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Let $\{X_n\}$ be a sequence of compact subsets of a metric space $M$ with $X_1\supset X_2\supset X_3\supset\dotsm$. Prove that if $U$ is an open set containing $\bigcap X_n$, then there exists $X_n\subset U$.

So suppose for contradiction that there doesn't exist $X_n\subset U$, i.e. every $X_n$ has a part outside $U$. Since $\bigcap X_n$ is a subset of an open set $U$, every element of $\bigcap X_n$ has an open ball within $U$. To use compactness, I want to find an open cover for $X_n$. But I don't see any open cover to use...

Edit: Okay, following the hint that someone posted (and deleted), I think I got it.

Compact sets are closed, so $X_n$ is closed for all $n$, so $X_n'$ is open for all $n$ (where $'$ denotes complement.) Note that $X_n$ has an open cover consisting of $U,X_{n+1}',X_{n+2}',\ldots$ since $X_{n+1}'\cup X_{n+2}'\cup\ldots = (X_{n+1}\cap X_{n+2}\cap \ldots)' = (\bigcap X_i)'$. Since $X_n$ is compact, there exists a finite subcover $U,X_{a_1}',X_{a_2}',\ldots,X_{a_k}'$ where $a_1<a_2<\ldots<a_k$. But this is a contradiction, since the part of $X_n$ which is the part of $X_{a_k}$ outside $U$ is not covered.

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The collection $\{U, X_2^c, X_3^c,... \}$ is an open cover of $X_1$, which is compact. To see this, suppose $x \in X_1$. Then either $x$ is in a finite number of the $X_n$, or $x \in \cap_n X_n \subset U$

Hence there is a finite subcover. We may take the subcover to be $U,X_{i_1}^c,...,X_{i_k}^c$, and since they are nested, the cover can be $U,X_{\max(i_1,...,i_k)}^c$. Since $X_{\max(i_1,...,i_k)} \subset X_1$, it follows that $X_{\max(i_1,...,i_k)} \subset U$.

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