Prove the following.
Theorem (Uniform boundedness principle). Let \(X\) be a complete metric space, and let \(\mathcal{F}\) be a subset of \(\mathcal{C}(X, \mathbb{R})\) such that for each \(a \in X\), the set
\[ \mathcal{F}_a = \ts{f(a) \mid f \in \mathcal{F}} \]
is bounded. Then there is a nonempty open set \(U\) of \(X\) on which the functions in \(\mathcal{F}\) are uniformly bounded, that is, there is a number \(M\) such that \(\abs{f(x)} \leq M\) for all \(x \in U\) and all \(f \in \mathcal{F}\). [Hint: Let \(A_N = \ts{x \mid \abs{f(x)} \leq N \text{ for all } f \in \mathcal{F}}\).]
Solution
For \(N\in\mathbb Z_+\) set \[ A_N=\{x\in X:|f(x)|\le N\text{ for every }f\in\mathcal F\}. \] Each \(A_N\) is closed, since \[ A_N=\bigcap_{f\in\mathcal F}f^{-1}([-N,N]). \] Pointwise boundedness says every \(x\) belongs to some \(A_N\), so \(X=\bigcup_NA_N\). A complete metric space is Baire; therefore some \(A_N\) has nonempty interior. Choose a nonempty open \(U\subset A_N\). Then for every \(x\in U\) and every \(f\in\mathcal F\), \(|f(x)|\le N\). Taking \(M=N\) proves the assertion.