Normed Division Ring Completions are Isometric and Isomorphic/Lemma 2

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 $\ds x = \lim_{n \mathop \to \infty} x_n$ for some sequence $\sequence {x_n} \subseteq R_1$

Then $\psi$ is a well-defined mapping.

Proof
Let $x \in S_1$.

By the definition of dense subset:
 * $\map \cl {R_1} = S_1$

By Closure of Subset of Metric Space by Convergent Sequence, there exists a sequence $\sequence {x_n} \subseteq R_1 $ that converges to $x$, that is:
 * $\ds \lim_{n \mathop \to \infty} x_n = x$

By Isometric Image of Cauchy Sequence is Cauchy Sequence, $\sequence {\map {\psi'} {x_n} }$ is a Cauchy sequence in $R_2 \subseteq S_2$.

Since $S_2$ is complete then the sequence $\sequence {\map {\psi'} {x_n} }$ has a limit, say $y$.

Let $\sequence {x_n}$ and $\sequence {x'_n}$ be sequences in $\map {\phi_1} R$ such that:
 * $\ds \lim_{n \mathop \to \infty} x_n = \lim_{n \mathop \to \infty} x'_n = x$

Then:

The result follows.