## 5.24 Limits of spectral spaces

Lemma 5.23.14 tells us that every spectral space is a cofiltered limit of finite sober spaces. Every finite sober space is a spectral space and every continuous map of finite sober spaces is a spectral map of spectral spaces. In this section we prove some lemmas concerning limits of systems of spectral topological spaces along spectral maps.

Lemma 5.24.1. Let $\mathcal{I}$ be a category. Let $i \mapsto X_ i$ be a diagram of spectral spaces such that for $a : j \to i$ in $\mathcal{I}$ the corresponding map $f_ a : X_ j \to X_ i$ is spectral.

1. Given subsets $Z_ i \subset X_ i$ closed in the constructible topology with $f_ a(Z_ j) \subset Z_ i$ for all $a : j \to i$ in $\mathcal{I}$, then $\mathop{\mathrm{lim}}\nolimits Z_ i$ is quasi-compact.

2. The space $X = \mathop{\mathrm{lim}}\nolimits X_ i$ is quasi-compact.

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). $\square$

Lemma 5.24.2. Let $\mathcal{I}$ be a cofiltered category. Let $i \mapsto X_ i$ be a diagram of spectral spaces such that for $a : j \to i$ in $\mathcal{I}$ the corresponding map $f_ a : X_ j \to X_ i$ is spectral.

1. Given nonempty subsets $Z_ i \subset X_ i$ closed in the constructible topology with $f_ a(Z_ j) \subset Z_ i$ for all $a : j \to i$ in $\mathcal{I}$, then $\mathop{\mathrm{lim}}\nolimits Z_ i$ is nonempty.

2. If each $X_ i$ is nonempty, then $X = \mathop{\mathrm{lim}}\nolimits X_ i$ is nonempty.

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. $\square$

Lemma 5.24.3. Let $\mathcal{I}$ be a cofiltered category. Let $i \mapsto X_ i$ be a diagram of spectral spaces such that for $a : j \to i$ in $\mathcal{I}$ the corresponding map $f_ a : X_ j \to X_ i$ is spectral. Let $X = \mathop{\mathrm{lim}}\nolimits X_ i$ with projections $p_ i : X \to X_ i$. Let $i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ and let $E, F \subset X_ i$ be subsets with $E$ closed in the constructible topology and $F$ open in the constructible topology. Then $p_ i^{-1}(E) \subset p_ i^{-1}(F)$ if and only if there is a morphism $a : j \to i$ in $\mathcal{I}$ such that $f_ a^{-1}(E) \subset f_ a^{-1}(F)$.

Proof. Observe that

$p_ i^{-1}(E) \setminus p_ i^{-1}(F) = \mathop{\mathrm{lim}}\nolimits _{a : j \to i} f_ a^{-1}(E) \setminus f_ a^{-1}(F)$

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. $\square$

Lemma 5.24.4. Let $\mathcal{I}$ be a cofiltered category. Let $i \mapsto X_ i$ be a diagram of spectral spaces such that for $a : j \to i$ in $\mathcal{I}$ the corresponding map $f_ a : X_ j \to X_ i$ is spectral. Let $X = \mathop{\mathrm{lim}}\nolimits X_ i$ with projections $p_ i : X \to X_ i$. Let $E \subset X$ be a constructible subset. Then there exists an $i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ and a constructible subset $E_ i \subset X_ i$ such that $p_ i^{-1}(E_ i) = E$. If $E$ is open, resp. closed, we may choose $E_ i$ open, resp. closed.

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 $E$. $\square$

Lemma 5.24.5. Let $\mathcal{I}$ be a cofiltered index category. Let $i \mapsto X_ i$ be a diagram of spectral spaces such that for $a : j \to i$ in $\mathcal{I}$ the corresponding map $f_ a : X_ j \to X_ i$ is spectral. Then the inverse limit $X = \mathop{\mathrm{lim}}\nolimits X_ i$ is a spectral topological space and the projection maps $p_ i : X \to X_ i$ are spectral.

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

$p_ i^{-1}(U_ i) = \mathop{\mathrm{lim}}\nolimits _{a : j \to i} f_ a^{-1}(U_ i)$

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. $\square$

Lemma 5.24.6. Let $\mathcal{I}$ be a cofiltered index category. Let $i \mapsto X_ i$ be a diagram of spectral spaces such that for $a : j \to i$ in $\mathcal{I}$ the corresponding map $f_ a : X_ j \to X_ i$ is spectral. Set $X = \mathop{\mathrm{lim}}\nolimits X_ i$ and denote $p_ i : X \to X_ i$ the projection.

1. Given any quasi-compact open $U \subset X$ there exists an $i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ and a quasi-compact open $U_ i \subset X_ i$ such that $p_ i^{-1}(U_ i) = U$.

2. Given $U_ i \subset X_ i$ and $U_ j \subset X_ j$ quasi-compact opens such that $p_ i^{-1}(U_ i) \subset p_ j^{-1}(U_ j)$ there exist $k \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ and morphisms $a : k \to i$ and $b : k \to j$ such that $f_ a^{-1}(U_ i) \subset f_ b^{-1}(U_ j)$.

3. If $U_ i, U_{1, i}, \ldots , U_{n, i} \subset X_ i$ are quasi-compact opens and $p_ i^{-1}(U_ i) = p_ i^{-1}(U_{1, i}) \cup \ldots \cup p_ i^{-1}(U_{n, i})$ then $f_ a^{-1}(U_ i) = f_ a^{-1}(U_{1, i}) \cup \ldots \cup f_ a^{-1}(U_{n, i})$ for some morphism $a : j \to i$ in $\mathcal{I}$.

4. Same statement as in (3) but for intersections.

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. $\square$

Lemma 5.24.7. Let $W$ be a subset of a spectral space $X$. The following are equivalent:

1. $W$ is an intersection of constructible sets and closed under generalizations,

2. $W$ is quasi-compact and closed under generalizations,

3. there exists a quasi-compact subset $E \subset X$ such that $W$ is the set of points specializing to $E$,

4. $W$ is an intersection of quasi-compact open subsets,

5. there exists a nonempty set $I$ and quasi-compact opens $U_ i \subset X$, $i \in I$ such that $W = \bigcap U_ i$ and for all $i, j \in I$ there exists a $k \in I$ with $U_ k \subset U_ i \cap U_ j$.

In this case we have (a) $W$ is a spectral space, (b) $W = \mathop{\mathrm{lim}}\nolimits U_ i$ as topological spaces, and (c) for any open $U$ containing $W$ there exists an $i$ with $U_ i \subset U$.

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 $i$. $\square$

Lemma 5.24.8. Let $X$ be a spectral space. Let $E \subset X$ be a constructible subset. Let $W \subset X$ be the set of points of $X$ which specialize to a point of $E$. Then $W \setminus E$ is a spectral space. If $W = \bigcap U_ i$ with $U_ i$ as in Lemma 5.24.7 (5) then $W \setminus E = \mathop{\mathrm{lim}}\nolimits (U_ i \setminus E)$.

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. $\square$

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