Normed Division Ring Operations are Continuous/Inversion

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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 $R^* = R \setminus \set 0$

Let $d^*$ be the subspace metric on $R^*$.


Then the mapping:

$\iota : \struct {R^* ,d^*} \to \struct {R, d} : \map \iota x = x^{-1}$

is continuous.


Proof

Let $x_0 \in R^*$.

Let $\epsilon > 0$ be given.


Let $\delta = \min \set {\dfrac {\norm {x_0} } 2, \dfrac {\norm {x_0}^2 \epsilon} 2 }$

Let $x \in R^*$ such that:

$\map {d^*} {x, x_0} < \delta$

By the definition of the subspace metric on $R^*$ and the definition of the metric induced by the norm on $R$:

$\map {d^*} {x, x_0} = \map d {x, x_0} = \norm {x - x_0} < \delta$

Then:

\(\ds \norm {x_0}\) \(\le\) \(\ds \norm {x - x_0} + \norm x\) Norm Axiom $\text N 3$: Triangle Inequality
\(\ds \) \(\) \(\ds \)
\(\ds \) \(<\) \(\ds \delta + \norm x\) by hypothesis: $\norm {x - x_0} < \delta$
\(\ds \) \(\) \(\ds \)
\(\ds \) \(\le\) \(\ds \dfrac {\norm {x_0} } 2 + \norm x\) by hypothesis: $\delta \le \dfrac {\norm {x_0} } 2$
\(\ds \) \(\) \(\ds \)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\norm {x_0} } 2\) \(<\) \(\ds \norm x\) subtracting $\dfrac {\norm {x_0} } 2$ from both sides
\(\ds \) \(\) \(\ds \)
\(\ds \leadsto \ \ \) \(\ds \dfrac 2 {\norm {x_0} }\) \(>\) \(\ds \dfrac 1 {\norm x}\) inverting both sides of the equation


Hence:

\(\ds \map d {x^{-1}, x_0^{-1} }\) \(=\) \(\ds \norm {x^{-1} - x_0^{-1} }\) Definition of Metric Induced by Norm on Division Ring
\(\ds \) \(\) \(\ds \)
\(\ds \) \(=\) \(\ds \dfrac 1 {\norm x} \paren {\norm x \norm {x^{-1} - x_0^{-1} } \norm {x_0} } \dfrac 1 {\norm {x_0} }\)
\(\ds \) \(\) \(\ds \)
\(\ds \) \(=\) \(\ds \dfrac 1 {\norm x \norm {x_0} } \paren {\norm {x \paren {x^{-1} - x_0^{-1} } x_0} }\) Norm Axiom $\text N 2$: Multiplicativity
\(\ds \) \(\) \(\ds \)
\(\ds \) \(=\) \(\ds \dfrac 1 {\norm x \norm {x_0} } \paren {\norm {x x^{-1} x_0 - x x_0^{-1} x_0} }\) Ring Axiom $\text D$: Distributivity of Product over Addition
\(\ds \) \(\) \(\ds \)
\(\ds \) \(=\) \(\ds \dfrac 1 {\norm x \norm {x_0} } \paren {\norm {x_0 - x} }\) Definition of Division Ring
\(\ds \) \(\) \(\ds \)
\(\ds \) \(<\) \(\ds \dfrac 2 {\norm {x_0}^2} \paren {\norm {x_0 - x} }\) from $\dfrac 1 {\norm x} < \dfrac 2 {\norm {x_0} }$ above
\(\ds \) \(\) \(\ds \)
\(\ds \) \(<\) \(\ds \dfrac {2 \delta} {\norm {x_0}^2}\) by hypothesis: $\norm {x - x_0} < \delta$
\(\ds \) \(\) \(\ds \)
\(\ds \) \(\le\) \(\ds \dfrac 2 {\norm {x_0}^2} \paren {\dfrac {\norm {x_0}^2 \epsilon} 2}\) by hypothesis: $\delta \le \dfrac {\norm {x_0}^2 \epsilon} 2$
\(\ds \) \(\) \(\ds \)
\(\ds \) \(=\) \(\ds \epsilon\) cancelling terms


Since $x_0$ and $\epsilon$ were arbitrary, by the definition of continuity then the mapping:

$\iota: \struct {R^*, d^*} \to \struct {R, d} : \map \iota x = x^{-1}$

is continuous.

$\blacksquare$


Sources

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