Equivalence of Definitions of Equivalent Division Ring Norms/Cauchy Sequence Equivalent implies Open Unit Ball Equivalent

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:
 * for all sequences $\sequence {x_n}$ in $R$: $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$ $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$

Then $\forall x \in R$:
 * $\norm x_1 < 1 \iff \norm x_2 < 1$

Proof
The contrapositive is proved.

Let there exist $x \in R$ such that $\norm x_1 < 1$ and $\norm x_2 \ge 1$.

Let $\sequence {x_n}$ be the sequence defined by: $\forall n: x_n = x^n$.

By Sequence of Powers of Number less than One in Normed Division Ring then $\sequence {x_n}$ is a null sequence in $\norm {\, \cdot \,}_1$.

By convergent sequence in normed division ring is a Cauchy sequence then $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$.

Let $0_R$ be the zero of $R$ and $1_R$ be the unit of $R$.

By Norm of Unity of Division Ring and the assumption $\norm x_1 < 1$:
 * $x \ne 1_R$

Hence:
 * $x - 1_R \ne 0_R$

By norm axiom $(\text N 1)$: Positive Definiteness:
 * $\norm {x - 1_R}_2 > 0$

Let $\epsilon = \dfrac {\norm {x - 1_R}_2} 2$.

Then $\norm {x - 1_R}_2 > \epsilon$.

We have that $\norm x_2 \ge 1$.

Hence for all $n \in \N$:

For all $n \in \N$:

So $\sequence {x_n}$ is not a Cauchy sequence in $\norm {\, \cdot \,}_2$.

The theorem now follows by the Rule of Transposition.