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Let $X$ be a noetherian scheme. The constructible sets are the smallest boolean algebra containing all of the open sets. It is easy to see that the constructible sets are exactly finite unions of locally closed sets. I have read several times that every constructible set is a finte disjoint union of locally closed sets. In all the examples I have seen this is indeed the case, but when I try to write down a proof of this fact I get stuck.

Does anyone know why this is true? I suspect that this is a standard trick which I have not seen.

DBr
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  • How do you show that contructible sets are finite union of locally closed sets? I don’t find this easy. – user5826 Sep 29 '21 at 16:18

2 Answers2

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Assume that $X$ is a noetherian scheme. Let $C \subseteq X $ be a constructible set. Then we can write $$ C = (U_1 \cap K_1) \cup (U_2 \cap K_2) \cup \dots \cup (U_s \cap K_s) $$ where each $ U_i $ is open and each $ K_i $ is closed. Assume that $ U_i \cap K_i \cap U_j \cap K_j \not= \emptyset $. Since we have the equation $$ U_j \cap K_j \backslash U_i \cap K_i = [U_j \cap (K_j \backslash U_i)] \cup [(U_j \backslash K_i) \cap K_j] $$ we can refine our decomposition for $C$ by adding in a new term with closed part $ K_j \backslash U_i \subset K_j$ (this is a strict inclusion). Since $X$ is noetherian, the infinite pigeon hole principle tells us that this process cannot go on forever. Therefore we must reach a point when all of the locally closed sets are disjoint.

user5826
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DBr
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  • What if $U_j \cap (K_j \setminus U_i)$ already appears in the decomposition of $C$? Then we must remove it and we don't get a smaller closed set... (We must remove it or else we can't claim the difference between two patches of the decomposition is nonempty) – Emolga Oct 27 '16 at 21:06
  • Moreover, what stops us from always refining locally-closed sets with the same closed part $K_j$? – Emolga Oct 27 '16 at 22:15
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Actually, the proof does not require the noetherian property. Here is a proof which I found here:

It suffices to prove that the family of sets which are finite disjoint unions of locally closed sets is a boolean algebra. It is obviously closed under finite intersections. Now the complement of such a set is a finite intersection of complements of locally closed sets, so it remains to show that the complement of a single locally closed set is a disjoint union of locally closed sets. Suppose $O\cap C$ is a locally closed set where $O$ is open, $C$ is closed. Then the claim follows immediately from $(O\cap C)^c=O^c\cup C^c=(O^c\cap C)\sqcup C^c$, since $O^c\cap C$ is closed and $C^c$ is open.

Potitov06
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Yuxiao Xie
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