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 $X \times X$.
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 ${X/\!\sim}\; = X$.