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 $y \in R$:
\(\ds \norm {y - x}_1 < \epsilon\) | \(\leadstoandfrom\) | \(\ds \norm {y - x}_2^\alpha < \epsilon\) | ||||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds \norm {y - x}_2 < \epsilon^{1 / \alpha}\) |
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 $y \in R$:
\(\ds \norm {y - x}_2 < \epsilon\) | \(\leadstoandfrom\) | \(\ds \norm {y - x}_2^\alpha < \epsilon^\alpha\) | ||||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds \norm {y - x}_1 < \epsilon^\alpha\) |
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,
$\blacksquare$
Sources
- 1997: Fernando Q. Gouvea: p-adic Numbers: An Introduction: $\S 3.1$ Absolute Values on $\Q$, Lemma $3.1.2$ and Problem $66$