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I have a simple statement about sequences I have proven. I am asking for help to see if this statement and proof are correct. If statement is correct, is there simpler way proving it? Huge thanks!!!! Here it is:

Statement: ${\{a_n}\}$ is a Cauchy sequence in R. Let $S_K = \{x=d(a_m,a_n)/m,n>K,K\in N\}$. Here $d(\cdot,\cdot)$ is a distance metric. Also $S_{K+1}\subset S_K $.

Prove: if ${\{a_n}\}$ diverges, then $\exists\epsilon>0\ni\forall K[\forall x\in S_K(x\geq\epsilon)]$. In words, if the sequence diverges then all the distances are larger then certain $\epsilon$.


(1) By Cauchy criterion: if $\{a_n\}$ diverges then $\exists\epsilon>0\ni \forall K \exists m,n\geq K \implies d(a_m,a_n)\geq\epsilon $. Rewriting this statement in terms of $S_K$: if $\{a_n\}$ diverges then $\exists\epsilon>0\ni \forall K \exists x\in S_K \implies x\geq\epsilon $. It follows then that $\exists\epsilon>0\ni \forall K (\sup S_K \geq\epsilon)$

(2)To prove the main statement we have to show that:

$\forall x \in S_1 \exists V\ni x\geq \sup S_V$. (aaa)

Take an arbitrary $x\in S_1$, first notice that $\exists U \ni x\in S_U\subset S_1$. There are two ways in which (aaa) is not true. (1) x is smallest element $\implies \bigcap{_{n\in N}}S_n = {\{x\}}$ which is not true since $n\rightarrow\infty $ and intersection should be empty. (2) $\forall S_K(S_K\subset S_U$ and $x\notin S_K)$ has no supremum which contradicts Least Upper Bound property of real numbers. Hence (aaa) is correct and main statement is true.

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There is a basic problem in your question: first you state that your sequence is Cauchy, and than that it diverges. – Martin Argerami Feb 18 '12 at 22:17
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The correct assertion is $$ \{a_n\} \mbox{ diverges if and only if there exists }\varepsilon>0 \mbox{ such that}\sup S_K>\varepsilon \mbox{ for all } K. $$

Note that since the sequence $\{\sup S_K\}$ is non-increasing and non-negative, $\displaystyle\lim_{K\to\infty}\sup S_K$ always exists. So the assertion to be proven is simply that the sequence is Cauchy if and only if the sequence $\{\sup S_K\}$ is convergent.

To prove this, notice that if $\{a_n\}$ diverges, then there exists $\varepsilon>0$ such that, for every $K\in\mathbb{N}$, there exist $m,n>K$ such that $d(a_m,a_n)>\varepsilon$; in other words, $\sup S_K>\varepsilon$. As $K$ was arbitrary, $\sup S_K\geq\varepsilon$ for all $K$.

Conversely, if $\{a_n\}$ converges, then it is Cauchy. Fix $\varepsilon>0$; then there exists $K$ such that $d(a_n,a_m)<\varepsilon$ for all $n,m>K$; in other words, $\sup S_K<\varepsilon$. As the sequence is monotone, this shows that $\sup S_{K'}<\varepsilon$ for all $K'>K$. That is, $\displaystyle\lim_{K\to\infty}\sup S_K=0$.

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No, that is not the assertion to be proved. The assertion is that if the sequence diverges, then there is $\epsilon>0$ such that every pair of terms are at least $\epsilon$ apart. This is obviously false. – Brian M. Scott Feb 19 '12 at 0:09
Ah, I totally agree with that. I read "$\sup S_K>\varepsilon$ for all $K$". – Martin Argerami Feb 19 '12 at 0:14
@Martin: thanks a lot for the explanation, one point I cant get my head around, if supSk>ε for all K, doesn't it mean every x in Sk is also > ε, since you can find 'u' such that x>supSu > ε? There must be wrong link in my logic, I cant find it! – Leon Feb 19 '12 at 11:53
@Leon: it might help you understand if you replace "sup" with "max". In any case, for example $\sup[0,1)=1$; no point in the set is $>1$. – Martin Argerami Feb 19 '12 at 14:05

It's not true that $\bigcap{_{n\in N}}S_n$ should be empty. For instance, every second term might be equal to the previous term; then $0$ would be in all $S_n$.

If I may offer some advice: You may want to focus slightly less on formal logical manipulations and slightly more on developing intuition and checking examples. It's easy to make an error in a formal derivation of this length; that happens to everyone once in a while. However, intuition and examples can help guide your formal work and help you check and correct it when it's gone astray. In the present case, it's easy to see that the statement can't be true because it's easy to come up with counterexamples. Even just a sequence that starts off with a distance of $0$ and then diverges any way it likes is a counterexample. By thinking about examples and tuning your intuition on them, you can avoid such mistakes much more easily than by just trying to ensure that all your formal steps are correct.

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I don't think there are counterexamples to the assertion in the question. You say "even just a sequence that starts off with a distance of 0 and then diverges any way it likes is a counterexample"; for such a sequence, it is true that there exists $\varepsilon>0$ such that, for every $K$, $\sup S_K\geq\varepsilon$. – Martin Argerami Feb 18 '12 at 23:01
@Martin: There certainly are counterexamples, and joriki has correctly described a large class of them. You misread the assertion in the problem. – Brian M. Scott Feb 19 '12 at 0:07
@Brian: yes, you are right. – Martin Argerami Feb 19 '12 at 0:15
@Martin:I admit, my love of formalism led me astray :-). Part (2) of the proof is incorrect, more then that, you can easily draw a pictorial representation of the sequence, value of the sequence member vs its count number in the sequence that will disprove the statement! The best you can do is make conclusions about suprema behaviour as you have shown! Thanks! – Leon Feb 20 '12 at 7:44

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