Is $\mathbb{R}$ complete under the metric $d(x,y) = \lvert x^3 - y^3 \rvert$?

Question

Looking at the graph of $y = x^3$ I feel like the answer is yes, but I can't figure out how to prove it. I've tried proving the contrapositive, that if a sequence doesn't converge it's not cauchy, but I can't figure it out.

Answer 1

Let $(x_n)$ be Cauchy in $(\mathbb{R},d).$ Then, $(x_n^3)$ is Cauchy in $(\mathbb{R},|\cdot|),$ so it converges to some $x'$ (as this space is complete). Let $x=(x')^{1/3}.$ Then, $$d(x_n,x)=|x_n^3-x'|\rightarrow 0.$$ So, yes, it's complete.

Answer 2

Your metric function $\ d\ $ is indeed a metric in $\ \mathbb R.\ $ Your metric space $\ \mathbf R' := (\mathbb R\ d)\ $ is isometric to the standard 1-dim Euclidean space $\ \mathbf R\ :=\ (\mathbb R\ \rho)\ $ where $\ \rho(x\ y):=|x-y|\ $ for every $\ x\ y\in\mathbb R $.

It follows that your space $\ \mathbf R'\ $ is complete.