a) Exist a not bounded subsequence of $a_{k}$ or b)There exist two subsequences of $a_{k}$ such that they converge to different limits. I have been trying with this proof as follow.
I know that not Cauchy sequence implies that: $\exists \epsilon>0$ for $\forall N \in\Bbb N$ and $\exists n,m>N$ such that $\Vert a_{n}-a_{m}\Vert\ge\epsilon$.
And I im giving a certain $N_{2}=1$ and getting two subsequences of $a_{k}$, and then using $N_{1}=1+max(n,m)$ and getting another subsequences, and over and over again, in a way that i could contract two subsequences that are monotonous.
Now, I can say that option a) doesn't happened and then I would happened that the original sequence is forced to be bounden by a certain M>0. My idea, but i don't know how, is that because I was creating a monotonous subsequence that is just getting greater for every element on the constructed subsequence, I could put an $\epsilon$ in the boundaries of M, because as much as I am make greater elements for those two monotonous subsequences, they should be force to accumulate on the boundaries of M, and analogously do it for the other subsequence, and finally conclude that there exist two different subsequences of $a_{k}$ such that both have different limits, (that are those made by M and -M).
Is this proof logical, or at least possibles, I would appreciate any help for the demonstration or tips to conclude that there should exist a convergent subsequences on the boundaries of M and -M
If you are dealing with ${\Bbb R}$ and have learned about $\limsup$ and $\liminf$ then you simply set: $$ a_+ = \limsup_{n\rightarrow \infty} a_n \ \ \ \mbox{and} \ \ \ \ a_- = \limsup_{n\rightarrow \infty} a_n.$$ Both $a_+$ and $a_-$ are limits of subsequences of $(a_n)$.
If either of them is infinite then the sequence was not bounded, whence case a. If they are finite and different we have case b. And if they are equal the sequence converges so it was Cauchy in the first place.