Tychonoff Theorem in the Realm of $\neg AC$

Question

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.

Answer 1

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.