Lemma 5.23.6. Let $X$ be a spectral space. Let $E \subset X$ be a subset closed in the constructible topology (for example constructible).

1. If $x \in \overline{E}$, then $x$ is the specialization of a point of $E$.

2. If $E$ is stable under specialization, then $E$ is closed.

3. If $E' \subset X$ is open in the constructible topology (for example constructible) and stable under generalization, then $E'$ is open.

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. $\square$

## Comments (2)

Comment #5060 by Laurent Moret-Bailly on

Suggested slogan: The spectral topology is determined by the constructible topology and the specialization relation.

In the same vein, here is a statement which contains Lemma 23.3(1) and improves on Lemma 23.10: Let $f:X\to Y$ be a map of spectral spaces. Then $f$ is spectral iff $f$ is constructibly continuous and takes secializations to specializations.

There are also:

• 4 comment(s) on Section 5.23: Spectral spaces

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