PRODUCTS OF COMPACT SPACES AND THE AXIOM OF CHOICE II

PRODUCTS OF COMPACT SPACES

AND THE AXIOM OF CHOICE II

Omar De la Cruz, Eric J. Hall, Paul Howard,

Kyriakos Keremedis, and Jean E. Rubin

March 4, 2002

Abstract. This is a continuation of [dhhkr]. We study the Tychonoff Compactness Theorem

for various definitions of compact and for various types of spaces, (first and second count-

able spaces, Hausdorff spaces, and subspaces of Rκ). We also study well ordered Tychonoffproducts 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 [dhhkr], where westudied the Theorem using the definitions of compactness studied in [dhhrs]. In this paperwe extend these results to various types of topological spaces, first countable, secondcountable, Hausdorff spaces and subsets of Rκ. We also consider the role the multiplechoice axiom plays when the products of topological spaces are well ordered.

It is well known (see [ty] or [k]) that the Tychonoff Compactness Theorem, “Products ofcompact spaces are compact” is equivalent to the Axiom of Choice (AC). In the last sectionwe prove that this equivalence does not hold when “compact” in Tychonoff’s Theorem isreplaced by certain weaker forms of compactness. We construct a Fraenkel-Mostowskimodel in which one of the product theorems is true, but AC is false.

We start by defining some of the terms we shall be using. Let S = (X, T ) be a topologicalspace.

1. B ⊆ T is called a base for T if for each x ∈ X and each neighborhood U of x there is aV ∈ B such that x ∈ V ⊆ U . (Thus, every element of T is a union of elements of B.)

2. S is first countable if each point has a countable base.

3. S is second countable if T has a countable base.

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

1991 Mathematics Subject Classification. 03E25, 04A25, 54D30.Key words and phrases. First Countable, Second Countable, Hausdorff, Products, Compactness, Choice,

Multiple Choice.

2 DE LA CRUZ, HALL, HOWARD, KEREMEDIS, AND RUBIN

5. S is said to be T0 if whenever x and y are distinct points there is a neighborhood of onethat does not contain the other.

6. S is said to be T1 if whenever x and y are distinct points, each has a neighborhood notcontaining the other. (Points are closed.)

7. S is said to be T2 (Hausdorff) if whenever x and y are distinct points, there is a neigh-borhood U of x and a neighborhood V of y such that U ∩ V = ∅.Clearly, every second countable space is first countable; every T2 space is T1; and every

T1 space is T0.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.

9. ACWO 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(ℵ0) is the axiom of choice for a countable number of sets.

12. ACfin(ℵ0) 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 X of non-empty sets,there is a function f such that for each x ∈ X , f(x) is a non-empty finite subset of X .

14. BPI is the Boolean Prime Ideal Theorem: Every Boolean algebra has a prime ideal.(The reader is referred to [hr] 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 ZF0, ZF without the foundation axiom. See [le] and [tr]. Unless otherwisestated, all proofs are in ZF0.

The types of compactness we will be studying are given in [dhhkr], but we shall repeatthose 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 {Xi : i ∈ k} be a set of topological spaces and let X =∏i∈kXi

be their product. A base for the product topology on X is all sets of the form∏i∈k Ui

where Ui is open in Xi and all but a finite number of the Ui = Xi.

PRODUCTS OF COMPACT SPACES II 3

P(X,Y) is the statement: Products of compact X spaces are compact Y. Pℵ0(X,Y) is

the statement: Countable products of compact X spaces are compact Y. Pr(X) is thestatement: 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 spacesof a certain type will have the same type. It is easy to show that a product of Ti spacesis Ti, for i = 0, 1, 2, but it is known, for example, that the product of two Lindelof spacesneed not be Lindelof. (Sorgenfrey line: R, where a base for the topology is the set of allhalf open intervals [a, b).)

We shall study two types of product theorems. Let P1t (X, Y ) be the following statement:

Every product of type t compact X spaces is compact Y , provided that the product itselfis of type t. The same statement but without the requirement that the product is of typet will be called P2

t (X, Y ). Clearly, P2t (X, Y ) implies P1

t (X, Y ), for all X, Y . If t is someproperty such as Hausdorff that is preserved under arbitrary products, then P1

t and P2t are

identical and the superscript will be omitted.

The following results for general topological spaces were obtained in [dhhkr]:(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)→ ACWO

(4) AC → P (F,A)→ P (F,D)→ P (F,H)→ ACWO

(5) AC → P (B,B)→ P (C,B)→ P (A,B)← P (D,B)← AC

It is easy to show that properties (1) through (5) also hold for P2t (X, Y ), where t is first

or second countable. However, in the proof of Theorem 1 in [dhhkr] that each of P(A,G)and P(A,H) implies AC, the topology used is not even T0. However, we show in section 4,that it is easy to extend the topology so that it is T1. In addition, it is shown in [lr2] thatPT2

(A,A) ↔ BPI and it is shown in [he] that PT2(B,B) is provable in ZF. We show in the

introduction that if BPI holds, then compact D implies compact G. Therefore, PT2(A,A)

+ PT2(D,D) → PT2

(D,G).


In [dhhrs], section 6, we give some relationships between the forms of compactness ifBPI holds. Here, we note that we can add to those relationships by showing that compactD implies compact G. (BPI implies that every set can be linearly ordered. Suppose Xis compact D and Y is an infinite subset of X that has no accumulation points. Thenevery subset of Y is closed. Let N = {Uy = X r {z ∈ Y : z < y} : z ∈ y} , where <is a linear ordering of Y which has no minimum element. Then N is a nest of open setswhich covers Y , but has no finite subcover.) Thus, the arrow diagram which gives therelationship between the forms when BPI holds can be changed to:

A, B, C −→D −→ H

↓F −→G

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 Pℵ0

fc (X, Y ) to stand for the statement “Countable products of first

countable compact X spaces are compact Y .” Likewise, we have Pℵ0sc (X, Y ) about second

countable spaces. Note that first and second countability are preserved under countableproducts.

It follows from [dhhrs] that the following are the arrow diagrams for first countablespaces.

E ←−−−− H ←−−−− D ←−−−− A −−−−→ C −−−−→ ByF −−−−→ G

Assuming AC,

A, B, C, D, F −→ E, H −→ G

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

F −−−−→ G ←−−−− E ←−−−− A,D,H −−−−→ C −−−−→ B

Consequently, we immediately have that the statements P1sc(X, Y ) for X, Y = A, D, or H,

are equivalent.With ZFC:

A,B,C,D,E, F,H −−−−→ G

(In addition in ZFC, if the space is second countable and T1, then all the forms of com-pactness are equivalent. See [dhhrs].)


In a topological space X , let us refer to a point that has only one neighborhood (whichmust be X) as a trivial point. A trivially compact topological space is space that has atrivial point (so a space X is trivially compact if every open cover contains X). A spaceX is indiscrete if its topology is {∅, X}. It is provable in ZFC that a product,

∏i∈I Xi, of

first countable spaces is first countable iff each Xi is first countable and all but countablymany of the Xi’s are indiscrete spaces, and that a product of T2 spaces is second countableiff each factor is second countable and all but countably many factors are one-point spaces,[w]. In ZF, we show in the next lemma that uncountable products are not first countableunless all but κ of the factors are trivially compact, where κ is a countable union of finitesets.

Lemma 1. Let Xi, i ∈ I be topological spaces such that∏i∈I Xi is first countable and

non-empty.

(a) Xi is first countable, for all i ∈ I.

(b) For each s ∈∏i∈I Xi, the set Js := {i ∈ I : πi(s) is not a trivial point in Xi} is a

countable union of finite sets.

Proof. Part (a) is clear.

For (b), let s ∈∏i∈I Xi, and let Bs = {un : n ∈ ω} be a neighborhood base for s. For

each i ∈ Js, define f(i) as the least n ∈ ω such that

un ⊆ π−1i v

for some neighborhood v of πi(s) in Xi such that v 6= Xi.

Given n ∈ ω, since un is an open set in∏i∈I Xi, it contains a basic neighborhood of the

form π−1i0vi0 ∩· · ·∩π−1

ikvik , where vi0 , . . . , vik are open sets in Xi0 , . . . , Xik , respectively. In

this case, f−1(n) ⊆ {i0, . . . , ik}. This means that f defines a partition of Js into countablymany finite pieces. �

Corollary 1. ACfin(ℵ0) implies that if∏i∈I Xi is first countable, then for each s ∈∏

i∈I Xi, the set Js (as in Lemma 1(b)) is countable.

Proof. ACfin(ℵ0) implies that countable unions of finite sets are countable. �

The following can be checked case by case:

Lemma 2. If S is a compact X space, and T is a trivially space, then S × T is compactX, where X is A, B, C, D, E, G, H. �

Notice that the previous lemma does not apply when X is F. For example, suppose Tis an infinite but Dedekind finite set given the indiscrete topology, and let S = {0, 1} withthe discrete topology. Then, S×T is an infinite set with no complete accumulation point,since a neighborhood of the form {i} × T does not have the same cardinality as S × T .


Theorem 1. Let X,Y be definitions of compactness, with Y different from F. If ACfin(ℵ0)

holds, then Pℵ0

fc (X, Y )→ P1fc(X, Y ) and Pℵ0

sc (X, Y )→ P1sc(X, Y )

Proof. We give the proof of Pℵ0

fc (X, Y )→ P1fc(X, Y ). The proof of Pℵ0

sc (X, Y )→ P1sc(X, Y )

is similar.Let Xi, for i ∈ I, be first countable compact X spaces such that

∏i∈I Xi is first count-

able. If∏i∈I Xi is empty, then it is immediately compact Y; otherwise, fix s ∈

∏i∈I Xi.

We have that ∏i∈I

Xi ≡∏i∈Js

Xi ×∏

i∈IrJs

Xi,

where Js = {i ∈ I : πi(s) is not a trivial point in Xi} and is countable by Corollary 1.Now

∏i∈IrJs Xi is trivially compact, since the projection of s into that space is a trivial

point there. And since Pℵ0

fc (X, Y ) implies that∏i∈Js Xi is compact Y, by Lemma 2 we

conclude that∏i∈I Xi is compact Y. �

The fact that ACfin(ℵ0) is needed for the results above is illustrated by the following

theorem. However, some statements of the form Pℵ0

fc (X, Y ) might automatically implyACfin(ℵ0), and make it redundant for some of the instances of Theorem 1.

Theorem 2. P1fc(X, Y ) cannot be proved in ZF, where X is any of the definitions of

compactness and Y is A, D, F, G, or H. Consequently, P2fc(X, Y ) cannot be proved in ZF

either, for the same values of X and Y.

Proof. Consider the second Fraenkel model (N 2 in [hr]), in which there is a set A whichhas a countable partition into pairs with no choice function. In this model we have that2A is a product of two-element spaces which is first countable, but not compact A, D, F,G, or H (see [dhhrs]). �

Note that it does not follow immediately from the definitions that P1fc(X, Y )→ P1

sc(X, Y ).The next theorem proves a result of this form.

Theorem 3. P1fc(A,A)→ P1

sc(A,A) and P1sc(A,A)↔ AC(ℵ0).

Proof. It is proved in [hkrs] that Pℵ0sc (A,A) ↔ AC(ℵ0), and thus both Pℵ0

sc (A,A) and

Pℵ0

fc (A,A) imply ACfin(ℵ0). It then follows from Theorem 1 that Pℵ0

fc (A,A) ↔ P1fc(A,A).

Likewise, it follows from Theorem 1 that Pℵ0sc (A,A)↔ P1

sc(A,A). The theorem follows easily

since Pℵ0

fc (A,A) clearly implies Pℵ0sc (A,A). �

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

Theorem 4.(a) P2

sc(X,E) is false in ZF for all definitions X of compactness.(b) P1

sc(G, Y ) is false in ZF for Y ∈ {A, B, C, D, E, F, H}.(c) P1

fc(X, Y ) is false in ZFC for X ∈ {E,G,H} and Y ∈ {A, B, C, D, F}.Proof. The two-element space 2 with the discrete topology is compact X for X = A, B, C,D, E, F, G, and H, while 22ω is not compact E (see [dhhrs]). This proves (a). Parts (b) and


(c) are consequences of counterexamples that are not even products (again see [dhhrs]).For part (b) note that the topology T = {(x,∞) : x ∈ Z} on the integers witnesses inZF that a second countable space can be compact G but not compact in any other sense.To prove (c) consider ω1 with the order topology. Assuming ω1 is regular, it is a firstcountable compact E, G, and H space that is not compact A, B, C, D, or F. �

From independence results found in [dhhrs] (Theorem 3(b) and Lemma 1(c)), it im-mediately follows that the following statements are not provable in ZF: P2

sc(B,X) forX ∈ {A,C,D,E, F,G,H} and P2

sc(X,F ) for X ∈ {A,B,C,D,E,G}.

Theorem 5. P2sc(A,C) implies AC.

Proof. We will refine the proof of P(A,A) → AC given in [rr].Assume P2

sc(A,C), and let X = {Xi : i ∈ I} be a collection of non-empty sets; wewill prove that X has a choice function. For each i ∈ I define the space 〈Yi, Ti〉 byYi = Xi ∪ {∞}, where ∞ stands for a fixed element not in

⋃X , and Ti = {∅, Yi, {∞}}.

These spaces are compact A (actually trivially compact).Let P =

∏i∈I Yi; clearly P is not empty. For every finite E ⊂ I, define PE =∏

i∈IrE Yi ×∏i∈E{∞}; every non-empty basic neighborhood in P is of the form PE

for some finite E ⊂ I. We have the following:Claim 1: s ∈ P is a trivial point in P if and only if s is a choice function for X . Indeed,a basic neighborhood PE is different from P iff E 6= ∅, and PE contains s iff s(i) /∈ Xi fori ∈ E.Claim 2: Every finite open cover of P contains P as an element. To prove this we willshow that if S, S′ 6= P are open sets, then S ∪ S′ 6= P ; an easy inductive argument canthen complete the proof.

Let S =⋃{PE : E ∈ E} and S′ =

⋃{PE : E ∈ E ′}, where E , E ′ are sets of finite subsets

of I. Take s, s′ ∈ P such that s /∈ S and s′ /∈ S′. Then for each E ∈ E there exists i ∈ Esuch that s(i) ∈ Xi, and for each E ∈ E ′ there exists i ∈ E such that s′(i) ∈ Xi. It is clearthat t ∈ P , defined by

t(i) = s(i), if s(i) ∈ Xi

t(i) = s′(i) otherwise

is neither in S nor in S′. This finishes the proof of Claim 2.Finally, in order to reach a contradiction, assume that X has no choice function; by

Claim 1, P has no trivial points. If B is a subbase for the product topology on P thenevery point has a neighborhood in B which is not P . This way, {u ∈ B : u 6= P} is an opencover of P with elements in B. By P2

sc(A,C), this cover has a finite subcover, but that isimpossible by Claim 2. �

Thus, it follows that each of P2sc(X, Y ), for X, Y = A or C, is equivalent to AC. Also,

it follows from [he] Theorem 4.2, that P2sc(F,F) is equivalent to AC. In addition, Theorem

1 of [dhhkr] holds for second countable spaces: Each of P2sc(A,G) and P2

sc(A,H) imply AC.


Consequently, (1) through (5) in the introduction also hold for P2sc(X, Y ). (In addition,

since P2fc(X, Y ) implies P2

sc(X, Y ), (1) through (5) hold for P2fc(X, Y ).)

Let Uω be the statement: Every non-principal filter on ω can be extended to an ultra-filter. Recall that Pr(A) is the statement: Projections in Tychonoff products of compactA spaces are closed. We close this section with a result showing that some weak forms ofAC imply Pr(A).

Theorem 6. Uω + AC(ℵ0) implies Pr(A) for first countable T2 spaces.

Proof. Let {(Xi, Ti) : i ∈ k} be a family of first countable compact T2 spaces and let (X, T )be their Tychonoff product. If X = ∅ then there is nothing to show. Let C be a closedset in X and i ∈ k. If πi(C) (the projection of C on its ith coordinate), is not closed thenπi(C) is an infinite subset of Xi and Xi \ πi(C) contains limit points from πi(C). Let x besuch a limit point in πi(C) and, by AC(ℵ0), let {xn : n ∈ ω} be a countably infinite subsetof πi(C) converging to x. By AC(ℵ0) again, we choose for each n ∈ ω, yn ∈ π−1

i (xn) ∩ C.Let F be a non-principal ultrafilter on {yn : n ∈ ω} extending the filterbase

{π−1i (V ) ∩ {yn : n ∈ ω} : V is a neighborhood of x}.

Using compactness and the fact that the spaces are T2, we can readily verify that for everyj ∈ k, j 6= i, there exists a unique xj ∈ Xj such that

{π−1j (V ) ∩ {yn : n ∈ ω} : V is a neighborhood of xj} ⊂ F .

Clearly, the element y ∈ X , πi(y) = x and πj(y) = xj for j 6= i is in the closure of C,hence, in C. On the other hand, since x /∈ πi(C), y /∈ C. This contradiction finishes theproof of the theorem. �

3 Hausdorff Spaces

Properties (1)-(5) in the introduction do hold for T1 spaces because Theorem 1 of[dhhkr] (Each of P(A,G) and P(A,H) implies AC) holds for T1 spaces, (add the set ofcofinite subsets to the topology on each Yi). However, these properties do not necessarilyhold for T2 or Hausdorff spaces. We shall add a diagram to indicate the relationships thatare known to hold for all spaces. First we have the relationships between the definitionsof compact.

H ←−−−− D ←−−−− A −−−−→ C −−−−→ ByF −−−−→ G

It is known that PT2(A,A) ↔ BPI. (See Theorem 7 below.) Also, it is known (see,

for example, [dhhrs]) that BPI implies that compact A, B, and C are equivalent. Thus,


PT2(A,A) implies PT2

(C,C). In addition, using the preceding diagram and that if X → Y ,then P(Z,X) → P(Z, Y ) and P(Y, Z) → P(X,Z), we obtain the following diagram (alsonote that products of T2 spaces are always T2):

PT2(D,H) ←−−−− PT2

(D,D)y yPT2(A,H) ←−−−− PT2(A,D) ←−−−− PT2(A,A) −−−−→ PT2(C,C) −−−−→ PT2(A,C)y

PT2(A,G)

It has been shown in [he] that PT2(B,B) is provable in ZF. Since PT2

(B,B) impliesPT2

(C,B) implies PT2(A,B), it follows that PT2

(C,B) and PT2(A,B) are also provable in ZF.

Also, PT2(X,E) is always false (a counterexample is 2R).

It is also shown in [he] that AC ↔ PT2(F,F). Since F implies G, it is also true that:

PT2(F, F ) −→ PT2

(F,G)

We show in section 6 that PT2(F,G) does not imply AC.

The next collection of results shows that, in contrast to (1) in the introduction, none ofthe arrows in the sequence of implications

AC→ PT2(A,A)→ PT2

(A,D)→ PT2(A,H)

are reversible.

Theorem 7. ( Los, Ryll-Nardzewski [lr2]) PT2(A,A) ↔ BPI.

Lemma 3. Assuming AC(LO), a T2 space X is compact D iff every well-ordered infinitesubset of X has a complete accumulation point.

Note that Lemma 3 is not very interesting in ZF, since ZF + AC(LO) implies AC.However, ZF0 + AC(LO) does not imply AC (see [tr]).

Proof. Suppose X is not compact D, so there is a covering nest {Ui : i ∈ I} of open setsthat does not have X as a member. Under AC(LO), every linearly ordered set can be well-ordered, so choose a well-ordering ≤ of I (which is linearly ordered by the subset relationon the Ui’s). Now we obtain a well-ordered nest by induction: For each ordinal α, letWα = Uj , where j is the ≤-least member of I such that Uj properly contains

⋃β<αWβ, if

such Wα exists. By throwing away all but some cofinal subset if necessary, we may assumethat the Wα’s are defined just for all α in some cardinal κ. Now using AC(LO) again, letxα ∈ Wα+1 rWα for each α ∈ κ. Assuming X is Hausdorff, this family of xα’s will haveno complete accumulation point.


Conversely, assume that (X, T ) is a compact D T2 space. We will show that everywell-ordered subset Y = {yj : j ∈ ℵ} has a complete accumulation point. We consider twocases:

(i) ℵ is a regular cardinal. For every j ∈ ℵ put Uj = {O ∈ T : O∩Y ⊂ {yi : i ∈ j}} andlet U = {

⋃Uj : j ∈ ℵ}. Assuming that Y has no c.a.p. (complete accumulation point) and

taking into consideration that every point x ∈ X has a neighborhood Ox meeting Y in aset of size < ℵ, we see that U is a well ordered open cover of X . Hence, for some j ∈ ℵ,X =

⋃Uj which is a contradiction.

(ii) ℵ is a singular cardinal. Let k = cf(ℵ) (the cofinality of k) and fix {kn : n ∈ k}a cofinal set of regular cardinals of ℵ. For every n ∈ k let Bn = {x ∈ X : x is a c.a.p.of {yi : i ∈ kn}}. By part (i) each Bn is non-empty. Let G = {gn : n ∈ k} be a choiceset of B = {Bn : n ∈ k}. Since k is a regular cardinal it follows by part (i) again, that Ghas a c.a.p. g, and this g must be a c.a.p. of Y . Indeed, if Vg is a neighborhood of g then|Vg ∩ {yi : kn ∈ i ∈ kn+1}| = kn+1 for all n ∈ k. Thus,

|Vg ∩ {yi : i ∈ ℵ}| =∑|Vg ∩ {yi : kn ∈ i ∈ kn+1}| =

∑kn = ℵ.

The equalities in the above display are justified because of AC(LO). �

As before, Uω is the statement: Every non-principal filter on ω can be extended toan ultrafilter. Similarly, let UWO 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 inpermutation models, where AC holds in the subuniverse of pure sets.

Theorem 8.

(a) Uω + AC(ℵ0) → PT2(A,H)

(b) UWO + AC(LO)→ PT2(A,D)

Proof. (a) Let {(Xi, Ti) : i ∈ k} be a family of T2 compact A spaces and let X be theirTychonoff product. In view of AC(ℵ0) it suffices to show that every countably infinitesubset S = {gn : n ∈ ω} of X has a limit point. Choose, by Uω, a non-trivial ultrafilter Fon S. It is straightforward to verify that for each j ∈ k there is a unique bj ∈ Xj such that

{π−1j (Vbj )∩S : Vbj is a neighborhood of bj} ⊂ F . (The existence follows from compactness

and the fact that F is an ultrafilter, and the uniqueness follows from T2.) Now it is clearthat the element b ∈ X given by b(i) = bi is a limit point of S finishing the proof of thetheorem.

(b) Choose a family of compact A Hausdorff spaces. By AC(LO) and Lemma 3, itsuffices to show that every well-ordered infinite subset S of their product has a limit point.If we choose, by UWO, an ultrafilter F on S such that each subset of S with cardinalitysmaller than that of S is not in F , then the proof proceeds just as in the proof of part(a). �


Corollary. PT2(A,D) does not imply PT2

(A,A).

Proof. The corollary follows from Theorem 7 and Theorem 8(b), since in ZF0, Uω +AC(LO) does not imply BPI.

(Construct a permutation model as follows: Let the set of atoms, A, be uncountable.The group of permutations, G, is the group of all permutations on A, and supports arecountable. (See N 12 in [hr].) BPI is false because BPI implies every set of pairs has achoice function, but the set of pairs of atoms in this model has no choice function. Trussshowed in [tr] that AC(LO) holds in this model.) �

Theorem 9. PT2(A,H) does not imply PT2

(A,D).

Proof. We first give the description of a permutation model N . The set of atoms A =⋃{Ai : i ∈ ω1}, where Ai = {aix : x ∈ C} and C is the set of points on the unit circle

centered at 0. The group of permutations G is the group of all permutations on A whichrotate the Ai’s by an angle θi ∈ R and supports are countable.

The following 2 claims are straightforward:

Claim 1. The family A = {Ai : i ∈ ω1} does not have a multiple choice function in N .

Claim 2. AC(ℵ0) is true in N .

Since Uω and AC(ℵ0) are true in N , it follows by Theorem 8(a) that PT2(A,H) also

holds in N . Furthermore, PT2(A,D) fails in N . Indeed, for every i ∈ ω1 let Xi = Ai ∪ {∗i}

carry the disjoint topology Ti induced from the (compact A) metric topology of Ai (Aiis identified with the unit circle) and the discrete topology on {∗i}. Clearly each Xi is acompact A T2 space. Let X be the Tychonoff product of the family {(Xi, Ti) : i ∈ ω1}.Let Ui = ∪{π−1

j ({∗i}) : j ∈ i}. Since AC(ℵ0) holds in N , it follows that no Ui covers X .

Since A has no choice set, we see that U = {Ui : i ∈ ω1} is a nested open cover of X whichhas no finite subcover. �

This completes the proof that none of the implications

AC→ PT2(A,A)→ PT2

(A,D)→ PT2(A,H)

are reversible.

Let U be the statement: Every infinite set has a non-trivial ultrafilter. It is known thatU is strictly weaker than BPI, so the next result shows that the implication PT2

(A,A) →PT2

(A,G) is not reversible in ZF.

Theorem 10. U implies PT2(A,G).

Proof. Let {(Xi, Ti) : i ∈ k} be a family of compact A, T2 spaces and let X be theirTychonoff product. Let S be an infinite subset of X and let F be a non-trivial ultrafilteron X . 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 compactA spaces. We shall use the notation PWO(X,Y) to mean that a well ordered product ofcompact X spaces is compact Y.

Theorem 11. PWO(A,D) if and only if PWO(A,A).

Proof. It suffices to show (→) as the other implication is clear. Fix {(Xn, Tn) : n ∈ ℵ}a well ordered family of compact A spaces and let (X, T ) be their Tychonoff product.Without loss of generality we may assume that for every v ∈ ℵ, Yv =

∏n∈vXn is compact

A. Let πv denote the projection of X to Xv. Let G be a family of closed subsets of X havingthe finite intersection property. Let Q = {π−1

v (Qv) : v ∈ ℵ}, Qv =⋂{πv(G) : G ∈ G}. It

can be readily verified that Q is a family of closed sets of X having the finite intersectionproperty. Since for every v ∈ ℵ, Yv is compact A, it follows thatW =

⋂{π−1

v (Qv) : v ∈ ℵ}is a nest of non-empty closed subsets of X . Since X is compact D, it follows that

⋂W 6= ∅.

Clearly, any point x ∈⋂W is in

⋂G and, consequently, X 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.

Proof. Let (X, T ) be a compact D space and U = {Ui : i ∈ k} be an open cover of X .Using Levy’s Lemma (form [67 B] in [hr]: Each set can be covered by a well ordered familyof finite sets) we express k as

⋃{kj : j ∈ ℵ}, where each kj is a finite subset of k and ℵ

is a well ordered cardinal. For every j ∈ ℵ put Oj =⋃{Ui : i ∈ kj}. Let m be the least

cardinal number for which there exists a subfamily F of {Oj : j ∈ ℵ} covering X . Weshow that m is finite. Assume on the contrary that m is infinite. For every j ∈ m putQj =

⋃{Oi : i ∈ j}. Since X is compact D, it follows that some Qj covers X . Thus,

{Oi : i ∈ j} is a subfamily of {Oj : j ∈ ℵ} of cardinality < m covering X . This is acontradiction. Hence, U has a finite subcover and X is compact A as required. �Theorem 13. MC + PWO(A,A) if and only if AC.

Proof. It suffices to show that MC + PWO(A,A) implies P(A,A), as P(A,A) is equivalentto AC and the implication in the other direction is clear.

Let {Xi : i ∈ k} be a family of compact A spaces and let X =∏i∈kXi be the Tychonoff

product. Use Levy’s Lemma again (see the proof of Theorem 12) to express k as a wellordered union of finite disjoint subsets of k. That is, k =

⋃{kv : v ∈ ℵ}. For every v ∈ ℵ,

let Qv =∏j∈kv Yj . Clearly Qv is compact A. The product Q =

∏v∈ℵQv is a compact A

space by PWO(A,A), and Q is homeomorphic to X .

It now follows from Theorems 12 and 13 that

Corollary. MC + PWO(D,D) if and only if AC.

It follows from Theorem 13 and its Corollary that neither PWO(A,A) nor PWO(D,D) isprovable in ZF0.


5 Products of Subsets of Rκ and Choice for Closed Sets

In this section we study products of subsets of Rκ; the results are different for wellorderable products and general products.

Theorem 14. Let {Xi : i ∈ I} be a collection of compact C topological spaces such that{c ⊂ Xi : i ∈ I, c 6= ∅, c is closed} has a choice function. Assume also that there is asequence {Si : i ∈ I} such that for each i ∈ I, Si is a subbase for Xi that witnesses theC-compactness of Xi. Then

∏i∈I Xi is compact C.

Proof. Consider the collection T = {Ui,s : i ∈ I, s ∈ Si} of open sets in∏i∈I Xi, where

Ui,s =∏j∈I Nj with Ni = s and Nj = Xj for j 6= i. It is easy to see that the set of

finite intersections of elements of T is a base for the Tychonoff topology of∏i∈I Xi, and

therefore T is a subbase for that topology.

Suppose now that C ⊂ T is a cover for∏i∈I Xi. Then there exists i0 ∈ I such that⋃

{s : Ui0,s ∈ C} = Xi0 ; otherwise, we have that for every i ∈ I, Di = Xir⋃{s : Ui,s ∈ C}

is a non-empty closed subset of X , and by hypothesis there is an element in∏i∈I Di, which

cannot be covered by any element of C. Since {s : Ui0,s ∈ C} is a cover of Xi by elementsof Si, it has a finite subcover F . Clearly, {Ui0,s : s ∈ F} is a finite subcover of C. �

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

For well orderable products, we have:

Theorem 15. Let {Xα : α < λ} be a collection of compact A topological spaces such that{c ⊂ Xα : α < λ, c 6= ∅, c is closed} has a choice function. Then

∏α<λXα is compact A.

Proof. Let C be a choice function for {c ⊂ Xα : α < λ, c 6= ∅, c is closed}. To see thatX =

∏α<λXα is compact A, let F be a family of closed subsets of X with the finite

intersection property (fA .i.p.); we will show that⋂F 6= ∅.

By simultaneous recursion on α we define a family {Fα : α < λ} of filters on X , and asequence 〈xα ∈ Xα : α < λ〉, satisfying:

(i) F0 is the filter generated by F .

(ii) Fα ⊂ Fβ for all α < β < λ.

(iii) For all β < λ and all α < β, π−1α (u) ∈ Fβ , for every neighborhood u of xα.

Once this construction is completed, the filter G generated by⋃{Fα : α < λ} will converge

to the point x = 〈xα ∈ Xα : α < λ〉 ∈ X . Then every neighborhood v of x intersects everyelement of F ; since each element of F is closed, x ∈

⋂F .

F0 is given by (i). Assuming that the filter Fδ is defined, we take xδ = C(⋂{πδ(y) : y ∈

Fδ}) (we have that⋂{πδ(y) : y ∈ Fδ} 6= ∅ because each Xα is compact A). We also define

Fδ+1 as the filter generated by the collection Fδ ∪ {π−1δ (v) : v is a neighborhood of x}. If

γ is a limit ordinal and Fδ is defined for all δ < γ, we define Fγ as the filter generated by⋃{Fδ : δ < γ}. This finishes the recursion; clearly conditions (ii) and (iii) are satisfied. �


Corollary. (a) Any product∏i∈I Xi of closed compact C sets of reals, with a witnessing

family of subbases {Si : i ∈ I}, is compact C.(b) Well ordered products of compact A sets of reals are compact A.(c) More generally, any product of compact C linearly ordered spaces with the order

topology (and a family of witnessing subbases) which are conditionally complete (any subsetwith upper bounds has a least upper bound) is compact C. If the product is well orderableand the spaces are compact A (equivalently, if they have a maximum and a minimumelement), then the product is compact A.

Proof. Part (c) implies both parts (a) and (b). For (c), if the product is empty, then it iscompact A. Assume then that the product is not empty; because of the previous theorems,we only need to prove that the collection of all closed subsets from all the factor spaceshas a choice function.

Fix one element x in the product; we will give a rule for choosing an element from agiven closed subset c of the i-th factor space using only the element xi. If xi is an upperbound for c, we pick the least upper bound of c. If not, we pick the greatest lower boundof (xi,∞) ∩ c. In either case the element we picked is in c, since c is closed. �

Notice that arbitrary products of compact A sets of reals are not necessarily compactA: consider the space 2A in the model N 2 in [hr], where A is the set of atoms.

Theorem 16. If κ is a well-orderable cardinal then the collection

{c ⊂ Rκ : c 6= ∅, c is compact A}

has a choice function.

Proof. Let c be a fixed compact A subset of Rκ. We will describe a rule for choosing anelement of c.

For each α < κ, let cα be the projection by πα of c on R. Since projections arecontinuous, cα is compact A. Therefore, cα ⊂ R is bounded. Let rα = inf(cα) and sα =sup(cα). Then Y =

∏α<κ[rα, sα] contains c, and Y is compact A by the previous corollary.

Also, for each α < κ, let Bα be the restriction to [rα, sα] of the base B for the topologyof R which contains all open intervals with rational endpoints. Clearly, Bα is a countablebase for [rα, sα].

Let G be a subfamily of {c}∪ {π−1α (b) : α < κ, b ∈ Bα} which contains c and is maximal

with respect to the f.i.p.; such a family can be easily constructed by transfinite induction,since {c} ∪ {π−1

α (b) : α < κ, b ∈ Bα} has cardinality κ.Let F be the filter generated by G, and for each α < κ, define

Aα = {x ∈ [rα, sα] : π−1α (b) ∈ F , for every neighborhood b ∈ Bα of x}.

We have that Aα 6= ∅. Since [rα, sα] is compact A, the filter Fα generated by the setsb ∈ Bα such that π−1

α (b) ∈ F has an accumulation point x. If b ∈ Bα is a neighborhood


of x, then it has a non-empty intersection with every member of Fα; consequently π−1α (b)

meets every member of F and therefore π−1α (b) ∈ F (by maximality).

Furthermore, Aα is a singleton. Otherwise, if b1, b2 ∈ Bα are disjoint neighborhoodsof two points in Aα, we would have π−1

α (b1), π−1α (b1) ∈ F , which is not possible because

these sets are disjoint. Now, for each α < κ, let xα be the unique element of Aα, and letxc = 〈xα : α < κ〉 ∈ Y . Then every neighborhood of xc in Y contains a neighborhoodwhich is a finite intersection of sets of the form π−1

α (b1), which are in F . Therefore, everyneighborhood of xc intersects c; since c is closed, then xc ∈ c. �

Notice that in the proof we used no arbitrary parameters. Therefore we can uniformlychoose elements from the compact A subsets of many different spaces Rκi at the sametime.

Corollary. Arbitrary products of compact A spaces which are a subspace of Rκ for somewell-orderable cardinal κ, are compact C. Also, if the index set is well orderable, then theproduct is compact A.

Proof. Let {Xi : i ∈ I} be a family of compact A spaces, such that for each i ∈ I, Xi is asubspace of Rκi . Then for each i ∈ I, the collection of closed subsets of Xi is contained inthe collection of compact A subsets of Rκi . Therefore, by the previous theorem, we have achoice function for the closed subsets of all the factor spaces. Applying Theorems 14 and15 we obtain the desired conclusions. �

We turn now to the study of compact F sets in this context.

Theorem 17. Assume that X has a well-ordered base and a choice function for its closedsubsets. If X is compact F, then it is compact A.

Proof. Let B be a well ordered base for X , and let F be a choice function for the closedsubsets of X , and suppose that X is not compact A.

From all the well-ordered open covers from B without a finite subcover, choose C ={uα : α < λ} with minimum λ, where λ is a cardinal. Let κ = cof(λ) and let f : κ→ λ bean increasing cofinal sequence. For each α < κ let

xα = F (X r⋃

β<f(α)

uβ).

(Each set X r⋃β<f(α) uβ is closed, and it is non-empty by the minimality of κ.) Then,

since κ is a regular cardinal, Y = {xα : α < κ} has cardinality κ, and for every γ < κ,|{xα : α < γ}| < κ.

Now, the set Y has no complete accumulation points because for every x ∈ X thereexists a neighborhood uα of x, and we have that |uα∩Y | < κ. Therefore, X is not compactF . �Corollary. For closed sets of reals, compact F implies compact A.

Proof. From our previous results, closed sets of reals clearly satisfy the hypotheses ofTheorem 17. �


Corollary. Well-ordered products of closed compact F sets of reals are compact A. �However, it is not true in general that even well-ordered products of compact F sets of

reals are compact F: consider the product space 2ω in the model M1 in [hr]; 2 = {0, 1} isclearly compact F, but 2ω is not, because it is Hausdorff and contains an infinite Dedekindfinite set.

The results in this section suggest the following questions: Are compact C (or F) setsof reals necessarily closed in R?

Since choice for closed sets played an important role in the previous results, the studyof the following principles becomes natural:

Definition. Let CSC(Ti), i = 1, 2 stand for the statements: If (X, τ) is a compact A Tispace then the family G of all non-empty closed subsets of X has a choice function.

Theorem 18. CSC(T1) is equivalent to AC.

Proof. We only need to prove that CSC(T1) implies AC. Fix A = {Ai : i ∈ I} a disjointfamily of non-empty sets and let Z be the one point compactification of the disjoint unionY of the family of spaces {(Ai, τi) : i ∈ I}, where τi is the cofinite topology on Ai. ByCSC(T1), let f be a choice function on the family of all non-empty closed subsets of Z.Clearly, the restriction h of f to the family A is a choice function, finishing the proof ofthe theorem. �

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

FORM 154. Tychonoff’s Compactness Theorem for Countably Many T2 Spaces: Theproduct of countably many compact A T2 spaces is compact A.

FORM 343. A product of non-empty, compact A T2 spaces is non-empty.

Theorem 19. (Keremedis–Tachtsis [kt]) CSC(T2) ↔ form 343.

Theorem 20. CSC(T2) ↔ P∗T2(A, C), where P∗T2

(A,C) is the principle: If {(Xi, τi) : i ∈I} is a family of non-empty compact A T2 spaces then the Tychonoff product X =

∏i∈I Xi

is compact C with respect to the standard subbase H = {π−1i (O) : O ∈ τi, i ∈ I} of X.

Proof. (→). Fix {(Xi, τi) : i ∈ I} a family of non-empty compact A spaces and letX =

∏i∈I Xi. We will show that X is compact A with respect to the standard subbase

H. Let U ⊂ H be an open cover of X . For every i ∈ I put Ui = U|{π−1i (O) : O ∈ τi}.

If some Ui covers X then it is easy to see that Ui, hence U , has a finite subcover. Soassume than no Ui covers X and arrive at a contradiction. Put G = {Gi : i ∈ I}, whereGi = Xi r

⋃{O ∈ τi : π−1

i (O) ∈ Ui}. Clearly each Gi with the subspace topology is ancompact A T2 space. Thus, by Theorem 19, G has a choice function f . It is straightforwardto see that f is not covered by U , contradicting the fact that U covers X .

(←). By Theorem 19, it suffices to prove form 343. Fix {(Xi, τi) : i ∈ I} a family ofnon-empty compact A T2 spaces. Let Yi = Xi ∪ {∗i}, ∗i /∈ Xi, carry the disjoint uniontopology of Xi and the discrete space {∗i}. Clearly Yi is an compact A T2 space. Hence,by P∗T2

(A, C), Y =∏i∈I Yi, the Tychonoff product of the family {Yi : i ∈ I}, is compact C


with respect to the standard subbase H. If∏i∈I Xi = ∅, then U = {π−1

i (∗i) : i ∈ I} ⊂ His an open cover of Y . It follows that U has a finite subcover, which is a contradiction. �Theorem 21. CSC(T2) implies PWO

T2(A, A), where PWO

T2(A, A) stands for: the Tychonoff

product of every well ordered family {(Xi, τi) : i ∈ ℵ} of compact A T2 spaces is compactA. In particular, CSC(T2) implies form 154.

Proof. Fix {(Xi, τi) : i ∈ k} a well ordered family of compact A T2 spaces and let X =∏i∈kXi be their Tychonoff product. Let G = {Gj : j ∈ J} be a family of closed sets

having the f.i.p. We will show that ∩G 6= ∅. Let f be a choice function of the familyof all non-empty closed sets of the one point compactification of the disjoint topologicalunion of the family {(Xi, τi) : i ∈ k}. It follows that the restriction h of f to the family{H : ∅ 6= H ⊂ Xi, (X r H) ∈ τi, i ∈ k} is a choice function. Now we can complete theproof of the theorem by mimicing the proof of Theorem 20. �

The status of the relationship between forms 343 and 154 is indicated as unknown inTable 1 of [hr]. By Theorems 19 and 21 we have the following:

Corollary. Form 343 implies form 154.

6 An Independence Result

Our last result shows that P(F,G) does not imply AC in ZF0. We do this by showingthat P(F,G) is true in Mostowski’s linearly ordered model, N . (This is the model N 3 in[hr].) The set of atoms A is ordered so that it is isomorphic to the set of rational numbers.The group G is the group of order preserving isomorphisms and the ideal of supports is theset of all finite subsets of A. If E is a finite subset of A, we let GE = {φ ∈ G : φ fixes E}.Then GE is a subgroup of G. (Note that if φ ∈ GE then φ fixes E pointwise.) We willalso use the following facts about N .

(1) Every element x of N has a smallest support supp(x) with the property that ∀φ ∈ G,φ(x) = x⇔ φ(supp(x)) =supp(x).

(2) If E and F 6= ∅ are disjoint finite subsets of A, then the set S = {φ(F ) : φ ∈ GE} is aninfinite, Dedekind finite set in the model.(If we choose t ∈ F , then the set {φ(t) : φ ∈ GE} is infinite. By (1), since supp(F ) = F ,if φ and ψ are in GE and φ(t) 6= ψ(t) then φ(F ) 6= ψ(F ). It follows that S is infinite.If S were Dedekind infinite, then there would be a finite subset G of A which was asupport of infinitely many elements of S. But the only finite subsets of A supported byG are the finitely many subsets of G. It follows that S is Dedekind finite.)

(3) IfE = {e0, e1, . . . , en} is any finite subset of A and ifW = {(−∞, e0), (e0, e1), . . . , (en,∞)}is the set of open intervals determined by E, then for any two subsets F and F ′ of Aboth disjoint from E, if |F ∩W | = |F ′ ∩W | for all W ∈ W, then there is a φ ∈ GE suchthat φ(F ) = F ′.(The proof of (3) uses the following fact about the ordering on A: If B = {b1, b2, . . . , bk}and C = {c1, c2, . . . , ck} are two subsets of A both with the same number of elementsand b1 < b2 < · · · < bk and c1 < c2 < · · · ck then there is an order isomorphism φ


of A such that for all i, 1 ≤ i ≤ k, φ(bi) = ci). In order to see that this is true wemay assume that A is the set of rational numbers with the usual ordering and let φbe the union of the functions φ1, φ2, . . . , φk−1 where φ1 is the linear function whosedomain is (−∞, b2] and whose graph goes through the points (b1, c1) and (b2, c2), for2 ≤ i < k − 1, φi is the linear function whose domain is [bi, bi+1) and whose graphgoes through the points (bi, ci) and (bi+1, ci+1) and φk−1 is the linear function whosedomain is [bk−1,∞) whose graph goes through the points (bk−1, ck−1) and (bk, ck). Now(3) follows by taking B = E ∪ F and C = E ∪ F ′.

Theorem 22. P(F,G) is true in the Mostowski linearly ordered model, N .

Proof. Assume that {(Xi, Ti) : i ∈ I} ∈ N is a collection of ordered pairs each of which is

an compact F topological space in N and let X =(∏

i∈I Xi

)Nwhere the superscript N

means that the product is taken in the model. Assume that Z ∈ N is an infinite subset ofX . Let E be a support of the ordered pair (X,Z). For each i ∈ I, let πi : X → Xi be theprojection map. We first claim that it suffices to prove that

(#)Z ⊆ X, has a cluster point (in N ) under the assumption

(∀i ∈ I)(|πi(Z)| > 1)

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

To prove the claim, let I ′ = {i ∈ I : |πi(Z)| > 1} , and let πI′ : X →∏i∈I′ Xi be

the projection map. Then (∀i ∈ I ′)(|πi(πI′(Z))| > 1). It is also easy to see that πI′(Z)is infinite since Z is. Therefore, if we assume (#), πI′(Z) has a cluster point in

∏i∈I′ Xi,

call it f ′. If we then define f ∈ X by

f(i) =

{f ′(i) i ∈ I ′

the unique element of πi(Z) otherwise

we can show that f is a cluster point of Z.For the remainder of the proof we shall assume

(∗) (∀i ∈ I)(|πi(Z)| > 1)

and we shall show that Z has a cluster point.

Lemma. If for any i ∈ I, there is an element t ∈ Xi for which supp(t) * supp(Xi), thenZ has a cluster point.

Proof. Assume i satisfies the hypothesis, then the set W = {φ(t) : φ ∈ Gsupp(Xi)} is aninfinite, Dedekind finite subset of Xi since it can be put in a one to one correspondencein the model with the set {φ(supp(t) r supp(Xi)) : φ ∈ Gsupp(Xi)}. (This latter set isinfinite and Dedekind finite in the model by (2).)


By our assumption (∗), there are two elements f and g of Z such that f(i) 6= g(i).Since Xi is compact F, the set W ′ = W ∪ {f(i), g(i)} has a complete accumulation pointz ∈ Xi. We assume without loss of generality that z 6= f(i). We first note that f(i) is inevery neighborhood of z. For assume that N is a neighborhood of z in Xi. Then since W ′

is Dedekind finite and |W ′ ∩N | = |W ′| we have W ′ ∩N = W ′. Hence, f(i) ∈ N . Defineh ∈ X by

h(j) =

{z if j = i

f(j) otherwise.

Then f 6= h and f is in every neighborhood of h since f(i) is in every neighborhood ofz = h(i). Therefore, h is an accumulation point of Z.

Using the lemma, we will assume for the remainder of the proof that

(∗∗) (∀i ∈ I)(∀t ∈ Xi)(supp(t) ⊆ supp(Xi)).

Lemma. If for some f ∈ Z, supp(f) * E then Z has a cluster point.

Proof. Assume that f0 is an element of Z whose support is not contained in E. Wemay also assume that the function i 7→ Xi is one to one and therefore, for any φ ∈ GE ,φ(i) = i⇔ φ(Xi) = Xi.

Let E = {e0, e1, . . . , en} and let W be the set of open intervals determined by E. Thatis, W = {(−∞, e0), (e0, e1), . . . , (en,∞)}. (See (3) above.) For the proof of the lemma wewill need to consider the GE orbits of I. For each i ∈ I, let ob(i) = {φ(i) : φ ∈ GE} andlet Ob= {ob(i) : i ∈ I}. For each Q ∈ Ob choose iQ ∈ Q so that

(†) for every open interval W determined by E, every element of supp(f0)∩W is less thanevery element of supp(iQ) ∩W .

The function Q 7→ iQ may not be in N , however the set f1 defined by

f1 = {(φ(iQ), φ(f0(iQ))) : φ ∈ GE and Q ∈ Ob}

is in the model since it has support E. We also make the following claims about f1.

Claim A. f1 is a function.

To prove claim A, assume that (φ1(iQ), φ1(f0(iQ))) and (φ2(iP ), φ2(f0(iP ))) are in f1

where φ1 and φ2 are in GE and Q and P are in Ob and assume that φ1(iQ) = φ2(iP ).

Then iQ = φ−11 φ2(iP ). From this we may draw two conclusions: First that iQ and iP are in

the same orbit and, hence, iQ = iP . Secondly (using (∗∗)) that f0(iQ) = φ−11 φ2(f0(iP )) =

φ−11 φ2(f0(iQ)) and therefore, φ1(f0(iQ)) = φ2(f0(iP )). Claim A follows.

Claim B. f1 ∈∏i∈I Xi

This claim follows from the facts that the domain of f1 is {φ(iQ) : φ ∈ GE and Q ∈ Ob} =I and for each φ(iQ) in the domain of f1, f1(φ(iQ)) = φ(f0(iQ)) ∈ Xφ(iQ).


Claim C. f1 is a cluster point of Z

We first note that proving claim C will complete the proof of the lemma. To prove theclaim, let N be a basic neighborhood of f1. Then N =

∏i∈I Ki where for some finite

subset I0 of I, Ki = Xi for all i /∈ I0. We will show that N contains an element f2 of Zdifferent from f1.

For each j ∈ I0, let Qj be the unique element of Ob such that j ∈ Qj and let φj ∈ GEbe chosen so that φj(iQj ) = j. Choose an η ∈ GE so that

(‡) for each of the intervals W determined by E and every element a of supp(f0) in W ,η(a) < c for every element c ∈ supp(j) ∩W .

Let f2 = η(f0). Then f2 6= f1 since E is a support of f1 but supp(f2) = η(supp(f0)) 6= Esince supp(f0) 6= E and η fixes E.

We also note that f2 = η(f0) ∈ Z since f0 ∈ Z, Z has support E and η fixes E pointwise.Therefore, to complete the proof of the claim, it suffices to show that f2 ∈ N . We will dothis by arguing that ∀j ∈ I0, f2(j) = f1(j).

Assume that j ∈ I0 then, using the notation introduced above, φj(iQj ) = j whereφj ∈ GE where j ∈ Qj . It follows that for each open interval W determined by E,|supp(iQj ) ∩W | = |supp(j) ∩W |. Similarly |supp(f0) ∩W | = |supp(η(f0)) ∩W |. By (†),supp(f0) ∩ supp(iQj ) ∩W = ∅ and by (‡), supp(η(f0)) ∩ supp(j) ∩W = ∅ hence,

|(supp(f0) ∪ supp(iQj )) ∩W | = |(supp(η(f0)) ∪ supp(j)) ∩W |.

Therefore, letting F = supp(f0)∪supp(iQj ) and F ′ = supp(η(f0))∪supp(j) we conclude byproperty 3 (in our list of properties ofN ) that there is a β ∈ GE such that β(F ) = F ′. Sinceβ is order preserving (and using (†) and (‡)) we conclude that β(supp(f0)) = supp(η(f0))and η(supp(iQj )) = supp(j). Therefore, β(f0) = η(f0) and β(iQj ) = j. Hence,

f2(j) = (η(f0))(j) = (β(f0))(j) = β(f0(β−1(j))) = β(f0(iQj )) = f1(j).

This completes the proof of claim C and therefore, the lemma is proved.

We therefore assume for the remainder of the proof that

(∗ ∗ ∗) ∀f ∈ Z, supp(f) ⊆ E.

It follows from (∗∗) that each Xi is well orderable in N and therefore, the power set ofXi in N (which we denote by P(Xi)

N ) is the same as the power set of Xi in the ground

model. Similarly (P(P(Xi)))N

is the same as P(P(Xi)) in the ground model. We denotethe ground model (that is, the model of ZF+AC from which N was constructed) by N0. Itfollows that (Xi, Ti) is an compact F topological space inN0. Therefore, sinceN0 is a model

of AC,∏i∈I Xi (taken in N0) is compact G in N0. Since Z ⊆

(∏i∈I Xi

)N ⊆ ∏i∈I Xi,Z has a cluster point in

∏i∈I Xi. Let h be such a cluster point. If h ∈ N , then it is not


hard to verify that h is a cluster point of Z in N in which case the proof is complete.Assume therefore, that h /∈ N . Then there is a permutation σ ∈ GE and a j ∈ Isuch that σ(h(j)) 6= h(σ(j)). (Otherwise h has support E.) One consequence of thisassumption is that σ(j) 6= j. For if σ(j) = j then by (∗∗), σ(h(j)) = h(j) and, hence,σ(h(j)) = h(j) = h(σ(j)).

Now, as in the proof of the previous lemma, we choose one representative iQ from eachQ ∈ Ob but rather than require (†) we require only that iob(j) = j. Again as in the proofof the previous lemma (see the definition of f1) we define

h1 = {(φ(iQ), φ(h(iQ))) : φ ∈ GE and Q ∈ Ob}.

The proof that h1 is in(∏

i∈I Xi

)Nis almost identical to the proof that f1 ∈

(∏i∈I Xi

)N.

We also note

(♥) ∀Q ∈ Ob, h1(iQ) = h(iQ).

We define h2 ∈∏i∈I Xi by

h2(i) =

{h1(i) i 6= σ(j)

h(i) i = σ(j).

It is clear that h2 ∈ N since h1 is. We will show that h2 is a cluster point of Z.First note thatσ(h2(j)) = σ(h1(j)) since j 6= σ(j) andσ(h1(j)) = σ(h(j)) since j = iob(j) by (♥) andσ(h(j)) 6= h(σ(j)) = h2(σ(j)) soσ(h2(j)) 6= h2(σ(j))and since σ ∈ GE it follows that supp(h2) * E. By (∗ ∗ ∗) h2 /∈ Z. Therefore, to show

that h2 is a cluster point of Z it suffices to show that every neighborhood of h2 containsan element of Z.

Let N be a neighborhood of h2, then N =∏i∈I Ki where for some finite subset I0 of

I, Ki = Xi for all i /∈ I0. For each k ∈ I0, choose ηk ∈ GE so that ηk(k) = iob(k). Foreach Q ∈ Ob, let NQ =

⋂{ηk(Kk) : k ∈ Q ∧ k 6= σ(j)}. Since Kk is a neighborhood

of h2(k) = h1(k) for all k ∈ Q such that k 6= σ(j) and since η fixes h1, ηk(Kk) is aneighborhood of h1(iQ). Further, since Kk = Xk for all but finitely many k, ηk(Kk) = XiQ

for all but finitely many k in Q. It follows that NQ is a neighborhood of h1(iQ) = h2(iQ).Let M =

∏i∈IMi where

Mi =

Nob(i) if i ∈ I0, i = iob(i), i 6= σ(j)

Kσ(j) if i = σ(j)

Xi otherwise

.


Then M is a neighborhood of h and therefore there is an element f of Z such that f ∈M .We complete the proof by showing that f ∈ N .

It follows from the choice of f that f(j) ∈ Kj and we also have for i /∈ I0, f(i) ∈ Xi = Ki.

Suppose i ∈ I0 and i 6= σ(j). Then f(iob(i)) ∈ Nob(i) ⊆ ηi(Ki). So η−1i (f(iob(i))) ∈ Ki.

That is, η−1i (f)(η−1

i (iob(i))) ∈ Ki. But by (∗ ∗ ∗), η−1i (f) = f and by the definition of ηi,

η−1i (iob(i)) = i. Therefore, f(i) ∈ Ki and, hence, f is in N . �

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Omar De la CruzDepartment of MathematicsPurdue UniversityWest Lafayette, IN [email protected]

Eric J. HallDepartment of MathematicsPurdue UniversityWest Lafayette, IN [email protected]

Paul HowardDepartment of MathematicsEastern Michigan UniversityYpsilanti, MI [email protected]

Kyriakos KeremedisDepartment of MathematicsUniversity of the AegeanKarlovasi, Samos, 83200, [email protected]

Jean E. RubinDepartment of MathematicsPurdue UniversityWest Lafayette, IN [email protected]

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PRODUCTS OF COMPACT SPACES AND THE AXIOM OF CHOICE II