The Stacks project

Proof. The equivalence of (1), (2), and (3) follows from Decent Spaces, Lemma 68.20.1 and the fact that a quasi-separated algebraic space is decent (Decent Spaces, Section 68.6). However in the next paragraph we will give a more elementary proof of the equivalence.

Note that (1) and (2) are equivalent by definition (Properties of Spaces, Definition 66.10.2). To prove the equivalence of (1) and (3) we may assume $X$ is quasi-compact. Choose

and $f_ i : V_ i \to U_ i$ as in Decent Spaces, Lemma 68.8.6. Say $x \in U_ i$, $x \not\in U_{i + 1}$. Then $x = f_ i(y)$ for a unique $y \in V_ i$. If (1) holds, then $y$ is a generic point of an irreducible component of $V_ i$ (Properties of Spaces, Lemma 66.11.1). Since $f_ i^{-1}(U_{i + 1})$ is a quasi-compact open of $V_ i$ not containing $y$, there is an open neighbourhood $W \subset V_ i$ of $y$ disjoint from $f_ i^{-1}(V_ i)$ (see Properties, Lemma 28.2.2 or more simply Algebra, Lemma 10.26.4). Then $f_ i|_ W : W \to X$ is an isomorphism onto its image and hence $x = f_ i(y)$ is a generic point of $|X|$. Conversely, assume (3) holds. Then $f_ i$ maps $\overline{\{ y\} }$ onto the irreducible component $\overline{\{ x\} }$ of $|U_ i|$. Since $|f_ i|$ is bijective over $\overline{\{ x\} }$, it follows that $\overline{\{ y\} }$ is an irreducible component of $U_ i$. Thus $x$ is a point of codimension $0$.

The final statement of the lemma is Properties of Spaces, Proposition 66.13.3. $\square$