hcju/proofwikiqa huggingface dataset:
=== Sufficient Condition === That $\mathbf 0 \in \map \ker T$ follows from Kernel of Linear Transformation contains Zero Vector. That $\map \ker T$ is a singleton follows from the definition of injection. {{qed|lemma}} === Necessary Condition === Let $\map \ker T = \set {\mathbf 0}$. Consider: :$\map T {\mathbf x} = \mathbf b$ where $\mathbf b$ is in the codomain of $T$. Let this equation have a solution: :$\mathbf x = \mathbf x_1 \in \mathbf V$ Suppose $\mathbf x = \mathbf x_2 \in \mathbf V$ is also a solution. Clearly: :$\map T {\mathbf x_1} = \map T {\mathbf x_2}$ Observe that: {{begin-eqn}} {{eqn | l = \map T {\mathbf x_1}       | r = \mathbf b }} {{eqn | ll= \land       | l = \map T {\mathbf x_2}       | r = \mathbf b }} {{eqn | ll= \leadsto       | l = \map T {\mathbf x_1} - \map T {\mathbf x_2}       | r = \mathbf b - \mathbf b }} {{eqn | r = \mathbf 0' }} {{eqn | ll= \leadsto       | l = \map T {\mathbf x_1 - \mathbf x_2}       | r = \mathbf 0'       | c = {{Defof|Linear Transformation on Vector Space}} }} {{eqn | ll= \leadsto       | l = \paren {\mathbf x_1 - \mathbf x_2}       | o = \in       | r = \map \ker T       | c = {{Defof|Kernel of Linear Transformation}} }} {{eqn | ll= \leadsto       | l = \mathbf x_1 - \mathbf x_2       | r = \mathbf 0       | c = {{Defof|Set Equality}}: recall $\map \ker T = \set {\mathbf 0}$ }} {{eqn | ll= \leadsto       | l = \mathbf x_1       | r = \mathbf x_2 }} {{end-eqn}} As $\mathbf x_1, \mathbf x_2$ were arbitrary: :$\forall \mathbf x_1,\mathbf x_2 \in \mathbf V: \map T {\mathbf x_1} = \map T {\mathbf x_2} \implies \mathbf x_1 = \mathbf x_2$ and the result follows from the definition of injectivity. {{qed}} Category:Linear Transformations n7es60exp1uazpyjxotyj4ili2emz1h
From Diagonal Relation is Ordering and Equivalence, we have that $\Delta_R$ is actually an ordering on $R$. From the definition of the diagonal relation: :$\tuple {x, y} \in \Delta_R \iff x = y$ Thus: {{begin-eqn}} {{eqn | l = \tuple {x, y}       | o = \in       | r = \Delta_R       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = x + z       | r = y + z       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {x + z, y + z}       | o = \in       | r = \Delta_R       | c =  }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = \tuple {x, y}       | o = \in       | r = \Delta_R       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = z + x       | r = z + y       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {z + x, z + y}       | o = \in       | r = \Delta_R       | c =  }} {{end-eqn}} So $\Delta_R$ is compatible with $+$. Then note that: {{begin-eqn}} {{eqn | l = \tuple {0_R, x}       | o = \in       | r = \Delta_R       | c =  }} {{eqn | lo= \land       | l = \tuple {0_R, y}       | o = \in       | r = \Delta_R       | c =  }} {{eqn | ll= \leadsto       | l = 0_R       | r = x       | c =  }} {{eqn | lo= \land       | l = 0_R       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = 0_R       | r = x \circ y       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {0_R, x \circ y}       | o = \in       | r = \Delta_R       | c =  }} {{end-eqn}} Hence the result, from the definition of an ordering compatible with the ring structure of $R$. {{qed}}
The main result Identity Mapping is Automorphism holds directly. {{qed}} Category:Semigroup Automorphisms Category:Identity Mappings ndb9aunvdsojqh52tr61ybp4zijnvjv
We have that: :an identity mapping is an order isomorphism :an identity mapping is a group automorphism :an identity mapping is a semigroup automorphism Hence the result by definition of ordered ring automorphism. {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = S }} {{eqn | ll= \implies       | l = \left({x, x}\right)       | o = \in       | r = \mathcal R       | c = {{Defof|Reflexive Relation}} }} {{eqn | ll= \implies       | l = \left({x, x}\right)       | o = \in       | r = \mathcal R^{-1}       | c = {{Defof|Inverse Relation}} }} {{end-eqn}} Hence the result by definition of reflexive relation. {{qed}}
{{begin-eqn}} {{eqn | l=x       | o=\in       | r=S       | c= }} {{eqn | ll=\implies       | l=\left({x, x}\right)       | o=\notin       | r=\mathcal R       | c=by definition of antireflexive relation }} {{eqn | ll=\implies       | l=\left({x, x}\right)       | o=\notin       | r=\mathcal R^{-1}       | c=by definition of inverse relation }} {{end-eqn}} Hence the result by definition of antireflexive relation. {{qed}}
Let $\mathcal R$ be non-reflexive. Then: {{begin-eqn}} {{eqn | l=\exists x \in S: \left({x, x}\right)       | o=\in       | r=\mathcal R       | c=as $\mathcal R$ is not antireflexive }} {{eqn | ll=\implies       | l=\exists x \in S: \left({x, x}\right)       | o=\in       | r=\mathcal R^{-1}       | c=by definition of inverse relation }} {{end-eqn}} Thus $\mathcal R^{-1}$ is not antireflexive. Also: {{begin-eqn}} {{eqn | l=\exists x \in S: \left({x, x}\right)       | o=\notin       | r=\mathcal R       | c=as $\mathcal R$ is not reflexive }} {{eqn | ll=\implies       | l=\exists x \in S: \left({x, x}\right)       | o=\notin       | r=\mathcal R^{-1}       | c=by definition of inverse relation }} {{end-eqn}} Thus $\mathcal R^{-1}$ is not reflexive. Hence the result, by definition of non-reflexive relation. {{qed}}
Let $\mathcal R$ be symmetric. Then from Relation equals Inverse iff Symmetric it follows that $\mathcal R^{-1}$ is also symmetric. {{qed}}
Let $\mathcal R$ be asymmetric. Then: : $\left({x, y}\right) \in \mathcal R \implies \left({y, x}\right) \notin \mathcal R$ Thus if $\left({x, y}\right) \in \mathcal R$ then: : $\left({y, x}\right) \in \mathcal R^{-1}$ and $\left({x, y}\right) \notin \mathcal R^{-1}$ Thus it follows that $\mathcal R^{-1}$ is also asymmetric. {{qed}}
Let $\RR$ be antisymmetric. Then: :$\tuple {x, y} \land \tuple {y, x} \in \RR \implies x = y$ It follows that: :$\tuple {y, x} \land \tuple {x, y} \in \RR^{-1} \implies x = y$ Thus it follows that $\RR^{-1}$ is also antisymmetric. {{qed}}
Let $\mathcal R$ be non-symmetric. Then: : $\exists \left({x_1, y_1}\right) \in \mathcal R \implies \left({y_1, x_1}\right) \in \mathcal R$ and also: : $\exists \left({x_2, y_2}\right) \in \mathcal R \implies \left({y_2, x_2}\right) \notin \mathcal R$ Thus: : $\exists \left({y_1, x_1}\right) \in \mathcal R^{-1} \implies \left({x_1, y_1}\right) \in \mathcal R^{-1}$ and also: : $\exists \left({y_2, x_2}\right) \in \mathcal R^{-1} \implies \left({x_2, y_2}\right) \notin \mathcal R^{-1}$ and so $\mathcal R^{-1}$ is non-symmetric. {{qed}}
Let $\RR$ be transitive. Then: :$\tuple {x, y}, \tuple {y, z} \in \RR \implies \tuple {x, z} \in \RR$ Thus: :$\tuple {y, x}, \tuple {z, y} \in \RR^{-1} \implies \tuple {z, x} \in \RR^{-1}$ and so $\RR^{-1}$ is transitive. {{qed}}
Let $\RR$ be transitive. Thus by definition: :$\RR \circ \RR \subseteq \RR$ Thus: {{begin-eqn}} {{eqn | l = \RR^{-1} \circ \RR^{-1}       | r = \paren {\RR \circ \RR}^{-1}       | c = Inverse of Composite Relation }} {{eqn | o = \subseteq       | r = \RR^{-1}       | c = Inverse of Subset of Relation is Subset of Inverse }} {{end-eqn}} {{qed}} Hence the result by definition of transitive relation.
Let $\RR$ be antitransitive. Then: :$\tuple {x, y}, \tuple {y, z} \in \RR \implies \tuple {x, z} \notin \RR$ Thus: :$\tuple {y, x}, \tuple {z, y} \in \RR^{-1} \implies \tuple {z, x} \notin \RR^{-1}$ and so $\RR^{-1}$ is antitransitive. {{qed}}
Let $\mathcal R$ be non-transitive. Then: : $\exists x_1, y_1, z_1 \in S: \left({x_1, y_1}\right), \left({y_1, z_1}\right) \in \mathcal R, \left({x_1, z_1}\right) \in \mathcal R$ : $\exists x_2, y_2, z_2 \in S: \left({x_2, y_2}\right), \left({y_2, z_2}\right) \in \mathcal R, \left({x_2, z_2}\right) \notin \mathcal R$ So: : $\exists x_1, y_1, z_1 \in S: \left({y_1, x_1}\right), \left({z_1, y_1}\right) \in \mathcal R^{-1}, \left({z_1, x_1}\right) \in \mathcal R^{-1}$ : $\exists x_2, y_2, z_2 \in S: \left({y_2, x_2}\right), \left({z_2, y_2}\right) \in \mathcal R^{-1}, \left({z_2, x_2}\right) \notin \mathcal R^{-1}$ So $\mathcal R^{-1}$ is non-transitive. {{qed}}
By definition, $\phi$ is a bijection. By Bijection iff Inverse is Bijection, $\phi^{-1}$ is also a bijection. By definition, an ordered ring isomorphism from $\phi: \left({S, +, \circ, \preceq}\right) \to \left({T, \oplus, *, \preccurlyeq}\right)$ is: :an order isomorphism from the ordered set $\left({S, \preceq}\right)$ to the ordered set $\left({T, \preccurlyeq}\right)$ :a group isomorphism from the group $\left({S, +}\right)$ to the group $\left({T, \oplus}\right)$ :a semigroup isomorphism from the semigroup $\left({S, \circ}\right)$ to the semigroup $\left({T, *}\right)$. From Inverse of Order Isomorphism is Order Isomorphism, $\phi^{-1}: \left({T, \preccurlyeq}\right) \to \left({S, \preceq}\right)$ is an order isomorphism. From Inverse of Algebraic Structure Isomorphism is Isomorphism: :$\phi^{-1}: \left({T, \oplus}\right) \to \left({S, +}\right)$ is a group isomorphism :$\phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$ is a semigroup isomorphism. From Isomorphism of Abelian Groups, $\phi^{-1}: \left({T, \oplus}\right) \to \left({S, +}\right)$ preserves the commutativity of $S$ and $T$. Hence the result. {{qed}}
From Unit Matrix is Orthogonal, the unit matrix $\mathbf I_n$ is orthogonal. Let $\mathbf A, \mathbf B \in \operatorname O \left({n, k}\right)$. Then, by definition, $\mathbf A$ and $\mathbf B$ are orthogonal. Then by Inverse of Orthogonal Matrix is Orthogonal: :$\mathbf B^{-1}$ is a orthogonal matrix. By Product of Orthogonal Matrices is Orthogonal Matrix: :$\mathbf A \mathbf B^{-1}$ is a orthogonal matrix. Thus by definition of orthogonal group: :$\mathbf A \mathbf B^{-1} \in \operatorname O \left({n, k}\right)$ Hence the result by One-Step Subgroup Test. {{qed}} Category:Orthogonal Groups Category:General Linear Group m22ama3xu94uxysoaj0ku2jn1iz1ck8
A direct corollary of Orthogonal Group is Subgroup of General Linear Group. {{qed}}
From Composite of Order Isomorphisms is Order Isomorphism, $\psi \circ \phi: \left({S_1, \preccurlyeq_1}\right) \to \left({S_3, \preccurlyeq_3}\right)$ is an order isomorphism. From Composite of Isomorphisms in Algebraic Structure is Isomorphism, $\psi \circ \phi$ is an algebraic structure isomorphism. From Isomorphism Preserves Groups, it follows that $\psi \circ \phi$ is a group isomorphism from $\left({S_1, +_1}\right)$ to $\left({S_3, +_3}\right)$. From Isomorphism Preserves Semigroups, it follows that $\psi \circ \phi$ is a semigroup isomorphism from $\left({S_1, \circ_1}\right)$ to $\left({S_3, \circ_3}\right)$. Hence the result. {{qed}}
From Composite of Order Embeddings is Order Embedding, $\psi \circ \phi: \left({S_1, \preceq_1}\right) \to \left({S_3, \preceq_3}\right)$ is an order embedding. From Composite of Monomorphisms is Monomorphism, $\psi \circ \phi$ is a monomorphism. From Group Monomorphism preserves Groups, it follows that $\psi \circ \phi$ is a group monomorphism from $\left({S_1, +_1}\right)$ to $\left({S_3, +_3}\right)$. From Semigroup Monomorphism preserves Semigroups, it follows that $\psi \circ \phi$ is a semigroup monomorphism from $\left({S_1, \circ_1}\right)$ to $\left({S_3, \circ_3}\right)$. Hence the result. {{qed}}
Let $v \in V$ and $s \in R$. Then: {{begin-eqn}} {{eqn|l = m_r \left({s \circ v}\right)      |r = r \circ \left({s \circ v}\right)      |c = Definition of rescaling }} {{eqn|r = \left({r \cdot s}\right) \circ v      |c = $V$ is an $R$-module }} {{eqn|r = \left({s \cdot r}\right) \circ v      |c = $R$ is a commutative ring }} {{eqn|r = s \circ \left({r \circ v}\right)      |c = $V$ is an $R$-module }} {{eqn|r = s \circ m_r \left({v}\right)      |c = Definition of rescaling }} {{end-eqn}} Next, for $v,w \in V$: {{begin-eqn}} {{eqn|l = m_r \left({v + w}\right)      |r = r \circ \left({v + w}\right)      |c = Definition of rescaling }} {{eqn|r = r \circ v + r \circ w      |c = $V$ is an $R$-module }} {{eqn|r = m_r \left({v}\right) + m_r \left({w}\right)      |c = Definition of rescaling }} {{end-eqn}} It follows that $m_r$ is a linear transformation. {{qed}} Category:Linear Transformations f21xhuuisueybdb7ar42npjhkjtfq0t
From Determinant of Diagonal Matrix, it follows directly that: :$\map \det {r \, \mathbf I_n} = \displaystyle \prod_{i \mathop = 1}^n r = r^n$ {{qed}} Category:Determinants 2xs8qedaiszlly34e7fjppgn0psrdr9
By definition, a rescaling matrix is also a diagonal matrix. Hence Inverse of Diagonal Matrix applies, and since $r$ is a unit, it gives the desired result. {{qed}} Category:Matrix Algebra rdzg8jav5ovuegzpvdyrnr0ul16ur42
Proof by induction: For all $n \in \N$, let $\map P n$ be the proposition: :$\paren {m \cdot x} \circ \paren {n \cdot x} = \paren {m n} \cdot \paren {x \circ x}$ In what follows, we make extensive use of Powers of Ring Elements: :$\forall n \in \Z: \forall x \in R: \paren {m \cdot x} \circ x = m \cdot \paren {x \circ x} = x \circ \paren {m \cdot x}$ First we verify $\map P 0$. When $n = 0$, we have: {{begin-eqn}} {{eqn | l = \paren {m \cdot x} \circ \paren {0 \cdot x}       | r = \paren {m \cdot x} \circ 0_R       | c =  }} {{eqn | r = 0_R       | c =  }} {{eqn | r = 0 \cdot \paren {x \circ x}       | c =  }} {{eqn | r = \paren {m 0} \cdot \paren {x \circ x}       | c =  }} {{end-eqn}} So $\map P 0$ holds. === Basis for the Induction === Next we verify $\map P 1$. When $n = 1$, we have: {{begin-eqn}} {{eqn | l = \paren {m \cdot x} \circ \paren {1 \cdot x}       | r = \paren {m \cdot x} \circ x       | c =  }} {{eqn | r = m \cdot \paren {x \circ x}       | c =  }} {{eqn | r = \paren {m 1} \cdot \paren {x \circ x}       | c =  }} {{end-eqn}} So $\map P 1$ holds. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\paren {m \cdot x} \circ \paren {k \cdot x} = \paren {m k} \cdot \paren {x \circ x}$ Then we need to show: :$\paren {m \cdot x} \circ \paren {\paren {k + 1} \cdot x} = \paren {m \paren {k + 1} } \cdot \paren {x \circ x}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \paren {m \cdot x} \circ \paren {\paren {k + 1} \cdot x}       | r = \paren {m \cdot x} \circ \paren {k \cdot x + x}       | c =  }} {{eqn | r = \paren {m \cdot x} \circ \paren {k \cdot x} + \paren {m \cdot x} \circ x       | c = {{Ring-axiom|D}} }} {{eqn | r = \paren {m k} \cdot \paren {x \circ x} + m \cdot \paren {x \circ x}       | c = Induction Hypothesis }} {{eqn | r = \paren {m k + k} \cdot \paren {x \circ x}       | c = {{Ring-axiom|D}} }} {{eqn | r = \paren {m \paren {k + 1} } \cdot \paren {x \circ x}       | c =  }} {{end-eqn}} So $\map P K \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall m \in \Z: \forall n \in \N: \paren {m \cdot x} \circ \paren {n \cdot x} = \paren {m n} \cdot \paren {x \circ x}$ {{qed|lemma}} The result for $n < 0$ follows directly from Powers of Group Elements. {{qed}} Category:Ring Theory Category:Proofs by Induction nh2n0mqnbv15b7l1leixn5etc5jip76
By definition, a division ring has no proper zero divisors. If $\struct {D, +, \circ}$ is finite, then from Characteristic of Finite Ring with No Zero Divisors, $\map {\operatorname {Char} } D$ is prime. On the other hand, suppose $\struct {D, +, \circ}$ is not finite. Then there are no $x, y \in D, x \ne 0 \ne y$ such that $x + y = 0$. Thus it follows that $\map {\operatorname {Char} } D$ is zero. {{qed}}
By definition, an integral domain has no proper zero divisors. If $\struct {D, +, \circ}$ is finite, then from Characteristic of Finite Ring with No Zero Divisors, $\operatorname{Char} \left({D}\right)$ is prime. On the other hand, suppose $\struct {D, +, \circ}$ is not finite. Then there are no $x, y \in D, x \ne 0 \ne y$ such that $x + y = 0$. Thus it follows that $\operatorname{Char} \left({D}\right)$ is zero. {{qed}}
By definition of generator: :$\forall G \in \mathcal G: G \in \Sigma$ The third axiom for a $\sigma$-algebra thus ensures that: :$\forall G \in \mathcal G: X \setminus G \in \Sigma$ Therefore, by definition of generated $\sigma$-algebra, $\sigma \left({\mathcal{G}'}\right) \subseteq \Sigma$. However, by the same argument (now applied to $\sigma \left({\mathcal{G}'}\right)$), also: :$\forall G \in \mathcal G: X \setminus \left({X \setminus G}\right) \in \sigma \left({\mathcal{G}'}\right)$ and by Set Difference with Set Difference and Intersection with Subset is Subset: :$X \setminus \left({X \setminus G}\right) = G \cap X = G$ Thus, it follows that $\Sigma = \sigma \left({\mathcal G}\right) \subseteq \sigma \left({\mathcal{G}'}\right)$. Hence the result, by definition of set equality. {{qed}} Category:Sigma-Algebras 3ib7boq4tglu1tojuzxop0s5xnys8wv
From Ring is Module over Itself, it follows that: :$\struct {R, +, \circ}_R$ is an $R$-module. :If $\struct {R, +, \circ}$ has a unity, then $\struct {R, +, \circ}_R$ is unitary. Now the theorem follows directly from Subring Module. {{qed}}
Checking the module axioms in turn: :$(\text M 1): \quad \lambda \circ \paren {x +_G y} = e_G = e_G +_G e_G = \paren {\lambda \circ x} +_G \paren {\lambda \circ y}$ :$(\text M 2): \quad \paren {\lambda +_R \mu} \circ x = e_G = e_G +_G e_G = \paren {\lambda \circ x} +_G \paren {\mu \circ x}$ :$(\text M 3): \quad \paren {\lambda \times_R \mu} \circ x = e_G = \lambda \circ e_G = \lambda \circ \paren {\mu \circ x}$ Thus the trivial module is indeed a module. {{qed}}
By definition, for a trivial module to be unitary, $R$ needs to be a ring with unity. For {{Module-axiom|4}} to apply, we require that: :$\forall x \in G: 1_R \circ x = x$ But for the trivial module: :$\forall x \in G: 1_R \circ x = e_G$ So {{Module-axiom|4}} can apply only when: :$\forall x \in G: x = e_G$ Thus for the trivial module to be unitary, it is necessary that $G$ be the trivial group, and thus to contain one element. {{qed}}
Follows from the fact that $\left({G, +_G, \circ}\right)_R$ has to be, by definition, a trivial module: $\circ$ can only be defined as: :$\forall \lambda \in R: \forall x \in G: \lambda \circ x = e_G$ {{qed}} Category:Module Theory 98hsbt8bc90fqn1tff9vjbo2blkvish
From Generated Sigma-Algebra Preserves Subset, it follows that: :$\sigma \left({\mathcal G}\right) \subseteq \sigma \left({\mathcal H}\right)$ Since $\sigma \left({\mathcal G}\right)$ is a $\sigma$-algebra containing $\mathcal H$: :$\sigma \left({\mathcal H}\right) \subseteq \sigma \left({\mathcal G}\right)$ from the definition of generated $\sigma$-algebra. Hence the result, from the definition of set equality. {{qed}} Category:Sigma-Algebras jdy4qnagp0ewlqivk12691qsj7fyqzv
By definition, $\Sigma$, being a $\sigma$-algebra, is closed under countable unions. From Sigma-Algebra Closed under Countable Intersection, it is also closed under countable intersections. Thence, by definition, $\Sigma$ is a monotone class. {{qed}} Category:Sigma-Algebras Category:Monotone Classes fzkm77bth2qkus1e99usmrmrjf455ur
=== Necessary Condition === By definition of linear combination of subset: :Every linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is a linear combination of $\set {a_k: 1 \mathop \le k \mathop \le n}$. {{qed|lemma}} === Sufficient Condition === Let $b$ be a linear combination of $\set {a_k: 1 \mathop \le k \mathop \le n} = \set {a_1, a_2, \ldots, a_n}$. Then there exists: :a sequence $\sequence {c_j}_{1 \mathop \le j \mathop \le m}$ of elements of $\set {a_1, a_2, \ldots, a_n}$ and: :a sequence $\sequence {\mu_j}_{1 \mathop \le j \mathop \le m}$ of scalars such that: ::$\displaystyle b = \sum_{j \mathop = 1}^m \mu_j c_j$ For each $k \in \closedint 1 n$, let $\lambda_k$ be defined as follows. If: :$a_k \in \set {c_1, c_2, \ldots, c_m}$ and: :$a_i \ne a_j$ for all indices $i$ such that $1 \le i < k$ let $\lambda_k$ be the sum of all scalars $\mu_j$ such that $c_j = a_k$. If: :$a_k \notin \set {c_1, c_2, \ldots, c_m}$ or: :$a_i = a_j$ for some index $i$ such that $1 \le i < k$ let $\lambda_k = 0$. It follows that: :$\displaystyle b = \sum_{j \mathop = 1}^m \mu_j c_j = \sum_{k \mathop = 1}^n \lambda_k a_k$ Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ and $\sequence {b_j}_{1 \mathop \le j \mathop \le m}$ be sequences of elements of $G$ such that $\set {a_1, a_2, \ldots, a_n}$ and $\set {b_1, b_2, \ldots, b_m}$ are identical. Then as a consequence of the above: :an element is a linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ {{iff}}: :it is a linear combination of $\set {a_k: 1 \mathop \le k \mathop \le n}$ {{qed}} {{explain|The above lacks coherence. Source is to be revisited and reinterpreted.}}
By definition of generated $\sigma$-algebra: :$\GG \subseteq \map \sigma \GG$ whence from Set Intersection Preserves Subsets: :$A \cap \GG \subseteq A \cap \map \sigma \GG$ and therefore, by definition of generated $\sigma$-algebra: :$\map \sigma {A \cap \GG} \subseteq A \cap \map \sigma \GG$ For the reverse inclusion, define $\Sigma$ by: :$\Sigma := \set {E \subseteq X: A \cap E \in \map \sigma {A \cap \GG} }$ We will show that $\Sigma$ is a $\sigma$-algebra on $X$. Since $A \in \map \sigma {A \cap \GG}$: :$A \cap X = A \in \map \sigma {A \cap \GG}$ and therefore $X \in \Sigma$. Suppose that $E \in \Sigma$. Then by Set Intersection Distributes over Set Difference and Intersection with Subset is Subset: :$\paren {X \setminus E} \cap A = \paren {X \cap A} \setminus \paren {E \cap A} = A \setminus \paren {E \cap A}$ Since $E \cap A \in \map \sigma {A \cap \GG}$ and this is a $\sigma$-algebra on $A$: :$A \setminus \paren {E \cap A} \in \map \sigma {A \cap \GG}$ Finally, let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence in $\Sigma$. Then by Intersection Distributes over Union: :$\displaystyle \paren {\bigcup_{n \mathop \in \N} E_n} \cap A = \bigcup_{n \mathop \in \N} \paren {E_n \cap A}$ The latter expression is a countable union of elements of $\map \sigma {A \cap \GG}$, hence again in $\map \sigma {A \cap \GG}$. Therefore, $\Sigma$ is a $\sigma$-algebra. It is also apparent that $\GG \subseteq \Sigma$ since: :$A \cap \GG \subseteq \map \sigma {A \cap \GG}$ by definition of generated $\sigma$-algebra. Thus, as $\Sigma$ is a $\sigma$-algebra: :$\map \sigma \GG \subseteq \Sigma$ and therefore: :$A \cap \map \sigma \GG \subseteq \map \sigma {A \cap \GG}$ Hence the result, by definition of set equality. {{qed}} Category:Sigma-Algebras imxcs2gyazaxw5soy70oxy6hh8xx2ly
From Union with Set Difference: :$T = T \setminus S \cup S$ and therefore by Set Difference is Right Distributive over Union: :$T \setminus R = \left({\left({T \setminus S}\right) \setminus R}\right) \cup \left({S \setminus R}\right)$ Now, by Set Difference with Union and Union with Superset is Superset: :$\left({T \setminus S}\right) \setminus R = T \setminus \left({S \cup R}\right) = T \setminus S$ Combining the above yields: :$T \setminus R = \left({T \setminus S}\right) \cup \left({S \setminus R}\right)$ {{qed}}
$(1):$ First we show that $x^\wedge \in G^{**}$: {{ProofWanted}} $(2):$ Then we show that $J: G \to G^{**}$ is a linear transformation: {{ProofWanted}}
Let $\left \langle {a_n} \right \rangle$ be an ordered basis of $G$. Then $\left \langle {J \left({a_n}\right)} \right \rangle$ is the ordered basis of $G^{**}$ dual to the ordered basis of $G^*$ dual to $\left \langle {a_n} \right \rangle$. {{ProofWanted}} From this it follows that $J$ is an isomorphism.
Let the line $L = \alpha_1 x_1 + \alpha_2 x_2 = \beta$ be homogeneous. Then the origin $\tuple {0, 0}$ lies on the line $L$. That is, $\alpha_1 0 + \alpha_2 0 = \beta \implies \beta = 0$. Let the equation of $L$ be $L = \alpha_1 x_1 + \alpha_2 x_2 = 0$. Then $0 = \alpha_1 0 + \alpha_2 0 \in L$ and so $\tuple {0, 0}$ lies on the line $L$. Hence $L$ is homogeneous. {{qed}}
Let the plane $P = \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$ be homogeneous. Then the origin $\left({0, 0, 0}\right)$ lies on the plane $P$. That is, $\alpha_1 0 + \alpha_2 0 + \alpha_3 0= \gamma \implies \gamma = 0$. Let the equation of $P$ be $P = \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = 0$. Then $0 = \alpha_1 0 + \alpha_2 0 + \alpha_3 0 \in P$ and so $\left({0, 0, 0}\right)$ lies on the plane $P$. Hence $P$ is homogeneous. {{qed}}