TITLE: Why are constructible sets a disjoint unions of locally closed sets
QUESTION [11 upvotes]: 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.

REPLY [2 votes]: 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 open and $C^c$ is closed.