Find a counter-expamle to $\lim_n \limsup_m d(a_n, a_m) =0 \implies (a_n)_n \ \text{is cauchy}.$ Let $(a_n)$ be a sequence in a $d$-metric space. I want to find a counter-example to the statement $$\lim_n \limsup_m d(a_n, a_m) =0 \implies (a_n) \ \text{is Cauchy}.$$
I know that $\lim_n \limsup_m d(a_n, a_m) =0$ is equivalent to $\limsup_n \limsup_m d(a_n, a_m) =0$, so the given condition is symmetric in $n$ and $m$. Any help is appreciated.

There is no counter-example.
Given $\epsilon>0$, pick $n$ such that $\limsup_m d(a_n,a_m)<\epsilon/2$. We may then take $N\in\mathbb{N}$ such that $d(a_n,a_m)<\epsilon/2$ for all $m\ge N$. This implies that for all $m_1, m_2\ge N$, $d(a_{m_1},a_{m_2})\le d(a_n,a_{m_1})+d(a_n,a_{m_2})<\epsilon$.
