Lemma 5.10.2 (0B7I)—The Stacks project

Lemma 5.10.2. Let $X$ be a topological space. Then $\dim (X) = \sup \dim _ x(X)$ where the supremum runs over the points $x$ of $X$.

Proof. It is clear that $\dim (X) \geq \dim _ x(X)$ for all $x \in X$ (see discussion following Definition 5.10.1). 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 $x \in Z_0$. 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. $\square$

Comments (2)

Comment #6799 by Xun on December 11, 2021 at 11:03

In the proof of the lemma, to conclude $\dim U\geq n$ , do we need to prove that $Z_i\cap U$ 's are all distinct? The proof seems forget to prove this.

Comment #6944 by Johan on January 20, 2022 at 20:09

Here is an exercise: let $U$ be an open subset of a topological space $X$ . Then the map $Z \mapsto U \cap Z$ is an inclusion preserving bijection between the set of irreducible closed subsets of $X$ which meet $U$ to the set of irreducible closed subsets of $U$ . Should we make this a lemma? What do you guys think?

Anyway, I tried to improve the wording of the proof. See changes here.

There are also:

3 comment(s) on Section 5.10: Krull dimension

Comments (2)

Post a comment