Find if the metric space $\mathbb R$ with $d(x,y) = |e^x - e^y|$ is complete

Question

Consider the metric space $(\mathbb R,d)$, where $d(x,y) = |e^x −e^y|$, $\forall x,y\in\mathbb R$. Is the metric space $(\mathbb R,d)$ complete?

I understand the definition of completeness but am unable to prove this?

Can you provide me with a counterexample, I understand it is not complete.

Answer 1

Consider the sequence $x_n = -n$. Compute that $$\lim_{n,m \rightarrow \infty} |e^{x_n} - e^{x_m}| = \lim_{n,m \rightarrow \infty}|e^{-n} - e^{-m}| \leq \lim_{n,m \rightarrow \infty} e^{-n} + e^{-m} = 0,$$ So $(x_n)$ is Cauchy. But $(x_n)$ is not convergent. To see this, suppose toward a contradiction that $\lim_{n \rightarrow \infty} |e^x - e^{-n}| = 0$. But then it follows that $e^x = 0$, which is absurd.

Answer 2

User Levi has answered this question.

A bit formal:

1) $x_n =\log n$ is Cauchy.

Let $m \ge n$.

$|e^{-\log m} -e^{-\log n}|=|1/m-1/n| <1/m+1/n \le 2/n;$

$\epsilon >0$;

Let $n_0>2/\epsilon.$

For $m \ge n \ge n_0 $ we have

$|e^{x_m}-e^{x_n}| < 2/n \le 2/n_0 <\epsilon$.

2) Assume $x \in \mathbb{R}$ is the limit.

For $\epsilon >0$

there is a $n_1$ s.t. for $n \ge n_1$

$|e^{-x}-1/n| < \epsilon.$

$\lim_{n \rightarrow \infty}|e^{-x}-1/n| =e^{-x} \lt \epsilon$;

Since $\epsilon$ is arbitrary it follows $e^{-x} = 0,$ a contradiction.