# 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,

$\blacksquare$

## Sources

- 1997: Fernando Q. Gouvea:
*p-adic Numbers: An Introduction*: $\S 3.1$ Absolute Values on $\Q$, Lemma $3.1.2$ and Problem $66$