If $(X,d)$ is a metric space then is Homeo$(X)$ (the group of homeomorphisms of $X$ with itself) endowed with the compact open topology metrizable?

At first I thought I could define a metric on Homeo$(X)$ using the metric $d$ but I can't find a good way to do that. I am not certain how to prove Homeo$(X)$ is regular and has a countable basis either.

If this question can't be answered in general then can it be done in the case when $X$ is connected, locally path connected and locally compact?

Thank you.

  • 2
    $\begingroup$ In general no: for example, space of homeomorphisms of the real line is not normal; and hence not metrisable. $\endgroup$ – Mozibur Ullah Oct 3 '15 at 13:34
  • 1
    $\begingroup$ This satisfies your additional conditions; so these restrictions don't allow a metric to be given there either. $\endgroup$ – Mozibur Ullah Oct 3 '15 at 13:35
  • 1
    $\begingroup$ where can I find a proof for the fact that Homeo$(\mathbb{R})$ is not normal? $\endgroup$ – R_D Oct 3 '15 at 13:37
  • 3
    $\begingroup$ If $X$ is compact it's just the sup metric. $\endgroup$ – user98602 Oct 3 '15 at 14:05

First of all, let $(X,d)$ be an uncountable set equipped with the discrete metric. The space $Homeo(X,d)$ is the set of bijections $X\to X$. The compact-open topology on this space is simply the topology of pointwise convergence, i.e., the subspace topology induced by the product topology on $X^X$. From this, it is easy to see that $Homeo(X,d)$ is not 1st countable and, hence, not metrizable. On the other hand, according to

Robert A. McCoy, Ibula Ntantu, Topological Properties of Spaces of Continuous Functions, Springer Lecture Notes in Math, Volume 1315 (1988), page 68,

for a metrizable space $X$, the space of continuous functions $C(X,X)$ (equipped with the compact-open topology) is metrizable iff $X$ is hemicompact, i.e. contains a countable collection of compact subsets $C_n\subset X$ such that every compact in $X$ is contained in some $C_n$. Thus, for example $Homeo({\mathbb R}, {\mathbb R})$ is metrizable (I am not sure where did the statement by Mozibur Ullah come from), since it is a subspace of a metrizable space.


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.