Let $S$ be a totally disconnected compact Hausdorff space and let $A\subset S$ be a closed subset. Let $S/A$ denote the space we get when collapsing $A$ to a point. Is this space still totally disconnected?
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$\begingroup$ Yes indeed! Great, I didn't know this space. Thanks. $\endgroup$– EchoOct 17 at 7:58
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$\begingroup$ On second thoughts, why is it a collapse by a closed set? I understand, it is a collapse $S/A$ of some t.d. space $S$, but I can't get $A$ closed. $\endgroup$– EchoOct 17 at 8:02
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3$\begingroup$ Yes, the quotient is totally disconnected. If you take two distinct points outside $A$, there are two clopen subsets disjoint of $A$ and separating them, hence at the quotient these map to clopen subsets, separating them. If you take a point outside $A$, there's a single clopen subset containing it and disjoint of $A$, and at the quotient this and its complement are two clopen subsets separating the two images. $\endgroup$– YCorOct 17 at 8:08
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$\begingroup$ @bof No clarification is needed: it is classical that in a totally disconnected compact Hausdorff space (totally disconnected being in the usual sense: connected components are singletons), distinct points are separated by clopen subsets $\endgroup$– YCorOct 17 at 8:53
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1$\begingroup$ If we allow the collapse of infinitely many different closed sets, i.e. if $A\subset S\times S$ is closed equivalence relation, then the equivalence relation on the cantor set that identifies the two endpoints of each removed interval is a counterexample. In this case, the quotient is again the unit inverval. $\endgroup$– HenrikRüpingOct 17 at 9:14
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Yes. If you take two distinct points outside $A$, there are two clopen subsets disjoint of $A$ and separating them, hence at the quotient these map to clopen subsets, separating them. If you take a point outside $A$, there's a single clopen subset containing it and disjoint of $A$, and at the quotient this and its complement are two clopen subsets separating the two images.
(I'm using that the image of a clopen subset is clopen as soon as it contains or is disjoint of $A$.)
A simple generalization is the following: if $X$ is totally disconnected compact Hausdorff, $A$ a closed subset, $A\to B$ a surjective continuous map to another totally disconnected compact Hausdorff space, then "collapsing $A$ to $B$" yields a totally disconnected compact Hausdorff space.
Above is when $B$ is a singleton.
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$\begingroup$ I think it's worth explicitly writing down that you use the facts that for compact Hausrdoff spaces, ⓐ “totally disconnected” and “totally separated” are equivalent, so we can separate distinct points by complementary clopen sets, and ⓑ regularity holds, so we can separate points from closed sets. This makes me wonder if the statement still holds when we remove “compact” (i.e., is the collapse of a closed set in a totally disconnected Hausdorff space still totally disconnected?). $\endgroup$– Gro-TsenOct 18 at 8:11
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$\begingroup$ (I feel like the answer to the last sentence in my previous comment had been answered in comments to the question itself which where then, sadly, deleted.) $\endgroup$– Gro-TsenOct 18 at 8:12
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$\begingroup$ Indeed. The idea (due to @bof) was to write the Knaster-Kuratowski fan as a quotient of a similar space where the top looks exaktly like the bottom,i.e. it is a Cantor set. and then one can divide out the top to get the fan. $\endgroup$ Oct 18 at 11:35