## 5.15 Constructible sets

Definition 5.15.1. Let $X$ be a topological space. Let $E \subset X$ be a subset of $X$.

We say $E$ is *constructible*^{1} in $X$ if $E$ is a finite union of subsets of the form $U \cap V^ c$ where $U, V \subset X$ are open and retrocompact in $X$.

We say $E$ is *locally constructible* in $X$ if there exists an open covering $X = \bigcup V_ i$ such that each $E \cap V_ i$ is constructible in $V_ i$.

Lemma 5.15.2. The collection of constructible sets is closed under finite intersections, finite unions and complements.

**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.
$\square$

Lemma 5.15.3. Let $f : X \to Y$ be a continuous map of topological spaces. If the inverse image of every retrocompact open subset of $Y$ is retrocompact in $X$, then inverse images of constructible sets are constructible.

**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.
$\square$

Lemma 5.15.4. Let $U \subset X$ be open. For a constructible set $E \subset X$ the intersection $E \cap U$ is constructible in $U$.

**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 $X$.
$\square$

Lemma 5.15.5. Let $U \subset X$ be a retrocompact open. Let $E \subset U$. If $E$ is constructible in $U$, then $E$ is constructible in $X$.

**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 $V \cap (U \setminus W)$.
$\square$

Lemma 5.15.6. Let $X$ be a topological space. Let $E \subset X$ be a subset. Let $X = V_1 \cup \ldots \cup V_ m$ be a finite covering by retrocompact opens. Then $E$ is constructible in $X$ if and only if $E \cap V_ j$ is constructible in $V_ j$ for each $j = 1, \ldots , m$.

**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.
$\square$

Lemma 5.15.7. Let $X$ be a topological space. Let $Z \subset X$ be a closed subset such that $X \setminus Z$ is quasi-compact. Then for a constructible set $E \subset X$ the intersection $E \cap Z$ is constructible in $Z$.

**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.
$\square$

Lemma 5.15.8. Let $X$ be a topological space. Let $T \subset X$ be a subset. Suppose

$T$ is retrocompact in $X$,

quasi-compact opens form a basis for the topology on $X$.

Then for a constructible set $E \subset X$ the intersection $E \cap T$ is constructible in $T$.

**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.
$\square$

Lemma 5.15.9. Let $Z \subset X$ be a closed subset whose complement is retrocompact open. Let $E \subset Z$. If $E$ is constructible in $Z$, then $E$ is constructible in $X$.

**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

\[ W \cap \tilde V = \left(V \cap W\right) \cup \left((X \setminus Z) \cap W\right). \]

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

\[ V_1 \cap (Z \setminus V_2) = \tilde V_1 \cap (X \setminus \tilde V_2) \]

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

Lemma 5.15.10. Let $X$ be a topological space. Every constructible subset of $X$ is retrocompact.

**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.
$\square$

Question: Does the following lemma also hold if we assume $X$ is a quasi-compact topological space? Compare with Lemma 5.15.7.

Lemma 5.15.11. Let $X$ be a topological space. Assume $X$ has a basis consisting of quasi-compact opens. For $E, E'$ constructible in $X$, the intersection $E \cap E'$ is constructible in $E$.

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

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.
$\square$

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 $U \setminus T = U \cap T^ c$ is retrocompact in $U$ as $T^ c$ is retrocompact in $X$. Hence the inclusion $T \subset X$ can be written as the composition of the inclusion $T \subset U$ of a closed subset with retrocompact complement and the inclusion $U \subset X$ which is retrocompact by our assumption on intersections of quasi-compact opens. Thus the lemma is a consequence of Lemmas 5.15.5 and 5.15.9.
$\square$

Lemma 5.15.14. Let $X$ be a topological space which has a basis for the topology consisting of quasi-compact opens. Let $E \subset X$ be a subset. Let $X = E_1 \cup \ldots \cup E_ m$ be a finite covering by constructible subsets. Then $E$ is constructible in $X$ if and only if $E \cap E_ j$ is constructible in $E_ j$ for each $j = 1, \ldots , m$.

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

Lemma 5.15.15. Let $X$ be a topological space. Suppose that $Z \subset X$ is irreducible. Let $E \subset X$ be a finite union of locally closed subsets (e.g. $E$ is constructible). The following are equivalent

The intersection $E \cap Z$ contains an open dense subset of $Z$.

The intersection $E \cap Z$ is dense in $Z$.

If $Z$ has a generic point $\xi $, then this is also equivalent to

We have $\xi \in E$.

**Proof.**
Write $E = \bigcup U_ i \cap Z_ i$ as the finite union of intersections of open sets $U_ i$ and closed sets $Z_ i$. Suppose that $E \cap Z$ is dense in $Z$. Note that the closure of $E \cap Z$ is the union of the closures of the intersections $U_ i \cap Z_ i \cap Z$. As $Z$ is irreducible we conclude that the closure of $U_ i \cap Z_ i \cap Z$ is $Z$ for some $i$. Fix such an $i$. It follows that $Z \subset Z_ i$ since otherwise the closed subset $Z \cap Z_ i$ of $Z$ would not be dense in $Z$. Then $U_ i \cap Z_ i \cap Z = U_ i \cap Z$ is an open nonempty subset of $Z$. Because $Z$ is irreducible, it is open dense. Hence $E \cap Z$ contains an open dense subset of $Z$. The converse is obvious.

Suppose that $\xi \in Z$ is a generic point. Of course if (1) $\Leftrightarrow $ (2) holds, then $\xi \in E$. Conversely, if $\xi \in E$, then $\xi \in U_ i \cap Z_ i$ for some $i = i_0$. Clearly this implies $Z \subset Z_{i_0}$ and hence $U_{i_0} \cap Z_{i_0} \cap Z = U_{i_0} \cap Z$ is an open not empty subset of $Z$. We conclude as before.
$\square$

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