# If $G$ is a group of isometries of $X$ then prove that $X/G$ and $X/\bar{G}$ are homotopically equvalent

Let $X$ be a connected, locally path connected, locally compact metric space. Let $G$ be a group of isometries of $X$ (that is a group of homeomorphisms of $X$ with itself that preserves distance). Let $\bar{G}$ be the closure of $G$ in Homeo$(X)$ endowed with the compact open topology.

Let $(P)$ be the property that for all $x,x' \in X$ and a neighbourhood $V$ of $x$ in $X$, if $x' \in \bar{G}x$ then $x' \in GV$

I want to prove that $X/G$ and $X/\bar{G}$ are homotopically equivalent.

I have shown that if $X$ and $G$ satisfy $(P)$ then they are homotopically equivalent (I have in fact shown this if $G$ is just a group of homeomorphisms of $X$ with itself).

So to prove this result I only need to show that property $(P)$ is true when $G$ is a group of isometries of $X$. But I don't know how to do so. Is there a way to prove it without using $(P)$?

Thank you.

Suppose we have $x, x' \in X$ such that $x' \in \bar{G}x$, and let $V$ be a neighbourhood of $x$ in $X$. Let $\epsilon$ denote the distance from $x$ to $X - V$. So $V$ contains the $\epsilon$ ball around $x$.
Now since $G$ is a group of isometries, for any $g\in G$ the $g$ translate of this $\epsilon$ - ball is the $\epsilon$ - ball around $g(x)$ which in turn is contained in $gV$. So we will be done if we can find a $g\in G$ such that $g(x)$ is within an $\epsilon$ distance of $x'$.
Since $x' \in \bar{G}x$ there exists a $\bar{g} \in \bar{G}$ such that $\bar{g}(x)=x'$. Since $\bar{g}\in G$ there is a net $(g_{\alpha})$ of points in $G$ converging to $\bar{g}$. So $g_{\alpha}(x) \longrightarrow \bar{g}(x)=x'$ So for the $\epsilon$ - ball around $x'$ I can find a $g_{\beta}$ from this net such that $g_{\beta}(x)$ is within $\epsilon$ of $x'$. Then I am done. $\square$