hcju/stacksqa huggingface dataset:
Let $S$ be a scheme and let $X$ be a locally separated algebraic space over $S$. We may assume $S = \Spec(\mathbf{Z})$, see Properties of Spaces, Definition \ref{spaces-properties-definition-separated}. Unadorned fibre products will be over $\mathbf{Z}$. Let $x \in |X|$. Choose a scheme $U$, an \'etale morphism $U \to X$, and a point $u \in U$ mapping to $x$ in $|X|$. As usual we identify $u = \Spec(\kappa(u))$. As $X$ is locally separated the morphism $$ u \times_X u \to u \times u $$ is an immersion (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-fibre-product-after-map}). Hence More on Groupoids, Lemma \ref{more-groupoids-lemma-locally-closed-image-is-closed} tells us that it is a closed immersion (use Schemes, Lemma \ref{schemes-lemma-immersion-when-closed}). As $u \times_X u \to u \times_X U$ is a monomorphism (base change of $u \to U$) and as $u \times_X U \to u$ is \'etale we conclude that $u \times_X u$ is a disjoint union of spectra of fields (see Remark \ref{remark-recall} and Schemes, Lemma \ref{schemes-lemma-mono-towards-spec-field}). Since it is also closed in the affine scheme $u \times u$ we conclude $u \times_X u$ is a finite disjoint union of spectra of fields. Thus $x$ can be represented by a monomorphism $\Spec(k) \to X$ where $k$ is a field, see Lemma \ref{lemma-R-finite-above-x}. \medskip\noindent Next, let $U = \Spec(A)$ be an affine scheme and let $U \to X$ be an \'etale morphism. To finish the proof it suffices to show that $F = U \times_X \Spec(k)$ is finite. Write $F = \coprod_{i \in I} \Spec(k_i)$ as the disjoint union of finite separable extensions of $k$. We have to show that $I$ is finite. Set $R = U \times_X U$. As $X$ is locally separated, the morphism $j : R \to U \times U$ is an immersion. Let $U' \subset U \times U$ be an open such that $j$ factors through a closed immersion $j' : R \to U'$. Let $e : U \to R$ be the diagonal map. Using that $e$ is a morphism between schemes \'etale over $U$ such that $\Delta = j \circ e$ is a closed immersion, we conclude that $R = e(U) \amalg W$ for some open and closed subscheme $W \subset R$. Since $j'$ is a closed immersion we conclude that $j'(W) \subset U'$ is closed and disjoint from $j'(e(U))$. Therefore $\overline{j(W)} \cap \Delta(U) = \emptyset$ in $U \times U$. Note that $W$ contains $\Spec(k_i \otimes_k k_{i'})$ for all $i \not = i'$, $i, i' \in I$. By Lemma \ref{lemma-infinite-number} we conclude that $I$ is finite as desired.
Let $T$ be a scheme and let $T \to Y$ be a morphism. Then $T \to Y$ is representable, hence the base change $T \times_Y X \to X$ is representable. Hence if $X$ is decent (or reasonable), then so is $T \times_Y X$, see Lemma \ref{lemma-representable-named-properties}. Similarly, for property $(\beta)$, see Lemma \ref{lemma-representable-properties}.
Let us prove the lemma in case $\omega = \beta$. In this case we have to show that any $x \in |X|$ is represented by a monomorphism from the spectrum of a field into $X$. Let $y = f(x) \in |Y|$. By assumption there exists a field $k$ and a monomorphism $\Spec(k) \to Y$ representing $y$. Then $x$ corresponds to a point $x'$ of $\Spec(k) \times_Y X$. By assumption $x'$ is represented by a monomorphism $\Spec(k') \to \Spec(k) \times_Y X$. Clearly the composition $\Spec(k') \to X$ is a monomorphism representing $x$. \medskip\noindent Let us prove the lemma in case $\omega = decent$. Let $x \in |X|$ and $y = f(x) \in |Y|$. By the result of the preceding paragraph we can choose a diagram $$ \xymatrix{ \Spec(k') \ar[r]_x \ar[d] & X \ar[d]^f \\ \Spec(k) \ar[r]^y & Y } $$ whose horizontal arrows monomorphisms. As $Y$ is decent the morphism $y$ is quasi-compact. As $f$ is decent the algebraic space $\Spec(k) \times_Y X$ is decent. Hence the monomorphism $\Spec(k') \to \Spec(k) \times_Y X$ is quasi-compact. Then the monomorphism $x : \Spec(k') \to X$ is quasi-compact as a composition of quasi-compact morphisms (use Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-base-change-quasi-compact} and \ref{spaces-morphisms-lemma-composition-quasi-compact}). As the point $x$ was arbitrary this implies $X$ is decent. \medskip\noindent Let us prove the lemma in case $\omega = reasonable$. Choose $V \to Y$ \'etale with $V$ an affine scheme. Choose $U \to V \times_Y X$ \'etale with $U$ an affine scheme. By assumption $V \to Y$ has universally bounded fibres. By Lemma \ref{lemma-base-change-universally-bounded} the morphism $V \times_Y X \to X$ has universally bounded fibres. By assumption on $f$ we see that $U \to V \times_Y X$ has universally bounded fibres. By Lemma \ref{lemma-composition-universally-bounded} the composition $U \to X$ has universally bounded fibres. Hence there exists sufficiently many \'etale morphisms $U \to X$ from schemes with universally bounded fibres, and we conclude that $X$ is reasonable.
Let $\omega \in \{\beta, decent, reasonable\}$. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of algebraic spaces over the scheme $S$. Assume $f$ and $g$ both have property $(\omega)$. Then we have to show that for any scheme $T$ and morphism $T \to Z$ the space $T \times_Z X$ has $(\omega)$. By Lemma \ref{lemma-base-change-relative-conditions} this reduces us to the following claim: Suppose that $Y$ is an algebraic space having property $(\omega)$, and that $f : X \to Y$ is a morphism with $(\omega)$. Then $X$ has $(\omega)$. This is the content of Lemma \ref{lemma-property-over-property}.
Namely, by Lemma \ref{lemma-property-for-morphism-out-of-property} the morphism $X \to Y$ has the property. Then the base change $X \times_Y Z \to Z$ has the property by Lemma \ref{lemma-base-change-relative-conditions}. And finally this implies $X \times_Y Z$ has the property by Lemma \ref{lemma-property-over-property}.
The implications (1) $\Rightarrow$ (2) $\Rightarrow$ (3) $\Rightarrow$ (4) are trivial. The implication (3) $\Rightarrow$ (1) can be seen as follows. Let $Z \to Y$ be a morphism whose source is a scheme over $S$. Consider the algebraic space $Z \times_Y X$. If we assume (3), then for any affine open $W \subset Z$, the open subspace $W \times_Y X$ of $Z \times_Y X$ has property $\mathcal{P}$. Hence by Lemma \ref{lemma-properties-local} the space $Z \times_Y X$ has property $\mathcal{P}$, i.e., (1) holds. A similar argument (omitted) shows that (4) implies (1). \medskip\noindent The implication (1) $\Rightarrow$ (5) is trivial. Let $V \to Y$ be an \'etale morphism from a scheme as in (5). Let $Z$ be an affine scheme, and let $Z \to Y$ be a morphism. Consider the diagram $$ \xymatrix{ Z \times_Y V \ar[r]_q \ar[d]_p & V \ar[d] \\ Z \ar[r] & Y } $$ Since $p$ is \'etale, and hence open, we can choose finitely many affine open subschemes $W_i \subset Z \times_Y V$ such that $Z = \bigcup p(W_i)$. Consider the commutative diagram $$ \xymatrix{ V \times_Y X \ar[d] & (\coprod W_i) \times_Y X \ar[l] \ar[d] \ar[r] & Z \times_Y X \ar[d] \\ V & \coprod W_i \ar[l] \ar[r] & Z } $$ We know $V \times_Y X$ has property $\mathcal{P}$. By Lemma \ref{lemma-representable-properties} we see that $(\coprod W_i) \times_Y X$ has property $\mathcal{P}$. Note that the morphism $(\coprod W_i) \times_Y X \to Z \times_Y X$ is \'etale and quasi-compact as the base change of $\coprod W_i \to Z$. Hence by Lemma \ref{lemma-descent-conditions} we conclude that $Z \times_Y X$ has property $\mathcal{P}$.
In Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-quasi-compact-existence-universally-closed} we proved that any quasi-compact morphism which satisfies the existence part of the valuative criterion is universally closed. To prove the other, assume that $f$ is universally closed. In the proof of Proposition \ref{proposition-characterize-universally-closed} we have seen that it suffices to show, for any valuation ring $A$, and any morphism $\Spec(A) \to Y$, that the base change $f_A : X_A \to \Spec(A)$ satisfies the existence part of the valuative criterion. By definition the algebraic space $X_A$ has property $(\gamma)$ and hence Proposition \ref{proposition-characterize-universally-closed} applies to the morphism $f_A$ and we win.
The morphism $q$ is quasi-finite at every $w \in F_{x, z}$, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-quasi-finite-locus}. Hence the lemma follows from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-quasi-finite-at-a-finite-number-of-points}.
By Lemma \ref{lemma-surjective-on-fibres} applied twice we may replace $Z$ by $Z_k = \Spec(k) \times_Y Z$ and $X$ by $X_k = \Spec(k) \times_Y X$. We may and do replace $Y$ by $\Spec(k)$ as well. Note that $Z_k \to \Spec(k)$ is quasi-finite at $z$ by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-quasi-finite-locus}. Choose a scheme $V$, a point $v \in V$, and an \'etale morphism $V \to Z_k$ mapping $v$ to $z$. Choose a scheme $U$, a point $u \in U$, and an \'etale morphism $U \to X_k$ mapping $u$ to $x$. Again by Lemma \ref{lemma-surjective-on-fibres} it suffices to show $F_{u, v}$ is finite for the diagram $$ \xymatrix{ U \times_{\Spec(k)} V \ar[r] \ar[d] & V \ar[d] \\ U \ar[r] & \Spec(k) } $$ The morphism $V \to \Spec(k)$ is quasi-finite at $v$ (follows from the general discussion in Morphisms of Spaces, Section \ref{spaces-morphisms-section-local-source-target} and the definition of being quasi-finite at a point). At this point the finiteness follows from Example \ref{example-schemes}. The parenthetical remark of the statement of the lemma follows from the fact that on decent spaces points are represented by monomorphisms from fields and from the fact that an \'etale morphism of algebraic spaces is locally quasi-finite.
We will use Properties of Spaces, Lemma \ref{spaces-properties-lemma-etale-open} and Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-monomorphism-injective-points} without further mention. Let $V \to Y$ be an \'etale morphism with $V$ affine such that there exists a $v \in V$ mapping to $y$. Since $\Spec(k) \to Y$ is quasi-compact there are a finite number of points of $V$ mapping to $y$ (Lemma \ref{lemma-UR-finite-above-x}). After shrinking $V$ we may assume $v$ is the only one. Choose a scheme $U$ and a surjective \'etale morphism $U \to X$. Consider the commutative diagram $$ \xymatrix{ U \ar[d] & U_V \ar[l] \ar[d] & U_v \ar[l] \ar[d] \\ X \ar[d] & X_V \ar[l] \ar[d] & X_v \ar[l] \ar[d] \\ Y & V \ar[l] & v \ar[l] } $$ Since $U_v \to U_V$ identifies $U_v$ with a subset of $U_V$ with the induced topology (Schemes, Lemma \ref{schemes-lemma-fibre-topological}), and since $|U_V| \to |X_V|$ and $|U_v| \to |X_v|$ are surjective and open, we see that $|X_v| \to |X_V|$ is a homeomorphism onto its image (with induced topology). On the other hand, the inverse image of $f^{-1}(\{y\})$ under the open map $|X_V| \to |X|$ is equal to $|X_v|$. We conclude that $|X_v| \to f^{-1}(\{y\})$ is open. The morphism $X_v \to X$ factors through $X_k$ and $|X_k| \to |X|$ is injective with image $f^{-1}(\{y\})$ by Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian}. Using $|X_v| \to |X_k| \to f^{-1}(\{y\})$ the lemma follows because $X_v \to X_k$ is surjective.
Parts (4) and (5) are equivalent for example by Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-locally-finite-type-quasi-finite-part} and \ref{spaces-morphisms-lemma-quasi-finite-at-point}. \medskip\noindent Let $U \to X$ be an \'etale morphism where $U$ is an affine scheme and let $u \in U$ be a point mapping to $x$. Moreover, if $x$ is a closed point, e.g., in case (2) or (3), then we may and do assume that $u$ is a closed point. Observe that $\dim_u(U) = \dim_x(X)$ by definition and that this is equal to $\dim(\mathcal{O}_{U, u})$ if $u$ is a closed point, see Algebra, Lemma \ref{algebra-lemma-dimension-closed-point-finite-type-field}. \medskip\noindent If $\dim_x(X) > 0$ and $u$ is closed, by the arguments above we can choose a nontrivial specialization $u' \leadsto u$ in $U$. Then the transcendence degree of $\kappa(u')$ over $k$ exceeds the transcendence degree of $\kappa(u)$ over $k$. It follows that the images $x$ and $x'$ in $X$ are distinct, because the transcendence degree of $x/k$ and $x'/k$ are well defined, see Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-dimension-fibre}. This applies in particular in cases (2) and (3) and we conclude that (2) and (3) imply (4). \medskip\noindent Conversely, if $X \to \Spec(k)$ is locally quasi-finite at $x$, then $U \to \Spec(k)$ is locally quasi-finite at $u$, hence $u$ is an isolated point of $U$ (Morphisms, Lemma \ref{morphisms-lemma-quasi-finite-at-point-characterize}). It follows that (5) implies (2) and (3) as $|U| \to |X|$ is continuous and open. \medskip\noindent Assume $X$ is decent and (1) holds. Then $\dim_x(X) = \dim_x(|X|)$ by Lemma \ref{lemma-dimension-decent-space} and the proof is complete.
Parts (4) and (5) are equivalent for example by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-locally-finite-type-quasi-finite-part}. \medskip\noindent Let $U \to X$ be a surjective \'etale morphism where $U$ is a scheme. \medskip\noindent If $\dim(U) > 0$, then choose a nontrivial specialization $u \leadsto u'$ in $U$ and the transcendence degree of $\kappa(u)$ over $k$ exceeds the transcendence degree of $\kappa(u')$ over $k$. It follows that the images $x$ and $x'$ in $X$ are distinct, because the transcendence degree of $x/k$ and $x'/k$ is well defined, see Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-dimension-fibre}. We conclude that (2) and (3) imply (4). \medskip\noindent Conversely, if $X \to \Spec(k)$ is locally quasi-finite, then $U$ is locally Noetherian (Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}) of dimension $0$ (Morphisms, Lemma \ref{morphisms-lemma-locally-quasi-finite-rel-dimension-0}) and hence is a disjoint union of spectra of Artinian local rings (Properties, Lemma \ref{properties-lemma-locally-Noetherian-dimension-0}). Hence $U$ is a discrete topological space, and since $|U| \to |X|$ is continuous and open, the same is true for $|X|$. In other words, (4) implies (2) and (3). \medskip\noindent Assume $X$ is decent and (1) holds. Then we may choose $U$ above to be affine. The fibres of $|U| \to |X|$ are finite (this is a part of the defining property of decent spaces). Hence $U$ is a finite type scheme over $k$ with finitely many points. Hence $U$ is quasi-finite over $k$ (Morphisms, Lemma \ref{morphisms-lemma-finite-fibre}) which by definition means that $X \to \Spec(k)$ is locally quasi-finite.
By Lemma \ref{lemma-conditions-on-point-on-space-over-field} conditions (\ref{item-isolated-in-k-fibre}), (\ref{item-no-specializations-in-k-fibre}), and (\ref{item-k-fibre-at-x-dim-0}) are equivalent to each other and to the condition that $X_k \to \Spec(k)$ is quasi-finite at $\tilde x$. Thus by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-quasi-finite-locus} they are also equivalent to (\ref{item-quasi-finite-at-x}). If $f$ is decent, then $X_k$ is a decent algebraic space and Lemma \ref{lemma-conditions-on-point-on-space-over-field} shows that (\ref{item-dimension-top-k-fibre}) implies (\ref{item-isolated-in-k-fibre}). \medskip\noindent If $Y$ is decent, then we can pick a quasi-compact monomorphism $\Spec(k') \to Y$ in the equivalence class of $y$. In this case Lemma \ref{lemma-topology-fibre} tells us that $|X_{k'}| \to F$ is a homeomorphism. Combined with the arguments given above this implies the remaining statements of the lemma; details omitted.
By Lemma \ref{lemma-conditions-on-space-over-field} conditions  (\ref{item-k-fibre-discrete}), (\ref{item-k-fibre-no-specializations}), and (\ref{item-k-fibre-dim-0}) are equivalent to each other and to the condition that $X_k \to \Spec(k)$ is locally quasi-finite. Thus by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-quasi-finite-locus} they are also equivalent to (\ref{item-quasi-finite-at-points-fibre}). If $f$ is decent, then $X_k$ is a decent algebraic space and Lemma \ref{lemma-conditions-on-space-over-field} shows that (\ref{item-k-fibre-finite}) implies (\ref{item-k-fibre-discrete}). \medskip\noindent The map $|X_k| \to F$ is surjective by Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian} and we see (\ref{item-k-fibre-finite}) $\Rightarrow$ (\ref{item-fibre-finite}). \medskip\noindent If $Y$ is decent, then we can pick a quasi-compact monomorphism $\Spec(k') \to Y$ in the equivalence class of $y$. In this case Lemma \ref{lemma-topology-fibre} tells us that $|X_{k'}| \to F$ is a homeomorphism. Combined with the arguments given above this implies the remaining statements of the lemma; details omitted.
This immediately reduces to the case $Y = \Spec(A)$ where $A$ is a zero dimensional local ring, i.e., $\Spec(A) = \{\mathfrak m_A\}$ is a singleton. If $X = \emptyset$, then there is nothing to prove. If not, choose a nonempty affine scheme $U = \Spec(B)$ and an \'etale morphism $U \to X$. As $|X|$ is a singleton (as a subset of $|Y|$, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-monomorphism-injective-points}) we see that $U \to X$ is surjective. Note that $U \times_X U = U \times_Y U = \Spec(B \otimes_A B)$. Thus we see that the ring maps $B \to B \otimes_A B$ are \'etale. Since $$ (B \otimes_A B)/\mathfrak m_A(B \otimes_A B) = (B/\mathfrak m_AB) \otimes_{A/\mathfrak m_A} (B/\mathfrak m_AB) $$ we see that $B/\mathfrak m_AB \to (B \otimes_A B)/\mathfrak m_A(B \otimes_A B)$ is flat and in fact free of rank equal to the dimension of $B/\mathfrak m_AB$ as a $A/\mathfrak m_A$-vector space. Since $B \to B \otimes_A B$ is \'etale, this can only happen if this dimension is finite (see for example Morphisms, Lemmas \ref{morphisms-lemma-etale-universally-bounded} and \ref{morphisms-lemma-locally-quasi-finite-qc-source-universally-bounded}). Every prime of $B$ lies over $\mathfrak m_A$ (the unique prime of $A$). Hence $\Spec(B) = \Spec(B/\mathfrak m_A)$ as a topological space, and this space is a finite discrete set as $B/\mathfrak m_A B$ is an Artinian ring, see Algebra, Lemmas \ref{algebra-lemma-finite-dimensional-algebra} and \ref{algebra-lemma-artinian-finite-length}. Hence all prime ideals of $B$ are maximal and $B = B_1 \times \ldots \times B_n$ is a product of finitely many local rings of dimension zero, see Algebra, Lemma \ref{algebra-lemma-product-local}. Thus $B \to B \otimes_A B$ is finite \'etale as all the local rings $B_i$ are henselian by Algebra, Lemma \ref{algebra-lemma-local-dimension-zero-henselian}. Thus $X$ is an affine scheme by Groupoids, Proposition \ref{groupoids-proposition-finite-flat-equivalence}.
Conditions (4) and (5) are equivalent for any algebraic space by definition, see Properties of Spaces, Definition \ref{spaces-properties-definition-dimension-local-ring}. Observe that any $Y$ as in (2) and (3) is decent by Lemma \ref{lemma-etale-named-properties}. Thus it suffices to prove the equivalence of (1) and (4) as then the equivalence with (2) and (3) follows since the dimension of the local ring of $Y$ at $y$ is equal to the dimension of the local ring of $X$ at $x$. Let $f : U \to X$ be an \'etale morphism from an affine scheme and let $u \in U$ be a point mapping to $x$. \medskip\noindent Assume (1). Let $u' \leadsto u$ be a specialization in $U$. Then $f(u') = f(u) = x$. By Lemma \ref{lemma-decent-no-specializations-map-to-same-point} we see that $u' = u$. Hence $u$ is a generic point of an irreducible component of $U$. Thus $\dim(\mathcal{O}_{U, u}) = 0$ and we see that (4) holds. \medskip\noindent Assume (4). The point $x$ is contained in an irreducible component $T \subset |X|$. Since $|X|$ is sober (Proposition \ref{proposition-reasonable-sober}) we $T$ has a generic point $x'$. Of course $x' \leadsto x$. Then we can lift this specialization to $u' \leadsto u$ in $U$ (Lemma \ref{lemma-decent-specialization}). This contradicts the assumption that $\dim(\mathcal{O}_{U, u}) = 0$ unless $u' = u$, i.e., $x' = x$.
Choose a scheme $U$, a point $u \in U$, and an \'etale morphism $U \to X$ sending $u$ to $\xi$. Then any sequence of nontrivial specializations $\xi_e \leadsto \ldots \leadsto \xi_0 = \xi$ can be lifted to a sequence $u_e \leadsto \ldots \leadsto u_0 = u$ in $U$ by Lemma \ref{lemma-decent-specialization}. Conversely, any sequence of nontrivial specializations $u_e \leadsto \ldots \leadsto u_0 = u$ in $U$ maps to a sequence of nontrivial specializations $\xi_e \leadsto \ldots \leadsto \xi_0 = \xi$ by Lemma \ref{lemma-decent-no-specializations-map-to-same-point}. Because $|X|$ and $U$ are sober topological spaces we conclude that the codimension of $T$ in $|X|$ and of $\overline{\{u\}}$ in $U$ are the same. In this way the lemma reduces to the schemes case which is Properties, Lemma \ref{properties-lemma-codimension-local-ring}.
Let $U$ be an affine scheme and let $a : U \to X$ be an \'etale morphism. We have to show that the fibres of $a$ are universally bounded. By assumption (1) the scheme $U$ has finitely many irreducible components. Let $u_1, \ldots, u_n \in U$ be the generic points of these irreducible components. Let $\{x_1, \ldots, x_m\} \subset |X|$ be the image of $\{u_1, \ldots, u_n\}$. Each $x_j$ is a point of codimension $0$. By assumption (2) we may choose a monomorphism $\Spec(k_j) \to X$ representing $x_j$. Then $$ U \times_X \Spec(k_j) = \coprod\nolimits_{a(u_i) = x_j} \Spec(\kappa(u_i)) $$ is finite over $\Spec(k_j)$ of degree $d_j = \sum_{a(u_i) = x_j} [\kappa(u_i) : k_j]$. Set $n = \max d_j$. \medskip\noindent Observe that $a$ is separated (Properties of Spaces, Lemma \ref{spaces-properties-lemma-separated-cover}). Consider the stratification $$ X = X_0 \supset X_1 \supset X_2 \supset \ldots $$ associated to $U \to X$ in Lemma \ref{lemma-stratify-flat-fp-lqf}. By our choice of $n$ above we conclude that $X_{n + 1}$ is empty. Namely, if not, then $a^{-1}(X_{n + 1})$ is a nonempty open of $U$ and hence would contain one of the $x_i$. This would mean that $X_{n + 1}$ contains $x_j = a(u_i)$ which is impossible. Hence we see that the fibres of $U \to X$ are universally bounded (in fact by the integer $n$).
In the proof we use Properties of Spaces, Lemma \ref{spaces-properties-lemma-codimension-0-points} without further mention. Assume (1). Then $X$ has a dense open subscheme $X'$ by Theorem \ref{theorem-decent-open-dense-scheme}. Since the closure of an irreducible component of $|X'|$ is an irreducible component of $|X|$, we see that $|X'|$ has finitely many irreducible components. Thus (3) holds. \medskip\noindent Assume $X' \subset X$ is as in (3). Let $\{x'_1, \ldots, x'_m\}$ be the generic points of the irreducible components of $X'$. Let $a : U \to X$ be an \'etale morphism with $U$ a quasi-compact scheme. It suffices to show that $U$ has finitely many irreducible components whose generic points lie over $\{x'_1, \ldots, x'_m\}$. It suffices to prove this for the members of a finite affine open cover of $U$, hence we may and do assume $U$ is affine. Note that $U' = a^{-1}(X') \subset U$ is a dense open. The generic points of irreducible components of $U'$ are the points lying over $\{x'_1, \ldots, x'_m\}$ and since $x'_j \to X$ is quasi-compact there are finitely many points of $U$ lying over $x'_j$ (Lemma \ref{lemma-UR-finite-above-x}). Hence $U'$ has finitely many irreducible components, which implies that the closures of these irreducible components are the irreducible components of $U$. Thus (2) holds. \medskip\noindent Assume (2). This implies (1) and the final statement by Lemma \ref{lemma-get-reasonable}. (We also use that a reasonable algebraic space is decent, see discussion following Definition \ref{definition-very-reasonable}.)