# Proof regarding nested compact subsets of a metric space

Let $\left\lbrace X_n \right\rbrace$ be a sequence of compact subsets of a metric space $M$ with $X_1 \supset X_2 \supset X_3 \supset \dots$. Prove that if $U$ is an open set containing $\cap X_n$, then there exists $X_n \subset U$.

This proof is tricky for me because I can't use many facts about compactness beyond its definition (a metric space $M$ is compact if every open cover of $M$ has a finite subcover). For example, I can't use the fact that every sequence in a compact metric space has a convergent subsequence, or the fact that a compact subset is closed and bounded. I do "know" that a finite union of compact subsets is compact and the intersection of compact subsets is compact.

My main idea has been a proof by contradiction in which I assume that $X_n \not\subset U$ for all $n$. Then I find a point in each $X_n$ that is not in $U$ and build a sequence $x_n$ that converges to a point in $U$. Since $x_n$ is in the closed subset $M \backslash U$, I would have a contradiction. But I can't figure out how to show that such a sequence exists.

• Well. $X_n$ are not empty and $U$ doesn't contain any $X_n$ , then you have at the very least a constant sequence. – Jacky Chong Oct 19 '16 at 4:04
• Okay. But a constant sequence in $M \backslash U$ could not converge to a point in $U$. – mr_gasbag Oct 19 '16 at 4:08
• If $x_1 = x_2 =x_3 = \ldots$ which belongs in all the $X_n$ then the limit is in $\bigcap X_n \subset U$. – Jacky Chong Oct 19 '16 at 4:09
• So I argue that if $X_n \not\subset U$ for all $n$, then there must exist a point $x \in X_n \cap M \backslash U$ for all $n$. Then $x \in \cap X_n \subset U$, a contradiction. I'm not so sure about this because it doesn't make use of the openness of $U$ or the compactness of $X_n$, and I know that the result doesn't hold without those restrictions. Am I missing something? – mr_gasbag Oct 19 '16 at 4:33

Given the tools available to you, sequences aren’t the best way to go.

HINT: Let $K=X_1\setminus U$; $K$ is a closed subset of the compact set $X_1$, so $K$ is compact. For each $n\in\Bbb Z^+$ let $V_n=M\setminus X_n$, and let $\mathscr{V}=\{V_n:n\in\Bbb Z^+\}$.

• Show that $\mathscr{V}$ is an open cover of $K$.
• Conclude that there is an $n\in\Bbb Z^+$ such that $K\subseteq V_n$.
• What does this tell you about $X_n$?
• This looks like it would work. But in order to show that $\mathscr{V}$ is an open cover of $K$, I need to show that $V_n = M \backslash X_n$ is open in $M$ for each $n$, which means (I think) using the theorem that the $X_n$ compact implies $X_n$ closed, which is not available to me. – mr_gasbag Oct 19 '16 at 15:13
• @mr_gasbag: Nothing keeps you from proving it as a step in the larger proof. If $x\notin X_n$, then for each $y\in X_n$ there are disjoint open sets $G_y$ and $H_y$ such that $x\in G_y$ and $y\in H_y$. $\{H_y:y\in X_n\}$ is an open cover of $X_n$, so it has a finite subcover, say $\{H_y:y\in F\}$ for some finite $F\subseteq X_n$. Then $\bigcap_{y\in F}G_y$ is an open nbhd of $x$ disjoint from $X_n$, and $X_n$ is therefore closed. – Brian M. Scott Oct 19 '16 at 15:18
• Great, thank you very much! – mr_gasbag Oct 19 '16 at 20:51
• @mr_gasbag: You’re very welcome. – Brian M. Scott Oct 19 '16 at 21:03

Suppose not. For each $n$ let $p_n\in X_n$ \ $U.$ Then $(p_n)_{n\in N}$ is a sequence in $X_1$ so there exists $q\in X_1$ and a strictly increasing $f:N\to N$ such that $(p_{f(n)})_{n\in N}$ converges to $q.$

And for $j>1,$ the sequence $(p_{f(n)})_{f(n)\geq j}$ is a sequence in $X_j,$ also converging to $q,$ so $q\in X_j.$

So $q\in \cap_{n\in N}X_n.$ So $q\in U.$

Now $(p_{f(n})_{n \in N}$ converges to $q ,$and $f$ is strictly increasing, so any nbhd of $q$ contains $p_{f(n)}$ for all but finitely many $n.$ And $U$ is a nbhd of $q$ (because $q\in U$). So $p_{f(n)}\in U$ for all but finitely many $n .$

This contradicts $\forall n\;(p_{f(n)}\not \in U).$