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. 



Metrizable implies hemicompact 
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. 


