# Compact group acting on regular space

Let $$G$$ be a compact topological group, $$X$$ be a regular topological space. Then the quotient space given by the continuos action of $$G$$ on $$X$$, $$X/G$$ is also regular.
Here's my attempt, though I feel it's wrong for some reason:
Let $$\bar{y} = [y] \in X/G, \bar{y}\not\in F\subset X/G$$ closed. Also, let $$p:X\rightarrow X/G$$ be the projection. Then: $$(p^{-1}(\bar{y}):=G\cdot y)\cap (p^{-1}(F):=G\cdot F)=\emptyset$$ Clearly, $$G\cdot F$$ is closed since $$p$$ is continuous. Since $$X$$ is regular, then, for each $$z\in G\cdot y$$, $$\exists U_z$$ neighbourhood of $$z$$, $$V_z$$ open set containing $$G\cdot F$$ such that $$U_z\cap V_z=\emptyset$$. Thus, $$G\cdot y\subset \bigcup_{z\in G\cdot y}U_z$$ Now, for each $$y\in X$$, define $$m_y:G\rightarrow X$$, by $$m_y(g)=g\cdot y$$. Clearly, $$m_y$$ is continuous for every $$y$$, and $$m_y(G)=G\cdot y$$. It follows that $$\{m_y^{-1}(U_z)\}_z$$ is an open cover for $$G$$, and, aince it is compact, admits a finite subcover $$G=\bigcup_{i=1}^nm_y^{-1}(U_{z_i})$$ From here it's pretty straightforward: It follows that $$G\cdot y\subset \bigcup_{i=1}^nU_{z_i}=U$$ And then $$G\cdot F\subset\bigcap_{i=1}^nV_{z_i}=V$$ So that, since $$p$$ is an open map and surjective, $$p(U),\,p(V)$$ are open subsets of $$X/G$$ such that $$\bar{y}\in p(U),\,F\subset p(V),\,p(U)\cap p(V)=\emptyset$$.
Are there any mistakes in this proof?

• Your notation is somewhat unpleasant. I guess ": =" means "is defined as". But $p^{-1}(\bar{y})$ is not defined as $G \cdot y$, it is a property (similarly for $p^{-1}(F)$). $(p^{-1}(\bar{y}):=G\cdot y)\cap (p^{-1}(F):=G\cdot F)=\emptyset$ is rather confusing, I suggest to write $\emptyset = p^{-1}(\{ \bar{y} \} \cap F) = p^{-1}(\bar{y}) \cap p^{-1}(F) = G \cdot y \cap G \cdot F$. – Paul Frost Feb 10 at 23:50