Cauchy sequence and uniform continuity on $\mathbb{R}$

Question

I was trying this T\F type question

A continuous function $f:\mathbb{R}\to \mathbb{R}$ is uniformly continuous if it maps Cauchy sequences into Cauchy sequences.

The answer given is False. I know that $x^2$ is not uniformly continuous on $\mathbb{R}$ so i thought let's try with this. $$|x_n^2-x_m^2|=|x_n-x_m||x_n+x_m|<\epsilon$$ if $$|x_n-x_m|<\frac{\epsilon}{|x_n+x_m|}$$ Now I'm stuck at what to take $\delta$. I don't know how to proceed. Thanks!!!

Answer 1

To show that $(x_n^2)$ is also cauchy note that a cauchy sequence is bounded in $\mathbb{R}$ so let $M$ be the bound, then $|x_n+x_m|< 2M$ So for $\epsilon/{2M}$, there is a $N$, st for $n,m>N$ $|x_n-x_m|<\epsilon/{2M}$, therefore $|x_n^2-x_m^2|<|x_n-x_m|(2M)<\epsilon$

Answer 2

Without division (else the case where $x_n\to 0$ becomes awkward). It helps to know that a Cauchy sequence is a bounded sequence: If $(x_n)_{n\in \mathbb N}$ is Cauchy then the set $\{x_n: n\in \mathbb N\}$ is a subset of $[-s,s]$ for some $s>0.$

Because there exists $n_1$ such that for all $n>n_1$ we have $|x_n-x_{n_1}|<1$, so for all $n>n_1$ we have $|x_n|<1+|x_{n_1}|$ . And let $A=\max \{x_m|:m\leq n_1\} .$ Let $s=\max (A,1+|x_{n_1}|).$ Then for all $n$ we have $|x_n|\leq s.$ $$ \text {Therefore }\;\;|x_m^2-x_n^2|=|x_m-x_n|\cdot |x_m+x_n|\leq |x_m-x_n|\cdot (|x_m|+|x_n|)\leq |x_m-x_n|\cdot 2s.$$ So if $|x_m-x_n|<\epsilon /2s$ then $|x_m^2-x_n^2|<\epsilon.$