Is $\mathbb{R}$ complete when endowed with the metric $d (x,y) = |e^x- e^y|$?

Question

Let $\mathbb{R}$ be endowed with the metric $d (x,y) = |e^x- e^y|$. Is $( \mathbb{R}, d)$ complete?

My attempt: Yes. Proof: $d(x_n, x_m) = |e^{-n} - e^{-m} | \le |e^{-n} | + |e^{-m}| = e^{-n} + e^{-m}$ Now we can choose N such that $e^{-N} < \frac{\epsilon}{2}.$ Then for all $ n, m\ge N . $

Now we have $e^{-n} + e^{-m} \le e^{-N} + e^{-N} < \frac{\epsilon}{2} + \frac{\epsilon}{2}= e $ that is we can say $x_n$ is cauchy so $( \mathbb{R}, d)$ is compelete

Answer 1

So here's a simple argument: consider

$$f:\mathbb{R}\to\mathbb{R}^+$$ $$f(x)=e^x$$

with $d$ on the left side and the standard Euclidean metric $|\cdot|$ on the right side. Note that $f$ is an isometry and so the left side is not complete because the right side isn't (sequences convergent to $0$ are Cauchy but the limit is not in $\mathbb{R}^+$).

---

An alternative approach is an extension of what you did so far. Consider $x_n=-n$. You've already established that $x_n$ is Cauchy. But it is not convergent! Indeed, assume that $x_n\to x$ in $d$. Then $d(x_n,x)=|e^{-n}-e^x|$ has to converg to $0$ (in the Euclidean metric). But $e^{-n}\to 0$ and so $d(x_n,x)\to e^x\neq 0$. Contradiction.