How do I prove that compact-open topology is metrizable? 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?
 A: I don't think this does induce the compact-open topology as stated.
Let $X = \{1,1/2, 1/3, \dots, 0\}$ with its usual Euclidean metric (so $X$ is a compact metric space).  Let $K_0 = \{0\}$ and $K_n = \{1/n\}$, so that $K_n$ is compact and $X = \bigcup_n K_n$.  Let $Y = \{0,1\}$ with the obvious metric.  Now the compact-open topology on $C(X,Y)$ should simply be the topology of uniform convergence.
Now $$\rho(f,g) = \frac{1}{2} \left( |f(0) - g(0)| + \sum_{n=1}^\infty 2^{-n} |f(1/n) - g(1/n)|\right)$$
where the $1/2$ comes from the fact that $\rho_n/(1+\rho_n)$ is either 0 or 1/2. 
Let $f_n(x) = 1$ if $x = 1/n$ and 0 otherwise.  Let $f = 0$.  Then $\rho(f_n, f) = 2^{-(n+1)} \to 0$ but $f_n$ does not converge to $f$ uniformly.
A: 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.
