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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