Equivalence of Definitions of Equivalent Division Ring Norms/Cauchy Sequence Equivalent implies Open Unit Ball 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 $\norm {\, \cdot \,}_1$ 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 $\forall x \in R$:

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


Proof

The contrapositive is proved.


Let there exist $x \in R$ such that $\norm x_1 < 1$ 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 of Division Ring and the assumption $\norm x_1 < 1$:

$x \ne 1_R$

Hence:

$x - 1_R \ne 0_R$

By norm axiom $(\text N 1)$: Positive Definiteness:

$\norm {x - 1_R}_2 > 0$


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

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


We have that $\norm x_2 \ge 1$.

Hence for all $n \in \N$:

\(\ds \norm {x_n}_2\) \(=\) \(\ds \norm {x^n}_2\) Definition of $x_n$
\(\ds \) \(=\) \(\ds \norm x_2^n\) Norm Axiom $(\text N 2)$: Multiplicativity
\(\ds \) \(\ge\) \(\ds 1\)


For all $n \in \N$:

\(\ds \norm {x_{n + 1} - x_n}_2\) \(=\) \(\ds \norm {x^{n + 1} - x^n}_2\)
\(\ds \) \(=\) \(\ds \norm {x^n x - x^n}_2\)
\(\ds \) \(=\) \(\ds \norm {x^n \paren {x - 1_R} }_2\)
\(\ds \) \(=\) \(\ds \norm {x^n}_2 \norm {x - 1_R}_2\) Norm Axiom $(\text N 2)$: Multiplicativity
\(\ds \) \(>\) \(\ds \epsilon\)

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

The theorem now follows by the Rule of Transposition.

$\blacksquare$


Sources

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