TITLE: Question about proof: Uniform cauchy $\Rightarrow$ Uniform convergence
QUESTION [7 upvotes]: I have one quick question regarding the proof of a theorem contained in here : https://www.math.ucdavis.edu/~hunter/m125a/intro_analysis_ch5.pdf

Theorem 5.13. A sequence $(f_n)$ of functions $f_n : A → R$ converges
  uniformly on $A$ if and only if it is uniformly Cauchy on $A$.


Question:
In the triangle inequality part
$|f_n(x) - f(x)| \leq |f_n(x) - f_m(x)| + |f_m(x) - f(x)|$
Why do we know that $f_m(x) \to f(x)$ as $m \to \infty$ so that $|f_m(x) - f(x)| < \frac{\epsilon}{2}$
I mean, isn't $|f_m(x) - f(x)|$ basically the same as $|f_n(x) - f(x)|$ on the left hand side? Just one is indexed by $m$ instead $n$? Why do we know that $|f_m(x) - f(x)| < \epsilon$ but we do not know immediately whether $|f_n(x) - f(x)| < \epsilon$

REPLY [7 votes]: Another way to look at it is as follows. This circumvents any consideration of the dependence of $m$ on $x$ which, clearly, is confusing.
It can be shown that for $n,m > N$ we have, for all $x \in A,$
$$|f_n(x) - f(x)| < \epsilon + |f_m(x) - f(x)|.$$
Hence,
$$|f_n(x) - f(x)| -\epsilon < |f_m(x) - f(x)|.$$
With $x$ fixed, we take the limit of both sides as $m \to \infty$.  Since $f_m(x)$ converges pointwise to $f(x)$ and the LHS does not depend on $m$ it follows that for all $x \in A$ and $n > N$,
$$ |f_n(x) - f(x)| -\epsilon = \lim_{m \to \infty}\left(|f_n(x) - f(x)| -\epsilon\right) \leqslant \lim_{m \to \infty}|f_m(x) - f(x)| = 0 \\ \implies |f_n(x) - f(x)|  \leqslant\epsilon .$$
Here we have used the following lemma which you can easily prove by contradiction:
$$a_m > a \, ,\forall m \in \mathbb{N} \implies \lim_{m \to \infty}a_m \geqslant a. $$