## Tag `0A31`

Chapter 5: Topology > Section 5.24: Limits of spectral spaces

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

- $W$ is an intersection of constructible sets and closed under generalizations,
- $W$ is quasi-compact and closed under generalizations,
- there exists a quasi-compact subset $E \subset X$ such that $W$ is the set of points specializing to $E$,
- $W$ is an intersection of quasi-compact open subsets,
- 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) $W = \mathop{\rm lim}\nolimits U_i$ 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 $W = \mathop{\rm lim}\nolimits U_i$ 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 $i$. $\square$

The code snippet corresponding to this tag is a part of the file `topology.tex` and is located in lines 4514–4534 (see updates for more information).

```
\begin{lemma}
\label{lemma-make-spectral-space}
Let $W$ be a subset of a spectral space $X$. The following are equivalent:
\begin{enumerate}
\item $W$ is an intersection of constructible sets and
closed under generalizations,
\item $W$ is quasi-compact and closed under generalizations,
\item there exists a quasi-compact subset $E \subset X$ such that
$W$ is the set of points specializing to $E$,
\item $W$ is an intersection of quasi-compact open subsets,
\item
\label{item-intersection-quasi-compact-open}
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 $U_k \subset U_i \cap U_j$.
\end{enumerate}
In this case we have (a) $W$ is a spectral space, (b) $W = \lim U_i$
as topological spaces, and (c) for any open $U$ containing $W$
there exists an $i$ with $U_i \subset U$.
\end{lemma}
\begin{proof}
Let $W \subset X$ satisfy (1). Then $W$ is closed in the constructible
topology, hence quasi-compact in the constructible topology (by
Lemmas \ref{lemma-constructible-hausdorff-quasi-compact} and
\ref{lemma-closed-in-quasi-compact}), hence quasi-compact in the topology
of $X$ (because opens in $X$ are open in the constructible topology). Thus
(2) holds.
\medskip\noindent
It is clear that (2) implies (3) by taking $E = W$.
\medskip\noindent
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.
\medskip\noindent
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).
\medskip\noindent
Let $I$ and $U_i$ be as in (5).
Since $W = \bigcap U_i$ we have $W = \lim U_i$ by the universal property
of limits. Then $W$ is a spectral space by
Lemma \ref{lemma-directed-inverse-limit-spectral-spaces}.
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 \ref{lemma-constructible-hausdorff-quasi-compact})
we conclude from
Lemma \ref{lemma-intersection-closed-in-quasi-compact}
that $E_i = \emptyset$ for some $i$.
\end{proof}
```

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