# Equivalence of Definitions of Equivalent Division Ring Norms/Norm is Power of Other Norm implies Cauchy 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:

$\exists \alpha \in \R_{\gt 0}: \forall x \in R: \norm{x}_1 = \norm{x}_2^\alpha$

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

## Proof

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

Let $\epsilon \gt 0$ be given.

Since $\sequence {x_n}$ is a Cauchy sequence then:

$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_1 \lt \epsilon^\alpha$

Then:

$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2^\alpha \lt \epsilon^\alpha$

Hence:

$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2 \lt \epsilon$

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

It follows that for all sequences $\sequence {x_n}$ in $R$:

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

$\Box$

Let $\sequence {x_n}$ be a Cauchy sequence in $\norm{\,\cdot\,}_2$.

Let $\epsilon \gt 0$ be given.

Since $\sequence {x_n}$ is a Cauchy sequence then:

$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2 \lt \epsilon^{1/\alpha}$

Then:

$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2^\alpha \lt \epsilon$

Hence:

$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_1 \lt \epsilon$

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

It follows that for all sequences $\sequence {x_n}$ in $R$:

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

The result follows

$\blacksquare$