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

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. $\square$

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