Confusion about the proof of uniform convergence

Question

Let $Y$ be a complete metric space. Then a uniformly Cauchy sequence $(f_n)$ of functions $f_n$ : $X → Y$ converges uniformly to a function $f : X → Y $.
I met this proof in the book, and I have confusion in the last step. "$d(f_n,f)\le d(f_n,f_m)+d(f_m,f) <\epsilon + d(f_m,f)$. Now let $m$ tends to infinity, we have $d(f_n,f)<\epsilon$" Why we can let $m$ tends to infinity here. It does not make sense to me. For each $x\in X$, $f_m(x)$ is Cauchy. And by completeness, we have $f_m(x)\to f(x)$ which means for given $\epsilon$, there exists $N(x)$ such that $d(f_m(x),f(x))< \epsilon$ for $n\ge N(x)$. However, we can not choose such a $N$ works for all $x$. Can anyone help me understand this?

Answer 1

Your criticism is fine. Here is how the proof can be completed.

Take $\varepsilon>0$; you want to prove that, for some $N\in\mathbb N$, you have $d(f_n,f)<\varepsilon$. Take $N\in\mathbb N$ such that$$m,n\geqslant N\implies d(f_m,f_n)<\frac\varepsilon2.$$Then, for each $x\in X$ and each $n\geqslant N$,\begin{align}d\bigl(f(x),f_n(x)\bigr)&=d\left(\lim_{m\to\infty}f_m(x),f_n(x)\right)\\&=\lim_{m\to\infty}\overbrace{d\bigl(f_m(x),f_n(x)\bigr)}^{\phantom{\varepsilon/2}<\varepsilon/2}\\&\leqslant\frac\varepsilon2\\&<\varepsilon.\end{align}