**Proof.**
We sketch the proof. By Properties of Spaces, Lemma 65.6.3 we have $X = U/R$ with $U$ affine. In this case, reasonable means $U \to X$ is universally bounded. Hence there exists an integer $N$ such that the “fibres” of $U \to X$ have degree at most $N$, see Definition 67.3.1. Denote $s, t : R \to U$ and $c : R \times _{s, U, t} R \to R$ the groupoid structural maps.

Claim: for every quasi-compact open $A \subset R$ there exists an open $R' \subset R$ such that

$A \subset R'$,

$R'$ is quasi-compact, and

$(U, R', s|_{R'}, t|_{R'}, c|_{R' \times _{s, U, t} R'})$ is a groupoid scheme.

Note that $e : U \to R$ is open as it is a section of the étale morphism $s : R \to U$, see Étale Morphisms, Proposition 41.6.1. Moreover $U$ is affine hence quasi-compact. Hence we may replace $A$ by $A \cup e(U) \subset R$, and assume that $A$ contains $e(U)$. Next, we define inductively $A^1 = A$, and

\[ A^ n = c(A^{n - 1} \times _{s, U, t} A) \subset R \]

for $n \geq 2$. Arguing inductively, we see that $A^ n$ is quasi-compact for all $n \geq 2$, as the image of the quasi-compact fibre product $A^{n - 1} \times _{s, U, t} A$. If $k$ is an algebraically closed field over $S$, and we consider $k$-points then

\[ A^ n(k) = \left\{ (u, u') \in U(k) : \begin{matrix} \text{there exist } u = u_1, u_2, \ldots , u_ n \in U(k)\text{ with}
\\ (u_ i , u_{i + 1}) \in A \text{ for all }i = 1, \ldots , n - 1.
\end{matrix} \right\} \]

But as the fibres of $U(k) \to X(k)$ have size at most $N$ we see that if $n > N$ then we get a repeat in the sequence above, and we can shorten it proving $A^ N = A^ n$ for all $n \geq N$. This implies that $R' = A^ N$ gives a groupoid scheme $(U, R', s|_{R'}, t|_{R'}, c|_{R' \times _{s, U, t} R'})$, proving the claim above.

Consider the map of sheaves on $(\mathit{Sch}/S)_{fppf}$

\[ \mathop{\mathrm{colim}}\nolimits _{R' \subset R} U/R' \longrightarrow U/R \]

where $R' \subset R$ runs over the quasi-compact open subschemes of $R$ which give étale equivalence relations as above. Each of the quotients $U/R'$ is an algebraic space (see Spaces, Theorem 64.10.5). Since $R'$ is quasi-compact, and $U$ affine the morphism $R' \to U \times _{\mathop{\mathrm{Spec}}(\mathbf{Z})} U$ is quasi-compact, and hence $U/R'$ is quasi-separated. Finally, if $T$ is a quasi-compact scheme, then

\[ \mathop{\mathrm{colim}}\nolimits _{R' \subset R} U(T)/R'(T) \longrightarrow U(T)/R(T) \]

is a bijection, since every morphism from $T$ into $R$ ends up in one of the open subrelations $R'$ by the claim above. This clearly implies that the colimit of the sheaves $U/R'$ is $U/R$. In other words the algebraic space $X = U/R$ is the colimit of the quasi-separated algebraic spaces $U/R'$.

Properties (1) and (2) follow from the discussion above. If $X$ is a scheme, then if we choose $U$ to be a finite disjoint union of affine opens of $X$ we will obtain (3). Details omitted.
$\square$

## Comments (2)

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