A first countable Hausdorff space is compactly generated

Question

I know that even a non-Hausdorff first countable space is compactly generated, but I assume that adding the property that the space is also Hausdorff, there is an easier proof. How would you prove that a first countable Hausdorff space is compactly generated? I assume using the fact that a compact subspace in a Hausdorff space is closed is to key to make the proof easier, but I don't see how.

I use the following definition for a compactly generated space: A space is compactly generated if (i) a subspace $ A $ is closed in $ X $ if and only if (ii) $ A\cap C $ is closed in $ C $ for all compact subspaces $ C\subseteq X $.

To show that (i) $ \Rightarrow $ (ii) is easy. Since $ X $ is a Hausdorff space, $ C $ is closed and the intersection $ A\cap C $ is an intersection between two closed sets and hence closed in both $ C $ and $ X $.

What about the converse?

Answer 1

Suppose that $A$ is not closed; then there is an $x\in(\operatorname{cl}A)\setminus A$. Since $X$ is first countable, there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $A$ converging to $x$. Let $C=\{x\}\cup\{x_n:n\in\Bbb N\}$; then $C$ is a compact subset of $X$, but $A\cap C=\{x_n:n\in\Bbb N\}$ is not closed in $X$.

First countability of $X$ is actually more than is needed: it suffices to assume that $X$ is sequential. If a subset of a sequential space is not closed, there is a sequence in it converging to a point not in it, which is precisely what we need here.

Answer 2

As Brian notes, sequential implies compactly generated in the sense OP descries, even without first-countability or $T_2$.

In this pi-Base pull request PatrickR points out that a slightly stronger property is shown in Proposition 1.6 of this paper: all sequential spaces are generated by maps from compact Hausdorff spaces. See this question for discussion on this variant.