# Metrization of spaces of functions

Let $M$ and $N$ be topological spaces. Are there necessary and sufficient conditions on the topological properties of the spaces such that $C(M,N)$ is metrizable?

For $M$ compact and $N$ a metric space, the space is obviously metrizable using the uniform convergence topology, $d(f,g)=\sup_{x\in M}d(f(x),g(x))$.

And also, if $N$ is a metric space, but $M$ is not necessarily compact the space of continuous bounded functions $C_0(M,N)=\{f\in C(M,N)\mid d(f(x),a)\leq K_f, \forall x\in M\}$ for a point $a\in N$ and $K_f>0$ is metrizable with the same distance.

But in general, which distances are usable in $C(M,N)$ in the context of a noncompact space $M$?

• For the metrization question to make sense, you first need to start with some topology on $C(M,N)$ (e.g., compact-open). Some metrization conditions are well known (en.wikipedia.org/wiki/Metrization_theorem, en.wikipedia.org/wiki/…) so if $C(M,N)$ fits them, it is metrizable. But perhaps this is not the kind of answer you are looking for... In that case, what is the motivation behind the question? – Igor Khavkine Sep 16 '13 at 20:09
• Im trying to obtain the less restrictive conditions on the topological properties on $M$, $N$, (Combinations of second- countable, lindelöf, hausdorff, paracompact or some othe properties) such that $C(M,N)$ is metrizable with some dynamically relevant topology in the sense that the stability notion of dynamical systems (en.wikipedia.org/wiki/Topological_conjugacy) is not vacuous. – user40076 Sep 16 '13 at 20:25
• I think one can repeat to some extent the construction of the compact-open topology in $C(M,N)$ by means of an ideal of sets $\mathcal{I}$ on $M$ (instead of the family of compacts of $X$). The metrizability of $N$ and the countable cofinality of $\mathcal{I}$ should be again the condition for metrizability; one gets the distance of uniform convergence on elements of $\mathcal{I}$. – Pietro Majer Sep 17 '13 at 10:41

As to the compact-open topology of $C(X,Y)$, it is metrizable if and only if $Y$ is metrizable, and $X$ is hemicompact.
Perhaps the constant functions should form a topological subspace of $$\ C(M , N)$$   which is canonically isomorphic with $$\ N$$. This would eliminate the non-metrizable spaces $$\ N$$.