The Stacks project

Lemma 5.23.5. 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 $X$ 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 $x$. 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 (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0903. Beware of the difference between the letter 'O' and the digit '0'.