# Uniform continuity and sequences negation

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?

• Check your negation again; there are no numbers $f(x)$ and $f(y)$ such that $|f(x) - f(y)| \ge \varepsilon$ for all $\varepsilon > 0$. Nov 13, 2020 at 15:00
• Your negation is incorrect : it should be "there exists $\varepsilon > 0$ such that for every $\delta$, etc. etc.". Now fix such an $\varepsilon$, call it $c$, and apply the non uniform continuity with $\delta = 1/n$. Nov 13, 2020 at 15:01
• @TheSilverDoe I edited the negation. How does it look now? Nov 13, 2020 at 15:02
• @ManCheese Take time to do it carefully : your $\delta$ is not introduced. Nov 13, 2020 at 15:03

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. $$|x_n-y_n|\leq \frac{1}{n}\quad \text{and}\quad |f(x_n)-f(y_n)|>\varepsilon .$$