Let \(X\) be a topological space and let \(Y\) be a metric space. Let \(f_n: X \to Y\) be a sequence of continuous functions. Let \(x_n\) be a sequence of points of \(X\) converging to \(x\). Show that if the sequence \((f_n)\) converges uniformly to \(f\), then \((f_n(x_n))\) converges to \(f(x)\).
Solution
Uniform convergence of continuous maps into a metric space implies that the limit map \(f\) is continuous. Let \(d\) be the metric on \(Y\). Then \[ d(f_n(x_n),f(x)) \le d(f_n(x_n),f(x_n))+d(f(x_n),f(x)). \] The first term tends to \(0\) because \[ \sup_{z\in X}d(f_n(z),f(z))\to0. \] The second tends to \(0\) because \(x_n\to x\) and \(f\) is continuous. Hence \[ f_n(x_n)\longrightarrow f(x). \]