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. $\square$

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 $X$ 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.

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