Prove or disprove that $(a_n)_{n=1}^{\infty}$ is Cauchy $\iff$ $\displaystyle\inf_{n \ge 1}{\sup_{k,l \ge n}{|a_k - a_l|}} = 0$ (How to state that a sequence is Cauchy in terms of $\limsup$ and $\liminf$? For the above post by me, I have this new claim and its unfinished proof, but I am not sure should I edit that old question to include the new claim and unfinished proof or should I answer by myself with the new claim and unfinished proof, so I ask a new question here, and editing this post to improve the format and content is welcome.) 
Claim. $(a_n)_{n=1}^{\infty}$ is Cauchy $\iff$ $\displaystyle\inf_{n \ge 1}{\sup_{k,l \ge n}{|a_k - a_l|}} = 0$

(Unfinished Proof)
Denote $\displaystyle a_n^+ := \sup_{k,l \ge n}{|a_k - a_l|}$
We have to show that: $(a_n)_{n=1}^{\infty}$ is Cauchy $\iff$ $\displaystyle\inf_{n \ge 1}{a_n^+} = 0$
($\Leftarrow$) Given $\epsilon \gt 0$, since $\epsilon$ is not a lower bound of  $(a_n^+)$, there is a $a_N^+$ such that $a_N^+ \lt \epsilon$, that is, $\displaystyle\sup_{k,l \ge N}{|a_k - a_l|} \lt \epsilon$. Hence $|a_k - a_l| \lt \epsilon$ for all $k,l \ge N$. Therefore $(a_n)_{n=1}^{\infty}$ is Cauchy
($\Rightarrow$) (i) Assume $\displaystyle\inf_{n \ge 1}{a_n^+} \lt 0$. Then $\exists N \ge 1$ such that $a_N^+ \lt 0$, that is $\displaystyle\sup_{k,l \gt N}{|a_k - a_l|} \lt 0$, hence $|a_k - a_l| \lt 0$ for all $k,l \gt N$. Particularly we have $0 \le |a_{N+1}-a_N| \lt 0$, hence $0 \lt 0$, a contradiction. (ii) Assume $\displaystyle\inf_{n \ge 1}{a_n^+} \gt 0$. This means $\displaystyle a_n^+ = \sup_{k,l \ge n}{|a_k - a_l|} \gt 0$ for all $n \ge 1$. Hence for arbitrary $n \ge 1$,......

So I am stuck here. Could someone help giving a formal proof or disprove this claim?
 A: An inf of numbers that are $\ge 0$ cannot be $<0$, as $0$ is a lower bound for all terms, and $\inf$ is the largest one, so at least $0$ as well.
And if $(a_n)$ is Cauchy, let $\varepsilon > 0$. Apply the definition of Cauchy to find $N$ such that $n,m \ge N$ implies $|a_n - a_m| < \frac{\varepsilon}{2}$.
Then $(a_N)^{+} \le \frac{\varepsilon}{2} < \varepsilon$ (all terms are bounded by it, so their sup is as well, as sup is the least upper bound). So $\varepsilon$ is not a lower bound for all $a_n^+$. As $\varepsilon > 0$ was arbitrary..  
A: For $(\Rightarrow)$, it is easier to argue directly. Assume that $\{a_n\}$ is Cauchy. Let $\epsilon >0$. Then there is $N\in \mathbb N$ so that 
$$|a_n -a_m|<\epsilon$$
for all $n, m \ge N$. By definition, this means $a_N^+ <\epsilon$ ans so 
$$\inf_{n\ge 1} a_n^+ \le \epsilon.$$
As $\epsilon >0$ is arbitrary, we have 
$$\inf_{n\ge 1} a_n^+ \le 0$$
but $a_n^+ \ge 0$ by definition, so 
$$\inf_{n\ge 1} a_n^+ = 0.$$
A: Put $\sup_{k,\> l\geq n}|a_k-a_l|=:\Delta_n\geq0$.
Assume that $\inf_n\Delta_n=0$, and that an $\epsilon>0$ is given. There is an $n$ such that $\Delta_n<\epsilon$, and this implies that $|a_k-a_l|\leq\Delta_n<\epsilon$ for all $k$, $l\geq n$. As $\epsilon>0$ was arbitrary this proves that the sequence $(a_k)_{k\geq0}$ is Cauchy. 
This part you had alright. Now for the other part:
Assume that the sequence $(a_k)_{k\geq0}$ is Cauchy, and that an $\epsilon>0$ is given. There is an $n$ such that $|a_k-a_l|\leq{\epsilon\over2}$ for all $k$, $l\geq n$, and this implies $\Delta_n\leq{\epsilon\over2}<\epsilon$. As $\epsilon>0$ was arbitrary this shows that no positive number $\epsilon$ can be a lower bound of the $\Delta_n$, from which we conclude that in fact $\inf_n\Delta_n=0$.
