# Cauchy Sequence is Bounded/Normed Division Ring/Proof 1

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## Theorem

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

Every Cauchy sequence in $R$ is bounded.

## Proof

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 $\forall n \ge n_1$:

\(\displaystyle \norm {x_n}\) | \(=\) | \(\displaystyle \norm {x_n - x_{n_1} + x_{n_1} }\) | |||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \norm {x_n - x_{n_1} } + \norm {x_{n_1} }\) | Norm axiom (N3) (Triangle Inequality). | ||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle 1 + \norm {x_{n_1} }\) | Since $n, n_1 \ge n_1$. |

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

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

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

$\blacksquare$

## Sources

- 2007: Svetlana Katok:
*p-adic Analysis Compared with Real*: $\S 1.2$: Normed Fields, Exercise $11$ $(2)$