Lemma 5.15.13. Let $X$ be a quasi-compact topological space having a basis consisting of quasi-compact opens such that the intersection of any two quasi-compact opens is quasi-compact. Let $T \subset X$ be a locally closed subset such that $T$ is quasi-compact and $T^ c$ is retrocompact in $X$. Then $T$ is constructible in $X$.

Proof. Note that $T$ is quasi-compact and open in $\overline{T}$. Using our basis of quasi-compact opens we can write $T = U \cap \overline{T}$ where $U$ is quasi-compact open in $X$. Then $V = U \setminus T = U \cap T^ c$ is retrocompact in $U$ as $T^ c$ is retrocompact in $X$. Hence $V$ is quasi-compact. Since the intersection of any two quasi-compact opens is quasi-compact any quasi-compact open of $X$ is retrocompact. Thus $T = U \cap V^ c$ with $U$ and $V = U \setminus T$ retrocompact opens of $X$. A fortiori, $T$ is constructible in $X$. $\square$

Comment #6471 by Owen on

I think the proof gives more: $T$ can be written as $A\cap B^c$ where $A$ and $B$ are retrocompact (i.e. quasi-compact) opens of $X$. Namely, $A=U$ and $B=U\smallsetminus T$. (Note that quasi-compact = retrocompact for opens of $X$ since $X$ is quasi-compact and quasi-separated.) The proof uses only what is written before the word 'Hence,' plus that a retrocompact open of $U$ is a retrocompact open of $X$ (5.12.2).

Comment #6549 by on

Very good. I have used your remark to simplify the proof but I didn't change the statement. Changes are here.

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