To prove that a sequence converges, it is sometimes easier to start by finding a subsequence that converges (or proving that such a subsequence exists). Then one can hope to deduce that the sequence itself converges. Another way of using subsequences is to exploit the following result: if every subsequence has a further subsequence that converges to a limit , then the whole sequence converges to .
Basic concepts of real analysis.
To prove that a Cauchy sequence of real numbers converges, assuming one of the other forms of the completeness axiom, one can argue as follows. First, one proves the Bolzano-Weierstrass theorem. Then one observes that a Cauchy sequence is bounded, and so by the Bolzano-Weierstrass theorem has a convergent subsequence. And finally, one proves that if a Cauchy sequence has a subsequence that tends to a limit , then the sequence itself tends to .
Here is a second way of proving that Cauchy sequences converge. Let be a Cauchy sequence. By the Cauchy condition we can choose a subsequence such that whenever . It follows that the closed intervals are nested, since if then , and . Therefore, their intersection is non-empty. Since their lengths tend to zero, the intersection is a singleton , and it is easy to check that the subsequence converges to . Therefore, since is Cauchy, so does the whole sequence.
The trick in the second approach was to replace a Cauchy sequence by a subsequence that was rapidly Cauchy. That gave us the inequality , which was crucial to the success of the proof.