Sequence is Bounded in Norm iff Bounded in Metric/Necessary Condition

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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 $R$.

Let $\sequence {x_n} $ be a bounded sequence in the normed division ring $\struct {R, \norm {\,\cdot\,}}$


Then:

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


Proof

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

\(\ds \map d { x_n , x_m }\) \(=\) \(\ds \norm {x_n - x_m}\) Definition of Metric Induced by Norm on Division Ring
\(\ds \) \(\le\) \(\ds \norm {x_n} + \norm {x_m}\) Norm of Difference
\(\ds \) \(\le\) \(\ds K + K\) Definition of Bounded Sequence in Normed Division Ring
\(\ds \) \(=\) \(\ds 2 K\)


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

$\blacksquare$


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