Prove that $d(x_n,x_{n+1})→0$ $\iff$ $(x_n)$ ia a Cauchy sequence.

Question

Let $X\neq\varnothing$ and let $d:X\times X→\mathbb{R}$ be a function such that

1. $$d(x,y)=0 \iff x=y,$$

2. $$d(x,y)=d(y,x),$$

3. $$d(x,z)≤\max⁡\{d(x,y),d(z,y)\}.$$

Let us say that $d$ is a metric. Let $(x_n)$ be a sequence in $X$. Prove that $$\lim_{n\to\infty}d(x_n,x_{n+1})=0$$ if and only if $(x_n)$ is a Cauchy sequence.

I proved that if $(x_n)$ is a Cauchy sequence then $d(x_n,x_{n+1})\to0$ as $n\to\infty$, but I could not prove that if $d(x_n,x_{n+1})\to0$ as $n\to\infty$ then $(x_n)$ is a Cauchy sequence. I know that it is not true for arbitrary spaces. I tried to prove that just for this special metric.

Answer 1

Let $\epsilon > 0$ be given. We wish to find $N$ such that $m, n > N$ implies $d(x_m, x_n) < \epsilon$.

We know we can find $N$ such that $n > N$ implies $d(x_n, x_{n+1}) < \epsilon$.

I claim this $N$ suffices. Indeed, if $m, n > N$ are arbitrary, WLOG $n > m$. Now \begin{align} d(x_n, x_m) \leq & \max\{d(x_n, x_{n+1}),\ d(x_{n+1}, x_m)\} \\ \leq & \max\{\epsilon,\ d(x_{n+1}, x_m)\} \\ \leq& \max\{\epsilon, \max\{d(x_{n+1}, x_{n+2}), d(x_{n+2}, x_m)\}\} \\ \leq& \cdots\\ \leq & \max\{\epsilon, \epsilon, \ldots, d(x_{m-1}, x_m)\} \\ \leq &\ \epsilon \end{align} as desired. (Note that I was sloppy about strict inequality; it doesn't matter because you can pick $N$ to make the distances less than $\epsilon/2$, or just track the strict inequalities more carefully.)