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May 16, 2018 at 22:24 history edited N. de Rancourt CC BY-SA 4.0
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May 16, 2018 at 12:55 comment added Will Brian @N.deRancourt: Go ahead and add it to your answer -- it's just a minor variation on your idea.
May 16, 2018 at 9:57 comment added N. de Rancourt Of course @WillBrian you are right! That's nice, maybe you should post it as an answer? Or if you prefer, I can add it on my answer.
May 15, 2018 at 20:45 comment added Dominic van der Zypen @N.deRancourt I like your answer so much that I have accepted it! Answers from other people are of course much welcome, too.
May 15, 2018 at 17:31 vote accept Dominic van der Zypen
May 15, 2018 at 14:14 comment added Will Brian Doesn't your argument apply equally well to the subspace of $X^\omega$ consisting of eventually constant sequences? It's still true that the closed sets can be expressed as trees, and it's still true that every open set is a union of disjoint cones. But this subspace has cardinality $|X|$, so this version of the argument would work for any infinite cardinal.
May 15, 2018 at 13:46 comment added N. de Rancourt However @DominicvanderZypen, maybe you should not accept the answer yet, the question is still interesting as we have no example for a lot of cardinals. It would be nice if other people could look at this question.
May 15, 2018 at 13:44 history edited N. de Rancourt CC BY-SA 4.0
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May 15, 2018 at 13:43 comment added N. de Rancourt Yes @NikWeaver, Will Brian was right: it was "empty interior". I edit to correct the mistake.
May 15, 2018 at 13:40 vote accept Dominic van der Zypen
May 15, 2018 at 14:33
May 15, 2018 at 13:01 comment added Will Brian @NikWeaver: I think it should say empty interior.
May 15, 2018 at 12:59 comment added Nik Weaver Every nonempty compact subset of $\omega^\omega$ has nonempty interior? What about singletons?
May 15, 2018 at 12:40 history answered N. de Rancourt CC BY-SA 4.0