Irreducibles and opens

Thanks to Xun
https://stacks.math.columbia.edu/tag/0B7I#comment-6799

topology.tex

@@ -1258,9 +1258,14 @@ \section{Krull dimension}
 Thus an inequality in one direction. For the converse, let $n \geq 0$
 and suppose that $\dim(X) \geq n$. Then we can find a chain of irreducible
 closed subsets $Z_0 \subset Z_1 \subset \ldots \subset Z_n \subset X$.
-Pick $x \in Z_0$. Then we see that every open neighbourhood $U$ of $x$
-has a chain of irreducible closed subsets
-$Z_0 \cap U \subset Z_1 \cap U \subset \ldots Z_n \cap U \subset U$.
+Pick $x \in Z_0$. For every open neighbourhood $U$ of $x$ we get
+a chain of irreducible closed subsets
+$$
+Z_0 \cap U \subset Z_1 \cap U \subset \ldots \subset Z_n \cap U
+$$
+in $U$. Namely, the sets $U \cap Z_i$ are irreducible closed in $U$ and the
+inclusions are strict (details omitted; hint: the closure of $U \cap Z_i$
+is $Z_i$).
 In this way we see that $\dim_x(X) \geq n$ which proves the other
 inequality.
 \end{proof}