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?



How do I prove that compact-open topology is metrizable?
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.


