Published Jun 24, 2021
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Definition 2.6.1 A sequence \((a_n)\) is called a Cauchy sequence if, for every \(\epsilon > 0\), there exists an \(N \in N\) such that whenever \(m, n \geqslant N\) it follows that \(|a_n - a_m| < \epsilon\).
Theorem 2.6.2 Every convergent sequence is a Cauchy sequence.
Lemma 2.6.3 Cauchy sequences are bounded.
Proof. Given \(\epsilon = 1\), there exists an \(N\) such that \(|x_m - x_n| < 1\)for all \(m, n \geqslant N\). Thus, we must have \(|x_n| < |x_N| + 1\) for all \(n \geqslant N\). It follows that $$ M = max\{|x_1|,|x_2|,|x_3|,\cdot \cdot \cdot, |x_{N-1}|,|x_N| + 1\} $$ is a bound for the sequence \((x_n)\)
Theorem 2.6.4 Cauchy Criterion. A sequence converges if and only if it is a Cauchy sequence
Proof. \((\Leftarrow)\) For this direction, we start with a Cauchy sequence \((x_n)\), which is bounded. So we may use the Bolzano-Weierstrass Theorem to produce a convergent subsequence \((x_{n_k})\). Set $$ x = \lim x_{n_k} $$ The idea is to show that the original sequence \((x_n)\) converges to this same limit. Once again, we will use a triangle inequality argument. We know the terms in the subsequence are getting close to the limit \(x\), and the assumption that \((x_n)\) is Cauchy implies the terms in the "tail" of the sequence ae close to each other. Thus, we want to make each of these distance less than half of the prescribed \(\epsilon\).
Let \(\epsilon > 0\). Because \((x_n)\) is Cauchy, there exists \(N\) such that $$ |x_n - x_m| < \frac{\epsilon}{2} $$ whenever \(m, n \geqslant N\). Now, we also know that \((x_{n_k}) \to x\), so choose a term in this subsequence, call it \(x_{n_K}\), with \(n_K \geqslant N\) and $$ |x_{n_K} - x| < \frac{\epsilon}{2} $$ To see that \(N\) has the desired property (for the original sequence \((x_n)\)), observe that if \(n \geqslant N\), then $$ \begin{align} |x_n - x| &= |x_n - x_{n_K} + x_{n_K} - x| \\ &\leqslant |x_n - x_{n_K}| + |x_{n_K} - x| \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align} $$