# When is Quotient topology by Group action Hausdorff?

Let $$G$$ be a topological group and $$X$$ be topological space, and $$\rho$$ be a continuous action of the topological group $$G$$ on $$X$$. We know that $$\pi$$ projection of topological space onto quotient space is open.

Let $$R=\{(x,y):x\sim y\}$$

We know $$X/\sim$$ is Hausdorff iff $$R$$ is closed in $$X\times X$$

Let $$O_x$$ be the orbit of $$x\in X$$.

Is it true that If each $$O_x$$ is closed in $$X$$ then $$X/\sim$$ is Hausdorff?

Or is there any other condition for Hausdorff using an Orbit of action? It's in my notes but I don't know why it has to be true.

My attempt:

here, $$R=\bigcup_{x\in X} \{x\}\times O_x$$ But I have no idea how to prove R is closed.

Edit :

If $$G$$ acts by homeomorphisms the quotient map $$p: X \to X / G$$ is always open (contrary to general quotient maps): this is because $$V \subset X/G$$ is open if and only if $$p^{-1}(V) \subset X$$ is open and $$p^{-1}(p(U)) = \bigcup_{g \in G}gU$$ is a union of open sets if $$U \subset X$$ is open. Therefore $$X/G$$ is Hausdorff if and only if the orbit equivalence relation is a closed subset of $$X \times X$$.

Could someone please explain $$X/G$$ is Hausdorff if and only if the orbit equivalence relation is a closed subset of $$X \times X$$.

• Check Tu's "Differentiable Manifolds", pages 73-76. – Bliipbluup Jan 21 at 14:18

Is it true that: If each $$O_x$$ is closed in $$X$$ then $$X/\!\sim$$ is Hausdorff?
If each $$O_x$$ is closed in $$X$$ then $$X/\!\sim$$ has the property that each point is closed. That does not imply that $$X/\!\sim$$ is Hausdorff.
Indeed, let $$X$$ be a non-Hausdorff topological space with every point closed (say, $$T_1$$ but not $$T_2$$). Let $$G = \{e\}$$ be the trivial group, and $$\pi$$ be the trivial action. Then $${X/\!\sim}\; = X$$.
• If $G$ acts by homeomorphisms the quotient map $p: X \to X / G$ is always open (contrary to general quotient maps): this is because $V \subset X/G$ is open if and only if $p^{-1}(V) \subset X$ is open and $p^{-1}(p(U)) = \bigcup_{g \in G}gU$ is a union of open sets if $U \subset X$ is open. Therefore $X/G$ is Hausdorff if and only if the orbit equivalence relation is a closed subset of $X \times X$. – Cloud JR K Feb 4 at 5:17