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

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 metric space $\struct {R, d}$

Then:
 * $\sequence {x_n} $ is a bounded sequence in the normed division ring $\struct {R, \norm {\,\cdot\,}}$

Proof
Let $\sequence {x_n} $ be a bounded sequence in the metric space $\struct {R, d}$ then:
 * $\exists K \in \R_{\gt 0} : \forall n, m : \map d { x_n, x_m } \le K$

By the definition of the Metric induced by a norm this is equivalent to:
 * $\exists K \in \R_{\gt 0} : \forall n, m : \norm { x_n, x_m } \le K$

Then $\forall n \in \N$:

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

Also See

 * Bounded sequence in a normed division ring
 * Metric induced by a norm on a division ring
 * Bounded sequence in a metric space