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.

$\blacksquare$