Cauchy sequence and completeness of $\mathbb{R}$ equipped with the following metric

Question

Let $d: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ such that $d(x,y) = |e^{-x} - e^{-y}|$.

a) Show that $x_n=n^2, n \in \mathbb{N}$ is Cauchy.

b) Is $(\mathbb{R},d)$ complete?

So for a), let $\varepsilon > 0$ and $n,m > N > \frac{2}{\varepsilon}$. Then $$d(x_n,x_m) = |e^{-n^2} - e^{-m^2}| \leq \frac{1}{n} - \frac{1}{m} \leq \frac{1}{n} + \frac{1}{m} \leq \frac{2}{N} < \varepsilon.$$

Correct me if I am wrong. I also think that this metric space is not complete because if $x_n$ would converge shouldn't the limit be the same one as in $\mathbb{R}$ (which we know does not exist)?

Answer 1

Yeah you're right the metric space isn't complete. Take $x_n = n^2$ for example. You've already proven that it's Cauchy. Now assume that $x_n \to L \in \mathbb{R}$. Then we have $\forall \epsilon >0$ $\exists n_0 \in \mathbb{N}$ s.t. $n \ge n_0 \implies d(x_n,L) < \epsilon \implies |e^{-n^2} - e^{-L}| < \epsilon$. Take the limit $n \to \infty$ from both sides. Then we have:

$$\epsilon > |\lim_{n \to \infty} e^{-n^2} - e^{-L}| = |0 - e^{-L}|$$

As this is true for all $\epsilon > 0$ we have that $e^{-L} = 0$. But this is impossible as such value for $L$ doesn't exist.

Answer 2

No. This is a different distance. What makes you think that the limit should be the same?

Note that if $D$ is the usual distance in $\mathbb R$, then the function $f\colon(\mathbb{R},d)\longrightarrow\bigl((0,+\infty),D\bigr)$ defined by $f(x)=e^{-x}$ is an isometry. Since $\bigl((0,+\infty),D\bigr)$ is not complete, it follows from this fact that $(\mathbb{R},d)$ is not complete.

Answer 3

The first point may be correct if you better explain the inequalities involved.

About the second part, the answer is not. Actually a metric space is complete if every Cauchy sequence converge to a point of the space. So, considering $n^2$, and assuming it converge to $x \in \mathbb{R}$, then we have $$ \lim_{n \to \infty} d(n^2,x) = 0 \iff \lim_{n \to \infty} |e^{-x} - e^{-n^2}| = 0$$ but we have that $$ |e^{-x} - e^{-n^2}| \geq |e^{-x}| + |e^{-n^2}| \geq |e^{-x}| $$ that is a contraddiction