TITLE: How do I prove that compact-open topology is metrizable?
QUESTION [5 upvotes]: Let $X$ be a $\sigma$-compact topological space and $(Y,d)$ be a metric space.
Let $\{K_n\}$ be a sequence of compact subsets of $X$ whose union is $X$.
Define $\rho_n(f,g):=\sup \{d(f(z),g(z)): z\in K_n\}$ and $\rho(f,g)=\sum_{n=0}^\infty (\frac{1}{2})^n \frac{\rho_n(f,g)}{1+\rho_n(f,g)}$ for all $f,g\in C(X,Y)$.
Then, it can be directly checked that $\rho$ is a metric on $C(X,Y)$.
Assuming $K_n\subset int(K_{n+1})$ for all $n$, it can be easily shown that $\rho$ induces the compact-open topology on $C(X,Y)$. However, this conditions seems too strong.
In Conway's functions of one complex variable text, it's written there that if $X$ is Baire in addition, then it can be still shown that $\rho$ induces the compact-open topology. However, I don't underatand why. How is it so? Is there any reference for it?

REPLY [5 votes]: Some more remarks: also note that you can embed $(Y,d)$ at zero price into a Banach space by the Fréchet-Kuratowski isometric embedding, so w.l.o.g. you can assume $Y$ itself is  a Banach space. The compact–open topology makes $C(X,Y)$ a LCTVS, with the family of seminorms $ \{  \|\cdot\|_{\infty,K}  \}   $ for all $K\Subset X$. It is metrizable iff it is first-countable iff $X$ is hemicompact, that is the family of its compact subsets has countable cofinality.