# Sequence is Bounded in Norm iff Bounded in Metric

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## Theorem

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

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

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

Then:

- $\sequence {x_n} $ is a bounded sequence in the normed division ring $\struct {R, \norm {\,\cdot\,} }$ if and only if $\sequence {x_n} $ is a bounded sequence in the metric space $\struct {R, d}$.

## Proof

### Necessary Condition

Let $\sequence {x_n} $ be a bounded sequence in $\struct {R, \norm {\,\cdot\,} }$.

Then:

- $\exists K \in \R_{\gt 0} : \forall n : \norm {x_n} \le K$

Then $\forall n, m \in \N$:

\(\displaystyle \map d { x_n , x_m }\) | \(=\) | \(\displaystyle \norm {x_n - x_m}\) | Definition of Metric Induced by Norm on Division Ring | ||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \norm {x_n} + \norm {x_m}\) | Norm of Difference | ||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle K + K\) | Definition of Bounded Sequence in Normed Division Ring | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2 K\) |

Hence the sequence $\sequence {x_n} $ is bounded by $2 K$ in the metric space $\struct {R, d}$.

$\Box$

### Sufficent Condition

Let $\sequence {x_n} $ be a bounded sequence in the metric space $\struct {R, d}$.

Then:

- $\exists K \in \R_{> 0} : \forall n, m : \map d {x_n , x_m} \le K$

By the definition of the metric induced by a norm this is equivalent to:

- $\exists K \in \R_{> 0} : \forall n, m : \norm {x_n - x_m} \le K$

Then $\forall n \in \N$:

\(\displaystyle \norm {x_n}\) | \(=\) | \(\displaystyle \norm {x_n - x_1 + x_1}\) | |||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \norm {x_n - x_1} + \norm {x_1}\) | Norm axiom (N3) (Triangle Inequality) | ||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle K + \norm {x_1}\) | Assumption of bounded sequence |

Hence the sequence $\sequence {x_n}$ is bounded by $K + \norm {x_1}$ in the normed division ring $\struct {R, \norm {\,\cdot\,} }$.

$\blacksquare$