# Is this sequence necessarily Cauchy?

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?

• I'm not so sure. I think we can manipulate the harmonic series to fit the criteria. At any rate if k,l > M, it think (but don't know) the best you can do is $|a_k - a_l| < 1/m + 1/(m+1) + .... 1/(m+k-l)$ for some m, and that isn't cauchy. – fleablood Nov 16 '15 at 6:13
• @fleablood Yes I came across the same thing, but as you've pointed out it doesn't lead anywhere... – user118224 Nov 16 '15 at 6:38