Alternative approach to prove that if a sequence is a Cauchy Sequence, then it converges.

Question

I am familiar with the traditional proof that if a sequence of real numbers is a Cauchy sequence, then it is convergent. However, I am stuck on a problem that asks me to prove the implication using the fact that if all the subsequences of a sequence converge to a real number then the sequence itself converges.

Question: (i) Let $(a_n)$ be a bounded sequence of real numbers and let $x \in \mathbb{R}$ have the property that every subsequence of $(a_n) \neq (a_n)$ converges to $x$. Show that $(a_n) \to x$. (ii) Furthermore, use this result to prove that if a sequence is a Cauchy sequence, then it converges.

Proving part (i) was fairly easy for me. Outlining a proof for part (ii) is tripping me up. I am having a hard time applying the properties of a Cauchy sequence to the proof other than the fact that this implies the sequence is bounded.

Any help is greatly appreciated.

Answer 1

For (i), simply let $b_n:=a_{n+1} ,\forall n\ge1.$ Then by the assumption, $(b_n)$ is a convergent subsequence of $(a_n)$ , and it implies obviously that $(a_n)$ is a convergent sequence by the definition of limits.

For (ii), try to argue by contradiction with the definitions of convergent sequences, limits, and Cauchy sequences.

The basic idea is that, if a sequence $(a_n)$ is divergent, then no matter how rear in the "tail"(i.e. $\{a_n\,\vert n\ge k, k$ large$\}$, there always exist two elements of it which are not close enough. However, every two terms with sufficiently large indexes of a Cauchy sequence can get sufficiently close. This is the contradiction.