# Convergent Sequence in Normed Division Ring is Bounded/Proof 1

## 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

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

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

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:

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

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

$\blacksquare$

## Sources

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