Tychonoff's theorem for completely regular spaces and the axiom of choice

Question

It is well-known that Tychonoff's theorem, i.e., that the product of any set-indexed family of compact spaces is compact, is equivalent to the axiom of choice. It is also the case that if the spaces are Hausdorff, then the full axiom of choice is no longer needed (I believe the claim becomes equivalent to the ultrafilter axiom). What if the spaces have even more structure? In particular, what sort of equivalency holds for completely regular spaces? In other words, if the product of a set-indexed family of compact completely regular spaces is always compact, which choice-like axiom is implied.

Answer 1

Going from Hausdorff to Tikhonov gains you nothing. Theorem $4.70$ in Horst Herrlich, Axiom of Choice, Springer Lecture Notes in Mathematics $1876$:

Equivalent are:

1. Products of compact Hausdorff spaces are compact.
2. Products of finite discrete spaces are compact.
3. Products of finite spaces are compact.
4. Hilbert cubes $[0,1]^I$ are compact.
5. Cantor cubes $2^I$ are compact.
6. PIT.
7. UFT.

The last two are the Boolean prime ideal theorem and the ultrafilter theorem. In particular, if $(4)$ implies UFT, then certainly the result for Tikhonov spaces must do so.