Corollary. Every closed subspace of \(\mathbb{R}^N\) has topological dimension at most \(N\).
Solution
1. \(\RR^N\) has covering dimension \(N\).
This is the standard computation of the Lebesgue covering dimension of \(\RR^N\).
2. A closed subspace \(A\) of a space \(X\) with \(\dim X \le N\) has \(\dim A \le N\).
Let \(\mathcal{U}\) be an open cover of \(A\). Each \(U \in \mathcal{U}\) is \(A \cap U'\) for some open \(U' \subseteq X\). The sets \(U'\) together with \(X \setminus A\) form an open cover of \(X\), which has an open refinement \(\mathcal{V}\) of order at most \(N + 1\). The sets \(V \cap A\), \(V \in \mathcal{V}\), form an open cover of \(A\) of order at most \(N + 1\); each \(V\) meeting \(A\) lies in some \(U'\), since \(X \setminus A\) misses \(A\), so \(V \cap A\) lies in the corresponding \(U\).
3. Q.E.D.
Apply step 2 to \(X = \RR^N\), using step 1.