Leigh.Samphier/Sandbox/Element of Completion is Limit of Sequence in Normed Division Ring
Theorem
Let $\struct {R_1, \norm {\, \cdot \,}_1 }, \struct {R_2, \norm {\, \cdot \,}_2 }$ be normed division rings.
Let $\struct {R_2, \norm {\, \cdot \,}_2 }$ be a completion of $\struct {R_1, \norm {\, \cdot \,}_1 }$ with distance-preserving ring monomorphism $\phi: R_1 \to R_2$.
Then for all $x \in R_2$, there exists a sequence $\sequence{x_n}$ in $R_1$:
- $x = \displaystyle \lim_{n \mathop \to \infty} \map \phi {x_n}$
Corollary
Let $\struct {R, \norm {\, \cdot \,} }$ be a complete normed division ring.
Let $\struct {S, \norm {\, \cdot \,}}$ be a dense normed division subring of $\struct {R, \norm {\, \cdot \,}}$.
Then for all $x \in R$, there exists a sequence $\sequence{x_n}$ in $S$:
- $x = \displaystyle \lim_{n \mathop \to \infty} x_n$
Proof
Let $x \in R_2$.
By definition of the completion $\struct {R_2, \norm {\, \cdot \,}_2 }$:
- $\map {\phi^\to} {R_1}$ is a dense subset of $\struct {R_2, \norm {\, \cdot \,}_2 }$.
By the definition of a dense subset:
- $\map \cl {\map {\phi^\to} {R_1}} = R_2$
By Closure of Subset of Metric Space by Convergent Sequence:
That is:
- $\displaystyle \lim_{n \mathop \to \infty} y_n = x$
From Injection to Image is Bijection, we can define:
- $\forall n \in \N : x_n = \map {\phi^{-1}} {y_n}$
Then $\sequence{x_n}$ is a sequence in $R_1$:
- $\displaystyle \lim_{n \mathop \to \infty} \map \phi {x_n} = \lim_{n \mathop \to \infty} y_n = x$
$\blacksquare$