Normed Division Ring Completions are Isometric and Isomorphic/Lemma 4

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:
 * $\forall x \in S_1: \psi \paren x = \displaystyle \lim_{n \to \infty} \psi’ \paren {x_n}$
 * where $x = \displaystyle \lim_{n \to \infty} x_n$ 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 $\displaystyle \lim_{n \to \infty} x_n = x, \lim_{n \to \infty} y_n = y$ then:

On the other hand:

By Convergent Sequence in Metric Space has Unique Limit then:
 * $\norm {x - y} = \norm {\psi \paren x - \psi \paren y}$

It follows that $\psi$ is distance-preserving.

By Distance-Preserving Surjection is Isometry of Metric Spaces then $\psi$ is an isometry.