1
$\begingroup$

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 $M$?

$\endgroup$
  • 3
    $\begingroup$ 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? $\endgroup$ – Igor Khavkine Sep 16 '13 at 20:09
  • 1
    $\begingroup$ 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. $\endgroup$ – user40076 Sep 16 '13 at 20:25
  • $\begingroup$ 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 $\mathcal{I}$. $\endgroup$ – Pietro Majer Sep 17 '13 at 10:41
9
$\begingroup$

As to the compact-open topology of $C(X,Y)$, it is metrizable if and only if $Y$ is metrizable, and $X$ is hemicompact.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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 $\ N$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.