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Lemma 5.18.6. A finite Jacobson space is discrete. A Jacobson space with finitely many closed points is discrete.

Proof. If $X$ is finite then the set $X_0 \subset X$ of closed points is finite. Assume $X_0$ is is finite and $X$ is Jacobson. Then $\overline{X_0} = X$ by the Jacobson property. Now $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

Is there something wrong with the following argument that a finite Jacobson space discrete without any separation axioms? If is finite Jacobson, the subset of closed points, then, on the one hand, . On the other hand, , and hence is finite, so is a finite union of closed sets, hence closed, so . Every point is closed, and by finiteness, every point is open.

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  • 8 comment(s) on Section 5.18: Jacobson spaces

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