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

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 TheLimitOfAUniformlyConvergentSequenceOfContinuousFunctionsIsContinuous 2013-03-22 15:21:58 2013-03-22 15:21:58 neapol1s (9480) neapol1s (9480) 13 neapol1s (9480) Theorem msc 40A30 LimitFunctionOfSequence