hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = H g_1       | r = H g_2       | c =  }} {{eqn | ll= \leadstoandfrom       | l = g_1 {g_2}^{-1}       | o = \in       | r = H       | c = Right Cosets are Equal iff Product with Inverse in Subgroup }} {{eqn | ll= \leadstoandfrom       | l = \paren { {g_1}^{-1} }^{-1} {g_2}^{-1}       | o = \in       | r = H       | c = Inverse of Group Inverse }} {{eqn | ll= \leadstoandfrom       | l = {g_1}^{-1} H       | r = {g_2}^{-1} H       | c = Left Cosets are Equal iff Product with Inverse in Subgroup }} {{end-eqn}} {{qed}}
From the Tower Law for Subgroups: :$\index G K = \index G H \index H K$ As $\index G K = p$ is prime, either $\index G H = p$ or $\index H K = p$. Thus either $\index G H = 1$ or $\index H K = 1$. The result follows from Index is One iff Subgroup equals Group. {{qed}}
Comparing the Cayley tables of the symmetry group of the rectangle with the Klein $4$-group the isomorphism can be seen:  {{:Symmetry Group of Rectangle/Cayley Table}} {{:Klein Four-Group/Cayley Table}} Thus the required isomorphism $\phi$ can be set up as: {{begin-eqn}} {{eqn | l = \map \phi e       | r = e }} {{eqn | l = \map \phi r       | r = a }} {{eqn | l = \map \phi h       | r = b }} {{eqn | l = \map \phi v       | r = c }} {{end-eqn}} {{qed}}
By inspection, the Cayley table is constructed: :$\begin{array}{c|cccc}   & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & e & a \\ c & c & b & a & e \\ \end{array}$ Again by inspection this can be seen to be the same as the Cayley table for the Klein $4$-group. {{qed}}
By the definition of the metric induced by a norm then: :a sequence $\sequence {x_n}$ is a Cauchy sequence in $\struct {R, \norm {\, \cdot \,} }$ {{iff}} $\sequence {x_n}$ is a Cauchy sequence in $\struct {R, d}$. So $\mathcal C$ is the set of Cauchy sequences in $\struct {R, d}$. Let $\sim$ be the equivalence relation on $\mathcal C$ defined by: :$\displaystyle \sequence {x_n} \sim \sequence {y_n} \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$ Let $\tilde {\mathcal C} = \mathcal C  / \sim$ denote the set of equivalence classes under $\sim$. For $\sequence {x_n} \in \mathcal C$, let $\eqclass {x_n} {}$ denote the equivalence class containing $\sequence {x_n}$. === Lemma 1 === {{:Quotient of Cauchy Sequences is Metric Completion/Lemma 1}}{{qed|lemma}} Let $\tilde d: \tilde {\mathcal C} \times \tilde {\mathcal C} \to \R_{\ge 0}$ be the metric defined by: :$\displaystyle \map {\tilde d} {\eqclass {x_n} {}, \eqclass {y_n} {} } = \lim_{n \mathop \to \infty} \map d {x_n, y_n}$ By Completion of a Metric Space then: :$\struct {\tilde {\mathcal C}, \tilde d}$ is the metric completion of $\struct {R, d}$. === Lemma 2 === {{:Quotient of Cauchy Sequences is Metric Completion/Lemma 2}}{{qed|lemma}} Let $\tilde \phi: R \to \tilde {\mathcal C}$ be the mapping defined by: :$\map {\tilde \phi} a = \eqclass {a, a, a, \dotsc} {}$ where $\eqclass {a, a, a, \dotsc} {}$ denotes the equivalence class containing the constant sequence $\sequence {a, a, a, \dotsc}$. By Completion of a Metric Space then: :$\map {\tilde \phi} R$ is a dense subset of $\tilde {\mathcal C}$. By Lemma 1 then: :$\forall \sequence {x_n} \in \mathcal C: \sequence {x_n} + \mathcal N = \eqclass {x_n} {}$ In particular: :$\forall a \in R: \sequence {a, a, a, \dotsc} + \mathcal N = \eqclass {a, a, a, \dotsc} {}$ That is: :$\forall a \in R: \map \phi a = \map {\tilde \phi} a$ Hence $\phi = \tilde \phi$. The result follows. {{qed}}
Let $d$ be the metric induced by $\struct {R, \norm {\, \cdot \,} }$. Let $\CC$ be the ring of Cauchy sequences over $R$. Let $\NN = \set {\sequence {x_n}: \displaystyle \lim_{n \mathop \to \infty} x_n = 0_R}$. Let $\norm {\, \cdot \,}: \CC \, \big / \NN \to \R_{\ge 0}$ be the norm on the quotient ring $\CC \, \big / \NN$ defined by: :$\displaystyle \forall \sequence {x_n} + \NN: \norm {\sequence {x_n} + \NN } = \lim_{n \mathop \to \infty} \norm{x_n}$ Let $d'$ be the metric induced by $\struct {\CC \, \big / \NN, \norm {\, \cdot \,} }$. By Quotient Ring of Cauchy Sequences is Normed Division Ring, $\struct {\CC \, \big / \NN, \norm {\, \cdot \,} }$ is a normed division ring. By Quotient of Cauchy Sequences is Metric Completion, $\struct {\CC \, \big / \NN, d' }$ is the metric completion of $\struct {R, d}$. Let $\phi: R \to \CC \, \big / \NN$ be the mapping from $R$ to the quotient ring $\CC \,\big / \NN$ defined by: :$\quad \quad \quad \forall a \in R: \map \phi a = \tuple {a, a, a, \ldots} + \NN$ where $\tuple {a, a, a, \ldots} + \NN$ is the left coset in $\CC \, \big / \NN$ that contains the constant sequence $\tuple {a, a, a, \ldots} $. By Quotient of Cauchy Sequences is Metric Completion, $\map \phi R$ is a dense subset of $\struct {\CC \, \big / \NN, d' }$. By Embedding Division Ring into Quotient Ring of Cauchy Sequences, $\phi$ is a distance-preserving ring monomorphism. By the definition of a normed division ring completion, $\struct {\CC \, \big / \NN, \norm {\, \cdot \,} }$ is a normed division ring completion of $\struct {R, \norm {\, \cdot \,} }$. {{qed}}
Let $x \in G \setminus \set e$. From Identity is Only Group Element of Order 1: :$\order x > 1$ where $\order x$ denotes the order of $x$. From Lagrange's Theorem, $\order x$ is $3$, $9$ or $27$. Thus one of the following applies: {{begin-eqn}} {{eqn | l = \order x       | r = 3 }} {{eqn | l = \order {x^3}       | r = 3 }} {{eqn | l = \order {x^9}       | r = 3 }} {{end-eqn}} Hence one of the following holds: {{begin-eqn}} {{eqn | l = \order {\gen x}       | r = 3 }} {{eqn | l = \order {\gen {x^3} }       | r = 3 }} {{eqn | l = \order {\gen {x^9} }       | r = 3 }} {{end-eqn}} where $\gen x$ denotes the subgroup generated by $x$. Thus one of these: $\gen x$, $\gen {x^3}$ or $\gen {x^9}$ is a subgroup of $G$ of order $3$. {{qed}}
Proof by Counterexample: Consider $S_5$, the symmetric group on $5$ letters. By Order of Symmetric Group, $\order {S_5} = 5! = 120$. We have that $120 = 8 \times 15$ and so $15$ is a divisor of $120$. However, Symmetric Group on 5 Letters has no Subgroup of Order 15. {{qed}}
Proof by Counterexample: Consider the symmetric group $S_4$. Then the order of the alternating group $A_4$ is $12$. We list the subgroups of $A_4$: {{:Alternating Group on 4 Letters/Subgroups}} Now $6$ divides $12$. But there is no subgroup of $A_4$ of order $6$. {{qed}}
Proof by Counterexample: Let $a \in G$ such that $a \ne a^{-1}$. Let $X = \set {a, a^{-1} }, Y = \set a, Z = \set {a^{-1} }$. Then: :$X \circ \paren {Y \cap Z} = X \circ \O = \O$ :$\paren {X \circ Y} \cap \paren {X \circ Z} = \set {a^2, e} \cap \set {e, a^{-2} } \ne \O$ so: :$X \circ \paren {Y \cap Z} \ne \paren {X \circ Y} \cap \paren {X \circ Z}$ {{qed}}
Suppose $x K \cap H \ne \O$. Then: {{begin-eqn}} {{eqn | l = x K \cap H       | o = \ne       | r = \O       | c =  }} {{eqn | ll= \leadsto       | lo= \exists y \in G       | l = y       | o = \in       | r = x K \cap H       | c =  }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = x K       | c =  }} {{eqn | ll= \leadsto       | l = x K       | r = y K       | c = Left Cosets are Equal iff Element in Other Left Coset }} {{end-eqn}} We have that: :$y \in H$ and: :$K \subseteq H$ As $H$ is a group, it is closed. Hence: :$\forall x \in K: y x \in H$ which means, by definition of subset product: :$y K \subseteq H$ Hence the result. {{qed}}
Let $x \in a H \cap b K$. Then: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = a H }} {{eqn | ll= \leadsto       | l = x H       | r = a H       | c = Left Cosets are Equal iff Element in Other Left Coset }} {{end-eqn}} and similarly: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = b K }} {{eqn | ll= \leadsto       | l = x K       | r = b K       | c = Left Cosets are Equal iff Element in Other Left Coset }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = a H \cap b K       | r = x H \cap x K }} {{eqn | r = x \paren {H \cap K}       | c = Corollary to Product of Subset with Intersection }} {{end-eqn}} Hence the result by definition of left coset. {{qed}}
{{begin-eqn}} {{eqn | l = H x       | r = H y       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x y^{-1}       | o = \in       | r = H       | c = Right Cosets are Equal iff Product with Inverse in Subgroup }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = H y       | c = Element in Right Coset iff Product with Inverse in Subgroup }} {{end-eqn}} {{Qed}}
As $K \subsetneq H$ and $K$ is a subgroups of $G$, it follows that $K$ is a proper subgroup of $H$. That is, $K \ne H$ Hence from Subgroup of Subgroup with Prime Index: :$H = G$ {{qed}}
From Order of Cyclic Group equals Order of Generator: :$\order x = \order {\gen x}$ where $\gen x$ denotes the subgroup of $G$ generated  by $x$. By definition, $\gen x$ is a cyclic group. By Cyclic Group is Abelian, all elements of $\gen x$ commute with $x$. Thus by definition of centralizer: :$\gen x \subseteq \map {C_G} x$ From Centralizer of Group Element is Subgroup, $\map {C_G} x$ is a subgroup of $G$. So $\gen x$ is a subgroup of $\map {C_G} x$. The result follows from Lagrange's theorem. {{qed}}
Let $x \in G$. Consider the left coset $x G_t$. Let $\map x t = s$. Then: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = x G_t       | c =  }} {{eqn | ll= \iff       | l = y^{-1} x       | o = \in       | r = G_t       | c = Element in Left Coset iff Product with Inverse in Subgroup }} {{eqn | ll= \iff       | l = \map {y^{-1} } {\map x t}       | r = t       | c = Definition of $G_t$ }} {{eqn | ll= \iff       | l = \map {y^{-1} } s       | r = t       | c = Definition of $s$ }} {{eqn | ll= \iff       | l = \map y t       | r = s       | c =  }} {{end-eqn}} {{qed}}
From Order of Cyclic Group equals Order of Generator: :$\order x = \order {\gen x}$ and: :$\order y = \order {\gen y}$ where $\order {\gen x}, \order {\gen y}$ denote the orders of $\gen x$ and $\gen y$ respectively. From Intersection of Subgroups is Subgroup: :$\gen x \cap \gen y$ is a subgroup of both $\gen x$ and $\gen y$. From Lagrange's Theorem: :$\order {\gen x \cap \gen y} \divides \order {\gen x}$ and: :$\order {\gen x \cap \gen y} \divides \order {\gen y}$ But as  $\order {\gen x} \perp \order {\gen y}$: :$\order {\gen x \cap \gen y} = 1$ Hence the result by definition of trivial subgroup. {{qed}}
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 $\map \cl {\phi \sqbrk R} = R'$. By 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, $\displaystyle \lim_{n \mathop \to \infty} x_n' = x'$ :there exists a sequence $\sequence {y_n'} \subseteq \phi \sqbrk R$ that converges to $y'$, that is, $\displaystyle \lim_{n \mathop \to \infty} y_n' = y'$ By product rule for convergent sequences: :$\displaystyle \lim_{n \mathop \to \infty} x_n' y_n' = x' y'$ :$\displaystyle \lim_{n \mathop \to \infty} y_n' x_n' = y' x'$ 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: :$\displaystyle x' y' = \lim_{n \mathop \to \infty} x_n' y_n' = \lim_{n \mathop \to \infty} y_n' x_n' = y' x'$ Since $x', y' \in R'$ were arbitrary, then $R'$ is a commutative ring. The result follows. {{qed|lemma}} === 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. {{qed}}
Let $e \in G$ be the identity of $G$. Let $g \in G$ be an arbitrary element of $G$ such that $g \ne e$. From Identity is Only Group Element of Order 1, only $e$ has order $1$. Thus from Order of Element Divides Order of Finite Group: :$\order g \in \set {2, 4, 8}$ Suppose $\order g = 8$. Then $G$ is cyclic. So by Cyclic Group is Abelian, $G$ would be abelian. So $g$ cannot be of order $8$. Suppose all elements of $G \setminus \set e$ are of order $2$. Then by definition $G$ is a Boolean group. By Boolean Group is Abelian, $G$ would be abelian.  So the only option left is for at least one element of $G$ to be of order $4$. {{qed}}
Let $\order g$ denote the order of an element $g$ of $G$. From Order of Element Divides Order of Finite Group: :$\order g \divides p$ where $\divides$ denotes divisibility. By definition of prime number, the only divisors of $p$ are $1$ and $p$. From Identity is Only Group Element of Order 1, only $e$ has order $1$ in $G$. That leaves $p - 1$ elements of order $p$. {{qed}}
Let $H \le G$ be a subgroup of $G$ of order $p$. From Prime Group is Cyclic, $H$ is a cyclic group. From Group of Prime Order p has p-1 Elements of Order p, $H$ has $p - 1$ elements of order $p$. From Intersection of Subgroups of Prime Order, each of the $m$ sets of $p - 1$ non-identity elements of the $m$ subgroups of order $p$ are pairwise disjoint. Each elements of $G$ order $p$ is the generator of a cyclic group of order $p$. Therefore there are no elements of order $p$ apart from the $m \paren {p - 1}$ such elements of the $m$ subgroups of order $p$. Hence the result. {{qed}}
By Order of Element Divides Order of Finite Group, all elements of $G$ have order in $\set {1, p, p^2}$. But as $G$ is non-cyclic, it can have no element of order $p^2$. By Identity is Only Group Element of Order 1, $G$ has $p^2 - 1$ elements of order $p$. Let $m$ denote the number of subgroups of $G$ of order $p$. From Number of Order p Elements in Group with m Order p Subgroups, $G$ has $m \paren {p - 1}$ elements of order $p$. Then: {{begin-eqn}} {{eqn | l = m \paren {p - 1}       | r = p^2 - 1       | c =  }} {{eqn | r = \paren {p + 1} \paren {p - 1}       | c =  }} {{eqn | ll= \leadsto       | l = m       | r = p + 1       | c =  }} {{end-eqn}} Together with the trivial subgroup and $G$ itself, that makes $p + 3$ subgroups. {{qed}}
Let $x \in G$ be of order $p$. Consider $\gen x$, the subgroup generated by $x$. By Group of Prime Order p has p-1 Elements of Order p, the elements of $\gen x$ are all of order $p$ except $e$. Thus, by hypothesis, there must exist another $y \in G$ of order $p$. Consider the subset of $G$: :$S := \set {x^i y^j: 0 \le i, j < p}$ By the Finite Subgroup Test, $S$ is a subgroup of $G$ which has $p^2$ elements. By Lagrange's theorem: :$\order S \divides \order G$ But $\order S = p^2$ and so $p^2 \divides \order G$. {{qed}}
By Order of Element Divides Order of Finite Group, the order of elements of $G$ are all in $\set {1, p, q, p q}$. We have that Identity is Only Group Element of Order 1. Suppose $G$ were to contain more than $p - 1$ elements of order $p$. Then by Order of Finite Abelian Group with $p+$ Order $p$ Elements is Divisible by $p^2$: :$p^2 \divides \order G$ where $\divides$ denotes divisibility. As $p^2 \nmid p q$ it follows that $G$ contains no more than $p - 1$ elements of order $p$. Similarly it follows that $G$ contains no more than $q - 1$ elements of order $q$. These, along with $e$, account for $\paren {p - 1} + \paren {q - 1} + 1 = p + q - 1$ elements. Thus the number of elements of order $p q$ is: :$p q - \paren {p + q - 1} = \paren {p - 1} \paren {q - 1}$ which is strictly positive. The result follows from Group whose Order equals Order of Element is Cyclic. {{qed}}
A group is a semigroup. The result then follows from the General Morphism Property for Semigroups. {{Qed}}
From General Linear Group to Determinant is Homomorphism: :$\det$ is a group homomorphism. The special linear group $\SL {n, \R}$ is the subset of $\GL {n, \R}$ such that: :$\forall \mathbf A \in \SL {n, \R}: \map \det {\mathbf A} = 1$ From Real Multiplication Identity is One: : $1$ is the identity of the multiplicative group of real numbers. It follows by definition that $\SL {n, \R}$ is the kernel of the $\det$ mapping. {{qed}}
Let $\det: \GL {n, \R} \to \struct {\R_{\ne 0}, \times}$ be the group homomorphism: :$\mathbf A \mapsto \map \det {\mathbf A}$ where $\map \det {\mathbf A}$ is the determinant of $\mathbf A$. This is demonstrated to be a homomorphism in General Linear Group to Determinant is Homomorphism From the the corollary to General Linear Group to Determinant is Homomorphism, the kernel of $\phi$ is $\SL {n, \R}$. Thus from Kernel is Normal Subgroup of Domain, $\SL {n, \R}$ is normal in $\GL {n, \R}$. From the First Isomorphism Theorem for Groups: :$\Img \phi \cong \GL {n, \R} / \SL {n, \R}$ By definition, the image of $\phi$ is the multiplicative group of real numbers. Hence the result. {{qed}}
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. {{qed}} Category:Completion of Normed Division Ring b6j5h8yb3sr3hlymhfcig8m9vaqko0w
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: :$\displaystyle \lim_{n \mathop \to \infty} x_n = x$ 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: :$\displaystyle \lim_{n \mathop \to \infty} x_n = \lim_{n \mathop \to \infty} x'_n = x$ Then: {{begin-eqn}} {{eqn | ll=        | l = \lim_{n \mathop \to \infty} x_n - x'_n       | r = \paren {\lim_{n \mathop \to \infty} x_n} - \paren {\lim_{n \mathop \to \infty} x'_n}       | c = Difference rule for convergent sequences }} {{eqn | ll=        | r = x - x }} {{eqn | ll=        | r = 0_{S_1} }} {{eqn | o = }} {{eqn | ll= \leadsto       | r = \lim_{n \mathop \to \infty} \norm {x_n - x'_n}_1       | l = 0       | c = {{Defof|Convergent Sequence in Normed Division Ring}} }} {{eqn | r = \lim_{n \mathop \to \infty} \norm {\map {\psi'} {x_n} - \map {\psi'} {x'_n} }_2       | c = {{Defof|Isometric Metric Spaces}} }} {{eqn | o = }} {{eqn | ll= \leadsto       | r = \lim_{n \mathop \to \infty} \map {\psi'} {x_n} - \map {\psi'} {x'_n}       | l = 0_{S_2}       | c = {{Defof|Convergent Sequence in Normed Division Ring}} }} {{eqn | r = \paren {\lim_{n \mathop \to \infty} \map {\psi'} {x_n} } - \paren {\lim_{n \mathop \to \infty} \map {\psi'} {x'_n} }       | c = Difference rule for Convergent Sequences }} {{eqn | o = }} {{eqn | ll= \leadsto       | l = \lim_{n \mathop \to \infty} \map {\psi'} {x_n}       | r = \lim_{n \mathop \to \infty} \map {\psi'} {x'_n} }} {{end-eqn}} The result follows. {{qed}} Category:Completion of Normed Division Ring rzmal498jrtaydzxec6a2xtuequuibx
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, $\displaystyle \lim_{n \mathop \to \infty} y_n = y$ 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 $\psi$: {{begin-eqn}} {{eqn | l = \map \psi x       | r = \lim_{n \mathop \to \infty} \map {\psi'} {\map {\psi'^{-1} } {y_n} } }} {{eqn | r = \lim_{n \mathop \to \infty} y_n }} {{eqn | r = y }} {{end-eqn}} {{qed}} Category:Completion of Normed Division Ring 6eiqrtc4nq88rnl0wwdcc4j7gy5om73
Let $x, y \in S_1$. Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $R_1$ such that $\displaystyle \lim_{n \mathop \to \infty} x_n = x, \lim_{n \mathop \to \infty} y_n = y$. Then: {{begin-eqn}} {{eqn | l = x - y       | r = \lim_{n \mathop \to \infty} x_n - y_n       | c = Difference Rule for Convergent Sequences }} {{eqn | o = }} {{eqn | ll= \leadsto       | l = \norm {x - y}       | r = \lim_{n \mathop \to \infty} \norm {x_n - y_n}       | c = Modulus of Limit }} {{eqn | r = \lim_{n \mathop \to \infty} \norm {\map {\psi'} {x_n} - \map {\psi'} {y_n} }       | c = {{Defof|Isometry (Metric Spaces)}} }} {{end-eqn}} On the other hand: {{begin-eqn}} {{eqn | l = \map \psi x - \map \psi y       | r = \paren {\lim_{n \mathop \to \infty} \map {\psi'} {x_n} } - \paren {\lim_{n \mathop \to \infty} \map {\psi'} {y_n} }       | c = Definition of $\psi$ }} {{eqn | r = \lim_{n \mathop \to \infty} \map {\psi'} {x_n} - \map {\psi'} {y_n}       | c = Difference Rule for Convergent Sequences }} {{eqn | o = }} {{eqn | ll= \leadsto       | l = \norm {\map \psi x - \map \psi y}       | r = \lim_{n \mathop \to \infty} \norm {\map {\psi'} {x_n} - \map {\psi'} {y_n} }       | c = Modulus of Limit }} {{end-eqn}} 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. {{qed}} Category:Completion of Normed Division Ring powpdrlvt5p4znr4rmcsoi3ipftmw7j