Local compactness and actions

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Let $X$ be a Hausdorff topological space and $G$ a group acting continuously on $X$. We denote the group action as $(g,x) rightarrow gx$. Let $xin X$ and $U$ a neighborhood of $x$ such that the set $P(x,U)=g in G: gx in U$ is relatively compact. Then, I claim, $G$ is locally compact.



Proof. It suffices to prove that $P(x,U)$ is a neighborhood of the identity $e in G$. The action map $f:G times X rightarrow X$ is continuous so the map $F:G rightarrow X: g rightarrow gx$ is continuous too, mapping $e$ to $x$. Thus, since $U$ is a neighborhood of $x$ there exists a neighborhood $V subset G$ of $e$ such that $F(V) subset U$, that is $Vx subset U$. So, if $g in V$, then $gx in U$, which gives $g in P$. Thus, $V subset P$, meaning $P$ is a neighborhood of $e$. Since it is relatively compact, we are done.



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group-actions locally-compact-groups




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    It seems right to me.
    – Javi
    Jul 20 at 19:29














up vote
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Let $X$ be a Hausdorff topological space and $G$ a group acting continuously on $X$. We denote the group action as $(g,x) rightarrow gx$. Let $xin X$ and $U$ a neighborhood of $x$ such that the set $P(x,U)=g in G: gx in U$ is relatively compact. Then, I claim, $G$ is locally compact.



Proof. It suffices to prove that $P(x,U)$ is a neighborhood of the identity $e in G$. The action map $f:G times X rightarrow X$ is continuous so the map $F:G rightarrow X: g rightarrow gx$ is continuous too, mapping $e$ to $x$. Thus, since $U$ is a neighborhood of $x$ there exists a neighborhood $V subset G$ of $e$ such that $F(V) subset U$, that is $Vx subset U$. So, if $g in V$, then $gx in U$, which gives $g in P$. Thus, $V subset P$, meaning $P$ is a neighborhood of $e$. Since it is relatively compact, we are done.



Is this correct?




group-actions locally-compact-groups




share|cite|improve this question

















  • 1




    It seems right to me.
    – Javi
    Jul 20 at 19:29












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Let $X$ be a Hausdorff topological space and $G$ a group acting continuously on $X$. We denote the group action as $(g,x) rightarrow gx$. Let $xin X$ and $U$ a neighborhood of $x$ such that the set $P(x,U)=g in G: gx in U$ is relatively compact. Then, I claim, $G$ is locally compact.



Proof. It suffices to prove that $P(x,U)$ is a neighborhood of the identity $e in G$. The action map $f:G times X rightarrow X$ is continuous so the map $F:G rightarrow X: g rightarrow gx$ is continuous too, mapping $e$ to $x$. Thus, since $U$ is a neighborhood of $x$ there exists a neighborhood $V subset G$ of $e$ such that $F(V) subset U$, that is $Vx subset U$. So, if $g in V$, then $gx in U$, which gives $g in P$. Thus, $V subset P$, meaning $P$ is a neighborhood of $e$. Since it is relatively compact, we are done.



Is this correct?




group-actions locally-compact-groups




share|cite|improve this question













Let $X$ be a Hausdorff topological space and $G$ a group acting continuously on $X$. We denote the group action as $(g,x) rightarrow gx$. Let $xin X$ and $U$ a neighborhood of $x$ such that the set $P(x,U)=g in G: gx in U$ is relatively compact. Then, I claim, $G$ is locally compact.



Proof. It suffices to prove that $P(x,U)$ is a neighborhood of the identity $e in G$. The action map $f:G times X rightarrow X$ is continuous so the map $F:G rightarrow X: g rightarrow gx$ is continuous too, mapping $e$ to $x$. Thus, since $U$ is a neighborhood of $x$ there exists a neighborhood $V subset G$ of $e$ such that $F(V) subset U$, that is $Vx subset U$. So, if $g in V$, then $gx in U$, which gives $g in P$. Thus, $V subset P$, meaning $P$ is a neighborhood of $e$. Since it is relatively compact, we are done.



Is this correct?





group-actions locally-compact-groups





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 20 at 19:28









Javi

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asked Jul 20 at 19:00









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  • 1




    It seems right to me.
    – Javi
    Jul 20 at 19:29












  • 1




    It seems right to me.
    – Javi
    Jul 20 at 19:29







1




1




It seems right to me.
– Javi
Jul 20 at 19:29




It seems right to me.
– Javi
Jul 20 at 19:29















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