Given a sequence $a_n$ s.t. forall $\varepsilon>0$ exists $m,n_{0} \in \mathbb{N}$ s.t. $\forall n \geq n_0$ $|a_n-a_m|<\varepsilon$

Question

So I have that question:

Suppose a sequence $a_n$ satisfies that forall $\varepsilon>0$ exists $m,n_{0} \in \mathbb{N}$ s.t. $\forall n \geq n_0$ $|a_n-a_m|<\varepsilon$ determine if that sequence is a cauchy sequence.

I think it's not necessarily cauchy because the constant $a_m$ depends on $\varepsilon$, but I really struggle to give a counter example, i will be glad if someone will help me to think about a counter example (or maybe tell me that I'm wrong)

thanks in advance.

Answer 1

It is correct that $m$ depends on $\varepsilon$, but that condition actually does imply that $(a_n)$ is a Cauchy sequence. Roughly speaking: If $a_n$ and $a_k$ are both close to some $a_m$ then they are close to each other. Formally, this is the triangle inequality:

Given $\varepsilon > 0$ there exist $m,n_{0} \in \mathbb{N}$ such that $|a_n-a_m|<\varepsilon/2$ for all $n \ge n_0$. Then $$ |a_n - a_k| \le |a_n-a_m| + |a_m-a_k| < \epsilon $$ for all $n, k \ge n_0$, so that the Cauchy condition is satisfied.