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Problem of the Week #242 - Jan 24, 2017

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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
Here is this week's POTW:

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Suppose $\Gamma$ is a finite group of homeomorphisms of a Hausdorff space $M$ such that every non-identity element of $\Gamma$ is fixed point free. Show that $\Gamma$ acts on $M$ properly discontinuously.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,892
No one solved this week's problem. You can read my solution below.


Fix $p\in M$. For each $g\in \Gamma\setminus\{1\}$, $p \neq gp$. Since $M$ is Hausdorff, for every $g\in \Gamma\setminus\{1\}$, there are disjoint open sets $U_g \ni p$ and $V_g \ni gp$. Set $$W = \bigcap_{g\in \Gamma\setminus\{1\}} (U_g \cap g^{-1}(V_g))$$ Since each of the sets $U_g \cap g^{-1}(V_g)$ is an open neighborhood of $p$ and $\Gamma\setminus\{1\}$ is finite, then $W$ is open neighborhood of $p$. Given $g\neq 1$, $gW\cap W = \emptyset$. Indeed, if $gW\cap W \neq \emptyset$, then there are $w,w'\in W$ for which $gw = w'$. As $w'\in U_g$ and $w\in g^{-1}(V_g)$, we have $w' = gw \in U_g \cap V_g$, a contradiction. Consequently, $\Gamma$ acts on $M$ properly discontinuously.
 
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