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

## Contents

## Theorem

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

Every convergent sequence in $R$ is a Cauchy sequence.

## Proof

Let $\sequence {x_n}$ be a sequence in $R$ that converges to the limit $l \in R$.

Let $\epsilon > 0$.

Then also $\dfrac \epsilon 2 > 0$.

Because $\sequence {x_n}$ converges to $l$, we have:

- $\exists N: \forall n > N: \norm {x_n - l} < \dfrac \epsilon 2$

So if $m > N$ and $n > N$, then:

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

\(\displaystyle \) | \(\le\) | \(\displaystyle \norm {x_n - l} + \norm {l - x_m}\) | Triangle Inequality | ||||||||||

\(\displaystyle \) | \(<\) | \(\displaystyle \frac \epsilon 2 + \frac \epsilon 2\) | (by choice of $N$) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \epsilon\) |

Thus $\sequence {x_n}$ is a Cauchy sequence.

$\blacksquare$

## Also see

## Sources

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