I am using the following definitions (from Wikipedia):
- A space $X$ is locally compact if every $x \in X$ has a compact neighborhood;
- A space $X$ is compactly generated if a subset $A \subseteq X$ is closed if and only if $A \cap K \subseteq K$ is closed in every compact subset $K \subseteq X.$
According to Wikipedia's article on compactly generated spaces,
Every locally compact space is compactly generated.
I proved this claim for locally compact Hausdorff spaces, where compact subsets are also closed. My proof doesn't work in a non-Hausdorff case. I would like help with this case.
Best Answer
Note that to show that a space $X$ is compactly generated, it suffices to show that given any non-closed $A \subseteq X$ there is a compact $K \subseteq X$ such that $A \cap K$ is not a closed subset of $K$.
Let $X$ be a locally compact space, and suppose that $A \subseteq X$ is not closed. Taking any $x \in \overline{A} \setminus A$, let $K$ be a compact neighbourhood of $x$. Note that if $V$ is any (open) neighbourhood of $x$, then so is $V \cap \operatorname{Int} (K)$, and so $\varnothing \neq A \cap (V \cap \operatorname{Int}(K)) \subseteq (A \cap K ) \cap V$. Therefore $x \in \overline{A \cap K} \cap K = \operatorname{cl}_K ( A \cap K )$, and since $x \notin A \cap K$ it follows that $A \cap K$ is not a closed subset of $K$.