The Stacks project

Proof. Note that if $U_1$, $U_2$ are open and retrocompact in $X$ then so is $U_1 \cup U_2$ because the union of two quasi-compact subsets of $X$ is quasi-compact. It is also true that $U_1 \cap U_2$ is retrocompact. Namely, suppose $U \subset X$ is quasi-compact open, then $U_2 \cap U$ is quasi-compact because $U_2$ is retrocompact in $X$, and then we conclude $U_1 \cap (U_2 \cap U)$ is quasi-compact because $U_1$ is retrocompact in $X$. From this it is formal to show that the complement of a constructible set is constructible, that finite unions of constructibles are constructible, and that finite intersections of constructibles are constructible. $\square$

Proof. This is true because $f^{-1}(U \cap V^ c) = f^{-1}(U) \cap f^{-1}(V)^ c$, combined with the definition of constructible sets. $\square$

Proof. Suppose that $V \subset X$ is retrocompact open in $X$. It suffices to show that $V \cap U$ is retrocompact in $U$ by Lemma 5.15.3. To show this let $W \subset U$ be open and quasi-compact. Then $W$ is open and quasi-compact in $X$. Hence $V \cap W = V \cap U \cap W$ is quasi-compact as $V$ is retrocompact in $X$. $\square$

Proof. Suppose that $V, W \subset U$ are retrocompact open in $U$. Then $V, W$ are retrocompact open in $X$ (Lemma 5.12.2). Hence $V \cap (U \setminus W) = V \cap (X \setminus W)$ is constructible in $X$. We conclude since every constructible subset of $U$ is a finite union of subsets of the form $V \cap (U \setminus W)$. $\square$

Proof. If $E$ is constructible in $X$, then by Lemma 5.15.4 we see that $E \cap V_ j$ is constructible in $V_ j$ for all $j$. Conversely, suppose that $E \cap V_ j$ is constructible in $V_ j$ for each $j = 1, \ldots , m$. Then $E = \bigcup E \cap V_ j$ is a finite union of constructible sets by Lemma 5.15.5 and hence constructible. $\square$

Proof. Suppose that $V \subset X$ is retrocompact open in $X$. It suffices to show that $V \cap Z$ is retrocompact in $Z$ by Lemma 5.15.3. To show this let $W \subset Z$ be open and quasi-compact. The subset $W' = W \cup (X \setminus Z)$ is quasi-compact, open, and $W = Z \cap W'$. Hence $V \cap Z \cap W = V \cap Z \cap W'$ is a closed subset of the quasi-compact open $V \cap W'$ as $V$ is retrocompact in $X$. Thus $V \cap Z \cap W$ is quasi-compact by Lemma 5.12.3. $\square$

Proof. Suppose that $V \subset X$ is retrocompact open in $X$. It suffices to show that $V \cap T$ is retrocompact in $T$ by Lemma 5.15.3. To show this let $W \subset T$ be open and quasi-compact. By assumption (2) we can find a quasi-compact open $W' \subset X$ such that $W = T \cap W'$ (details omitted). Hence $V \cap T \cap W = V \cap T \cap W'$ is the intersection of $T$ with the quasi-compact open $V \cap W'$ as $V$ is retrocompact in $X$. Thus $V \cap T \cap W$ is quasi-compact. $\square$

Proof. Suppose that $V \subset Z$ is retrocompact open in $Z$. Consider the open subset $\tilde V = V \cup (X \setminus Z)$ of $X$. Let $W \subset X$ be quasi-compact open. Then

The first part is quasi-compact as $V \cap W = V \cap (Z \cap W)$ and $(Z \cap W)$ is quasi-compact open in $Z$ (Lemma 5.12.3) and $V$ is retrocompact in $Z$. The second part is quasi-compact as $(X \setminus Z)$ is retrocompact in $X$. In this way we see that $\tilde V$ is retrocompact in $X$. Thus if $V_1, V_2 \subset Z$ are retrocompact open, then

is constructible in $X$. We conclude since every constructible subset of $Z$ is a finite union of subsets of the form $V_1 \cap (Z \setminus V_2)$. $\square$

Proof. Let $E = \bigcup _{i = 1, \ldots , n} U_ i \cap V_ i^ c$ with $U_ i, V_ i$ retrocompact open in $X$. Let $W \subset X$ be quasi-compact open. Then $E \cap W = \bigcup _{i = 1, \ldots , n} U_ i \cap V_ i^ c \cap W$. Thus it suffices to show that $U \cap V^ c \cap W$ is quasi-compact if $U, V$ are retrocompact open and $W$ is quasi-compact open. This is true because $U \cap V^ c \cap W$ is a closed subset of the quasi-compact $U \cap W$ so Lemma 5.12.3 applies. $\square$

Proof. Combine Lemmas 5.15.8 and 5.15.10. $\square$

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

Then we can apply Lemma 5.15.9 to the first inclusion and Lemma 5.15.5 to the second. $\square$

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 $X$. $\square$

Proof. Combine Lemmas 5.15.11 and 5.15.12. $\square$

Proof. The implication (1) $\Rightarrow $ (2) is clear. Assume (2). Note that $E \cap Z$ is a finite union of locally closed subsets $Z_ i$ of $Z$. Since $Z$ is irreducible, one of the $Z_ i$ must be dense in $Z$. Then this $Z_ i$ is dense open in $Z$ as it is open in its closure. Hence (1) holds.

Suppose that $\xi \in Z$ is a generic point. If the equivalent conditions (1) and (2) hold, then $\xi \in E$. Conversely, if $\xi \in E$ then $\xi \in E \cap Z$ and hence $E \cap Z$ is dense in $Z$. $\square$