Let $U\subseteq X$ be an open set such that $\bigcap\limits_{n=1}^\infty A_n\subseteq U$. Show that $A_m\subseteq U$ for some $m\geq 1$

Question

Let $X$ be a compact topological space and let $A_1,A_2,...$ be closed sets in $X$ with $A_1\supseteq A_2\supseteq...$. Let $U\subseteq X$ be an open set such that $\bigcap\limits_{n=1}^\infty A_n\subseteq U$. Show that $A_m\subseteq U$ for some $m\geq 1$.

My attempt: Since $X$ is compact we know that each $A_n$ is also compact. Suppose on the contrary that $A_m\nsubseteq U$ for all $m\geq 1$. Then $\bigcap\limits_{n=1}^\infty A_n\subset A_m$, but $A_m\nsubseteq\bigcap\limits_{n=1}^\infty A_n$. Since the countable intersection of closed sets is closed, $\bigcap\limits_{n=1}^\infty A_n$ is compact.

At this point I'm out of ideas to go further, and I'm not sure where to go from here.

Answer 1

All $A_n^\complement$ are open sets and if $x$ is not any of them, $x \in \bigcap_n A_n$ and then by assumption $x \in U$. Also, the $A_n^\complement$ are an increasing family of open sets, as the $A_n$ are decreasing.

The previous remarks show that $\{U\} \cup \{A_n^\complement: n \in \Bbb N\}$ form an open cover of $X$, and so has a finite subcover, which is (WLOG) of the form $\{U\} \cup \{A_m^\complement\}$ for some $m$ (because we can take the largest index among the finitely many we used and it already includes the ones with smaller indices, if there are any originally).

But $U \cup A_m^\complement = X$ implies $A_m \subseteq U$, as required (any $x \in A_m$ would not be covered otherwise).

Answer 2

Since $X$ is compact the open cover $\mathcal C:=\{U\}\cup\{X-A_i:i\in\Bbb N\}$, comprised of $U$ and each complement $X-A_i$, must admit a finite sub-cover say $\mathcal C':=\{U,X-A_N\}$ implying $A_N$ is covered by $U\cup X-A_N$ and therefore $A_N\subseteq U$.