General Topology – Intersection of Clopen Sets Containing x is the Connected Component of x

Question

Let $X$ be a topological space, $x\in X $, $C$ is a connected component of $x$.

Define $A$ to be the intersection of all the open-and-closed sets that contain $x$ (also called the pseudo-component sometimes).

I wish to show that $A=C$, if $X$ is also compact (without it I think there is a counter-example).

Obviously $C$ is contained in $A$ (always) since any clopen set that contains $x$ contains $C$.
Also, $A$ is closed in $X$ so if $X$ is compact, $A$ is also compact. Not sure on how to proceed (tried supposing that $A$ is not contained in $C$ and getting a cover of $A$, doesn't seem to work).

Any help will be appreciated!

Answer 1

This does not work without the assumption that $X$ is normal. For a counterexample, consider the

$$**\textbf{sequence with two limits}**$$ Let $X=\{0,0'\}\cup\{1/n\mid n\in\Bbb N\}$, where $\{1/n\mid n\in\Bbb N\}$ is discrete and a neighborhood of $0$ or $0'$ is a set containing almost all of the points $1/n$. This space is compact but not normal, and $\{0,0'\}$ is the quasi-component of $0$ but is not connected.

If we assume that your $X$ is normal, then we can prove that $A$ is connected:
Assume $A$ splits into two components $B,D$. Since $A$ is closed, $B$ and $D$ are both closed, so if $X$ is normal there are disjoint open neighborhoods $U$ and $V$ around $B$ and $D$, respectively. The open sets $U$ and $V$ cover the intersection of all clopen neighborhoods of $A$, so if $X$ is compact, there must exist a finite number of clopen sets around $A$, say $A_1,\dots,A_n$ such that $U\cup V$ covers $K=\bigcap_1^n A_i$. Can you finish the proof from here?

(If you want to see the solution, hover over the shaded area below)

Note that $K$ is clopen. We can assume that $x\in U$. It is not difficult to see that $K\cap U$ is clopen and does not contain all of $A$, contradicting the definition of $A$.