Normed Division Ring Completions are Isometric and Isomorphic/Lemma 5

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 ring isomorphism.

Proof
By lemma 4, $\psi$ is an isometry.

By the definition of an isometry, $\psi$ is a bijection.

By the definition of a ring isomorphism, all that remains is to show that $\psi$ is a ring homomorphism.

That is:
 * $(1): \quad \forall x, y \in S_1: \map \psi {x + y} = \map \psi x + \map \psi y$
 * $(2): \quad \forall x, y \in S_1: \map \psi {x y} = \map \psi x \map \psi y$

Let $x, y \in S_1$.

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

Let $y = \displaystyle \lim_{n \mathop \to \infty} y_n$ for some sequence $\sequence {y_n} \subseteq R_1$.

By Sum Rule for Sequences in Normed Division Ring then:
 * $x + y = \displaystyle \lim_{n \mathop \to \infty} \paren {x_n + y_n}$

By Product Rule for Sequences in Normed Division Ring then:
 * $x y = \displaystyle \lim_{n \mathop \to \infty} \paren {x_n y_n}$

Then:

and: