Lemma 5.15.12. Let X be a topological space. Assume X has a basis consisting of quasi-compact opens. Let E be constructible in X and F \subset E constructible in E. Then F is constructible in X.
Proof. Observe that any retrocompact subset T of X has a basis for the induced topology consisting of quasi-compact opens. In particular this holds for any constructible subset (Lemma 5.15.10). Write E = E_1 \cup \ldots \cup E_ n with E_ i = U_ i \cap V_ i^ c where U_ i, V_ i \subset X are retrocompact open. Note that E_ i = E \cap E_ i is constructible in E by Lemma 5.15.11. Hence F \cap E_ i is constructible in E_ i by Lemma 5.15.11. Thus it suffices to prove the lemma in case E = U \cap V^ c where U, V \subset X are retrocompact open. In this case the inclusion E \subset X is a composition
E = U \cap V^ c \to U \to X
Then we can apply Lemma 5.15.9 to the first inclusion and Lemma 5.15.5 to the second. \square
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