Proof of the Arzela-Ascoli's theorem

Question

I'm trying understand the proof of the Arzela-Ascoli's theorem given in this lecture notes on pages $4$ and $5$, but there is one step that I didn't understand in the first paragraph on page $5$:

$d(x_n,y_n) \rightarrow 0$ while $|f(x_n) - f(y_n)| \geq \varepsilon_0$

I would like to know why this is valid. I think this remain because of the hypothesis by absurd that there are, $\varepsilon_0$, $f_n \in \mathcal{F}$ and $x_n,y_n \in X$ such that $d(x_n,y_n) < \frac{1}{n}$, but $|f_n(x_n) - f(y_n)| \geq \varepsilon_0$ for each $n \in \mathbb{N} \backslash \{ 0 \}$, but I can't see how this imply that $|f(x_n) - f(y_n)| \geq \varepsilon_0$ (observe here that the inequality is applied for $f$ now and not for $f_n$).

Thanks in advance!

Answer 1

You use the uniform convergence of $f_n$ to derive this.

We have $f_n$ converging uniformly to $f$, so fix an arbitrary $k>0$ and $d(f_m, f_{m'})<\varepsilon_0/k$ for all $m,m'>N(k)$. Hence for $n>N(k)$, $$ \lvert f(x_n)-f(y_n)\rvert\geq \lvert f_n(x_n)-f_n(y_n)\rvert - \lvert f(x_n)-f_n(x_n)\rvert - \lvert f(y_n)-f_n(y_n)\rvert \geq \varepsilon_0 -\frac1k\varepsilon_0-\frac1k\varepsilon_0=\varepsilon_0(1-\frac2k) $$ but $k$ is arbitrary. So $\lvert f(x_n)-f(y_n)\rvert\geq\varepsilon_0$.