Let $G$ be a discrete group which acts continuously on a Stonean space $\Omega$. Consider the map $f\colon \Omega\to \{0,1\}^G$ sending $x\in \Omega$ to $\chi\_{G\_x}$, where $\chi\_{G\_x}$ denotes the characteristic function of the stabilizer $G\_x$. 


Why is $f$ continuous?


We can restrict to the subbasis-sets of the product topology and for $g\in G$ fixed, it is enough to consider the preimages of $\{0\}\times \{0,1\}^{G\setminus \{g\}}$ and $\{1\}\times \{0,1\}^{G\setminus \{g\}}$ under $f$, which must be open in $\Omega$. That $f^{-1}(\{0\}\times \{0,1\}^{G\setminus \{g\}}) $ is open in $\Omega$ can be seen using that in the Hausdorff space $\Omega$, the diagonal $\{(x,x)\in \Omega\times \Omega\mid x\in \Omega\}$ is closed in $\Omega\times \Omega$, but without using that $\Omega$ is Stonean.


My question therefore reduces to: Why is $ f^{-1}(\{1\}\times \{0,1\}^{G\setminus \{g\}}) $ open in $\Omega$ ? I think to prove this, it must be used that $\Omega$ is Stonean and probably it helps to know that in a Stonean space, the closure of disjoint open sets is again disjoint. 


The background of my question is the following: I am currently reading the paper <https://arxiv.org/pdf/1509.01870.pdf> and that $f$ is continuous is used in the proof of Theorem 4.1, direction $<=$. 



continuity of certain map which is defined on a Stonean space
It's indeed true: whenever a discrete group $G$ acts continuously on a Hausdorff, extremally disconnected space $X$, then the map $x\mapsto G\_x$ is continuous, where the set of subgroups of $G$ is endowed with its compact topology given by inclusion in $2^G$.


One has to show that for any $g\in G$, the map $u\_g:x\mapsto \chi\_{G\_x}(g)$, $X\to\{0,1\}$ is continuous. That is, that $u\_g^{-1}(\{0\})$ and $u\_g^{-1}(\{1\})$ are both closed. 


That $u\_g^{-1}(\{0\})=\{x:gx=x\}$ is closed holds for whenever $G$ acts continuously on a Hausdorff topological space. (It's false when $X$ is not assumed Hausdorff.)


The claim is that $u\_g^{-1}(\{1\})=\{x:gx\neq x\}$ is closed for every $g$. 


Assume otherwise: this means that there exists $x\_0\in X$ such that $gx\_0=x\_0$ and $x\_0$ is in the closure of $\{x:gx\neq x\}$. Let $V\subset X$ an open subset, maximal for the property that $V\cap gV=\emptyset$.


Claim: *if $X$ is Hausdorff, then $x\_0$ belongs to the closure of $V$*. Indeed, let $U$ be the complement of this closure. If by contradiction $x\_0\in U$, define $U'=U\cap g^{-1}U$: this is an open neighborhood of $x\_0$. By assumption, there exists $x\in U'$ with $gx\neq x$. Using that $X$ is Hausdorff, one can find an open neighborhood $V'$ of $x$, with $V'\subset U'$ and $V'\cap gV'=\emptyset$. Then taking $V\cup V'$ contradicts the maximality of $V$.


But then $x\_0$ also belongs to the closure of $gV$. Assuming that $X$ is extremally disconnected, this contradicts that $V$ and $gV$ are disjoint.


The latter argument shows more generally that for any continuous self-map $g$ of a Hausdorff extremally disconnected space $X$, the closed subset $\{x:g(x)=x\}$ is also open.


