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In the paper

R. Arens: A Topology for Spaces of Transformations, Ann. of Math. 47(1946), 480-495

the author states in the introduction that if $B$ is a metric space and the space of continuous functions $C(A,B)$ in the compact-open topology is metrizable, then $A$ is hemicompact.

Later on, Theorem 8 says: If $C(A,B)$ satisfies the first axiom of countability and if for all points $x,y \in A$ there is a continuous $f: A \to \mathbb{R}$ with $f(x) \neq f(y)$, then $A$ is hemicompact.

Now, if $C(A,B)$ is metrizable, it's first countable. But I don't know how to find the $f$.

Can anyone give a hint ?

Added: In light of David's comment, it's clear that Theorem 8 alone doesn't imply hemicompactness. Thus the question:

Is it true that if $B$ is a metric space and $C(A,B)$ is metrizable, then $A$ is hemicompact ?

From the quoted Theorem 8 it's clear that if $A$ has in addition the separation property by continuous functions, then it's hemicompact. But since the converse ($A$ hemicompact, $B$ metric $\Rightarrow$ $C(A;B)$ metrizable) is true without this separation property, I wonder if it's really necessary.

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Why do you think $f$ should exist? It seems the existence of $f$ is an assumption in the second result, rather than a conclusion? Or perhaps I'm too tired. – Mark Grant Feb 29 '12 at 22:44
You can't find the f without an additional hypothesis. Let A be an indiscrete space and B a metric space. Then C(A,B) is homeomorphic to B and thus metrizable, but distinct points in A are not separated by continuous real-valued functions. – David Milovich Feb 29 '12 at 22:45
@David: I see. Thanks for the example. But isn't A in your example compact ? That would mean that Arens' introductory statement could be nevertheless true. – Ralph Feb 29 '12 at 22:57
up vote 2 down vote accepted

For regular spaces the implication is false in general: let $A$ be regular and such that all continuous real-valued functions on it are constant (see Problem 2.7.17 in Engelking's General Topology). Then $C(A,B)$ consists of constant functions only whenever $B$ is metrizable (or even just completely regular) and therefore it is metrizable too. However $\sigma$-compact regular spaces are Lindelöf, hence normal, which a space like $A$ definitely is not.

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Thanks for the answer. I thought about trying such a space (Hewitt's construction to be explicit), but shied away from figuring out how the compact subsets might look like. But just using "regular + $\sigma$-compact $\Rightarrow$ normal" is striking. – Ralph Mar 18 '12 at 11:50

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