# Sequence is Bounded in Norm iff Bounded in Metric

## 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 $\struct {R, \norm {\,\cdot\,} }$.

$\blacksquare$