Equivalence of Definitions of Equivalent Division Ring Norms/Topologically Equivalent implies Convergently 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 $d_1$ and $d_2$ be the metrics induced by the norms $\norm{\,\cdot\,}_1$ and $\norm{\,\cdot\,}_2$ respectively. Let $d_1$ and $d_2$ be topologically equivalent metrics.

Then:
 * $d_1$ and $d_2$ Are convergently equivalent metrics.

Proof
Let $\sequence{x_n}$ converge to $l$ in $\norm{\,\cdot\,}_1$.

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

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

By Open Ball is Open Set then $B_\epsilon^2 \paren {l}$ is open set in $\struct{R, d_2}$.

Since $d_1$ and $d_2$ are topologically equivalent metrics then $B_\epsilon^2 \paren {l}$ is open set in $\struct{R, d_1}$.

By the definition of an open set in a metric space then:
 * $\exists \delta \in \R_{\gt 0}: B_\delta^1 \paren {l} \subseteq B_\epsilon^2 \paren {l}$

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

Since $\sequence{x_n}$ converges to $l$ in $\norm{\,\cdot\,}_1$ then:
 * $\exists N \in \N: \forall n \ge N: \norm{x_n - l}_1 < \delta$

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

Since $\sequence{x_n}$ and $\epsilon \gt 0$ were arbitrary then it has been shown that for all sequences $\sequence {x_n}$ in $R$:
 * $\sequence {x_n}$ converges to $l$ in $\norm{\,\cdot\,}_1 \implies \sequence {x_n}$ converges to $l$ in $\norm{\,\cdot\,}_2$.

By a similar argument it is shown that for all sequences $\sequence {x_n}$ in $R$:
 * $\sequence {x_n}$ converges to $l$ in $\norm{\,\cdot\,}_2\implies \sequence {x_n}$ converges to $l$ in $\norm{\,\cdot\,}_1$.

The result follows.