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

Theorem
Let $R$ be a division ring.

Let $\norm{\,\cdot\,}_1: R \to \R_{\ge 0}$ and $\norm{\,\cdot\,}_2: R \to \R_{\ge 0}$ be norms on $R$.

Let $\norm{\,\cdot\,}_1$ and $\norm{\,\cdot\,}_2$ satisfy:
 * $\forall x \in R:\norm{x}_1 \lt 1 \iff \norm{x}_2 \lt 1$

Then:
 * $\exists \alpha \in \R_{\gt 0}: \forall x \in R: \norm{x}_1 = \norm{x}_2^\alpha$

Case 1
Let for all $x \in R, x \neq 0_R$, satisfy $\norm{x}_1 \ge 1$.

Lemma 1
By assumption, for all $x \in R, x \neq 0_R$, then $\norm{x}_2 \ge 1$.

Similarly $\norm{\,\cdot\,}_2$ is the trivial norm.

Hence $\norm{\,\cdot\,}_1$ and $\norm{\,\cdot\,}_2$ are equal.

For $\alpha = 1$ the result follows.

Case 2
Let $x_0 \in R$ such that $x_0 \neq 0_R$ and $\norm{x_0}_1 \lt 1$.

By assumption then $\norm{x_0}_2 \lt 1$.

Let $\alpha = \dfrac {\log \norm {x_0}_1 } {\log \norm {x_0}_2 }$.

Then $\norm{x_0}_1 = \norm{x_0}_2^\alpha$.

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

Let $x \in R, x \neq 0_R, x \neq x_0$.

Case 2.1
Let $\norm {x}_1 = \norm {x_0}_1$.

Then:

and similarly:

Aiming for a contradiction, suppose that $\norm {x}_2 \neq \norm {x_0}_2$.

By Real Numbers form Totally Ordered Field either:
 * $\norm {x}_1 \lt \norm {x_0}_1$

or:
 * $\norm {x_0}_1 \lt \norm {x}_1$.

Hence either:
 * $\norm{x x_0^{-1}} = \norm{x} \norm{x_0^{-1}} = \dfrac {\norm {x}_1} {\norm {x_0}_1} \lt 1$

or:
 * $\norm{x_0 x^{-1}} = \norm{x_0} \norm{x^{-1}} = \dfrac {\norm {x}_1} {\norm {x_0}_1} \lt 1$.

This contradicts the premise:
 * $\forall x \in R:\norm{x}_1 \lt 1 \iff \norm{x}_2 \lt 1$

So $\norm {x}_2 = \norm {x_0}_2$.

Hence:
 * $\norm{x}_1 = \norm{x_0}_1 = \norm{x_0}_2^\alpha = \norm{x}_2^\alpha$

Case 2.2
Let $\norm {x}_1 \neq \norm {x_0}_1$, $\norm {x}_1 \lt 1$.

Case 2.3
Let $\norm {x}_1 \neq \norm {x_0}_1$, $\norm {x}_1 \gt 1$.

Case 2.4
Suppose $\norm {x}_1 \neq \norm {x_0}_1$, $\norm {x}_1 = 1$.