Prove the following.
Theorem (Arzela’s theorem). Let \(X\) be compact; let \(f_n \in \mathcal{C}(X, \mathbb{R}^k)\). If the collection \(\ts{f_n}\) is pointwise bounded and equicontinuous, then the sequence \(f_n\) has a uniformly convergent subsequence.
Solution
1. Uniform boundedness from equicontinuity and compactness:The sequence \((f_n)\) is uniformly bounded on \(X\): \(\sup_{n \in \mathbb{Z}_+} \sup_{x \in X} \|f_n(x)\| \le M < \infty\).Proof:
1.1. By equicontinuity at \(\varepsilon = 1\), for each \(x \in X\) there is an open neighborhood \(U_x \subseteq X\) such that \(\|f_n(y) - f_n(x)\| < 1\) for all \(y \in U_x\) and all \(n \in \mathbb{Z}_+\).
1.2. Since \(X\) is compact, the open cover \(\{U_x\}_{x \in X}\) has a finite subcover \(\{U_{x_1}, \dots, U_{x_m}\}\).
1.3. By pointwise boundedness, for each \(j \in \{1, \dots, m\}\), there exists \(M_j < \infty\) such that \(\|f_n(x_j)\| \le M_j\) for all \(n\).
1.4. Setting \(M = \max_{1 \le j \le m} M_j + 1\), for every \(y \in X\), choosing \(x_j\) with \(y \in U_{x_j}\) gives:\[\|f_n(y)\| \le \|f_n(x_j)\| + \|f_n(y) - f_n(x_j)\| < M_j + 1 \le M.\]
2. Extraction of a subsequence converging on a countable dense net:Proof:
2.1. For each integer \(m \ge 1\), equicontinuity with \(\varepsilon = 1/m\) yields a finite open cover \(\{U_{m, 1}, \dots, U_{m, p_m}\}\) centered at points \(\{x_{m, 1}, \dots, x_{m, p_m}\}\).
2.2. The union \(D = \bigcup_{m=1}^\infty \{x_{m, 1}, \dots, x_{m, p_m}\} = \{d_1, d_2, d_3, \dots\}\) is a countable subset of \(X\).
2.3. Since \((f_n(d_1))\) is bounded in \(\mathbb{R}^k\), by the Bolzano-Weierstrass theorem there exists a subsequence \((f_{1, n})\) converging at \(d_1\).
2.4. Inductively, for each \(j \ge 2\), choose \((f_{j, n})\) as a subsequence of \((f_{j-1, n})\) converging at \(d_j\).
2.5. The diagonal sequence \(g_n = f_{n, n}\) is a subsequence of \((f_n)\) that converges pointwise at every point of \(D\).
3. Uniform convergence of the diagonal subsequence \(g_n\):Proof:
3.1. Let \(\varepsilon > 0\).
3.2. By equicontinuity, choose an integer \(m > \frac{3}{\varepsilon}\) such that for every \(x \in X\), there exists \(d_j \in D \cap \{x_{m, 1}, \dots, x_{m, p_m}\}\) with:\[\|g_n(x) - g_n(d_j)\| < \frac{\varepsilon}{3} \quad \text{for all } n \in \mathbb{Z}_+.\]
3.3. Since \(g_n\) converges at each of the finitely many points \(d_j \in \{x_{m, 1}, \dots, x_{m, p_m}\}\), there exists an index \(N\) such that for all \(p, q \ge N\) and all \(j \in \{1, \dots, p_m\}\):\[\|g_p(d_j) - g_q(d_j)\| < \frac{\varepsilon}{3}.\]
3.4. For an arbitrary point \(x \in X\), select \(d_j\) with \(\|g_n(x) - g_n(d_j)\| < \varepsilon/3\). Then for all \(p, q \ge N\):\[\|g_p(x) - g_q(x)\| \le \|g_p(x) - g_p(d_j)\| + \|g_p(d_j) - g_q(d_j)\| + \|g_q(d_j) - g_q(x)\| < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon.\]
3.5. Thus \((g_n)\) is uniformly Cauchy on \(X\).
3.6. Since \(\mathbb{R}^k\) is complete, \((g_n)\) converges uniformly to a continuous function \(g: X \to \mathbb{R}^k\). Q.E.D.