Lemma 5.16.1. Let $X$ be a Noetherian topological space. The constructible sets in $X$ are precisely the finite unions of locally closed subsets of $X$.
5.16 Constructible sets and Noetherian spaces
Proof. This follows immediately from Lemma 5.12.13. $\square$
Lemma 5.16.2. Let $f : X \to Y$ be a continuous map of Noetherian topological spaces. If $E \subset Y$ is constructible in $Y$, then $f^{-1}(E)$ is constructible in $X$.
Proof. Follows immediately from Lemma 5.16.1 and the definition of a continuous map. $\square$
Lemma 5.16.3. Let $X$ be a Noetherian topological space. Let $E \subset X$ be a subset. The following are equivalent:
$E$ is constructible in $X$, and
for every irreducible closed $Z \subset X$ the intersection $E \cap Z$ either contains a nonempty open of $Z$ or is not dense in $Z$.
Proof. Assume $E$ is constructible and $Z \subset X$ irreducible closed. Then $E \cap Z$ is constructible in $Z$ by Lemma 5.16.2. Hence $E \cap Z$ is a finite union of nonempty locally closed subsets $T_ i$ of $Z$. Clearly if none of the $T_ i$ is open in $Z$, then $E \cap Z$ is not dense in $Z$. In this way we see that (1) implies (2).
Conversely, assume (2) holds. Consider the set $\mathcal{S}$ of closed subsets $Y$ of $X$ such that $E \cap Y$ is not constructible in $Y$. If $\mathcal{S} \not= \emptyset $, then it has a smallest element $Y$ as $X$ is Noetherian. Let $Y = Y_1 \cup \ldots \cup Y_ r$ be the decomposition of $Y$ into its irreducible components, see Lemma 5.9.2. If $r > 1$, then each $Y_ i \cap E$ is constructible in $Y_ i$ and hence a finite union of locally closed subsets of $Y_ i$. Thus $E \cap Y$ is a finite union of locally closed subsets of $Y$ too and we conclude that $E \cap Y$ is constructible in $Y$ by Lemma 5.16.1. This is a contradiction and so $r = 1$. If $r = 1$, then $Y$ is irreducible, and by assumption (2) we see that $E \cap Y$ either (a) contains an open $V$ of $Y$ or (b) is not dense in $Y$. In case (a) we see, by minimality of $Y$, that $E \cap (Y \setminus V)$ is a finite union of locally closed subsets of $Y \setminus V$. Thus $E \cap Y$ is a finite union of locally closed subsets of $Y$ and is constructible by Lemma 5.16.1. This is a contradiction and so we must be in case (b). In case (b) we see that $E \cap Y = E \cap Y'$ for some proper closed subset $Y' \subset Y$. By minimality of $Y$ we see that $E \cap Y'$ is a finite union of locally closed subsets of $Y'$ and we see that $E \cap Y' = E \cap Y$ is a finite union of locally closed subsets of $Y$ and is constructible by Lemma 5.16.1. This contradiction finishes the proof of the lemma. $\square$
Lemma 5.16.4. Let $X$ be a Noetherian topological space. Let $x \in X$. Let $E \subset X$ be constructible in $X$. The following are equivalent:
$E$ is a neighbourhood of $x$, and
for every irreducible closed subset $Y$ of $X$ which contains $x$ the intersection $E \cap Y$ is dense in $Y$.
Proof. It is clear that (1) implies (2). Assume (2). Consider the set $\mathcal{S}$ of closed subsets $Y$ of $X$ containing $x$ such that $E \cap Y$ is not a neighbourhood of $x$ in $Y$. If $\mathcal{S} \not= \emptyset $, then it has a minimal element $Y$ as $X$ is Noetherian. Suppose $Y = Y_1 \cup Y_2$ with two smaller nonempty closed subsets $Y_1$, $Y_2$. If $x \in Y_ i$ for $i = 1, 2$, then $Y_ i \cap E$ is a neighbourhood of $x$ in $Y_ i$ and we conclude $Y \cap E$ is a neighbourhood of $x$ in $Y$ which is a contradiction. If $x \in Y_1$ but $x \not\in Y_2$ (say), then $Y_1 \cap E$ is a neighbourhood of $x$ in $Y_1$ and hence also in $Y$, which is a contradiction as well. We conclude that $Y$ is irreducible closed. By assumption (2) we see that $E \cap Y$ is dense in $Y$. Thus $E \cap Y$ contains an open $V$ of $Y$, see Lemma 5.16.3. If $x \in V$ then $E \cap Y$ is a neighbourhood of $x$ in $Y$ which is a contradiction. If $x \not\in V$, then $Y' = Y \setminus V$ is a proper closed subset of $Y$ containing $x$. By minimality of $Y$ we see that $E \cap Y'$ contains an open neighbourhood $V' \subset Y'$ of $x$ in $Y'$. But then $V' \cup V$ is an open neighbourhood of $x$ in $Y$ contained in $E$, a contradiction. This contradiction finishes the proof of the lemma. $\square$
Lemma 5.16.5. Let $X$ be a Noetherian topological space. Let $E \subset X$ be a subset. The following are equivalent:
$E$ is open in $X$, and
for every irreducible closed subset $Y$ of $X$ the intersection $E \cap Y$ is either empty or contains a nonempty open of $Y$.
Proof. This follows formally from Lemmas 5.16.3 and 5.16.4. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)