Criterion equivalent to a sequence being Cauchy?

Question

Everyone knows the usual Cauchy criterion for metric spaces: $(x_n)_n$ is Cauchy if $$\forall \epsilon>0,\,\exists N \text{ such that }\forall n,m\geq N, d(x_n,x_m)\leq \epsilon.$$

Instead, I am interested in sequences $(x_n)_n$ that satisfy the following condition (C): for every sequences $\phi,\psi:\mathbb{N}\to\mathbb{N}$ satisfying $\lim_n \phi(n)=\lim_n\psi (n)=+\infty$, we have $$\lim_n d(x_{\phi(n)},x_{\psi(n)})=0.$$

Obviously, every Cauchy sequence satisfies (C). It is very natural to expect that (C) is actually equivalent to Cauchy criterion, but I haven't succeeded in proving it (except in some easy cases that I will mention bellow). My question: is this equivalence true or is there a wild counterexample?

For real sequences the equivalence holds, because we can just take $\phi$ and $\psi$ that realise $\liminf x_n$ and $\limsup x_n$. Also, it holds when the metric space is compact.

Answer 1

Suppose that $(x_n)_{n\in\mathbb N}$ is not a Cauchy sequence. There is then a $\varepsilon>0$ such that, for every $p\in\mathbb N$, there are $m,n\in\mathbb N$ such that $m,n\geqslant p$ and that $d(x_m,x_n)\geqslant\varepsilon$. Well, take $\phi(1),\psi(1)\in\mathbb N$ such that $d\bigl(x_{\phi(1)},x_{\psi(1)}\bigr)\geqslant\varepsilon$. Now, take $p_2>\max\{\phi(1),\psi(1)\}$. Then there are $\phi(2),\psi(2)\geqslant p_2$ such that $d\bigl(x_{\phi(2)},x_{\psi(2)}\bigr)\geqslant\varepsilon$. And so on. Like this, we can construct two strictly increasing functions from $\mathbb N$ into $\mathbb N$ for which it is clear that $\lim_n d\bigl(x_{\phi(n)},x_{\psi(n)}\bigr)$ cannot be $0$. Since they are strictly increasing, it is clear that $\lim_n\phi(n)=\lim_n\psi(n)=+\infty$.

Answer 2

Let it be that $(x_n)_n$ is not a Cauchy sequence.

Then some $\epsilon>0$ exists such that for every $n$ we can find $k,m\geq n$ with $d(x_k,x_m)\geq\epsilon$.

That gives possibilities to find sequences $\phi(n)$ and $\psi(n)$ with $d(x_{\phi(n)},x_{\psi(n)})\geq\epsilon>0$ for every $n$.