$f_n\to f$ uniformly if, and only if $M_n\to 0$ when $n\to\infty$

Question

Theorem: Let $\lim_{n\to\infty}{f_n(x)=f(x)}$ and ($x\in E$). Let $M_n=\sup_{x\in E}|{f_n(x)-f(x}|$. Then $f_n\to f$ uniformly if, and only if $M_n\to 0$ when $n\to\infty$

This is a theorem on Rudin's principles of mathematical analysis. He doesn't give a detailed proof so I wanted to give it a try. This is what I got.

First assume $f_n\to f$ uniformly. Then $f_n$ also converges pointwise on E and $\lim_{n\to\infty}{f_n(x)=f(x)}$. Then $\lim_{n\to\infty}{M_n}=\sup_{x\in E}|f(x)-f(x)|=0$.

For the converse, let's assume $M_n\to 0$ when $n\to\infty$. Then for all $\epsilon>0$ there is an integer $N$ such that when $n\geq N$ implies that $|M_n-0|=|M_n|\leq\epsilon$, but $|f_n(x)-f(x)|\leq\sup_{x\in E}|f_n(x)-f(x)|\leq\epsilon$ for all $x\in E$. We conclude that $f_n\to f$ uniformly.

Answer 1

The proof of the converse is good, but the first part is wrong. Pointwise convergence will not imply that the supremum goes to zero.