Question 3.T.9: (Cauchy’s Criterion) A sequence of real numbers is convergen...
(Cauchy’s Criterion)
A sequence of real numbers is convergent if, and only if, it is a Cauchy sequence.
Let (x_{n}) be a Cauchy sequence and A = \left\{x_{n} : n ∈ \mathbb{N}\right\}. We have two cases:
(i) The set A is finite. In this case one of the terms of the sequence, say x, is repeated infinitely often. We shall prove that x_{n} → x. Given ε > 0, there is an integer N such that
m, n ≥ N ⇒ |x_{n} − x_{m}| < ε.
Since the term x is repeated infinitely often in the sequence, there is an m > N such that x_{m} = x. Hence
|x_{n} − x| = |x_{n} − x_{m}| < ε for all n ≥ N,
which just means x_{n} → x.
(ii) A is infinite. In view of Lemma 3.1, the set A is bounded and therefore, by the Bolzano-Weierstrass Theorem, it has a cluster point x. We shall prove that x_{n} → x. Given ε > 0, there is an integer N such that
|x_{n} − x_{m}| < ε for all n, m ≥ N.
Since x ∈ \hat{A}, the interval (x − ε, x + ε) contains an infinite number of terms of the sequence (x_{n}). Hence there is an m ≥ N such that x_{m} ∈ (x − ε, x + ε), i.e., such that |x_{m} − x| < ε. Now, if n ≥ N, then
|x_{n} − x| ≤ |x_{n} − x_{m}| + |x_{m} − x| < ε + ε = 2ε,
which proves x_{n} → x.