Small changes in proof 005K

Thanks to Alex Scheffelin and Laurent Moret-Bailly
https://stacks.math.columbia.edu/tag/005K#comment-7400
https://stacks.math.columbia.edu/tag/005K#comment-7401

topology.tex

@@ -2627,26 +2627,16 @@ \section{Constructible sets}
 \end{lemma}
 
 \begin{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.
+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.
 
 \medskip\noindent
-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.
+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 $Z$.
 \end{proof}