Normed Division Ring Operations are Continuous/Multiplication

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Theorem

Let $\struct {R, +, *, \norm {\,\cdot\,} }$ be a normed division ring.

Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.

Let $p \in \R_{\ge 1} \cup \set \infty$.

Let $d_p$ be the $p$-Product Metric on $R \times R$.


Then the mapping:

$* : \struct {R \times R, d_p} \to \struct {R, d}$

is continuous.


Proof

By $p$-Product Metric Induces Product Topology and Continuous Mapping is Continuous on Induced Topological Spaces, it suffices to consider the case $p = \infty$.


Let $\tuple {x_0, y_0} \in R \times R$.

Let $\epsilon > 0$ be given.


Let $\delta = \min \set {\dfrac \epsilon {1 + \norm {y_0} + \norm {x_0} }, 1}$


Since $1 + \norm {y_0} + \norm {x_0} > 0$ then $\delta > 0$


Let $\tuple {x,y} \in R \times R$ such that:

$\map {d_\infty} {\tuple {x, y}, \tuple{x_0, y_0} } < \delta$

By the definition of the $p$-product metric $d_\infty$:

$\max \set {\map d {x, x_0}, \map d {y, y_0}} < \delta$

or equivalently:

$\map d {x, x_0} < \delta$
$\map d {y, y_0} < \delta$


Then:

\(\ds \norm y\) \(=\) \(\ds \norm {y - y_0 + y_0}\)
\(\ds \) \(\le\) \(\ds \norm {y - y_0} + \norm {y_0}\) Norm Axiom $\text N 3$: Triangle Inequality
\(\ds \) \(\le\) \(\ds \map d {y, y_0} + \norm {y_0}\) Definition of Metric Induced by Norm on Division Ring
\(\ds \) \(<\) \(\ds \delta + \norm {y_0}\)
\(\ds \) \(\le\) \(\ds 1 + \norm {y_0}\)

Hence:

\(\ds \map d {x y, x_0 y_0}\) \(=\) \(\ds \norm {x y - x_0 y_0}\) Definition of Metric Induced by Norm on Division Ring
\(\ds \) \(=\) \(\ds \norm {x y - x_0 y + x_0 y - x_0 y_0}\)
\(\ds \) \(\le\) \(\ds \norm {x y - x_0 y} + \norm {x_0 y - x_0 y_0}\) Norm Axiom $\text N 3$: Triangle Inequality
\(\ds \) \(\le\) \(\ds \norm {x - x_0} \norm y + \norm {x_0} \norm {y - y_0}\) Norm Axiom $\text N 2$: Multiplicativity
\(\ds \) \(\le\) \(\ds \map d {x, x_0} \norm y + \norm {x_0} \map d {y, y_0}\) Definition of Metric Induced by Norm on Division Ring
\(\ds \) \(<\) \(\ds \delta \norm y + \norm {x_0} \delta\)
\(\ds \) \(\le\) \(\ds \delta \paren {\norm y + \norm {x_0} }\)
\(\ds \) \(<\) \(\ds \delta \paren {1 + \norm {y_0} + \norm {x_0} }\)
\(\ds \) \(\le\) \(\ds \dfrac \epsilon {1 + \norm {y_0} + \norm {x_0} } \paren {1 + \norm {y_0} + \norm {x_0} }\)
\(\ds \) \(=\) \(\ds \epsilon\)


We have that $\tuple {x_0, y_0}$ and $\epsilon$ are arbitrary.

Hence, by the definition of continuity, the mapping:

$* : \struct {R \times R, d_\infty} \to \struct {R, d}$

is continuous.

$\blacksquare$


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