Corollary

a) The dimension of the topological space $X$ is the least upper bound of the dimensions of its irreducible components (II, § 4, No. 1, Def. 2).

b) Let $x$ be a point of $X$. Then $\dim_x(X) \geq \sup \dim_x(X_i)$, where $X_i$ runs through the family of irreducible components of $X$ that contain $x$; there is equality if $x$ has a neighbourhood $V$ which has only a finite number of irreducible components (which is the case for example if $V$ is noetherian).

The first assertion is immediate since the chains of irreducible closed subsets of $X$ are the chains of irreducible closed subsets of the irreducible components of $X$ (II, § 4, No. 1, Prop. 5). The inequality $\dim_x(X) \geq \sup_{i} \dim_x(X_i)$ follows from Prop. 1, a). Let $V$ be a neighbourhood of $x$ which has only a finite number of irreducible components, and let $(V_j)_{j \in J}$ be the family of irreducible components of $V$ which contain $x$. It follows from Prop. 1, b) and c) that one has

one concludes by remarking that each of the $V_j$ is contained in one of the $X_i$, $i \in J_x$, and that one therefore has $\sup_{j \in J} \dim_x(V_j) \leq \sup_{i \in J_x} \dim_x(X_i)$.