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 $y$ .

Title	the limit of a uniformly convergent sequence of continuous functions is continuous
Canonical name	TheLimitOfAUniformlyConvergentSequenceOfContinuousFunctionsIsContinuous
Date of creation	2013-03-22 15:21:58
Last modified on	2013-03-22 15:21:58
Owner	neapol1s (9480)
Last modified by	neapol1s (9480)
Numerical id	13
Author	neapol1s (9480)
Entry type	Theorem
Classification	msc 40A30
Related topic	LimitFunctionOfSequence