Let $d(x,y) = |\phi(x) - \phi(y)|$, where $\phi(x) = x / (1+|x|).$ Show that $d$ is a metric on $\mathbb{R}$ that is not complete.

Question

Question: Let $d(x,y) = |\phi(x) - \phi(y)|$, where $\phi(x) = x / (1+|x|).$ Show that $d$ is a metric on $\mathbb{R}$ that is not complete.

There is another similar question here, but no proof is given. Here I provide a full proof to the question.

My attempt:

we wish to obtain a Cauchy sequence $(x_n)_{n \in \mathbb{N}}$ that does not converge in $\mathbb{R}.$

For all $n \in \mathbb{N},$ let $x_n = n.$

For any $\varepsilon>0,$ choose $N \in \mathbb{N}$ such that $\frac{1}{N} < \varepsilon.$
So, for any $m>n\geq N,$ we have $$d(x_n,x_m) = \left| \frac{x_n}{1+x_n} - \frac{x_m}{1+x_m} \right| = \frac{|x_n - x_m|}{(1+x_n)(1+x_m)} \leq \frac{1}{n} - \frac{1}{m} \leq \frac{1}{n} \leq \frac{1}{N} < \varepsilon.$$ Therefore, $(x_n)_{n \in \mathbb{N}}$ is Cauchy.

We claim that $(x_n)_{n \in \mathbb{N}}$ 'converges' to $\infty.$ For any $\varepsilon>0,$ choose $N \in \mathbb{N}$ such that $\frac{1}{N} < \varepsilon.$ For any $n \geq N,$ we have $$d(x_n,\infty) = \left| \frac{x_n}{1+x_n} - 1 \right| = \left| \frac{1}{1+n} \right| \leq \frac{1}{n} \leq \frac{1}{N} < \varepsilon.$$ It follows that $(x_n)_{n \in \mathbb{N}}$ 'converges' to $\infty.$ Since $\mathbb{R}$ is Hausdorff, limit of a sequence is unique. However, $\infty \not\in \mathbb{R}.$ Therefore, $d$ is not complete on $\mathbb{R}.$

Is my proof correct?

Answer 1

If you have to prove that a Cauchy sequence $(x_n)_n\subset (X,d)$ has no limit in $X$, than you have to argue that $d(x_n,x)\not\to 0$ for all $x\in X$.

You cannot compare the sequence to an outside point of $X$, because your metric is defined on $X\times X$. Since $\infty\notin\mathbb{R}$, your computation of $d(x_n,\infty)$ is not defined.

(Although you could complete $(\mathbb{R},d)$, the additional points are abstract in first place and it is not nessessary.)

Here you have to prove, that each $x\in\mathbb{R}$ cannot be the limit point of $(x_n)_n$. This can be seen by $$ d(x_n,x)=\left|\frac{x_n}{1+|x_n|}-\frac{x}{1+|x|}\right|\to \left|1-\frac{x}{1+|x|}\right|>0. $$ Since $x$ was arbitrary, no point of $\mathbb{R}$ is the limit of $(x_n)_n$ and $(x_n)_n$ doesn't converge in $(\mathbb{R},d)$.