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[In the question below, I use "compact" to mean "compact Hausdorff" and I use "quasi-compact" for what is commonly called "compact" in wikipedia and other sources.]

The problem is, let $X$ be a compact space and define $x\sim y$ if the points are in the same connected component. Show that $X/\sim$ is totally disconnected and compact with quotient topology.

My problem is not with the totally disconnected, it is with the compact, I know that the projection gives a quasi-compact quotient, but not compact because I have not proven that it is a Hausdorff space.

One idea is that compact spaces are normal and use it on two connected components of $X$ (which are closed sets), so you could separate points on $X/\sim$ with open sets, but it's not clear to me , because I think that open sets could intersect with another same component.

I know that totally disconnected spaces might not be hausdorff so I must think that the space is a quotient with this relation, the same goes for saying that a quotient of a hausdorff space might not be hausdorff.

I don't know if using the intersection of clopen sets containing a point $x \in X$ to coincide with the connected component of $x$ in a compact space works for this.

Does any of this work?

PatrickR
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Franco Gómez
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    The continuous image of a compact space is compact, the quotient map is surjective so the quotient is compact. – Emilio V. Nov 16 '23 at 06:23
  • The preimage of open sets is open under the quotient map. So for an open cover of the quotient space consider the preimages of the open sets which will be an open cover of $X$. Now use the compactness of $X$ to extract a finite subcover and pass to the quotient map. – CyclotomicField Nov 16 '23 at 09:31
  • @CyclotomicField we are using the definitiom that space its compact iff every open coverment has finite subcover and its hausdorff space – Franco Gómez Nov 16 '23 at 09:53
  • @EmilioV. In that proof why quotient its hausdorff? – Franco Gómez Nov 16 '23 at 09:55
  • To prove the quotient is Haussdorff separate the connected components and pass the open sets that separate them to the quotient. Now all the point are separated in the quotient space. – CyclotomicField Nov 16 '23 at 10:06
  • @CyclotomicField extactly, but how. What a really want its for any classes [x] and [y] find open disjoint sets A and B in quotient. This means find two unions of connected component in X, that be open and disjoint. What I could do its using that X its normal, for find open disjoint sets A and B that A covers [x] and B covers [y] as set points in X, but nothing its saying that A_0 and B_0 are union of connected component – Franco Gómez Nov 16 '23 at 11:11
  • All the points in the quotient map are the connected components of $X$ so by separating the connected components in $X$ you're separating the points in $X / \sim$ after you pass to the quotient map. This is the property that makes it totally disconnected too. – CyclotomicField Nov 16 '23 at 11:26
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    @CyclotomicField You have to be a little careful when you separate - need to separate by saturated open sets in order to guarantee their images are still disjoint in the quotient. – M W Nov 16 '23 at 11:30
  • Are you using the same definition of compact as in wikipedia? That is the usual definition here. If not (i.e. if you are using the French convention), you need to make that clear in the body of your question. – PatrickR Nov 17 '23 at 06:38
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    @PatrickR If you read the question, it becomes very clear: "I know that the projection gives a quasi-compact quotient, but not compact because I have not proven that it is a Hausdorff space." – Paul Frost Nov 17 '23 at 08:57
  • @PaulFrost I know and I noticed exactly what you mentioned. What I was trying to get at is to help the OP understand that for the benefit of readers he should specify up front what he means (in case his usage is not entirely standard) instead of relying to people inferring things later on. So the OP can generate better questions in the future. – PatrickR Nov 18 '23 at 04:10
  • @FrancoGómez Unless you are a topologist, most people here will not know what you mean by "French convention". If you don't mind, I'll edit your post to make things clear. – PatrickR Nov 18 '23 at 04:13
  • Up to you, but I would still recommend that you use "compact Hausdorff" in general when that's what you mean. – PatrickR Nov 18 '23 at 04:18

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You will need to use the fact that in a compact Hausdorff space, quasicomponents coincide with components, so two distinct components may be separated by a clopen (closed and open) set.

Then for components $C_1\neq C_2$ in $X$, let $U\subset X$ be clopen with $U\supseteq C_1$ and $U\cap C_2=\emptyset$.

Note that $U$, as a clopen set, is saturated. That is, if $x\sim y$ then $x\in U$ if and only if $y\in U$. This follows from the fact that otherwise, $U$ and $X\backslash U$ would separate the component containing $x$ and $y$.

Since $U$ is saturated, if $\pi\colon X\to X/\sim$ is the quotient map, we have $\pi^{-1}(\pi(U))=U$, so that $\pi(U)$ is clopen in $X/\sim$, contains $\pi(C_1)$, and is disjoint from $\pi(C_2)$.

Then the disjoint open sets $\pi(U)$ and $(X/\sim)\backslash \pi(U)$ separate $\pi(C_1)$ and $\pi(C_2)$. Since this can be done for arbitrary components, $X/\sim$ is Hausdorff.

M W
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  • How is that first implies the existence of thats clopen sets? I know that there are open sets that separate closed sets – Franco Gómez Nov 16 '23 at 12:10
  • @FrancoGómez Two points are in the same quasicomponent if and only if they cannot be separated by a clopen set. So given two distinct components, we know they are distinct quasicomponents (from the linked result), so there is a clopen set separating them – M W Nov 16 '23 at 12:13
  • The idea its: If I have A and B connected component in X, fix x in A, and so for every y in B there is a clopen C_y cointaining y but not x, the union of this sets cover B, this implies that there is a clopen set containing B but not x (because X its compact so finite union its clopen). Now how construct clopen set containing set A? – Franco Gómez Nov 16 '23 at 12:31
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    @FrancoGómez choose a single $x\in A$ and a single $y\in B$. From the linked result there is a clopen set $U$ containing $x$ but not $y$. Since components are connected, a given component is always either contained in a given clopen set or disjoint from it. Therefore $U$ contains the component $A$ and is disjoint from the component $B$. – M W Nov 16 '23 at 12:39
  • Can you explain how to find a clopen with $U\supseteq C_1$ and $U\cap C_2=\emptyset$ if $C_1\neq C_2$ ? – Kritiker der Elche Nov 17 '23 at 11:32
  • @KritikerderElche in the linked post, it shows that quasi components are the same as components, so as in my previous comment, you just choose one point from each component, the linked result gives a clopen set separating the two points, and then by connectedness it separates the two components. – M W Nov 17 '23 at 18:17