Normed Division Ring Completions are Isometric and Isomorphic/Lemma 3

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:
 * $\forall x \in S_1: \map \psi x = \displaystyle \lim_{n \mathop \to \infty} \map {\psi'} {x_n}$

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

Then:
 * $\psi$ is a surjective mapping.

Proof
Let $y \in S_2$.

By the definition of dense subset:
 * $\map \cl {R_2} = S_2$

By Closure of Subset of Metric Space by Convergent Sequence:
 * there exists a sequence $\sequence {y_n} \subseteq R_2 $ that converges to $y$, that is, $\displaystyle \lim_{n \mathop \to \infty} y_n = y$

By Isometric Image of Cauchy Sequence is Cauchy Sequence, $\sequence {\map {\psi'^{-1} } {y_n} }$ is a Cauchy sequence in $R_1 \subseteq S_1$.

Because $S_1$ is complete, the sequence $\sequence {\map {\psi’^{-1} } {y_n} }$ has a limit, say $x$.

By the definition of $\psi$: