Characterisation of Non-Archimedean Division Ring Norms/Corollary 2

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

\(\displaystyle \dfrac 1 {\norm{y}^n}\) \(=\) \(\displaystyle \norm {y^{-1} }^n\) $\quad$ Norm of Inverse $\quad$
\(\displaystyle \) \(\) \(\displaystyle \) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \norm {x}^n\) $\quad$ Definition of $y$ $\quad$
\(\displaystyle \) \(\) \(\displaystyle \) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \norm {x^n}\) $\quad$ Norm axiom (N2) {Multiplicativity) $\quad$
\(\displaystyle \) \(\) \(\displaystyle \) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \norm {\paren {m \cdot 1_R}^n}\) $\quad$ Definition of $x$ $\quad$
\(\displaystyle \) \(\) \(\displaystyle \) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \norm {m^n \cdot 1_R}\) $\quad$ $\quad$


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 $C \lt +\infty$

$\Box$


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 $C = 1$.

$\blacksquare$

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