Exercise 6.2.5

In the forward direction, suppose $(f_n)$ converges uniformly to $f$ and set $N$ large enough that $n\ge N$ has $|f_n(x)-f(x)|<𝜖∕2$ for all $x$, then use the triangle inequality (where $m\ge N$ as well)

In the reverse direction, suppose we can find an $N$ so that every $n,m\ge N$ has $|f_n(x)-f_m(x)|<𝜖∕2$ for all $x$. Fix $x$ and apply Theorem 2.6.4 to conclude the sequence $(f_n(x))$ converges to some limit $L$, and define $f(x)=L$. Doing this for all $x$ gives us the pointwise limit $f$. Now we show $(f_n)\to f$ uniformly using the fact that $|f_n(x)-f_m(x)|<𝜖∕2$for all $x$. Let $n\ge N$, notice that for all $m$

For all $m\ge N$ we have $|f_n(x)-f_m(x)|<𝜖∕2$ and

For any $x$ we can choose $m$ large enough to ensure $|f_m(x)-f(x)|<𝜖∕2$ (pointwise convergence), and since the inequality is for all $m$ this implies $|f_n(x)-f(x)|\le 𝜖$ for all $x$ as desired.