Showing that a sequence is Cauchy w.r.t. $d(x, y) = \left|e^{x} - e^{y}\right|$

Question

I am trying to demonstrate that a metric space $X$ is not complete with the metric $d(x, y) = \left|e^x - e^y\right|$. To this end I am first trying to show that a sequence $x_n = \frac{1}{n}$ is Cauchy w.r.t. the said metric. My current mental block is with the "final touch", where one shows that the distance between any two elements of the sequence is at most $\epsilon$. Specifically, let $n_\epsilon > \left|\frac{1}{\ln(\epsilon)}\right|$. Then, for $l, k \geq n_\epsilon$ it holds that $\left|e^{x_l} - e^{x_k}\right| = \left|e^{\frac{1}{l}} - e^{\frac{1}{k}}\right| < \left|e^{\frac{1}{\min\{l, k\}}}\right| \leq \left|e^{\frac{1}{n_\epsilon}}\right| < \left|e^{\left|\ln(\epsilon)\right|}\right|$.

This might seem as an elementary step to most, but how to you go about with the absolute value around the logarithm in the exponent?

Answer 1

Your approach won't work since $e^{1/n} > e^0 = 1$ for any $n \in \mathbb{N}$, and we can find no bound $e^{1/n_\epsilon} < \epsilon <1$.

Instead, for $l > k > n$, we have

$$\left|e^{\frac{1}{l}} - e^{\frac{1}{k}} \right|= e^{\frac{1}{k}}(1 - e^{\frac{1}{l}- \frac{1}{k}}) \leqslant e \cdot \left(1 - e^{-\left(\frac{1}{k}- \frac{1}{l}\right)} \right)$$

Since $e^{-x} \geqslant 1 - x $ for $0 \leqslant x < 1$, it follows that with $n > 1/(e\epsilon)$,

$$\left|e^{\frac{1}{l}} - e^{\frac{1}{k}} \right|\leqslant e \left(\frac{1}{k} - \frac{1}{l}\right) < \frac{e}{k}< \frac{e}{n}< \epsilon$$

Answer 2

Actually, the sequence $(1/n)_{n \geq 1}$ is Cauchy in $\mathbb{R}$, and $\exp$ is uniformly continuous over $[0,1]$, so the sequence $(\exp(1/n))_{n \geq 1}$ is also Cauchy in $\mathbb{R}$, which means that $(1/n)_{n \geq 1}$ is Cauchy for your metric $d$.