the limit of a uniformly convergent sequence of continuous functions is continuous

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 $y$ .