# Is every continuous action of a compact topological group closed?

I am reading Bredon's Introduction to compact transformation groups, and came across the following result and proof on page 34: Even though he writes "Recall our standing assumption that $$X$$ is Hausdorff," I cannot see where any properties of Hausdorffness of $$X$$ are used in this proof. I suspect it would be used to ensure that convergent nets in $$X$$ converge uniquely, but I do not see why this fact would be needed for this proof.

Is it true that the continuous action of a compact topological group on a topological space $$X$$ is always closed, regardless of whether $$X$$ is Hausdorff? If not, then where in the above proof is Hausdorffness used?

• There's also the question of whether one assumes a topological group to be $T_0$, and hence Hausdorff, by default. Jul 13, 2021 at 7:24
• Compact/ Locally compact groups are usually assumed Hausdorff by default, even when topological groups are not.
– YCor
Jul 13, 2021 at 8:03
• Hausdorffness of $X$ doesn't seem to be used at any point in the proof.
– YCor
Jul 13, 2021 at 8:05
• The only critical point seems to be the existence of a subnet, but this is well known, see f.i. Kelley, General Topology. Jul 13, 2021 at 9:16

Yes, it is true, and the following argument will surely convince you so. Recall that a topological space $$K$$ is compact iff it is universally closed: for every topological space $$X$$, the coordinate projection map $$X\times K \to X$$ is closed. Now, use the obvious fact that $$G\times X\to G\times X$$, $$(g,x)\mapsto (g,gx)$$, is a homoemorphism and realize the action map $$G\times X \to X$$ as a composition of this homeo with the coordinate projection map.
• Thanks for the great answer. It appears we do not even need $G$ to be Hausdorff (a question raised by David Roberts in the comments to my question).
• @Ben, right, but this is not a big deal, really. If $G$ is not Hausdorff then, denoting $G_1=\overline{\{1\}}$, we have that $G_1$ is a closed normal subgroup of $G$ on which the topology is trivial, thus each $G_1$ orbit in $X$ is inseparable. Then the action map $G\times X \to X$ fibers naturally over the action map of $G/G_1$ on $X/G_1$, thus everything reduces to the separable case. That is, the theory of dynamical systems reduces easily to the Hausdorff world. Jul 13, 2021 at 16:34
Let $$C\subseteq G\times X$$ be closed. Note that $$x_0\notin \Theta(C)$$ if and only if $$\{(g,g^{-1}x_0)\mid g\in G\}\cap C = \emptyset$$ which is if and only if $$G\times \{(e,x_0)\}\subseteq \Phi^{-1}((G\times X)\setminus C)$$ where $$\Phi: (g,(h,x))\mapsto (hg^{-1}, gx)$$. Now, let $$x_0\notin \Theta(C)$$. By our observation, $$G\times \{(e,x_0)\}\subseteq \Phi^{-1}((G\times X)\setminus C)$$ and since $$C$$ is closed, $$\Phi^{-1}((G\times X)\setminus C)$$ is a neighborhood of $$G\times \{(e,x_0)\}$$. Hence, (by the tube lemma), we can find neighborhoods $$U$$ of $$e$$ and $$V$$ of $$x_0$$ such that $$G\times (U\times V) \subseteq \Phi^{-1}((G\times X)\setminus C)$$. It is easy to see that $$\Theta(U\times V)$$ is a neighborhood of $$x_0$$ which does not intersect $$\Theta(C)$$.