Equicontinuous on a compact set implies uniform equicontinuous

Question

Theorem

$(M,d), (N,\rho)$ metric spaces, $K \subset M$ compact. If $\mathcal{F} \subset F(M,N)$, where $F(M,N) = \{ f \mid f:M \rightarrow N\}$, is equicontinuous on $K$, then $\mathcal{F}$ is uniformly equicontinuous on $K$.

Proof from class

Suppose not. Then $\exists \; \epsilon_0>0 $ s.t. $\forall \; \delta >0 , \exists \;x' \in K, y' \in M$ s.t. $d(x',y') < \delta$ but $\rho(f(x'),f(y') ) \geq \epsilon_0$ for some $f \in \mathcal{F}$.

1)

Let $\delta = 1/n$, then $x_n \in K, y_n \in M$ s.t. $d(x_n,y_n) < 1/n$ but $\rho(f(x_n),f(y_n) ) \geq \epsilon_0$ for some $f \in \mathcal{F}$.

$\{x_N\} _{n\in \mathbb{N}}$ had a convergent subsequence ($K$ compact.) i.e., $\exists x_{n_k} \rightarrow l \in K$ as $k \rightarrow \infty$.

Using a $3$ epsilon argument, there is a sequence $y_{n_k}$ that converges to $l$ as well.

2)

since $\mathcal{F}$ is equicontinuous at $l$, this implies for every $\epsilon>0$, (in particular, for $\epsilon_0$ that was fixed above) if $d(x,l) < \delta$ and $d(y<l) < \delta$, then $\rho(f(x),f(y)) < \epsilon \; \forall \; f \in \mathcal{F}.$

Then, for $k$ greater than some integer $N$, $d(x_{n_k},l) < \delta$ and $d(y_{n_k},l) < \delta$, then $\rho(f(x_{n_k}),f(y_{n_k})) < \epsilon \; \forall \; f \in \mathcal{F}.$

This is a contradiction.

I am confused about why this proof works. We proved that for the points $x,y$ that were in the particular sequence converging to $l$, $\mathcal{F}$ is unifomrly continuous. But suppose there are $x',y' \in K$ s.t. $d(x',y')< \delta$ but $d(x',l) \not < \delta$ or $d(l,y') \not< \delta$ (i.e., they are $\delta$-close but not in the sequence converging to $l$). Then we don't know if $\rho(f(x'),f(y')) < \epsilon$, do we?

Answer 1

You found a sequence $(x_n)$ that is $\delta$ close such that $(f(x_n)$) is $\epsilon$ close for all $f\in \mathcal F$, for large enough $n$. This is a contradiction of the original hypothesis.

There is an easier, direct proof that uses the Lebesgue number lemma, the proof of the existence of which is not hard and can be found in most topology texts.

By equicontinuity for each $x$ there is $\delta_x>0$ such that $d(x,y)<\delta\Rightarrow d(f(x),f(y))<\epsilon/2$ for all $f\in \mathcal F$. The balls $\{B(x,\delta_x)\}_{x\in M}$ cover $M$. Let $\lambda$ be the Lebesgue number for the cover. Then, if $d(x,y)<\lambda,$ there is a ball $B(z,\delta_z)$ such that $x,y\in B(z,\delta_z).$ It follows now from the triangle inequality that $d(f(x),f(y))<\epsilon$ for all $f\in \mathcal F$.