Lemma 5.23.14. A topological space is spectral if and only if it is a directed inverse limit of finite sober topological spaces.

Proof. One direction is given by Lemma 5.23.12. For the converse, assume $X$ is spectral. Then we may assume $X \subset \prod _{i \in I} W$ is a subset closed in the constructible topology where $W = \{ 0, 1\}$ as in Lemma 5.23.13. We can write

$\prod \nolimits _{i \in I} W = \mathop{\mathrm{lim}}\nolimits _{J \subset I\text{ finite }} \prod \nolimits _{j \in J} W$

as a cofiltered limit. For each $J$, let $X_ J \subset \prod _{j \in J} W$ be the image of $X$. Then we see that $X = \mathop{\mathrm{lim}}\nolimits X_ J$ as sets because $X$ is closed in the product with the constructible topology (detail omitted). A formal argument (omitted) on limits shows that $X = \mathop{\mathrm{lim}}\nolimits X_ J$ as topological spaces. $\square$

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