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I'm doing the following exercise from Just/Weese:

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Some thoughts: To show that the statement is not provable from $ZF$ I could either show that it implies the axiom of choice or Tychonoff or I could assume that it is provable from $ZF$ and produce a contradiction. Since I'm give a statement that is known not to imply $AC$ and I'm told to use it, I will do so. The idea is that if the second statement in the exercise is provable from $ZF$ then the first statement implies Tychonoff's theorem which is the desired contradiction.

Proof: Let (*)"For every indexed family of non-empty sets there is a family of compact Hausdorff topologies." be provable from $ZF$. Assume that (**)"The product of compact Hausdorff families is compact.". Let $X_i$ be an arbitrary collection of sets compact spaces . Then by (*) we can turn them into a compact Hausdorff family and by (**) the product is compact. Hence Tychonoff's theorem holds which is equivalent to $AC$. But then (**) would be equivalent to $AC$.$\Box$

Some more thoughts: Given a single arbitrary set it is always possible to endow it with a compact Hausdorff topology: First put the discrete topology on it and then use one-point compactification to make it into a compact space.

The reason why this cannot be applied to an infinite family of sets is that for each set one needs to choose a bijection between the set and the set union one point. And to do so one needs $AC$.

Can you tell me if I have it right? Thank you.

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‘Hence Tychonoff’s theorem holds’: How do you justify this? You haven’t shown that an arbitrary product of compact spaces is compact. –  Brian M. Scott Dec 1 '12 at 8:10
    
@BrianM.Scott I thought I had: If I have a compact family I turn it into a compact HD family and then apply (**) to get that the product is compact. No? –  Rudy the Reindeer Dec 1 '12 at 8:13
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No: that just shows that a different product of spaces is compact, since you’ve changed the topologies. Hang on: I’ve almost written up an answer. $-$ I take that last bit back: I need to think a bit more. –  Brian M. Scott Dec 1 '12 at 8:14
    
Maybe I could apply the construction from the TY implies AC proof. –  Rudy the Reindeer Dec 1 '12 at 8:19
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Yes; in fact, that’s what I’ve just done and am about to post. –  Brian M. Scott Dec 1 '12 at 8:25

3 Answers 3

up vote 5 down vote accepted

You’ve not actually derived the Tikhonov product theorem from $(*)$ and $(**)$: your argument does not show that the arbitrary product of compact spaces is compact. However, you can derive AC itself. Let $\{A_i:i\in I\}$ be a non-empty set of non-empty sets. Without loss of generality assume that the $A_i$ are pairwise disjoint and disjoint from $I$. For $i\in I$ let $X_i=A_i\cup\{i\}$. By $(*)$ there are topologies $\tau_i$ for $i\in I$ such that each $\langle X_i,\tau_i\rangle$ is a compact Hausdorff space, and there is no harm in assuming that $\{i\}\in\tau_i$.

By $(**)$ $X=\prod\{X_i:i\in I\}$ is a compact Hausdorff space. For $i\in I$ let $\pi_i:X\to X_i$ be the projection map, and let $F_i=\pi_i^{-1}[A_i]$; $A_i$ is closed in $X_i$, and $\pi_i$ is continuous, so $F_i$ is closed in $X$. Let $\mathscr{F}=\{F_i:i\in I\}$; it’s not hard to check that $\mathscr{F}$ is a centred family of closed sets in $X$. The key is that if $J$ is a finite subset of $I$, one need only choose an $a_i\in A_i$ for each $i\in J$ in order to define a point $x_J$ of $\bigcap\{F_i:i\in J\}$, since one can set $\pi_i(x_J)=i$ for $i\in I\setminus J$. Thus, $\bigcap\mathscr{F}\ne\varnothing$, and any point of $\bigcap\mathscr{F}\ne\varnothing$ is a choice function for $\{A_i:i\in I\}$.

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This seems to assume that we already know not only that $(**)$ is "not equivalent to AC", but specifically that it is "weaker than AC". –  Henning Makholm Dec 1 '12 at 10:50
    
@Henning: Presumably we do, since we know that AC implies the full Tikhonov product theorem (and Matt’s remarks indicate that we actually know that the two are equivalent). –  Brian M. Scott Dec 1 '12 at 10:53
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@Matt: No. Let $\mathscr{A}$ be any set of sets. Then $\big\{A\times\{A\}:A\in\mathscr{A}\big\}$ is a pairwise disjoint family of sets that are clearly just copies of the members of $\mathscr{A}$. This has nothing to do the axiom DIS of one of your earlier questions. –  Brian M. Scott Dec 3 '12 at 9:33
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@Matt: Disjointedly? Oops, no: that’s how, not what. ;-) –  Brian M. Scott Dec 3 '12 at 9:39
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@Matt: In drawing the conclusion that $\bigcap\mathscr{F}$\ne\varnothing$. –  Brian M. Scott Dec 3 '12 at 21:06

If given a family of non-empty sets we can endow them (uniformly!) with compact Hausdorff topologies, then by assuring that their product is compact we can show that the axiom of choice holds.

Indeed this would be enough to derive Tychonoff's theorem, but this is not Tychonoff's theorem per se. Tychonoff theorem does not permit you to replace the topologies given to you.

Now if we could have proved that every collection of sets can be given Hausdorff compact topologies in a uniform way then indeed the restriction of the Tychonoff theorem to Hausdorff spaces would imply the axiom of choice, but it is weaker. Interestingly enough the Tychonoff theorem for Hausdorff spaces is equivalent to the ultrafilter lemma, and the myriad of useful equivalents which comes with it.

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Rephrasing Brian's argument in terms of the proof (TY) $\rightarrow$ (AC) preceding exercise 34 won't hurt:

Assume (*) and (**). Let $A_i$ be a arbitrary family of non-empty sets. We may assume they are pairwise disjoint. The goal is to prove that there exists a choice function $f \in \prod_i A_i$. To this end, define $X_i = A_i \cup \{i\}$ and use (*) to endow $X_i$ with compact Hausdorff topologies $\tilde{\tau_i}$. Let $\tau_i$ be the topology generated by $\tilde{\tau_i}$ and $\{i\}$. Then $\tau_i$ are compact and Hausdorff hence by (**) the product $\prod_i X_i$ is compact. Define $U_i = \pi_i^{-1}(\{i\})$. Then $U_i$ are open. If $U_i$ were an open cover of $\prod_i X_i$ then there would be a finite subcover $(U_j)_J$. Define $f$ to be the map $j \mapsto a_j$ for some $a_j \in A_j$ and $i \mapsto i$ for $i \notin J$. Then $f$ is not in $\bigcup_J U_j$. Hence $U_i$ cannot be a cover of $\prod_i X_i$. Hence there is $g \in \prod_i X_i \setminus \bigcup_i U_i$ which means that $g$ is a choice function for $\prod_i A_i$.

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