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?

  • 1
    $\begingroup$ 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 $\emptyset = p^{-1}(\{ \bar{y} \} \cap F) = p^{-1}(\bar{y}) \cap p^{-1}(F) = G \cdot y \cap G \cdot F$. $\endgroup$ – Paul Frost Feb 10 at 23:50

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