hcju/stacksqa huggingface dataset:
The equivalence of (1) and (2) follows from Lemma \ref{lemma-conditions-on-fibre-and-qf} (and the fact that a quasi-separated morphism is decent by Lemma \ref{lemma-properties-trivial-implications}). \medskip\noindent Assume the equivalent conditions of (1) and (2). Choose an affine scheme $V$ and an \'etale morphism $V \to Y$ mapping a point $v \in V$ to $y$. Then $v$ is a generic point of an irreducible component of $V$ by Properties of Spaces, Lemma \ref{spaces-properties-lemma-codimension-0-points}. Choose an affine scheme $U$ and a surjective \'etale morphism $U \to V \times_Y X$. Then $U \to V$ is of finite type. The morphism $U \to V$ is quasi-finite at every point lying over $v$ by (2). It follows that the fibre of $U \to V$ over $v$ is finite (Morphisms, Lemma \ref{morphisms-lemma-quasi-finite-at-a-finite-number-of-points}). By Morphisms, Lemma \ref{morphisms-lemma-generically-finite} after shrinking $V$ we may assume that $U \to V$ is finite. Let $$ R = U \times_{V \times_Y X} U $$ Since $f$ is quasi-separated, we see that $V \times_Y X$ is quasi-separated and hence $R$ is a quasi-compact scheme. Moreover the morphisms $R \to V$ is quasi-finite as the composition of an \'etale morphism $R \to U$ and a finite morphism $U \to V$. Hence we may apply Morphisms, Lemma \ref{morphisms-lemma-generically-finite} once more and after shrinking $V$ we may assume that $R \to V$ is finite as well. This of course implies that the two projections $R \to V$ are finite \'etale. It follows that $V/R = V \times_Y X$ is an affine scheme, see Groupoids, Proposition \ref{groupoids-proposition-finite-flat-equivalence}. By Morphisms, Lemma \ref{morphisms-lemma-image-proper-is-proper} we conclude that $V \times_Y X \to V$ is proper and by Morphisms, Lemma \ref{morphisms-lemma-finite-proper} we conclude that $V \times_Y X \to V$ is finite. Finally, we let $Y' \subset Y$ be the open subspace of $Y$ corresponding to the image of $|V| \to |Y|$. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-integral-local} we conclude that $Y' \times_Y X \to Y'$ is finite as the base change to $V$ is finite and as $V \to Y'$ is a surjective \'etale morphism. \medskip\noindent If $Y$ is decent and $f$ is quasi-separated, then we see that $X$ is decent too; use Lemmas \ref{lemma-properties-trivial-implications} and \ref{lemma-property-over-property}. Hence Lemma \ref{lemma-conditions-on-fibre-and-qf} applies to show that (4) implies (1) and (2). On the other hand, we see that (2) implies (4) by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-quasi-finite-at-a-finite-number-of-points}.
Since quasi-separated algebraic spaces are decent, the equivalence of (1) and (2) follows from Lemma \ref{lemma-conditions-on-fibre-and-qf}. To prove that (1) and (2) imply (3) we may and do replace $Y$ by a quasi-compact open containing $y$. Since $f^{-1}(\{y\})$ is finite, we can find a quasi-compact open subspace of $X' \subset X$ containing the fibre. The restriction $f|_{X'} : X' \to Y$ is quasi-compact and quasi-separated by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-quasi-compact-quasi-separated-permanence} (this is where we use that $Y$ is quasi-separated). Applying Lemma \ref{lemma-generically-finite} to $f|_{X'} : X' \to Y$ we see that (3) holds. We omit the proof that (3) implies (2).
Let $V$ be a scheme and let $V \to Y$ be a surjective \'etale morphism. Let $U$ be a scheme and let $U \to V \times_Y X$ be a surjective \'etale morphism. Then $f$ is quasi-finite at the image $x$ of a point $u \in U$ if and only if $U \to V$ is quasi-finite at $u$. Moreover, $x \in X^0$ if and only if $u$ is the generic point of an irreducible component of $U$ (Properties of Spaces, Lemma \ref{spaces-properties-lemma-codimension-0-points}). Thus the lemma reduces to the case of the morphism $U \to V$, i.e., to Morphisms, Lemma \ref{morphisms-lemma-quasi-finiteness-over-generic-point}.
By Lemmas \ref{lemma-finitely-many-irreducible-components} and \ref{lemma-decent-generic-points} we may assume $Y$ is a scheme with finitely many irreducible components. Shrinking further we may assume $Y$ is an irreducible affine scheme with generic point $y$. Then the fibre of $f$ over $y$ is finite. \medskip\noindent Assume $f$ is quasi-compact and $Y$ affine irreducible. Then $X$ is quasi-compact and we may choose an affine scheme $U$ and a surjective \'etale morphism $U \to X$. Then $U \to Y$ is of finite type and the fibre of $U \to Y$ over $y$ is the set $U^0$ of generic points of irreducible components of $U$ (Properties of Spaces, Lemma \ref{spaces-properties-lemma-codimension-0-points}). Hence $U^0$ is finite (Morphisms, Lemma \ref{morphisms-lemma-quasi-finite-at-a-finite-number-of-points}) and after shrinking $Y$ we may assume that $U \to Y$ is finite (Morphisms, Lemma \ref{morphisms-lemma-generically-finite}). Next, consider $R = U \times_X U$. Since the projection $s : R \to U$ is \'etale we see that $R^0 = s^{-1}(U^0)$ lies over $y$. Since $R \to U \times_Y U$ is a monomorphism, we conclude that $R^0$ is finite as $U \times_Y U \to Y$ is finite. And $R$ is separated (Properties of Spaces, Lemma \ref{spaces-properties-lemma-separated-cover}). Thus we may shrink $Y$ once more to reach the situation where $R$ is finite over $Y$ (Morphisms, Lemma \ref{morphisms-lemma-finite-over-dense-open}). In this case it follows that $X = U/R$ is finite over $Y$ by exactly the same arguments as given in the proof of Lemma \ref{lemma-generically-finite} (or we can simply apply that lemma because it follows immediately that $X$ is quasi-separated as well). \medskip\noindent Assume $f$ is separated and $Y$ affine irreducible. Choose $V \subset Y$ and $U \subset X$ as in Lemma \ref{lemma-generically-finite-reprise}. Since $f|_U : U \to V$ is finite, we see that $U \subset f^{-1}(V)$ is closed as well as open (Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-universally-closed-permanence} and \ref{spaces-morphisms-lemma-finite-proper}). Thus $f^{-1}(V) = U \amalg W$ for some open subspace $W$ of $X$. However, since $U$ contains all the codimension $0$ points of $X$ we conclude that $W = \emptyset$ (Properties of Spaces, Lemma \ref{spaces-properties-lemma-codimension-0-points-dense}) as desired.
Let $X' \subset X$ and $Y' \subset Y$ be the maximal open subspaces which are representable, see Lemma \ref{lemma-finitely-many-irreducible-components}. By Lemma \ref{lemma-quasi-finiteness-over-generic-point} the fibre of $f$ over $y$ is consists of points of codimension $0$ of $X$ and is therefore contained in $X'$. Hence $X \times_Y \Spec(\mathcal{O}_{Y, y}) = X' \times_{Y'} \Spec(\mathcal{O}_{Y', y})$ and the result follows from Morphisms, Lemma \ref{morphisms-lemma-birational-generic-fibres}.
By Lemma \ref{lemma-finitely-many-irreducible-components} we may assume $Y$ is a scheme. By Lemma \ref{lemma-finite-over-dense-open} we may assume that $f$ is finite. Then $X$ is a scheme too and the result follows from Morphisms, Lemma \ref{morphisms-lemma-birational-isomorphism-over-dense-open}.
The algebraic space $U = X \times_Y V$ is decent (Lemma \ref{lemma-etale-named-properties}). The generic points of $V$ and $U$ are the elements of $|V|$ and $|U|$ which lie over generic points of $|Y|$ and $|X|$ (Lemma \ref{lemma-decent-generic-points}). Since $Y$ is decent we conclude there are finitely many generic points on $V$. Let $\xi \in |X|$ be a generic point of an irreducible component. By the discussion following Definition \ref{definition-birational} we have a cartesian square $$ \xymatrix{ \Spec(\mathcal{O}_{X, \xi}) \ar[d] \ar[r] & X \ar[d] \\ \Spec(\mathcal{O}_{Y, f(\xi)}) \ar[r] & Y } $$ whose horizontal morphisms are monomorphisms identifying local rings and where the left vertical arrow is an isomorphism. It follows that in the diagram $$ \xymatrix{ \Spec(\mathcal{O}_{X, \xi}) \times_X U \ar[d] \ar[r] & U \ar[d] \\ \Spec(\mathcal{O}_{Y, f(\xi)}) \times_Y V \ar[r] & V } $$ the vertical arrow on the left is an isomorphism. The horizonal arrows have image contained in the schematic locus of $U$ and $V$ and identify local rings (some details omitted). Since the image of the horizontal arrows are the points of $|U|$, resp.\ $|V|$ lying over $\xi$, resp.\ $f(\xi)$ we conclude.
By Lemma \ref{lemma-finitely-many-irreducible-components} we see that $X$ and $Y$ satisfy the equivalent conditions of Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-prepare-normalization} and the normalizations are defined. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-normalization-normal} the algebraic space $X^\nu$ is normal and maps codimension $0$ points to codimension $0$ points. Since $f$ maps codimension $0$ points to codimension $0$ points (this is the same as generic points on decent spaces by Lemma \ref{lemma-decent-generic-points}) we obtain from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-normalization-normal} a factorization of the composition $X^\nu \to X \to Y$ through $Y^\nu$. \medskip\noindent Observe that $X^\nu$ and $Y^\nu$ are decent for example by Lemma \ref{lemma-representable-named-properties}. Moreover the maps $X^\nu \to X$ and $Y^\nu \to Y$ induce bijections on irreducible components (see references above) hence $X^\nu$ and $Y^\nu$ both have a finite number of irreducible components and the map $X^\nu \to Y^\nu$ induces a bijection between their generic points. To prove that $X^\nu \to Y^\nu$ is birational, it therefore suffices to show it induces an isomorphism on local rings at these points. To do this we may replace $X$ and $Y$ by open neighbourhoods of their generic points, hence we may assume $X$ and $Y$ are affine irreducible schemes with generic points $x$ and $y$. Since $f$ is birational the map $\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}$ is an isomorphism. Let $x^\nu \in X^\nu$ and $y^\nu \in Y^\nu$ be the points lying over $x$ and $y$. By construction of the normalization we see that $\mathcal{O}_{X^\nu, x^\nu} = \mathcal{O}_{X, x}/\mathfrak m_x$ and similarly on $Y$. Thus the map $\mathcal{O}_{X^\nu, x^\nu} \to \mathcal{O}_{Y^\nu, y^\nu}$ is an isomorphism as well.
Let $V \to Y$ be an \'etale morphism with $V$ affine. It suffices to show that $U = X \times_Y V \to V$ is an isomorphism. By Lemma \ref{lemma-birational-etale-localization} and its proof we see that $U$ and $V$ are decent and have finitely many irreducible components and that $U \to V$ is birational. By Properties, Lemma \ref{properties-lemma-normal-locally-finite-nr-irreducibles} $V$ is a finite disjoint union of integral schemes. Thus we may assume $V$ is integral. As $f$ is birational, we see that $U$ is irreducible and reduced, i.e., integral (note that $U$ is a scheme as $f$ is integral, hence representable). Thus we may assume that $X$ and $Y$ are integral schemes and the result follows from the case of schemes, see Morphisms, Lemma \ref{morphisms-lemma-finite-birational-over-normal}.
Let $U \to X$ be a surjective \'etale morphism where $U$ is a scheme. Then $U$ is Jacobson (by definition) and for a morphism of schemes $V \to U$ which is locally of finite type we see that $V$ is Jacobson by the corresponding result for schemes (Morphisms, Lemma \ref{morphisms-lemma-Jacobson-universally-Jacobson}). Thus if $Y \to X$ is a morphism of algebraic spaces which is locally of finite type, then setting $V = U \times_X Y$ we see that $Y$ is Jacobson by definition.
Let $U \to X$ be an \'etale morphism with $U$ a scheme and with $u \in U$ closed mapping to $x$, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-identify-finite-type-points}. Observe that $W$, $W \times_X U$, and $U$ are Jacobson schemes by Lemma \ref{lemma-Jacobson-universally-Jacobson}. Hence finite type points on these schemes are the same thing as closed points by Morphisms, Lemma \ref{morphisms-lemma-jacobson-finite-type-points}. The inverse image $T \subset W \times_X U$ of $u$ is a nonempty (as $x$ in the image of $W \to X$) closed subset. By Morphisms, Lemma \ref{morphisms-lemma-enough-finite-type-points} there is a closed point $t$ of $W \times_X U$ which maps to $u$. As $W \times_X U \to W$ is locally of finite type the image of $t$ in $W$ is closed by Morphisms, Lemma \ref{morphisms-lemma-jacobson-finite-type-points}.
If $x \in |X|$ is closed, then we can represent $x$ by a closed immersion $\Spec(k) \to X$, see Lemma \ref{lemma-decent-space-closed-point}. Hence $x$ is certainly a finite type point. \medskip\noindent Conversely, let $x \in |X|$ be a finite type point. We know that $x$ can be represented by a quasi-compact monomorphism $\Spec(k) \to X$ where $k$ is a field (Definition \ref{definition-very-reasonable}). On the other hand, by definition, there exists a morphism $\Spec(k') \to X$ which is locally of finite type and represents $x$ (Morphisms, Definition \ref{morphisms-definition-finite-type-point}). We obtain a factorization $\Spec(k') \to \Spec(k) \to X$. Let $U \to X$ be any \'etale morphism with $U$ affine and consider the morphisms $$ \Spec(k') \times_X U \to \Spec(k) \times_X U \to U $$ The quasi-compact scheme $\Spec(k) \times_X U$ is \'etale over $\Spec(k)$ hence is a finite disjoint union of spectra of fields (Remark \ref{remark-recall}). Moreover, the first morphism is surjective and locally of finite type (Morphisms, Lemma \ref{morphisms-lemma-permanence-finite-type}) hence surjective on finite type points (Morphisms, Lemma \ref{morphisms-lemma-finite-type-points-surjective-morphism}) and the composition (which is locally of finite type) sends finite type points to closed points as $U$ is Jacobson (Morphisms, Lemma \ref{morphisms-lemma-jacobson-finite-type-points}). Thus the image of $\Spec(k) \times_X U \to U$ is a finite set of closed points hence closed. Since this is true for every affine $U$ and \'etale morphism $U \to X$, we conclude that $x \in |X|$ is closed.
Assume $X$ is Jacobson and that $T \subset |X|$ is a closed subset. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-enough-finite-type-points} we see that $T \cap X_{\text{ft-pts}}$ is dense in $T$. By Lemma \ref{lemma-decent-Jacobson-ft-pts} we see that $X_{\text{ft-pts}}$ are the closed points of $|X|$. Thus $|X|$ is indeed Jacobson. \medskip\noindent Assume $|X|$ is Jacobson. Let $f : U \to X$ be an \'etale morphism with $U$ an affine scheme. We have to show that $U$ is Jacobson. If $x \in |X|$ is closed, then the fibre $F = f^{-1}(\{x\})$ is a finite (by definition of decent) closed (by construction of the topology on $|X|$) subset of $U$. Since there are no specializations between points of $F$ (Lemma \ref{lemma-decent-no-specializations-map-to-same-point}) we conclude that every point of $F$ is closed in $U$. If $U$ is not Jacobson, then there exists a non-closed point $u \in U$ such that $\{u\}$ is locally closed (Topology, Lemma \ref{topology-lemma-non-jacobson-Noetherian-characterize}). We will show that $f(u) \in |X|$ is closed; by the above $u$ is closed in $U$ which is a contradiction and finishes the proof. To prove this we may replace $U$ by an affine open neighbourhood of $u$. Thus we may assume that $\{u\}$ is closed in $U$. Let $R = U \times_X U$ with projections $s, t : R \to U$. Then $s^{-1}(\{u\}) = \{r_1, \ldots, r_m\}$ is finite (by definition of decent spaces). After replacing $U$ by a smaller affine open neighbourhood of $u$ we may assume that $t(r_j) = u$ for $j = 1, \ldots, m$. It follows that $\{u\}$ is an $R$-invariant closed subset of $U$. Hence $\{f(u)\}$ is a locally closed subset of $X$ as it is closed in the open $|f|(|U|)$ of $|X|$. Since $|X|$ is Jacobson we conclude that $f(u)$ is closed in $|X|$ as desired.
We may replace by any open subspace containing $x$. Thus we may assume that $X$ is quasi-compact. Then $|X|$ is a Noetherian topological space (Properties of Spaces, Lemma \ref{spaces-properties-lemma-Noetherian-topology}). Thus $W$ is a Noetherian topological space (Topology, Lemma \ref{topology-lemma-Noetherian}). \medskip\noindent Combining Lemma \ref{lemma-locally-Noetherian-decent-quasi-separated} with Properties of Spaces, Lemma \ref{spaces-properties-lemma-quasi-compact-quasi-separated-spectral} we see that $|X|$ is a spectral toplogical space. By Topology, Lemma \ref{topology-lemma-make-spectral-space} we see that $W \cup \{x\}$ is a spectral topological space. Now $W$ is a quasi-compact open of $W \cup \{x\}$ and hence $W$ is spectral by Topology, Lemma \ref{topology-lemma-spectral-sub}. \medskip\noindent Let $E \subset W$ be an irreducible closed subset. Then if $Z \subset |X|$ is the closure of $E$ we see that $x \in Z$. There is a unique generic point $\eta \in Z$ by Proposition \ref{proposition-reasonable-sober}. Of course $\eta \in W$ and hence $\eta \in E$. We conclude that $E$ has a unique generic point, i.e., $W$ is sober. \medskip\noindent Let $x' \in W$ be a point such that $\{x'\}$ is locally closed in $W$. To finish the proof we have to show that $x'$ is a closed point of $W$. If not, then there exists a nontrivial specialization $x' \leadsto x'_1$ in $W$. Let $U$ be an affine scheme, $u \in U$ a point, and let $U \to X$ be an \'etale morphism mapping $u$ to $x$. By Lemma \ref{lemma-decent-specialization} we can choose specializations $u' \leadsto u'_1 \leadsto u$ mapping to $x' \leadsto x'_1 \leadsto x$. Let $\mathfrak p' \subset \mathcal{O}_{U, u}$ be the prime ideal corresponding to $u'$. The existence of the specializations implies that $\dim(\mathcal{O}_{U, u}/\mathfrak p') \geq 2$. Hence every nonempty open of $\Spec(\mathcal{O}_{U, u}/\mathfrak p')$ is infinite by Algebra, Lemma \ref{algebra-lemma-Noetherian-local-domain-dim-2-infinite-opens}. By Lemma \ref{lemma-decent-no-specializations-map-to-same-point} we obtain a continuous map $$ \Spec(\mathcal{O}_{U, u}/\mathfrak p') \setminus \{\mathfrak m_u/\mathfrak p'\} \longrightarrow W $$ Since the generic point of the LHS maps to $x'$ the image is contained in $\overline{\{x'\}}$. We conclude the inverse image of $\{x'\}$ under the displayed arrow is nonempty open hence infinite. However, the fibres of $U \to X$ are finite as $X$ is decent and we conclude that $\{x'\}$ is infinite. This contradiction finishes the proof.
The equivalence of (1) and (2) follows from the fact that irreducible components of $U$ passing through $u$ are in $1$-$1$ correspondence with minimal primes of the local ring of $U$ at $u$. The ring $\mathcal{O}_{X, x}^h$ is the henselization of $\mathcal{O}_{U, u}$, see discussion following Definition \ref{definition-henselian-local-ring}. In particular (3) and (4) are equivalent by More on Algebra, Lemma \ref{more-algebra-lemma-unibranch}. The equivalence of (2) and (3) follows from More on Morphisms, Lemma \ref{more-morphisms-lemma-nr-branches}.
The equivalence of (1) and (2) follows from the fact that irreducible components of $U$ passing through $u$ are in $1$-$1$ correspondence with minimal primes of the local ring of $U$ at $u$. The ring $\mathcal{O}_{X, x}$ is the henselization of $\mathcal{O}_{U, u}$, see discussion following Definition \ref{definition-henselian-local-ring}. In particular (3) and (4) are equivalent by More on Algebra, Lemma \ref{more-algebra-lemma-unibranch}. The equivalence of (2) and (3) follows from More on Morphisms, Lemma \ref{more-morphisms-lemma-nr-branches}.
Let $\varphi : U \to X$ be a surjective \'etale morphism where $U$ is a scheme. Then the similarly defined function $\delta_U$ is a dimension function on $U$ by Morphisms, Lemma \ref{morphisms-lemma-dimension-function-propagates}. On the other hand, by the definition of relative transcendence degree in (Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-dimension-fibre}) we see that $\delta_U(u) = \delta_X(\varphi(u))$. \medskip\noindent Let $x \leadsto x'$ be a specialization of points in $|X|$. by Lemma \ref{lemma-decent-specialization} we can find a specialization $u \leadsto u'$ of points of $U$ with $\varphi(u) = x$ and $\varphi(u') = x'$. Moreover, we see that $x = x'$ if and only if $u = u'$, see Lemma \ref{lemma-decent-no-specializations-map-to-same-point}. Thus the fact that $\delta_U$ is a dimension function implies that $\delta_X$ is a dimension function, see Topology, Definition \ref{topology-definition-dimension-function}.
The question is local on $S$ (use Topology, Lemma \ref{topology-lemma-catenary}). Thus we may assume that $S$ has a dimension function, see Topology, Lemma \ref{topology-lemma-locally-dimension-function}. Then we conclude that $|X|$ has a dimension function by Lemma \ref{lemma-scheme-with-dimension-function}. Since $|X|$ is sober (Proposition \ref{proposition-reasonable-sober}) we conclude that $|X|$ is catenary by Topology, Lemma \ref{topology-lemma-dimension-function-catenary}.
Let $x \mapsto x'$, $x \not = x'$ be a specialization in $|X|$. Choose $y \in |Y|$ with $|f|(y) = x$. Since $|f|$ is closed (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-finite-proper}) we find a specialization $y \leadsto y'$ with $|f|(y') = x'$. Thus we conclude that $\delta(x) = \delta(|f|(y)) > \delta(|f|(y')) = \delta(x')$ (see Topology, Definition \ref{topology-definition-dimension-function}). If $x \leadsto x'$ is an immediate specialization, then $y \leadsto y'$ is an immediate specialization too: namely if $y \leadsto y'' \leadsto y'$, then $|f|(y'')$ must be either $x$ or $x'$ and there are no nontrivial specializations between points of fibres of $|f|$ by Lemma \ref{lemma-conditions-on-fibre-and-qf}.
By Properties of Spaces, Lemma \ref{spaces-properties-lemma-subscheme} the question is local on $X$. Hence we may assume there exists an affine scheme $U$ and a surjective \'etale morphism $U \to X$ (Properties of Spaces, Lemma \ref{spaces-properties-lemma-cover-by-union-affines}). Let $n$ be an integer bounding the degrees of the fibres of $U \to X$ which exists as $X$ is reasonable, see Definition \ref{definition-very-reasonable}. We will argue by induction on $n$ that whenever \begin{enumerate} \item $U \to X$ is a surjective \'etale morphism whose fibres have degree $\leq n$, and \item $U$ is isomorphic to a locally closed subscheme of an affine scheme \end{enumerate} then the schematic locus is dense in $X$. \medskip\noindent Let $X_n \subset X$ be the open subspace which is the complement of the closed subspace $Z_{n - 1} \subset X$ constructed in Lemma \ref{lemma-quasi-compact-reasonable-stratification} using the morphism $U \to X$. Let $U_n \subset U$ be the inverse image of $X_n$. Then $U_n \to X_n$ is finite locally free of degree $n$. Hence $X_n$ is a scheme by Properties of Spaces, Proposition \ref{spaces-properties-proposition-finite-flat-equivalence-global} (and the fact that any finite set of points of $U_n$ is contained in an affine open of $U_n$, see Properties, Lemma \ref{properties-lemma-ample-finite-set-in-affine}). \medskip\noindent Let $X' \subset X$ be the open subspace such that $|X'|$ is the interior of $|Z_{n - 1}|$ in $|X|$ (see Topology, Definition \ref{topology-definition-nowhere-dense}). Let $U' \subset U$ be the inverse image. Then $U' \to X'$ is surjective \'etale and has degrees of fibres bounded by $n - 1$. By induction we see that the schematic locus of $X'$ is an open dense $X'' \subset X'$. By elementary topology we see that $X'' \cup X_n \subset X$ is open and dense and we win.
We have seen in Lemma \ref{lemma-kolmogorov} that $|X|$ is Kolmogorov. Hence it remains to show that every irreducible closed subset $T \subset |X|$ has a generic point. By Properties of Spaces, Lemma \ref{spaces-properties-lemma-reduced-closed-subspace} there exists a closed subspace $Z \subset X$ with $|Z| = |T|$. By definition this means that $Z \to X$ is a representable morphism of algebraic spaces. Hence $Z$ is a decent algebraic space by Lemma \ref{lemma-representable-properties}. By Theorem \ref{theorem-decent-open-dense-scheme} we see that there exists an open dense subspace $Z' \subset Z$ which is a scheme. This means that $|Z'| \subset T$ is open dense. Hence the topological space $|Z'|$ is irreducible, which means that $Z'$ is an irreducible scheme. By Schemes, Lemma \ref{schemes-lemma-scheme-sober} we conclude that $|Z'|$ is the closure of a single point $\eta \in T$ and hence also $T = \overline{\{\eta\}}$, and we win.