I know that when $X$ is a normal and totally disconnected space, the Stone-Cech compactification $\beta(X)$ is totally disconnected. But I can't find a counterexample when considering $X$ totally disconnected only.
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Eric Wofsey
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Is $\beta$ a continuous function? – Giacomo Parolin Apr 01 '20 at 18:51
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3No. Is a functor. $\beta(X) is the Stone-Cech compactification for X. Excuse me – Santiago Angarita García Apr 01 '20 at 18:59
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Are there any particular examples, you have checked? or suspect might work? – G. Chiusole Apr 01 '20 at 19:45
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For example when $X$ is a discrete space, $\beta(X)$ is totally disconnected – Santiago Angarita García Apr 01 '20 at 20:18
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1Do you assume $X$ is at least completely regular, so it embeds in $\beta(X)$? – Eric Wofsey Apr 01 '20 at 21:16
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No. I'm thinking $X$ like a general topological space – Santiago Angarita García Apr 05 '20 at 20:35
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No. Any totally disconnected compact Hausdorff space is zero-dimensional (clopen sets generate the topology), and any subspace of a zero-dimensional space is zero-dimensional. So, if $X$ is completely regular and $\beta(X)$ is totally disconnected, then $X$ must be zero-dimensional. This means any completely regular space which is totally disconnected but not zero-dimensional is a counterexample. Such a space is complicated to construct but one well-known example is Cantor's teepee (also known as the deleted Knaster-Kuratowski fan). Note that this example is also normal (it is a subspace of $\mathbb{R}^2$), so your statement about the case that $X$ is normal is incorrect.
Steven Clontz
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Eric Wofsey
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2I believe that $\beta X$ is zero-dimensional iff $X$ is strongly zero-dimensional. See Engelking. – Henno Brandsma Apr 01 '20 at 21:57
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Is possible to remove the assumption of X completely regular? – Santiago Angarita García Apr 05 '20 at 21:50
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1Seems very unlikely. I haven't checked the details but I suspect the construction I describe here gives a counterexample: if you take a Cantor set and enlarge its topology to make a countable dense set open, it will no longer be zero-dimensional, but I think its Stone-Cech compactification will still be the original Cantor set. – Eric Wofsey Apr 05 '20 at 22:11
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I'll work about your hint, thank you so much. Can I do an other question? I'd like ask you about hint for to prove your afirmation : Any totally disconnected compact Hausdorff space is zero-dimensional (clopen sets generate the topology) – Santiago Angarita García Apr 06 '20 at 23:33
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It suffices to show the topology generated by the clopen sets is Hausdorff, by compactness (you cannot have a compact topology that is strictly finer than a Hausdorff topology). In other words, you need to show that clopen sets separate points; for that see https://math.stackexchange.com/questions/11412/any-two-points-in-a-stone-space-can-be-disconnected-by-clopen-sets – Eric Wofsey Apr 07 '20 at 01:19
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Erdős' space $E = {(x_n)_n \in \ell^2\mid \forall n: x_n \in \Bbb Q}$ in the subspace topology from $\ell^2$ is separable metric, $\dim(E)=1$ (and $E \times E \simeq E$, so dimensions don't add up in products in general) and $E$ is totally disconnected. A bit of a late addition... – Henno Brandsma May 14 '21 at 13:15