In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). Your email address will not be published. This follows formally from Lemmas 5.16.3 and 5.16.4. The following are equivalent:įor every irreducible closed subset $Y$ of $X$ the intersection $E \cap Y$ is either empty or contains a nonempty open of $Y$. Let $X$ be a Noetherian topological space. This contradiction finishes the proof of the lemma. But then $V' \cup V$ is an open neighbourhood of $x$ in $Y$ contained in $E$, a contradiction. By minimality of $Y$ we see that $E \cap Y'$ contains an open neighbourhood $V' \subset Y'$ of $x$ in $Y'$. If $x \not\in V$, then $Y' = Y \setminus V$ is a proper closed subset of $Y$ containing $x$. If $x \in V$ then $E \cap Y$ is a neighbourhood of $x$ in $Y$ which is a contradiction. Thus $E \cap Y$ contains an open $V$ of $Y$, see Lemma 5.16.3. By assumption (2) we see that $E \cap Y$ is dense in $Y$. We conclude that $Y$ is irreducible closed. 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. 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. Suppose $Y = Y_1 \cup Y_2$ with two smaller nonempty closed subsets $Y_1$, $Y_2$. If $E \subset Y$ is constructible in $Y$, then $f^ \not= \emptyset $, then it has a minimal element $Y$ as $X$ is Noetherian. Let $f : X \to Y$ be a continuous map of Noetherian topological spaces. This follows immediately from Lemma 5.12.13. The constructible sets in $X$ are precisely the finite unions of locally closed subsets of $X$. 5.16 Constructible sets and Noetherian spaces
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