cauchy sequence on $\mathbb{R}$

Question

i want to show that $\mathbb{R}$ with the following metric : $d_1(x,y)=|x^3-y^3|$ is complete. I think a good way to show it is to show that a sequence which is Cauchy for $d_1$ will also be Cauchy for the usual metric $d(x,y)=|x-y|$ but i'm not able to write it properly. More precisely i want to write $$|x^3-y^3|=|x-y||x^2+xy+xy^2|$$

and thus get $d(x,y)=\frac{d_1(x,y)}{|x^2+xy+y^2|}$ but i'm not able to ensure that if $d_1(x_m,x_n)\leq \varepsilon$ for $n$ and $m$ sufficiently large, then $d(x_m,x_n)\leq \varepsilon$ also.

Answer 1

If $\left ( x_n \right )_{n\geq 1} $ is Cauchy in $d_1$ then $\left ( x_n ^{\frac {1}{3}} \right )_{n\geq 1}$ is Cauchy in $d$, and that's basically all you need.

Answer 2

Let $\epsilon >0$.

We have $d_1(x,y)=|x^3-y^3|$. Suppose $\left \{ x_n \right \}_{n\geq 0}$ is Cauchy in $d_1$. Then, $|x_n^3-x_m^3|<\epsilon $ for all $n,m$ large enough. But this says now that the sequence $\left \{ x^3_n \right \}_{n\geq 0}$ is Cauchy in $d$, the usual metric. Therefore, there is a $y\in \mathbb R$ such that $x^3_n\rightarrow y$. Write $y$ as $x^3$ for some $x\in \mathbb R$ (it is always possible to do this since $t\mapsto t^3$ is bijective) and now note that we have $d_1(x_n,x)=|x_n^3-x^3|\rightarrow 0$ which is what we want.

Answer 3

A sequence that is Cauchy in $d_1$ is Cauchy in $d$. See here An Algebraic Proof that $|y^3 - x^3| \ge |(y - x)|^3/4 $ for a proof that $|x_n^3 - x_m^3| \lt \epsilon \implies |x_n - x_m| \lt (4\epsilon)^{1/3}$.

Therefore the sequence has a limit $x$ in $d$ (because $\mathbb{R}$ is complete with $d$) and if $|x_n - x| \to 0$ then so does $|x_n^3 - x^3| $, so the Cauchy sequence has a limit in $d_1$.