Is a topological group action continuous if and only if all the stabilizers are open?

Let $G$ be a topological group and $(X,\mu)$ be a $G$-set, i.e. $\mu$ defines an action $X \times G \rightarrow X$.

Is it then true that $\mu$ is continuous if and only if for every $x \in X$ the stabilizer subgroup $G_x$ is open in $G$?

If yes, how does one prove this?

Solution 1:

Neither implication is true.

1) Let $X$ be a topological space admitting a discontinuous bijection $f: X \rightarrow X$. (E.g. we may take $X = \mathbb{R}$.) Let $G = \mathbb{Z}$ with the discrete topology. There is a unique set-theoretic action of $G$ on $X$ such that $1 \cdot x = f(x)$. Because $1 \cdot = f$ is discontinuous, the action is not continuous. However, since $\mathbb{Z}$ is discrete, all subgroups are open. In particular the point stabilizers $G_x$ are open.

2) Let $G$ be a nondiscrete topological group [Pierre-Yves Gaillard suggests $\mathbb{R}$ in his comment above], and view the group law $G \times G \rightarrow G$ as a left action of $G$ on itself. By definition this action is continuous, but the point stabilizers are $\{e\}$, which is not open since $G$ is not discrete.

Added: Wait! Upon closer reading, it was not explicitly said that $X$ is a topological space, only a $G$-set. In fact, if we view $X$ as a "naked set" and endow it with the discrete topology, then the equivalence becomes true.