TITLE: Isotropy subgroup is closed
QUESTION [5 upvotes]: I am studying a book and I am asked to prove the following:
Show that the isotropy subgroup (for a certain right action of a topological group $G$) of $y\in Y$ is closed for $Y$ some topological space.
I have a proof for the case that $Y$ is Hausdorrf:
$ \sigma^{-1}(y)\cap (\{y\}\times G) = \{y\}\times \text{Iso}(y)$
and since $\sigma$ is continuous and $\{y\}$ and $\{y\}\times G$ are closed, so is Iso($y$). But this relies crucially on the fact that the point is closed, which is only true for Hausdorff spaces. So my question is:
Do we need Hausdorff to prove the result?
Thanks.

REPLY [3 votes]: If you endow $Y$ with the cofinite topology then the points are closed, and obviuosly $Y$ with this topology is not Hausdorff. Now, let $G$ be a topological group acting continuously on the right of $Y$, i.e., there is a continuous map $\phi:Y\times G\to Y, (y,g)\to y\cdot g$, satisfaying $(y\cdot h)\cdot g=y\cdot(hg)$ and $y\cdot 1=y$. Then, for a fixed $y_o$, the induced map $\phi_{y_o}:G\to Y$, given by $g\mapsto y_o\cdot g$, is also continuous (because $\phi$ is!). Now $Iso(y_o)=\phi_{y_o}^{-1}(y_o)$ is closed.