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May 25, 2018 at 13:35 vote accept Dominic van der Zypen
May 25, 2018 at 9:34 comment added Henno Brandsma @RamirodelaVega And if $X$ is compact and has exactly two non-homeomorphic "types", $X$ is homogeneous.
May 24, 2018 at 18:50 answer added Henno Brandsma timeline score: 7
May 24, 2018 at 16:13 vote accept Dominic van der Zypen
May 25, 2018 at 13:35
May 24, 2018 at 13:31 comment added Ramiro de la Vega @David, any infinite compact space $X$ has non-homeomorphic open subsets (e.g. $X$ and $X \setminus \{p\}$ for a non-isolated $p \in X$).
May 24, 2018 at 9:33 history edited Dominic van der Zypen CC BY-SA 4.0
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May 24, 2018 at 9:31 comment added Dominic van der Zypen @PietroMajer thanks for the hint, will do
May 24, 2018 at 8:34 comment added HenrikRüping @bof: Thanks of course it is homogeneous.There are points which are endpoints of some removed intervals and points which are not. I was mistakenly thinking that a homeomorphism can't map one of the first kind to one of the second kind. However that idea at least shows that there are no order preserving (thinking of the Cantor set as a subset of the real line) homeomorphisms that map a point of the first kind to one of the second kind.
May 24, 2018 at 8:28 comment added bof @HenrikRüping Why do say that the Cantor set is not homogeneous? It is, after all, the product of countably many $2$-point discrete spaces. Isn't a product of homogeneous spaces homogeneous?
May 24, 2018 at 8:17 comment added HenrikRüping Related: In the Cantor set all clopen sets are homeomorphic (I believe) and it is not homogeneous.
May 24, 2018 at 7:44 comment added Pietro Majer Don't forget to link preceding relevant questions of yours like mathoverflow.net/questions/300253/…
May 24, 2018 at 6:45 history asked Dominic van der Zypen CC BY-SA 4.0