Equivalence of Definitions of Equivalent Division Ring Norms/Norm is Power of Other Norm implies Topologically Equivalent

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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 $\norm {\,\cdot\,}_1$ and $\norm {\,\cdot\,}_2$ satisfy:

$\exists \alpha \in \R_{\gt 0}: \forall x \in R: \norm x_1 = \norm x_2^\alpha$


Then $d_1$ and $d_2$ are topologically equivalent metrics


Proof

Let $x \in R$ and $\epsilon \in \R_{\gt 0}$

Then for $y \in R$:

\(\displaystyle \norm {y - x}_1 < \epsilon\) \(\leadstoandfrom\) \(\displaystyle \norm {y - x}_2^\alpha < \epsilon\)
\(\displaystyle \) \(\leadstoandfrom\) \(\displaystyle \norm {y - x}_2 < \epsilon^{1 / \alpha}\)

Hence:

$\map {B^1_\epsilon} x = \map {B^2_{\epsilon^{1 / \alpha} } } x$

where:

$\map {B^1_\epsilon} x$ is the open ball in $d_1$ centered on $x$ with radius $\epsilon$
$\map {B^2_{\epsilon^{1 / \alpha} } } x$ is the open ball in $d_2$ centered on $x$ with radius $\epsilon^{1 / \alpha}$


Since $x$ and $\epsilon$ were arbitrary then:

every $d_1$-open ball is a $d_2$-open ball.


Similarly, for $y \in R$:

\(\displaystyle \norm {y - x}_2 < \epsilon\) \(\leadstoandfrom\) \(\displaystyle \norm {y - x}_2^\alpha < \epsilon^\alpha\)
\(\displaystyle \) \(\leadstoandfrom\) \(\displaystyle \norm {y - x}_1 < \epsilon^\alpha\)

So:

every $d_2$-open ball is a $d_1$-open ball.


By the definition of an open set of a metric space it follows that $d_1$ and $d_2$ are topologically equivalent metrics,


$\blacksquare$

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