Consider a sequence $(p_n)^{\infty}_{n=1}$ and its range after index $N$, $E_{N} = \{ p \in X \mid p \in range( (p_n)^{\infty}_{n=N} )\}$ and $S_N = \{ d(p,q) \in \mathbb R \mid p,q \in E_N \}$. Note that $E_N$ are just the terms in the sequence after index $N$ all collected in a set. I want to show the following is true:
$$ (p_n)^{\infty}_{n=1} \text{ Cauchy } \iff \lim_{N \rightarrow \infty} \left[ \sup_{p,q \in E_N} d(p,q)\right] = 0$$
we want to show:
$$ \forall \epsilon >0, \exists N_S: N \geq N_S \implies \sup E_N < \epsilon $$
So I started by writing the definition of Cauchy:
$$ \forall \epsilon >0, \exists N_C: n,m \geq N_C \implies d(p_n,p_m) < \epsilon $$
then since there is an UB the completeness axiom guarantees a sup exists:
$$ d(p_n,q_m) \leq \sup_{p,q \in E_{N_C}} d(p,q) \leq \epsilon$$
where the last inequality is justified because $\sup_{p,q \in E_{N_C}} d(p,q)$ the diameter of $E_{N_C}$ is a least upper bound the other upper bound can only be larger or equal. However, the definition of convergence requires me to show a strict inequality which is where I am stuck. Since we have an inequality we know that its either equal xOR less than, but not which one. How does one proceed to remove the equality part?
I tried a proof but it just feels incorrect. I will write it but I'm not its right nor does it feel "standard".
Proof:
We want to show that for every epsilon greater than zero we can choose some $N_S$ such that the sup is bounded after $N$ i.e. $ \sup {S_N} < \epsilon$.
Lets consider $N_C$ s.t $d(p_n,p_m) < \epsilon$. The claim is that there must exist some $N_c + K$ s.t.
$$ \sup S_{N_c + k} < \sup S_{N_c } \leq \epsilon $$
if there wasn't then I claim $(p_n)_n$ wouldn't be Cauchy. The idea is that if that were true then for every $k \in \mathbb N $ we have:
$$ \sup S_{N_C + k} \geq \sup S_{N_C } $$
in the case where $\sup S_{N_C }$ is not zero it means that there is some epsilon where the Cauchy criterion doesn't hold, in particular for $\epsilon = \sup S_{N_C}$. Why doesn't it hold? Well it doesn't matter how far we go in the sequence of supremums we are always at least $\sup S_{N_C}$. I guess this is the last part of the proof I am struggling to complete.
Notice that if $\sup S_{N_C } = 0$ then $\sup S_{N_C + k} = 0 $ since $S_{N_C + K} \subset S_{N_C}$ so:
$$ \sup S_{N_C } = 0 = \sup S_{N_C + k} < \epsilon$$
becomes a hard inequality since epsilon is chosen to be positive.
Note that I have seen the proof:
Cauchy sequences and diameter of a set
which answer how to do the $\implies$ right direction of the proof, but I guess I am curious to know if the proof I suggested is correct of if it can be made to work. Perhaps not, idk.
Proof2:
I couldn't figure out if my original proof is correct so I want to propose a new one:
We want $\epsilon > 0, \exists N_{\epsilon'} : N>N_{\epsilon'} \implies \sup S_{N} < \epsilon '$
Consider any $\epsilon ' > 0$. Choose some $ \epsilon < \epsilon ' $. By Cauchy property of the sequence we know:
$$ \exists N_{\epsilon}: n_{N_{\epsilon}}, m_{N_{\epsilon}} \geq N_{\epsilon}: d( p_{ n_{N_{\epsilon}} }, q_{ m_{N_{\epsilon}} }) < \epsilon $$
Consider any $N \geq N_{\epsilon}$. For such $N$ we have that:
$$ n_{N}, m_{N} \geq N \geq N_{\epsilon } \implies d(p_{n_N},p_{m_N}) < \epsilon$$
by properties of chosen $N_{\epsilon}$ and the fact $S_N \subset S_{N_\epsilon} $:
$$ \forall d(p_{n_N},p_{n_M}) \in S_{N}, d(p_{n_N},p_{n_M}) \leq \sup S_N \leq \sup S_{N_{\epsilon}} \leq \epsilon $$
by choise of $\epsilon'$ we have $ \sup S_N < \epsilon$ since $\epsilon < \epsilon ' $. So we just let $N_{\epsilon'} = N_{\epsilon}$ to complete the proof. Since $N \geq N_{\epsilon}$ we complete the proof.
Alternatively, one can jut see this proof as choosing some new $\epsilon$ s.t.$ \epsilon < \epsilon'$ then adding $k$ to $\epsilon$ so that $\epsilon+k \approx \epsilon'$ to get $N \geq N_{\epsilon'} \implies \sup S_{N_{\epsilon}} \leq \epsilon \leq \epsilon + k < \epsilon '$. We just need any value of $k$ s.t $0 < k < \epsilon' - \epsilon$ e.g. $k = \frac{\epsilon' - \epsilon}{2}$
Don't you think that the claim you want to prove is obvious from the definitions?
Let $X$ be a metric space, and for arbitrary $A\subset X$ define ${\rm diam}(A):=\sup\{d(x,y)\,|\,x, y\in A\}$. Let $(p_n)_{n\geq1}$ be a sequence of points in $X$, and let $E_n:=\{p_k\,|\, k\geq n\}\subset X$ $(n\geq1)$ denote the tails of this sequence. Then the following are in turn equivalent:
The sequence $(p_n)_{n\geq1}$ is Cauchy.
For any $\epsilon>0$ there is an $N$ such that $d(p_n,p_m)\leq\epsilon$ whenever $m$, $n\geq N$.
For any $\epsilon>0$ there is an $N$ such that for all $n\geq N$ we have $d(x,y)\leq\epsilon$ for all $x$, $y\in E_n$.
For any $\epsilon>0$ there is an $N$ such that ${\rm diam}(E_n)\leq\epsilon$ for all $n\geq N$.