nested compact set question Suppose $A \subset \mathbb R^n$ is not compact. Show that there exists a sequence $F_1 \supset F_2\supset F_3\supset\cdots$ of closed sets such that $F_k \cap A\ne\emptyset$ for all $k$ and $\left(\bigcap_{k=1}^\infty F_k\right)\cap A = \emptyset$.
I remember a similar theorem that assume $A$ is compact and concludes the intersection is not empty (not if and only if), this is the opposite version of the questions, how to achieve this?
 A: Hint: Since $A$ is not compact there is some open cover of $A$ which do not admit a finite subcover. Show that we may assume that this open cover is countable $\{U_n\mid n\in\mathbb N\}$, that $U_n\subseteq U_k$ for $n<k$, and that $A\nsubseteq U_n$ for all $n$.
Let $F_n=\mathbb R^n\setminus U_n$, and show that this sequence has the wanted properties.
A: This is not an answer, it just elaborates a point in Asaf's answer below.
This shows that any infinite open cover of $A$ can be taken to be a countable open cover.
Let ${\cal O}$ be a collection of open sets in $\mathbb{R}^n$, and let $\Omega = \cup_{U \in {\cal O}} U$.
$\mathbb{R}^n$ is second countable, so it has a countable base ${\cal B} = \{B_k\}$. Let ${\cal C} = \{B \in {\cal B} | \exists U \in {\cal O}, B \subset U\}$. Since ${\cal C} \subset {\cal B}$, it is countable. Let $C_k$ be an enumeration of ${\cal C}$. For each $k$, choose $U_k \in {\cal O}$ such that $C_k \subset U_k$. By renumbering, we may take the $U_k$ to be distinct.
By construction, $\{U_k\} \subset {\cal O}$ and is countable. Furthermore, $\Omega = \cup_k U_k$. It is immediate that $\cup_k U_k \subset \Omega$, and if $x \in \Omega$, then $x \in U$ for some $U \in {\cal O}$. Since $U$ is open, and ${\cal B}$ is a base, we have $x \in B \subset U$ for some $B \in {\cal B}$, and by construction, $B \in {\cal C}$, and hence $B \subset U_k$ for some $k$. Hence $x \in \cup_k U_k$.
If ${\cal O}$ is an (infinite) open cover of $A$, then the above shows that there is a countable open cover (composed of members of ${\cal O}$) that also covers $A$. Clearly if the original cover does not admit a finite subcover, then the countable open cover cannot either.
