Products of Compact Spaces and the Axiom of Choice II ££ £££ Omar De La Cruz1,Erichall£ 1, Paul Howard2, Kyriakos Keremedis3, and Jean E

Math. Log. Quart. 49, No. 1, 57 – 71 (2003) / DOI 10.1002/malq.200310004

Products of compact spaces and the axiom of choice II ££ £££ Omar De la Cruz1,EricHall£ 1, Paul Howard2, Kyriakos Keremedis3, and Jean E. Rubin 1 Department of Mathematics, Purdue University, West Lafayette, IN 47907, U. S. A. 2 Department of Mathematics, Eastern Michigan University, Ypsilanti, MI 48197, U, S. A. 3 Department of Mathematics, University of the Aegean, Karlovasi, Samos, 83200, Greece

Received 19 January 2002, revised 4 March 2002, accepted 6 March 2002 Published online 26 December 2002

Key words First countable, second countable, Hausdorff space, Tychonoff-product, compact spaces, axiom of choice, axiom of multiple choice.

MSC (2000) 03E25, 04A25, 54D30 This paper is dedicated to the memory of Jean Rubin, our friend, colleague and mentor. This is a continuation of [2]. We study the Tychonoff Compactness Theorem for various deﬁnitions of com-

pactness and for various types of spaces (ﬁrst and second countable spaces, Hausdorff spaces, and subspaces

of Ê ). We also study well ordered Tychonoff products and the effect that the multiple choice axiom has on such products.

1 Introduction and deﬁnitions

We started our study of the Tychonoff Compactness Theorem in [2], where we studied the Theorem using the

deﬁnitions of compactness studied in [3]. In this paper we extend these results to various types of topological spaces, ﬁrst countable spaces, second countable spaces, Hausdorff spaces and subsets of Ê . We also consider the role the multiple choice axiom plays when the products of topological spaces are well ordered. It is well known (see [18] or [10]) that the Tychonoff Compactness Theorem, “Products of compact spaces are compact” is equivalent to the Axiom of Choice (AC). In the last section we prove that this equivalence does not hold when “compact” in Tychonoff’s Theorem is replaced by certain weaker forms of compactness. We construct

a Fraenkel-Mostowski model in which one of the product theorems is true, but AC is false.

´ Ì µ

We start by deﬁning some of the terms we shall be using. Let be a topological space.

Ì ¾ ¾

1. Ì is called a base for if for each and each neighborhood of there is a such that

¾ Ì . (Thus, every element of is a union of elements of .)

2. is ﬁrst countable if each point has a countable base. Ì 3. is second countable if has a countable base.

4. is Lindelof¨ if each open cover has a countable subcover.

5. is said to be ¼ if whenever and are distinct points, there is a neighborhood of one that does not

contain the other.

6. is said to be ½ if whenever and are distinct points, each has a neighborhood not containing the other.

(Points are closed.)

7. is said to be ¾ (or Hausdorff ) if whenever and are distinct points, there is a neighborhood of

and a neighborhood of such that .

Ì Ì Ì Ì

½ ½ ¼ Clearly, every second countable space is ﬁrst countable; every ¾ space is ;every space is .

£ Corresponding author: e-mail: [email protected]

££ Jean Rubin died on October 25, 2002.

£££ e-mail: [email protected] (De la Cruz), [email protected] (Howard), and [email protected] (Keremedis)

c 2003 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim 0942-5616/03/0101-0057 $ 17.50+.50/0 58 O. De la Cruz, E. Hall, P. Howard, K. Keremedis, and J. E. Rubin: Products of compact spaces and AC II

Some of the weak forms of AC that we shall be using include the following:

8. AC(WO) is the axiom of choice for a well ordered family of sets. Ç 9. ACÏ is the axiom of choice for a family of well orderable sets.

10. AC(LO) is the axiom of choice for a linearly ordered family of sets.

11. AC( ¼ ) is the axiom of choice for a countable number of sets.

´ µ ¼

12. ¬Ò is the axiom of choice for a countable number of ﬁnite sets.

13. MC is the the Multiple Choice Axiom: For every non-empty set of non-empty sets, there is a function

¾ ´µ such that for each , is a non-empty ﬁnite subset of . 14. BPI is the Boolean Prime Ideal Theorem: Every Boolean algebra has a prime ideal. (The reader is referred to [7] for more information about the statements given in items 8 – 14 above.)

It is known that each of AC(LO) and MC are equivalent to AC in ZF. However, they do not imply AC in ZF¼ ,

ZF without the foundation axiom (see [12] and [17]). Unless otherwise stated, all proofs are in ZF¼ . The types of compactness we will be studying are given in [2], but we shall repeat those definitions for convenience. Definitions of Compact Spaces. 1. A space is compact A if every open covering has a finite subcovering. 2. A space is compact B if every ultrafilter has an accumulation point. 3. A space is compact C if there is a subbase for the topology such that every open covering by elements of the subbase has a finite subcovering. 4. A space is compact D if every covering by elements of a nest of open sets has a finite subcovering. 5. A space is compact E if every sequence has a convergent subsequence. 6. A space is compact F if every infinite subset has a complete accumulation point. 7. A space is compact G if every infinite subset has an accumulation point.

8. A space is compact H if every countable open covering has a ﬁnite subcovering.

Product Topology. Let be a set of topological spaces and let be their

product. A base for the product topology on is the family of all sets of the form ,where is open

in and for all but a ﬁnite number of coordinates, .

P(X,Y) is the statement: Products of compact X spaces are compact Y. P (X,Y) is the statement: Count- ¼ able products of compact X spaces are compact Y. Pr(X) is the statement: Projections in Tychonoff-products of compact X spaces are closed.

One of the questions we will be concerned with is when the product of topological spaces of a certain type

Ì Ì ¼ ½ ¾ will have the same type. It is easy to show that a product of spaces is ,for , but it is known, for

example, that the product of two Lindel¨of spaces need not be Lindel¨of. (Sorgenfrey line: Ê , where a base for the

µ topology is the set of all half open intervals .)

We shall study two types of product theorems. Let P½ (X,Y) be the following statement: Every product of

type compact X spaces is compact Y, provided that the product itself is of type . The same statement but

¾ ¾ ½

without the requirement that the product is of type will be called P (X,Y). Clearly, P (X,Y) implies P (X,Y),

½ ¾

for all X, Y. If is some property such as Hausdorff that is preserved under arbitrary products, then P and P are

identical and the superscript will be omitted. The following results for general topological spaces were obtained in [2]: (1) Each of the following statements is equivalent to AC: P(A,A), P(A,C), P(A,D), P(A,F), P(A,G), P(A,H), P(B,A), P(B,C), P(B,D), P(B,F), P(B,G), P(B,H), P(C,A), P(C,C), P(C,D), P(C,F), P(C,G), P(C,H), P(D,A),

P(D,C), P(D,D), P(D,F), P(D,G), P(D,H), and P(F,F).

(2) AC P(F,A) P(F,C) P(F,B).

(3) AC P(F,A) P(F,G) AC Ï .

(4) AC P(F,A) P(F,D) P(F,H) AC Ï .

(5) AC P(B,B) P(C,B) P(A,B) P(D,B) AC.

Math. Log. Quart. 49, No. 1 (2003) 59 ¾

It is easy to show that properties (1) through (5) also hold for P (X,Y), where is ﬁrst or second countable.

However, in the proof of [2, Theorem 1] that each of P(A,G) and P(A,H) implies AC, the topology used is not

Ì Ì ½

even ¼ . However, we show in Section 4, that it is easy to extend the topology so that it is . In addition, it

È È Ì

isshownin[14]that Ì (A,A) is equivalent to BPI and it is shown in [4] that (B,B) is provable in ZF.We

¾ ¾

È ·È Ì

show in the introduction that if BPI holds, then compact D implies compact G. Therefore, Ì (A,A) (D,D)

¾ ¾ È

implies Ì (D,G). ¾ In [3, Section 6], we give some relationships between the forms of compactness if BPI holds. Here, we note

that we can add to those relationships by showing that compact D implies compact G. (BPI implies that every

set can be linearly ordered. Suppose is compact D and is an inﬁnite subset of that has no accumulation

Ò ¾ ¾

points. Then every subset of is closed. Let ,where is a

linear ordering of which has no minimum element. Then is a nest of open sets which covers , but has no ﬁnite subcover.) Thus, the arrow diagram which gives the relationship between the forms when BPI holds can be

changed to: ¹

A, B, C ¹ DH

FG¹

2 First and second countable spaces

In this section we shall study product theorems for ﬁrst and second countable spaces. We shall use the notation

È (X,Y) to stand for the statement “Countable products of ﬁrst countable compact X spaces are compact Y”.

Likewise, we have È (X,Y) about second countable spaces. Note that ﬁrst and second countability are preserved × under countable products.

It follows from [3] that the following are the arrow diagrams for ﬁrst countable spaces.

¹ ¹

EHDACB

F ¹ G

Assuming AC, we have ¹ A, B, C, D, F ¹ E, H G

For second countable spaces we have the following relationships between the forms of compactness:

¹ ¹

FGE¹ A, D, H CB

Consequently, we immediately have that the statements È (X,Y) for X, Y A, D, or H, are equivalent. In ZFC × we have:

A, B, C, D, E, F, H ¹ G Ì

(In addition in ZFC, if the space is second countable and ½ , then all the forms of compactness are equivalent –

see [3].) In a topological space , let us refer to a point that has only one neighborhood (which must be )asatrivial

point.Atrivially compact topological space is a space that has a trivial point (so a space is trivially compact

if every open cover contains ). A space is indiscrete if its topology is .ItisprovableinZFC that

a product of first countable spaces is first countable iff each is first countable and all but countably

Ì ¾ many of the ’s are indiscrete spaces, and that a product of spaces is second countable iff each factor is second countable and all but countably many factors are one-point spaces (see [19]). In ZF, we show in the next

lemma that uncountable products are not ﬁrst countable unless all but of the factors are trivially compact, where

is a countable union of ﬁnite sets.

60 O. De la Cruz, E. Hall, P. Howard, K. Keremedis, and J. E. Rubin: Products of compact spaces and AC II

Lemma 1 Let , , be topological spaces such that is ﬁrst countable and non-empty.

(a) is ﬁrst countable, for all .

¾ ¾ ´µ

(b) For each ,theset is not a trivial point in is a countable union of

ﬁnite sets.

¾ ¾

P r o o f. Part (a) is clear. For (b), let ,andlet be a neighborhood base

¾ ´µ ¾ ´µ

for . For each deﬁne as the least such that for some neighborhood of

in such that .Given ,since is an open set in , it contains a basic neighborhood of

½ ½

¡¡¡

the form ,where are open sets in , respectively. In this case,

¼ ¼ ¼

´µ

¼ . This means that deﬁnes a partition of into countably many ﬁnite pieces.

É É

´ µ ¾

¬Ò ¼

Corollary 1 implies that if is ﬁrst countable, then for each ,theset

¾ ¾

(as in Lemma 1(b)) is countable.

´ µ ¼ Proof. ¬Ò implies that countable unions of ﬁnite sets are countable.

The following can be checked case by case:

¢ Lemma 2 If is a compact X space and is a trivially compact space, then is compact X, where X

is A, B, C, D, E, G,orH. £

Notice that the previous lemma does not apply when X is compact F. For example, suppose is an inﬁnite

¼ ½ ¢

but Dedekind ﬁnite set given the indiscrete topology, and let with the discrete topology. Then

is an inﬁnite set with no complete accumulation point, since a neighborhood of the form does not have

the same cardinality as .

´ µ ¼

Theorem 1 Let X, Y be deﬁnitions of compactness, with Y different from F.If ¬Ò holds, then

½ ½

È ´ µ È È ´ µ È ´ µ µ

(a) implies ´ and (b) implies .

× ×

P r o o f. We give the proof of (a). The proof of (b) is similar. Let for be ﬁrst countable com-

É É

pact X spaces such that is ﬁrst countable. If is empty, then it is immediately compact Y.

¾ ¾

É É É É

¾ ¢

Otherwise, ﬁx a point . Then we have that ,where

¾ ¾ ¾ ¾ Ò

× ×

¾ ´µ

is not a trivial point in and is countable by Corollary 1. Now the space is

¾ Ò

´ µ È

trivially compact, since the projection of into that space is a trivial point there. And since implies

É É

that is compact Y, by Lemma 2 we conclude that is compact Y.

¾ ¾

´ µ ¼

The fact that ¬Ò is needed for the results above is illustrated by the following theorem. However, some

´ µ È ´ µ ¼

statements of the form might automatically imply ¬Ò , and make it redundant for some of the

instances of Theorem 1.

È µ

Theorem 2 ´ cannot be proved in ZF,whereX is any of the deﬁnitions of compactness and Y is

È µ

A, D, F, G,orH. Consequently, ´ cannot be proved in ZF either, for the same values of X and Y.

P r o o f. Consider the second Fraenkel model (Æ in [7]), in which there is a set which has a countable partition into pairs with no choice function. In this model we have that ¾ is a product of two-element spaces

which is ﬁrst countable, but not compact X for X A, D, F, G, or H (see [3]).

½ ½

´ µ ´ µ È

Note that it does not follow immediately from the deﬁnitions that È implies . The next

theorem proves a result of this form.

½ ½ ½

È ´ µ È ´ µ È ´ µ ´ µ

Theorem 3 implies , and is equivalent to ¼ .

× ×

¼ ¼

È È È

Proof. Itisprovedin[5]that (A,A) is equivalent to AC( ¼ ), and thus both (A,A) and (A,A)

× ×

´ µ È È ¼

imply ¬Ò . It then follows from Theorem 1 that (A,A) is equivalent to (A,A). Likewise, it follows

È È

from Theorem 1 that È (A,A) is equivalent to (A,A). The theorem follows easily since (A,A) clearly

× ×

implies È (A,A). × Math. Log. Quart. 49, No. 1 (2003) 61

We mention some Tychonoff-type product assertions that are provably false.

Theorem 4

´ µ

(a) È is false in ZF for all deﬁnitions X of compactness.

´ µ ¾ À

(b) È is false in ZF for .

´ µ ¾ À ¾

P r o o f. The two-element space ¾ with the discrete topology is compact X for X A, B, C, D, E, F, G, and H,

while ¾ is not compact E (see [3]). This proves (a). Parts (b) and (c) are consequences of counterexamples that

´ ½µ ¾ are not even products (again see [3]). For part (b) note that the topology Ì on the integers

witnesses in ZF that a second countable space can be compact G but not compact in any other sense. To prove

¾ À ¾ that is not compact Y for .

From independence results found in [3, Theorem 3(b) and Lemma 1(c)], it immediately follows that the

¾ ¾

´ µ ¾ À È ´ µ

following statements are not provable in ZF: È for and for

× ×

´ µ

Theorem 5 È implies AC. ×

P r o o f. We will reﬁne the proof in [16] that P(A,A) implies AC.

È ¾

Assume (A,C), and let be a collection of non-empty sets; we will prove that has a

¾ ½ ½

choice function. For each deﬁne the space by ,where stands for a ﬁxed element

Ë É

not in ,and . These spaces are compact A (actually trivially compact). Let ;

É É

¢ ½

clearly is not empty. For every ﬁnite ,deﬁne ; every non-empty basic

¾ Ò ¾

neighborhood in is of the form for some ﬁnite . We have the following facts:

Claim 1. is a trivial point in if and only if is a choice function for .

´µ ¾ ¾

Indeed, a basic neighborhood is different from iff ,and contains iff for .

Claim 2. Every ﬁnite open cover of contains as an element.

¼ ¼

To prove this we will show that if are open sets, then ; an easy inductive argument can

Ë Ë

¼ ¼ ¼

¾ ¾

then complete the proof. Let and ,where are sets of ﬁnite

¼ ¼ ¼

¾ ¾ ¾ ¾ ¾

subsets of .Take such that and . Then for each there exists such that

¼ ¼

´µ ¾ ¾ ¾ ´µ ¾ ¾

, and for each there exists such that . It is clear that ,deﬁnedby

´µ ´µ ¾

if ,

´µ

otherwise

is neither in nor in . This ﬁnishes the proof of Claim 2.

Finally, in order to reach a contradiction, assume that has no choice function; by Claim 1, has no trivial

points. If is a subbase for the product topology on , then every point has a neighborhood in which is not .

¾ È

This way, is an open cover of with elements in .By (A,C), this cover has a ﬁnite ×

subcover, but that is impossible by Claim 2.

Thus, it follows that each of È (X,Y), for X, Y A or C, is equivalent to AC. Also, it follows from

× ¾

[4, Theorem 4.2], that È (F,F) is equivalent to AC. In addition, Theorem 1 of [2] holds for second countable

¾ ¾ È

spaces: Each of È (A,G) and (A,H) imply AC. Consequently, (1) – (5) in the introduction also hold for

× ×

¾ ¾ ¾ ¾

È È È

È (X,Y). (In addition, since (X,Y) implies (X,Y), (1) – (5) hold for (X,Y).)

× ×

Let be the statement: Every non-principal ﬁlter on can be extended to an ultraﬁlter. Recall that Pr(A) is the statement: Projections in Tychonoff-products of compact A spaces are closed. We close this section with a

result showing that some weak forms of AC imply Pr(A).

·´ µ Ì

¼ ¾

Theorem 6 implies Pr(A) for ﬁrst countable spaces.

´ µ ¾ Ì ´ µ

Proof. Let be a family of ﬁrst countable compact spaces and let be their Ty-

¾ ´ µ chonoff-product. If , then there is nothing to show. Let be a closed set in and .If

62 O. De la Cruz, E. Hall, P. Howard, K. Keremedis, and J. E. Rubin: Products of compact spaces and AC II

´ µ Ò ´ µ

(the projection of on its coordinate) is not closed, then is an inﬁnite subset of and

´ µ ´ µ µ ¾

contains limit points from .Let be such a limit point in and, by AC( ,let

´ µ ¾ ¼

be a countably inﬁnite subset of converging to .ByAC() again, we choose for each ,

¾ ´ µ ¾

.Let be a non-principal ultraﬁlter on extending the ﬁlterbase

´ µ ¾

is a neighborhood of

Ì ¾

Using compactness and the fact that the spaces are ¾ , we can readily verify that for every , ,there

¾ ´ µ ¾

exists a unique such that is a neighborhood of . Clearly, the

¾ ´ µ ´ µ

element with and for is in the closure of , hence, in . On the other hand,

¾ ´ µ ¾ since , . This contradiction ﬁnishes the proof of the theorem.

3 Hausdorff spaces Ì

Properties (1) – (5) in the introduction do hold for ½ spaces because Theorem 1 of [2] (“Each of P(A,G) and

P(A,H) implies AC”) holds for ½ spaces (add the set of coﬁnite subsets to the topology on each ). However, Ì these properties do not necessarily hold for ¾ or Hausdorff spaces. We shall add a diagram to indicate the relationships that are known to hold for all spaces. First we have the relationships between the deﬁnitions of

compact.

¹ ¹

HDACB

F ¹ G È

It is known that Ì (A,A) is equivalent to BPI (see Theorem 7 below). Also, it is known (see, for example, [3])

È È Ì

that BPI implies that compact A, compact B, and compact C are equivalent. Thus, Ì (A,A) implies (C,C). In

¾ ¾

addition, using the preceding diagram and that if compact X implies compact Y, then P(Z,X) implies P(Z,Y) and

Ì Ì ¾

P(Y,Z) implies P(X,Z), we obtain the following diagram (also note that products of ¾ spaces are always ):

È È Ì

Ì (D,H) (D,D)

¾ ¾

¹ ¹

È È È È È

Ì Ì Ì Ì

Ì (A,H) (A,D) (A,A) (C,C) (A,C)

¾ ¾ ¾ ¾ ¾

Ì (A,G)

È È È È

Ì Ì Ì

It has been shown in [4] that Ì (B,B) is provable in ZF.Since (B,B) implies (C,B) and (C,B)

¾ ¾ ¾ ¾

È È È È

Ì Ì Ì

implies Ì (A,B), it follows that (C,B) and (A,B) are also provable in ZF.Also, (X,E)isalwaysfalse

¾ ¾ ¾ ¾ Ê

(a counterexample is ¾ ).

È È È

Ì Ì

It is also shown in [4] that AC iff Ì (F,F). Since F implies G, it is also true that (F,F) implies (F,G).

¾ ¾ ¾ È

We show in Section 6 that Ì (F,G) does not imply AC. ¾ The next collection of results shows that, in contrast to (1) in the introduction, none of the following arrows

are reversible:

¹ ¹ ¹

È È È

Ì Ì

Ì (A,A) (A,D) (A,H)

¾ ¾

( ) AC ¾

´ µ È

Theorem 7 (Ło´s and Ryll-Nardzewski [14]) Ì is equivalent to BPI.

Lemma 3 Assuming AC(LO),a ¾ space is compact D iff every well-ordered inﬁnite subset of has a complete accumulation point.

Note that Lemma 3 is not very interesting in ZF,sinceZF + AC(LO) implies AC. However, ZF¼ +AC(LO) does not imply AC (see [17]).

Math. Log. Quart. 49, No. 1 (2003) 63

P r o o f. Suppose is not compact D, so there is a covering nest of open sets that does not have

as a member. By AC(LO), every linearly ordered set can be well-ordered, so choose a well-ordering of

(which is linearly ordered by the subset relation on the ’s). Now we obtain a well-ordered nest by induction:

For each ordinal ,let ,where is the -least member of such that properly contains ,

if such exists. By throwing away all but some coﬁnal subset if necessary, we may assume that the ’s

¾ Ò ¾

·½

are deﬁned just for all in some cardinal . Now using AC(LO) again, let for each .

Assuming is Hausdorff, this family of ’s will have no complete accumulation point.

´ µ Ì

Conversely, assume that is a compact D ¾ space. We will show that every well-ordered subset

has a complete accumulation point. We consider two cases:

¾ ¾ ¾

(i) is a regular cardinal. For every ordinal put and let

. Assuming that has no complete accumulation point and taking into consideration

that every point has a neighborhood meeting in a set of size , we see that is a well ordered

open cover of . Hence, for some , which is a contradiction.

´µ ¾

(ii) is a singular cardinal. Let (the cofinality of ) and fix a cofinal set of regular

¾ ¾ ¾

cardinals in .Forevery let is a complete accumulation point of .By(i)

¾ ¾

each is non-empty. Let be a choice set of the set .Since is a regular

cardinal it follows by part (i) again, that has a complete accumulation point , and this must be a complete

¾ ¾

·½ ·½

accumulation point of . Indeed, if is a neighborhood of ,then for all

È È

¾ ¾ ¾ ¾

·½ . Thus, . The equalities in the above

display are justiﬁed because of AC(LO).

As before, is the statement: “Every non-principal ﬁlter on can be extended to an ultraﬁlter”. Similarly,

Ç let Ï be the statement: “Every non-principal ﬁlter on a well-ordered set can be extended to an ultraﬁlter”.

Note that these principles are always true in permutation models, where AC holds in the subuniverse of pure sets.

·´ µ È ·´ÄÇµ È ´ Àµ ´ µ

¼ Ì ÏÇ Ì

Theorem 8 (a) implies . (b) implies .

¾ ¾

Proof.

´ µ ¾ Ì

(a) Let be a family of compact A spaces and let be their Tychonoff-product. In view

of AC( ¼ ) it sufﬁces to show that every countably inﬁnite subset of has a limit point.

Choose, by , a non-trivial ultraﬁlter on . It is straightforward to verify that for each there is a unique

¾ ´ µ

such that is a neighborhood of . (The existence follows from compactness

Ì ¾

and the fact that is an ultraﬁlter, and the uniqueness follows from ¾ .) Now it is clear that the element

´µ given by is a limit point of ﬁnishing the proof of the theorem.

(b) Choose a family of compact A Hausdorff spaces. By AC(LO) and Lemma 3, it sufﬁces to show that every

well-ordered inﬁnite subset of their product has a limit point. If we choose, by Ï an ultraﬁlter on such

that each subset of with cardinality smaller than that of is not in , then the proof proceeds just as in the

proof of part (a).

È ´ µ È ´ µ Ì

Corollary 2 Ì does not imply .

¾ ¾

P r o o f. The corollary follows from Theorem 7 and Theorem 8(b), since in ZF , + AC(LO) does not

imply BPI. (Construct a permutation model as follows: Let the set of atoms be uncountable. The group of

Æ ½¾ permutations is the group of all permutations on , and supports are countable. (See in [7].) BPI is false because BPI implies that every set of pairs has a choice function, but the set of pairs of atoms in this model has

no choice function. Truss showed in [17] that AC(LO) holds in this model.)

È ´ Àµ È ´ µ Ì

Theorem 9 Ì does not imply .

¾ ¾

Æ ¾ ½

P r o o f. We ﬁrst give the description of a permutation model . The set of atoms is ,

¾ ¼

where and is the set of points on the unit circle centered at . The group of permutations

¾ Ê is the group of all permutations on which rotate the ’s by an angle and supports are countable. The

following two claims are straightforward:

¾ Æ ½

Claim 1. The family does not have a multiple choice function in .

Claim 2. AC( ¼ ) is true in .

64 O. De la Cruz, E. Hall, P. Howard, K. Keremedis, and J. E. Rubin: Products of compact spaces and AC II

Æ È Æ

¼ Ì

Since U and AC( ) are true in , it follows by Theorem 8(a) that (A,H) also holds in .Furthermore,

È Æ ¾ £

Ì (A,D) fails in . Indeed, for every let carry the disjoint topology induced from

the (compactA) metric topology of ( is identiﬁed with the unit circle) and the discrete topology on .

Ì ´ µ ¾

¾ ½

Clearly each is a compact A space. Let be the Tychonoff-product of the family .Let

´£ µ ¾ Æ

.SinceAC( ) holds in , it follows that no covers .Since has no choice

¾ ½ set, we see that is a nested open cover of which has no ﬁnite subcover.

This completes the proof that none of the implications in (£) are reversible.

Let be the statement: “Every inﬁnite set has a non-trivial ultraﬁlter”. It is known that is strictly weaker

È È Ì

than BPI, so the next result shows that the implication Ì (A,A) (A,G) is not reversible in ZF.

¾ ¾

È ´ µ

Theorem 10 implies Ì .

´ µ ¾ Ì

Proof. Let be a family of compact A spaces and let be their Tychonoff-product. Let

be an infinite subset of and let be a non-trivial ultrafilter on . Continuing as in the proof Theorem 8(a) we can finish the proof of this theorem.

4 Well ordered products and the multiple choice axiom

Our ﬁrst theorem in this section gives a property of well ordered products of compact A spaces. We shall use the Ç

notation PÏ (X,Y) to mean that a well ordered product of compact X spaces is compact Y.

ÏÇ ÏÇ

´ µ È ´ µ

Theorem 11 È if and only if .

µ ´ µ ¾

Proof. Itsufﬁcestoshow( ) as the other implication is clear. Fix a well ordered family

of compact A spaces and let ´ be their Tychonoff-product. Without loss of generality we may assume

that for every , is compact A. Let denote the projection of to .Let be a

É ´ µ ¾

family of closed subsets of having the ﬁnite intersection property. Let ,where

´µ ¾ É

. It can be readily veriﬁed that is a family of closed sets of having the ﬁnite

¾ Ï ´ µ ¾

intersection property. Since for every , is compact A, it follows that is

a nest of non-empty closed subsets of .Since is compact D, it follows that . Clearly, any point

Ì Ì

¾ Ï is in and, consequently, is compact A as required, ﬁnishing the proof of the theorem. Our next result gives a property of the Multiple Choice Axiom MC.

Theorem 12 MC implies that compact D spaces are compact A.

´ µ Í ¾

Proof. Let be a compact D space and be an open cover of . Using Levy’s

Lemma (Form [67 B] in [7]: “Each set can be covered by a well ordered family of ﬁnite sets”) we express

¾ ¾

as , where each is a ﬁnite subset of and is a well ordered cardinal. For every

put .Let be the least cardinal number for which there exists a subfamily of

¾ ¾

covering . We show that is ﬁnite. Assume on the contrary that is inﬁnite. For every

¾ ¾

put .Since is compact D, it follows that some covers . Thus, is

¾ Í a subfamily of of cardinality covering . This is a contradiction. Hence, has a ﬁnite

subcover and is compact A as required.

ÏÇ

´ µ

Theorem 13 Å·È if and only if AC. Ç

P r o o f. It sufﬁces to show that MC + PÏ (A,A) implies P(A,A), as P(A,A) is equivalent to AC, and the

implication in the other direction is clear. Let be a family of compact A spaces and let

be the Tychonoff-product. Use Levy’s Lemma again (see the proof of Theorem 12) to express as a well ordered

Ë É

¾ ¾

union of ﬁnite disjoint subsets of .Thatis, .Forevery ,let . Clearly

ÏÇ

is compact A. Then is a compact A space by P (A,A), and is homeomorphic to .

It now follows from Theorems 12 and 13:

ÏÇ

´ µ

Corollary 3 Å · È if and only if AC.

Ç ÏÇ ¼ It follows from Theorem 13 and Corollary 3 that neither PÏ (A,A) nor P (D,D) is provable in ZF .

Math. Log. Quart. 49, No. 1 (2003) 65

5 Products of subsets of Ê and choice for closed sets

In this section we study products of subsets of Ê ; the results are different for well orderable products and general

products.

Theorem 14 Let be a collection of compact C topological spaces such that the family

and is closed has a choice function. Assume also that there exists a sequence

¾ ¾

such that for each , is a subbase for that witnesses the fact that is compact C.

Then is compact C.

¾ ¾

P r o o f. Let us consider the collection and of open sets in ,where

with and for . It is easy to see that the set of ﬁnite intersections

of elements of is a base for the Tychonoff-topology of , and therefore is a subbase for that topology.

É Ë

¾ ¾

Suppose now that is a cover for . Then there exists such that ;

¼ ¼

¾ Ò ¾

otherwise, we have that for every , is a non-empty closed subset of , and by

hypothesis there is an element in , which cannot be covered by any element of .Since

is a cover of by elements of , it has a ﬁnite subcover . Clearly, is a ﬁnite subcover ¼

of .

The existence of the witnessing family of subbases is satisﬁed, for example, in the case of a power of a ﬁxed space.

For well orderable products, we have:

Theorem 15 Let be a collection of compact A topological spaces such that the family

, , and is closed has a choice function. Then is compact A.

Proof. Let be a choice function for , ,and is closed . To see that

is compact A, let be a family of closed subsets of with the ﬁnite intersection property (f. i. p.); we will show

that .

By simultaneous recursion on we deﬁne a family of ﬁlters on , and a sequence

, satisfying

(i) ¼ is the ﬁlter generated by ;

(ii) for all ;

´µ ¾

(iii) for all and all , , for every neighborhood of .

Once this construction is completed, the ﬁlter generated by will converge to the point

¾ ¾

. Then every neighborhood of intersects every element of ; since each

element of is closed, .

´ ´ µ ¾ µ

Æ Æ Æ Æ

¼ is given by (i). Assuming that the ﬁlter is deﬁned, we take (we have

´ µ ¾

Æ Æ ·½

that Æ because each is compact A). We also deﬁne as the ﬁlter generated by the

´ µ Æ Æ

collection Æ is a neighborhood of .If is a limit ordinal and is deﬁned for all ,we

Æ Æ define as the filter generated by . This finishes the recursion; clearly conditions (ii) and (iii) are satisfied.

Corollary 4

(a) Any product of closed compact C sets of reals with a witnessing family of subbases

¾ is compact C. (b) Well ordered products of compact A sets of reals are compact A.

´ µ witnessing subbasesµ which are conditionally complete any subset with upper bounds has a least upper bound is

compact C. If the product is well orderable and the spaces are compact A ´equivalently, if they have a maximum

and a minimum elementµ, then the product is compact A.

P r o o f. Part (c) implies both parts (a) and (b). For (c), if the product is empty, then it is compact A. Assume then that the product is not empty; because of the previous theorems, we only need to prove that the collection of 66 O. De la Cruz, E. Hall, P. Howard, K. Keremedis, and J. E. Rubin: Products of compact spaces and AC II

all closed subsets from all the factor spaces has a choice function. Fix one element in the product; we will give

a rule for choosing an element from a given closed subset of the -th factor space using only the element .

If is an upper bound for , we pick the least upper bound of . If not, we pick the greatest lower bound of

´ ½µ

. In either case the element we picked is in ,since is closed.

Notice that arbitrary products of compact A sets of reals are not necessarily compact A: consider the space ¾

in the model Æ in [7], where is the set of atoms.

Ê Theorem 16 If is a well-orderable cardinal, then the collection , is compact A has a

choice function.

Proof. Let be a ﬁxed compact A subset of . We will describe a rule for choosing an element of .

For each ,let be the projection by of on . Since projections are continuous, is compact A.

Ê Ò ´ µ ×ÙÔ´ µ

Therefore, is bounded. Let and .Then contains ,

and is compact A by the previous corollary. Also, for each ,let be the restriction to of

the base for the topology of which contains all open intervals with rational endpoints. Clearly, is a

´µ ¾

countable base for .Let be a subfamily of which contains and

is maximal with respect to the f. i. p.; such a family can be easily constructed by transﬁnite induction, since

´µ ¾

has cardinality .Let be the ﬁlter generated by , and for each ,

¾ ´µ ¾ ¾

deﬁne , for every neighborhood of .Wehavethat .Since

¾ ´µ ¾

is compact A, the ﬁlter generated by the sets such that has an accumulation point

.If is a neighborhood of , then it has a non-empty intersection with every member of ; consequently

½ ½

´µ ´µ ¾

meets every member of and therefore (by maximality). Furthermore, is a singleton.

½ ½

¾ ´ µ ´ µ ¾

¾ ½ ½

Otherwise, if ½ are disjoint neighborhoodsof two points in , we would have ,

which is not possible because these sets are disjoint. Now, for each ,let be the unique element of ,

and let . Then every neighborhood of in contains a neighborhood which is a ﬁnite

´ µ

intersection of sets of the form ½ , which are in . Therefore, every neighborhood of intersects ;since

¾ is closed, then .

Notice that in the proof we used no arbitrary parameters. Therefore we can uniformly choose elements from

the compact A subsets of many different spaces Ê at the same time.

Corollary 5 Arbitrary products of compact A spaces which are a subspace of Ê for some well-orderable

cardinal , are compact C. Also, if the index set is well orderable, then the product is compact A.

¾ ¾ Ê

Proof. Let be a family of compact A spaces such that for each , is a subspace of .

Then for each , the collection of closed subsets of is contained in the collection of compact A subsets

of Ê . Therefore, by the previous theorem, we have a choice function for the closed subsets of all the factor spaces. Applying Theorems 14 and 15 we obtain the desired conclusions.

We turn now to the study of compact F sets in this context. Theorem 17 Assume that has a well-ordered base and a choice function for its closed subsets. If is

compact F, then it is compact A.

Proof. Let be a well ordered base for ,let be a choice function for the closed subsets of ,and

suppose that is not compact A. From all the well-ordered open covers from without a ﬁnite subcover, choose

Ó´µ

with minimum ,where is a cardinal. Let and let be an increasing

Ë Ë

µ ´ Ò Ò

coﬁnal sequence. For each let . (Each set is closed, and it is

´µ ´µ

non-empty by the minimality of .) Then, since is a regular cardinal, has cardinality ,and

for every , .Now,theset has no complete accumulation points because for every

¾ there exists a neighborhood of , and we have that . Therefore, is not compact F. Corollary 6 For closed sets of reals, compact F implies compact A. P r o o f. From our previous results, closed sets of reals satisfy the hypotheses of Theorem 17. Corollary 7 Well-ordered products of closed compact F sets of reals are compact A. Math. Log. Quart. 49, No. 1 (2003) 67

However, it is not true in general that even well-ordered products of compact F sets of reals are compact F:

Å½ ¾¼ ½ ¾ Consider the product space ¾ in the model in [7]; is clearly compact F, but is not, because it is Hausdorff and contains an inﬁnite Dedekind ﬁnite set. The results in this section suggest the following questions: Are compact C (or compact F) sets of reals neces-

sarily closed in Ê ? Since choice for closed sets played an important role in the previous results, the study of the following princi-

ples becomes natural:

Ì ½ ¾ ´ µ Ì

Deﬁnition Let CSC( ), , stand for the statements: “If is a compact A space, then the

family of all non-empty closed subsets of has a choice function”.

Ë´Ì µ

Theorem 18 ½ is equivalent to AC.

Ì ¾

P r o o f. We only need to prove that CSC( ½ ) implies AC. Fix a disjoint family of

non-empty sets and let be the one point compactiﬁcation of the disjoint union of the family of spaces

´ µ ¾ Ì

,where is the coﬁnite topology on .ByCSC( ), let be a choice function on the family

of all non-empty closed subsets of . Clearly, the restriction of to the family is a choice function, ﬁnishing the proof of the theorem.

In the last part of this section we use the following statements from [7]: Ì

Form 154. Tychonoff’s Compactness Theorem for Countably Many ¾ Spaces: “The product of countably Ì

many compact A ¾ spaces is compact A”. Ì

Form 343. “Every product of non-empty compact A ¾ spaces is non-empty”.

Ë´Ì µ

Theorem 19 (Keremedis and Tachtsis [11]) ¾ is equivalent to Form 343.

£ £

´ µ ´ µ Ë´Ì µ È È

Theorem 20 ¾ is equivalent to ,where is the following principle: “If

Ì Ì

¾ ¾

´ µ ¾ Ì

is a family of non-empty compact A spaces, then the Tychonoff-product is

À ´ µ ¾ ¾

compact C with respect to the standard subbase of ”.

Proof.

µ ´ µ ¾

( ). Fix a family of non-empty compact A spaces and let be . We will

À Í À

show that is compact A with respect to the standard subbase .Let be an open cover of .For

¾ Í Í ´ µ ¾ Í Í Í

every put .Ifsome covers , then it is easy to see that , hence ,

Í ¾

has a ﬁnite subcover. So assume that no covers and arrive at a contradiction. Put ,

Ò ¾ ´ µ ¾Í Ì

where . Clearly each with the subspace topology is a compact A

space. Thus, by Theorem 19, has a choice function . It is straightforward to see that is not covered by ,

contradicting the fact that Í covers .

´ Ì

( ). By Theorem 19, it sufﬁces to prove Form 343. Fix a family of non-empty compact A ¾ spaces

´ µ ¾ £ £ ¾

.Let , , carry the disjoint union topology of and the dis-

£ Ì

crete space . Clearly is a compact A space. Hence, by P (A, C), , the Tychonoff-

¾ À

product of the family , is compact C with respect to the standard subbase .If ,

Í ´£ µ ¾ À Í

then is an open cover of . It follows that has a ﬁnite subcover, which is a

contradiction.

ÏÇ ÏÇ

Ë´Ì µ È È ´ µ ´ µ

Theorem 21 ¾ implies ,where stands for: “The Tychonoff-product of every

Ì Ì

¾ ¾

´ µ ¾ Ì Ë´Ì µ

¾ ¾ well ordered family of compact A spaces is compact A”. In particular, implies

Form 154.

´ µ ¾ Ì

P r o o f. Fix a well ordered family of compact A spaces, and let be

their Tychonoff-product. Let be a family of closed sets having the f. i. p. We will show that

.Let be a choice function of the family of all non-empty closed sets of the one point compactiﬁcation

´ µ ¾

of the disjoint topological union of the family . It follows that the restriction of to the

´ Ò µ ¾ ¾ family is a choice function. Now we can complete the proof of the theorem by mimicing the proof of Theorem 20. 68 O. De la Cruz, E. Hall, P. Howard, K. Keremedis, and J. E. Rubin: Products of compact spaces and AC II

The status of the relationship between Form 343 and Form 154 is indicated as unknown in Table 1 of [7]. By Theorems 19 and 21 we have the following: Corollary 8 Form 343 implies Form 154.

6 An independence result

Our last result shows that P(F,G) does not imply AC in ZF¼ . We do this by showing that P(F,G) is true in

Æ ¿ Mostowski’s linearly ordered model Æ . (This is the model in [7].) The set of atoms is ordered so that it

is isomorphic to the set of rational numbers. The group is the group of order preserving isomorphisms and the

ideal of supports is the set of all finite subsets of .If is a finite subset of ,welet fixes .

¾ Then is a subgroup of . (Note that if ,then ﬁxes pointwise.) We will also use the following

facts about Æ .

Æ ´µ ¾ ´µ

(1) Every element of has a smallest support supp with the property that for all , iff

´×ÙÔÔ ´µµ ×ÙÔÔ´µ

´ µ ¾

(2) If and are disjoint finite subsets of ,then is an infinite, Dedekind finite

¾ ´µ ¾ ´ µ

set in the model. (If we choose , then the set is inﬁnite. By (1), since supp ,

´µ ´µ ´ µ ´ µ

if and are in and ,then . It follows that is inﬁnite. If were Dedekind

infinite, then there would be a finite subset of which was a support of infinitely many elements of .Butthe

only finite subsets of supported by are the finitely many subsets of . It follows that is Dedekind finite.)

Ï ´ ½ µ ´ µ ´ ½µ

½ ¼ ¼ ½

(3) If ¼ is any ﬁnite subset of and if is

the set of open intervals determined by , then for any two subsets and of both disjoint from ,if

¼ ¼

¾Ï ¾ ´ µ

for all , then there is a such that . (The proof of (3) uses the

¾ ½ ¾

following fact about the ordering on :If ½ and are two subsets of

¡¡¡ ¡¡¡

¾ ½ ¾

both with the same number of elements and ½ and , then there is an order

½ ´ µ µ

isomorphism of such that for all , , . In order to see that this is true we may assume that

¾ ½

is the set of rational numbers with the usual ordering and let be the union of the functions ½ ,

´ ½ ´ µ

¾ ½ ½

where ½ is the linear function whose domain is and whose graph goes through the points and

´ µ ¾ ½ µ

¾ ·½

¾ ,for , is the linear function whose domain is and whose graph goes through

´ µ ´ µ ½µ

·½ ·½ ½ ½

the points and ,and is the linear function whose domain is whose graph goes

´ µ ´ µ

½ ½

through the points and . Now (3) follows by taking and .

µ Æ

Theorem 22 È´ is true in the Mostowski linearly ordered model .

´ µ ¾ ¾ Æ

Proof. Assumethat is a collection of ordered pairs each of which is an compact F

Æ Æ

topological space in and let , where the superscript means that the product is taken in

¾Æ ´ µ

the model. Assume that is an inﬁnite subset of .Let be a support of the ordered pair .For

each ,let be the projection map. We ﬁrst claim that it sufﬁces to prove that

Æ ¾ ´ µ ½

(I) has a cluster point (in ) under the assumption that for all , .

(Note that is a cluster point of if every neighborhood of contains a point in different from .)

¾ ´ µ ½

To prove the claim, let ,andlet be the projection map.

¼ ¼

¾ ´ ´ µµ ½ ´ µ

Then for all , . It is also easy to see that is inﬁnite since is. Therefore, if we

´ µ ¾

assume (I), has a cluster point in , call it . If we then deﬁne by

¼ ¼

´µ ¾

if ,

´µ

´ µ

the unique element of otherwise, we can show that is a cluster point of .

For the remainder of the proof we shall assume

¾ ´ µ ½

(II) for all , ,

and we shall show that has a cluster point.

Math. Log. Quart. 49, No. 1 (2003) 69

¾ ¾ ×ÙÔÔ´µ ¶ ×ÙÔÔ´ µ

Lemma 4 If for any ,thereisa for which ,then has a cluster point.

¾ ´µ ¾

Proof. Assume satisﬁes the hypothesis, then ×ÙÔÔ´ is an inﬁnite, Dede-

kind ﬁnite subset of since it can be put in a one to one correspondence in the model with the set

´×ÙÔÔ ´µ Ò ×ÙÔÔ´ µµ ¾

´ µ

×ÙÔÔ . (This latter set is inﬁnite and Dedekind ﬁnite in the model by (2).)

´µ ´µ

By our assumption (II), there are two elements and of such that .Since is compact F, the set

´µ´µ ¾

has a complete accumulation point . We assume without loss of generality that

´µ ´µ

. We ﬁrst note that is in every neighborhood of . For, assume that is a neighborhood of in .

¼ ¼ ¼ ¼ ¼

´µ ¾

Then since is Dedekind ﬁnite and we have . Hence, .Deﬁne

if ,

¾ ´ µ ´µ

by Then and is in every neighborhood of since is in every

´ µ

otherwise.

´µ neighborhood of . Therefore, is an accumulation point of .

Using Lemma 4, we will assume for the remainder of the proof that

¾ ¾ ×ÙÔÔ ´µ ×ÙÔÔ´ µ

(III) for all and for all , .

¾ ×ÙÔÔ´ µ ¶

Lemma 5 If for some , ,then has a cluster point.

Proof. Assumethat ¼ is an element of whose support is not contained in . We may also assume

% ¾ ´µ ´ µ

that the function is one to one and therefore, for any , iff .Let

¼ and let be the set of open intervals determined by . That is (see (3) above),

Ï ´ ½ µ ´ µ ´ ½µ

¼ ½

¼ . For the proof of the lemma we will need to consider the orbits

¾ Ó#´µ´µ ¾ Ç# Ó#´µ ¾ ¾ Ç#

of . For each ,let and let . For each choose

so that

´ µ

(IV) for every open interval determined by , every element of supp ¼ is less than

´ µ

every element of supp .

% Æ ½

The function may not be in , however the set deﬁned by

´´ µ´ ´ µµµ ¾ ¾ Ç#

½ and

is in the model since it has support . We also make the following claims about ½ .

Claim A. ½ is a function.

´ ´ µ ´ ´ µµµ ´ ´ µ ´ ´ µµµ

½ ¼ ¾ ¾ ¼ ½ ½ ¾

To prove Claim A, assume that ½ and are in ,where and

´ µ ´ µ ´ µ

½ ¾ ¾

are in and and are in Ob and assume that .Then .Fromthiswemay

draw two conclusions: First that and are in the same orbit, hence, . Secondly (using (III)) that

½ ½

´ µ ´ ´ µµ ´ ´ µµ ´ ´ µµ ´ ´ µµ

¼ ½ ¼ ¾ ¼ ¾ ¼ ¼

¾ and therefore, . Claim A follows.

½ ½

Claim B. ½ .

$ÓÑ´ µ ´ µ ¾ ¾ Ç#

This claim follows from the facts that ½ and and, for each

´ µ ¾ $ÓÑ´ µ ´´ µµ ´ ´ µµ ¾

½ ½ ¼

´ µ

, .

Claim C. ½ is a cluster point of .

We ﬁrst note that proving Claim C will complete the proof of the lemma. To prove the claim, let beabasic

¼ ¼

neighborhood of ½ .Then , where for some ﬁnite subset of , for all . We will

show that contains an element ¾ of different from .

¾ ¾ ¾

For each ¼ ,let be the unique element of Ob such that and let be chosen so that

´ ¾ ¾Ï µ

. Choose an so that for every

´ µ

(V) for each of the intervals determined by and every element of supp ¼ in ,

´µ ¾ ×ÙÔÔ´ µ for every element .

70 O. De la Cruz, E. Hall, P. Howard, K. Keremedis, and J. E. Rubin: Products of compact spaces and AC II

´ µ ´ µ ´×ÙÔÔ ´ µµ ´ µ

¼ ¾ ½ ½ ¾ ¼ ¼

Let ¾ .Then as is a support of but supp since supp and

´ µ ¾ ¾

¼ ¼

ﬁxes . We also note that ¾ since , has support and ﬁxes pointwise. Therefore,

¾ ¾ ¼

to complete the proof of the claim, it sufﬁces to show that ¾ . We will do this by arguing that for all ,

´ µ ´ µ ¾ ´ ¾ µ

½ ¼

¾ . Assume that , then, using the notation introduced above, ,where

¾ ×ÙÔÔ´ µ ×ÙÔÔ´ µ

for . It follows that for each open interval determined by , .

×ÙÔÔ´ µ ×ÙÔÔ´ ´ µµ ´ µ ×ÙÔÔ´ µ

¼ ¼

Similarly ¼ . By (IV), supp and by (V),

´ ´ µµ ×ÙÔÔ ´ µ

supp ¼ , hence

µµ ´×ÙÔÔ ´ ´ µµ ×ÙÔÔ´ µµ ´×ÙÔÔ´ µ ×ÙÔÔ´

¼ ¼

×ÙÔÔ´ µ ×ÙÔÔ´ ×ÙÔÔ ´ ´ µµ ×ÙÔÔ´ µ µ

Therefore, letting ¼ and , we conclude by property (3)

Æ ¾ ´ µ

in our list of properties of that there is a such that .Since is order preserving (and

µµ ×ÙÔÔ´ µ ´×ÙÔÔ ´ µµ ×ÙÔÔ´ ´ µµ ´×ÙÔÔ ´

using (IV) and (V)) we conclude that ¼ and . Therefore,

µ ´ µ ´ µ ´

¼ and . Hence,

´ µ´ ´ µµ´ µ´ ´ µµ´ µ ´ ´ ´ µµµ ´ ´ µµ ´ µ

¾ ¼ ¼ ¼ ¼ ½

This completes the proof of Claim C and therefore, Lemma 5 is proved.

We therefore assume for the remainder of the proof that

¾ ×ÙÔÔ ´ µ

(VI) for all , .

Æ Æ

It follows from (III) that each is well orderable in and therefore, the power set of in (which we

È ´ µ ´È ´È ´ µµµ

denote by ) is the same as the power set of in the ground model. Similarly is the same

È ´È ´ µµ · Æ

as in the ground model. We denote the ground model (that is, the model of from which

Æ ´ µ Æ Æ

¼ ¼

was constructed) by ¼ . It follows that is an compact F topological space in . Therefore, since is

É É É

Æ Æ

¼ ¼

a model of AC, (taken in ) is compact G in .Since , has a cluster

¾ ¾ ¾

¾Æ

point in .Let be such a cluster point. If , then it is not hard to verify that is a cluster point of

Æ ¾Æ ¾

in in which case the proof is complete. Assume therefore, that . Then there is a permutation

¾ ´´ µµ ´ ´ µµ

and a such that .(Otherwise has support .) One consequence of this assumption is

´ µ ´ µ ´´ µµ ´ µ ´´ µµ ´ µ´ ´ µµ

that .For,if , then by (III), and, hence, .

¾ Ç#

Now, as in the proof of the previous lemma, we choose one representative from each but rather

´ µ

than require (IV) we require only that Ó . Again as in the proof of the previous lemma (see the deﬁnition

´´ µ´´ µµµ ¾ ¾ Ç# ¾

½ ½

of ½ )wedeﬁne and . The proof for is almost

¾ ½

identical to the proof that . We also note

¾ Ç# ´ µ´ µ

(VII) for all , ½ .

´µ ´ µ

½ if ,

¾ ´µ ¾Æ

¾ ¾ ¾ ½

We deﬁne by It is clear that since is. We will show

´µ ´ µ

if .

that ¾ is a cluster point of .

First note that

´ ´ µµ ´ ´ µµ ´ µ ½

¾ since ,

´ ´ µµ ´´ µµ

´ µ

since Ó by (VII),

´´ µµ ´ ´ µµ ´ ´ µµ

´ ´ µµ ´ ´ µµ ¾ ´ µ ¶ ¾

¾ ¾ ¾

so ¾ , and since it follows that supp .By(VI) . Therefore, to show

that ¾ is a cluster point of , it sufﬁces to show that every neighborhood of contains an element of .

Let be a neighborhood of ¾ ,then , where for some ﬁnite subset of , for

¾ ¾ ¾ ´ µ ¾ Ç#

¼ ¼

´ µ

all . For each , choose so that Ó . For each ,let

´ µ ¾ ´ µ

and

Math. Log. Quart. 49, No. 1 (2003) 71

´ µ ´ µ ¾ ´ µ ´ µ

¾ ½ ½

Since is a neighborhood of for all such that and since ﬁxes , is

´ µ ´ µ

a neighborhood of ½ . Further, since for all but ﬁnitely many , for all but ﬁnitely

´ µ ´ µ

½ ¾

many in . Thus is a neighborhood of .Let ,where

¾ ´ µ

Ó´µ ´µ

Ó if , ,and ,

´ µ

if ,

otherwise.

Then is a neighborhood of and therefore there is an element of such that . We complete the

proof by showing that .

´ µ ¾ ¾ ´µ ¾ ¾

¼ ¼

It follows from the choice of that and we also have for , . Suppose

½ ½ ½

´ µ ´ µ ¾ ´ µ ´ ´ µµ ¾ ´ µ´ ´ µµ ¾

µ Ó´µ Ó´µ Ó´µ

and .Then Ó´ .So .Thatis, .

½ ½

´ µ ´ µ ´µ ¾

´µ

But by (VI), and by the deﬁnition of , Ó . Therefore, and, hence,

is in .

References

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