# 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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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
• 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