Proposition 2

Let $X$ be a topological space. Then $\dim(X) = \sup_{x \in X} \dim_x(X)$.

In fact, by definition one has $\dim(X) \geq \dim_x(X)$ for every $x \in X$. On the other hand, if $Z_0 \subset ... \subset Z_n$ is a chain of irreducible closed subsets of $X$, for every $x \in Z_0$ and every open neighbourhood $U$ of $x$, the sets $Z_0 \cap U, ..., Z_n \cap U$ form a chain of irreducible closed subsets of $U$ (II, § 4, No. 1, Prop. 7). Hence $\dim_x(X) \geq n$, whence $\dim(X) \leq \sup_{x \in X} \dim_x(X)$.