Normed Division Ring Completions are Isometric and Isomorphic/Lemma 4
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Theorem
Let $\struct {S_1, \norm {\, \cdot \,}_1 }$ and $\struct {S_2, \norm {\, \cdot \,}_2 }$ be complete normed division rings.
Let $R_1$ be a dense subring of $S_1$.
Let $R_2$ be a dense subring of $S_2$.
Let $\psi': R_1 \to R_2$ be an isometric ring isomorphism.
Let $\psi:S_1 \to S_2$ be defined by:
- $\forall x \in S_1: \map \psi x = \ds \lim_{n \mathop \to \infty} \map {\psi'}{x_n}$
- where $x = \ds \lim_{n \mathop \to \infty} x_n$ for some sequence $\sequence {x_n} \subseteq R_1$
Then:
- $\psi$ is an isometry.
Proof
Let $x, y \in S_1$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $R_1$ such that $\ds \lim_{n \mathop \to \infty} x_n = x, \lim_{n \mathop \to \infty} y_n = y$.
Then:
\(\ds x - y\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} x_n - y_n\) | Difference Rule for Sequences in Normed Division Ring | |||||||||||
\(\ds \) | \(\) | \(\ds \) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \norm {x - y}\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \norm {x_n - y_n}\) | Modulus of Limit | ||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \norm {\map {\psi'} {x_n} - \map {\psi'} {y_n} }\) | Definition of Isometry (Metric Spaces) |
On the other hand:
\(\ds \map \psi x - \map \psi y\) | \(=\) | \(\ds \paren {\lim_{n \mathop \to \infty} \map {\psi'} {x_n} } - \paren {\lim_{n \mathop \to \infty} \map {\psi'} {y_n} }\) | Definition of $\psi$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \map {\psi'} {x_n} - \map {\psi'} {y_n}\) | Difference Rule for Sequences in Normed Division Ring | |||||||||||
\(\ds \) | \(\) | \(\ds \) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \norm {\map \psi x - \map \psi y}\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \norm {\map {\psi'} {x_n} - \map {\psi'} {y_n} }\) | Modulus of Limit |
By Convergent Sequence in Metric Space has Unique Limit:
- $\norm {x - y} = \norm {\map \psi x - \map \psi y}$
It follows that $\psi$ is distance-preserving.
By Distance-Preserving Surjection is Isometry of Metric Spaces then $\psi$ is an isometry.
$\blacksquare$