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 $y \in R$:

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

Hence:

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

where:

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


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

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


Similarly, for $y \in R$:

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

So:

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


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


$\blacksquare$

Sources