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Let $(X,d)$ be a metric space. Show that $F \subseteq X$ is closed iff $F \cap K$ is closed for every compact set $K\subseteq X$.

If $F\subseteq X$ is closed then $K\subseteq X$ compact implies $K$ closed in $X$. Then $F\cap K$ is closed (finite interseccion of closed subsets.)

However trying to show the other direction, I'm having some problems. I've thought about the following. Given a sequence $(x_n)_{n \in \mathbb{N}}\subseteq F$ such that $x_n \to x$, I want to see that $x \in F$. I consider the subset $A=\{x_1,x_2,...,x_n,...,x\}\subseteq X$. If $A$ is compact, then $F\cap A$ is closed by hypothesis, and therefore since $(x_n)_{n \in \mathbb{N}}\subseteq F \cap A$ we have that $x_n \to x \in F\cap A$ and therefore, $x_n \to x \in F$ which implies $F$ closed.

However, I am not being able to prove that $A$ is compact. I thought about the following. Given an open cover $(U_i)_{i \in I}$ we have that $A\subseteq (U_i)_{i \in I}$, I want to see that there exist a finite subcover. Since $x_n \to x$, given $\epsilon > 0$ there exist $n_0 \in \mathbb{N}$ such that $d(x_n,x) < \epsilon$ if $n \geq n_0$. So if I take $x_{n_0}$, there exist $U_{i_0}$ with $i_{0}\in I$ such that $x_{n_0} \in U_{i_0}$, and since $U_{i_0}$ is open, there exist $\epsilon_0 > 0$ such that $B(x,\epsilon_0) \subset U_{i_0}$. The problem is that if $ \epsilon_0< \epsilon$ Then I can't be sure that all the $x_n$ with $n \geq n_0$ will be in $U_{i_0}$, and of course the problem is that the $\epsilon_{0}$ of the $U_{i_0}$ depends on the $x_{n_0}$ and this one depends on the $\epsilon$ given. Am I on the correct track? How can I solve this?

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  • $\begingroup$ If $x_n \to x$, then the set $\{x_n\} \cup \{x\}$ is compact. $\endgroup$ – copper.hat May 26 '15 at 2:02
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To show that $A = \{ x, x_1, x_2, \ldots \}$ is compact, you can do the following: Let $\{ U_i \}_{i \in I}$ be an open cover of $A$. Now there exists a $i_0 \in I$, such that $x \in U_{i_0}$. Since $x_k \to x$, there exists a $N \in \Bbb N^\times$, such that $x_k \in U_{i_0}$ for all $k \geq N$. For each $k \in \{1, \ldots, N-1\}$ we find a $i_k \in I$, such that $x_k \in U_{i_k}$, so $\{U_{i_k}\}_{k=0}^{N-1}$ is a finite subcover of $\{U_i\}_{i \in I}$.

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Here's one idea. Let $x_n$ be a sequence of points in $F$ which converges to $x \in X$ (and we do not yet know whether $x \in F$). Consider the set $K=\left \{ x_n : n \in \mathbb{N} \right \} \cup \{ x \}$. $K$ is compact (why?), so by assumption $F \cap K$ is closed. Hence $x \in F \cap K$ and hence $x \in F$.

I think this is easier than trying to play with open covers, partly because $F$ needn't be compact.

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  • $\begingroup$ @Jose27 Good catch, fixed. $\endgroup$ – Ian May 26 '15 at 11:46

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