Is this sequence necessarily Cauchy?

Question

Let $(X,d)$ be a metric space and $(x_n)$ a sequence in it such that for all $i,j,n\in\mathbb{N}_{>0}:$

$$d(x_i,x_j)<n+1\Rightarrow d(x_{i+1},x_{j+1})<n$$ $$d(x_i,x_j)<\frac{1}{n}\Rightarrow d(x_{i+1},x_{j+1})<\frac{1}{n+1}$$ Must $(x_n)$ be Cauchy?

I think it is but I am having trouble proving it. I have proved the following things:

• $d(x_{i+n},x_{j+n})\to 0\quad \forall i,j$. This is a consequence of $d(x_n,x_{n+1})\to 0$. In other words for any $k$, "gaps of length $k$ approach $0$".

• for sufficiently large $i$, $d(x_i,x_{i+n})$ is bounded (and hence has a convergent subsequence). In other words, arbitrarily large gaps cannot increase without bound provided we are "far enough in the sequence".

Is it possible to use any of this to show $(x_n)$ is Cauchy?

Answer 1

Here is a little table I drew

I am sorry for the quality.

The order is irrelevant because $d(x_i, x_j)=d(x_j,x_i)$. Basically, every element in the column $i$, is bounded by the sequence $\left \lceil d(x_1,x_i) \right \rceil + 1$, $\left \lceil d(x_1,x_i) \right \rceil$, $\left \lceil d(x_1,x_i) \right \rceil - 1$, ..., $1$, $\frac{1}{2}$, ... which goes to $0$ as long as $\left \lceil d(x_1,x_i) \right \rceil$ is bounded. And here is the result ...

Sequence $\left \{ x_n \right \}$ is Cauchy iff sequence $\left \{d(x_1,x_n) \right \}_{n=1}^{\infty }$ is bounded.

Well, basically every Cauchy sequence is bounded (https://proofwiki.org/wiki/Cauchy_Sequence_is_Bounded), and so is $\left \{ d(x_1,x_n) \right \}_{n=1}^{\infty }$.

And if $\left \{ d(x_1,x_n) \right \}_{n=1}^{\infty }$ is bounded, then for each column $i$, $$\lim_{j \to \infty } d(x_{1+j},x_{i+j})=0$$ or replace $j$ with $n$ and $$\lim_{n \to \infty } d(x_{1+n},x_{i+n})=0$$ and replace $i$ with $p$ and $\forall p$ $$\lim_{n \to \infty } d(x_{1+n},x_{p+n})=0$$ just to make it looking in the classical form. Then $$d(x_{n},x_{p+n}) \leq d(x_{n},x_{1+n}) + d(x_{1+n},x_{p+n}) = d(x_{1+(n-1)},x_{2+(n-1)}) + d(x_{1+n},x_{p+n})$$

• $d(x_{1+(n-1)},x_{2+(n-1)})$ is from the first column.
• $d(x_{1+n},x_{p+n})$ is from the p-th column.

both small enough. And the final step, because $\left \{ d(x_1,x_n) \right \}_{n=1}^{\infty }$ is bounded, there $\exists M \in \mathbb{N}$ such that: $$d(x_{1},x_{2}) < M, \space d(x_{1},x_{p}) < M$$ $$d(x_{1+1},x_{2+1}) < M-1, \space d(x_{1+1},x_{p+1}) < M-1$$ $$...$$ $$d(x_{1+M-1},x_{2+M-1}) < 1, \space d(x_{1+M-1},x_{p+M-1}) < 1$$ $$d(x_{1+M},x_{2+M}) < \frac{1}{2}, \space d(x_{1+M},x_{p+M}) < \frac{1}{2}$$ $$d(x_{1+M+1},x_{2+M+1}) < \frac{1}{3}, \space d(x_{1+M+1},x_{p+M+1}) < \frac{1}{3}$$ $$d(x_{1+M+k},x_{2+M+k}) < \frac{1}{2+k}, \space d(x_{1+M+k},x_{p+M+k}) < \frac{1}{2+k}$$ $$...$$ $$d(x_{1+n-1},x_{2+n-1}) < \frac{1}{2+n-1-M}, \space d(x_{1+n-1},x_{p+n-1}) < \frac{1}{2+n-1-M}$$ $$d(x_{1+n},x_{2+n}) < \frac{1}{2+n-M}, \space d(x_{1+n},x_{p+n}) < \frac{1}{2+n-M}$$ Which basically means $$d(x_{n},x_{p+n}) < \frac{1}{2+n-1-M} + \frac{1}{2+n-M} < \frac{2}{2+n-1-M}$$ which is independent of $p$.