The Stacks project

Proof. Let $x, y \in X$ with $x \not= y$. Since $X$ is sober, there is an open subset $U$ containing exactly one of the two points $x, y$. Say $x \in U$. We may replace $U$ by a quasi-compact open neighbourhood of $x$ contained in $U$. Then $U$ and $U^ c$ are open and closed in the constructible topology. Hence $X$ is Hausdorff in the constructible topology because $x \in U$ and $y \in U^ c$ are disjoint opens in the constructible topology. The existence of $U$ also implies $x$ and $y$ are in distinct connected components in the constructible topology, whence $X$ is totally disconnected in the constructible topology.

Let $\mathcal{B}$ be the collection of subsets $B \subset X$ with $B$ either quasi-compact open or closed with quasi-compact complement. If $B \in \mathcal{B}$ then $B^ c \in \mathcal{B}$. It suffices to show every covering $X = \bigcup _{i \in I} B_ i$ with $B_ i \in \mathcal{B}$ has a finite refinement, see Lemma 5.12.15. Taking complements we see that we have to show that any family $\{ B_ i\} _{i \in I}$ of elements of $\mathcal{B}$ such that $B_{i_1} \cap \ldots \cap B_{i_ n} \not= \emptyset $ for all $n$ and all $i_1, \ldots , i_ n \in I$ has a common point of intersection. We may and do assume $B_ i \not= B_{i'}$ for $i \not= i'$.

To get a contradiction assume $\{ B_ i\} _{i \in I}$ is a family of elements of $\mathcal{B}$ having the finite intersection property but empty intersection. An application of Zorn's lemma shows that we may assume our family is maximal (details omitted). Let $I' \subset I$ be those indices such that $B_ i$ is closed and set $Z = \bigcap _{i \in I'} B_ i$. This is a closed subset of $X$ which is nonempty by Lemma 5.12.6. If $Z$ is reducible, then we can write $Z = Z' \cup Z''$ as a union of two closed subsets, neither equal to $Z$. This means in particular that we can find a quasi-compact open $U' \subset X$ meeting $Z'$ but not $Z''$. Similarly, we can find a quasi-compact open $U'' \subset X$ meeting $Z''$ but not $Z'$. Set $B' = X \setminus U'$ and $B'' = X \setminus U''$. Note that $Z'' \subset B'$ and $Z' \subset B''$. If there exist a finite number of indices $i_1, \ldots , i_ n \in I$ such that $B' \cap B_{i_1} \cap \ldots \cap B_{i_ n} = \emptyset $ as well as a finite number of indices $j_1, \ldots , j_ m \in I$ such that $B'' \cap B_{j_1} \cap \ldots \cap B_{j_ m} = \emptyset $ then we find that $Z \cap B_{i_1} \cap \ldots \cap B_{i_ n} \cap B_{j_1} \cap \ldots \cap B_{j_ m} = \emptyset $. However, the set $B_{i_1} \cap \ldots \cap B_{i_ n} \cap B_{j_1} \cap \ldots \cap B_{j_ m}$ is quasi-compact hence we would find a finite number of indices $i'_1, \ldots , i'_ l \in I'$ with $B_{i_1} \cap \ldots \cap B_{i_ n} \cap B_{j_1} \cap \ldots \cap B_{j_ m} \cap B_{i'_1} \cap \ldots \cap B_{i'_ l} = \emptyset $, a contradiction. Thus we see that we may add either $B'$ or $B''$ to the given family contradicting maximality. We conclude that $Z$ is irreducible. However, this leads to a contradiction as well, as now every nonempty (by the same argument as above) open $Z \cap B_ i$ for $i \in I \setminus I'$ contains the unique generic point of $Z$. This contradiction proves the lemma. $\square$

Proof. Let $X'$ and $Y'$ denote $X$ and $Y$ endowed with the constructible topology which are quasi-compact Hausdorff spaces by Lemma 5.23.2. Part (1) says $X' \to Y'$ is continuous and follows immediately from the definitions. Part (3) follows as $f(X')$ is a quasi-compact subset of the Hausdorff space $Y'$, see Lemma 5.12.4. We have a commutative diagram

of continuous maps of topological spaces. Since $Y'$ is Hausdorff we see that the fibres $X'_ y$ are closed in $X'$. As $X'$ is quasi-compact we see that $X'_ y$ is quasi-compact (Lemma 5.12.3). As $X'_ y \to X_ y$ is a surjective continuous map we conclude that $X_ y$ is quasi-compact (Lemma 5.12.7). $\square$

Proof. The only if part of this is Lemma 5.23.3. Assume $f$ is continuous in the constructible topology. Let $V \subset Y$ be quasi-compact open. Then $V$ is open and closed in the constructible topology. Hence $f^{-1}(V)$ is open and closed in the constructible topology. Hence $f^{-1}(V)$ is quasi-compact in the constructible topology as $X$ is quasi-compact in the constructible topology by Lemma 5.23.2. Since the identity $f^{-1}(V) \to f^{-1}(V)$ is surjective and continuous from the constructible topology to the usual topology, we conclude that $f^{-1}(V)$ is quasi-compact in the topology of $X$ by Lemma 5.12.7. This finishes the proof. $\square$

Proof. Let $Z \subset E$ be a closed irreducible subset. Let $\eta $ be the generic point of the closure $\overline{Z}$ of $Z$ in $X$. To prove that $E$ is sober, we show that $\eta \in E$. If not, then since $E$ is closed in the constructible topology, there exists a constructible subset $F \subset X$ such that $\eta \in F$ and $F \cap E = \emptyset $. By Lemma 5.15.15 this implies $F \cap \overline{Z}$ contains a nonempty open subset of $\overline{Z}$. But this is impossible as $\overline{Z}$ is the closure of $Z$ and $Z \cap F = \emptyset $.

Since $E$ is closed in the constructible topology, it is quasi-compact in the constructible topology (Lemmas 5.12.3 and 5.23.2). Hence a fortiori it is quasi-compact in the topology coming from $X$. If $U \subset X$ is a quasi-compact open, then $E \cap U$ is closed in the constructible topology, hence quasi-compact (as seen above). It follows that the quasi-compact open subsets of $E$ are the intersections $E \cap U$ with $U$ quasi-compact open in $X$. These form a basis for the topology. Finally, given two $U, U' \subset X$ quasi-compact opens, the intersection $(E \cap U) \cap (E \cap U') = E \cap (U \cap U')$ and $U \cap U'$ is quasi-compact as $X$ is spectral. This finishes the proof. $\square$

Proof. Proof of (1). Let $x \in \overline{E}$. Let $\{ U_ i\} $ be the set of quasi-compact open neighbourhoods of $x$. A finite intersection of the $U_ i$ is another one. The intersection $U_ i \cap E$ is nonempty for all $i$. Since the subsets $U_ i \cap E$ are closed in the constructible topology we see that $\bigcap (U_ i \cap E)$ is nonempty by Lemma 5.23.2 and Lemma 5.12.6. Since $\{ U_ i\} $ is a fundamental system of open neighbourhoods of $x$, we see that $\bigcap U_ i$ is the set of generalizations of $x$. Thus $x$ is a specialization of a point of $E$.

Part (2) is immediate from (1).

Proof of (3). Assume $E'$ is as in (3). The complement of $E'$ is closed in the constructible topology (Lemma 5.15.2) and closed under specialization (Lemma 5.19.2). Hence the complement is closed by (2), i.e., $E'$ is open. $\square$

Proof. Let $\{ U_ i\} $ be the set of quasi-compact open neighbourhoods of $x$. A finite intersection of the $U_ i$ is another one. Let $\{ V_ j\} $ be the set of quasi-compact open neighbourhoods of $y$. A finite intersection of the $V_ j$ is another one. If $U_ i \cap V_ j$ is empty for some $i, j$ we are done. If not, then the intersection $U_ i \cap V_ j$ is nonempty for all $i$ and $j$. The sets $U_ i \cap V_ j$ are closed in the constructible topology on $X$. By Lemma 5.23.2 we see that $\bigcap (U_ i \cap V_ j)$ is nonempty (Lemma 5.12.6). Since $X$ is a sober space and $\{ U_ i\} $ is a fundamental system of open neighbourhoods of $x$, we see that $\bigcap U_ i$ is the set of generalizations of $x$. Similarly, $\bigcap V_ j$ is the set of generalizations of $y$. Thus any element of $\bigcap (U_ i \cap V_ j)$ specializes to both $x$ and $y$. $\square$

Proof. Lemma 5.22.2 shows the implication (1) $\Rightarrow $ (3). Irreducible components are closed, so if $X$ is totally disconnected, then every point is closed. So (3) implies (6). The equivalence of (6) and (5) is immediate, and (6) $\Leftrightarrow $ (7) holds because $X$ is sober. Assume (5). Then all constructible subsets of $X$ are closed (Lemma 5.23.6), in particular all quasi-compact opens are closed. So (5) implies (4). Since $X$ is sober, for any two points there is a quasi-compact open containing exactly one of them, hence (4) implies (2). Parts (4) and (8) are equivalent by the definition of the constructible topology. It remains to prove (2) implies (1). Suppose $X$ is Hausdorff. Every quasi-compact open is also closed (Lemma 5.12.4). This implies $X$ is totally disconnected. Hence it is profinite, by Lemma 5.22.2. $\square$

Proof. Combine Lemmas 5.12.10 and 5.22.5. $\square$

Proof. Let $X$, $Y$ be spectral spaces. Denote $p : X \times Y \to X$ and $q : X \times Y \to Y$ the projections. Let $Z \subset X \times Y$ be a closed irreducible subset. Then $p(Z) \subset X$ is irreducible and $q(Z) \subset Y$ is irreducible. Let $x \in X$ be the generic point of the closure of $p(X)$ and let $y \in Y$ be the generic point of the closure of $q(Y)$. If $(x, y) \not\in Z$, then there exist opens $x \in U \subset X$, $y \in V \subset Y$ such that $Z \cap U \times V = \emptyset $. Hence $Z$ is contained in $(X \setminus U) \times Y \cup X \times (Y \setminus V)$. Since $Z$ is irreducible, we see that either $Z \subset (X \setminus U) \times Y$ or $Z \subset X \times (Y \setminus V)$. In the first case $p(Z) \subset (X \setminus U)$ and in the second case $q(Z) \subset (Y \setminus V)$. Both cases are absurd as $x$ is in the closure of $p(Z)$ and $y$ is in the closure of $q(Z)$. Thus we conclude that $(x, y) \in Z$, which means that $(x, y)$ is the generic point for $Z$.

A basis of the topology of $X \times Y$ are the opens of the form $U \times V$ with $U \subset X$ and $V \subset Y$ quasi-compact open (here we use that $X$ and $Y$ are spectral). Then $U \times V$ is quasi-compact as the product of quasi-compact spaces is quasi-compact. Moreover, any quasi-compact open of $X \times Y$ is a finite union of such quasi-compact rectangles $U \times V$. It follows that the intersection of two such is again quasi-compact (since $X$ and $Y$ are spectral). This concludes the proof. $\square$

Proof. Since $f$ is spectral it defines a continuous map between $X$ and $Y$ in the constructible topology. By Lemmas 5.23.2 and 5.17.8 it follows that $X \to Y$ is a homeomorphism in the constructible topology. Let $U \subset X$ be quasi-compact open. Then $f(U)$ is constructible in $Y$. Let $y \in Y$ specialize to a point in $f(U)$. By the last assumption we see that $f^{-1}(y)$ specializes to a point of $U$. Hence $f^{-1}(y) \in U$. Thus $y \in f(U)$. It follows that $f(U)$ is open, see Lemma 5.23.6. Whence $f$ is a homeomorphism. To prove the lemma in case specializations lift along $f$ one shows instead that $f(Z)$ is closed if $X \setminus Z$ is a quasi-compact open of $X$. $\square$

Proof. Let $I$ be a directed set. Let $X_ i$ be an inverse system of finite sober spaces over $I$. Let $X = \mathop{\mathrm{lim}}\nolimits X_ i$ which exists by Lemma 5.14.1. As a set $X = \mathop{\mathrm{lim}}\nolimits X_ i$. Denote $p_ i : X \to X_ i$ the projection. Because $I$ is directed we may apply Lemma 5.14.2. A basis for the topology is given by the opens $p_ i^{-1}(U_ i)$ for $U_ i \subset X_ i$ open. Since an open covering of $p_ i^{-1}(U_ i)$ is in particular an open covering in the profinite topology, we conclude that $p_ i^{-1}(U_ i)$ is quasi-compact. Given $U_ i \subset X_ i$ and $U_ j \subset X_ j$, then $p_ i^{-1}(U_ i) \cap p_ j^{-1}(U_ j) = p_ k^{-1}(U_ k)$ for some $k \geq i, j$ and open $U_ k \subset X_ k$. Finally, if $Z \subset X$ is irreducible and closed, then $p_ i(Z) \subset X_ i$ is irreducible and therefore has a unique generic point $\xi _ i$ (because $X_ i$ is a finite sober topological space). Then $\xi = \mathop{\mathrm{lim}}\nolimits \xi _ i$ is a generic point of $Z$ (it is a point of $Z$ as $Z$ is closed). This finishes the proof. $\square$

Proof. Write $W = \{ 0, 1\} $ where $0$ is a specialization of $1$ but not vice versa. Let $I$ be a set. The space $\prod _{i \in I} W$ is spectral by Lemma 5.23.12. Thus we see that a subspace of $\prod _{i \in I} W$ closed in the constructible topology is a spectral space by Lemma 5.23.5.

For the converse, let $X$ be a spectral space. Let $U \subset X$ be a quasi-compact open. Consider the continuous map

which maps every point in $U$ to $1$ and every point in $X \setminus U$ to $0$. Taking the product of these maps we obtain a continuous map

By construction the map $f : X \to Y$ is spectral. By Lemma 5.23.3 the image of $f$ is closed in the constructible topology. If $x', x \in X$ are distinct, then since $X$ is sober either $x'$ is not a specialization of $x$ or conversely. In either case (as the quasi-compact opens form a basis for the topology of $X$) there exists a quasi-compact open $U \subset X$ such that $f_ U(x') \not= f_ U(x)$. Thus $f$ is injective. Let $Y = f(X)$ endowed with the induced topology. Let $y' \leadsto y$ be a specialization in $Y$ and say $f(x') = y'$ and $f(x) = y$. Arguing as above we see that $x' \leadsto x$, since otherwise there is a $U$ such that $x \in U$ and $x' \not\in U$, which would imply $f_ U(x') \not\leadsto f_ U(x)$. We conclude that $f : X \to Y$ is a homeomorphism by Lemma 5.23.11. $\square$

Proof. One direction is given by Lemma 5.23.12. For the converse, assume $X$ is spectral. Then we may assume $X \subset \prod _{i \in I} W$ is a subset closed in the constructible topology where $W = \{ 0, 1\} $ as in Lemma 5.23.13. We can write

as a cofiltered limit. For each $J$, let $X_ J \subset \prod _{j \in J} W$ be the image of $X$. Then we see that $X = \mathop{\mathrm{lim}}\nolimits X_ J$ as sets because $X$ is closed in the product with the constructible topology (detail omitted). A formal argument (omitted) on limits shows that $X = \mathop{\mathrm{lim}}\nolimits X_ J$ as topological spaces. $\square$

Proof. Let $U \subset X$ be open and let $U' \subset X'$ be the corresponding open, i.e., the open such that $c^{-1}(U') = U$. Then $U$ is quasi-compact if and only if $U'$ is quasi-compact, as pulling back by $c$ is a bijection between the opens of $X$ and $X'$ which commutes with unions. This in particular proves (1).

Proof of (2). It follows from the above that $X'$ has a basis of quasi-compact opens. Since $c^{-1}$ also commutes with intersections of pairs of opens, we see that the intersection of two quasi-compact opens $X'$ is quasi-compact. Finally, $X'$ is quasi-compact by (1) and sober by construction. Hence $X'$ is spectral.

Proof of (3). It is immediate that $X'$ is Noetherian as this is defined in terms of the acc for open subsets which holds for $X$. We have already seen in (2) that $X'$ is spectral. $\square$