,CPs be a list of all subformulas of cp (including, of course, cp itself). For each Cpj on this list that is of the form 3VijCPi for some f ::::; s, we define a functional class G j as follows: Fix mj > i j such that the free variables of CPi are among va, ... ,vmj . The domain of G j will be the class of all mrtuples (an, ... ,aij-l> aij+l," . ,amJ (we will use the shorthand ii to denote such tuples). If 3Vij CPi(ii), then we define Gj(ii) = min{a : 3aij EVa CPi(aO, ... ,aij-I,·aij,aj;+I,.·. ,amJ}. If -dVij cpf.(ii), then we define Gj(ii) = O. EXERCISE 24.16 (PG). Convince yourself that G j really is a functional class, i.e., convince yourself that there is a formula '¢j of Ls such that G j = {(ii, a) ,¢j(ii, an.
24.2. APPLICATIONS OF ELEMENTARY SUB MODELS IN SET THEORY
171
For every ordinal ,,(, the restriction Gjr(V,)mj is a function. By the Axiom of Replacement, the range of the above restriction is a set, and hence there exists f3 such that rng(Gjr(V'Y)m j ) ~ V/3. Thus, we can define a functional class F j : ON --+ ON as follows: F j (() = max{(,min{'17: rng(Gjr(V., K is a cardinal less than A, and K Then for every x E M, if H>. F Ixl = K, then x ~ M.
~
M.
PROOF. First note that although in general it is not the case that K ~ M implies K E M,12 our assumptions do imply that K E M: If there exists x E M such that H>. F Ixl = K, then in particular, H>. F 3J.L (lxl = J.L). Byelementarity, M F 3J.L (Ixl = J.L); and hence there exists J.L E M such that M F Ixl = J.L. Again by elementarity, H>. F Ixl = J.L, hence J.L = K, and thus K E M. Now M F 3f E "'x f[K] = x; hence there exists f E M such that M F f E "'x /\ f[K] = X. But then H>. F f E "'x /\ f[K] = x; hence f really maps K onto x. If y E x, then there is some E K such that f(O = y. But K ~ M implies that ~ E M, and hence M F f(e) = ii for some y. Since H>. F f(O = y, this ii must be y. 0
e
COROLLARY 24.22. If M
-< H>. and x is a countable set such that x E M, then
x~M.
PROOF. This follows immediately from Exercise 24.23 and Lemma 24.21.
0
Now suppose a E M n WI' Then a is a countable set of ordinals; hence by Corollary 24.22, a ~ M. In other words, f3 E M for every f3 E a. A similar argument applies if a < K+ and K ~ M. We have proved the following: CLAIM 24.23. Suppose M -< H>.. (a) If M is countable, then M nWl = 8 for some countable ordinal 8. (b) If IMI = K and K ~ M, then M n K+ = 8 for some ordinal 8 < K+. Now we are ready to present some applications of elementary submodels. Our first example will be a solution to Exercise 16.2. LEMMA 24.24. Let B be an uncountable family of finite subsets of WI' Then there exists a countable subset N ~ WI such that every b E B which is not contained in N is a member of an uncountable Il-system A ~ B with kernel r ~ N. PROOF. Let B be as in the assumptions. Fix a "sufficiently large" A, and let M be a countable elementary submodel of H>. such that B E M (recall that the existence of such M follows from Corollary 24.13). Let N = M n WI' By Claim 24.23, N is equal to a countable ordinal 8. We show that N is as required. Consider bE B such that b is not contained in N. Let r = bnN. By Exercise 24.22, b is not an element of M, but r is. Now let a be an arbitrary countable ordinal in M. Then a < 8, and thus
H>.
(24.34)
Fr
~
b /\ b\r =I- 0/\ min(b\r) > a.
In particular, (24.35)
H>.
F 3x E B (r ~ x /\ x\r =I- 0/\ min(x\r) > a).
12For example, if M = VI< then
I
a).
By elementarity, (24.38)
Since H>. 2 WI, (24.38) allows us to recursively construct a transfinite sequence (a€)€<Wl of countable ordinals and a sequence (b€)~<Wl such that r is a proper subset of b~ and a < a~ < min(b~\r) :::; maxb~ < a,., for all € < TJ < WI. It is easy to see that the set {b~ : € < wd is a a-system with kernel r, as desired. 0 EXERCISE 24.24 (G). Prove the following generalization of Lemma 24.24: LEMMA 24.25. Let K be an uncountable cardinal, and let B be a family of finite subsets of K+ with IBI = K+. Then there exists a subset N ~ K+ of cardinality K such that every b E B which is not contained in N is a member of a tl-system A E [BJI. such that B E M. Let N = M. We show that N is as required. Consider b E B such that b is not contained in M, and let r = b n N. By Exercise 24.22, b is not an element of M, but r is. In the proof of Lemma 24.24, we proceeded by considering countable ordinals
in M. This won't help us here, since b may not be a set of ordinals. We need to consider something more general: Let D be an arbitrary countable set such that DE M and r ~ D. By Corollary 24.22, D ~ M, and thus bnD = rand b\D i= 0. It follows that (24.39)
M
F 'VD 2
H>.
F 'VD 2 r (IDI
r (IDI :::; No
-+
3a E B (a
i= r /\ anD =
r)).
-+
3a E B (a
i= r /\ anD =
r)).
By elementarity, (24.40)
:::; No
EXERCISE 24.25 (G). Complete the proof of Lemma 24.26.
o
Now let us consider a situation where the kernels of tl-systems may be infinite. THEOREM 24.27. Let J.L and v be infinite cardinals such that vI' = v. Let B be a family of sets such that IBI > v and Ibl :::; J.L for all b E B. Then there exists a set N of cardinality v such that every b E B that is not contained in N is a member of a tl-system A ~ B with kernel r ~ N such that IAI > v. EXERCISE 24.26 (G). Convince yourself that Theorem 24.27 implies Theorem 16.3 for the case when both K and A are successor cardinals. Hint: Let K = J.L+ and A = v+.
24.2.
APPLICATIONS OF ELEMENTARY SUBMODELS IN SET THEORY
175
PROOF OF THEOREM 24.27. Let J.I., v, B be as in the assumptions. Fix a "sufficiently large" A, and let M be an elementary submodel of H>. such that J.I., v, B EM. We'd like to show that N = M is as required. Thus, in particular, M must be of cardinality v. Moreover, if bE B is such that b is not contained in M and r = bnN, then b should not be an element of M, but r should be. How can we make sure that this is the case? The first of these two requirements is relatively easy to handle: Let us choose M in such a way that v ~ M. By Corollary 24.13, it is possible to find M -< H>. of cardinality v such that vU{v,B} C M.13 Note that the assumptions of Theorem 24.2 imply J.I. < v, and hence we have {! ~ M and {! EM for every {! :S J.I.. If M is as above and b E M is a set of cardinality {! :S J.I., then by elementarity, M F Ibl = {!. By Lemma 24.21, b ~ M. In other words, if bE B is not a subset of M, then b ~ M. Making sure that r = b n M is an element of M is trickier. In contrast to the proofs of Lemmas 24.24- 24.26, r does not need to be finite, and Exercise 24.22 no longer applies. We need a new concept. Let K, be an infinite cardinal. We say that M is closed under subsets of size less than K, if [MJ. is closed" under subsets of size less than No. If A is regular, then H>. itself is closed under subsets of size less than A. Now let K, = J.I.+, and suppose we have an elementary submodel M -< H>. such that v U {v, B} ~ M and M is closed under subsets of size less than K,. Let b, r be as above. Then r is a subset of M of size :S J.I. < K" and hence rEM. Now we can reason as in the proof of Lemma 24.26: Let D be an arbitrary set of cardinality :S v such that D E M and r ~ D. Since v ~ M, Lemma 24.21 implies that D ~ M, and thus b n D = rand b\D =1= 0. Hence (24.41)
M
F VD 2 r(IDI
:S v
---+
3a E B (a
=1=
r /I. anD
= r».
Byelementarity, (24.42)
H>.
F VD 2 r (IDI :S v ---+ 3a E B (a =1= r /I. anD = r».
As in Exercise 24.25, we can recursively construct transfinite sequences (D~)~..
F U' is not an open cover of X.
This contradicts the choice of M as an elementary submodel of H>...
o
For dessert, let us show that some of the large cardinals you have encountered in this text can be characterized in terms of elementary embeddings. Let us say that a cardinal K, has the Reflection Property14 if for every R ~ VI< there exists a < K, such that (Va, E, R) -< (VI.+ for each
MATHOGRAPHICAL REMARKS
PROOF.
(24.91)
Let
K
185
be as in the assumptions, and let j,M be as in (**). Since
F V>" < K (>.. is an infinite cardinal
-+
2), = >.. +),
F V>" < j(K) (>.. is an infinite cardinal
-+
2),
VI..+).
How realistic are M's notions of infinite cardinals?
>.. really is an infinite cardinal, Hint: Argue as in Exercise 24.14.
EXERCISE 24.34 (G). Convince your-self that if
then M
F >.. is an infinite cardinal.
By Exercise 24.34,
(24.93)
M
FK
is an infinite cardinal.
Hence (24.92) implies in particular that
(24.94) This means, there exist
M
f
F 21< =
K+.
E M and an ordinal {3 E M such that
M F f is a bijection from P(K) onto {3, and for every ordinal a: such that K < a: < {3, (24.95)
(24.96)
MFa: is not a cardinal.
By Claim 12.19(j), there exists a set x E M such that onto {3 and (24.97)
M
f
really is a bijection from x
FX =P(K).
Since M witnesses (**) and P(K) ~ VI 0, and Ij3(A) has already been defined for all /3 < a. If a = /3 + 1 for some /3, then we let I",(A) =
'If;;l I(Aj3).
198
25.
BOOLEAN ALGEBRAS
If a is a limit ordinal, then we let
I,,(A)
= U Ifj(A). fj 1/. For a < J.L' consider Dn = {cn . a: a E D, Cn ' a ~ O},
and let E = U{Dn : a < J.L / }. On the one hand, E is a disjoint family with lEI ~ IDI > 1/. But on the other hand, lEI = 2:n. ~ 11:, the implication IZ(a"() I = 11:+ particular, !Ya(a,e) I = 11:+.
-+
!Ya(a"() I = 11:+ holds. In 00
Let B be a Boolean algebra, X, Y ~ B, and z E B. We write X ..L Y if x ..L y (i.e., x· y = 0) for all x E X and y E Y. Instead of {z} ..L Y we simply write z ..L Y. Similarly, we write X ~ Y if x ~ y for all x E X and y E Y. An element b of B separates X and Y if X ~ band Y ..L b, or vice versa. Note that if some b separates the sets X and Y, then X ..L Y. We say that B has the countable separation property if for all X, Y E [B]~No such that X ..L Y there exists b E B that separates X and Y. EXERCISE 25.32 (G). Let B be a Boolean algebra. Show that the following are equivalent: (i) B has the countable separation property; (ii) For all disjoint sets X, Y E [B]~No such that X ..L Y there exists b E B separating X and Y; (iii) ' 2'\, the Pigeonhole Principle implies that there exist io, il E I such that io #- i1 and fio = fi Consider the set W = iliEI Wi, where Wio = Ui o , Wil = Vi, , and Wi = Xi for i "E I\{io,id. Then W is a nonempty open subset of iliEIXi with W n D = 0. 0 The last two theorems of this appendix illustrate the use of cardinal invariants of the continuum in topology. More material on this topic can be found in the articles The integers and topology, by E. K. van Douwen, in Handbook of Set-Theoretical Topology, K. Kunen and J. E. Vaughan, eds., North-Holland, 1984, 111-167; and
26. APPENDIX: SOME GENERAL TOPOLOGY
215
Small Uncountable Cardinals and Topology, by J. E. Vaughan, in Open Problems in Topology, J. van Mill and G. M. Reed, eds., North-Holland, 1990, 195-216. A topological space X is called sequentially compact if every sequence of points in X has a convergent subsequence. Every compact metrizable space is sequentially compact. In particular, the space D(2) is sequentially compact. THEOREM 26.8. Suppose K. < t and {(Xo, To) : a < K.} is a family of sequentially compact topological spaces. Then the space I1o')~, 49 K f+ (>')~, 50 n ~ (m);, 51 K -+ (>,);;W, 60 K -+ (>'i)kO" 60 >. -+ ({3i)kO" 62 >. -+ ({3) 'd, 62 >. -+ ({3);;W, 62 w -+ (F)~, 114