I'm working off of the definition that $\mathcal{G}$ is a groupoid if it has a partial product which is associative (when applicable) and an involutive inverse map so that $xx^{-1}$ is always defined and $xx^{-1}y=y=yzz^{-1}$ when the products all exist. A groupoid is locally compact if it has a locally compact topology such that the multiplication and inversion maps are continuous. The range map $r$ is given by $r(x)=xx^{-1}$ and source map $s: x\mapsto x^{-1}x$
Following this, a groupoid is called $r$-discrete if the unit space $\mathcal{G}^{(0)}:=r(\mathcal{G})$ is open. Several sources cite a result originally due to Renault saying that if $\mathcal{G}$ is a locally compact $r$-discrete groupoid then, for each $u\in \mathcal{G}^{(0)}$, the set $\mathcal{G}^u:=r^{-1}$({$u$}) is discrete in the relative topology. For notation, $\mathcal{G}_u:=s^{-1}$({$u$})
The provided proof goes as follows: an $x\in \mathcal{G}_v^u=\mathcal{G}^u\cap \mathcal{G}_v$ defines a homeomorphism from $\mathcal{G}^u$ to $\mathcal{G}^v$. Since {$v$} is open in $\mathcal{G}^v$, we must have {$u$} is open in $\mathcal{G}^u$.
I have a few issues with this:
- This assumes that $\mathcal{G}_v^u$ is nonempty. What happens if the set is empty? I can easily write an r-discrete groupoid with $\mathcal{G}_v^u$ empty when $u\neq v$.
- Why does this assume that {$v$} is open in $\mathcal{G}^v$? Isn't this exactly what we wish to show?
Any help is appreciated! For a reference, this is Lemma I 2.7(i) in Renault's "A Groupoid Approach to $C^*$ Algebras.
I believe I've actually resolved the issues:
Since $\mathcal{G}^{(0)}$ is open, we have that $\mathcal{G}^{(0)}\cap \mathcal{G}^u$ is open in the relative topology. This set is exactly {$u$} and resolves my second question.
Secondly, we simply need $\mathcal{G}^u_v$ to be nonempty for some $v$. This is always true at least in the case that $u=v$.