Let $X$ be a topological space, $(Y, d)$ a metric space and $C(X, Y)$ the space of continuous maps with the topology of compact convergence.

Question: What is a minimal topological condition on $X$ such that $C(X, Y)$ is a sequential space?

The motivation of the question is the following:

Let $(f_n)$ be a sequence in $C(X, Y)$ that is equicontinuous and $ \overline{ \{ f_n(x) : n \in \mathbb{N} \} } $ is compact in $(Y, d)$ for every $x \in X$. By Ascoli theorem, the closure $ \overline{ \{ f_n : n \in \mathbb{N} \} } $ in $C(X, Y)$ is compact. Unfortunately, in general $(f_n)$ may not have any convergent subsequence (in the topology of compact convergence).

To guarantee a convergent subsequence, it is enough for $C(X, Y)$ to be a sequential space. Then $ \overline{ \{ f_n : n \in \mathbb{N} \} } $ would be a compact subspace that is also sequential, which implies that it is sequentially compact.

It is well known that if $X$ is hemicompact, then $C(X, Y)$ is metrizable. Of course $C(X, Y)$ would be sequential in this case. But the hemicompactness of $X$ is too strong a condition. We only need $C(X, Y)$ to be sequential, and metrizability is a big overkill.

On the other hand, it is not enough for $X$ to be a locally compact metric space. See this Math.SE page for a counter-example.

  • 1
    $\begingroup$ Here are some remarks on the special case that $Y=\mathbb{R}$ and $X$ is Tychonoff. Then $C(X)$ is (strongly) Fréchet-Urysohn iff it is sequential iff it is compactly generated. On the other hand, if $X$ is first-countable, then $C(X)$ is Fréchet-Urysohn iff $X$ is hemicompact. In general I think even the case that $Y=$ seperable metric is not too well understood. $\endgroup$
    – Tyrone
    Jun 23 at 11:11
  • 2
    $\begingroup$ The reference for the above is a paper by E.G. Pytkeev. A recent refernce to his paper is from Topological properties of some function spaces by Gabriyelyana and Osipov, which contains some statements you might be interested in. $\endgroup$
    – Tyrone
    Jun 23 at 11:23

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