Lemma 5.23.15. Let $X$ be a topological space and let $c : X \to X'$ be the universal map from $X$ to a sober topological space, see Lemma 5.8.16.
If $X$ is quasi-compact, so is $X'$.
If $X$ is quasi-compact, has a basis of quasi-compact opens, and the intersection of two quasi-compact opens is quasi-compact, then $X'$ is spectral.
If $X$ is Noetherian, then $X'$ is a Noetherian spectral space.
Proof. Let $U \subset X$ be open and let $U' \subset X'$ be the corresponding open, i.e., the open such that $c^{-1}(U') = U$. Then $U$ is quasi-compact if and only if $U'$ is quasi-compact, as pulling back by $c$ is a bijection between the opens of $X$ and $X'$ which commutes with unions. This in particular proves (1).
Proof of (2). It follows from the above that $X'$ has a basis of quasi-compact opens. Since $c^{-1}$ also commutes with intersections of pairs of opens, we see that the intersection of two quasi-compact opens $X'$ is quasi-compact. Finally, $X'$ is quasi-compact by (1) and sober by construction. Hence $X'$ is spectral.
Proof of (3). It is immediate that $X'$ is Noetherian as this is defined in terms of the acc for open subsets which holds for $X$. We have already seen in (2) that $X'$ is spectral. $\square$
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