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May 15, 2018 at 2:06 vote accept Claudia Correa
Mar 21, 2018 at 21:55 comment added Not Mike @YCor effectively yes, that's what I'm saying, see Theorem 3.2 of projecteuclid.org/euclid.ijm/1255454110
Mar 21, 2018 at 21:49 comment added YCor But only theoretically... e.g., if I have "given" spaces $X,Y$, I don't see if the problem whether $X'$ and $Y'$ are homeomorphic is trivial...
Mar 21, 2018 at 21:47 comment added YCor @NotMike I'm not sure what you're proving (I'm not familiar with all notation). Are you saying that to every compact space $X$ one can canonically associate some totally disconnected space $X'$ with some quotient map $X'\to X$ (this Gleason extremally disconnected space of $X$?) such that $X$ is a quotient of $bN-N$ iff $X'$ is a quotient of $bN-N$? Indeed I'd then agree that this more precise assertion makes the problems at least theoretically equivalent.
Mar 21, 2018 at 21:33 comment added Not Mike @YCor whoops meant $CO(G)$ instead of $G$.
Mar 21, 2018 at 21:08 comment added Not Mike @YCor let $B$ be the completion of $A=\mathcal{P}(\omega)/fin$ then any continuous map taking $S(A)$ onto $X$ can be pulled-back to a map of $S(B)$ onto $X$. Let $G$ be the extermally disconnected Gleason space associated with $X$; since extremally disconnected compact spaces are projective, either $G$ maps onto $S(B)$ or vice versa; without loss of generality, $S(B)$ maps onto $G$, in this case we can find some $A_0 \subset A$ which generates a sub-algebra isomorphic to a dense sub-algebra of $G$.
Mar 21, 2018 at 20:48 comment added YCor @NotMike I don't think that the fact you mention make the problems equivalent.
Mar 21, 2018 at 20:11 comment added Not Mike @YCor Since every compact space is the continuous image of an extremally disconnected compact space, it is enough to only consider totally disconnected images; the two problems are equivalent.
Mar 21, 2018 at 17:11 comment added YCor In restriction to totally disconnected spaces, this is equivalent, by Stone duality, to classifying Boolean subalgebras of $2^\mathbf{N}/(\mathrm{fin})$ up to isomorphism. Here $(\mathrm{fin})$ is the ideal of finite subsets.
Mar 21, 2018 at 13:38 history edited Will Brian CC BY-SA 3.0
deleted 1 character in body; edited title
Mar 21, 2018 at 2:19 history edited Johannes Hahn CC BY-SA 3.0
TeXified title and question
Mar 21, 2018 at 1:49 history edited Martin Sleziak
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Mar 21, 2018 at 1:40 answer added Will Brian timeline score: 23
Mar 21, 2018 at 0:49 review First posts
Mar 21, 2018 at 2:19
Mar 21, 2018 at 0:46 history asked Claudia Correa CC BY-SA 3.0