コーシーの収束条件

Deffinition

Real sequence $(a_n)_{n\in\mathbb{N}}$ is Cauchy sequence if it fills bellow.
$$(\forall\epsilon\gt 0)(\exists n_0\in\mathbb{N})(\forall n,m \in \mathbb{N})(m,n \ge n_0 \Rightarrow |a_m - a_n| \lt \epsilon)$$

Statement

Suppose $(a_n)_{n\in\mathbb{N}}$ is Cauchy sequence.
1. $(a_n)_{n\in\mathbb{N}}$ is bounded(有界).
2. if $(a_n)_{n\in\mathbb{N}}$ has a partial sequence $(a_{n(k)})_{k\in\mathbb{N}}$ which is $\lim_{k\rightarrow\infty}a_{n(k)} = a$, then $\lim_{n\rightarrow\infty}a_n = a$

Proof

1.
for $1(\gt 0)$, $\exists n_1$, $(m,n \ge n_1 \Rightarrow |a_m - a_n| \lt 1)$.
So
$$|a_m - a_{n_1}| \lt 1 \\ \Leftrightarrow a_{n_1} - 1 \lt a_m \lt a_{n_1} + 1$$ therefore,
If
$$M = \max\{|a_0|,|a_1|,\dots,|a_{n_1}|,|a_{n_1} - 1|, |a_{n_1} + 1|\}$$ then
$$|a_n| \lt M (\forall n \in \mathbb{N})$$ 2. Given any small number of $\epsilon$,
$$\exists n_0,(m,n \ge n_0 \Rightarrow |a_m - a_n| \lt \frac{\epsilon}{2})$$ and
$$\exists k_0,(k \ge k_0 \Rightarrow |a_n(k) - a| \lt \frac{\epsilon}{2})$$ So
$m_0 = \max\{n_0,k_0\}$
then
$$k \gt m_0 \Rightarrow |a - a_k| \le |a - a_{n(k)}| + |a_{n(k)} - a_k| \lt \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$ whitch means
$$\lim_{n \rightarrow \infty} a_n = a$$

Statement

1. $(a_n)_{n\in\mathbb{N}}$ is convergent. ($\lim_{n\rightarrow \infty} a_n = a$)
$$\Leftrightarrow$$ 2. Real sequence $(a_n)_{n\in\mathbb{N}}$ is Cauchy sequence.

Proof

$1.\Rightarrow 2.$
$\forall\epsilon\gt 0,\exists n_0 \in \mathbb{N}, (n \gt n_0 \Rightarrow |a_n - a| \lt \frac{\epsilon}{2})$
So
$m,n \gt n_0 \Rightarrow |a_m - a_n| \le |a_m - a| + |a - a_n| \lt \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$
$2.\Rightarrow 1.$

 ボルツァーノ・ワイヤストラスの定理より有界無限数列は収束する部分列をもつ