The intersection of all stabilizers in a continuous action

Question

Let $G$ be a topological group (locally compact and Hausdorff) continuously acting on a topological space $M$. Does this implies that the normal subgroup of $G$ made by the elements fixing each $m\in M$ is closed? If so, any hint on how to prove it?

Answer 1

We could use nets to see it (often handy in topological group arguments, I find):

so the action $a: G \times X \to X$ where $a(g,x) = g \cdot x$ is a continuous map.

Let $I \to g_i$ be a net in $G$ in the said normal subgroup, and suppose $g_i \to g$.

Let $x \in X$, then $g \cdot x = a(g,x) = a(\lim_i g_i, x) = \lim_i a(g_i,x)$ (by continuity of $a$ in the first coordinate) $= \lim_i g_i \cdot x = \lim_i x = x$ as each $g_i$ fixes $x$. As $x \in X$ is arbitrary, so $g$ is also in the intersection of all stabilisers. So that set is closed under net-limits, hence closed.

No assumptions on $G$ are needed, and we use Hausdorffness on $X$ because we want the limits of nets to be uniquely defined. Eric Wofsey points out in his answer that something on $X$ is needed here..

Answer 2

This is false in general. For instance, if $M$ has the indiscrete topology, then any action of $G$ on $M$ is continuous, so the normal subgroup fixing all elements of $M$ could be any normal subgroup at all (given any normal subgroup $N$, let $M=G/N$ with the usual action).

It is true assuming $M$ is $T_1$. Indeed, in that case the subgroup in question is just the intersection of the closed sets $\mu_m^{-1}(\{m\})$ for each $m\in M$, where $\mu_m:G\to M$ is the map $\mu_m(g)=g\cdot m$.