Real Analysis/Uniform Convergence

Definition: A sequence of real-valued functions $$f_n{(x)}$$ is uniformly convergent if there is a function f(x) such that for every $$\epsilon>0$$ there is an $$N > 0$$ such that when $$n > N$$ for every x in the domain of the functions f, then $$|f_n(x)-f(x)| < \epsilon$$

Theorem (Uniform Convergence Theorem))
Let $$f_n$$ be a series of continuous functions that uniformly converges to a function $$f$$. Then $$f$$ is continuous.

Proof
There exists an N such that for all n>N, $$|f_n(x) - f(x)|<\frac{\epsilon}{3}$$ for any x. Now let n>N, and consider the continuous function $$f_n$$. Since it is continuous, there exists a $$\delta$$ such that if $$|x'-x|<\delta$$, then $$|f_n(x)-f_n(x')|<\frac{\epsilon}{3}$$. Then $$|f(x')-f(x)|\le |f(x')-f_n(x')|+|f_n(x')-f_n(x)|+ |f_n(x)-f(x)| < \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3} = \epsilon$$ so the function f(x) is continuous.