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.
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asked Jun 15, 2016 at 13:58
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$\begingroup$ Are sober spaces completely regular? $\endgroup$– Asaf Karagila ♦Jun 15, 2016 at 14:03
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$\begingroup$ Not necessarily. $\endgroup$– Ittay WeissJun 15, 2016 at 14:36
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$\begingroup$ It's not mentioned separately in the book "Consequences of the axiom of choice" (sober ones are, but they seem to use choice too). I saw no separation axioms beyond Hausdorff. $\endgroup$– Henno BrandsmaJun 15, 2016 at 17:52
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1 Answer
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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:
- Products of compact Hausdorff spaces are compact.
- Products of finite discrete spaces are compact.
- Products of finite spaces are compact.
- Hilbert cubes $[0,1]^I$ are compact.
- Cantor cubes $2^I$ are compact.
- PIT.
- 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.
answered Jun 15, 2016 at 19:53
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$\begingroup$ Thanks. Decisive answers are the best kind. $\endgroup$ Jun 16, 2016 at 7:19
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$\begingroup$ @Ittay: My pleasure. (They are indeed.) $\endgroup$ Jun 16, 2016 at 7:26
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