Stabilisers of group action open imply the action is continuous Let $\mu \colon X \times G \longrightarrow X$ be the action of a topological group on a set $X$. We consider $X$ to be a topological space with the discrete topology. Suppose that for all $x \in X$, the stabiliser $I_x =\{ g \in G \mid xg = x \}$ is open in $G$. Why does this mean that the function $\mu$ is continuous?

Because for a given $x \in X$ we have
$$
\mu^{-1}(\{ x \}) = \{ (y, g) : y g = x \}
= \bigcup \{ z \} \times I_{z} g_{zx},
$$
where the union ranges over all $z$ in the orbit of $x$, and for each such $z$, we choose $g_{zx}$ to be a fixed element such that $z g_{zx} = x$.
Clearly the translate $I_{z} g_{zx}$ of $I_{z}$ is open in $G$, and so is $\{ z \}$ in $X$, so their product is open in $X \times G$.
