Cauchy Sequence is Bounded/Normed Division Ring

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Theorem

Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring.

Every Cauchy sequence in $R$ is bounded.


Proof 1

Let $\sequence {x_n} $ be a Cauchy sequence in $R$ then:

$\forall \epsilon \in \R_{\gt 0}: \exists N \in \N : \forall n, m \ge N: \norm {x_n - x_m} \lt \epsilon$.


Let $n_1$ satisfy:

$\forall n, m \ge n_1: \norm {x_n - x_m} \lt 1$


Then $\forall n \ge n_1$:

\(\displaystyle \norm {x_n}\) \(=\) \(\displaystyle \norm {x_n - x_{n_1} + x_{n_1} }\) $\quad$ $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \norm {x_n - x_{n_1} } + \norm {x_{n_1} }\) $\quad$ Norm axiom (N3) (Triangle Inequality). $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle 1 + \norm {x_{n_1} }\) $\quad$ Since $n, n_1 \ge n_1$. $\quad$


Let $K = \max \set {\norm {x_1}, \norm {x_2}, \dots, \norm {x_{n_1 - 1}}, 1 + \norm {x_{n_1} } }$ then:

$\forall n \lt n_1: \norm {x_n} \le K$
$\forall n \ge n_1: \norm {x_n} \le 1 + \norm {x_{n_1} } \le K$

By the definition of a bounded sequence in a normed division ring then $\sequence {x_n}$ is bounded.

$\blacksquare$


Proof 2

Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$.

Let $\sequence {x_n} $ be a Cauchy sequence in $\struct {R, \norm {\,\cdot\,}}$.

By the definition of a Cauchy sequence in a normed division ring then $\sequence {x_n} $ is a Cauchy sequence in $\struct {R, d}$.

By Cauchy sequence is bounded in metric space then $\sequence {x_n} $ is a bounded sequence in $\struct {R, d}$.

By Sequence is Bounded in Norm iff Bounded in Metric then $\sequence {x_n} $ is a bounded sequence in $\struct {R, \norm {\,\cdot\,}}$.

$\blacksquare$


Proof 3

Let $\sequence {x_n} $ be a Cauchy sequence in $R$.

By Norm Sequence of Cauchy Sequence has Limit, $\sequence {\norm {x_n} }$ is a convergent sequence in $\R$.

By Convergent Real Sequence is Bounded, $\sequence {\norm {x_n} }$ is bounded.

That is:

$\exists M \in \R_{\gt 0}: \forall n \in \N: \norm {x_n} = \size {\norm {x_n} } \le M$

By the definition of a bounded sequence in a normed division ring, $\sequence {x_n}$ is bounded.

$\blacksquare$


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