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I know that if a set is closed then it can be expressed as a countable union of compact sets. But is the converse true?

I am asking because while reading a real analysis book, the following comment is remarked: Let $\bar{B_k}$ the the closure of a ball with radius $k$ centered at 0. If $F\cap\bar{B_k}$ is closed for any $k \in \mathbb{Z}^+$ then $F$ is closed. Or is there another way of showing the passing remark?

Tomas Jorovic
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    in the real line, let each compact set be a single point at $x=1/n.$ – Will Jagy Nov 02 '13 at 20:18
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    Or take the sets $[1/n, 1-1/n]$ which are clearly closed but their union is an open set (in the usual topology). – Old John Nov 02 '13 at 20:20
  • Alright, so how do I go about proving my problem then? – Tomas Jorovic Nov 02 '13 at 20:22
  • The remark is true because the interiors of the balls cover the entire space. So if there was a boundary point $p \in \overline{F}\setminus F$, it would lie in some $B_k$. Then it would also be in $\overline{F\cap\overline{B}_k}\setminus (F\cap \overline{B}_k)$. – Daniel Fischer Nov 02 '13 at 20:24

2 Answers2

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As the comments showed, it is easy to construct counterexamples showing that the converse is false.

For your second paragraph, you can just prove $F$ is closed by proving it contains all its limit points: let $x$ be a limit point of $F$, with $(x_n) \subset F$ and $x_n \to x$ as $n \to \infty$. Any convergent sequence in a metric space is bounded, so there exists $k > 0$ with $x_n \in F \cap {\overline B_k}$ for all $n$. $F \cap {\overline B_k}$ is closed, so $x \in F \cap {\overline B_k} \subset F$.

Stefan Smith
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In $\mathbb{R}^d$, every open subset is a countable union of compacta.

ncmathsadist
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