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Dec 6, 2018 at 15:33 comment added Emil Jeřábek 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.
Dec 6, 2018 at 15:21 comment added Will Brian 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).
Dec 6, 2018 at 14:31 comment added LCO Yes, that's why I wonder if by adding conditions on X, it's still possible to construct compactifications with DC.
Dec 6, 2018 at 14:25 comment added Asaf Karagila 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).
Dec 6, 2018 at 14:25 comment added Emil Jeřábek Right. BPI and DC are quite tangential to each other, hence I would expect DC not to be of much help for constructing compactifications.
Dec 6, 2018 at 14:21 comment added godelian 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.
Dec 6, 2018 at 14:05 history asked LCO CC BY-SA 4.0