3
$\begingroup$

If $X$ is a Tychonoff space, then using the Tychonoff theorem and thus the full axiom of choice, it follows that $X$ admits a Hausdorff compactification.

My question is : what remains true if we do not have the axiom of choice, but still allow weaker versions like Dependent choice (DC) or countable choice ?

More concretely, would this result still be true under (DC) if we add a stronger condition on $X$ like being metrizable, or second-countable ? In those cases, could we obtain a metrizable compactification ?

Improve this question
$\endgroup$
6
  • 4
    $\begingroup$ Actually the full strength of AC is not needed for having the Stone-Cech compactification, and the weaker Boolean prime ideal principle (BPI) is enough. This is because BPI alone implies that the product of compact Hausdorff spaces is compact (is in fact equivalent to that statement). However BPI and DC do not imply each other. $\endgroup$
    – godelian
    Commented Dec 6, 2018 at 14:21
  • $\begingroup$ Right. BPI and DC are quite tangential to each other, hence I would expect DC not to be of much help for constructing compactifications. $\endgroup$
    – Emil Jeřábek
    Commented Dec 6, 2018 at 14:25
  • $\begingroup$ What @godelian said. Also DC, while being very useful on its own, is a statement about making countably many choices. Tychonoff's theorem is about making arbitrary products (and thus choices). $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 6, 2018 at 14:25
  • 2
    $\begingroup$ I suspect DC is enough to prove that every separable metrizable space has a metrizable compactification (but I don't have the time to check the details today -- hence a comment and not an answer). The idea is that you can embed $X$ into $[0,1]^\omega$ without using any choice at all, and then you can use DC to prove that $[0,1]^\omega$ is compact. At least, I think DC should suffice for the second part (but I'm not sure). $\endgroup$
    – Will Brian
    Commented Dec 6, 2018 at 15:21
  • 1
    $\begingroup$ I think $[0,1]^\omega$ is provably compact in ZF. You have a countable, hence well ordered basis, and all the magic happens in there. So, I believe ZF proves the existence of compactifications for all second-countable $T_{3\frac12}$ spaces. I still doubt there is a natural class of spaces where DC can help. $\endgroup$
    – Emil Jeřábek
    Commented Dec 6, 2018 at 15:33

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .