Theorem 19.1
Theorem 19.1. Let \(\{a_n\}\) be a convergent sequence. Then for every \(\varepsilon > 0\), there exists a positive integer \(N\) such that if \(m, n \geq N\), then \[ |a_m - a_n| < \varepsilon. \]
Proof. Let $\{ a_n \}$ be a sequence with the limit $L$ and let $\varepsilon > 0$. Then $\exists N \in \mathbb{P}$ such that $|a_n - L| < \frac{\varepsilon}{2}, \forall n \geq N$. Thus $$|a_m - a_n| = |a_m - L - a_n + L| \leq |a_m - L| + |a_n - L| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon, \forall n, m \geq N. \blacksquare$$
Definition 19.2
Definition 19.2. If \(\{a_n\}\) is a sequence such that for every \(\varepsilon > 0\), there exists a positive integer \(N\) such that if \(m, n \geq N\), we have \[ |a_m - a_n| < \varepsilon, \] then we call \(\{a_n\}\) a Cauchy sequence.
어떤 수열이 수렴함을 증명하고자 할 때, 극한의 정의인 Definition 10.2에 의해서 보이려고 하면 반드시 수렴값이 주어져야한다. 그러나 값을 구체적으로 모르더라도 어떤 수열이 코시 수열임을 보이기만 하면 수렴성은 반드시 보장되어 있다.
Theorem 19.3
Theorem 19.3. Let \(\{a_n\}\) be a real sequence. Then \(\{a_n\}\) is convergent if and only if \(\{a_n\}\) is a Cauchy sequence.
Proof. $(\Longrightarrow)$ We have shown this in Theorem 19.1.
$(\Longleftarrow)$ Let $\varepsilon = 1$. By Definition 19.2, $\exists N \in \mathbb{P}$ such that $|a_n - a_m| < 1, \forall n, m \geq \mathbb{P}$. Then $|a_n - a_N| < 1, \forall n \geq N$, and we have $$|a_n| = |a_n - a_N + a_N| \leq |a_n - a_N| + |a_N| \leq 1 + |a_N|, \forall n \geq N.$$ Thus $$|a_n| \leq \max \{ |a_1|, \cdots, |a_{N_1}|, |a_{N}| + 1 \}, \forall n \in \mathbb{P},$$ whiche means that $\{ a_n \}$ is bounded.
By Bolzano-Weierstrass Theorem, $\{ a_n \}$ has a subsequence $\{ a_{n_k} \}$ with the limit $L$. Let $\varepsilon > 0$. Then $\exists N \in \mathbb{P}$ such that $|a_n - a_m| < \frac{\varepsilon}{2}, \forall n, m \geq N$, and $\exists N' \in \mathbb{P}$ such that $|a_{n_k} - L| < \frac{\varepsilon}{2}, \forall k \geq N'$. Choose $K \geq N'$ and $n_K \geq N$. Then $$|a_n - L| = |a_n - a_{n_K} + a_{n_K} - L| \leq |a_n - a_{n_K}| + |a_{n_K} - L| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon, \forall n \geq N.$$ Thus $\{ a_n \}$ converges to $L$. $\blacksquare$