Leigh.Samphier/Sandbox/Normed Division Ring Determines Norm on Completion

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Theorem

Let $\struct {R_1, \norm {\, \cdot \,}_1 }$ be a normed division ring.

Let $\struct {R_2, \norm {\, \cdot \,}_2 }$ be a normed division ring 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 $R_1$:

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

and

$\norm x_2 = \displaystyle \lim_{n \mathop \to \infty} \norm {x_n}_1$

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 $S$:

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

and

$\norm x = \displaystyle \lim_{n \mathop \to \infty} \norm {x_n}$


Proof

$\blacksquare$