# Convergent Sequence in Normed Division Ring is Bounded

## Theorem

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

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

Let $\sequence {x_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the following limit:

- $\displaystyle \lim_{n \mathop \to \infty} x_n = l$

Then $\sequence {x_n}$ is bounded.

## Proof 1

Let $\sequence {x_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the limit $l$, then:

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

Let $n_1$ satisfy:

- $\forall n \ge n_1: \norm {x_n - l} < 1$

Then $\forall n \ge n_1$:

\(\displaystyle \norm {x_n}\) | \(=\) | \(\displaystyle \norm {x_n - l + l}\) | |||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \norm {x_n - l} + \norm l\) | Norm axiom (N3) (Triangle Inequality) | ||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle 1 + \norm l\) | as $n \ge n_1$ |

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

Then:

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

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

$\blacksquare$

## Proof 2

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

Let $\sequence {x_n}$ be convergent to the limit $l$ in $\struct {R, \norm {\,\cdot\,}}$.

By the definition of a convergent sequence in a normed division ring, $\sequence {x_n} $ is convergent to the limit $l$ in $\struct {R, d}$.

By Convergent Sequence in Metric Space is Bounded, $\sequence {x_n} $ is a bounded sequence in $\struct {R, d}$.

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

$\blacksquare$

## Proof 3

Let $\sequence {x_n}$ be convergent to the limit $l$ in $\struct {R, \norm {\,\cdot\,} }$.

By modulus of limit in normed division ring, $\sequence {\norm {x_n} }$ is a convergent sequence in $\R$.

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

That is:

- $\exists M \in \R_{> 0}: \forall n, \norm {x_n} = \size {\norm {x_n} } \le M$

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

$\blacksquare$

## Proof 4

Let $\sequence {x_n}$ be convergent to the limit $l$ in $\struct {R, \norm {\,\cdot\,}}$.

By Convergent Sequence is Cauchy Sequence in Normed Division Ring, $\sequence {x_n}$ is a Cauchy sequence in $\struct {R, \norm {\,\cdot\,}}$.

By Cauchy Sequence in Normed Division Ring is Bounded, $\sequence {x_n}$ is a bounded sequence in $\struct {R, \norm {\,\cdot\,}}$.

$\blacksquare$

## Sources

- 2007: Svetlana Katok:
*p-adic Analysis Compared with Real*: $\S 1.2$: Normed Fields, Exercise $11 \ (1)$