Uniform continuity and sequences negation

Question

Show that if $f$ is not uniformly continuous on an interval $[a,b]$ then there are sequences $\{x_n\}$ and $\{y_n\}$ chosen from $[a,b]$ so that $x_n-y_n\to0$ but $|f(x_n)-f(y_n)|>c$ for some $c>0$

I've only barely started the proof for this one.

---

Proof

Suppose $f:[a,b]\to \mathbb{R}$ is not uniformly continuous

Then $\exists c >0: \forall \delta >0,$ there are $x_ ,y \in [a,b]: |x-y|\leq \delta \quad \text{and}\quad |f(x)-f(y)|\geq c$

Let $\{x_n\}, \{y_n\} \subseteq \mathbb{R}$

...

---

I'm not sure where to go from here. I get stuck because we don't know what $\{x_n\}$ and $\{y_n\}$ converge to, which makes using the definition of convergence tricky here. Any ideas?

Answer 1

The negation of $$\forall \varepsilon >0, \exists \delta >0: \forall x,y\in [a,b]: |x-y|\leq \delta \implies |f(x)-f(y)|<\varepsilon $$

Is $$\exists \varepsilon >0: \forall \delta >0, \exists x_\delta ,y_\delta \in [a,b]: |x-y|\leq \delta \quad \text{and}\quad |f(x)-f(y)|\geq \varepsilon .$$ In particular, there is $\varepsilon >0$ s.t. for all $n\in\mathbb N$, there are $x_n,y_n\in[a,b]$ s.t. $$|x_n-y_n|\leq \frac{1}{n}\quad \text{and}\quad |f(x_n)-f(y_n)|>\varepsilon .$$