[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?
Best Answer
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.