Theorem. The limit of a uniformly convergent sequence of continuous functions is continuous.

Proof. Let $f_n,f:X\rightarrow Y$ , where $(X,\rho)$ and $(Y,d)$ are metric spaces. Suppose $f_n \rightarrow f$ uniformly and each $f_n$ is continuous. Then given any $\epsilon>0$ , there exists $N$ such that $n>N$ implies $d(f(x),f_{n}(x)) < \frac{\epsilon}{3}$ for all $x$ . Pick an arbitrary $n$ larger than $N$ . Since $f_n$ is continuous, given any point $x_0$ , there exists $\delta>0$ such that $0<\rho(x,x_0)<\delta$ implies $d(f_n(x), f_n(x_0))<\frac{\epsilon}{3}$ . Therefore, given any $x_0$ and $\epsilon>0$ , there exists $\delta>0$ such that $$ 0<\rho(x,x_0)<\delta\Rightarrow d(f(x),f(x_0)) \leq d(f(x),f_n(x)) + d(f_n(x),f_n(x_0)) + d(f_n(x_0),f(x_0)) < \epsilon $$ Therefore, $f$ is continuous.

The theorem also generalizes to when $X$ is an arbitrary topological space. To generalize it to $X$ an arbitrary topological space, note that if $d(f_n(x),f(x))< \epsilon/3$ for all $x$ , then $x_0\in f_n^{-1}(B_{\epsilon/3}(f_n(x_0)))\subseteq f^{-1}(B_\epsilon(f(x_0)))$ , so $f^{-1}(B_\epsilon(f(x_0)))$ is a neighbourhood of $x_0$ . Here $B_{\epsilon}(y)$ denote the open ball of radius $\epsilon$ , centered at $y$ .