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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?

Martin Sleziak
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Barbara
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2 Answers2

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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.

Brian M. Scott
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  • So $X$ being Hausdorff does not add anything of interest to the proof? – Barbara Nov 22 '16 at 21:30
  • @Gjermund: Not that I can see at the moment, anyway. – Brian M. Scott Nov 22 '16 at 21:33
  • @BrianMScott: Your second $X$ should be $x$. I can not edit it because the edit is less then 6 characters. – Barbara Nov 24 '16 at 17:24
  • What is $C$ compact? Is it a one-point-compactification (Alexandroff compactification)? – Barbara Nov 24 '16 at 17:39
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    @Gjermund: Thanks for catching the typo. $C$ is compact because it’s a convergent sequence together with its limit: any open set that contains $x$ automatically contains all but finitely many of the $x_n$. (And yes, that does make it in effect the one-point compactification of a countably infinite discrete space.) – Brian M. Scott Nov 24 '16 at 18:19
  • In the end, you mean that $A \cap C \subset C$ closed instead of being closed in $X$, right? – MPB94 Nov 29 '17 at 10:16
  • @MPB94: No, I meant that it is not closed in $X$. Of course it is also not closed in $C$. – Brian M. Scott Oct 17 '22 at 17:43
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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.

Steven Clontz
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