Equivalence of Definitions of Equivalent Division Ring Norms

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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.


The following definitions of the concept of Equivalent Division Ring Norms are equivalent:

Topologically Equivalent

$\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equivalent if and only if $d_1$ and $d_2$ are topologically equivalent metrics.

Convergently Equivalent

$\norm {\,\cdot\,}_1$ and $\norm {\,\cdot\,}_2$ are equivalent if and only if for all sequences $\sequence {x_n}$ in $R$:

$\sequence {x_n}$ converges to $l$ in $\norm{\,\cdot\,}_1 \iff \sequence {x_n}$ converges to $l$ in $\norm {\,\cdot\,}_2$

Null Sequence Equivalent

$\norm {\,\cdot\,}_1$ and $\norm {\,\cdot\,}_2$ are equivalent if and only if for all sequences $\sequence {x_n}$ in $R$:

$\sequence {x_n}$ is a null sequence in $\norm{\,\cdot\,}_1 \iff \sequence {x_n}$ is a null sequence in $\norm {\,\cdot\,}_2$

Open Unit Ball Equivalent

$\norm {\,\cdot\,}_1$ and $\norm {\,\cdot\,}_2$ are equivalent if and only if $\forall x \in R: \norm x_1 < 1 \iff \norm x_2 < 1$

Norm is Power of Other Norm

$\norm{\,\cdot\,}_1$ and $\norm{\,\cdot\,}_2$ are equivalent if and only if $\exists \alpha \in \R_{\gt 0}: \forall x \in R: \norm{x}_1 = \norm{x}_2^\alpha$

Cauchy Sequence Equivalent

$\norm {\,\cdot\,}_1$ and $\norm {\,\cdot\,}_2$ are equivalent if and only if 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

Topologically Equivalent implies Convergently Equivalent

Let $d_1$ and $d_2$ be topologically equivalent metrics.


Then:

$d_1$ and $d_2$ are convergently equivalent metrics.

$\Box$


Convergently Equivalent implies Null Sequence Equivalent

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

for all sequences $\sequence {x_n}$ in $R:\sequence {x_n}$ converges to $l$ in $\norm{\, \cdot \,}_1 \iff \sequence {x_n}$ is a converges to $l$ in $\norm {\, \cdot \,}_2$


Then for all sequences $\sequence {x_n}$ in $R$:

$\sequence {x_n}$ is a null sequence in $\norm {\, \cdot \,}_1 \iff \sequence {x_n}$ is a null sequence in $\norm {\, \cdot \,}_2$

$\Box$


Null Sequence Equivalent implies Open Unit Ball Equivalent

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

for all sequences $\sequence {x_n}$ in $R:\sequence {x_n}$ is a null sequence in $\norm {\, \cdot \,}_1 \iff \sequence {x_n}$ is a null sequence in $\norm {\, \cdot \,}_2$


Then $\forall x \in R$:

$\norm x_1 < 1 \iff \norm x_2 < 1$

$\Box$


Open Unit Ball Equivalent implies Norm is Power of Other Norm

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

$\forall x \in R: \norm x_1 < 1 \iff \norm x_2 < 1$


Then:

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

$\Box$


Norm is Power of Other Norm implies Topologically Equivalent

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.

$\Box$


Norm is Power of Other Norm implies Cauchy Sequence Equivalent

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

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


Then for all sequences $\sequence {x_n}$ in $R$:

$\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$ if and only if $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$

$\Box$


Cauchy Sequence Equivalent implies Open Unit Ball Equivalent

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:

for all sequences $\sequence {x_n}$ in $R$: $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$ if and only if $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$


Then $\forall x \in R$:

$\norm x_1 < 1 \iff \norm x_2 < 1$

$\blacksquare$


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