Uniform Limits of Continuous Functions

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Now we show that a uniformly convergent sequence of continuous functions has to converge to a continuous function.

Proof: Let $\epsilon >0$ and let $x_0\in I$. Since $(f_n{)}_{n\in \mathbb{N}}$ converges uniformly to $f:I\to \mathbb{K}$, there exists some $N$ such that for all $n\ge N$ and $x\in I$ holds $$|f(x)-f_n(x)|<\frac{\epsilon }{3}.$$ Since $f_n$ is continuous on $I$, there exists some $\delta >0$ such that for all $x\in I$ with $|x-x_0|<\delta$ holds $$|f_n(x_0)-f_n(x)|<\frac{\epsilon }{3}.$$ Altogether, we then have $$\begin{array}{rl}|f(x)-f(x_0)|=& |f(x)-f_n(x)+f_n(x)-f_n(x_0)+f_n(x_0)-f(x_0)|\\ \le & |f(x)-f_n(x)|+|f_n(x)-f_n(x_0)|+|f_n(x_0)-f(x_0)|\\ <& \frac{\epsilon }{3}+\frac{\epsilon }{3}+\frac{\epsilon }{3}=\epsilon .\end{array}$$ Therefore, $f$ is continuous by the $\epsilon$-$\delta$ criterion. $◻$