Show that the set \(\mathcal{B}(\mathbb{R}, \mathbb{R})\) of bounded functions \(f: \mathbb{R} \to \mathbb{R}\) is closed in \(\mathbb{R}^{\mathbb{R}}\) in the uniform topology, but not in the topology of compact convergence.
Solution
1. \(\mathcal B(\RR,\RR)\) is closed in the uniform topology.
The uniform topology is induced by a metric, so it suffices to show that a uniform limit \(f\) of a sequence \(f_n \in \mathcal B(\RR,\RR)\) is bounded. Choose \(N\) with \(\sup_x \abs{f_N(x) - f(x)} < 1\), and let \(M_N\) bound \(\abs{f_N}\). Then \(\abs{f(x)} \le \abs{f_N(x)} + 1 \le M_N + 1\) for all \(x\).
2. \(\mathcal B(\RR,\RR)\) is not closed in the topology of compact convergence.
Let \(f_n(x) = x\) for \(\abs{x} \le n\) and \(f_n(x) = 0\) otherwise, and let \(f(x) = x\). Each \(f_n\) is bounded by \(n\), and \(f\) is unbounded. For a compact \(K \subseteq \RR\) and \(n \ge \sup_{x \in K} \abs{x}\), \(f_n = f\) on \(K\), so \(f_n \to f\) uniformly on every compact set. Hence \(f\) lies in the closure of \(\mathcal B(\RR,\RR)\) in the topology of compact convergence but not in \(\mathcal B(\RR,\RR)\).
3. Q.E.D.
Steps 1 and 2 prove the two claims.