Prove that $f$ is not uniformly continuous if and only if there is a ε > 0 and sequences Prove that $f : \mathbb{R}\rightarrow \mathbb{R}$ is not uniformly continuous if and only if there is a $\varepsilon > 0$ and sequences $\{x_n\}$ and $\{y_n\}$ in $\Bbb{R}$ such that $|x_n-y_n|<\dfrac{1}{n}$ and $|f(x_n)-f(y_n)| \geqslant \varepsilon$ for any $n\in \mathbb{N}$
I do not know how to get the suceciones...
 A: A function $f$ is uniformly continuous on $\mathbb{R}$ iff $$\forall \epsilon >0\, \exists  \delta>0\, \forall u\in\mathbb{R}\, \forall v\in\mathbb{R}\Big[|u-v|<\delta \space \implies \space |f(u)-f(v)|<\epsilon\Big].$$
Hence the negation, a function $f$ is not uniformly continuous on $\mathbb{R}$ iff $$\exists \epsilon >0\, \forall  \delta>0\, \exists u\in\mathbb{R}\, \exists v\in\mathbb{R}\Big[|u-v|<\delta \space \& \space |f(u)-f(v)|\ge\epsilon\Big].$$
Notice that this should be true for all $\delta>0$ (There is an epsilon such that you give me any positive real number and I'll be able to give you $u\, \& v$ with the satisfying property). So you choose $\delta_n:= \frac{1}{n}$ which is positive for all $n.$ So now we have that a function $f$ is not uniformly continuous on $\mathbb{R}$ if $$\exists \epsilon >0\, \forall  \frac{1}{n},n\in\mathbb{N}\,\, \exists u\in\mathbb{R}\, \exists v\in\mathbb{R}\Big[|u-v|<\frac{1}{n} \space \& \space |f(u)-f(v)|\ge\epsilon\Big].$$
But for each $n$ you are getting a pair. For $n=1$ you get $u_1,v_1$....for $n=2$ you get $u_2,v_2$.... and so forth and you end up getting sequences $\{u_n\}$ and $\{v_n\}.$ Hence $f : \mathbb{R}\rightarrow \mathbb{R}$ is not uniformly continuous if there is an $\varepsilon > 0$ and sequences $\{u_n\}$ and $\{v_n\}$ in $\Bbb{R}$ such that $|u_n-v_n|<\dfrac{1}{n}$ and $|f(u_n)-f(v_n)| \geqslant \varepsilon$ for any $n\in \mathbb{N}$
