Proposition 1

Let X be a topological space. a) If Y is a subspace of X, then $\dim(Y) \leq \dim(X)$, and $\dim_y(Y) \leq \dim_y(X)$ for every point y of Y. b) Let x be a point of X and V a neighbourhood of x in X. Then $\dim_x(X) = \dim_x(V)$. c) Let $(X_i)_{i \in I}$ be a finite family of closed subsets of X such that $X = \bigcup_{i \in I} X_i$. Then $\dim(X) = \sup_{i \in I} \dim(X_i)$ and, for every point x of X, one has $\dim_x(X) = \sup_{i \in J_x} \dim_x(X_i)$, where $J_x$ denotes the set of $i \in I$ such that $x \in X_i$.

Let us prove a). If Z is an irreducible closed subset of Y, its closure $\overline{Z}$ in X is irreducible (II, § 4, No. 1, Prop. 2), and one has $\overline{Z} \cap Y = Z$. Thus every chain c of irreducible closed subsets of Y defines, by passing to the closure in X, a chain of irreducible closed subsets of X, of the same length as c. The inequality $\dim(Y) \leq \dim(X)$ follows from this. If U is an open subset of X containing a point y of Y, then therefore $\dim(U \cap Y) \leq \dim(U)$, whence $\dim_y(Y) \leq \dim_y(X)$. Let us prove b). By definition one has $\dim_x(X) \leq \dim_x(V)$, and the opposite inequality follows from a). Let us prove c). Let $Z_0 \subset \ldots \subset Z_n$ be a chain of irreducible closed subsets of X. One has $Z_n = \bigcup_{i \in I} (Z_n \cap X_i)$ and each of the sets $Z_n \cap X_i$ is closed in $Z_n$;

since I is finite, $Z_n$ is contained in one of the $X_i$. Consequently, one has $\dim(X) \leq \sup_{i} \dim(X_i)$, whence equality by a).

Now let $x$ be a point of $X$ and $n = \sup_{i \in J_x} \dim_x(X_i)$, where $J_x$ is as in the statement. One has $\dim_x(X) \geq n$ by a), and, to establish equality, one may suppose $n$ finite. For every $i \in J_x$, let $U_i$ be an open neighbourhood of $x$ in $X$, such that $\dim(U_i \cap X_i) \leq n$. Put $U = (\bigcap_{i \in J_x} U_i) \cap (\bigcap_{i \in I - J_x} \mathscr{C} X_i)$; the set $U$ is open in $X$. Moreover, one has $\dim(U) = \sup_{i \in J_x} \dim(U \cap X_i) \leq n$ by the preceding paragraph, hence $\dim_x(X) \leq n$.