Lemma 76.9.4. In Situation 76.9.2 assume $R$ is a decent space. Then $R$ is a separated algebraic space. In Situation 76.9.1 assume that $G$ is a decent algebraic space. Then $G$ is separated algebraic space.
Next, we prove the first assertion. To do this we may replace $B$ by $S$. By the paragraph above the stabilizer group scheme $G \to U$ is separated. By Groupoids in Spaces, Lemma 75.28.2 the morphism $j = (t, s) : R \to U \times _ S U$ is separated. As $U$ is the spectrum of a field the scheme $U \times _ S U$ is affine (by the construction of fibre products in Schemes, Section 26.17). Hence $R$ is separated, see Morphisms of Spaces, Lemma 64.4.9. $\square$
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