Corollary

If $(X_\alpha)_{\alpha \in A}$ is an open covering, or a locally finite closed covering, of a topological space $X$, one has

It is enough to prove that, for every point $x$ of $X$, one has $\dim_x(X) = \sup_{\alpha \in A_x} \dim_x(X_\alpha)$, where $A_x$ is the set of $\alpha \in A$ such that $x \in X_\alpha$. This is clear in the case of an open covering, and follows from Prop. 1, c), in the case of a locally finite closed covering.