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 $\norm{\,\cdot\,}_1 \iff \sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_2$

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

Proof
The contrapositive is proved.

Let there exist $x \in R$ such that $\norm{x}_1 \lt 1$ 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 and the assumption $\norm{x}_1 \lt 1$ then $x \neq 1_R$.

Then $x - 1_R \neq 0_R$.

By norm axiom (N1) (Positive Definiteness) then $\norm {x - 1_R}_2 \gt 0$.

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

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

Since $\norm{x}_2 \ge 1$ then for all $n \in \N$ then:

For all $n \in \N$ then:

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

The theorem now follows by the Rule of Transposition.