Lemma 28.2.1. Let $X$ be a scheme. A subset $E$ of $X$ is locally constructible in $X$ if and only if $E \cap U$ is constructible in $U$ for every affine open $U$ of $X$.
28.2 Constructible sets
Constructible and locally constructible sets are introduced in Topology, Section 5.15. We may characterize locally constructible subsets of schemes as follows.
Proof. Assume $E$ is locally constructible. Then there exists an open covering $X = \bigcup U_ i$ such that $E \cap U_ i$ is constructible in $U_ i$ for each $i$. Let $V \subset X$ be any affine open. We can find a finite open affine covering $V = V_1 \cup \ldots \cup V_ m$ such that for each $j$ we have $V_ j \subset U_ i$ for some $i = i(j)$. By Topology, Lemma 5.15.4 we see that each $E \cap V_ j$ is constructible in $V_ j$. Since the inclusions $V_ j \to V$ are quasi-compact (see Schemes, Lemma 26.19.2) we conclude that $E \cap V$ is constructible in $V$ by Topology, Lemma 5.15.6. The converse implication is immediate. $\square$
Lemma 28.2.2. Let $X$ be a scheme and let $E \subset X$ be a locally constructible subset. Let $\xi \in X$ be a generic point of an irreducible component of $X$.
If $\xi \in E$, then an open neighbourhood of $\xi $ is contained in $E$.
If $\xi \not\in E$, then an open neighbourhood of $\xi $ is disjoint from $E$.
Proof. As the complement of a locally constructible subset is locally constructible it suffices to show (2). We may assume $X$ is affine and hence $E$ constructible (Lemma 28.2.1). In this case $X$ is a spectral space (Algebra, Lemma 10.26.2). Then $\xi \not\in E$ implies $\xi \not\in \overline{E}$ by Topology, Lemma 5.23.6 and the fact that there are no points of $X$ different from $\xi $ which specialize to $\xi $. $\square$
Lemma 28.2.3. Let $X$ be a quasi-separated scheme. The intersection of any two quasi-compact opens of $X$ is a quasi-compact open of $X$. Every quasi-compact open of $X$ is retrocompact in $X$.
Proof. If $U$ and $V$ are quasi-compact open then $U \cap V = \Delta ^{-1}(U \times V)$, where $\Delta : X \to X \times X$ is the diagonal. As $X$ is quasi-separated we see that $\Delta $ is quasi-compact. Hence we see that $U \cap V$ is quasi-compact as $U \times V$ is quasi-compact (details omitted; use Schemes, Lemma 26.17.4 to see $U \times V$ is a finite union of affines). The other assertions follow from the first and Topology, Lemma 5.27.1. $\square$
Lemma 28.2.4. Let $X$ be a quasi-compact and quasi-separated scheme. Then the underlying topological space of $X$ is a spectral space.
Proof. By Topology, Definition 5.23.1 we have to check that $X$ is sober, quasi-compact, has a basis of quasi-compact opens, and the intersection of any two quasi-compact opens is quasi-compact. This follows from Schemes, Lemma 26.11.1 and 26.11.2 and Lemma 28.2.3 above. $\square$
Lemma 28.2.5. Let $X$ be a quasi-compact and quasi-separated scheme. Any locally constructible subset of $X$ is constructible.
Proof. As $X$ is quasi-compact we can choose a finite affine open covering $X = V_1 \cup \ldots \cup V_ m$. As $X$ is quasi-separated each $V_ i$ is retrocompact in $X$ by Lemma 28.2.3. Hence by Topology, Lemma 5.15.6 we see that $E \subset X$ is constructible in $X$ if and only if $E \cap V_ j$ is constructible in $V_ j$. Thus we win by Lemma 28.2.1. $\square$
Lemma 28.2.6. Let $X$ be a scheme. A subset $E$ of $X$ is retrocompact in $X$ if and only if $E \cap U$ is quasi-compact for every affine open $U$ of $X$.
Proof. Immediate from the fact that every quasi-compact open of $X$ is a finite union of affine opens. $\square$
Lemma 28.2.7. A partition $X = \coprod _{i \in I} X_ i$ of a scheme $X$ with retrocompact parts is locally finite if and only if the parts are locally constructible.
Proof. See Topology, Definitions 5.12.1, 5.28.1, and 5.28.4 for the definitions of retrocompact, partition, and locally finite.
If the partition is locally finite and $U \subset X$ is an affine open, then we see that $U = \coprod _{i \in I} U \cap X_ i$ is a finite partition (more precisely, all but a finite number of its parts are empty). Hence $U \cap X_ i$ is quasi-compact and its complement is retrocompact in $U$ as a finite union of retrocompact parts. Thus $U \cap X_ i$ is constructible by Topology, Lemma 5.15.13. It follows that $X_ i$ is locally constructible by Lemma 28.2.1.
Assume the parts are locally constructible. Then for any affine open $U \subset X$ we obtain a covering $U = \coprod X_ i \cap U$ by constructible subsets. Since the constructible topology is quasi-compact, see Topology, Lemma 5.23.2, this covering has a finite refinement, i.e., the partition is locally finite. $\square$
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