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Feb 22, 2020 at 21:37 answer added Pierre PC timeline score: 3
Sep 5, 2019 at 21:08 history edited YCor CC BY-SA 4.0
edited title
Sep 5, 2019 at 18:15 answer added Goldstern timeline score: 6
Mar 5, 2019 at 16:28 comment added Dominic van der Zypen @YCor Thanks for your encouraging comment!
Mar 4, 2019 at 21:46 history edited YCor CC BY-SA 4.0
added the topological interpretation
Mar 4, 2019 at 21:44 comment added YCor By the way, this is an excellent basic question, I didn't know the answer when I read it and I was unable to solve it myself (my answer below only reflects by Google skills). Don't know why it's been downvoted (the other interpretation of the question, before being edited, led to an interesting answer too).
Feb 23, 2019 at 17:46 history edited Dominic van der Zypen CC BY-SA 4.0
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Feb 23, 2019 at 17:46 comment added Dominic van der Zypen Apologies for the unclear question - @YCor perfectly answers it, but also thanks to Juan for his interesting answer!
Feb 23, 2019 at 17:44 vote accept Dominic van der Zypen
Feb 23, 2019 at 17:34 answer added YCor timeline score: 10
Feb 23, 2019 at 16:49 comment added YCor A remark: let $K$ be a Cantor space. There's an equivalence between (a) the number of metrizable countable Stone spaces mod isomorphism is $c$, and (b) the number of closed subsets of $K$ modulo $\mathrm{Homeo}(K)$ is $c$. Indeed, clearly (a) implies (b) since every metrizable countable Stone space is homeomorphic to a closed subset of a Cantor space. (b) implies (a): indeed given a closed subset $F$ of $K$, one can produce a compact space $K_F$ as $K$ union a discrete open countable set, accumulating exactly on $F$. Then $K_F$ and $K_{F'}$ are homeo only if $F,F'$ are in the same orbit.
Feb 23, 2019 at 16:27 answer added juan timeline score: 3
Feb 23, 2019 at 16:24 history edited YCor
edited tags
Feb 23, 2019 at 16:23 comment added YCor I guess you mean "Boolean algebra structures on $\omega$". Equivalently are there $c$ homeomorphism types of metrizable Stone spaces. At least there are $\aleph_1$ (namely metrizable countable Stone spaces): the number of superatomic countable Boolean algebras modulo isomorphism is exactly $\aleph_1$. In particular the answer is yes under CH.
Feb 23, 2019 at 16:09 history asked Dominic van der Zypen CC BY-SA 4.0