Can someone tell me if what I have done is correct?
Proposition : Let $X$ be a compact metric space and $G$ a finite group acting on it. Let $p : X \rightarrow X/G$ be the orbit map. Let $B = \{ U_{\alpha}\}_{\alpha}$ be a $\sigma$ - locally finite basis of $X$ and let $B' = \{p(U_{\alpha})\}$. Then $B'$ is $\sigma$ - locally finite basis of $X/G$.
My attempt at a proof :
- $B'$ is a basis -
First of all since $p$ is an open map $B'$ is a collection of open sets.
Let $\tilde{x}$ be a point in $X/G$ and $\tilde{U}$ an open neighbourhood of $\tilde{x}$ in $X/G$. We need to find an element in $B'$ containing $\tilde{x}$ and contained in $\tilde{U}$. Let $x\in p^{-1}(\tilde{x})$. Then $p^{-1}(\tilde{U})$ is an open neighbourhood of $x$ and hence contains a basis element $U_{\alpha}$ that contains $x$. Then $p(U_{\alpha})$ is my required element of $B'$.
- B' is $\sigma$ - locally finite -
If we write $B = \bigcup _{n=1}^{\infty} B_n$ where $B_n$'s are locally finite subcollections of $B$ then we can write $$B' = \bigcup_{n=1}^{\infty}B'_n \text{ where }B'_n = \{ p(U_{\alpha}) : U_{\alpha} \in B_n \}$$ Now it remains to show that $B'_n$ is a locally finite collection -
If $\tilde{x} \in X/G$ we need to show that there is an open set $\tilde{U}$ containing $\tilde{x}$ that intersects only finitely many $p(U_{\alpha})$'s in $B'_n$.
Since $G$ is finite, $p^{-1}(\tilde{x}) = \{x_1, ... , x_m\}$ and each $x_i$ has an open neighbourhood $V_i$ that intersects finitely many members of $B_n$.
Let $V = \bigcap_{i=1}^{m} p(V_i)$. Then $V$ is an open neighbourhood of $\tilde{x}$. My claim is that that this is the required $\tilde{U}$.
Suppose $V\cap p(U_{\alpha}) \neq \emptyset$ for some $U_{\alpha}$ in $B_n$. This means that, $$p^{-1}(V\cap p(U_{\alpha}))=p^{-1}(V)\cap p^{-1}(p(U_{\alpha})) \neq \emptyset$$
Now, $$p^{-1}(V)=\bigcap_{i=1}^{m}p^{-1}(p(V_i))=\bigcap_{i=1}^{m}GV_i \subseteq GV_i \text{ for all } i$$ And similarly, $$p^{-1}(p(U_{\alpha}))=GU_{\alpha}$$
So, $$GV_i\cap GU_{\alpha} = G(V_i\cap U_{\alpha}) = \bigcup_\limits{g\in G}g(V_i\cap U_{\alpha})\neq \emptyset $$
So, $V_i \cap U_{\alpha} \neq \emptyset $ for all $i$, meaning that there are only finitely many $U_{\alpha}$ for which $V\cap p(U_{\alpha}) \neq \emptyset$
Done$\square$
Remark: The definitions I am using are as follows -
A $\sigma$ - locally finite basis of a topological space $X$ is a basis that can be written as a countable union of locally finite collections.
A collection of subsets of a topological space $X$ is called locally finite if every point of $X$ has an open neighbourhood that intersects only finitely many members of the collection.
Thank you.