Characterisation of Non-Archimedean Division Ring Norms/Corollary 2

Theorem
Let $\struct{R, \norm{\,\cdot\,}}$ be a normed division ring with unity $1_R$.

Let $\,\,\sup \set {\norm{n \cdot 1_R}: n \in \N_{\gt 0}} = C \lt +\infty$. where $n \cdot 1_R = \underbrace {1_R + 1_R + \dots + 1_R}_{n \, times}$

Then $\norm{\,\cdot\,}$ is non-Archimedean and $C = 1$.

Proof
Aiming for a contradiction, let $C \gt 1$.

By Characterizing Property of Supremum of Subset of Real Numbers then:
 * $\exists m \in \N_{\gt 0}: \norm{m \cdot 1_R} \gt 1$.

Let
 * $x = m \cdot 1_R$
 * $y = x^{-1}$

By Norm of Inverse then:
 * $\norm {y} \lt 1$.

By Sequence of Powers of Number less than One then:
 * $\displaystyle \lim_{n \to \infty} \norm{y}^n = 0$

By Reciprocal of Null Sequence then:
 * $\displaystyle \lim_{n \to \infty} \frac 1 {\norm{y}^n} = +\infty$

For all $n \in \N_{\gt 0}$:

{{eqn|r= \norm {x^n} |c= Norm axiom (N2) {Multiplicativity) }}

So:
 * $\displaystyle \lim_{n \to \infty} \norm {m^n \cdot 1_R} = +\infty$

Hence:
 * $\sup \set {\norm{n \cdot 1_R}: n \in \N_{\gt 0}} = +\infty$

This contradicts the assumption that $C \lt +\infty$

It follows that $C \le 1$.

Then  $\forall n \in \N_{>0}: \norm{n \cdot 1_R} \le 1$.

By Characterisation of Non-Archimedean Division Ring Norms then $\norm{\,\cdot\,}$ is non-Archimedean.

By Corollary 1 then  $\sup \set {\norm{n \cdot 1_R}: n \in \N_{\gt 0}} = 1$.

So $C = 1$.