Show that a compact Hausdorff space $X$ is totally disconnected if and only if for every two distinct $x, y ∈ X$, there exists a clopen set containing $x$ but not $y$.
My attempt:
Let $X$ be a compact Hausdorff Space which is totally disconnected. Since $X$ is compact Hausdorff, equivalent classes coincide with components. $X$ is totally disconnected, so the equivalence class of a point $x∈X$ is {$x$}. If $y∈X$ is a different point, then $y$ isn't in the equivalence class of $x$. Thus there exist disjoint open sets $U,V$ containing a,b, respectively, such that $X=U∪V$.
How to prove the other inclusion?
Best Answer
It makes no sense to talk about equivalence classes until you tell us what the equivalence relation is. From context I can guess that you’re talking about the quasicomponents of $X$, but you really need to make that explicit. Assuming that I’ve guessed correctly, your argument is fine otherwise.
The other direction is actually the easier one, since it’s true whether or not $X$ is compact. Suppose that $C\subseteq X$ has at least two points. Let $x,y\in C$ with $x\ne y$; by hypothesis there is a clopen set $H$ containing $x$ but not $y$. Then $C\cap H$ and $C\setminus H$ form a separation of $C$, so $C$ is not connected. Thus, the components of $X$ are the singletons $\{x\}$ for $x\in X$, and $X$ is therefore totally disconnected.