Leigh.Samphier/Sandbox/Element of Completion is Limit of Sequence in Normed Division Ring

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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:

there exists a sequence $\sequence {y_n} \subseteq \map {\phi^\to} {R_1}$ that converges to $x$

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 $R_1$:

$\displaystyle \lim_{n \mathop \to \infty} \map \phi {x_n} = \lim_{n \mathop \to \infty} y_n = x$

$\blacksquare$