nyuuzyou/wikis huggingface dataset:
Normed Division Ring Completions are Isometric and Isomorphic

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\struct {S_1, \norm {\, \cdot \,}_1 }$ and $\struct {S_2, \norm {\, \cdot \,}_2 }$ be normed division ring completions of $\struct {R, \norm {\, \cdot \,} }$
Then there exists an isometric isomorphism:
:$\psi: \struct {S_1, \norm {\, \cdot \,}_1 } \to \struct {S_2, \norm {\, \cdot \,}_2 }$
== Proof ==
By the definition of a normed division ring completion then:
:there exists a distance-preserving ring monomorphisms $\phi_1: R \to S_1$
:$R_1 = \map {\phi_1} R$ is a dense subring of $S_1$
:$S_1$ is a complete metric space
:there exists a distance-preserving ring monomorphisms $\phi_2: R \to S_2$
:$R_2 = \map {\phi_2} R$ is a dense subring of $S_2$
:$S_2$ is a complete metric space
=== Lemma 1 ===
=== Lemma 2 ===
=== Lemma 3 ===
=== Lemma 4 ===
=== Lemma 5 ===
Category:Normed Division Ring Completions are Isometric and Isomorphic
Category:Completion of Normed Division Ring
Category:Normed Division Rings
Category:Complete Metric Spaces
Normed Division Ring Completions are Isometric and Isomorphic/Lemma 1

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }, \struct {S_1, \norm {\, \cdot \,}_1 }$ and $\struct {S_2, \norm {\, \cdot \,}_2 }$ be normed division rings.
Let $\phi_1: R \to S_1, \phi_2: R \to S_2$ be distance-preserving ring monomorphisms.
Let $\psi' = \phi_2 \circ \phi_1^{-1}:\phi_1 \paren R \to \phi_2 \paren R$ be the composition of $\phi_1^{-1}$ with $\phi_2$. 
Then $\psi': \struct {\map {\phi_1} R, \norm {\, \cdot \,}_1 } \to \struct {\map {\phi_2} R, \norm {\, \cdot \,}_2 }$ is an isometric ring isomorphism.
== Proof ==
By Monomorphism Image is Isomorphic to Domain, $\phi_1:R \to \map {\phi_1} R$ and $\phi_2:R \to \map {\phi_2} R$ are ring isomorphisms.
By Distance-Preserving Image Isometric to Domain for Metric Spaces, $\phi_1:R \to \map {\phi_1} R$ and $\phi_2:R \to \map {\phi_2} R$ are isometries.
By Inverse of Algebraic Structure Isomorphism is Isomorphism, $\phi_1^{-1}:\map {\phi_1} R \to R$ is a ring isomorphism.
By Inverse of Isometry of Metric Spaces is Isometry, $\phi_1^{-1}:\map {\phi_1} R \to R$ is a ring isomorphism is an isometry.
By Composition of Ring Isomorphisms is Ring Isomorphism, $\psi'$ is a ring isomorphism.
By Composition of Isometries is Isometry, $\psi'$ is an isometry.
Category:Normed Division Ring Completions are Isometric and Isomorphic
Normed Division Ring Completions are Isometric and Isomorphic/Lemma 2

== 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:
:$\forall x \in S_1: \map \psi x = \ds \lim_{n \mathop \to \infty} \map {\psi'} {x_n}$
where $\ds x = \lim_{n \mathop \to \infty} x_n$ for some sequence $\sequence {x_n} \subseteq R_1$
Then $\psi$ is a well-defined mapping.
== Proof ==
Let $x \in S_1$.
By the definition of dense subset:
:$\map \cl {R_1} = S_1$
By Closure of Subset of Metric Space by Convergent Sequence, there exists a sequence $\sequence {x_n} \subseteq R_1 $ that converges to $x$, that is:
:$\ds \lim_{n \mathop \to \infty} x_n = x$
By Isometric Image of Cauchy Sequence is Cauchy Sequence, $\sequence {\map {\psi'} {x_n} }$ is a Cauchy sequence in $R_2 \subseteq S_2$.
Since $S_2$ is complete then the sequence $\sequence {\map {\psi'} {x_n} }$ has a limit, say $y$.
Let $\sequence {x_n}$ and $\sequence {x'_n}$ be sequences in $\map {\phi_1} R$ such that:
:$\ds \lim_{n \mathop \to \infty} x_n = \lim_{n \mathop \to \infty} x'_n = x$
Then:
The result follows.
Category:Normed Division Ring Completions are Isometric and Isomorphic
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:
:$\forall x \in S_1: \map \psi x = \ds \lim_{n \mathop \to \infty} \map {\psi'} {x_n}$
where $x = \ds \lim_{n \mathop \to \infty} x_n$ 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, $\ds \lim_{n \mathop \to \infty} y_n = y$
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 $\psi$:
Category:Normed Division Ring Completions are Isometric and Isomorphic
Normed Division Ring Completions are Isometric and Isomorphic/Lemma 4

== 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:
:$\forall x \in S_1: \map \psi x = \ds \lim_{n \mathop \to \infty} \map {\psi'}{x_n}$
:where $x = \ds \lim_{n \mathop \to \infty} x_n$ for some sequence $\sequence {x_n} \subseteq R_1$
Then:
:$\psi$ is an isometry.
== Proof ==
Let $x, y \in S_1$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $R_1$ such that $\ds \lim_{n \mathop \to \infty} x_n = x, \lim_{n \mathop \to \infty} y_n = y$.
Then:
On the other hand:
By Convergent Sequence in Metric Space has Unique Limit:
:$\norm {x - y} = \norm {\map \psi x - \map \psi y}$
It follows that $\psi$ is distance-preserving.
By Distance-Preserving Surjection is Isometry of Metric Spaces then $\psi$ is an isometry.
Category:Normed Division Ring Completions are Isometric and Isomorphic
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:
:$\forall x \in S_1: \map \psi x = \ds \lim_{n \mathop \to \infty} \map {\psi'} {x_n}$
:where $x = \ds \lim_{n \mathop \to \infty} x_n$ for some sequence $\sequence {x_n} \subseteq R_1$.
Then:
:$\psi$ is a ring isomorphism.
== Proof ==
By , $\psi$ 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 $x = \ds \lim_{n \mathop \to \infty} x_n$ for some sequence $\sequence {x_n} \subseteq R_1$.
Let $y = \ds \lim_{n \mathop \to \infty} y_n$ for some sequence $\sequence {y_n} \subseteq R_1$.
By Sum Rule for Sequences in Normed Division Ring then:
:$x + y = \ds \lim_{n \mathop \to \infty} \paren {x_n + y_n}$
By Product Rule for Sequences in Normed Division Ring then:
:$x y = \ds \lim_{n \mathop \to \infty} \paren {x_n y_n}$
Then:
and:
Category:Normed Division Ring Completions are Isometric and Isomorphic
Normed Division Ring Operations are Continuous

== Theorem ==
Let $\struct {R, +, *, \norm {\,\cdot\,} }$ be a normed division ring.
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
Let $p \in \R_{\ge 1} \cup \set \infty$.
Let $d_p$ be the $p$-product metric on $R \times R$.
Let $R^* = R \setminus \set 0$
Let $d^*$ be the restriction of $d$ to $R^*$.
Then the following results hold:
=== Addition is Continuous ===
The mapping:
=== Negation is Continuous ===
The mapping:
=== Multiplication is Continuous ===
The mapping:
=== Inversion is Continuous ===
The mapping:
== Corollary ==
.
== Sources ==
* : $\S 2.3$: Topology: Problem $43$
Category:Normed Division Rings
Category:Topological Division Rings
Normed Division Ring Operations are Continuous/Addition

== Theorem ==
Let $\struct {R, +, *, \norm {\,\cdot\,} }$ be a normed division ring.
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
Let $p \in \R_{\ge 1} \cup \set \infty$.
Let $d_p$ be the $p$-product metric on $R \times R$.
Then the mapping:
:$+ : \struct {R \times R, d_p} \to \struct{R,d}$ 
is continuous.
== Proof ==
By $p$-Product Metric Induces Product Topology and Continuous Mapping is Continuous on Induced Topological Spaces, it suffices to consider the case $p = \infty$.
Let $\tuple {x_0, y_0} \in R \times R$.
Let $\epsilon > 0$ be given.
Let $\tuple {x, y} \in R \times R$ such that:
:$\map {d_\infty} {\tuple {x, y}, \tuple{x_0, y_0} } < \dfrac \epsilon 2$
By the definition of the product metric $d_\infty$ then:
:$\max \set {\map d {x, x_0}, \map d {y, y_0} } < \dfrac \epsilon 2$
or equivalently:
:$\map d {x, x_0} < \dfrac \epsilon 2$
:$\map d {y, y_0} < \dfrac \epsilon 2$
Then:
Since $\tuple {x_0, y_0}$ and $\epsilon$ were arbitrary, by the definition of continuity then the mapping:
:$+ : \struct {R \times R, d_\infty} \to \struct {R, d}$ 
is continuous.
== Sources ==
* : $\S 2.3$: Topology, Problem $43$
Category:Normed Division Rings
Category:Topological Division Rings
Normed Division Ring Operations are Continuous/Corollary

== Theorem ==
Let $\struct {R, +, \times, \norm {\,\cdot\,} }$ be a normed division ring.
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
Let $\tau$ be the topology induced by the metric $d$.
Then:
:$\struct {R, \tau}$ is a topological division ring.
== Proof ==
Let $d_\infty$ be the Chebyshev distance metric on $R \times R$.
Let $\tau^\times$ be the product topology on $R \times R$.
By $p$-Product Metric Induces Product Topology, $\tau^\times$ is the topology induced by the metric $d_\infty$.
Let $R^* = R \setminus \set 0$.
Let $d^*$ be the restriction of $d$ to $R^*$.
Let $\tau^*$ be the subspace topology on $R^*$.
By Metric Subspace Induces Subspace Topology then $\tau^*$ is the topology induced by the metric $d^*$
By Normed Division Ring Operations are Continuous and Continuous Mapping is Continuous on Induced Topological Spaces, the mappings:
::$\phi : \struct {R \times R, \tau^\times} \to \struct {R, \tau} : \map \phi {x, y} = x + y$ 
::$\theta : \struct {R ,\tau} \to \struct {R, \tau} : \map \theta x = -x$ 
::$\psi : \struct {R \times R, \tau^\times} \to \struct {R, \tau} : \map \psi {x, y} = x y$ 
::$\xi : \struct {R^* ,\tau^*} \to \struct {R, \tau} : \map \xi x = x^{-1}$ 
are continuous.
By the definition of a topological division ring then the result follows.
== Sources ==
* : $\S 2.3$: Topology, Problem $43$
Category:Normed Division Rings
Category:Topological Division Rings
Normed Division Ring Operations are Continuous/Inversion

== Theorem ==
Let $\struct {R, +, *, \norm {\,\cdot\,} }$ be a normed division ring.
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
Let $R^* = R \setminus \set 0$
Let $d^*$ be the subspace metric on $R^*$.
Then the mapping:
:$\iota : \struct {R^* ,d^*} \to \struct {R, d} : \map \iota x = x^{-1}$ 
is continuous.
== Proof ==
Let $x_0 \in R^*$.
Let $\epsilon > 0$ be given.
Let $\delta = \min \set {\dfrac {\norm {x_0} } 2, \dfrac {\norm {x_0}^2 \epsilon} 2 }$
Let $x \in R^*$ such that:
:$\map {d^*} {x, x_0} < \delta$
By the definition of the subspace metric on $R^*$ and the definition of the metric induced by the norm on $R$:
:$\map {d^*} {x, x_0} = \map d {x, x_0} = \norm {x - x_0} < \delta$
Then:
Hence:
Since $x_0$ and $\epsilon$ were arbitrary, by the definition of continuity then the mapping:
:$\iota: \struct {R^*, d^*} \to \struct {R, d} : \map \iota x = x^{-1}$ 
is continuous.
== Sources ==
* : $\S 2.3$: Topology, Problem $43$
Category:Normed Division Rings
Category:Topological Division Rings
Normed Division Ring Operations are Continuous/Multiplication

== Theorem ==
Let $\struct {R, +, *, \norm {\,\cdot\,} }$ be a normed division ring.
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
Let $p \in \R_{\ge 1} \cup \set \infty$.
Let $d_p$ be the $p$-Product Metric on $R \times R$.
Then the mapping:
:$* : \struct {R \times R, d_p} \to \struct {R, d}$ 
is continuous.
== Proof ==
By $p$-Product Metric Induces Product Topology and Continuous Mapping is Continuous on Induced Topological Spaces, it suffices to consider the case $p = \infty$.
Let $\tuple {x_0, y_0} \in R \times R$.
Let $\epsilon > 0$ be given.
Let $\delta = \min \set {\dfrac \epsilon {1 + \norm {y_0} + \norm {x_0} }, 1}$
Since $1 + \norm {y_0} + \norm {x_0} > 0$ then $\delta > 0$
Let $\tuple {x,y} \in R \times R$ such that:
:$\map {d_\infty} {\tuple {x, y}, \tuple{x_0, y_0} } < \delta$
By the definition of the $p$-product metric $d_\infty$:
:$\max \set {\map d {x, x_0}, \map d {y, y_0}} < \delta$
or equivalently:
:$\map d {x, x_0} < \delta$
:$\map d {y, y_0} < \delta$
Then:
Hence:
We have that $\tuple {x_0, y_0}$ and $\epsilon$ are arbitrary.
Hence, by the definition of continuity, the mapping:
:$* : \struct {R \times R, d_\infty} \to \struct {R, d}$ 
is continuous.
== Sources ==
* : $\S 2.3$: Topology, Problem $43$
Category:Normed Division Rings
Category:Topological Division Rings
Normed Division Ring Operations are Continuous/Negation

== Theorem ==
Let $\struct {R, +, *, \norm {\,\cdot\,} }$ be a normed division ring.
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
Then the mapping:
:$\eta: \struct {R, d} \to \struct {R, d}: \map \eta x = -x$ 
is continuous.
== Proof ==
Let $x_0 \in R$.
Let $\epsilon > 0$ be given.
Let $x \in R$ such that:
:$\map d {x, x_0} < \epsilon$
Then:
Since $x_0$ and $\epsilon$ were arbitrary, by the definition of continuity then the mapping:
:$\eta: \struct {R, d} \to \struct {R, d} : \map \eta x = -x$ 
is continuous.
== Sources ==
* : $\S 2.3$: Topology, Problem $43$
Category:Normed Division Rings
Category:Topological Division Rings
Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit

== Theorem ==
Let $\struct {R, \norm{\,\cdot\,}_R}$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
Let $Q = \CC / \NN$ where $\CC / \NN$ denotes a quotient ring.
Let $\norm {\, \cdot \,}_Q: Q \to \R_{\ge 0}$ be the norm on the quotient ring $Q$ defined by:
:$\ds \forall \sequence {x_n} + \NN: \norm {\sequence {x_n} + \NN }_Q = \lim_{n \mathop \to \infty} \norm{x_n}_R$
Let $\phi: R \to Q$ be the mapping from $R$ to the quotient ring $Q$ defined by:
:$\forall a \in R: \map \phi a = \tuple {a, a, a, \ldots} + \NN$
where $\sequence {a, a, a, \ldots} + \NN$ is the left coset in $Q$ that contains the constant sequence $\sequence {a, a, a, \ldots} $.
Let $\sequence{x_n}$ be a sequence in $R$.
Let $y \in Q$.
Then:
:$\sequence{\map \phi {x_n}}$ converges to $y$ $\sequence{x_n} \in y$
== Proof ==
=== Lemma 1 ===
=== Lemma 2 ===
=== Necessary Condition ===
=== Sufficient Condition ===
Category:Normed Division Rings
Category:Complete Metric Spaces
Category:Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit
Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit/Lemma 1

== Theorem ==
Let $\struct {R, \norm{\,\cdot\,}_R}$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
Let $Q = \CC / \NN$ where $\CC / \NN$ denotes a quotient ring.
Let $\norm {\, \cdot \,}_Q: Q \to \R_{\ge 0}$ be the norm on the quotient ring $Q$ defined by:
:$\ds \forall \sequence {x_n} + \NN: \norm {\sequence {x_n} + \NN }_Q = \lim_{n \mathop \to \infty} \norm{x_n}_R$
Let $\phi: R \to Q$ be the mapping from $R$ to the quotient ring $Q$ defined by:
:$\forall a \in R: \map \phi a = \tuple {a, a, a, \ldots} + \NN$
where $\sequence {a, a, a, \ldots} + \NN$ is the left coset in $Q$ that contains the constant sequence $\sequence {a, a, a, \ldots} $.
Let $\sequence{x_n}$ be a sequence in $R$.
Then:
:$\sequence{x_n}$ is a Cauchy sequence $\sequence{\map \phi {x_n}}$ is a Cauchy sequence 
== Proof ==
From Embedding Division Ring into Quotient Ring of Cauchy Sequences:
:the mapping $\phi: R \to Q$ is a distance-preserving monomorphism
We have
Category:Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit
Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit/Lemma 2

== Theorem ==
Let $\struct {R, \norm{\,\cdot\,}_R}$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
Let $Q = \CC / \NN$ where $\CC / \NN$ denotes a quotient ring.
Let $\norm {\, \cdot \,}_Q: Q \to \R_{\ge 0}$ be the norm on the quotient ring $Q$ defined by:
:$\ds \forall \sequence {x_n} + \NN: \norm {\sequence {x_n} + \NN }_Q = \lim_{n \mathop \to \infty} \norm{x_n}_R$
Let $\phi: R \to Q$ be the mapping from $R$ to the quotient ring $Q$ defined by:
:$\forall a \in R: \map \phi a = \tuple {a, a, a, \ldots} + \NN$
where $\tuple {a, a, a, \ldots} + \NN$ is the left coset in $Q$ that contains the constant sequence $\sequence {a, a, a, \ldots} $.
Let $\sequence{x_n}$ be a sequence in $R$.
Let $y$ be the left coset in $Q$ that contains $\sequence{x_n}$.
For each $n \in \N$:
:$\norm{\map \phi {x_n} - y}_Q = \ds \lim_{m \mathop \to \infty} \norm{x_n - x_m}_R$
== Proof ==
Let $n \in \N$.
We have:
:$\tuple {x_n, x_n, x_n, \ldots} \in \map \phi {x_n}$
:$\sequence{x_m} \in y$
From Element of Group is in Unique Coset of Subgroup:
:$\tuple {x_n, x_n, x_n, \ldots} + \NN = \map \phi {x_n}$
:$\sequence{x_m} + \NN = y$
Then:
From Element of Group is in its own Coset:
:$\sequence{x_n - x_m}_{m \in \N} \in \map \phi {x_n} - y$
By definition of induced norm:
:$\norm{\map \phi {x_n} - y}_Q = \ds \lim_{m \mathop \to \infty} \norm{x_n - x_m}_R$
Category:Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit
Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit/Necessary Condition

== Theorem ==
If $\sequence{\map \phi {x_n}}$ converges to $y$ then $\sequence{x_n} \in y$
== Proof ==
=== Lemma 1 ===
Let $\sequence{\map \phi {x_n}}$ converge to $y$.
From Convergent Sequence in Normed Division Ring is Cauchy Sequence:
:$\sequence{\map \phi {x_n}}$ is a Cauchy Sequence
From Lemma 1:
:$\sequence{x_n}$ is a Cauchy Sequence
Let $y'$ be the left coset that contains $\sequence{x_n}$.
From sufficient condition:
:$\ds \lim_{n \mathop \to \infty} \norm{\map \phi {x_n} }_Q = y'$
From Convergent Sequence in Metric Space has Unique Limit:
:$y = y'$
Category:Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit
Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit/Sufficient Condition

== Theorem ==
If $\sequence{x_n} \in y$ then $\sequence{\map \phi {x_n}}$ converges to $y$
== Proof ==
=== Lemma 2 ===
Let $\sequence{x_n} \in y$.
Then $\sequence{x_n}$ is a Cauchy Sequence by definition of $y$.
Let $\epsilon > 0$ be arbitrary.
By definition of a Cauchy sequence:
:$\exists N \in \N: \forall n, m \ge N : \norm{x_n - x_m}_R < \dfrac \epsilon 2$
Let $n \ge N$ be arbitrary.
From Difference Rule for Cauchy Sequences in Normed Division Ring, the sequence $\sequence{x_n - x_m}_{m \in \N}$ is a Cauchy sequence.
Hence:
:$\forall m \ge N : \norm{x_n - x_m}_R < \dfrac \epsilon 2$
From Norm Sequence of Cauchy Sequence has Limit:
:$\ds \lim_{m \mathop \to \infty} \norm{x_n - x_m}_R$ exists
From Inequality Rule for Real Sequences:
:$\ds \lim_{m \mathop \to \infty} \norm{x_n - x_m}_R \le \dfrac \epsilon 2 < \epsilon$
From Lemma 2:
:$\norm{\map \phi {x_n} - y}_Q < \epsilon$
By definition of convergent sequence:
:$\ds \lim_{n \mathop \to \infty} \norm{\map \phi {x_n} }_Q = y$
Category:Normed Division Ring Sequence Converges in Completion iff Sequence Represents Limit
Normed Division Ring is Dense Subring of Completion

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\struct {R', \norm {\, \cdot \,}' }$ be a normed division ring completion of $\struct {R, \norm {\, \cdot \,} }$
Then:
:$\struct {R, \norm {\, \cdot \,} }$ is isometrically isomorphic to a dense normed division subring of $\struct {R', \norm {\, \cdot \,}' }$.
== Proof ==
By the definition of a normed division ring completion then:
:$(1): \quad$ there exists a distance-preserving ring monomorphism $\phi: R \to R'$.
:$(2): \quad \struct {R', \norm {\, \cdot \,}' }$ is a complete metric space.
:$(3): \quad \phi \sqbrk R$ is a dense subspace in $\struct {R', \norm {\, \cdot \,}' }$.
By image of a ring homomorphism is a subring then $\phi \sqbrk R$ is a subring of $R'$ and $\phi: R \to \phi \sqbrk R$ is an isomorphism.
By Epimorphism from Division Ring to Ring then $\phi \sqbrk R$ is a division subring of $R'$.
By Division Subring of Normed Division Ring then $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ is a normed division subring of $\struct {R', \norm {\, \cdot \,}' }$.
By Distance-Preserving Surjection is Isometry of Metric Spaces then $\phi: R \to \phi \sqbrk R$ is an isometry.
The result follows.
Category:Normed Division Rings
Category:Complete Metric Spaces
Category:Completion of Normed Division Ring
Normed Division Ring is Field iff Completion is Field

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\struct {R', \norm {\, \cdot \,}' }$ be a normed division ring completion of $\struct {R, \norm {\, \cdot \,} }$
Then:
:$R$ is a field $R'$ is a field.
== Proof ==
By the definition of a normed division ring completion then:
:$(1): \quad$ there exists a distance-preserving ring monomorphism $\phi: R \to R'$.
:$(2): \quad$ $\struct {R', \norm {\, \cdot \,}' }$ is a complete metric space.
:$(3): \quad$ The image $\phi \sqbrk R$ of $\phi$ is a dense subspace in $\struct {R', \norm {\, \cdot \,}' }$.
By image of a ring homomorphism is a subring then $\phi \sqbrk R$ is a subring of $R'$ and $\phi: R \to \phi \sqbrk R$ is an isomorphism.
=== Necessary Condition ===
Let $x', y' \in R'$.
By the definition of a dense subset then $\map \cl {\phi \sqbrk R} = R'$.
From Closure of Subset of Metric Space by Convergent Sequence:
:there exists a sequence $\sequence {x_n'} \subseteq \phi \sqbrk R$ that converges to $x'$, that is, $\ds \lim_{n \mathop \to \infty} x_n' = x'$
:there exists a sequence $\sequence {y_n'} \subseteq \phi \sqbrk R$ that converges to $y'$, that is, $\ds \lim_{n \mathop \to \infty} y_n' = y'$
From Product Rule for Sequences in Normed Division Ring:
:$\ds \lim_{n \mathop \to \infty} x_n' y_n' = x' y'$
:$\ds \lim_{n \mathop \to \infty} y_n' x_n' = y' x'$
By the definition of a field, $R$ is a commutative ring.
Because $\phi: R \to \phi \sqbrk R$ is an isomorphism, by Isomorphism Preserves Commutativity, $\phi \sqbrk R$ is a commutative ring.
Because $\sequence {x_n'} \subseteq \phi \sqbrk R$ and $\sequence {y_n'} \subseteq \phi \sqbrk R$:
:$\forall n: x_n' y_n' = y_n' x_n'$
Hence:
:$\ds x' y' = \lim_{n \mathop \to \infty} x_n' y_n' = \lim_{n \mathop \to \infty} y_n' x_n' = y' x'$
Since $x', y' \in R'$ were arbitrary, then $R'$ is a commutative ring.
The result follows.
=== Sufficient Condition ===
By Restriction of Commutative Operation is Commutative, $\phi \sqbrk R$ is a commutative ring.
By Monomorphism Image is Isomorphic to Domain, $\phi: R \to \phi \sqbrk R$ is a ring isomorphism.
By Inverse of Algebraic Structure Isomorphism is Isomorphism, $\phi^{-1}: \phi \sqbrk R \to R$ is a ring isomorphism.
By Isomorphism Preserves Commutativity, $R$ is a commutative ring.
The result follows.
== Sources ==
* : $\S 3.2$: Completions
Category:Normed Division Rings
Category:Complete Metric Spaces
Category:Completion of Normed Division Ring