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.


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?

  • $\begingroup$ Check your negation again; there are no numbers $f(x)$ and $f(y)$ such that $|f(x) - f(y)| \ge \varepsilon$ for all $\varepsilon > 0$. $\endgroup$
    – user847970
    Nov 13, 2020 at 15:00
  • 2
    $\begingroup$ 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 $\delta = 1/n$. $\endgroup$ Nov 13, 2020 at 15:01
  • $\begingroup$ @TheSilverDoe I edited the negation. How does it look now? $\endgroup$ Nov 13, 2020 at 15:02
  • $\begingroup$ @ManCheese Take time to do it carefully : your $\delta$ is not introduced. $\endgroup$ Nov 13, 2020 at 15:03

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


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