# Cauchy Sequence is Bounded/Normed Division Ring

## 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 by definition:

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

Let $n_1$ satisfy:

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

Then $\forall n \ge n_1$:

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

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

Then:

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

It follows by definition that $\sequence {x_n}$ 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:

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

Thus, by definition, $\sequence {x_n}$ is bounded.

$\blacksquare$