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I want to show that a subset of a metric space $X$ is closed iff its intersection with every compact subset of $X$ is closed

Stefan Hamcke
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Sherif Gamal Salem
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2 Answers2

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Let $(X,d)$ be a metric space and $Y \subset X$ be such that $Y \cap F$ is closed for any closed set $F$.

Let $(x_n)$ be a sequence in $Y$ converging to some $x \in X$. Then $A=\{x_n: n \geq 0\} \cup \{x\}$ is compact. By assumption, $A \cap Y$ is closed, so $x \in \overline{A \cap Y} =A \cap Y \subset Y$. You deduce that $Y$ is closed.

Showing that if $Y$ is closed, then its intersection with every compact subset of $X$ is closed is a consequence of the following fact (in metric spaces): a closed subset of a compact set is compact, hence closed in $X$.

Seirios
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    So we see that we need not use all compact sets, only these special ones: a convergent sequence together with its limit. – GEdgar Mar 06 '13 at 13:40
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If $F\subset{X}$ is closed and $C\subset{X}$ is compact, then $C$ is also closed (compact in Hausdorff is closed), and hence the intersection $F\cap{C}$ is closed as well.

Axiom
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