The Stacks project

Proof. The limit $Z = \mathop{\mathrm{lim}}\nolimits Z_ i$ exists by Lemma 5.14.1. Denote $X'_ i$ the space $X_ i$ endowed with the constructible topology and $Z'_ i$ the corresponding subspace of $X'_ i$. Let $a : j \to i$ in $\mathcal{I}$ be a morphism. As $f_ a$ is spectral it defines a continuous map $f_ a : X'_ j \to X'_ i$. Thus $f_ a|_{Z_ j} : Z'_ j \to Z'_ i$ is a continuous map of quasi-compact Hausdorff spaces (by Lemmas 5.23.2 and 5.12.3). Thus $Z' = \mathop{\mathrm{lim}}\nolimits Z_ i$ is quasi-compact by Lemma 5.14.5. The maps $Z'_ i \to Z_ i$ are continuous, hence $Z' \to Z$ is continuous and a bijection on underlying sets. Hence $Z$ is quasi-compact as the image of the surjective continuous map $Z' \to Z$ (Lemma 5.12.7). $\square$

Proof. Denote $X'_ i$ the space $X_ i$ endowed with the constructible topology and $Z'_ i$ the corresponding subspace of $X'_ i$. Let $a : j \to i$ in $\mathcal{I}$ be a morphism. As $f_ a$ is spectral it defines a continuous map $f_ a : X'_ j \to X'_ i$. Thus $f_ a|_{Z_ j} : Z'_ j \to Z'_ i$ is a continuous map of quasi-compact Hausdorff spaces (by Lemmas 5.23.2 and 5.12.3). By Lemma 5.14.6 the space $\mathop{\mathrm{lim}}\nolimits Z'_ i$ is nonempty. Since $\mathop{\mathrm{lim}}\nolimits Z'_ i = \mathop{\mathrm{lim}}\nolimits Z_ i$ as sets we conclude. $\square$

Proof. Observe that

Since $f_ a$ is a spectral map, it is continuous in the constructible topology hence the set $f_ a^{-1}(E) \setminus f_ a^{-1}(F)$ is closed in the constructible topology. Hence Lemma 5.24.2 applies to show that the LHS is nonempty if and only if each of the spaces of the RHS is nonempty. $\square$

Proof. Assume $E$ is a quasi-compact open of $X$. By Lemma 5.14.2 we can write $E = p_ i^{-1}(U_ i)$ for some $i$ and some open $U_ i \subset X_ i$. Write $U_ i = \bigcup U_{i, \alpha }$ as a union of quasi-compact opens. As $E$ is quasi-compact we can find $\alpha _1, \ldots , \alpha _ n$ such that $E = p_ i^{-1}(U_{i, \alpha _1} \cup \ldots \cup U_{i, \alpha _ n})$. Hence $E_ i = U_{i, \alpha _1} \cup \ldots \cup U_{i, \alpha _ n}$ works.

Assume $E$ is a constructible closed subset. Then $E^ c$ is quasi-compact open. So $E^ c = p_ i^{-1}(F_ i)$ for some $i$ and quasi-compact open $F_ i \subset X_ i$ by the result of the previous paragraph. Then $E = p_ i^{-1}(F_ i^ c)$ as desired.

If $E$ is general we can write $E = \bigcup _{l = 1, \ldots , n} U_ l \cap Z_ l$ with $U_ l$ constructible open and $Z_ l$ constructible closed. By the result of the previous paragraphs we may write $U_ l = p_{i_ l}^{-1}(U_{l, i_ l})$ and $Z_ l = p_{j_ l}^{-1}(Z_{l, j_ l})$ with $U_{l, i_ l} \subset X_{i_ l}$ constructible open and $Z_{l, j_ l} \subset X_{j_ l}$ constructible closed. As $\mathcal{I}$ is cofiltered we may choose an object $k$ of $\mathcal{I}$ and morphism $a_ l : k \to i_ l$ and $b_ l : k \to j_ l$. Then taking $E_ k = \bigcup _{l = 1, \ldots , n} f_{a_ l}^{-1}(U_{l, i_ l}) \cap f_{b_ l}^{-1}(Z_{l, j_ l})$ we obtain a constructible subset of $X_ k$ whose inverse image in $X$ is $E$. $\square$

Proof. The limit $X = \mathop{\mathrm{lim}}\nolimits X_ i$ exists (Lemma 5.14.1) and is quasi-compact by Lemma 5.24.1.

Denote $p_ i : X \to X_ i$ the projection. Because $\mathcal{I}$ is cofiltered we can apply Lemma 5.14.2. Hence a basis for the topology on $X$ is given by the opens $p_ i^{-1}(U_ i)$ for $U_ i \subset X_ i$ open. Since a basis for the topology of $X_ i$ is given by the quasi-compact open, we conclude that a basis for the topology on $X$ is given by $p_ i^{-1}(U_ i)$ with $U_ i \subset X_ i$ quasi-compact open. A formal argument shows that

as topological spaces. Since each $f_ a$ is spectral the sets $f_ a^{-1}(U_ i)$ are closed in the constructible topology of $X_ j$ and hence $p_ i^{-1}(U_ i)$ is quasi-compact by Lemma 5.24.1. Thus $X$ has a basis for the topology consisting of quasi-compact opens.

Any quasi-compact open $U$ of $X$ is of the form $U = p_ i^{-1}(U_ i)$ for some $i$ and some quasi-compact open $U_ i \subset X_ i$ (see Lemma 5.24.4). Given $U_ i \subset X_ i$ and $U_ j \subset X_ j$ quasi-compact open, then $p_ i^{-1}(U_ i) \cap p_ j^{-1}(U_ j) = p_ k^{-1}(U_ k)$ for some $k$ and quasi-compact open $U_ k \subset X_ k$. Namely, choose $k$ and morphisms $k \to i$ and $k \to j$ and let $U_ k$ be the intersection of the pullbacks of $U_ i$ and $U_ j$ to $X_ k$. Thus we see that the intersection of two quasi-compact opens of $X$ is quasi-compact open.

Finally, let $Z \subset X$ be irreducible and closed. Then $p_ i(Z) \subset X_ i$ is irreducible and therefore $Z_ i = \overline{p_ i(Z)}$ has a unique generic point $\xi _ i$ (because $X_ i$ is a spectral space). Then $f_ a(\xi _ j) = \xi _ i$ for $a : j \to i$ in $\mathcal{I}$ because $\overline{f_ a(Z_ j)} = Z_ i$. Hence $\xi = \mathop{\mathrm{lim}}\nolimits \xi _ i$ is a point of $X$. Claim: $\xi \in Z$. Namely, if not we can find a quasi-compact open containing $\xi $ disjoint from $Z$. This would be of the form $p_ i^{-1}(U_ i)$ for some $i$ and quasi-compact open $U_ i \subset X_ i$. Then $\xi _ i \in U_ i$ but $p_ i(Z) \cap U_ i = \emptyset $ which contradicts $\xi _ i \in \overline{p_ i(Z)}$. So $\xi \in Z$ and hence $\overline{\{ \xi \} } \subset Z$. Conversely, every $z \in Z$ is in the closure of $\xi $. Namely, given a quasi-compact open neighbourhood $U$ of $z$ we write $U = p_ i^{-1}(U_ i)$ for some $i$ and quasi-compact open $U_ i \subset X_ i$. We see that $p_ i(z) \in U_ i$ hence $\xi _ i \in U_ i$ hence $\xi \in U$. Thus $\xi $ is a generic point of $Z$. We omit the proof that $\xi $ is the unique generic point of $Z$ (hint: show that a second generic point has to be equal to $\xi $ by showing that it has to map to $\xi _ i$ in $X_ i$ since by spectrality of $X_ i$ the irreducible $Z_ i$ has a unique generic point). This finishes the proof. $\square$

Proof. Part (1) is a special case of Lemma 5.24.4. Part (2) is a special case of Lemma 5.24.3 as quasi-compact opens are both open and closed in the constructible topology. Parts (3) and (4) follow formally from (1) and (2) and the fact that taking inverse images of subsets commutes with taking unions and intersections. $\square$

Proof. Let $W \subset X$ satisfy (1). Then $W$ is closed in the constructible topology, hence quasi-compact in the constructible topology (by Lemmas 5.23.2 and 5.12.3), hence quasi-compact in the topology of $X$ (because opens in $X$ are open in the constructible topology). Thus (2) holds.

It is clear that (2) implies (3) by taking $E = W$.

Let $X$ be a spectral space and let $E \subset W$ be as in (3). Since every point of $W$ specializes to a point of $E$ we see that an open of $W$ which contains $E$ is equal to $W$. Hence since $E$ is quasi-compact, so is $W$. If $x \in X$, $x \not\in W$, then $Z = \overline{\{ x\} }$ is disjoint from $W$. Since $W$ is quasi-compact we can find a quasi-compact open $U$ with $W \subset U$ and $U \cap Z = \emptyset $. We conclude that (4) holds.

If $W = \bigcap _{j \in J} U_ j$ then setting $I$ equal to the set of finite subsets of $J$ and $U_ i = U_{j_1} \cap \ldots \cap U_{j_ r}$ for $i = \{ j_1, \ldots , j_ r\} $ shows that (4) implies (5). It is immediate that (5) implies (1).

Let $I$ and $U_ i$ be as in (5). Since $W = \bigcap U_ i$ we have $W = \mathop{\mathrm{lim}}\nolimits U_ i$ by the universal property of limits. Then $W$ is a spectral space by Lemma 5.24.5. Let $U \subset X$ be an open neighbourhood of $W$. Then $E_ i = U_ i \cap (X \setminus U)$ is a family of constructible subsets of the spectral space $Z = X \setminus U$ with empty intersection. Using that the spectral topology on $Z$ is quasi-compact (Lemma 5.23.2) we conclude from Lemma 5.12.6 that $E_ i = \emptyset $ for some $i$. $\square$

Proof. Since $E$ is constructible, it is quasi-compact and hence Lemma 5.24.7 applies to $W$. If $E$ is constructible, then $E$ is constructible in $U_ i$ for all $i \in I$. Hence $U_ i \setminus E$ is spectral by Lemma 5.23.5. Since $W \setminus E = \bigcap (U_ i \setminus E)$ we have $W \setminus E = \mathop{\mathrm{lim}}\nolimits U_ i \setminus E$ by the universal property of limits. Then $W \setminus E$ is a spectral space by Lemma 5.24.5. $\square$