Lemma 59.82.3. Let $Z \subset X$ be a closed subset of a topological space $X$. Assume

$X$ is a spectral space (Topology, Definition 5.23.1), and

for $x \in X$ the intersection $Z \cap \overline{\{ x\} }$ is connected (in particular nonempty).

If $Z = Z_1 \amalg Z_2$ with $Z_ i$ closed in $Z$, then there exists a decomposition $X = X_1 \amalg X_2$ with $X_ i$ closed in $X$ and $Z_ i = Z \cap X_ i$.

**Proof.**
Observe that $Z_ i$ is quasi-compact. Hence the set of points $W_ i$ specializing to $Z_ i$ is closed in the constructible topology by Topology, Lemma 5.24.7. Assumption (2) implies that $X = W_1 \amalg W_2$. Let $x \in \overline{W_1}$. By Topology, Lemma 5.23.6 part (1) there exists a specialization $x_1 \leadsto x$ with $x_1 \in W_1$. Thus $\overline{\{ x\} } \subset \overline{\{ x_1\} }$ and we see that $x \in W_1$. In other words, setting $X_ i = W_ i$ does the job.
$\square$

## Comments (2)

Comment #3640 by Brian Conrad on

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