Prove that $\mathbb{R}$ with this metric is complete

Question

Prove that $\mathbb{R}$ endowed with the metric $d(x,\ y)=|x^3-y^3|$ is complete.

My attempt:

Let $\{x_n\}$ be a Cauchy sequence in $(\mathbb{R},\ d)$. We must prove that $x_n\to x$ where $x\in\mathbb{R}$.

The main problem I face is that I don't know what $x$ is. If I did, I could try proving that $d(x_n,\ x)\to 0$.

Answer 1

Let $(x_n)_{n \in \mathbb{N}}$ be a Cauchy sequence for the metric $d$, and let's consider the sequence $(y_n)_{n \in \mathbb{N}}$ defined by $y_n = x_n^3$.

$(x_n)_{n \in \mathbb{N}}$ is a Cauchy sequence, so you have $$\forall \varepsilon > 0, \exists N \in \mathbb{N}, \forall p,q \geq N, |x_p^3 - x_q^3| \leq \varepsilon$$

i.e. $$\forall \varepsilon > 0, \exists N \in \mathbb{N}, \forall p,q \geq N, |y_p - y_q| \leq \varepsilon$$

So $(y_n)_{n \in \mathbb{N}}$ is a Cauchy sequence for the usual metric on $\mathbb{R}$, therefore it converges. Let $y$ be its limit.

We have then that $$\forall \varepsilon > 0, \exists N \in \mathbb{N}, \forall n \geq N, |y_n-y| \leq \varepsilon$$

so $$\forall \varepsilon > 0, \exists N \in \mathbb{N}, \forall n \geq N, |x_n^3-\left(y^{\frac{1}{3}}\right)^3| \leq \varepsilon$$

So you get the definition of the convergence, for the metric $d$, of the sequence $(x_n)_{n \in \mathbb{N}}$ to its limit $y^{\frac{1}{3}}$.

So your topological space is complete, as this proof.

Answer 2

Suppose that $(x_n)_n$ is a Cauchy sequence. It's bounded in the usual metric, so you may find some closed and bounded interval (in usual sense) $[-M,M]$ so that $(x_n)_n \subseteq [-M,M]$, from which you may find a convergent subsequence (still in the usual Euclidean metric) $(x_{n_k})_k$ so that $x_{n_k} \to x$. Check that $x_{n_k} \overset{d}{\to} x$.

Fix $M' = \max_{(x,y) \in [-M,M]^2} |x^2+xy+y^2|$. (Well defined since $(x,y) \mapsto |x^2+xy+y^2|$ is continuous.) Then $$d(x_{n_k},x) = |x_{n_k}^3-x^3| = |x_{n_k}-x||x_{n_k}^2+x_{n_k}x+x^2| \le M'|x_{n_k}-x|\xrightarrow[k \to \infty]{} 0.$$

Since $(x_n)_n$ is Cauchy, the fluctuation of the tail of $(x_n)_n$ can be made arbitrarily small. This forbids $(x_n)_n$ from having two limit points. Therefore, $x_n \overset{d}{\to} x$.