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 $\forall x \in R: \norm{x}_1 \lt 1 \iff \norm{x}_2 \lt 1$

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.

Let $\sequence {x_n}$ converge to $l$ in $\norm {\,\cdot\,}_1$.

Let $\epsilon \in \R_{\gt 0}$ be given.


Let $\map {B_\epsilon^2} i$ denote the open ball centered on $l$ of radius $\epsilon$ in $\struct {R, \norm{\,\cdot\,}_2}$.

By Open Ball is Open Set then $\map {B_\epsilon^2} l$ is open set in $\struct{R, d_2}$.

Since $d_1$ and $d_2$ are topologically equivalent metrics then $\map {B_\epsilon^2} l$ is open set in $\struct {R, d_1}$.

By the definition of an open set in a metric space then:

$\exists \delta \in \R_{\gt 0}: \map {B_\delta^1} l \subseteq \map {B_\epsilon^2} l$

Hence:

$\forall x \in R: \norm {x - l}_1 < \delta \implies \norm {x - l}_2 < \epsilon$


Since $\sequence {x_n}$ converges to $l$ in $\norm{\,\cdot\,}_1$ then:

$\exists N \in \N: \forall n \ge N: \norm {x_n - l}_1 < \delta$

Hence:

$\exists N \in \N: \forall n \ge N: \norm {x_n - l}_2 < \epsilon$


Since $\sequence {x_n}$ and $\epsilon \gt 0$ were arbitrary then it has been shown that for all sequences $\sequence {x_n}$ in $R$:

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


By a similar argument it is shown that for all sequences $\sequence {x_n}$ in $R$:

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

The result follows.

$\Box$


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 $\norm{\,\cdot\,}_2$


Let $0_R$ be the zero of $R$, then:

$\sequence {x_n}$ converges to $0_R$ in $\norm{\,\cdot\,}_1 \iff \sequence {x_n}$ converges to $0_R$ in $\norm{\,\cdot\,}_2$

Hence:

$\sequence {x_n}$ 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 $\norm{\,\cdot\,}_2$


Let $x \in R$.

Let $\sequence {x_n}$ be the sequence defined by: $\forall n: x_n = x^n$.


\(\displaystyle \norm{x}_1 \lt 1 \quad\) \(\iff\) \(\displaystyle \) $\sequence {x_n}$ is a null sequence in $\norm{\,\cdot\,}_1$ Sequence of Powers of Number less than One in Normed Division Ring
\(\displaystyle \) \(\iff\) \(\displaystyle \) $\sequence {x_n}$ is a null sequence in $\norm{\,\cdot\,}_2$ Assumption
\(\displaystyle \) \(\iff\) \(\displaystyle \) $\norm{x}_2 \lt 1$ Sequence of Powers of Number less than One in Normed Division Ring


$\Box$


Open Unit Ball Equivalent implies Norm is Power of Other Norm

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

$\forall x \in R:\norm{x}_1 \lt 1 \iff \norm{x}_2 \lt 1$


Case 1

Let for all $x \in R, x \neq 0_R$, satisfy $\norm{x}_1 \ge 1$.

Lemma 1

Then:

$\norm{\,\cdot\,}_1$ is the trivial norm.


By assumption, for all $x \in R, x \neq 0_R$, then $\norm{x}_2 \ge 1$.

Similarly $\norm{\,\cdot\,}_2$ is the trivial norm.

Hence $\norm{\,\cdot\,}_1$ and $\norm{\,\cdot\,}_2$ are equal.

For $\alpha = 1$ the result follows.

$\Box$

Case 2

Let $x_0 \in R$ such that $x_0 \neq 0_R$ and $\norm{x_0}_1 \lt 1$.

By assumption then $\norm{x_0}_2 \lt 1$.

Let $\alpha = \dfrac {\log \norm {x_0}_1 } {\log \norm {x_0}_2 }$.

Then $\norm{x_0}_1 = \norm{x_0}_2^\alpha$.


Since $\norm{x_0}_1, \norm{x_0}_2 \lt 1$ then:

$\log \norm {x_0}_1 < 0$
$\log \norm {x_0}_2 < 0$

So $\alpha \gt 0$.

Lemma 2

Then:

$\forall x \in R: \norm{x}_1 = \norm{x}_2^\alpha$


$\Box$


Norm is Power of Other Norm implies Topologically Equivalent

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

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

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,

$\Box$


Norm is Power of Other Norm implies Cauchy Sequence Equivalent

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

$\exists \alpha \in \R_{\gt 0}: \forall x \in R: \norm{x}_1 = \norm{x}_2^\alpha$

Let $\sequence {x_n}$ be a Cauchy sequence in $\norm{\,\cdot\,}_1$.

Let $\epsilon \gt 0$ be given.


Since $\sequence {x_n}$ is a Cauchy sequence then:

$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_1 \lt \epsilon^\alpha$

Then:

$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2^\alpha \lt \epsilon^\alpha$

Hence:

$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2 \lt \epsilon$


So $\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_2$


It follows that for all sequences $\sequence {x_n}$ in $R$:

$\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_1 \implies \sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_2$

$\Box$

Let $\sequence {x_n}$ be a Cauchy sequence in $\norm{\,\cdot\,}_2$.

Let $\epsilon \gt 0$ be given.


Since $\sequence {x_n}$ is a Cauchy sequence then:

$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2 \lt \epsilon^{1/\alpha}$

Then:

$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2^\alpha \lt \epsilon$

Hence:

$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_1 \lt \epsilon$


So $\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_1$


It follows that for all sequences $\sequence {x_n}$ in $R$:

$\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_2 \implies \sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_1$


The result follows

$\Box$


Cauchy Sequence Equivalent implies Open Unit Ball Equivalent

The contrapositive is proved.


Let there exist $x \in R$ such that $\norm{x}_1 \lt 1$ and $\norm{x}_2 \ge 1$.

Let $\sequence {x_n}$ be the sequence defined by: $\forall n: x_n = x^n$.

By Sequence of Powers of Number less than One in Normed Division Ring then $\sequence {x_n}$ is a null sequence in $\norm{\,\cdot\,}_1$.

By convergent sequence in normed division ring is a Cauchy sequence then $\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_1$.


Let $0_R$ be the zero of $R$ and $1_R$ be the unit of $R$.

By norm of unity and the assumption $\norm{x}_1 \lt 1$ then $x \neq 1_R$.

Then $x - 1_R \neq 0_R$.

By norm axiom (N1) (Positive Definiteness) then $\norm {x - 1_R}_2 \gt 0$.


Let $\epsilon = \dfrac {\norm {x - 1_R}_2} 2$.


Then $\norm {x - 1_R}_2 \gt \epsilon$.


Since $\norm{x}_2 \ge 1$ then for all $n \in \N$ then:

\(\displaystyle \norm{x_n}_2\) \(=\) \(\displaystyle \norm{x^n}_2\) Definition of $x_n$
\(\displaystyle \) \(=\) \(\displaystyle \norm{x}_2^n\) norm axiom (N2) (Multiplicativity)
\(\displaystyle \) \(\ge\) \(\displaystyle 1\)


For all $n \in \N$ then:

\(\displaystyle \norm {x_{n+1} - x_n}_2\) \(=\) \(\displaystyle \norm {x^{n+1} - x^n}_2\)
\(\displaystyle \) \(=\) \(\displaystyle \norm {x^n x - x^n}_2\)
\(\displaystyle \) \(=\) \(\displaystyle \norm {x^n \paren {x - 1_R} }_2\)
\(\displaystyle \) \(=\) \(\displaystyle \norm {x^n}_2 \norm {x - 1_R}_2\) norm axiom (N2) (Multiplicativity)
\(\displaystyle \) \(\gt\) \(\displaystyle \epsilon\)

So $\sequence {x_n}$ is not a Cauchy sequence in $\norm{\,\cdot\,}_2$.

The theorem now follows by the Rule of Transposition.

$\blacksquare$


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