# Is a fixed point set of a compact group action on a compact space also compact? [closed]

Let $G$ be a compact group acting on a compact topological space $X$, that is the function $$G\times X\rightarrow X$$ $$(g,x)\mapsto g\cdot x$$ is continuous and satisfies $$g_1\cdot(g_2\cdot x )=(g_1g_2)\cdot x$$ $$e\cdot x=x$$ for any $g_1,g_2\in G$ and $x\in X$, where $e\in G$ is the neutral element.

Does this imply that the fixed point set of this action, $$X^G:=\{x\in X|g\cdot x=x \text{ for any g\in G}\},$$ is compact?

Remark. We consider the subset topology on $X^G$.

## closed as off-topic by user99914, clark, Xander Henderson, Claude Leibovici, Cyclohexanol.Oct 21 '17 at 4:30

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## 1 Answer

Assuming $X$ is Hausdorff, yes (and you don't need $G$ to be compact). Since $X$ is Hausdorff, for any $g\in G$, the set $\{x\in X:g\cdot x=x\}$ is closed in $X$. Since $X^G$ is just the intersection of these sets, it is also closed, and hence compact.

(If $X$ is not Hausdorff, then all bets are off. For instance, let $A$ be an infinite set and let $X=A\cup\{x,y\}$, topologized by saying a set is open iff it is either contained in $A$ or is all of $X$. Then $X$ is compact, and a cyclic group $G$ of order $2$ acts on $X$ by swapping $x$ and $y$ and fixing $A$. But then $X^G=A$ is an infinite set with the discrete topology, which is not compact.)

• Won't it be the case that on the $X^G$ you will have also indescrete topology and hence it will be compact? – piotrmizerka Oct 20 '17 at 22:59
• In addition to piotr's comment, your argument also does not assume the action of $G$ is continuous! – Qiaochu Yuan Oct 20 '17 at 23:04
• Right, I've given an actual counterexample now. My argument does not assume the entire action map $G\times X\to X$ is continuous but it does assume that for each $g\in G$, $x\mapsto g\cdot x$ is continuous. – Eric Wofsey Oct 20 '17 at 23:12
• In fact this argument adapts without much difficulty to show that compact Hausdorff spaces are closed under all limits, not just taking fixed points. This is necessary in order for the inclusion of compact Hausdorff spaces into spaces to have a left adjoint, namely the Stone-Cech compactification. – Qiaochu Yuan Oct 20 '17 at 23:24
• hmm,... I think that $X=A\cup\{x,y\}$ is not compact. You can pick out an infinite open cover $\{\{a_{\alpha},x,y\}\}$ where $\alpha$ iterates over all elements of $A$. Then if we remove even one of the elements, say $\{a_{\alpha_0},x,y\}$, of the cover it won't be a cover anymore since the new "cover" won't contain this element. – piotrmizerka Oct 20 '17 at 23:31