# Given $X$ compact and Hausdorff, $C$ a component of $X$, and $U$ open set containing $C$, show that $\exists V$ clopen such that $C\subset V\subset U$

Question

I'm struggling with the following:

Let $C$ be a connected component of a compact Hausdorff space $X$ and let $U$ be an open set containing $C$. Prove that there exists a clopen set $V$ such that $C\subset V\subset U$.

I think I have a solution, but what I've "proved" is stronger: that $C$ must be itself clopen, so $C$ itself can function as the desired $V$.

Attempt

Let $\{C_{\alpha}\,|\,\alpha\in A\}$ be the collection of components in $X$. Since $X$ is normal (compact and Hausdorff implies normal) and each $C_{\alpha}$ is closed, we can find a collection of disjoint open sets $\mathcal{U}:=\{U_{\alpha}\,|\,\alpha\in A\}$ such that $C_{\alpha}\subseteq U_{\alpha}$ for each $\alpha\in A$. Now $\mathcal{U}$ is an open cover of $X$, so via compactness there exists a finite subcover, $\mathcal{U}_0=\{U_{\alpha_1},U_{\alpha_2},\ldots \}\subseteq\mathcal{U}$. It follows that there are finitely many components of $X$ and hence each component is open: given $C_{\alpha_i}$, we see that $$X-C_{\alpha_i}=C_{\alpha_1}\cup\cdots\cup C_{\alpha_{i-1}}\cup C_{\alpha_{i+1}}\cup\cdots\cup C_{\alpha_n},$$ which is the finite union of closed sets.

Is this correct or have I made a grave mistake? If I've indeed made a mistake, could anyone point me in the right direction? Also, apologies for the frequent postings, I have a qualifying exam in a week and I've studying quite a lot lately.

Edit 1: I should add that this is the second part of a two part question. The first part was showing that the components and quasi-components coincide.

Edit 2: It's also worth noting that I am not quite sure whether "$\subset$" means a proper subset or not in the statement of the problem (this was a question taken from a old qualifying exam and I'm not quite sure who wrote it). However, if it were a proper subset, then I believe I could think of a counterexample to the statement. (I'm thinking of two disjoint closed disks in $\mathbb{R}^2$ where the $C$ is open of the disks and $U$ is one of the disks in union with some open set strictly contained in the other closed disk.)

• Why the $C_{\alpha_{i}}$ are open? – BBVM Aug 7 '16 at 2:13
• @BBVM Well, since there are finitely many components, and the components partition $X$, it's clear that $$X-C_{\alpha_i}=C_{\alpha_1}\cup\cdots\cup C_{\alpha_{i-1}}\cup C_{\alpha_{i+1}}\cup\cdots\cup C_{\alpha_n}.$$ Since the right-hand side is a finite union of closed sets, $X-C_{\alpha_i}$ is closed, and hence its complement, $C_{\alpha_i}$, is open. Is there any issue with this argument? – Blake Aug 7 '16 at 4:33
• The problem with your argument is that the disjoint open sets $U_\alpha$ need not exist; normality only applies to pairs of closed subspaces, not an infinite amount of them. For example, if $X$ is the closure of the sequence $\{ 1/n \}$ in $\mathbb{R}$, then $\{0\}$ is a connected component which isn't open. – Alex Provost Aug 7 '16 at 4:42
• @Alex Provost Okay, that makes sense. I think I understand now. Thank you. That was just the observation I needed. – Blake Aug 7 '16 at 4:47
• $\subset$ and $\subseteqq$ are synonymous. If $X$ is connected then $C=X$ and we must have $C=U=V=X$. – DanielWainfleet Aug 10 '16 at 1:03

The answer by user254665 shows where you went astray.

HINT: You know from the first part that $C$ is the quasicomponent of $x$. (There is also a proof here.) Now use the fact that the quasicomponent of a point is the intersection of all clopen sets containing $x$. (If this isn’t your definition of quasicomponent, you’ll have to show that it follows from your definition.) Thus, if $\mathscr{H}$ is the family of all clopen sets containing $C$, we have $C=\bigcap\mathscr{H}$.

For each $H\in\mathscr{H}$ let $\widehat H=X\setminus H$. For each $y\in X\setminus U$ there is an $H_y\in\mathscr{H}$ such that $y\in \widehat{H_y}$. Clearly $\{\widehat{H_y}:y\in X\setminus U\}$ is an open cover of the compact set $X\setminus U$, so ... ? In case you get completely stuck, I’ve completed the answer in the spoiler-protected block below.

So there is a finite $\mathscr{F}\subseteq\mathscr{H}$ such that $\{\widehat H:H\in\mathscr{F}\}$ covers $X\setminus U$. But then $\bigcap\mathscr{F}$ is a clopen nbhd of $C$ contained in $U$.

(The inclusions in the question are non-strict, i.e., what I would write $\subseteq$.)

• @Blake: You’re welcome! – Brian M. Scott Aug 8 '16 at 15:01

The real subspace $X=\{0\}\cup\{1/n: n\in N\}$ is compact Hausdorff . The maximal connected subspaces of $X$ are $\{\{p\}: p\in X\}$ and there are infinitely many of them, and one of them, $\{0\},$ is not open.

Your error is assuming that any family $F= \{C_a\}_{a\in A}$ of pairwise-disjoint closed subsets can be completely separated. The problem is that some $C_a$ may fail to be disjoint from the closure of the set $\cup (F$ \ $\{C_a\}).$

I am working on a proof. Thought I had it but I didn't.