Lemma 5.18.6 (07JU)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 5: Topology
Section 5.18: Jacobson spaces
Lemma 5.18.6 (cite)

Lemma 5.18.6. A finite Jacobson space is discrete.

Proof. If $X$ is finite Jacobson, $X_0 \subset X$ the subset of closed points, then, on the one hand, $\overline{X_0} = X$. On the other hand, $X$, and hence $X_0$ is finite, so $X_0 =\{ x_1, \ldots , x_ n\} = \bigcup _{i = 1, \ldots , n} \{ x_ i\} $ is a finite union of closed sets, hence closed, so $X = \overline{X_0} = X_0$. Every point is closed, and by finiteness, every point is open. $\square$

Comments (2)

Comment #778 by Keenan Kidwell on June 30, 2014 at 23:37

Is there something wrong with the following argument that a finite Jacobson space discrete without any separation axioms? If $X$ is finite Jacobson, $X^0\subseteq X$ the subset of closed points, then, on the one hand, $\overline{X^0}=X$ . On the other hand, $X$ , and hence $X^0$ is finite, so $X^0=\{x_1,\ldots,x_n\}=\bigcup_{i=1}^n\{x_i\}$ is a finite union of closed sets, hence closed, so $X=\overline{X^0}=X^0$ . Every point is closed, and by finiteness, every point is open.

Comment #795 by Johan on July 09, 2014 at 12:48

Wonderful! Thanks for this and your other comments which I fixed in this commit.

There are also:

8 comment(s) on Section 5.18: Jacobson spaces

Comments (2)

Post a comment