Lemma 61.2.1. Let $X$ be a spectral space. Let $X_0 \subset X$ be the set of closed points. The following are equivalent

Every open covering of $X$ can be refined by a finite disjoint union decomposition $X = \coprod U_ i$ with $U_ i$ open and closed in $X$.

The composition $X_0 \to X \to \pi _0(X)$ is bijective.

Moreover, if $X_0$ is closed in $X$ and every point of $X$ specializes to a unique point of $X_0$, then these conditions are satisfied.

**Proof.**
We will use without further mention that $X_0$ is quasi-compact (Topology, Lemma 5.12.9) and $\pi _0(X)$ is profinite (Topology, Lemma 5.23.9). Picture

\[ \xymatrix{ X_0 \ar[rd]_ f \ar[r] & X \ar[d]^\pi \\ & \pi _0(X) } \]

If (2) holds, the continuous bijective map $f : X_0 \to \pi _0(X)$ is a homeomorphism by Topology, Lemma 5.17.8. Given an open covering $X = \bigcup U_ i$, we get an open covering $\pi _0(X) = \bigcup f(X_0 \cap U_ i)$. By Topology, Lemma 5.22.4 we can find a finite open covering of the form $\pi _0(X) = \coprod V_ j$ which refines this covering. Since $X_0 \to \pi _0(X)$ is bijective each connected component of $X$ has a unique closed point, whence is equal to the set of points specializing to this closed point. Hence $\pi ^{-1}(V_ j)$ is the set of points specializing to the points of $f^{-1}(V_ j)$. Now, if $f^{-1}(V_ j) \subset X_0 \cap U_ i \subset U_ i$, then it follows that $\pi ^{-1}(V_ j) \subset U_ i$ (because the open set $U_ i$ is closed under generalizations). In this way we see that the open covering $X = \coprod \pi ^{-1}(V_ j)$ refines the covering we started out with. In this way we see that (2) implies (1).

Assume (1). Let $x, y \in X$ be closed points. Then we have the open covering $X = (X \setminus \{ x\} ) \cup (X \setminus \{ y\} )$. It follows from (1) that there exists a disjoint union decomposition $X = U \amalg V$ with $U$ and $V$ open (and closed) and $x \in U$ and $y \in V$. In particular we see that every connected component of $X$ has at most one closed point. By Topology, Lemma 5.12.8 every connected component (being closed) also does have a closed point. Thus $X_0 \to \pi _0(X)$ is bijective. In this way we see that (1) implies (2).

Assume $X_0$ is closed in $X$ and every point specializes to a unique point of $X_0$. Then $X_0$ is a spectral space (Topology, Lemma 5.23.5) consisting of closed points, hence profinite (Topology, Lemma 5.23.8). Let $x, y \in X_0$ be distinct. By Topology, Lemma 5.22.4 we can find a disjoint union decomposition $X_0 = U_0 \amalg V_0$ with $U_0$ and $V_0$ open and closed and $x \in U_0$ and $y \in V_0$. Let $U \subset X$, resp. $V \subset X$ be the set of points specializing to $U_0$, resp. $V_0$. Observe that $X = U \amalg V$. By Topology, Lemma 5.24.7 we see that $U$ is an intersection of quasi-compact open subsets. Hence $U$ is closed in the constructible topology. Since $U$ is closed under specialization, we see that $U$ is closed by Topology, Lemma 5.23.6. By symmetry $V$ is closed and hence $U$ and $V$ are both open and closed. This proves that $x, y$ are not in the same connected component of $X$. In other words, $X_0 \to \pi _0(X)$ is injective. The map is also surjective by Topology, Lemma 5.12.8 and the fact that connected components are closed. In this way we see that the final condition implies (2).
$\square$

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