Lemma 5.23.13. Let $W$ be the topological space with two points, one closed, the other not. A topological space is spectral if and only if it is homeomorphic to a subspace of a product of copies of $W$ which is closed in the constructible topology.

**Proof.**
Write $W = \{ 0, 1\} $ where $0$ is a specialization of $1$ but not vice versa. Let $I$ be a set. The space $\prod _{i \in I} W$ is spectral by Lemma 5.23.12. Thus we see that a subspace of $\prod _{i \in I} W$ closed in the constructible topology is a spectral space by Lemma 5.23.5.

For the converse, let $X$ be a spectral space. Let $U \subset X$ be a quasi-compact open. Consider the continuous map

which maps every point in $U$ to $1$ and every point in $X \setminus U$ to $0$. Taking the product of these maps we obtain a continuous map

By construction the map $f : X \to Y$ is spectral. By Lemma 5.23.3 the image of $f$ is closed in the constructible topology. If $x', x \in X$ are distinct, then since $X$ is sober either $x'$ is not a specialization of $x$ or conversely. In either case (as the quasi-compact opens form a basis for the topology of $X$) there exists a quasi-compact open $U \subset X$ such that $f_ U(x') \not= f_ U(x)$. Thus $f$ is injective. Let $Y = f(X)$ endowed with the induced topology. Let $y' \leadsto y$ be a specialization in $Y$ and say $f(x') = y'$ and $f(x) = y$. Arguing as above we see that $x' \leadsto x$, since otherwise there is a $U$ such that $x \in U$ and $x' \not\in U$, which would imply $f_ U(x') \not\leadsto f_ U(x)$. We conclude that $f : X \to Y$ is a homeomorphism by Lemma 5.23.11. $\square$

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