Uniform Cauchy implies uniform convergence

Uniform Cauchy implies uniform convergence: Let $X$ be a set and let $(Y,d)$ be a complete metric space . If $(f_n)$ is a uniformly Cauchy sequence of functions $f_n:X\to Y$, then there exists a function $f:X\to Y$ such that $f_n\to f$ uniformly on $X$.

This lemma is the completeness principle for uniform convergence: the space of bounded functions with the sup metric is complete when the target is complete.

Proof sketch: Fix $x\in X$. Since $(f_n)$ is uniformly Cauchy, the sequence $(f_n(x))$ is Cauchy in $Y$, hence converges to some value $f(x)\in Y$. This defines $f:X\to Y$. To show uniform convergence, use the uniform Cauchy property: given $\varepsilon>0$, choose $N$ so that $d(f_n(x),f_m(x))<\varepsilon$ for all $x$ and $m,n\ge N$, then let $m\to\infty$ to obtain $d(f_n(x),f(x))\le \varepsilon$ uniformly in $x$.