Tychonoff Theorem in the Realm of $\neg AC$

It's widely know that the Tychonoff Theorem is equivalent to the Axiom of Choice; thus, assuming the negation of the axiom of choice, I'd like to know if there is a canonical example of a collection of compact spaces whose product fails to be compact.

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@ArthurFischer this fact does not imply the possibility of the existence of such collection? – Paulo Henrique Feb 19 '13 at 16:14
It guarantees that there is such a collection. – Brian M. Scott Feb 19 '13 at 16:14
Are you asking about a canonical example of a collection of compact spaces whose product fails to be compact (under the assumption of $\neg$AC)? – arjafi Feb 19 '13 at 16:20
Fëanor, do you mean you want to see a model $\mathcal{M}$ of ZF in which Tychonoff’s Theorem fails, together with a product of compact spaces in $\mathcal{M}$ that witnesses the failure? – Haskell Curry Feb 19 '13 at 16:21
Then the question should be edited to reflect this fact. (This is a much more interesting question than the tautology I originally assumed.) – arjafi Feb 19 '13 at 16:26

I suspect what you want is to be given an example of a collection of compact spaces whose product is not compact, under the assumption that choice fails.

Choice is equivalent to: The product of non-empty sets is non-empty. Thus, assuming its negation, there is a collection $(A_i\mid i\in I)$ of non-empty sets whose product is empty. Let $*$ be a point not in any of the $A_i$, and let $X_i=A_i\cup\{*\}$ for all $i$. Make $X_i$ into a compact space by requiring that neighborhoods of $*$ are cofinite. Then $$\prod_i X_i$$ is not compact. Otherwise, from its compactness one would easily deduce that $\prod_i A_i\ne\emptyset$. (I suspect Jech's "The axiom of choice" should give a few more details, if needed.)

If you rather state choice as "there is a set that is not well-orderable", we can get back to the example above as follows: Let $A$ be not well-orderable, and let $(A_i\mid i\in I)$ list all nonempty subsets of $A$. Then $\prod_i A_i$ is empty, or else any element of the product (a choice function on the subsets of $A$) would easily induce by transfinite recursion a well-ordering of $A$.

Now, if what you are asking for is a "canonical" example, so that I tell you "explicitly" what the $A_i$ are, that is not possible. The universe of sets is stratified via a notion of "rank", indexed by the ordinals. For any ordinal $\alpha$, we can exhibit models where choice fails in general, but all Tychonoff products of compact spaces of rank $\alpha$ or lower are compact. So no "canonical" example of what you want exist.

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This construction can be "simplified" somewhat in a way that the op might like: In any model where choice fails for countable families of two element sets, all of your $X_{i}$ can be taken to be three element discrete spaces and the index set $I$ can be $\mathbb{N}$. – Shawn Henry Feb 19 '13 at 16:57
@Shawn: Luckily, or unluckily, this is not always the case. In many of the models considered, countable choice is present; or at least choice from collections of finite sets is present. – Asaf Karagila Feb 19 '13 at 17:00
Andres, the part where you suspect Jech's book will have more details intrigued me. And it turns out that Jech gives this as an exercise (Ch. 2, Ex. 8) with a hint which includes your given construction and another step along the way. – Asaf Karagila Feb 19 '13 at 19:52
@Asaf: I know that there are lots of models with various amounts of choice. My comment was intended to provide a an example which is "minimal" in some sense (I posted this earlier, then deleted it because I wasn't sure, but now I have proofs): It is provable in ZF that the countable product of two element discrete spaces is compact, and AC for countable families of n element sets is equivalent to the Tychonoff theorem for countable products of n+1 element discrete spaces. – Shawn Henry Feb 19 '13 at 22:30