Quasi-components and components coincide for compact Hausdorff spaces

Question

I am attempting to solve exercise 1.10.1 from Topology and Geometry by Bredon:

If $X$ is a compact Hausdorff space then show that its quasicomponents are connected (and hence that its quasi-components coincide with its components).

The definition of the relation used for quasi-components is "$d(p) = d(q)$ for every discrete valued map $d$ on $X$."

He gives the hint:
If $C$ is a quasi-component, let $C = \cap C_{\alpha}$ where the $C_\alpha$ are the clopen sets containing $C$. If $C$ is disconnected, then $C = A \cup B, A \cap B = \emptyset, A,B$ closed. Let $f:X \to [0,1]$ be $0$ on $A$ and 1 on $B$. Put $U = f^{-1}([0,\frac{1}{2}))$ and apply the following result:
Let $X$ be a compact space and let $\{C_\alpha \mid \alpha \in A\}$ be a collection of closed sets, closed with respect to finite intersections. Let $C = \cap C_\alpha$ and suppose that $C \subset U$ with $U$ open. Then $C_\alpha \subset U$ for some $\alpha$.

I don't understand which collection of closed sets to use. If we use the clopen sets $C_\alpha$, and we have that $C \subset U$ but $U \subset X - B$, so $B = \emptyset$, and we don't need some $C_\alpha \subset U$. Furthermore, how would we show that $C \subset U$, it seems to more or less assumes $B = \emptyset$.

Answer 1

I think the hint is somewhat misleading. In fact, the clopen sets $C_\alpha$ containing $C$ are the correct choice, but the choice of $U$ is inadequate. So let us do it properly.

The quasi-component $C = C(x)$ of a point $x \in X$ is the intersection of all clopen subsets of $X$ containing $x$. In particular $C$ is a closed set. Let $\mathfrak C = \{ C_{\alpha} \mid \alpha \in A\}$ denote the set of clopen subsets of $X$containing $x$.

If $C$ is disconnected, then $C$ is the union of two non-empty disjoint closed sets $A, B \subset C$. W.l.o.g. assume $x \in A$. Note that $A,B$ are closed in $X$ since they are closed subsets of the closed $C$.

Since $X$ is normal, we find open neighborhoods $V$ of $A$ and $W$ of $B$ such that $V \cap W = \emptyset$. We have $C = \bigcap_\alpha C_\alpha \subset U = V \cup W$. We claim that $C_\alpha \subset U$ for some $\alpha$. In fact, $\bigcap_\alpha (C_\alpha \cap (X \setminus U)) = (\bigcap_\alpha C_\alpha) \cap (X \setminus U) = C \cap (X \setminus U) = \emptyset$. The sets $C_\alpha \cap (X \setminus U)$ are compact, hence the finite intersection property applies to show that there are finitely many $\alpha_i \in A$ such that $\bigcap_i (C_{\alpha_i} \cap (X \setminus U)) = (\bigcap_i C_{\alpha_i}) \cap (X \setminus U) = \emptyset$. But clearly $C^* = \bigcap_i C_{\alpha_i} \in \mathfrak C$ and $C^* \subset U$.

We conclude $$\overline {C_\alpha \cap V} \subset \overline C_\alpha \cap \overline V = C_\alpha \cap \overline V = (C_\alpha \cap U) \cap \overline V = C_\alpha \cap (U \cap \overline V) = C_\alpha \cap V, $$ thus $\overline {C_\alpha \cap V} = C_\alpha \cap V$. Hence $C_\alpha \cap V$ is also clopen. It contains $A$ and a fortiori $x$. Thus $C \subset C_\alpha \cap V \subset V$ which is impossible.