Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 1

Theorem

Let $R$ be a division ring.

Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ be a norm on $R$.

Let $\norm {\, \cdot \,}_1$ satisfy:

$\forall x \in R, x \ne 0_R:\norm x_1 \ge 1$

Then:

$\norm {\, \cdot \,}_1$ is the trivial norm.

Proof

We prove the contrapositive.

Let $\norm {\, \cdot \,}_1$ be a nontrivial norm.

Then:

$\exists y \in R: \norm y_1 \ne 0, \norm y_1 \ne 1$.

By Real Numbers form Ordered Field either $\norm y_1 < 1$ or $\norm y_1 > 1$.

Suppose $\norm y_1 > 1$.

By Norm axiom $(\text N 1)$: Positive Definiteness:

$y \ne 0_R$

By Norm of Inverse in Division Ring:

$\norm {y^{-1} }_1 = \dfrac 1 {\norm y_1} < 1$

So either $\norm y_1 < 1$ or $\norm {y^{-1} }_1 < 1$

So:

$\exists x \in R, x \ne 0_R:\norm x_1 < 1$

The theorem now follows by the Rule of Transposition.

$\blacksquare$