# continuity of certain map which is defined on a Stonean space

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 $$<=$$.

• @YCor he said the space is Stonean en.m.wikipedia.org/wiki/Extremally_disconnected_space which is a very strong property that I don't think your example has. Nov 13, 2018 at 23:30
• @BenjaminSteinberg thanks for fixing the confusion.
– YCor
Nov 14, 2018 at 4:51

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.

• Nice, I understand. Thank you very much Nov 14, 2018 at 10:56