Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2

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

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

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

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

Proof
Since $\norm{x_0}_1 \lt 1$ then $\norm{x_0}_2 \lt 1$ and:
 * $\log \norm {x_0}_1 \lt 0$
 * $\log \norm {x_0}_2 \lt 0$

Hence $\alpha \gt 0$

Since $\forall x \in R:\norm{x}_1 \lt 1 \iff \norm{x}_2 \lt 1$:

Lemma 4
Let $x \in R, x \neq 0_R$.

Case 1
Let $\norm {x}_1 \neq 1$.

By Lemma 4 then:
 * $\norm {x}_2 \neq 1$

Let $\beta = \dfrac {\log \norm {x}_1 } {\log \norm {x}_2 }$.

Then $\norm{x}_1 = \norm{x}_2^\beta$.

Lemma 5
Hence $\norm{x}_1 = \norm{x}_2^\alpha$.

Case 2
Let $\norm {x}_1 = 1$.

By Lemma 4 then:
 * $\norm {x}_2 = 1$

Hence:
 * $\norm {x}_1 = 1 = 1^\alpha = \norm {x}_2^\alpha$