5
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ Why the $C_{\alpha_{i}}$ are open? $\endgroup$ – BBVM Aug 7 '16 at 2:13
  • $\begingroup$ @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? $\endgroup$ – Blake Aug 7 '16 at 4:33
  • 1
    $\begingroup$ 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. $\endgroup$ – Alex Provost Aug 7 '16 at 4:42
  • $\begingroup$ @Alex Provost Okay, that makes sense. I think I understand now. Thank you. That was just the observation I needed. $\endgroup$ – Blake Aug 7 '16 at 4:47
  • $\begingroup$ $\subset$ and $\subseteqq$ are synonymous. If $X$ is connected then $C=X$ and we must have $C=U=V=X$. $\endgroup$ – DanielWainfleet Aug 10 '16 at 1:03
2
$\begingroup$

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 $\subseteq$.)

$\endgroup$
  • $\begingroup$ @Blake: You’re welcome! $\endgroup$ – Brian M. Scott Aug 8 '16 at 15:01
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.