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Timeline for Subsets of the Cantor set

Current License: CC BY-SA 4.0

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Mar 7, 2022 at 18:00 history edited LSpice CC BY-SA 4.0
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Mar 7, 2022 at 17:41 history edited gaam2296 CC BY-SA 4.0
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Mar 7, 2022 at 17:40 vote accept gaam2296
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Mar 7, 2022 at 11:46 review Close votes
Mar 13, 2022 at 3:11
Mar 7, 2022 at 11:25 answer added KP Hart timeline score: 3
Mar 5, 2022 at 21:17 comment added YCor @SaúlRodríguezMartín oops, my mistake, yes of course: if closed (=clopen) nonempty it's homeomorphic to Cantor, and if not closed it's homeomorphic to Cantor minus singleton. Anyway this leaves the reasoning unchanged: it works in any Hausdorff $U'$ with no isolated point.
Mar 5, 2022 at 21:11 comment added Saúl RM @YCor just wanted to point out that $U$ can also be homeomorphic to the Cantor set. Also, the same reasoning can be applied without knowing how open subsets of the Cantor set are up to homeomorphism: it is enough to notice that $U$ is uncountable if it is non empty (because every nonempty open set of the Cantor set contains a small copy of the Cantor set), and every point of $U$ is an accumulation point of $D\cap U$.
Mar 5, 2022 at 21:01 comment added YCor If $U$ is assumed non-empty, just being a nonempty open subset of a Cantor set, it is homeomorphic to Cantor minus one point (or equivalently to Cantor $\times$ discrete countable). So the point is to check that every dense subset $D$ of $U$ has infinitely many accumulation points inside $D$. Indeed every element of $D$ is an accumulation point of $D$ (this just follows from the fact that $U$ has no isolated point).
Mar 5, 2022 at 20:59 comment added YCor If $U$ is empty the assertion is false. So it should be assumed that $U$ is nonempty.
S Mar 5, 2022 at 20:45 review First questions
Mar 5, 2022 at 21:05
S Mar 5, 2022 at 20:45 history asked gaam2296 CC BY-SA 4.0