# Under what conditions is the compact-open topology compactly generated?

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?

• 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. Commented Apr 17, 2023 at 1:08

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.
• 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. Commented Apr 14, 2023 at 18:15
• @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. Commented Apr 14, 2023 at 18:57
• @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. Commented Apr 14, 2023 at 19:00