Equivalence of Definitions of Equivalent Division Ring Norms/Topologically Equivalent implies Convergently Equivalent

Theorem
Let $\struct{R, \norm{\,\cdot\,}_1}$ and $\struct{R, \norm{\,\cdot\,}_2}$ be normed division rings on the same underlying division ring $R$. Let $\norm{\,\cdot\,}_1$ and $\norm{\,\cdot\,}_2$ be topologically equivalent norms.

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 $R$.

Let $\sequence{x_n}$ be a null sequence in $\norm{\,\cdot\,}_1$.

Let $\epsilon \in \R_{\gt 0}$ be given.

Let $B_\epsilon^2 \paren {0_R}$ denote the open ball centred on $0_R$ of radius $\epsilon$ in $\struct{R, \norm{\,\cdot\,}_2}$.

By Open Ball is Open then $B_\epsilon^2 \paren {0_R}$ is open in $\struct{R, \norm{\,\cdot\,}_2}$.

Since $\norm{\,\cdot\,}_1$ and $\norm{\,\cdot\,}_2$ are topologically equivalent norms then $B_\epsilon^2 \paren {0_R}$ is open in $\struct{R, \norm{\,\cdot\,}_1}$.

By Open Balls are local base then:
 * $\exists \delta \in \R_{\gt 0}: B_\delta^1 \paren {0_R} \subseteq B_\epsilon^2 \paren {0_R}$

Hence:
 * $\forall x \in R: \norm{x}_1 < \delta \implies \norm{x}_2 < \epsilon$

Since $\sequence{x_n}$ is a null sequence in $\norm{\,\cdot\,}_1$ then:
 * $\exists N \in \N: \forall n \ge N: \norm{x_n}_1 < \delta$

Hence:
 * $\exists N \in \N: \forall n \ge N: \norm{x_n}_2 < \epsilon$

Since $\sequence{x_n}$ and $\epsilon \gt 0$ were arbitrary then it has been shown that:
 * For all sequences $\sequence {x_n} \in R: \sequence {x_n}$ is a null sequence in $\norm{\,\cdot\,}_1 \implies \sequence {x_n}$ is a null sequence in $\norm{\,\cdot\,}_2$

By a similar argument it is shown that:
 * For all sequences $\sequence {x_n} \in R: \sequence {x_n}$ is a null sequence in $\norm{\,\cdot\,}_2 \implies \sequence {x_n}$ is a null sequence in $\norm{\,\cdot\,}_1$

The result follows.