Specifically, I'm wondering, if X and Y are Hausdorff, and Y is compactly generated, does it follow that C(X,Y), with the compact-open topology, is compactly generated?

Edit: answered as written, but curious about other conditions that do imply the compact-open topology is compactly generated. E.g. if we strengthen the assumption to Y being locally compact?

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    $\begingroup$ If $Y$ is metrisable and $X$ is hemicompact, then $C(X,Y)$ is metrisable, hence compactly generated. The hemicompactness of $X$ is essential. First-countability of $Y$ is not sufficient for $C(X,Y)$ to be compactly generated (even when $X$ is compact metric). Neither is local compactness of $Y$ sufficient: neither $C(\mathbb{Q},\mathbb{R})$ nor $C(\mathbb{Q},I)$ is compactly generated. $\endgroup$
    – Tyrone
    Commented Apr 17, 2023 at 1:08

1 Answer 1


Not necessarily: consider the compactly generated space $Y=\mathbb R^\infty=\varinjlim \mathbb R^n$, which is the direct limit of Euclidean spaces. Then for the countable discrete space $X=\omega$ the function space $C(X,Y)$ is homeomorphic to $(\mathbb R^\infty)^\omega$ and hence is not sequential and so is not compactly generated. To see that the space $(\mathbb R^\infty)^\omega$ is not sequential, one should apply the known fact that the product $\mathbb R^\infty\times\mathbb R^\omega$ is not sequential.

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    $\begingroup$ I was confused by this because I'd been under the impression that both X and Y being compactly generated was sufficient, but it looks like I was just wrong about that. Thanks. $\endgroup$ Commented Apr 14, 2023 at 16:56
  • $\begingroup$ I think different definitions are being used, since $(\mathbb R^\infty)^\omega$ should be compactly generated under the algebraic topologist's definition (ncatlab.org/nlab/show/compactly+generated+topological+space). With that definition the compact-open topology is always CG when X and Y are CG and X is Hausdorff. $\endgroup$ Commented Apr 14, 2023 at 18:15
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    $\begingroup$ @MarcHoyois I looked at the definition of compactly-generated in nLab. It is just the definition of a k-space, well-known in General Topology. If all compact subsets of a topological space $X$ are metrizable, then it is a $k$-space if and only if it is sequential. And the space $(\mathbb R^\infty)^\omega$ has all compact subsets metrizable. Since this space is not sequential, it is not compactly generated. $\endgroup$ Commented Apr 14, 2023 at 18:57
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    $\begingroup$ @MarcHoyois The compact-open topology does not preserve the compact-generacy. If you will change the compact-open topology on $C(X,Y)$ by the stronger topology induced by compact subsets, then on will loose many important properties of $C(X,Y)$. For example, if $Y$ is a topological group, then $C(X,Y)$ is a topological group in the compact-open topology but not necessarily in the stronger compact-generated topology on $C(X,Y)$. This problem (with failure of compact generacy in function spaces) does not have good positive solution. $\endgroup$ Commented Apr 14, 2023 at 19:00
  • $\begingroup$ @TarasBanakh Thanks for the follow-up, of course you are right! Some of the statements on the nLab page and the sources therein are easily misinterpreted... $\endgroup$ Commented Apr 15, 2023 at 8:15

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