# 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 $\norm{\,\cdot\,}_2$

$\blacksquare$