Let $(X,d)$ be a compact metric space and $G$ be a finite group of homeomorphisms of $X$. Let $p:X\rightarrow X/G$ be the orbit map.
Then we can define a (psuedo) metric on $X/G$ as follows - $$D(\tilde{x},\tilde{y})=\inf\{d(gx,q_1)+d(g_1q_1,q_2)+\cdots+d(g_{n-1}q_{n-1},g'y)\}$$ where the infimum ranges over all finite sequences of points $(q_1,\cdots ,q_{n-1})$ in $X$ and $(g,g',g_1,\cdots g_{n-1})$ in $G$. This is just a reformulation of this definition from Wikipedia.
I have two questions -
- Is this proof correct for showing $D(\tilde{x},\tilde{y})=0\Rightarrow \tilde{x}=\tilde y$?
Proof - If $D(\tilde{x},\tilde{y})=0$ then since $X$ is compact and $G$ is finite, infimum is attained at some finite sequence of points. So $$D(\tilde{x},\tilde{y})=d(gx,q_1)+d(g_1q_1,q_2)+\cdots+d(g_{n-1}q_{n-1},g'y) =0$$ Since each term in the sum is non-negative, each term is 0 and we get, $$y=(g')^{-1}g_{n-1}g_{n-2}\cdots g_1g\cdot x \Longrightarrow \tilde{x}=\tilde{y}$$
- Is the quotient topology on $X/G$ the same as the metric topology on $X/G$ induced by $D$?
My attempt - I have to show that a basis for the metric topology is also a basis for the quotient topology. For that I need to first show any open ball $B_D(\tilde{x},r)$ in $X/G$ with metric topology is open in the quotient topology. So it is enough to show that $U=p^{-1}(B_D(\tilde{x},r))$ is open in $X$. I tried to show that for any $y\in U$ the ball $B_d (y,r)\subseteq U$ but I wasn't successful. And next I need to prove that for any $\tilde{x}\in X/G$ and an open set $\tilde{U}$ of $\tilde{x}$ in $X/G$ there is a ball around $\tilde{x}$ contained in $\tilde{U}$. I haven't been able to show this either.
Is the way I am going about everything right? Can someone please help out.
Thank you.