hcju/proofwikiqa huggingface dataset:
Let $\O \subset \mathbb S \subseteq \mathbb K$. By Intersection of Ring Ideals is Largest Ideal Contained in all Ideals: :$\bigcap \mathbb S$ is the largest ideal of $K$ contained in each of the elements of $\mathbb S$. By Intersection of Ring Ideals Containing Subset is Smallest: :The intersection of the set of all ideals of $K$ containing $\bigcup \mathbb S$ is the smallest ideal of $K$ containing $\bigcup \mathbb S$. Thus, not only is $\bigcap \mathbb S$ a lower bound of $\mathbb S$, but also the largest, and therefore an infimum. The supremum of $\mathbb S$ is the join of the set of all ideals of $\mathbb S$. From Sum of Ideals is Ideal: General Result, this supremum is: :$\displaystyle \sum_{S \mathop \in \mathbb S} \struct {S, +, \circ}$ That is: :$\struct {S_1, +, \circ} + \struct {S_2, +, \circ} + \dotsb$ where addition of ideals is as defined in subset product. Therefore $\struct {\mathbb K, \subseteq}$ is a complete lattice. {{qed}}
From Subgroup of Abelian Group is Normal, $H_1$ and $H_2$ are normal. The result follows from Subset Product with Normal Subgroup is Subgroup. {{qed}} Category:Abelian Groups Category:Subset Products 42ddr5761fjsy89srlm2c6znwzueywi
By definition, $\struct {R, +}$ is an abelian group. So from Subset Product of Abelian Subgroups, we have that: :$\struct{J, +} = \struct{J_1, +} + \struct{J_2, +}$ is itself a subgroup of $R$. Now consider $a \circ b$ where $a \in J, b \in R$. Then: {{begin-eqn}} {{eqn | l = a \circ b       | r = \paren {a_1 + a_2} \circ b       | c = by definition of $a$ }} {{eqn | r = \paren {a_1 \circ b} + \paren {a_2 \circ b}       | c = as $\circ$ is distributive over $+$ }} {{eqn | o = \in       | r = J_1 + J_2       | c = as $\paren {a_i \circ b} \in J_i$ because $J_i$ is an ideal of $R$ }} {{end-eqn}} Similarly, $b \circ a \in J_1 + J_2$ So by definition $J_1 + J_2$ is an ideal of $R$. {{qed}}
Let $J = R \circ b$ The center of $R$ is defined as: :$Z \left({R}\right) = \left\{{x \in R: \forall s \in R: s \circ x = x \circ s}\right\}$ Therefore: :$R \circ b = b \circ R = \left\{{x \circ b: x \in R}\right\} = \left\{{b \circ x: x \in R}\right\}$ and so: :$x \in J \implies x \circ b \in J \land b \circ x \in J$ Thus $J$ is an ideal of $R$ and so $J = \left({b}\right)$. {{qed}}
Consider the structures $\left({R_1, +_1}\right), \left({R_2, +_2}\right), \ldots, \left({R_n, +_n}\right)$. By the definition of a ring, these are all groups. From External Direct Product of Groups is Group we have that the their external direct product: : $\displaystyle \left({R, +}\right) = \prod_{k \mathop = 1}^n \left({R_k, +_k}\right)$ is a group. Similarly, consider the structures $\left({R_1, \circ_1}\right), \left({R_2, \circ_2}\right), \ldots, \left({R_n, \circ_n}\right)$. By the definition of a ring, these are all semigroups. From External Direct Product of Semigroups we have that the their external direct product: : $\displaystyle \left({R, \circ}\right) = \prod_{k \mathop = 1}^n \left({R_k, \circ_k}\right)$ is a semigroup. Finally we note that from External Direct Product Distributivity, $\circ$ as defined here is distributive over $+$. Hence the result, by definition of ring. {{qed}}
As all the $\left({R_k, +_k, \circ_k}\right)$ are ringoids, $\circ_k$ distributes over $+_k$ for all $k$. Let $x, y, z \in R$. Then: {{begin-eqn}} {{eqn | l = x \circ \left({y + z}\right)       | r = \left({x_1, x_2, \ldots, x_n}\right) \circ \left({\left({y_1, y_2, \ldots, y_n}\right) + \left({z_1, z_2, \ldots, z_n}\right)}\right)       | c =  }} {{eqn | r = \left({x_1, x_2, \ldots, x_n}\right) \circ \left({y_1 + z_1, y_2 + z_2, \ldots, y_n + z_n}\right)       | c =  }} {{eqn | r = \left({x_1 \circ \left({y_1 + z_1}\right), x_2 \circ \left({y_2 + z_2}\right), \ldots, x_n \circ \left({y_n + z_n}\right)}\right)       | c =  }} {{eqn | r = \left({\left({x_1 \circ z_1}\right) + \left({y_1 \circ z_1}\right), \left({x_2 \circ z_2}\right) + \left({y_2 \circ z_2}\right), \ldots, \left({x_n \circ z_n}\right) + \left({y_n \circ z_n}\right)}\right)       | c =  }} {{eqn | r = \left({\left({x_1 \circ z_1}\right), \left({x_2 \circ z_2}\right), \ldots, \left({x_n \circ z_n}\right)}\right) + \left({\left({y_1 \circ z_1}\right) + \left({y_2 \circ z_2}\right) + \ldots + \left({y_n \circ z_n}\right)}\right)       | c =  }} {{eqn | r = \left({\left({x_1, x_2, \ldots, x_n}\right) \circ \left({z_1, z_2, \ldots, z_n}\right)}\right) + \left({\left({x_1, x_2, \ldots, x_n}\right) \circ \left({z_1, z_2, \ldots, z_n}\right)}\right)       | c =  }} {{eqn | r = \left({x \circ y}\right) + \left({x \circ z}\right)       | c =  }} {{end-eqn}} In the same way: :$\left({y + z}\right) \circ x = \left({y \circ x}\right) + \left({z \circ x}\right)$ {{qed}} Category:External Direct Products Category:Distributive Operations 0wp7cqiw919zcvcopuv7m87fb8j34dw
Let $y = \left({y_1, y_2, \ldots, y_n}\right) \in R$. Let $x = \left({x_1, x_2, \ldots, x_n}\right) \in R_k$ By definition of direct product, we have: :$x \circ y = \left({x_1 \circ y_1, x_2 \circ y_2, \ldots, x_n \circ y_n}\right)$ :$y \circ x = \left({y_1 \circ x_1, y_2 \circ x_2, \ldots, y_n \circ x_n}\right)$ But we have, $\forall j \ne k: x_j \circ y_j = 0 = y_j \circ x_j$ as $x_j = 0$. So it follows that $x \circ y \in R_k$ and $y \circ x \in R_k$ Hence $R\,'_k$ is an ideal of $R$. {{qed}}
From Ideal of External Direct Sum of Rings we have that $R'_k$ is an ideal of $R$, and thus a subring of $R$. The result follows by application of Canonical Injection is Monomorphism. {{qed}}
From Ideal of External Direct Sum of Rings we already have that $R'_k$ is an ideal of $R$. The result follows by application of Projection is Epimorphism. {{qed}}
Let $S$ be the cartesian product of $\sequence {\struct {S_k, +} }$ First note that $\phi$ is a group homomorphism from $\struct {S, +}$ to $\struct {R, +}$, as: :$\displaystyle \sum_{j \mathop = 1}^n \paren {x_j + y_j} = \sum_{j \mathop = 1}^n x_j + \sum_{j \mathop = 1}^n y_j$ from Associativity on Indexing Set. So $\phi$ is a ring homomorphism {{iff}}: :$\displaystyle \paren {\sum_{j \mathop = 1}^n x_j} \circ \paren {\sum_{j \mathop = 1}^n y_j} = \sum_{j \mathop = 1}^n \paren {x_j \circ y_j}$ Now, let: :$\struct {S, +, \circ}$ be the cartesian product of $\sequence {\struct {S_k, +, \circ} }$ :$\phi: S \to R$ be the mapping defined as: ::$\map \phi {x_1, x_2, \ldots, x_n} = x_1 + x_2 + \cdots x_n$ Clearly $\phi$ is a surjection {{iff}} $(1)$ holds. By Internal Direct Product Generated by Subgroups, $\phi$ is a ring isomorphism {{iff}} $(1)$ and $(2)$ hold. Let $\phi: S \to R$ be a ring isomorphism. By Canonical Injection from Ideal of External Direct Sum of Rings, $\map {\inj_k} {S_k}$ is an ideal of $S$. So $\map \phi {\map {\inj_k} {S_k} }$ is an ideal of $R$. But $\phi$ and $\pr_k$ coincide on $\map {\inj_k} {S_k}$. So: :$\map \phi {\map {\inj_k} {S_k} } = \map {\pr_k} {\map {\inj_k} {S_k} } = S_k$ and so $(3)$ holds. Now suppose $(3)$ holds, and the $S_k$ are all ideals of $R$. By $(2)$ and by definition of independent subgroups: :$i \ne j \implies S_i \cap S_j = \set {0_R}$ So for all $\tuple {x_1, x_2, \ldots, x_n}, \tuple {y_1, y_2, \ldots, y_n} \in S$: {{begin-eqn}} {{eqn | l = \map \phi {x_1, x_2, \ldots, x_n} \circ \map \phi {y_1, y_2, \ldots, y_n}       | r = \paren {\sum_{i \mathop = 1}^n x_i} \circ \paren {\sum_{j \mathop = 1}^n y_j}       | c =  }} {{eqn | r = \sum_{i \mathop = 1}^n \paren {\sum_{j \mathop = 1}^n x_i \circ y_j}       | c =  }} {{eqn | r = \sum_{i \mathop = 1}^n x_i \circ y_i       | c =  }} {{eqn | r = \map \phi {x_1, x_2, \ldots, x_n} \circ \map \phi {y_1, y_2, \ldots, y_n}       | c =  }} {{end-eqn}} because as $S_i, S_j$ are ideals, we have: :$x_i \circ j_j = S_i \cap S_j = \set {0_R}$ Hence the three conditions are sufficient for $\phi$ to be a ring isomorphism. {{qed}}
As $R$ is a ring, both $+$ and $\circ$ are closed on $R$ by definition. From Closure of Pointwise Operation on Algebraic Structure, it follows that both $+'$ and $\circ'$ are closed on $R^S$: :$\forall f, g \in R^S: f +' g \in R^S$ :$\forall f, g \in R^S: f \circ' g \in R^S$ By Structure Induced by Abelian Group Operation is Abelian Group, $\struct {R^S, +'}$ is an abelian group. By Structure Induced by Associative Operation is Associative, $\struct {R^S, \circ'}$ is a semigroup. From Pointwise Operation on Distributive Structure is Distributive, $\circ'$ is distributive over $+'$. The result follows by definition of ring. {{qed}}
=== Necessary Condition === Suppose $\struct {L, +, \circ}$ is a division subring of $\struct {K, +, \circ}$. The conditions $(1)$ to $(3)$ hold by virtue of the Subring Test. Then $(4)$ also holds by the definition of a division ring: :$\forall x \in L^*: \exists ! x^{-1} \in L^*: x^{-1} \circ x = x \circ x^{-1} = 1_L$ {{qed|lemma}} === Sufficient Condition === Suppose the conditions $(1)$ to $(4)$ hold. By $(1)$ to $(3)$, it follows from Subring Test that $\struct {L, +, \circ}$ is a subring of $\struct {K, +, \circ}$. By $(4)$, every element of $L^*$ has a product inverse. Thus, from the Two-Step Subgroup Test, $\struct {L^*, \circ}$ is a group. Therefore, $\struct {L, +, \circ}$ is a ring in which every element has a product inverse, which makes $\struct {L, +, \circ}$ a division ring. {{qed}} Category:Subrings 1o60mugxbp4aeldykaapiwlbxt3ytex
Let $\O \subset \mathbb S \subseteq \mathbb K$. By Intersection of Division Subrings is Largest Division Subring Contained in all Division Subrings: :$\bigcap \mathbb S$ is the largest division subring of $K$ contained in each of the elements of $\mathbb S$. By Intersection of Division Subrings Containing Subset is Smallest: :The intersection of the set of all division subrings of $K$ containing $\bigcup \mathbb S$ is the smallest division subring of $K$ containing $\bigcup \mathbb S$. Thus: :Not only is $\bigcap \mathbb S$ a lower bound of $\mathbb S$, but also the largest, and therefore an infimum. :The supremum of $\mathbb S$ is the intersection of the set of all division subrings of $K$. Therefore $\struct {\mathbb K, \subseteq}$ is a complete lattice. {{qed}}
Let $\O \subset \mathbb S \subseteq \mathbb F$. By Intersection of Subfields is Largest Subfield Contained in all Subfields: :$\bigcap \mathbb S$ is the largest subfield of $F$ contained in each of the elements of $\mathbb S$. By Intersection of Subfields Containing Subset is Smallest: :The intersection of the set of all subfields of $F$ containing $\bigcup \mathbb S$ is the smallest subfield of $F$ containing $\bigcup \mathbb S$. Thus: :Not only is $\bigcap \mathbb S$ a lower bound of $\mathbb S$, but also the largest, and therefore an infimum. :The supremum of $\mathbb S$ is the intersection of the set of all subfields of $R$. Therefore $\struct {\mathbb F, \subseteq}$ is a complete lattice. {{qed}}
The fact that $g_a$ is a group homomorphism follows directly from Index Laws for Monoids: {{begin-eqn}} {{eqn | l = \map {g_a} m + \map {g_a} n       | r = m \cdot a + n \cdot a }} {{eqn | r = \paren {m + n} \cdot a }} {{eqn | r = \map {g_a} {m + n} }} {{end-eqn}} By Multiple of Ring Product, we have that: :$\forall n \in \Z_{>0}: \paren {n \cdot x} \circ y = n \cdot \paren {x \circ y} = x \circ \paren {n \cdot y}$ So: :$\forall n \in \Z_{>0}: n \cdot a = \paren {n \cdot a} \circ 1_R = a \circ \paren {n \cdot 1_R}$ so when $n \cdot 1_R = 0$ we have $n \cdot a = 0$. For $n \in \Z_{<0}$, we have $-n \in \Z_{>0}$. So: :$n \cdot a = -\paren {-n} \cdot a = -\paren {\paren {-n \cdot a} \circ 1_R} = -\paren {a \circ \paren {-n \cdot 1_R}}$ so when $-n \cdot 1_R = 0$ we have $n \cdot a = 0$. For $n = 0$, we trivially have $n \cdot a = 0$. By definition of characteristic, we have: :$p \divides n \iff n \cdot 1_R = 0$ by the above, this implies $n \cdot a = 0$. Therefore: :$p \divides n \implies n \cdot a = 0$ We also have: {{begin-eqn}} {{eqn | l = n       | o = \in       | r = \ideal p }} {{eqn | ll = \leadstoandfrom       | l = p       | o = \divides       | r = n       | c = {{Defof|Ideal of Ring}} }} {{eqn | ll = \leadsto       | l = n \cdot a       | r = 0 }} {{eqn | ll = \leadstoandfrom       | l = n       | o = \in       | r = \map \ker {g_a}       | c = {{Defof|Kernel of Group Homomorphism}} }} {{end-eqn}} By definition of subset: :$\ideal p \subseteq \map \ker {g_a}$ {{qed}}
From Field of Characteristic Zero has Unique Prime Subfield, if $\Char F = 0$, then its prime subfield is isomorphic to $\Q$, the field of rational numbers. From Field of Prime Characteristic has Unique Prime Subfield, if $\Char F = p$, then its prime subfield is isomorphic to $\Z_p$, the Ring of Integers Modulo $p$. From Characteristic of Field is Zero or Prime, $p$ is prime. {{Qed}}
Let $H$ be a subgroup of $\struct {\Z_m, +_m}$ Suppose: : $(1): \quad h + \ideal m \in H$, where $\ideal m$ is a principal ideal of $\struct {\Z_m, +_m, \times_m}$ and  : $(2): \quad n \in \N_{>0}$. Then by definition of multiplication on integers and Homomorphism of Powers as applied to integers: {{begin-eqn}} {{eqn | l = \paren {n + \ideal m} \times \paren {h + \ideal m}       | r = \map {q_m} n \times \map {q_m} h       | c = where $q_m$ is the quotient mapping }} {{eqn | r = \map {q_m} {n \times h}       | c =  }} {{eqn | r = \map {q_m} {n \cdot h}       | c =  }} {{eqn | r = n \cdot \map {q_m} h       | c =  }} {{end-eqn}} But: :$n \cdot \map {q_m} h \in \gen {\map {q_m} h}$ where $\gen {\map {q_m} h}$ is the group generated by $\map {q_m} h$. Hence by Epimorphism from Integers to Cyclic Group, $n \cdot \map {q_m} h \in H$. The result follows. {{qed}}
From Subgroup of Finite Cyclic Group is Determined by Order, there exists exactly one subgroup $\left \langle {a^{n/m}} \right \rangle$ of $G$ with $a$ elements. So the mapping as defined is indeed a bijection by definition. {{qed}}
From Integers under Addition form Infinite Cyclic Group, the identity element $1_\Z$ of the ring of integers $\struct {\Z, +, \times}$ is a generator of the group $\struct {\Z, +}$. Thus from Quotient Group of Cyclic Group, the identity element $1_{\Z_m}$ of the ring $\struct {\Z_m, +_m, \times_m}$ is a generator of the group $\struct {\Z_m, +_m}$. Let $a \in \Z_m$. Suppose $1_{\Z_m} \in \gen a$, where $\gen a$ signifies the group generated by $a$. Then the smallest subgroup of $\struct {\Z_m, +_m}$ containing $1_{\Z_m}$, that is $\struct {\Z_m, +_m}$ itself, is contained in $\gen a$. Thus $\gen a = \struct {\Z_m, +_m}$ iff $1_{\Z_m} \in \gen a$. However, from Subgroup of Additive Group Modulo m is Ideal of Ring, $\gen a$ is an ideal of $\struct {\Z_m, +_m, \times_m}$, and hence is the principal ideal $\ideal a$ generated by $a$. But from Principal Ideal from Element in Center of Ring, $1_{\Z_m} \in \gen a$ {{iff}} $a$ is a unit of the ring $\struct {\Z_m, +_m, \times_m}$. Hence the result. {{qed}}
Note that by definition, $\map \DD {\mathbb J} \subseteq \R^{\mathbb J}$.  Let $f, g \in \map \DD {\mathbb J}$.  Let $\lambda \in \R$. From Linear Combination of Derivatives, we have that:  :$f + \lambda g$ is differentiable on $\mathbb J$.  That is:  :$f + \lambda g \in \map \DD {\mathbb J}$ So, by One-Step Vector Subspace Test: :$\struct {\map \DD {\mathbb J}, +, \times}_\R$ is a subspace of $\R^{\mathbb J}$.  {{qed}}
Note that by definition, $\map {\CC^{\paren m} } {\mathbb J} \subseteq \R^{\mathbb J}$. Let $f, g \in \map {\CC^{\paren m} } {\mathbb J}$. Let $\lambda \in \R$. Applying Linear Combination of Derivatives $m$ times we have:  :$f + \lambda g$ is $m$-times differentiable on $\mathbb J$ with $m$th derivative $f^{\paren m} + \lambda g^{\paren m}$. Since both $f$ and $g$ are of differentiability class $m$: :$f^{\paren m}$ and $g^{\paren m}$ are continuous on $\mathbb J$.  From Combination Theorem for Continuous Functions: Combined Sum Rule: :$f^{\paren m} + \lambda g^{\paren m} = \paren {f + \lambda g}^{\paren m}$ is continuous on $\mathbb J$. So: :$f + \lambda g \in \map {\CC^{\paren m} } {\mathbb J}$ Therefore, by One-Step Vector Subspace Test: :$\struct {\map {\CC^{\paren m} } {\mathbb J}, +, \times}_\R$ is a subspace of $\struct {\R^{\mathbb J}, +, \times}_\R$. {{qed}}
Note that by definition, $\map {\mathcal D} {\mathbb J} \subseteq \R^{\mathbb J}$.  Let $f, g \in \map \RR {\mathbb J}$. Let $\lambda \in \R$. By Linear Combination of Definite Integrals: :$f + \lambda g$ is Riemann integrable on $\mathbb J$. That is:  :$f + \lambda g \in \map \RR {\mathbb J}$ So by One-Step Vector Subspace Test: :$\struct {\map \RR {\mathbb J}, +, \times}_\R$ is a subspace of $\struct {\R^{\mathbb J}, +, \times}_\R$. {{qed}}
By definition, every real polynomial function is a linear combination of $B$. Suppose: :$\displaystyle \sum_{k \mathop = 0}^m \alpha_k I^k = 0, \alpha_m \ne 0$ Then by differentiating $m$ times, we obtain from Nth Derivative of Nth Power: :$m! \alpha_m = 0$ whence $\alpha_m = 0$ which is a contradiction. Hence $B$ is linearly independent and therefore is a basis for $\map P \R$. {{qed}}
:500pxrightthumb Let $P$ be any point in the plane for which we want to provide a linear combination of $a_1$ and $a_2$. Let the distance from $O$ to the point determined by $a_1$ be defined as being $1$ unit of length on the line $L_1$. Let the distance from $O$ to the point determined by $a_2$ be defined as being $1$ unit of length on the line $L_2$. Draw lines parallel to $L_1$ and $L_2$ through $P$. Then the coordinates $\lambda_1$ and $\lambda_2$ of $P$ are given by: :$P = \lambda_1 a_1 + \lambda_2 a_2$ by the Parallelogram Law. {{ProofWanted}}
Apply Taylor's Theorem to the transformations $X$, $\mathbf Y$ at the point $\epsilon = 0$: {{begin-eqn}} {{eqn | l = X       | r = \map \Phi {x, \mathbf y, \mathbf y'; \epsilon} }} {{eqn | r = \map \Phi {x, \mathbf y, \mathbf y'; 0} + \epsilon \frac {\partial \map \Phi {x, \mathbf y, \mathbf y'; \epsilon} } {\partial \epsilon} \Bigg\rvert_{\epsilon \mathop = 0} + \map \OO \epsilon }} {{eqn | r = x + \epsilon \phi + \map \OO \epsilon }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \mathbf Y       | r = \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; \epsilon} }} {{eqn | r = \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; 0} + \epsilon \frac {\partial \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; \epsilon} } {\partial \epsilon} \Bigg \rvert_{\epsilon \mathop = 0} + \map \OO \epsilon }} {{eqn | r = y + \epsilon \boldsymbol \psi + \map \OO \epsilon }} {{end-eqn}} Use the general variation formula, and suppose that the curve $\mathbf y = \map {\mathbf y} x$ is an extremal of $J \sqbrk {\mathbf y}$: :$\delta x = \epsilon \phi, \quad \delta y = \epsilon\psi$ :$\delta J = \epsilon \bigintlimits {\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y' \cdot \nabla_{\mathbf y'} F }\phi} {x \mathop = x_0} {x \mathop = x_1}$ Since $J \sqbrk {\mathbf y}$ is invariant under the transformation, the variation vanishes. Then for any $\epsilon \ne 0$ we have: $\bigintlimits {\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y'\cdot \nabla_{\mathbf y'} F} \phi} {x \mathop = x_0} {} = \bigintlimits {\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y' \cdot \nabla_{\mathbf y'} F }\phi} {x \mathop = x_1} {}$ This has to hold for arbitrary $x_0$ and $x_1$. Since only a constant mapping produces the same result for any input, the term in brackets has to be a constant. {{qed}} {{Namedfor|Emmy Noether|cat = Noether E}}
By definition of evaluation linear transformation: :$\forall x \in G: y' \in H^*: \sequence {x, \map {u^t} {y'} } = \sequence {\map u x, y'}$ Since we have: {{begin-eqn}} {{eqn | l = \sequence {x, \map {u^t} {y' + z'} }       | r = \sequence {\map u x, y' + z'}       | c =  }} {{eqn | r = \sequence {\map u x, y'} + \sequence {\map u x, z'}       | c =  }} {{eqn | r = \sequence {x, \map {u^t} {y'} } + \sequence {x, \map {u^t} {z'} }       | c =  }} {{eqn | r = \sequence {x, \map {u^t} {y'} + \map {u^t} {z'} }       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \sequence {x, \map {u^t} {\lambda y'} }       | r = \sequence {\map u x, \lambda y'}       | c =  }} {{eqn | r = \lambda \sequence {\map u x, y'}       | c =  }} {{eqn | r = \lambda \sequence {x, \map {u^t} {y'} }       | c =  }} {{eqn | r = \sequence {x, \lambda \map {u^t} {y'} }       | c =  }} {{end-eqn}} it follows that $u^t: H^* \to G^*$ is a linear transformation. {{qed}}
=== Definition 1 implies Definition 2 === Let $K \in \R$ be a supremum of $f$ by definition 1. Then from the definition: : $\text{(a)}: \quad K$ is an upper bound of $f \left({x}\right)$ in $\R$. : $\text{(b)}: \quad K \le M$ for all upper bounds $M$ of $f \left({S}\right)$ in $\R$. As $K$ is an upper bound it follows that: :$(1): \quad \forall x \in S: f \left({x}\right) \le K$ Now let $\epsilon \in \R_{>0}$. {{AimForCont}} $(2)$ were false: :$\forall x \in S: f \left({x}\right) \le K - \epsilon$ Then by definition, $K - \epsilon$ is an upper bound of $f$. But by definition that means $K \le K + \epsilon$. So by Real Plus Epsilon: : $K < K$ From this contradiction we conclude that: : $(2): \quad \exists x \in S: \forall \epsilon \in \R_{>0}: K - \epsilon < f \left({x}\right)$ Thus $K$ is a supremum of $f$ by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $K$ be a supremum of $f$ by definition 2: : $(1) \quad \forall x \in S: f \left({x}\right) \le K$ : $(2) \quad \exists x \in S: \forall \epsilon \in \R_{>0}: K - \epsilon < f \left({x}\right)$ From $(1)$ we have that $K$ is an upper bound of $f$. {{AimForCont}} that $K$ is ''not'' a supremum of $f$ by definition 1. Then: : $\exists M \in \R, M < K: \forall x \in S: f \left({x}\right) \le M$ Then: :$\exists \epsilon \in \R_{>0}: M = K - \epsilon$ Hence: :$\exists \epsilon \in \R_{>0}: \forall x \in S: f \left({x}\right) \le K - \epsilon$ This contradicts $(2)$. Thus $K$ is a supremum of $f$ by definition 1. {{qed}} Category:Real Analysis dhw6ddzg7b8ps4uhwpmz8l3ed78h72m
{{wtd|Use Characterizing Property of Infimum of Subset of Real Numbers}} === Definition 1 implies Definition 2 === Let $k \in \R$ be an infimum of $f$ by definition 1. Then from the definition: :$\text{(a)}: \quad k$ is a lower bound of $\map f x$ in $\R$. :$\text{(b)}: \quad k \ge m$ for all lower bounds $m$ of $f \sqbrk S$ in $\R$. As $k$ is a lower bound it follows that: :$(1): \quad \forall x \in S: k \le \map f x$ Now let $\epsilon \in \R_{>0}$. {{AimForCont}} $(2)$ were false: :$\forall x \in S: k + \epsilon \le \map f x$ Then by definition, $k + \epsilon$ is a lower bound of $f$. But by definition that means $k \ge k + \epsilon$. So by Real Plus Epsilon: :$k > k$ From this contradiction we conclude that: :$(2): \quad \exists x \in S: \forall \epsilon \in \R_{>0}: \map f x < k + \epsilon$ Thus $k$ is an infimum of $f$ by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $k$ be an infimum of $f$ by definition 2: :$(1): \quad \forall x \in S: k \le \map f x$ :$(2): \quad \exists x \in S: \forall \epsilon \in \R_{>0}: \map f x < k + \epsilon$ From $(1)$ we have that $k$ is a lower bound of $f$. {{AimForCont}} that $k$ is ''not'' an infimum of $f$ by definition 1. Then: :$\exists m \in \R, m > k: \forall x \in S: \map f x \ge m$ Then: :$\exists \epsilon \in \R_{>0}: m = k + \epsilon$ Hence: :$\exists \epsilon \in \R_{>0}: \forall x \in S: k + \epsilon \le \map f x$ This contradicts $(2)$. Thus $k$ is an infimum of $f$ by definition 1. {{qed}} Category:Infima k5t6anxwz58e3xa9n3e2rss9dstbjlm
By definition of supremum: :$\forall a \in T: a \preceq M$ By definition of infimum: :$\forall a \in T: m \preceq a$ The result follows from transitivity of ordering. {{qed}}