# Quasi-components and components coincide for compact Hausdorff spaces

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 $$B = \emptyset$$.

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.

• why are $A,B$ closed in $X$? @PaulFrost Jun 6 '20 at 16:22
• @RubenBaeza $A,B$ are closed in $C$ and $C$ is closed in $X$. Jun 6 '20 at 16:52
• @PaulFrost Maybe it is a dumb question but why $\bigcap_{\alpha} C_{\alpha} \subset U$ implies that $C_\alpha \subset U$ for some $\alpha$? May 1 '21 at 17:00
• @Zanzag See my update. May 1 '21 at 23:00