# Is Homeo$(X)$ metrizable?

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.

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