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

$\blacksquare$


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