hcju/proofwikiqa huggingface dataset:
=== $(3)$ implies $(3')$ === Let $X$ be a set, and let $\SS$ be a collection of subsets of $X$. Suppose that for all $A, A_1 \in \SS$ such that $A_1 \subseteq A$, there exists a finite sequence of sets $A_2, A_3, \ldots, A_n \in \SS$ such that: :$A_1, A_2, \ldots, A_n$ are pairwise disjoint :$\displaystyle A = \bigcup_{k \mathop = 1}^n A_k$ Let $B \in \SS$, and let $A_1 = A \cap B$. It follows that $A_1 \in \SS$, by definition. Also, $A_1 \subseteq A$ by Intersection is Subset. Then: {{begin-eqn}} {{eqn | l = A \setminus B       | r = A \setminus \left({A \cap B}\right)       | c = Set Difference with Intersection is Difference }} {{eqn | r = A \setminus A_1 }} {{eqn | r = \left({\bigcup_{k \mathop = 1}^n A_k}\right) \setminus A_1 }} {{eqn | r = \bigcup_{k \mathop = 1}^n \, \left({A_k \setminus A_1}\right)       | c = Set Difference is Right Distributive over Union }} {{eqn | r = \bigcup_{k \mathop = 2}^n \, \left({A_k \setminus A_1}\right)       | c = Set Difference with Self is Empty Set and Union with Empty Set }} {{eqn | r = \bigcup_{k \mathop = 2}^n A_k       | c = Set Difference with Disjoint Set }} {{end-eqn}} as required. {{qed|lemma}} === $(3')$ implies $(3)$ === Now suppose that for all $A, B \in \SS$, there exists a finite sequence of pairwise disjoint sets $A_1, A_2, \ldots, A_n \in \SS$ such that $\displaystyle A \setminus B = \bigcup_{k \mathop = 1}^n A_k$. Then $B$ is disjoint with each of the sets $A_k$. Let $B \subseteq A$. Then: {{begin-eqn}} {{eqn | l = B \cup \bigcup_{k \mathop = 1}^n A_k       | r = \left({A \setminus B}\right) \cup B }} {{eqn | r = A \cup B       | c = Set Difference Union Second Set is Union }} {{eqn | r = A       | c = Union with Superset is Superset }} {{end-eqn}} as required. {{qed}} Category:Semirings of Sets pfw64woccuy4o9v9j9bdj3kaaclv85c
From Imaginary Part as Mapping is Surjection, $f$ is a surjection. Let $z_1, z_2 \in \C$. Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$. Then: {{begin-eqn}} {{eqn | l = \map f {z_1 + z_2}       | r = \map \Im {z_1 + z_2}       | c = Definition of $f$ }} {{eqn | r = \map \Im {x_1 + i y_1 + x_2 + i y_2}       | c = Definition of $z_1$ and $z_2$ }} {{eqn | r = y_1 + y_2       | c = {{Defof|Imaginary Part}} }} {{eqn | r = \map \Im {z_1} + \map \Im {z_2}       | c = {{Defof|Imaginary Part}} }} {{eqn | r = \map f {z_1} + \map f {z_2}       | c = Definition of $f$ }} {{end-eqn}} So $f$ is a group homomorphism. Thus $f$ is a surjective group homomorphism and therefore by definition a group epimorphism. Finally: :$\forall y \in \R: \map \Im {x + 0 i} = 0 i = 0$ It follows from Complex Addition Identity is Zero that:  :$\map \ker f = \set {x: x \in \R} = \R$ {{qed}}
=== Proof of Existence === Proved in Matrix is Row Equivalent to Reduced Echelon Matrix. {{qed|lemma}} === Proof of Uniqueness === {{ProofWanted}} Category:Echelon Matrices tqbd2jjcoeedgo03370m94xifgo20hg
{{begin-eqn}} {{eqn | l = T       | r = \paren {T \setminus S} \cup \paren {T \cap S}       | c = Set Difference Union Intersection }} {{eqn | r = \paren {T \setminus S} \cup S       | c = Intersection with Subset is Subset }} {{eqn | ll= \leadsto       | l = \mu \paren T       | r = \mu \paren {T \setminus S} + \mu \paren S       | c = Measure is Finitely Additive Function, Set Difference Intersection with Second Set is Empty Set }} {{eqn | ll= \leadsto       | l = \mu \paren T - \mu \paren S       | r = \mu \paren {T \setminus S})       | c = Subtraction defined as $\mu \paren S < +\infty$ }} {{end-eqn}} {{qed}} Category:Measure Theory aqnz065pz56fykosrzns46w9htorram
From Image of Preimage under Mapping: : $B \subseteq T \implies \left({f \circ f^{-1} }\right) \left[{B}\right] = B \cap f \left[{S}\right]$ The result follows from Intersection is Subset. {{qed}}
Let $y \in B$. Then: :$\exists x \in S: y = f \left({x}\right)$ Therefore by definition of preimage of subset: :$\exists x \in f^{-1} \left[{B}\right]$ It follows by definition of image of subset that: :$y \in f \left[{f^{-1} \left[{B}\right]}\right]$ Thus by definition of composition $f$ with $f^{-1}$: :$y \in \left({f \circ f^{-1}}\right) \left[{B}\right]$ The result follows by definition of subset. {{qed}}
Let $B \subseteq T$. Then: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = \paren {f \circ f^{-1} } \sqbrk B }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = f \sqbrk {f^{-1} \sqbrk B}       | c = {{Defof|Composition of Mappings}} }} {{eqn | ll= \leadsto       | lo= \exists x \in f^{-1} \sqbrk B:       | l = \map f x       | r = y       | c = {{Defof|Image of Subset under Mapping}} }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = B       | c = {{Defof|Preimage of Subset under Mapping}} }} {{end-eqn}} So by definition of subset: :$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B \subseteq B$ {{qed}}
=== Transitive === Let $a, b, c \in \overline \R$. Suppose that $a \le b$ and $b \le c$. If $c = +\infty$, then by the definition of $\le$, $a \le c$. If $b = +\infty$, then by Positive Infinity is Maximal, $b = c$, so $a \le c$. If $a = +\infty$, then applying Positive Infinity is Maximal twice yields $a \le c$. The cases for $a$, $b$, or $c$ being $-\infty$ are similar. The only remaining case is that $a, b, c \in \R$. Since $\le$ is transitive on $\R$, $a \le c$. {{qed|lemma}} === Antisymmetric === Let $a, b \in \overline{\R}$. Suppose that $a \le b$ and $b \le a$. If $a \in \R$ and $b \in \R$, then since $\le$ is antisymmetric on $\R$, $a = b$. If $a = +\infty$, then since $a \le b$ and Positive Infinity is Maximal, $b = +\infty$. Thus $a = b$. If $a = -\infty$, then since $b \le a$ and Negative Infinity is Minimal, $b = -\infty$. Similarly $b = +\infty \implies a = +\infty$ and $b = -\infty \implies a = -\infty$. {{qed|lemma}} === Reflexive === Let $a \in \overline \R$. If $a \in \R$, then since $\le$ is reflexive on $\R$, $a \le a$. If $a = +\infty$, then since $\tuple {+\infty, +\infty} \in \set {\tuple {x, +\infty}: x \in \overline \R}$, $a \le a$. Similarly, if $a = -\infty$, then $a \le a$. {{qed}} Category:Usual Ordering on Extended Real Numbers 5hb6v34dla40bkoqe30ujv6f1dudwat
By definition of isomorphism, each $\phi_k$ is a homomorphism which is a bijection. From Cartesian Product of Bijections is Bijection: General Result, $\phi$ is a bijection. From Homomorphism of External Direct Products: General Result, $\phi$ is a homomorphism. Hence the result. {{qed}}
Let $\left({G_1, \circ_1}\right), \left({G_2, \circ_2}\right), \ldots, \left({G_n, \circ_n}\right)$ be groups. Let $\displaystyle \left({G, \circ}\right) = \prod_{k \mathop = 1}^n G_k$ be the external direct product of $\left({G_1, \circ_1}\right), \left({G_2, \circ_2}\right), \ldots, \left({G_n, \circ_n}\right)$. Taking the group axioms in turn: === G0: Closure === From External Direct Product Closure: General Result it follows that $\left({G, \circ}\right)$ is closed. {{qed|lemma}} === G1: Associativity === From External Direct Product Associativity: General Result it follows that $\left({G, \circ}\right)$ is associative. {{qed|lemma}} === G2: Identity === Let $e_1, e_2, \ldots, e_n$ be the identity elements of $\left({G_1, \circ_1}\right), \left({G_2, \circ_2}\right), \ldots, \left({G_n, \circ_n}\right)$ respectively. From External Direct Product Identity: General Result it follows that $\left({e_1, e_2, \ldots, e_n}\right)$ is the identity element of $\left({G, \circ}\right)$. {{qed|lemma}} === G3: Inverses === Let $g_1, g_2, \ldots, g_n$ be arbitrary elements of $G_1, G_2, \ldots, G_n$ Let $g_1^{-1}, g_2^{-1}, \ldots, g_n^{-1}$ be the inverse elements of $g_1, g_2, \ldots, g_n$ in $\left({G_1, \circ_1}\right), \left({G_2, \circ_2}\right), \ldots, \left({G_n, \circ_n}\right)$ respectively. From External Direct Product Inverses: General Result it follows that $\left({g_1^{-1}, g_2^{-1}, \ldots, g_n^{-1}}\right)$ is the inverse element of $\left({g_1, g_2, \ldots, g_n}\right)$ in $\left({G, \circ}\right)$. {{qed|lemma}} All group axioms are fulfilled, hence $\left({G, \circ}\right)$ is a group. {{qed}}
Let $e_1, e_2, \ldots, e_n$ be the identity elements of $\left({S_1, \circ_1}\right), \left({S_2, \circ_2}\right), \ldots, \left({S_n, \circ_n}\right)$ respectively. Let $x := \left({x_1, x_2, \ldots, x_n}\right)$. Let $y := \left({x_1, x_2, \ldots, x_n}\right)$. From External Direct Product Identity, $e := \left({e_1, e_2, \ldots, e_n}\right)$ is the identity element of $S^n$. Then: {{begin-eqn}} {{eqn | l = x \circ y       | r = \left({x_1, x_2, \ldots, x_n}\right) \circ \left({y_1, y_2, \ldots, y_n}\right)       | c =  }} {{eqn | r = \left({x_1 \circ_1 y_1, x_2 \circ_2 y_2, \ldots, x_n \circ_n y_n}\right)       | c = Definition of External Direct Product }} {{eqn | r = \left({e_1, e_2, \ldots, e_n}\right)       | c = {{Defof|Inverse Element}} }} {{eqn | r = e       | c = External Direct Product Identity }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = y \circ x       | r = \left({y_1, y_2, \ldots, y_n}\right) \circ \left({x_1, x_2, \ldots, x_n}\right)       | c =  }} {{eqn | r = \left({y_1 \circ_1 x_1, y_2 \circ_2 x_2, \ldots, y_n \circ_n x_n}\right)       | c = Definition of External Direct Product }} {{eqn | r = \left({e_1, e_2, \ldots, e_n}\right)       | c = {{Defof|Inverse Element}} }} {{eqn | r = e       | c = External Direct Product Identity }} {{end-eqn}} {{qed}}
Let $s := \left({s_1, s_2, \ldots, s_n}\right)$ be an arbitrary element of $\left({S_1, \circ_1}\right) \times \left({S_2, \circ_2}\right) \times \cdots \times \left({S_n, \circ_n}\right)$. Let $e := \left({e_1, e_2, \ldots, e_n}\right)$. Then: {{begin-eqn}} {{eqn | l = s \circ e       | r = \left({s_1, s_2, \ldots, s_n}\right) \circ \left({e_1, e_2, \ldots, e_n}\right)       | c =  }} {{eqn | r = \left({s_1 \circ_1 e_1, s_2 \circ_2 e_2, \ldots, s_n \circ_n e_n}\right)       | c = Definition of External Direct Product }} {{eqn | r = \left({s_1, s_2, \ldots, s_n}\right)       | c = {{Defof|Identity Element}} }} {{eqn | r = s       | c = Definition of $s$ }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = e \circ s       | r = \left({e_1, e_2, \ldots, e_n}\right) \circ \left({s_1, s_2, \ldots, s_n}\right)       | c =  }} {{eqn | r = \left({e_1 \circ_1 s_1, e_2 \circ_2 s_2, \ldots, e_n \circ_n s_n}\right)       | c = Definition of External Direct Product }} {{eqn | r = \left({s_1, s_2, \ldots, s_n}\right)       | c = {{Defof|Identity Element}} }} {{eqn | r = s       | c = Definition of $s$ }} {{end-eqn}} {{qed}}
Suppose that, for all $k \in \N^*_n$, $\circ_k$ is commutative. Let $\left({s_1, s_2, \ldots, s_n}\right)$ and $\left({t_1, t_2, \ldots, t_n}\right)$ be elements of $\left({S_1, \circ_1}\right) \times \left({S_2, \circ_2}\right) \times \cdots \times \left({S_n, \circ_n}\right)$. {{begin-eqn}} {{eqn | l = \left({s_1, s_2, \ldots, s_n}\right) \circ \left({t_1, t_2, \ldots, t_n}\right)       | r = \left({s_1 \circ_1 t_1, s_2 \circ_2 t_2, \ldots, s_n \circ_n t_n}\right)       | c =  }} {{eqn | r = \left({t_1 \circ_1 s_1, t_2 \circ_2 s_2, \ldots, t_n \circ_n s_n}\right)       | c =  }} {{eqn | r = \left({t_1, t_2, \ldots, t_n}\right) \circ \left({s_1, s_2, \ldots, s_n}\right)       | c =  }} {{end-eqn}} Hence the result. {{qed}}
Proof by Counterexample: Let $\mathbf a = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$, $\mathbf b = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$, $\mathbf c = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ be vectors in $\R^3$. {{begin-eqn}} {{eqn | l = \mathbf a \times \left({\mathbf b \times \mathbf c}\right)       | r = \mathbf a \times \left({\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \times \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} }\right) }} {{eqn | r = \mathbf a \times \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} }} {{eqn | r = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \times \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} }} {{eqn | r = \begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix} }} {{eqn | l = \left({\mathbf a \times \mathbf b}\right) \times \mathbf c       | r =  \left({\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \times \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} }\right) \times \mathbf c }} {{eqn | r = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \times \mathbf c }} {{eqn | r = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \times \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} }} {{eqn | r = \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} }} {{end-eqn}} {{qed}} Category:Vector Cross Product 7bzmkkoao1qhhs5bmyk2u1rrudeclby
{{begin-eqn}} {{eqn | l = \paren {\mathbf a + \mathbf b} \times \paren {\mathbf a + \mathbf b}       | r = \mathbf a \times \mathbf a + \mathbf a \times \mathbf b + \mathbf b \times \mathbf a + \mathbf b \times \mathbf b       | c = Vector Cross Product Operator is Bilinear }} {{eqn | l = 0       | r = 0 + \mathbf a \times \mathbf b + \mathbf b \times \mathbf a + 0       | c = Cross Product of Vector with Itself is Zero }} {{eqn | l = \mathbf a \times \mathbf b       | r = -\paren {\mathbf b \times \mathbf a}       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \mathbf a \times \mathbf b       | r = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ a_i & a_j & a_k \\ b_i & b_j & b_k \\ \end{vmatrix}       | c = {{Defof|Vector Cross Product}} }} {{eqn | r = -\begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ b_i & b_j & b_k \\ a_i & a_j & a_k \\ \end{vmatrix}       | c = Determinant with Rows Transposed }} {{eqn | r = -\left({\mathbf b \times \mathbf a}\right)       | c = {{Defof|Vector Cross Product}} }} {{end-eqn}} {{qed}}
Consider the $j$th projection. As long as none of $S_1, S_2, \ldots, S_n$ is the empty set, then: :$\displaystyle \forall x \in S_j: \exists \left({s_1, s_2, \ldots, s_{j-1}, x, s_{j+1}, \ldots, s_n}\right) \in \prod_{k \mathop = 1}^n S_k: \operatorname{pr}_j \left({\left({s_1, s_2, \ldots, s_{j-1}, x, s_{j+1}, \ldots, s_n}\right)}\right) = x$ Hence the result. {{qed}} Category:Surjections Category:Projections ijq8vf7kihzko82t9y6fuflloiv2t1s
From Canonical Injection is Injection we have that the canonical injections are in fact injective. It remains to prove the morphism property. Let $x, y \in \struct {S_j, \circ_j}$. Then: {{begin-eqn}} {{eqn | l = \map {\inj_j} {x \circ_j y}       | r = \tuple {e_1, e_2, \dotsc, e_{j - 1}, x \circ_j y, e_{j + 1}, \dotsc, e_n} }} {{eqn | r = \tuple {e_1 \circ_1 e_1, e_2 \circ_2 e_2, \dotsc, e_{j - 1} \circ_{j - 1} e_{j - 1}, x \circ_j y, e_{j + 1} \circ_{j + 1} e_{j + 1}, \dotsc, e_n \circ_n e_n} }} {{eqn | r = \tuple {e_1, e_2, \dotsc, e_{j - 1}, x, e_{j + 1}, \dotsc, e_n} \circ \tuple {e_1, e_2, \dotsc, e_{j - 1}, y, e_{j + 1}, \dotsc, e_n} }} {{eqn | r = \map {\inj_j} x \circ \map {\inj_j} y }} {{end-eqn}} and the morphism property has been demonstrated to hold. Thus $\displaystyle \inj_j: \struct {S_j, \circ_j} \to \prod_{i \mathop = 1}^n \struct {S_i, \circ_i}$ has been shown to be an injective homomorphism and therefore a monomorphism. {{Qed}}
By the definition of null space: :$\mathbf x \in \map {\mathrm N} {\mathbf A} \iff \mathbf A \mathbf x = \mathbf 0$ From the corollary to Row Equivalent Matrix for Homogeneous System has same Solutions: :$\mathbf A \mathbf x = \mathbf 0 \iff \map {\mathrm {rref} } {\mathbf A} \mathbf x = \mathbf 0$ Hence the result, by the definition of set equality. {{qed}}
Let $\mathbf x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \in \R^m$. We have that: {{begin-eqn}} {{eqn | l = \mathbf x       | o = \in        | r = \map {\mathrm N} {\mathbf A} }} {{eqn | ll= \leadstoandfrom       | l = \mathbf A \mathbf x_{n \times 1}       | r = \mathbf 0_{m \times 1} }} {{eqn | ll= \leadstoandfrom       | l = \begin{bmatrix} \mathbf a_1 & \mathbf a_2 & \cdots & \mathbf a_n \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}       | r = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix} }} {{eqn | ll= \leadstoandfrom       | l = \sum_{k \mathop = 1}^n x_k \mathbf a_k       | r = \mathbf 0 }} {{end-eqn}} === Sufficient Condition === Let $\set {\mathbf a_1, \mathbf a_2, \cdots, \mathbf a_n}$ be linearly independent. Then by definition: :$\forall k: 1 \le k \le n: x_k = 0 \iff \mathbf x = \mathbf 0_{n \times 1}$ By the definition of null space: :$\map {\mathrm N} {\mathbf A} = \set {\mathbf 0_{n \times 1} }$ {{qed|lemma}} === Necessary Condition === Let $\map {\mathrm N} {\mathbf A} = \set {\mathbf 0_{n \times 1} }$. Then by the definition of null space: :$\mathbf x = \mathbf 0_{n \times 1}$ This means that: :$\forall k: 1 \le k \le n: x_k = 0$ from which it follows that $\set {\mathbf a_1, \mathbf a_2, \cdots, \mathbf a_n}$ is linearly independent. {{qed}}
Let $\left\langle{S_n}\right\rangle_{n \in \N}$ be a sequence of non-empty sets. Define: :$\displaystyle S = \bigsqcup_{n \mathop \in \N} S_n$ where $\bigsqcup$ denotes disjoint union. Let $\mathcal R$ be the binary endorelation on $S$ defined by: :$\left({x, m}\right) \ \mathcal R \ \left({y, n}\right) \iff n = m + 1$ Note that $\mathcal R$ satisfies: :$\forall a \in S : \exists b \in S : a \ \mathcal R \ b$ Using the axiom of dependent choice, there exists a sequence $\left\langle{y_n}\right\rangle_{n \in \N}$ in $S$ such that $y_n \ \mathcal R \ y_{n + 1}$ for all $n \in \N$. Letting $y_n = \left({s_n, N_n}\right)$ for all $n \in \N$, it follows by the definition of $\mathcal R$ that $N_{n + 1} = N_n + 1$. A straightforward application of mathematical induction shows that $N_n = n + N$ for some $N \in \N$. So $s_n \in S_{n + N}$ for all $n \in \N$. The cartesian product $S_0 \times S_1 \times \cdots \times S_{N-1}$ is non-empty. So there exists a finite sequence $x_0, x_1, \ldots, x_{N-1}$ with $x_n \in S_n$ for all natural numbers $n < N$. Now, define $x_n = s_{n - N}$ for all natural numbers $n \ge N$. Then $x_n \in S_n$ for all $n \in \N$. {{qed}} {{ADC||3}}
Let $\mathcal R$ be a binary endorelation on a non-empty set $S$ such that: :$\forall a \in S: \exists b \in S: a \mathrel{\mathcal R} b$ For an element $x \in S$, define: :$R \left({x}\right) = \left\{{y \in S: x \mathrel{\mathcal R} y}\right\}$ By assumption, $R \left({x}\right)$ is non-empty for all $x \in S$. Now, consider the indexed family of sets: :$\left\langle{R \left({x}\right)}\right\rangle_{x \mathop \in S}$ Using the axiom of choice, there exists a mapping $f: S \to S$ such that: :$\forall x \in S: f \left({x}\right) \in R \left({x}\right)$ That is: :$x \mathrel{\mathcal R} f \left({x}\right)$ So, for any $x \in S$, the sequence: :$\left\langle{x_n}\right\rangle_{n \mathop \in \N} = \left\langle{f^n \left({x}\right)}\right\rangle_{n \mathop \in \N}$ where $f^n$ denotes the composition of $f$ with itself $n$ times, is a sequence such that: :$x_n \mathrel{\mathcal R} x_{n+1}$ for all $n \in \N$, as desired. {{qed}}
Observe that, vacuously, any $s \in S$ is a lower bound for $\O$. But for any lower bound $s$ of $\O$, $s \preceq \map \inf \O$ by definition of infimum. Hence: :$\forall s \in S: s \preceq \map \inf \O$ {{qed}}
Observe that, vacuously, any $s \in S$ is an upper bound for $\O$. === Necessary Condition === Let $\map \sup \O$ exist. For any upper bound $s$ of $\O$, $\map \sup \O \preceq s$ by definition of supremum. Hence: :$\forall s \in S: \map \sup \O \preceq s$ {{qed|lemma}} === Sufficient Condition === Let $t$ be the smallest element of $S$. Then $t$ is an upper bound of $\O$. For any upper bound $s$ of $\O$, $t \preceq s$ by definition of the smallest element. By definition of the supremun: :$t = \map \sup \O$ {{qed}}
Let $\left\langle{S_n}\right\rangle_{n \in \N}$ be a sequence of countable sets. Define: :$\displaystyle S = \bigcup_{n \mathop \in \N} S_n$ For all $n \in \N$, let $\mathcal F_n$ denote the set of all injections from $S_n$ to $\N$. Since $S_n$ is countable, $\mathcal F_n$ is non-empty. Using the axiom of countable choice, there exists a sequence $\left\langle{f_n}\right\rangle_{n \in \N}$ such that $f_n \in \mathcal F_n$ for all $n \in \N$. Let $\phi: S \to \N \times \N$, where $\times$ denotes the cartesian product, be the mapping defined by: :$\phi \left({x}\right) = \left({n, f_n \left({x}\right)}\right)$ where $n$ is the (unique) smallest natural number such that $x \in S_n$. From the Well-Ordering Principle, such an $n$ exists; hence, the mapping $\phi$ exists. Since each $f_n$ is an injection, it (trivially) follows that $\phi$ is an injection. Since $\N \times \N$ is countable, there exists an injection $\alpha: \N \times \N \to \N$. From Composite of Injections is Injection, the mapping $\alpha \circ \phi: S \to \N$ is an injection. Hence, $S$ is countable. {{qed}}
Consider the countable sets $S_0, S_1, S_2, \ldots$ where $\displaystyle S = \bigcup_{i \mathop \in \N} {S_i}$. Assume that none of these sets have any elements in common. Otherwise, we can consider the sets $S_0' = S_0, S_1' = S_1 \setminus S_0, S_2' = S_2 \setminus \paren {S_0 \cup S_1}, \ldots$ All of these are countable by Subset of Countable Set is Countable, and they have the same union $\displaystyle S = \bigcup_{i \mathop \in \N} {S_i'}$. Now we write the elements of $S_0', S_1', S_2', \ldots$ in the form of a (possibly infinite) table: :$\begin{array} {*{4}c}  {a_{00}} & {a_{01}} & {a_{02}} & \cdots \\  {a_{10}} & {a_{11}} & {a_{12}} & \cdots \\  {a_{20}} & {a_{21}} & {a_{22}} & \cdots \\  \vdots  & \vdots  & \vdots & \ddots \\ \end{array}$ where $a_{ij}$ is the $j$th element of set $S_i$. This table clearly contains all the elements of $\displaystyle S = \bigcup_{i \mathop \in \N} {S_i}$. Furthermore, we have that $\phi: S \to \N \times \N, a_{ij} \mapsto \tuple {i, j}$ is an injection. By Cartesian Product of Countable Sets is Countable, there exists an injection $\psi: \N \times \N \to \N$. Then $\psi \circ \phi: S \to \N$ is also an injection by Composite of Injections is Injection. That is to say, $S$ is countable. {{qed}}
From Projection is Surjection, $\operatorname{pr}_j$ is a surjection for all $j$. We now need to show it is a homomorphism. Let $s, t \in \left({S, \circ}\right)$ where $s = \left({s_1, s_2, \ldots, s_j, \ldots, s_n}\right)$ and $t = \left({t_1, t_2, \ldots, t_j, \ldots, t_n}\right)$. Then: {{begin-eqn}} {{eqn | l = \operatorname{pr}_j \left({s \circ t}\right)       | r = \operatorname{pr}_j \left({\left({s_1, s_2, \ldots, s_j, \ldots, s_n}\right) \circ \left({t_1, t_2, \ldots, t_j, \ldots, t_n}\right)}\right)       | c =  }} {{eqn | r = \operatorname{pr}_j \left({\left({s_1 \circ_1 t_1, s_2 \circ_2 t_2, \ldots, s_j \circ_j t_j, \ldots, s_n \circ_n t_n}\right) }\right)       | c =  }} {{eqn | r = s_j \circ_j t_j       | c =  }} {{eqn | r = \operatorname{pr}_j \left({s}\right) \circ_j \operatorname{pr}_j \left({t}\right)       | c =  }} {{end-eqn}} ... and thus the morphism property is demonstrated. {{Qed}}
Let $\displaystyle \tuple {s_1, s_2, \dotsc, s_{j - 1}, s_j, s_{j + 1}, \dotsc, s_n} \in \prod_{i \mathop = 1}^n \struct {S_i, \circ_i}$. So: :$s_j \in S_j$ From the definition of the canonical injection: :$\map {\inj_j} {s_j} = \tuple {e_1, e_2, \dotsc, e_{j - 1}, s_j, e_{j + 1}, \dotsc, e_n}$ So from the definition of the $j$th projection: :$\map {\pr_j} {e_1, e_2, \dotsc, e_{j - 1}, s_j, e_{j + 1}, \dotsc, e_n} = s_j$ Thus: :$\map {\pr_j \circ \inj_j} {s_j} = s_j$ and the result follows from the definition of the identity mapping. {{qed}}
When $x,y \in \R$, then $x +_{\overline{\R}} y = y +_{\overline{\R}} x$ follows from Real Addition is Commutative. The remaining cases where the expressions are defined, are already imposed in the definition of $+_{\overline{\R}}$. {{qed}} Category:Extended Real Numbers re0bgbsugf5tynnbcvx6y0icrntr5fa