Uniform convergence implies uniform Cauchy

Uniform convergence implies uniform Cauchy: Let $X$ be a set and let $(f_n)$ be functions $f_n:X\to (Y,d)$ into a metric space . If $f_n\to f$ uniformly on $X$, then $(f_n)$ is uniformly Cauchy : $\forall\varepsilon>0\;\exists N\;\forall m,n\ge N:\ \sup_{x\in X} d(f_n(x),f_m(x))<\varepsilon.$

This is the Cauchy criterion direction for uniform convergence and is often the easiest way to prove uniform convergence: show uniform Cauchy in a complete codomain.

Proof sketch: Fix $\varepsilon>0$. Choose $N$ so that $\sup_x d(f_n(x),f(x))<\varepsilon/2$ for all $n\ge N$. Then for $m,n\ge N$, $d(f_n(x),f_m(x))\le d(f_n(x),f(x))+d(f(x),f_m(x))<\varepsilon,$ uniformly in $x$.