Equivalence of Definitions of Equivalent Division Ring Norms/Convergently Equivalent implies Null Sequence 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 $R:\sequence {x_n}$ converges to $l$ in $\norm{\, \cdot \,}_1 \iff \sequence {x_n}$ is a converges to $l$ in $\norm {\, \cdot \,}_2$

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

Proof
Let $0_R$ be the zero of $R$, then:
 * $\sequence {x_n}$ converges to $0_R$ in $\norm {\, \cdot \,}_1 \iff \sequence {x_n}$ converges to $0_R$ in $\norm {\, \cdot \,}_2$

Hence:
 * $\sequence {x_n}$ is a null sequence in $\norm{\,\cdot\,}_1 \iff \sequence {x_n}$ is a null sequence in $\norm{\,\cdot\,}_2$