id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-19200 | Isomorphism between Algebraic Structures induces Isomorphism between Induced Structures | Let $A$ be a set.
Let $\struct {S, \odot}$ and $\struct {T, \otimes}$ be algebraic structures.
Let:
:$S^A$ denote the set of mappings from $A$ to $S$
:$T^A$ denote the set of mappings from $A$ to $T$.
Let:
:$\struct {S^A, \odot}$ denote the algebraic structure on $S^A$ induced by $\odot$
:$\struct {T^A, \otimes}$ denot... | Let $f: A \to S$ and $g: A \to S$ be arbitrary elements of $S^A$.
We have that $\phi$ is an isomorphism.
Hence $\phi$ is {{afortiori}} a homomorphism which is a bijection.
Hence, from Bijection iff exists Mapping which is Left and Right Inverse, $\phi$ has an inverse mapping $\phi^{-1}: T \to S$ such that:
:$\phi \circ... | Let $A$ be a [[Definition:Set|set]].
Let $\struct {S, \odot}$ and $\struct {T, \otimes}$ be [[Definition:Algebraic Structure|algebraic structures]].
Let:
:$S^A$ denote the [[Definition:Set of All Mappings|set of mappings]] from $A$ to $S$
:$T^A$ denote the [[Definition:Set of All Mappings|set of mappings]] from $A$ t... | Let $f: A \to S$ and $g: A \to S$ be arbitrary [[Definition:Element|elements]] of $S^A$.
We have that $\phi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
Hence $\phi$ is {{afortiori}} a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] which is a [[Definition:Bijection|bijection]].
Henc... | Isomorphism between Algebraic Structures induces Isomorphism between Induced Structures | https://proofwiki.org/wiki/Isomorphism_between_Algebraic_Structures_induces_Isomorphism_between_Induced_Structures | https://proofwiki.org/wiki/Isomorphism_between_Algebraic_Structures_induces_Isomorphism_between_Induced_Structures | [
"Composite Mappings",
"Isomorphisms (Abstract Algebra)",
"Pointwise Operations"
] | [
"Definition:Set",
"Definition:Algebraic Structure",
"Definition:Set of All Mappings",
"Definition:Set of All Mappings",
"Definition:Pointwise Operation/Induced Structure",
"Definition:Pointwise Operation/Induced Structure",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Mapping",
"Definiti... | [
"Definition:Element",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Bijection",
"Bijection iff exists Mapping which is Left and Right Inverse",
"Definition:Inverse Mapping",
"Definition:Identity Mapping",
"Inverse of Bijection is Bijection",
"... |
proofwiki-19201 | Vitali Set Existence Theorem/Lemma | For all real numbers in the closed unit interval $\mathbb I = \closedint 0 1$, define the relation $\sim$ such that:
:$\forall x, y \in \mathbb I: x \sim y \iff x - y \in \Q$
where $\Q$ is the set of rational numbers.
That is, $x \sim y$ {{iff}} their difference is rational.
Then $\sim$ is an equivalence relation. | Checking in turn each of the criteria for equivalence: | For all [[Definition:Real Number|real numbers]] in the [[Definition:Closed Unit Interval|closed unit interval]] $\mathbb I = \closedint 0 1$, define the [[Definition:Relation|relation]] $\sim$ such that:
:$\forall x, y \in \mathbb I: x \sim y \iff x - y \in \Q$
where $\Q$ is the set of [[Definition:Rational Number|rati... | Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Vitali Set Existence Theorem/Lemma | https://proofwiki.org/wiki/Vitali_Set_Existence_Theorem/Lemma | https://proofwiki.org/wiki/Vitali_Set_Existence_Theorem/Lemma | [
"Vitali Set Existence Theorem"
] | [
"Definition:Real Number",
"Definition:Real Interval/Unit Interval/Closed",
"Definition:Relation",
"Definition:Rational Number",
"Definition:Subtraction/Real Numbers",
"Definition:Rational Number",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-19202 | Induced Structure from Doubleton is Isomorphic to External Direct Product with Self | Let $A$ be a set.
Let $\struct {S, \odot}$ be an algebraic structure.
Let $S^A$ denote the set of mappings from $A$ to $S$.
Let $\struct {S^A, \odot}$ denote the algebraic structure on $S^A$ induced by $\odot$.
Then $\struct {S^A, \odot}$ is isomorphic to the external direct product of $\struct {S, \odot}$ with itself. | Let $A = \set {a, b}$ where $a$ and $b$ are arbitrary.
Let $\phi: S^A \to S^2$ be defined as:
:$\forall f \in S^A: \map \phi f = \tuple {\map f a, \map f b}$
We are to show that $\phi$ is an isomorphism.
First we demonstrate that $\phi$ is a homomorphism.
So, let $f, g \in S^A$.
Recall that:
:by definition of pointwise... | Let $A$ be a [[Definition:Set|set]].
Let $\struct {S, \odot}$ be an [[Definition:Algebraic Structure|algebraic structure]].
Let $S^A$ denote the [[Definition:Set of All Mappings|set of mappings]] from $A$ to $S$.
Let $\struct {S^A, \odot}$ denote the [[Definition:Induced Structure|algebraic structure on $S^A$ induce... | Let $A = \set {a, b}$ where $a$ and $b$ are arbitrary.
Let $\phi: S^A \to S^2$ be defined as:
:$\forall f \in S^A: \map \phi f = \tuple {\map f a, \map f b}$
We are to show that $\phi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
First we demonstrate that $\phi$ is a [[Definition:Homomorphism ... | Induced Structure from Doubleton is Isomorphic to External Direct Product with Self | https://proofwiki.org/wiki/Induced_Structure_from_Doubleton_is_Isomorphic_to_External_Direct_Product_with_Self | https://proofwiki.org/wiki/Induced_Structure_from_Doubleton_is_Isomorphic_to_External_Direct_Product_with_Self | [
"External Direct Products",
"Isomorphisms (Abstract Algebra)",
"Pointwise Operations",
"Doubletons"
] | [
"Definition:Set",
"Definition:Algebraic Structure",
"Definition:Set of All Mappings",
"Definition:Pointwise Operation/Induced Structure",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:External Direct Product"
] | [
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Pointwise Operation",
"Definition:External Direct Product",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Bijection",
"Equality of Ordered Pairs",
"Equality of Mappings",
"Definition:In... |
proofwiki-19203 | Structure Induced on Set of Self-Maps on Entropic Structure is Entropic | Let $\struct {S, \odot}$ be a magma.
Let $\struct {S, \odot}$ be an entropic structure.
Let $S^S$ be the set of all mappings from $S$ to itself.
Let $\struct {S^S, \oplus}$ denote the algebraic structure on $S^S$ induced by $\odot$.
Then $\struct {S^S, \oplus}$ is an entropic structure. | Recall the definition of algebraic structure on $S^S$ induced by $\odot$:
Let $f: S \to S$ and $g: S \to S$ be self-maps on $S$, and thus elements of $S^S$.
The pointwise operation on $S^S$ induced by $\odot$ is defined as:
:$\forall x \in S: \map {\paren {f \oplus g} } x = \map f x \odot \map g x$
Let $f, g, p, q \in ... | Let $\struct {S, \odot}$ be a [[Definition:Magma|magma]].
Let $\struct {S, \odot}$ be an [[Definition:Entropic Structure|entropic structure]].
Let $S^S$ be the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to itself.
Let $\struct {S^S, \oplus}$ denote the [[Definition:Induced Structure|algebraic st... | Recall the definition of [[Definition:Induced Structure|algebraic structure on $S^S$ induced by $\odot$]]:
Let $f: S \to S$ and $g: S \to S$ be [[Definition:Self-Map|self-maps]] on $S$, and thus [[Definition:Element|elements]] of $S^S$.
The [[Definition:Pointwise Operation|pointwise operation on $S^S$ induced by $\od... | Structure Induced on Set of Self-Maps on Entropic Structure is Entropic | https://proofwiki.org/wiki/Structure_Induced_on_Set_of_Self-Maps_on_Entropic_Structure_is_Entropic | https://proofwiki.org/wiki/Structure_Induced_on_Set_of_Self-Maps_on_Entropic_Structure_is_Entropic | [
"Entropic Structures",
"Pointwise Operations",
"Magmas"
] | [
"Definition:Magma",
"Definition:Entropic Structure",
"Definition:Set of All Mappings",
"Definition:Pointwise Operation/Induced Structure",
"Definition:Entropic Structure"
] | [
"Definition:Pointwise Operation/Induced Structure",
"Definition:Self-Map",
"Definition:Element",
"Definition:Pointwise Operation",
"Equality of Mappings",
"Definition:Entropic Structure"
] |
proofwiki-19204 | Set of Endomorphisms on Entropic Structure is Closed in Induced Structure on Set of Self-Maps | Let $\struct {S, \odot}$ be a magma.
Let $\struct {S, \odot}$ be an entropic structure.
Let $S^S$ be the set of all mappings from $S$ to itself.
Let $\struct {S^S, \oplus}$ denote the algebraic structure on $S^S$ induced by $\odot$.
Let $T \subseteq S^S$ denote the set of endomorphisms on $\struct {S, \odot}$.
Then $\s... | Recall the definition of algebraic structure on $S^S$ induced by $\odot$:
Let $f: S \to S$ and $g: S \to S$ be self-maps on $S$, and thus elements of $S^S$.
The pointwise operation on $S^S$ induced by $\odot$ is defined as:
:$\forall x \in S: \map {\paren {f \oplus g} } x = \map f x \odot \map g x$
Let $f, g \in T$ be ... | Let $\struct {S, \odot}$ be a [[Definition:Magma|magma]].
Let $\struct {S, \odot}$ be an [[Definition:Entropic Structure|entropic structure]].
Let $S^S$ be the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to itself.
Let $\struct {S^S, \oplus}$ denote the [[Definition:Induced Structure|algebraic st... | Recall the definition of [[Definition:Induced Structure|algebraic structure on $S^S$ induced by $\odot$]]:
Let $f: S \to S$ and $g: S \to S$ be [[Definition:Self-Map|self-maps]] on $S$, and thus [[Definition:Element|elements]] of $S^S$.
The [[Definition:Pointwise Operation|pointwise operation on $S^S$ induced by $\od... | Set of Endomorphisms on Entropic Structure is Closed in Induced Structure on Set of Self-Maps | https://proofwiki.org/wiki/Set_of_Endomorphisms_on_Entropic_Structure_is_Closed_in_Induced_Structure_on_Set_of_Self-Maps | https://proofwiki.org/wiki/Set_of_Endomorphisms_on_Entropic_Structure_is_Closed_in_Induced_Structure_on_Set_of_Self-Maps | [
"Set of Endomorphisms on Entropic Structure is Closed in Induced Structure on Set of Self-Maps",
"Entropic Structures",
"Pointwise Operations",
"Endomorphisms",
"Self-Maps"
] | [
"Definition:Magma",
"Definition:Entropic Structure",
"Definition:Set of All Mappings",
"Definition:Pointwise Operation/Induced Structure",
"Definition:Set",
"Definition:Endomorphism",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] | [
"Definition:Pointwise Operation/Induced Structure",
"Definition:Self-Map",
"Definition:Element",
"Definition:Pointwise Operation",
"Definition:Endomorphism",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] |
proofwiki-19205 | Replicative Function of x minus Floor of x is Replicative/Lemma | Let $x \in \R$.
Suppose $x - \floor x < \dfrac 1 n$.
Then:
:$\floor {x + \dfrac k n} = \dfrac {\floor {n x} } n$
for any $0 \le k \le n - 1$. | We have $n x < n \floor x + 1$.
By Number less than Integer iff Floor less than Integer:
:$\floor {n x} < n \floor x + 1$
Thus $\floor {n x} \le n \floor x$.
From definition of floor function:
:$n x \ge n \floor x$
By Number not less than Integer iff Floor not less than Integer:
:$\floor {n x} \ge n \floor x$
Therefore... | Let $x \in \R$.
Suppose $x - \floor x < \dfrac 1 n$.
Then:
:$\floor {x + \dfrac k n} = \dfrac {\floor {n x} } n$
for any $0 \le k \le n - 1$. | We have $n x < n \floor x + 1$.
By [[Number less than Integer iff Floor less than Integer]]:
:$\floor {n x} < n \floor x + 1$
Thus $\floor {n x} \le n \floor x$.
From definition of [[Definition:Floor Function|floor function]]:
:$n x \ge n \floor x$
By [[Number not less than Integer iff Floor not less than Integer]]... | Replicative Function of x minus Floor of x is Replicative/Lemma | https://proofwiki.org/wiki/Replicative_Function_of_x_minus_Floor_of_x_is_Replicative/Lemma | https://proofwiki.org/wiki/Replicative_Function_of_x_minus_Floor_of_x_is_Replicative/Lemma | [
"Replicative Function of x minus Floor of x is Replicative"
] | [] | [
"Number less than Integer iff Floor less than Integer",
"Definition:Floor Function",
"Number not less than Integer iff Floor not less than Integer",
"Number less than Integer iff Floor less than Integer",
"Category:Replicative Function of x minus Floor of x is Replicative"
] |
proofwiki-19206 | Set of Endomorphisms on Entropic Structure is Closed in Induced Structure on Set of Self-Maps/Converse | Let $T \subseteq S^S$ denote the set of endomorphisms on $\struct {S, \odot}$.
Let $\struct {T, \oplus_T}$ be closed in $\struct {S^S, \oplus}$.
Then it is not necessarily the case that $\struct {S, \odot}$ is an entropic structure. | Proof by Counterexample:
Let $S = \set {a, b}$.
Let $\odot$ be the operation on $S$ defined by Cayley table as:
:<nowiki>$\begin{array} {c|cc} \odot & a & b \\
\hline
a & b & b \\
b & a & b
\end{array}$</nowiki>
First we note that:
{{begin-eqn}}
{{eqn | l = \paren {a \odot a} \odot \paren {b \odot a}
| r = b \odo... | Let $T \subseteq S^S$ denote the [[Definition:Set|set]] of [[Definition:Endomorphism|endomorphisms]] on $\struct {S, \odot}$.
Let $\struct {T, \oplus_T}$ be [[Definition:Closed Algebraic Structure|closed]] in $\struct {S^S, \oplus}$.
Then it is not necessarily the case that $\struct {S, \odot}$ is an [[Definition:En... | [[Proof by Counterexample]]:
Let $S = \set {a, b}$.
Let $\odot$ be the [[Definition:Binary Operation|operation]] on $S$ defined by [[Definition:Cayley Table|Cayley table]] as:
:<nowiki>$\begin{array} {c|cc} \odot & a & b \\
\hline
a & b & b \\
b & a & b
\end{array}$</nowiki>
First we note that:
{{begin-eqn}}
{{eqn... | Set of Endomorphisms on Entropic Structure is Closed in Induced Structure on Set of Self-Maps/Converse | https://proofwiki.org/wiki/Set_of_Endomorphisms_on_Entropic_Structure_is_Closed_in_Induced_Structure_on_Set_of_Self-Maps/Converse | https://proofwiki.org/wiki/Set_of_Endomorphisms_on_Entropic_Structure_is_Closed_in_Induced_Structure_on_Set_of_Self-Maps/Converse | [
"Set of Endomorphisms on Entropic Structure is Closed in Induced Structure on Set of Self-Maps",
"Entropic Structures",
"Pointwise Operations",
"Endomorphisms"
] | [
"Definition:Set",
"Definition:Endomorphism",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Entropic Structure"
] | [
"Proof by Counterexample",
"Definition:Operation/Binary Operation",
"Definition:Cayley Table",
"Definition:Entropic Structure",
"Definition:Set",
"Definition:Endomorphism",
"Definition:Endomorphism",
"Definition:Mapping",
"Cardinality of Set of All Mappings",
"Definition:Identity Mapping",
"Iden... |
proofwiki-19207 | Equivalent Conditions for Entropic Structure/Pointwise Operation is Homomorphism | Let $\struct {T, \circledast}$ be an arbitrary algebraic structure.
Let $f$ and $g$ be mappings from $\struct {T, \circledast}$ to $\struct {S, \odot}$.
Let $f \odot g$ denote the pointwise operation on $S^T$ induced by $\odot$.
Then:
:If $f$ and $g$ are homomorphisms, then $f \odot g$ is also a homomorphism
{{iff}}:
:... | === Sufficient Condition ===
Let $\struct {S, \odot}$ be such that if $f$ and $g$ are homomorphisms, then $f \odot g$ is also a homomorphism.
So, let $f: T \to S$ and $g: T \to S$ be arbitrary homomorphisms.
Let $a, b \in T$ be arbitrary.
Because $T$ is arbitrary, and $f$ and $g$ are arbitrary, it follows that:
{{begin... | Let $\struct {T, \circledast}$ be an arbitrary [[Definition:Algebraic Structure|algebraic structure]].
Let $f$ and $g$ be [[Definition:Mapping|mappings]] from $\struct {T, \circledast}$ to $\struct {S, \odot}$.
Let $f \odot g$ denote the [[Definition:Pointwise Operation|pointwise operation]] on $S^T$ induced by $\odo... | === Sufficient Condition ===
Let $\struct {S, \odot}$ be such that if $f$ and $g$ are [[Definition:Homomorphism (Abstract Algebra)|homomorphisms]], then $f \odot g$ is also a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
So, let $f: T \to S$ and $g: T \to S$ be arbitrary [[Definition:Homomorphism (Abst... | Equivalent Conditions for Entropic Structure/Pointwise Operation is Homomorphism | https://proofwiki.org/wiki/Equivalent_Conditions_for_Entropic_Structure/Pointwise_Operation_is_Homomorphism | https://proofwiki.org/wiki/Equivalent_Conditions_for_Entropic_Structure/Pointwise_Operation_is_Homomorphism | [
"Equivalent Conditions for Entropic Structure",
"Pointwise Operations",
"Homomorphisms (Abstract Algebra)"
] | [
"Definition:Algebraic Structure",
"Definition:Mapping",
"Definition:Pointwise Operation",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Entropic Structure"
] | [
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Entropic Structure",
"Definition:Entropic Structure",
"Defin... |
proofwiki-19208 | Composite with Constant Mapping is Constant Mapping | Let $f_c: S \to T$ be the constant mapping defined as:
:$\forall x \in S: \map {f_c} x = c$
where $c \in T$.
Then for all mappings $g: \Dom g \to S$:
:$f_c \circ g$ is a constant mapping
and for all mappings $h: T \to \Cdm h$:
:$h \circ f_c$ is a constant mapping
where:
:$\Dom g$ denotes the domain of $g$
:$\Cdm h$ den... | {{begin-eqn}}
{{eqn | q = \forall x \in \Dom g
| l = \map {\paren {f_c \circ g} } x
| r = \map {f_c} {\map g x}
| c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = c
| c = {{Defof|Constant Mapping}}
}}
{{end-eqn}}
{{qed|lemma}}
{{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map {\pare... | Let $f_c: S \to T$ be the [[Definition:Constant Mapping|constant mapping]] defined as:
:$\forall x \in S: \map {f_c} x = c$
where $c \in T$.
Then for all [[Definition:Mapping|mappings]] $g: \Dom g \to S$:
:$f_c \circ g$ is a [[Definition:Constant Mapping|constant mapping]]
and for all [[Definition:Mapping|mappings]]... | {{begin-eqn}}
{{eqn | q = \forall x \in \Dom g
| l = \map {\paren {f_c \circ g} } x
| r = \map {f_c} {\map g x}
| c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = c
| c = {{Defof|Constant Mapping}}
}}
{{end-eqn}}
{{qed|lemma}}
{{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map {\pa... | Composite with Constant Mapping is Constant Mapping | https://proofwiki.org/wiki/Composite_with_Constant_Mapping_is_Constant_Mapping | https://proofwiki.org/wiki/Composite_with_Constant_Mapping_is_Constant_Mapping | [
"Constant Mappings",
"Composite Mappings"
] | [
"Definition:Constant Mapping",
"Definition:Mapping",
"Definition:Constant Mapping",
"Definition:Mapping",
"Definition:Constant Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Composition of Mappings"
] | [
"Definition:Constant",
"Category:Constant Mappings",
"Category:Composite Mappings"
] |
proofwiki-19209 | Equivalent Conditions for Entropic Structure/Pointwise Operation of Homomorphisms from External Direct Product is Homomorphism | Let $\struct {S \times S, \otimes}$ denote the external direct product of $\struct {S, \odot}$ with itself:
:$\forall \tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S: \tuple {x_1, y_1} \otimes \tuple {x_2, y_2} = \tuple {x_1 \odot x_2, y_1 \odot y_2}$
Let $f$ and $g$ be mappings from $\struct {S \times S, \otimes}$... | === Sufficient Condition ===
Let $\struct {S, \odot}$ be such that if $f$ and $g$ are homomorphisms, then $f \odot g$ is also a homomorphism.
So, let $f: S \times S \to S$ and $g: S \times S \to S$ be arbitrary homomorphisms.
Let $w$, $x$, $y$ and $d$ in $S$ be arbitrary.
Then as $f$ and $g$ are arbitrary:
{{begin-eqn}... | Let $\struct {S \times S, \otimes}$ denote the [[Definition:External Direct Product|external direct product]] of $\struct {S, \odot}$ with itself:
:$\forall \tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S: \tuple {x_1, y_1} \otimes \tuple {x_2, y_2} = \tuple {x_1 \odot x_2, y_1 \odot y_2}$
Let $f$ and $g$ be [[De... | === Sufficient Condition ===
Let $\struct {S, \odot}$ be such that if $f$ and $g$ are [[Definition:Homomorphism (Abstract Algebra)|homomorphisms]], then $f \odot g$ is also a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
So, let $f: S \times S \to S$ and $g: S \times S \to S$ be arbitrary [[Definition:... | Equivalent Conditions for Entropic Structure/Pointwise Operation of Homomorphisms from External Direct Product is Homomorphism | https://proofwiki.org/wiki/Equivalent_Conditions_for_Entropic_Structure/Pointwise_Operation_of_Homomorphisms_from_External_Direct_Product_is_Homomorphism | https://proofwiki.org/wiki/Equivalent_Conditions_for_Entropic_Structure/Pointwise_Operation_of_Homomorphisms_from_External_Direct_Product_is_Homomorphism | [
"Equivalent Conditions for Entropic Structure",
"Pointwise Operations",
"Homomorphisms (Abstract Algebra)",
"External Direct Products"
] | [
"Definition:External Direct Product",
"Definition:Mapping",
"Definition:Pointwise Operation",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Entropic Structure"
] | [
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Entropic Structure",
"Definition:Entropic Structure",
"Defin... |
proofwiki-19210 | Equivalent Conditions for Entropic Structure/Mapping from External Direct Product is Homomorphism | Let $\struct {S \times S, \otimes}$ denote the external direct product of $\struct {S, \odot}$ with itself:
:$\forall \tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S: \tuple {x_1, y_1} \otimes \tuple {x_2, y_2} = \tuple {x_1 \odot x_2, y_1 \odot y_2}$
Consider the operation $\odot$ as a mapping from $S \times S$ to... | === Sufficient Condition ===
Let $\struct {S, \odot}$ be such that $\odot: S \times S \to S$ is a homomorphism.
Let $\tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S$ be arbitrary.
We have:
{{begin-eqn}}
{{eqn | o =
| r = \paren {x_1 \odot x_2} \odot \paren {y_1 \odot y_2}
| c =
}}
{{eqn | r = \paren {... | Let $\struct {S \times S, \otimes}$ denote the [[Definition:External Direct Product|external direct product]] of $\struct {S, \odot}$ with itself:
:$\forall \tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S: \tuple {x_1, y_1} \otimes \tuple {x_2, y_2} = \tuple {x_1 \odot x_2, y_1 \odot y_2}$
Consider the operation $... | === Sufficient Condition ===
Let $\struct {S, \odot}$ be such that $\odot: S \times S \to S$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
Let $\tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S$ be arbitrary.
We have:
{{begin-eqn}}
{{eqn | o =
| r = \paren {x_1 \odot x_2} \odot \paren ... | Equivalent Conditions for Entropic Structure/Mapping from External Direct Product is Homomorphism | https://proofwiki.org/wiki/Equivalent_Conditions_for_Entropic_Structure/Mapping_from_External_Direct_Product_is_Homomorphism | https://proofwiki.org/wiki/Equivalent_Conditions_for_Entropic_Structure/Mapping_from_External_Direct_Product_is_Homomorphism | [
"Equivalent Conditions for Entropic Structure",
"Homomorphisms (Abstract Algebra)",
"External Direct Products"
] | [
"Definition:External Direct Product",
"Definition:Mapping",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Entropic Structure"
] | [
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Entropic Structure",
"Definition:Entropic Structure",
"Definition:Homomorphism (Abstract Algebra)"
] |
proofwiki-19211 | Condition for Algebra Loop to be Abelian Group | Let $\struct {S, \odot}$ be an algebra loop.
Let $\struct {S \times S, \otimes}$ denote the external direct product of $\struct {S, \odot}$ with itself:
:$\forall \tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S: \tuple {x_1, y_1} \otimes \tuple {x_2, y_2} = \tuple {x_1 \odot x_2, y_1 \odot y_2}$
Let $f$ and $g$ be ... | Recall that an '''algebra loop''' $\struct {S, \circ}$ is a quasigroup with an identity element:
:$\exists e \in S: \forall x \in S: x \circ e = x = e \circ x$
From Equivalent Conditions for Entropic Structure: Pointwise Operation of Homomorphisms from External Direct Product is Homomorphism:
:$\struct {S, \odot}$ is a... | Let $\struct {S, \odot}$ be an [[Definition:Algebra Loop|algebra loop]].
Let $\struct {S \times S, \otimes}$ denote the [[Definition:External Direct Product|external direct product]] of $\struct {S, \odot}$ with itself:
:$\forall \tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S: \tuple {x_1, y_1} \otimes \tuple {x_... | Recall that an '''algebra loop''' $\struct {S, \circ}$ is a [[Definition:Quasigroup|quasigroup]] with an [[Definition:Identity Element|identity element]]:
:$\exists e \in S: \forall x \in S: x \circ e = x = e \circ x$
From [[Equivalent Conditions for Entropic Structure/Pointwise Operation of Homomorphisms from Externa... | Condition for Algebra Loop to be Abelian Group | https://proofwiki.org/wiki/Condition_for_Algebra_Loop_to_be_Abelian_Group | https://proofwiki.org/wiki/Condition_for_Algebra_Loop_to_be_Abelian_Group | [
"Algebra Loops",
"Abelian Groups"
] | [
"Definition:Algebra Loop",
"Definition:External Direct Product",
"Definition:Mapping",
"Definition:Pointwise Operation",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Abelian Group"
] | [
"Definition:Quasigroup",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Equivalent Conditions for Entropic Structure/Pointwise Operation of Homomorphisms from External Direct Product is Homomorphism",
"Definition:Entropic Structure",
"Entropic Structure with Identity is Commutative Monoid",
... |
proofwiki-19212 | External Direct Product of Congruence Relations | Let $\struct {A, \odot}$ and $\struct {B, \circledast}$ be algebraic structures.
Let $\RR$ and $\SS$ be congruence relations on $\struct {A, \odot}$ and $\struct {B, \circledast}$ respectively.
Let $\struct {A / \RR, \odot_\RR}$ and $\struct {B / \SS, \circledast_\SS}$ denote the quotient structures defined by $\RR$ an... | We have that $\RR$ and $\SS$ are congruence relations on $\struct {A, \odot}$ and $\struct {B, \circledast}$ respectively.
Hence {{afortiori}} $\RR$ and $\SS$ are equivalence relations on $A$ and $B$ respectively.
Thus from Cartesian Product of Equivalence Relations we have that $\TT$ is an equivalence relation.
Let $\... | Let $\struct {A, \odot}$ and $\struct {B, \circledast}$ be [[Definition:Algebraic Structure|algebraic structures]].
Let $\RR$ and $\SS$ be [[Definition:Congruence Relation|congruence relations]] on $\struct {A, \odot}$ and $\struct {B, \circledast}$ respectively.
Let $\struct {A / \RR, \odot_\RR}$ and $\struct {B / \... | We have that $\RR$ and $\SS$ are [[Definition:Congruence Relation|congruence relations]] on $\struct {A, \odot}$ and $\struct {B, \circledast}$ respectively.
Hence {{afortiori}} $\RR$ and $\SS$ are [[Definition:Equivalence Relation|equivalence relations]] on $A$ and $B$ respectively.
Thus from [[Cartesian Product of ... | External Direct Product of Congruence Relations | https://proofwiki.org/wiki/External_Direct_Product_of_Congruence_Relations | https://proofwiki.org/wiki/External_Direct_Product_of_Congruence_Relations | [
"Congruence Relations",
"External Direct Products",
"Quotient Mappings"
] | [
"Definition:Algebraic Structure",
"Definition:Congruence Relation",
"Definition:Quotient Structure",
"Definition:External Direct Product",
"Definition:Relation",
"Definition:Congruence Relation",
"Definition:Mapping",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Definition:Congruence Relation",
"Definition:Equivalence Relation",
"Cartesian Product of Equivalence Relations",
"Definition:Equivalence Relation",
"Definition:Congruence Relation",
"Definition:Isomorphism (Abstract Algebra)",
"Cartesian Product of Equivalence Relations",
"Definition:Equivalence Cla... |
proofwiki-19213 | Cartesian Product of Equivalence Relations | Let $A$ and $B$ be sets.
Let $\RR$ and $\SS$ be equivalence relations on $A$ and $B$ respectively.
Let $\TT$ be the relation on $A \times B$ defined as:
:$\forall \tuple {u, v}, \tuple {x, y} \in A \times B: \tuple {u, v} \mathrel \TT \tuple {x, y} \iff u \mathrel \RR x \land v \mathrel \SS y$
Then $\TT$ is an equivale... | We have that $\RR$ and $\SS$ are equivalence relations on $A$ and $B$ respectively.
Thus they are both reflexive, symmetric and transitive.
Checking in turn each of the criteria for equivalence: | Let $A$ and $B$ be [[Definition:Set|sets]].
Let $\RR$ and $\SS$ be [[Definition:Equivalence Relation|equivalence relations]] on $A$ and $B$ respectively.
Let $\TT$ be the [[Definition:Relation|relation]] on $A \times B$ defined as:
:$\forall \tuple {u, v}, \tuple {x, y} \in A \times B: \tuple {u, v} \mathrel \TT \tu... | We have that $\RR$ and $\SS$ are [[Definition:Equivalence Relation|equivalence relations]] on $A$ and $B$ respectively.
Thus they are both [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
Checking in turn each of the criteria ... | Cartesian Product of Equivalence Relations | https://proofwiki.org/wiki/Cartesian_Product_of_Equivalence_Relations | https://proofwiki.org/wiki/Cartesian_Product_of_Equivalence_Relations | [
"Equivalence Relations",
"Cartesian Product"
] | [
"Definition:Set",
"Definition:Equivalence Relation",
"Definition:Relation",
"Definition:Equivalence Relation",
"Definition:Equivalence Class",
"Definition:Element"
] | [
"Definition:Equivalence Relation",
"Definition:Reflexive Relation",
"Definition:Symmetric Relation",
"Definition:Transitive Relation",
"Definition:Equivalence Relation",
"Definition:Reflexive Relation",
"Definition:Reflexive Relation",
"Definition:Reflexive Relation",
"Definition:Symmetric Relation"... |
proofwiki-19214 | Condition for Mapping between Structures to be Homomorphism | Let $\struct {A, \odot}$ and $\struct {B, \circledast}$ be magmas.
Let $\struct {A \times B, \otimes}$ be the external direct product of $\struct {A, \odot}$ and $\struct {B, \circledast}$.
Let $\phi: A \to B$ be a mapping.
Let $\phi$ be considered as a subset of the Cartesian product $A \times B$.
Then:
:$\phi$ is a h... | Let $\phi$ be a homomorphism
Let $\tuple {a, b}, \tuple {c, d} \in A \times B$ such that:
{{begin-eqn}}
{{eqn | l = \tuple {a, b}
| o = \in
| r = \phi
}}
{{eqn | l = \tuple {c, d}
| o = \in
| r = \phi
| c =
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \map \phi a
| r = b
| c... | Let $\struct {A, \odot}$ and $\struct {B, \circledast}$ be [[Definition:Magma|magmas]].
Let $\struct {A \times B, \otimes}$ be the [[Definition:External Direct Product|external direct product]] of $\struct {A, \odot}$ and $\struct {B, \circledast}$.
Let $\phi: A \to B$ be a [[Definition:Mapping|mapping]].
Let $\phi... | Let $\phi$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]
Let $\tuple {a, b}, \tuple {c, d} \in A \times B$ such that:
{{begin-eqn}}
{{eqn | l = \tuple {a, b}
| o = \in
| r = \phi
}}
{{eqn | l = \tuple {c, d}
| o = \in
| r = \phi
| c =
}}
{{end-eqn}}
Then:
{{begin-eq... | Condition for Mapping between Structures to be Homomorphism | https://proofwiki.org/wiki/Condition_for_Mapping_between_Structures_to_be_Homomorphism | https://proofwiki.org/wiki/Condition_for_Mapping_between_Structures_to_be_Homomorphism | [
"Homomorphisms (Abstract Algebra)",
"External Direct Products"
] | [
"Definition:Magma",
"Definition:External Direct Product",
"Definition:Mapping",
"Definition:Subset",
"Definition:Cartesian Product",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Algebraic Structure",
"Definition:Submagma"
] | [
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Subset",
"Definition:Submagma",
"Definition:Submagma",
"Definition:Homomorphism (Abstract Algebra)"
] |
proofwiki-19215 | Condition for Semigroup to be Internal Direct Product of Subgroup and Subsemigroup with Right Operation | Let $\struct {S, \odot}$ be a semigroup.
Then:
:$\struct {S, \odot}$ is the internal direct product of a subgroup $\struct {G, \odot_G}$ and subsemigroup $\struct {H, \odot_H}$ such that $\odot_H$ is the right operation
{{iff}}:
:for all $x \in S$ there exists a left identity $a$ such that $x \odot a = x$, and element ... | === Sufficient Condition ===
Let $\struct {S, \odot}$ be the internal direct product of a subgroup $\struct {G, \odot_G}$ and subsemigroup $\struct {H, \odot_H}$ such that $\odot_H$ is the right operation.
Hence, by definition, the mapping $\phi: G \times H \to S$ defined as:
:$\forall g \in H, h \in H: \map \phi {g, h... | Let $\struct {S, \odot}$ be a [[Definition:Semigroup|semigroup]].
Then:
:$\struct {S, \odot}$ is the [[Definition:Internal Direct Product|internal direct product]] of a [[Definition:Subgroup|subgroup]] $\struct {G, \odot_G}$ and [[Definition:Subsemigroup|subsemigroup]] $\struct {H, \odot_H}$ such that $\odot_H$ is th... | === Sufficient Condition ===
Let $\struct {S, \odot}$ be the [[Definition:Internal Direct Product|internal direct product]] of a [[Definition:Subgroup|subgroup]] $\struct {G, \odot_G}$ and [[Definition:Subsemigroup|subsemigroup]] $\struct {H, \odot_H}$ such that $\odot_H$ is the [[Definition:Right Operation|right oper... | Condition for Semigroup to be Internal Direct Product of Subgroup and Subsemigroup with Right Operation | https://proofwiki.org/wiki/Condition_for_Semigroup_to_be_Internal_Direct_Product_of_Subgroup_and_Subsemigroup_with_Right_Operation | https://proofwiki.org/wiki/Condition_for_Semigroup_to_be_Internal_Direct_Product_of_Subgroup_and_Subsemigroup_with_Right_Operation | [
"Semigroups",
"Internal Direct Products",
"Right Operation"
] | [
"Definition:Semigroup",
"Definition:Internal Direct Product",
"Definition:Subgroup",
"Definition:Subsemigroup",
"Definition:Right Operation",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Element"
] | [
"Definition:Internal Direct Product",
"Definition:Subgroup",
"Definition:Subsemigroup",
"Definition:Right Operation",
"Definition:Mapping",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:External Direct Product",
"Definition:Bijection",
"Condition for Mapping between Structure and Cartesia... |
proofwiki-19216 | Pointwise Addition on Continuous Real Functions on Closed Unit Interval forms Group | Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $\R^J$ denote the set of all mappings from $J$ to $\R$.
Let $\struct {\R^J, +}$ denote the algebraic structure on $\R^J$ induced by addition:
:$\forall f... | Taking the group axioms in turn: | Let $J \subseteq \R$ denote the [[Definition:Closed Unit Interval|closed unit interval]] $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the [[Definition:Set|set]] of all [[Definition:Continuous Real Function|continuous real functions]] from $J$ to $\R$.
Let $\R^J$ denote the [[Definition:Set of All Mappings|set o... | Taking the [[Axiom:Group Axioms|group axioms]] in turn: | Pointwise Addition on Continuous Real Functions on Closed Unit Interval forms Group | https://proofwiki.org/wiki/Pointwise_Addition_on_Continuous_Real_Functions_on_Closed_Unit_Interval_forms_Group | https://proofwiki.org/wiki/Pointwise_Addition_on_Continuous_Real_Functions_on_Closed_Unit_Interval_forms_Group | [
"Examples of Groups",
"Pointwise Addition",
"Continuous Real Functions"
] | [
"Definition:Real Interval/Unit Interval/Closed",
"Definition:Set",
"Definition:Continuous Real Function",
"Definition:Set of All Mappings",
"Definition:Pointwise Operation/Induced Structure",
"Definition:Addition/Real Numbers",
"Definition:Subgroup"
] | [
"Axiom:Group Axioms",
"Axiom:Group Axioms"
] |
proofwiki-19217 | Pointwise Addition on Differentiable Real Functions on Closed Unit Interval forms Group | Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr D} J$ denote the set of all differentiable real functions from $J$ to $\R$.
Let $\R^J$ denote the set of all mappings from $J$ to $\R$.
Let $\struct {\R^J, +}$ denote the algebraic structure on $\R^J$ induced by addition:
:$\fora... | Taking the group axioms in turn: | Let $J \subseteq \R$ denote the [[Definition:Closed Unit Interval|closed unit interval]] $\closedint 0 1$.
Let $\map {\mathscr D} J$ denote the [[Definition:Set|set]] of all [[Definition:Differentiable Real Function|differentiable real functions]] from $J$ to $\R$.
Let $\R^J$ denote the [[Definition:Set of All Mappin... | Taking the [[Axiom:Group Axioms|group axioms]] in turn: | Pointwise Addition on Differentiable Real Functions on Closed Unit Interval forms Group | https://proofwiki.org/wiki/Pointwise_Addition_on_Differentiable_Real_Functions_on_Closed_Unit_Interval_forms_Group | https://proofwiki.org/wiki/Pointwise_Addition_on_Differentiable_Real_Functions_on_Closed_Unit_Interval_forms_Group | [
"Examples of Groups",
"Pointwise Addition",
"Differentiable Real Functions"
] | [
"Definition:Real Interval/Unit Interval/Closed",
"Definition:Set",
"Definition:Differentiable Mapping/Real Function",
"Definition:Set of All Mappings",
"Definition:Pointwise Operation/Induced Structure",
"Definition:Addition/Real Numbers",
"Definition:Subgroup"
] | [
"Axiom:Group Axioms",
"Axiom:Group Axioms"
] |
proofwiki-19218 | Integral of Compactly Supported Function | Let $f : \R \to \R$ be a continuous real function.
Let $K \subset \R$ be a compact subset, say, $\closedint a b$.
Let $K$ be the support of $f$:
:$\map \supp f = K$.
Then:
:$\ds \int_{- \infty}^\infty \map f x \rd x = \int_a^b \map f x \rd x$ | {{begin-eqn}}
{{eqn | l = \int_{-\infty}^\infty \map f x \rd x
| r = \int_{\overline \R} \map f x \rd x
| c = {{Defof|Extended Real Number Line}}
}}
{{eqn | r = \int_{K \cup \paren { {\overline \R} \setminus K} } \map f x \rd x
| c = {{Defof|Set Difference}}, {{Defof|Set Complement}}
}}
{{eqn | r = \i... | Let $f : \R \to \R$ be a [[Definition:Continuous Real Function|continuous real function]].
Let $K \subset \R$ be a [[Definition:Compact Subset of Real Euclidean Space|compact subset]], say, $\closedint a b$.
Let $K$ be the [[Definition:Support of Continuous Mapping|support]] of $f$:
:$\map \supp f = K$.
Then:
:$\... | {{begin-eqn}}
{{eqn | l = \int_{-\infty}^\infty \map f x \rd x
| r = \int_{\overline \R} \map f x \rd x
| c = {{Defof|Extended Real Number Line}}
}}
{{eqn | r = \int_{K \cup \paren { {\overline \R} \setminus K} } \map f x \rd x
| c = {{Defof|Set Difference}}, {{Defof|Set Complement}}
}}
{{eqn | r = \i... | Integral of Compactly Supported Function | https://proofwiki.org/wiki/Integral_of_Compactly_Supported_Function | https://proofwiki.org/wiki/Integral_of_Compactly_Supported_Function | [
"Calculus"
] | [
"Definition:Continuous Real Function",
"Definition:Compact Space/Euclidean Space",
"Definition:Support of Continuous Mapping"
] | [
"Riemann Integral Operator is Linear Mapping",
"Category:Calculus"
] |
proofwiki-19219 | Value of Compactly Supported Function outside its Support | Let $f : \R \to \R$ be a continuous real function.
Let $K \subseteq \R$ be a compact subset.
Let $K$ be the support of $f$:
:$\map \supp f = K$.
Then:
:$\forall x \notin K : \map f x = 0$ | We have that:
:$\R = K \cup \paren {\R \setminus K}$.
By definition of the support:
:$x \in \map \supp f \iff \map f x \ne 0$
By Biconditional Equivalent to Biconditional of Negations:
:$\neg \paren {x \in \map \supp f} \iff \neg \paren {\map f x \ne 0}$
That is:
:$x \notin K \iff \map f x = 0$
or
:$x \in \R \setminus... | Let $f : \R \to \R$ be a [[Definition:Continuous Real Function|continuous real function]].
Let $K \subseteq \R$ be a [[Definition:Compact Subset of Real Euclidean Space|compact subset]].
Let $K$ be the [[Definition:Support of Continuous Mapping|support]] of $f$:
:$\map \supp f = K$.
Then:
:$\forall x \notin K : \... | We have that:
:$\R = K \cup \paren {\R \setminus K}$.
By definition of the [[Definition:Support of Continuous Mapping|support]]:
:$x \in \map \supp f \iff \map f x \ne 0$
By [[Biconditional Equivalent to Biconditional of Negations]]:
:$\neg \paren {x \in \map \supp f} \iff \neg \paren {\map f x \ne 0}$
That is:
... | Value of Compactly Supported Function outside its Support | https://proofwiki.org/wiki/Value_of_Compactly_Supported_Function_outside_its_Support | https://proofwiki.org/wiki/Value_of_Compactly_Supported_Function_outside_its_Support | [
"Real Analysis"
] | [
"Definition:Continuous Real Function",
"Definition:Compact Space/Euclidean Space",
"Definition:Support of Continuous Mapping"
] | [
"Definition:Support of Continuous Mapping",
"Biconditional Equivalent to Biconditional of Negations",
"Category:Real Analysis"
] |
proofwiki-19220 | Dirichlet's Principle for Harmonic Functions/Riemannian Manifold | Let $\struct {M, g}$ be a compact connected $n$-dimensional Riemannian manifold with nonempty boundary.
Let $\map {C^\infty} M$ be the smooth function space.
Let $u \in \map {C^\infty} M$ be a smooth real function.
Let $\rd V_g$ be the Riemannian volume form.
Let $\grad$ be the gradient operator.
Let $\size {\, \cdot ... | {{ProofWanted}}
{{Namedfor|Johann Peter Gustav Lejeune Dirichlet|cat = Dirichlet}} | Let $\struct {M, g}$ be a [[Definition:Compact Manifold|compact]] [[Definition:Connected Manifold|connected]] [[Definition:Dimension of Riemannian Manifold|$n$-dimensional]] [[Definition:Riemannian Manifold|Riemannian manifold]] with [[Definition:Non-Empty Set|nonempty]] [[Definition:Boundary (Topology)|boundary]].
L... | {{ProofWanted}}
{{Namedfor|Johann Peter Gustav Lejeune Dirichlet|cat = Dirichlet}} | Dirichlet's Principle for Harmonic Functions/Riemannian Manifold | https://proofwiki.org/wiki/Dirichlet's_Principle_for_Harmonic_Functions/Riemannian_Manifold | https://proofwiki.org/wiki/Dirichlet's_Principle_for_Harmonic_Functions/Riemannian_Manifold | [
"Dirichlet's Principle for Harmonic Functions",
"Riemannian Geometry"
] | [
"Definition:Compact Manifold",
"Definition:Connected Manifold",
"Definition:Riemannian Manifold/Dimension",
"Definition:Riemannian Manifold",
"Definition:Non-Empty Set",
"Definition:Boundary (Topology)",
"Definition:Space of Continuous Mappings on Manifolds of Differentiability Class k",
"Definition:S... | [] |
proofwiki-19221 | Hasse Diagram of Dual Ordering | Let $D$ be the Hasse diagram of an ordering $\RR$.
Then the Hasse diagram $D'$ of the dual ordering $\RR^{-1}$ of $\RR$ is obtained by turning $D$ upside down. | Let $\tuple {x, y}$ be an ordered pair in $\RR$ which is represented by a line in $D$ joining $x$ and $y$ such that $x$ is below $y$.
Then $x \mathrel \RR y$ such that:
:$\not \exists z: x \mathrel \RR z \land \mathrel \RR y$
As $\RR^{-1}$ is the dual of $\RR$, we have that $y \mathrel {\RR^{-1} } x$ such that:
:$\not ... | Let $D$ be the [[Definition:Hasse Diagram|Hasse diagram]] of an [[Definition:Ordering|ordering]] $\RR$.
Then the [[Definition:Hasse Diagram|Hasse diagram]] $D'$ of the [[Definition:Dual Ordering|dual ordering]] $\RR^{-1}$ of $\RR$ is obtained by turning $D$ upside down. | Let $\tuple {x, y}$ be an [[Definition:Ordered Pair|ordered pair]] in $\RR$ which is represented by a [[Definition:Line Segment|line]] in $D$ joining $x$ and $y$ such that $x$ is below $y$.
Then $x \mathrel \RR y$ such that:
:$\not \exists z: x \mathrel \RR z \land \mathrel \RR y$
As $\RR^{-1}$ is the [[Definition:Du... | Hasse Diagram of Dual Ordering | https://proofwiki.org/wiki/Hasse_Diagram_of_Dual_Ordering | https://proofwiki.org/wiki/Hasse_Diagram_of_Dual_Ordering | [
"Hasse Diagrams",
"Dual Orderings"
] | [
"Definition:Hasse Diagram",
"Definition:Ordering",
"Definition:Hasse Diagram",
"Definition:Dual Ordering"
] | [
"Definition:Ordered Pair",
"Definition:Line/Segment",
"Definition:Dual Ordering",
"Definition:Line/Segment"
] |
proofwiki-19222 | Supremum of Elements of Sublattice not necessarily Same as for Lattice | Let $\struct {S, \preceq}$ be a lattice.
Let $\struct {T, \preceq_T}$ be a sublattice of $S$.
Let $a, b \in T$.
Then it is not necessarily the case that:
:$\sup_S \set {a, b}$
is the same as:
:$\sup_T \set {a, b}$ | Proof by Counterexample:
Let $\struct {G, \circ}$ be a group.
Let $\mathbb G$ be the set of all subgroups of $G$.
Let $\powerset G$ denote the power set of $G$.
Let $\struct {\powerset G, \subseteq}$ be the complete lattice formed by $\powerset G$ and $\subseteq$.
From Power Set is Complete Lattice, $\struct {\powerset... | Let $\struct {S, \preceq}$ be a [[Definition:Lattice (Ordered Set)|lattice]].
Let $\struct {T, \preceq_T}$ be a [[Definition:Sublattice|sublattice]] of $S$.
Let $a, b \in T$.
Then it is not necessarily the case that:
:$\sup_S \set {a, b}$
is the same as:
:$\sup_T \set {a, b}$ | [[Proof by Counterexample]]:
Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $\mathbb G$ be the [[Definition:Set|set]] of all [[Definition:Subgroup|subgroups]] of $G$.
Let $\powerset G$ denote the [[Definition:Power Set|power set]] of $G$.
Let $\struct {\powerset G, \subseteq}$ be the [[Definition:Co... | Supremum of Elements of Sublattice not necessarily Same as for Lattice | https://proofwiki.org/wiki/Supremum_of_Elements_of_Sublattice_not_necessarily_Same_as_for_Lattice | https://proofwiki.org/wiki/Supremum_of_Elements_of_Sublattice_not_necessarily_Same_as_for_Lattice | [
"Suprema",
"Sublattices"
] | [
"Definition:Lattice (Ordered Set)",
"Definition:Sublattice"
] | [
"Proof by Counterexample",
"Definition:Group",
"Definition:Set",
"Definition:Subgroup",
"Definition:Power Set",
"Definition:Complete Lattice",
"Power Set is Complete Lattice",
"Definition:Complete Lattice",
"Definition:Complete Lattice",
"Set of Subgroups forms Complete Lattice",
"Definition:Com... |
proofwiki-19223 | Generated Subsemigroup is not necessarily Same as Generated Group | Let $\struct {G, \circ}$ be a group.
Let $A \subseteq G$ be a subset of $G$.
Let $\struct {S, \circ}$ be the subsemigroup of $\struct {G, \circ}$ generated by $S$.
Let $\struct {H, \circ}$ be the subgroup of $\struct {G, \circ}$ generated by $S$.
Then it is not necessarily the case that $\struct {S, \circ}$ is the same... | Proof by Counterexample:
Let $\struct {\Z, +}$ be the additive group of integers.
Let $A$ be the set of positive odd integers.
From Generator of Subsemigroup: Positive Odd Numbers:
:the subsemigroup of $\struct {\Z, +}$ generated by $A$ is the semigroup of strictly positive integers under addition
while from Generator ... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $A \subseteq G$ be a [[Definition:Subset|subset]] of $G$.
Let $\struct {S, \circ}$ be the [[Definition:Generator of Subsemigroup|subsemigroup of $\struct {G, \circ}$ generated by $S$]].
Let $\struct {H, \circ}$ be the [[Definition:Generated Subgroup|subg... | [[Proof by Counterexample]]:
Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
Let $A$ be the [[Definition:Set|set]] of [[Definition:Positive Integer|positive]] [[Definition:Odd Integer|odd integers]].
From [[Generator of Subsemigroup/Examples/Positive Odd Numbers|Ge... | Generated Subsemigroup is not necessarily Same as Generated Group | https://proofwiki.org/wiki/Generated_Subsemigroup_is_not_necessarily_Same_as_Generated_Group | https://proofwiki.org/wiki/Generated_Subsemigroup_is_not_necessarily_Same_as_Generated_Group | [
"Generators of Groups",
"Subsemigroups"
] | [
"Definition:Group",
"Definition:Subset",
"Definition:Generator of Subsemigroup",
"Definition:Generated Subgroup"
] | [
"Proof by Counterexample",
"Definition:Additive Group of Integers",
"Definition:Set",
"Definition:Positive/Integer",
"Definition:Odd Integer",
"Generator of Subsemigroup/Examples/Positive Odd Numbers",
"Definition:Generator of Subsemigroup",
"Definition:Semigroup",
"Definition:Strictly Positive/Inte... |
proofwiki-19224 | Cardinality of Set of Relations | Let $S$ and $T$ be finite sets.
Let $\card S = n$ and $\card T = m$, where $\card {\, \cdot \,}$ denotes cardinality (that is, the number of elements).
Let $\RR$ denote the set of all relations from $S$ to $T$.
Then the cardinality of $\RR$ is given by:
:$\card \RR = 2^{n m}$ | By definition, a relation from $S$ to $T$ is a subset of the cartesian product $S \times T$ of $S$ and $T$.
From Cardinality of Cartesian Product of Finite Sets, we have:
:$\card {S \times T} = n m$
The number of subsets of $S \times T$ is the cardinality of the power set $\powerset {S \times T}$ of $S \times T$.
From ... | Let $S$ and $T$ be [[Definition:Finite Set|finite sets]].
Let $\card S = n$ and $\card T = m$, where $\card {\, \cdot \,}$ denotes [[Definition:Cardinality|cardinality]] (that is, the number of [[Definition:Element|elements]]).
Let $\RR$ denote the [[Definition:Set|set]] of all [[Definition:Relation|relations]] from ... | By definition, a [[Definition:Relation|relation]] from $S$ to $T$ is a [[Definition:Subset|subset]] of the [[Definition:Cartesian Product|cartesian product]] $S \times T$ of $S$ and $T$.
From [[Cardinality of Cartesian Product of Finite Sets]], we have:
:$\card {S \times T} = n m$
The number of [[Definition:Subset|su... | Cardinality of Set of Relations | https://proofwiki.org/wiki/Cardinality_of_Set_of_Relations | https://proofwiki.org/wiki/Cardinality_of_Set_of_Relations | [
"Combinatorics",
"Cardinality",
"Relations"
] | [
"Definition:Finite Set",
"Definition:Cardinality",
"Definition:Element",
"Definition:Set",
"Definition:Relation",
"Definition:Cardinality"
] | [
"Definition:Relation",
"Definition:Subset",
"Definition:Cartesian Product",
"Cardinality of Cartesian Product of Finite Sets",
"Definition:Subset",
"Definition:Cardinality",
"Definition:Power Set",
"Cardinality of Power Set of Finite Set",
"Category:Combinatorics",
"Category:Cardinality",
"Categ... |
proofwiki-19225 | Cardinality of Set of Endorelations | Let $S$ be a finite set.
Let $\card S = n$, where $\card {\, \cdot \,}$ denotes cardinality (that is, the number of elements).
Let $\RR$ denote the set of all endorelations on $S$.
Then the cardinality of $\RR$ is given by:
:$\card \RR = 2^{\paren {n^2} }$ | By definition, an endorelation on $S$ is a relation from $S$ to itself.
From Cardinality of Set of Relations, the number of relations from $S$ to $T$ is equal to $2^{\card S \times \card T}$.
In this case $S = T$ and the result follows.
{{qed}}
Category:Combinatorics
Category:Cardinality
Category:Endorelations
dgfdai5a... | Let $S$ be a [[Definition:Finite Set|finite set]].
Let $\card S = n$, where $\card {\, \cdot \,}$ denotes [[Definition:Cardinality|cardinality]] (that is, the number of [[Definition:Element|elements]]).
Let $\RR$ denote the [[Definition:Set|set]] of all [[Definition:Endorelation|endorelations]] on $S$.
Then the [[D... | By definition, an [[Definition:Endorelation|endorelation]] on $S$ is a [[Definition:Relation|relation]] from $S$ to itself.
From [[Cardinality of Set of Relations]], the number of [[Definition:Relation|relations]] from $S$ to $T$ is equal to $2^{\card S \times \card T}$.
In this case $S = T$ and the result follows.
{... | Cardinality of Set of Endorelations | https://proofwiki.org/wiki/Cardinality_of_Set_of_Endorelations | https://proofwiki.org/wiki/Cardinality_of_Set_of_Endorelations | [
"Combinatorics",
"Cardinality",
"Endorelations"
] | [
"Definition:Finite Set",
"Definition:Cardinality",
"Definition:Element",
"Definition:Set",
"Definition:Endorelation",
"Definition:Cardinality"
] | [
"Definition:Endorelation",
"Definition:Relation",
"Cardinality of Set of Relations",
"Definition:Relation",
"Category:Combinatorics",
"Category:Cardinality",
"Category:Endorelations"
] |
proofwiki-19226 | Number of Total Orderings on Finite Set | Let $S$ be a finite set with $n$ elements.
Then there are $n!$ different total orderings that can be applied to $S$. | A total ordering on $S$ is by definition a permutation on $S$ in the sense of an ordered selection.
The result follows from Number of Permutations.
{{qed}} | Let $S$ be a [[Definition:Finite Set|finite set]] with $n$ [[Definition:Element|elements]].
Then there are $n!$ different [[Definition:Total Ordering|total orderings]] that can be applied to $S$. | A [[Definition:Total Ordering|total ordering]] on $S$ is by definition a [[Definition:Permutation (Ordered Selection)|permutation]] on $S$ in the sense of an [[Definition:Ordered Selection|ordered selection]].
The result follows from [[Number of Permutations]].
{{qed}} | Number of Total Orderings on Finite Set | https://proofwiki.org/wiki/Number_of_Total_Orderings_on_Finite_Set | https://proofwiki.org/wiki/Number_of_Total_Orderings_on_Finite_Set | [
"Total Orderings"
] | [
"Definition:Finite Set",
"Definition:Element",
"Definition:Total Ordering"
] | [
"Definition:Total Ordering",
"Definition:Permutation/Ordered Selection",
"Definition:Permutation/Ordered Selection",
"Number of Permutations"
] |
proofwiki-19227 | Isomorphism Class of Total Orderings | Let $S$ be a finite set with $n$ elements.
There is exactly one isomorphism class containing the total orderings on $S$.
That is, every total ordering on $S$ is (order) isomorphic to every other total ordering. | {{ProofWanted|Intuitively obvious, probably needs to be hammered out in a proof by induction. I want to move on to other stuff today.}} | Let $S$ be a [[Definition:Finite Set|finite set]] with $n$ [[Definition:Element|elements]].
There is exactly one [[Definition:Isomorphism Class (Ordered Structures)|isomorphism class]] containing the [[Definition:Total Ordering|total orderings]] on $S$.
That is, every [[Definition:Total Ordering|total ordering]] on $... | {{ProofWanted|Intuitively obvious, probably needs to be hammered out in a proof by induction. I want to move on to other stuff today.}} | Isomorphism Class of Total Orderings | https://proofwiki.org/wiki/Isomorphism_Class_of_Total_Orderings | https://proofwiki.org/wiki/Isomorphism_Class_of_Total_Orderings | [
"Total Orderings"
] | [
"Definition:Finite Set",
"Definition:Element",
"Definition:Isomorphism Class (Ordered Structures)",
"Definition:Total Ordering",
"Definition:Total Ordering",
"Definition:Order Isomorphism",
"Definition:Total Ordering"
] | [] |
proofwiki-19228 | Reflexive Reduction of Transitive Relation is Transitive | Let $S$ be a set.
Let $\RR$ be a transitive relation on $S$.
Let $\RR^\ne$ be the reflexive reduction of $\RR$.
Then $\RR^\ne$ is transitive. | Let $a, b, c \in S$.
Let $a \mathrel {\RR^\ne} b$ and $b \mathrel {\RR^\ne} c$.
By the definition of reflexive reduction:
:$a \ne b$ and $a \mathrel \RR b$
:$b \ne c$ and $b \mathrel \RR c$
Since $\mathrel \RR$ is transitive:
:$a \mathrel \RR c$
{{qed}} | Let $S$ be a [[Definition:Set|set]].
Let $\RR$ be a [[Definition:Transitive Relation|transitive relation]] on $S$.
Let $\RR^\ne$ be the [[Definition:Reflexive Reduction|reflexive reduction]] of $\RR$.
Then $\RR^\ne$ is [[Definition:Transitive Relation|transitive]]. | Let $a, b, c \in S$.
Let $a \mathrel {\RR^\ne} b$ and $b \mathrel {\RR^\ne} c$.
By the definition of [[Definition:Reflexive Reduction|reflexive reduction]]:
:$a \ne b$ and $a \mathrel \RR b$
:$b \ne c$ and $b \mathrel \RR c$
Since $\mathrel \RR$ is [[Definition:Transitive Relation|transitive]]:
:$a \mathrel \RR c$... | Reflexive Reduction of Transitive Relation is Transitive | https://proofwiki.org/wiki/Reflexive_Reduction_of_Transitive_Relation_is_Transitive | https://proofwiki.org/wiki/Reflexive_Reduction_of_Transitive_Relation_is_Transitive | [
"Reflexive Reductions"
] | [
"Definition:Set",
"Definition:Transitive Relation",
"Definition:Reflexive Reduction",
"Definition:Transitive Relation"
] | [
"Definition:Reflexive Reduction",
"Definition:Transitive Relation"
] |
proofwiki-19229 | Composite of Increasing Mappings is Increasing | Let $\struct {S_1, \preceq_1}$, $\struct {S_2, \preceq_2}$ and $\struct {S_3, \preceq_3}$ be ordered sets.
Let:
:$\phi: \struct {S_1, \preceq_1} \to \struct {S_2, \preceq_2}$
and:
:$\psi: \struct {S_2, \preceq_2} \to \struct {S_3, \preceq_3}$
be increasing mappings.
Then $\psi \circ \phi: \struct {S_1, \preceq_1} \to \... | By definition of composition of mappings:
:$\map {\paren {\psi \circ \phi} } x = \map \psi {\map \phi x}$
As $\phi$ is an increasing mapping, we have:
:$\forall x_1, y_1 \in S_1: x_1 \preceq_1 y_1 \implies \map \phi {x_1} \preceq_2 \map \phi {y_1}$
As $\psi$ is an increasing mapping, we have:
:$\forall x_2, y_2 \in S_2... | Let $\struct {S_1, \preceq_1}$, $\struct {S_2, \preceq_2}$ and $\struct {S_3, \preceq_3}$ be [[Definition:Ordered Set|ordered sets]].
Let:
:$\phi: \struct {S_1, \preceq_1} \to \struct {S_2, \preceq_2}$
and:
:$\psi: \struct {S_2, \preceq_2} \to \struct {S_3, \preceq_3}$
be [[Definition:Increasing Mapping|increasing map... | By definition of [[Definition:Composition of Mappings|composition of mappings]]:
:$\map {\paren {\psi \circ \phi} } x = \map \psi {\map \phi x}$
As $\phi$ is an [[Definition:Increasing Mapping|increasing mapping]], we have:
:$\forall x_1, y_1 \in S_1: x_1 \preceq_1 y_1 \implies \map \phi {x_1} \preceq_2 \map \phi {y_... | Composite of Increasing Mappings is Increasing | https://proofwiki.org/wiki/Composite_of_Increasing_Mappings_is_Increasing | https://proofwiki.org/wiki/Composite_of_Increasing_Mappings_is_Increasing | [
"Increasing Mappings"
] | [
"Definition:Ordered Set",
"Definition:Increasing/Mapping",
"Definition:Increasing/Mapping"
] | [
"Definition:Composition of Mappings",
"Definition:Increasing/Mapping",
"Definition:Increasing/Mapping",
"Category:Increasing Mappings"
] |
proofwiki-19230 | Order-Preserving Identity Mapping between Ordered Structures not necessarily Isomorphism | Let $A$ be a set.
Let $\RR$ and $\SS$ be orderings on $A$ such that:
:$\forall a, b \in A: a \mathrel \RR b \implies a \mathrel \SS b$
Let $I_A$ denote the identity mapping on $A$.
Then it is not necessarily the case that $I_A$ is an order isomorphism from the ordered structures $\struct {A, \RR}$ and $\struct {A, \SS}... | Proof by Counterexample
Let $\RR: \N \to \N$ be the ordering on the natural numbers $\N$ defined as:
:$\forall a, b \in \N: a \mathrel \RR b \iff a \le b \text { and } \map P a = \map P b$
where $\map P x$ is the parity of $x$.
That is:
:$0 \mathrel \RR 2 \mathrel \RR 4 \mathrel \RR \cdots$
and:
:$1 \mathrel \RR 3 \mat... | Let $A$ be a [[Definition:Set|set]].
Let $\RR$ and $\SS$ be [[Definition:Ordering|orderings]] on $A$ such that:
:$\forall a, b \in A: a \mathrel \RR b \implies a \mathrel \SS b$
Let $I_A$ denote the [[Definition:Identity Mapping|identity mapping]] on $A$.
Then it is not necessarily the case that $I_A$ is an [[Defin... | [[Proof by Counterexample]]
Let $\RR: \N \to \N$ be the [[Definition:Ordering|ordering]] on the [[Definition:Natural Numbers|natural numbers]] $\N$ defined as:
:$\forall a, b \in \N: a \mathrel \RR b \iff a \le b \text { and } \map P a = \map P b$
where $\map P x$ is the [[Definition:Parity of Integer|parity]] of $x... | Order-Preserving Identity Mapping between Ordered Structures not necessarily Isomorphism | https://proofwiki.org/wiki/Order-Preserving_Identity_Mapping_between_Ordered_Structures_not_necessarily_Isomorphism | https://proofwiki.org/wiki/Order-Preserving_Identity_Mapping_between_Ordered_Structures_not_necessarily_Isomorphism | [
"Ordered Structures",
"Order Isomorphisms"
] | [
"Definition:Set",
"Definition:Ordering",
"Definition:Identity Mapping",
"Definition:Order Isomorphism",
"Definition:Ordered Structure"
] | [
"Proof by Counterexample",
"Definition:Ordering",
"Definition:Natural Numbers",
"Definition:Parity of Integer",
"Definition:Usual Ordering",
"Definition:Natural Numbers",
"Definition:Identity Mapping",
"Definition:Order Isomorphism"
] |
proofwiki-19231 | Relation Isomorphism preserves Ordering | Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be relational structures which are relationally isomorphic.
Let $\struct {A, \RR}$ be an ordered set.
Then $\struct {B, \SS}$ is also an ordered set. | Let $\struct {A, \RR}$ be an ordered set.
Recall the definition:
{{:Definition:Ordering/Definition 1}}
From Relation Isomorphism Preserves Reflexivity:
:$\SS$ is reflexive.
From Relation Isomorphism Preserves Antisymmetry:
:$\SS$ is antisymmetric.
From Relation Isomorphism Preserves Transitivity:
:$\SS$ is transitive.
... | Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be [[Definition:Relational Structure|relational structures]] which are [[Definition:Relation Isomorphism|relationally isomorphic]].
Let $\struct {A, \RR}$ be an [[Definition:Ordered Set|ordered set]].
Then $\struct {B, \SS}$ is also an [[Definition:Ordered Set|ordered se... | Let $\struct {A, \RR}$ be an [[Definition:Ordered Set|ordered set]].
Recall the [[Definition:Ordering/Definition 1|definition]]:
{{:Definition:Ordering/Definition 1}}
From [[Relation Isomorphism Preserves Reflexivity]]:
:$\SS$ is [[Definition:Reflexive Relation|reflexive]].
From [[Relation Isomorphism Preserves Anti... | Relation Isomorphism preserves Ordering | https://proofwiki.org/wiki/Relation_Isomorphism_preserves_Ordering | https://proofwiki.org/wiki/Relation_Isomorphism_preserves_Ordering | [
"Relation Isomorphisms",
"Orderings"
] | [
"Definition:Relational Structure",
"Definition:Relation Isomorphism",
"Definition:Ordered Set",
"Definition:Ordered Set"
] | [
"Definition:Ordered Set",
"Definition:Ordering/Definition 1",
"Relation Isomorphism Preserves Reflexivity",
"Definition:Reflexive Relation",
"Relation Isomorphism Preserves Antisymmetry",
"Definition:Antisymmetric Relation",
"Relation Isomorphism Preserves Transitivity",
"Definition:Transitive Relatio... |
proofwiki-19232 | Relation Isomorphism preserves Total Ordering | Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be relational structures which are relationally isomorphic.
Let $\struct {A, \RR}$ be a totally ordered set.
Then $\struct {B, \SS}$ is also a totally ordered set. | Let $\struct {A, \RR}$ be a totally ordered set.
Recall the definition:
{{:Definition:Total Ordering/Definition 1}}
From Relation Isomorphism preserves Ordering:
:$\SS$ is an ordering.
Let $\phi: \struct {A, \RR} \to \struct {B, \SS}$ be a relation isomorphism.
Let $x, y \in A$.
Then either $x \mathrel \RR y$ or $y \ma... | Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be [[Definition:Relational Structure|relational structures]] which are [[Definition:Relation Isomorphism|relationally isomorphic]].
Let $\struct {A, \RR}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Then $\struct {B, \SS}$ is also a [[Definition:Totally ... | Let $\struct {A, \RR}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Recall the [[Definition:Total Ordering/Definition 1|definition]]:
{{:Definition:Total Ordering/Definition 1}}
From [[Relation Isomorphism preserves Ordering]]:
:$\SS$ is an [[Definition:Ordering|ordering]].
Let $\phi: \struct {A, \R... | Relation Isomorphism preserves Total Ordering | https://proofwiki.org/wiki/Relation_Isomorphism_preserves_Total_Ordering | https://proofwiki.org/wiki/Relation_Isomorphism_preserves_Total_Ordering | [
"Relation Isomorphisms",
"Total Orderings"
] | [
"Definition:Relational Structure",
"Definition:Relation Isomorphism",
"Definition:Totally Ordered Set",
"Definition:Totally Ordered Set"
] | [
"Definition:Totally Ordered Set",
"Definition:Total Ordering/Definition 1",
"Relation Isomorphism preserves Ordering",
"Definition:Ordering",
"Definition:Relation Isomorphism",
"Definition:Relation Isomorphism",
"Definition:Total Ordering",
"Definition:Totally Ordered Set"
] |
proofwiki-19233 | Relation Isomorphism preserves Well-Ordering | Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be relational structures which are relationally isomorphic.
Let $\struct {A, \RR}$ be a well-ordered set.
Then $\struct {B, \SS}$ is also a well-ordered set. | Let $\struct {A, \RR}$ be a well-ordered set.
Recall the definition:
{{:Definition:Well-Ordering/Definition 1}}
By Well-Ordering is Total Ordering, $\RR$ is a total ordering.
From Relation Isomorphism preserves Total Ordering:
:$\SS$ is a total ordering on $B$.
Let $\phi: \struct {A, \RR} \to \struct {B, \SS}$ be a rel... | Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be [[Definition:Relational Structure|relational structures]] which are [[Definition:Relation Isomorphism|relationally isomorphic]].
Let $\struct {A, \RR}$ be a [[Definition:Well-Ordered Set|well-ordered set]].
Then $\struct {B, \SS}$ is also a [[Definition:Well-Ordered S... | Let $\struct {A, \RR}$ be a [[Definition:Well-Ordered Set|well-ordered set]].
Recall the [[Definition:Well-Ordering/Definition 1|definition]]:
{{:Definition:Well-Ordering/Definition 1}}
By [[Well-Ordering is Total Ordering]], $\RR$ is a [[Definition:Total Ordering|total ordering]].
From [[Relation Isomorphism preser... | Relation Isomorphism preserves Well-Ordering | https://proofwiki.org/wiki/Relation_Isomorphism_preserves_Well-Ordering | https://proofwiki.org/wiki/Relation_Isomorphism_preserves_Well-Ordering | [
"Relation Isomorphisms",
"Well-Orderings"
] | [
"Definition:Relational Structure",
"Definition:Relation Isomorphism",
"Definition:Well-Ordered Set",
"Definition:Well-Ordered Set"
] | [
"Definition:Well-Ordered Set",
"Definition:Well-Ordering/Definition 1",
"Equivalence of Definitions of Well-Ordering/Definition 1 implies Definition 2",
"Definition:Total Ordering",
"Relation Isomorphism preserves Total Ordering",
"Definition:Total Ordering",
"Definition:Relation Isomorphism",
"Defini... |
proofwiki-19234 | Relation Isomorphism preserves Lattice Structure | Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be relational structures which are relationally isomorphic.
Let $\struct {A, \RR}$ be a lattice.
Then $\struct {B, \SS}$ is also a lattice. | Let $\struct {A, \RR}$ be a lattice.
Recall the definition:
{{:Definition:Lattice (Ordered Set)}}
From Relation Isomorphism preserves Ordering:
:$\SS$ is an ordering on $B$.
Let $\phi: \struct {A, \RR} \to \struct {B, \SS}$ be a relation isomorphism.
We need to show that for all $x, y \in A$, the ordered set $\struct {... | Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be [[Definition:Relational Structure|relational structures]] which are [[Definition:Relation Isomorphism|relationally isomorphic]].
Let $\struct {A, \RR}$ be a [[Definition:Lattice (Ordered Set)|lattice]].
Then $\struct {B, \SS}$ is also a [[Definition:Lattice (Ordered S... | Let $\struct {A, \RR}$ be a [[Definition:Lattice (Ordered Set)|lattice]].
Recall the [[Definition:Lattice (Ordered Set)|definition]]:
{{:Definition:Lattice (Ordered Set)}}
From [[Relation Isomorphism preserves Ordering]]:
:$\SS$ is an [[Definition:Ordering|ordering]] on $B$.
Let $\phi: \struct {A, \RR} \to \struct ... | Relation Isomorphism preserves Lattice Structure | https://proofwiki.org/wiki/Relation_Isomorphism_preserves_Lattice_Structure | https://proofwiki.org/wiki/Relation_Isomorphism_preserves_Lattice_Structure | [
"Relation Isomorphisms",
"Lattice Theory"
] | [
"Definition:Relational Structure",
"Definition:Relation Isomorphism",
"Definition:Lattice (Ordered Set)",
"Definition:Lattice (Ordered Set)"
] | [
"Definition:Lattice (Ordered Set)",
"Definition:Lattice (Ordered Set)",
"Relation Isomorphism preserves Ordering",
"Definition:Ordering",
"Definition:Relation Isomorphism",
"Definition:Ordered Set",
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Supremum of Set",
"Definitio... |
proofwiki-19235 | Ordered Set with Greatest Element whose Subsets have Infimum is Complete Lattice | Let $\struct {S, \preccurlyeq}$ be an ordered set such that:
:$\struct {S, \preccurlyeq}$ has a greatest element $u$
:every non-empty subset of $S$ admits an infimum.
Then $\struct {S, \preccurlyeq}$ is a complete lattice. | For $\struct {S, \preccurlyeq}$ to be a complete lattice, it has to be such that:
:every non-empty subset of $S$ admits both a supremum and an infimum.
Let $T \subseteq S$ be an arbitrary subset of $S$.
We already have {{hypothesis}} that $T$ admits an infimum.
We are to show that $T$ admits a supremum.
By definition o... | Let $\struct {S, \preccurlyeq}$ be an [[Definition:Ordered Set|ordered set]] such that:
:$\struct {S, \preccurlyeq}$ has a [[Definition:Greatest Element|greatest element]] $u$
:every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$ admits an [[Definition:Infimum of Set|infimum]].
Then $\str... | For $\struct {S, \preccurlyeq}$ to be a [[Definition:Complete Lattice|complete lattice]], it has to be such that:
:every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$ admits both a [[Definition:Supremum of Set|supremum]] and an [[Definition:Infimum of Set|infimum]].
Let $T \subseteq S$ be... | Ordered Set with Greatest Element whose Subsets have Infimum is Complete Lattice | https://proofwiki.org/wiki/Ordered_Set_with_Greatest_Element_whose_Subsets_have_Infimum_is_Complete_Lattice | https://proofwiki.org/wiki/Ordered_Set_with_Greatest_Element_whose_Subsets_have_Infimum_is_Complete_Lattice | [
"Complete Lattices",
"Greatest Elements"
] | [
"Definition:Ordered Set",
"Definition:Greatest Element",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Infimum of Set",
"Definition:Complete Lattice"
] | [
"Definition:Complete Lattice",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Subset",
"Definition:Infimum of Set",
"Definition:Supremum of Set",
"Definition:Greatest Element",
"Definition:Upper Bound of Set",
"Definition:S... |
proofwiki-19236 | Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice | Let $A$ be a set.
Let $\SS$ be a set of subsets of $A$ such that:
:$A \in \SS$
:for every non-empty subset $\TT$ of $\SS$, $\ds \bigcap \TT \in \SS$, where $\ds \bigcap \TT$ denotes the intersection of $\TT$.
Then:
:the ordered set $\struct {\SS, \subseteq}$ is a complete lattice
where:
:$\ds \bigcap \TT$ is the infimu... | From Subset Relation is Ordering, $\struct {\SS, \subseteq}$ is indeed an ordered set.
We have by definition of $\SS$ that:
:$\forall H \in \SS: H \subseteq A$
Hence $A$ is the greatest element of $\SS$ with respect to the ordered set $\struct {A, \subseteq}$.
Then from Intersection is Largest Subset:
:$\forall K \in \... | Let $A$ be a [[Definition:Set|set]].
Let $\SS$ be a [[Definition:Set of Sets|set]] of [[Definition:Subset|subsets]] of $A$ such that:
:$A \in \SS$
:for every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] $\TT$ of $\SS$, $\ds \bigcap \TT \in \SS$, where $\ds \bigcap \TT$ denotes the [[Definition... | From [[Subset Relation is Ordering]], $\struct {\SS, \subseteq}$ is indeed an [[Definition:Ordered Set|ordered set]].
We have by definition of $\SS$ that:
:$\forall H \in \SS: H \subseteq A$
Hence $A$ is the [[Definition:Greatest Element|greatest element]] of $\SS$ with respect to the [[Definition:Ordered Set|ordered... | Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice | https://proofwiki.org/wiki/Set_of_Subsets_which_contains_Set_and_Intersection_of_Subsets_is_Complete_Lattice | https://proofwiki.org/wiki/Set_of_Subsets_which_contains_Set_and_Intersection_of_Subsets_is_Complete_Lattice | [
"Complete Lattices",
"Subsets"
] | [
"Definition:Set",
"Definition:Set of Sets",
"Definition:Subset",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Set Intersection/Set of Sets",
"Definition:Ordered Set",
"Definition:Complete Lattice",
"Definition:Infimum of Set"
] | [
"Subset Relation is Ordering",
"Definition:Ordered Set",
"Definition:Greatest Element",
"Definition:Ordered Set",
"Intersection is Largest Subset",
"Definition:Infimum of Set",
"Ordered Set with Greatest Element whose Subsets have Infimum is Complete Lattice"
] |
proofwiki-19237 | Intersection of Submagmas is Largest Submagma | Let $\HH$ be a set of submagmas of $\struct {S, \odot}$, where $\HH \ne \O$.
Then the intersection $\bigcap \HH$ of the elements of $\HH$ is the largest submagma of $\struct {S, \odot}$ contained in each element of $\HH$. | Let $K = \bigcap \HH$.
Let $K_i$ be an arbitrary element of $\HH$.
Then:
{{begin-eqn}}
{{eqn | l = a, b
| o = \in
| r = K
| c =
}}
{{eqn | ll= \leadsto
| q = \forall i
| l = a, b
| o = \in
| r = K_i
| c = {{Defof|Intersection of Set of Sets}}
}}
{{eqn | ll= \leadsto
... | Let $\HH$ be a [[Definition:Set|set]] of [[Definition:Submagma|submagmas]] of $\struct {S, \odot}$, where $\HH \ne \O$.
Then the [[Definition:Intersection of Set of Sets|intersection]] $\bigcap \HH$ of the [[Definition:Element|elements]] of $\HH$ is the largest [[Definition:Submagma|submagma]] of $\struct {S, \odot}$... | Let $K = \bigcap \HH$.
Let $K_i$ be an arbitrary [[Definition:Element|element]] of $\HH$.
Then:
{{begin-eqn}}
{{eqn | l = a, b
| o = \in
| r = K
| c =
}}
{{eqn | ll= \leadsto
| q = \forall i
| l = a, b
| o = \in
| r = K_i
| c = {{Defof|Intersection of Set of Sets}}
}}... | Intersection of Submagmas is Largest Submagma | https://proofwiki.org/wiki/Intersection_of_Submagmas_is_Largest_Submagma | https://proofwiki.org/wiki/Intersection_of_Submagmas_is_Largest_Submagma | [
"Submagmas",
"Set Intersection"
] | [
"Definition:Set",
"Definition:Submagma",
"Definition:Set Intersection/Set of Sets",
"Definition:Element",
"Definition:Submagma",
"Definition:Element"
] | [
"Definition:Element",
"Definition:Magma",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Submagma",
"Definition:Submagma",
"Definition:Element",
"Definition:Submagma",
"Definition:Submagma",
"Definition:Element"... |
proofwiki-19238 | Set of Submagmas of Magma under Subset Relation forms Complete Lattice | Let $\struct {A, \odot}$ be a magma.
Let $\SS$ be the set of submagmas of $A$.
Then:
:the ordered set $\struct {\SS, \subseteq}$ is a complete lattice
where for every subset $\TT$ of $\SS$:
:the infimum of $\TT$ necessarily admitted by $\TT$ is $\bigcap \TT$. | From Magma is Submagma of Itself:
:$\struct {A, \odot} \in \SS$
Let $\TT$ be a non-empty subset of $\SS$.
From Intersection of Submagmas is Largest Submagma:
:$\bigcap \TT \in \SS$
Hence, from Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice:
:$\struct {\SS, \subseteq}$ is a complete la... | Let $\struct {A, \odot}$ be a [[Definition:Magma|magma]].
Let $\SS$ be the [[Definition:Set|set]] of [[Definition:Submagma|submagmas]] of $A$.
Then:
:the [[Definition:Ordered Set|ordered set]] $\struct {\SS, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]]
where for every [[Definition:Subset|subset... | From [[Magma is Submagma of Itself]]:
:$\struct {A, \odot} \in \SS$
Let $\TT$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $\SS$.
From [[Intersection of Submagmas is Largest Submagma]]:
:$\bigcap \TT \in \SS$
Hence, from [[Set of Subsets which contains Set and Intersection of Subsets i... | Set of Submagmas of Magma under Subset Relation forms Complete Lattice | https://proofwiki.org/wiki/Set_of_Submagmas_of_Magma_under_Subset_Relation_forms_Complete_Lattice | https://proofwiki.org/wiki/Set_of_Submagmas_of_Magma_under_Subset_Relation_forms_Complete_Lattice | [
"Examples of Complete Lattices",
"Submagmas"
] | [
"Definition:Magma",
"Definition:Set",
"Definition:Submagma",
"Definition:Ordered Set",
"Definition:Complete Lattice",
"Definition:Subset",
"Definition:Infimum of Set"
] | [
"Magma is Submagma of Itself",
"Definition:Non-Empty Set",
"Definition:Subset",
"Intersection of Submagmas is Largest Submagma",
"Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice",
"Definition:Complete Lattice",
"Definition:Infimum of Set"
] |
proofwiki-19239 | Intersection of Subsemigroups/General Result/Proof 1 | Let $\struct {S, \circ}$ be a semigroup.
{{:Intersection of Subsemigroups/General Result}} | Let $T = \bigcap \mathbb S$.
Then:
{{begin-eqn}}
{{eqn | l = a, b
| o = \in
| r = T
| c =
}}
{{eqn | ll= \leadsto
| q = \forall K \in \mathbb S
| l = a, b
| o = \in
| r = K
| c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
| q = \forall K \in \mathbb S
... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
{{:Intersection of Subsemigroups/General Result}} | Let $T = \bigcap \mathbb S$.
Then:
{{begin-eqn}}
{{eqn | l = a, b
| o = \in
| r = T
| c =
}}
{{eqn | ll= \leadsto
| q = \forall K \in \mathbb S
| l = a, b
| o = \in
| r = K
| c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
| q = \forall K \in \mathbb S
... | Intersection of Subsemigroups/General Result/Proof 1 | https://proofwiki.org/wiki/Intersection_of_Subsemigroups/General_Result/Proof_1 | https://proofwiki.org/wiki/Intersection_of_Subsemigroups/General_Result/Proof_1 | [
"Intersection of Subsemigroups"
] | [
"Definition:Semigroup"
] | [
"Definition:Subsemigroup",
"Subsemigroup Closure Test",
"Definition:Subsemigroup",
"Definition:Subsemigroup",
"Definition:Subsemigroup"
] |
proofwiki-19240 | Intersection of Subsemigroups/General Result/Proof 2 | Let $\struct {S, \circ}$ be a semigroup.
{{:Intersection of Subsemigroups/General Result}} | From Set of Subsemigroups forms Complete Lattice:
:$\struct {\mathbb S, \subseteq}$ is a complete lattice.
where for every set $\mathbb H$ of subsemigroups of $S$:
:the infimum of $\mathbb H$ necessarily admitted by $\mathbb H$ is $\ds \bigcap \mathbb H$.
Hence the result, by definition of infimum.
{{qed}} | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
{{:Intersection of Subsemigroups/General Result}} | From [[Set of Subsemigroups forms Complete Lattice]]:
:$\struct {\mathbb S, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]].
where for every [[Definition:Set|set]] $\mathbb H$ of [[Definition:Subsemigroup|subsemigroups]] of $S$:
:the [[Definition:Infimum of Set|infimum]] of $\mathbb H$ necessarily ad... | Intersection of Subsemigroups/General Result/Proof 2 | https://proofwiki.org/wiki/Intersection_of_Subsemigroups/General_Result/Proof_2 | https://proofwiki.org/wiki/Intersection_of_Subsemigroups/General_Result/Proof_2 | [
"Intersection of Subsemigroups"
] | [
"Definition:Semigroup"
] | [
"Set of Subsemigroups forms Complete Lattice",
"Definition:Complete Lattice",
"Definition:Set",
"Definition:Subsemigroup",
"Definition:Infimum of Set",
"Definition:Infimum of Set"
] |
proofwiki-19241 | Elements of Semigroup with Equal Images under Homomorphisms form Subsemigroup | Let $\struct {A, \circ}$ and $\struct {B, *}$ be semigroups.
Let $f: A \to B$ and $g: A \to B$ be semigroup homomorphisms.
Then the set:
:$S = \set {x \in A: \map f x = \map g x}$
is a subsemigroup of $A$. | Let $x, y \in A$. Then:
{{begin-eqn}}
{{eqn | l = \map f {x \circ y}
| r = \map f x * \map f y
| c = Morphism Property
}}
{{eqn | r = \map g x * \map g y
| c = Definition of $A$
}}
{{eqn | r = \map g {x \circ y}
| c = Morphism Property
}}
{{end-eqn}}
Thus $x \circ y \in A$.
So, by the Subsemigro... | Let $\struct {A, \circ}$ and $\struct {B, *}$ be [[Definition:Semigroup|semigroups]].
Let $f: A \to B$ and $g: A \to B$ be [[Definition:Semigroup Homomorphism|semigroup homomorphisms]].
Then the [[Definition:Set|set]]:
:$S = \set {x \in A: \map f x = \map g x}$
is a [[Definition:Subsemigroup|subsemigroup]] of $A$. | Let $x, y \in A$. Then:
{{begin-eqn}}
{{eqn | l = \map f {x \circ y}
| r = \map f x * \map f y
| c = [[Definition:Morphism Property|Morphism Property]]
}}
{{eqn | r = \map g x * \map g y
| c = Definition of $A$
}}
{{eqn | r = \map g {x \circ y}
| c = [[Definition:Morphism Property|Morphism Prop... | Elements of Semigroup with Equal Images under Homomorphisms form Subsemigroup | https://proofwiki.org/wiki/Elements_of_Semigroup_with_Equal_Images_under_Homomorphisms_form_Subsemigroup | https://proofwiki.org/wiki/Elements_of_Semigroup_with_Equal_Images_under_Homomorphisms_form_Subsemigroup | [
"Semigroup Homomorphisms"
] | [
"Definition:Semigroup",
"Definition:Semigroup Homomorphism",
"Definition:Set",
"Definition:Subsemigroup"
] | [
"Definition:Morphism Property",
"Definition:Morphism Property",
"Subsemigroup Closure Test",
"Definition:Subsemigroup"
] |
proofwiki-19242 | Set of Congruence Relations on Algebraic Structure forms Complete Lattice | Let $\struct {S, \odot}$ be an algebraic structure.
Let $\map \RR \odot$ be the set of all congruence relations on $\struct {S, \odot}$.
Then $\struct {\map \RR \odot, \subseteq}$ is a complete lattice. | Elements of $\map \RR \odot$ are subsets of $S \times S$.
First we consider the trivial relation $S \times S$ itself.
From Trivial Relation is Universally Congruent, $S \times S$ is a congruence relation on $\struct {S, \odot}$.
Let $\TT$ be a subset of $\map \RR \odot$.
Consider the intersection $\HH = \ds \bigcap \TT... | Let $\struct {S, \odot}$ be an [[Definition:Algebraic Structure|algebraic structure]].
Let $\map \RR \odot$ be the [[Definition:Set|set]] of all [[Definition:Congruence Relation|congruence relations]] on $\struct {S, \odot}$.
Then $\struct {\map \RR \odot, \subseteq}$ is a [[Definition:Complete Lattice|complete latt... | [[Definition:Element|Elements]] of $\map \RR \odot$ are [[Definition:Subset|subsets]] of $S \times S$.
First we consider the [[Definition:Trivial Relation|trivial relation]] $S \times S$ itself.
From [[Trivial Relation is Universally Congruent]], $S \times S$ is a [[Definition:Congruence Relation|congruence relation]... | Set of Congruence Relations on Algebraic Structure forms Complete Lattice | https://proofwiki.org/wiki/Set_of_Congruence_Relations_on_Algebraic_Structure_forms_Complete_Lattice | https://proofwiki.org/wiki/Set_of_Congruence_Relations_on_Algebraic_Structure_forms_Complete_Lattice | [
"Congruence Relations",
"Examples of Complete Lattices"
] | [
"Definition:Algebraic Structure",
"Definition:Set",
"Definition:Congruence Relation",
"Definition:Complete Lattice"
] | [
"Definition:Element",
"Definition:Subset",
"Definition:Trivial Relation",
"Trivial Relation is Universally Congruent",
"Definition:Congruence Relation",
"Definition:Subset",
"Definition:Set Intersection/Set of Sets",
"Definition:Congruence Relation",
"Set of Subsets which contains Set and Intersecti... |
proofwiki-19243 | Group Epimorphism Preserves Generator | Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: G \to H$ be a group epimorphism.
Let $A$ be a generator for $\struct {G, \circ}$.
Then $\phi \sqbrk A$ is a generator for $\struct {H, *}$. | By definition of generator:
:$A$ is the intersection of all subgroups of $G$ containing $A$.
{{ProofWanted}} | Let $\struct {G, \circ}$ and $\struct {H, *}$ be [[Definition:Group|groups]].
Let $\phi: G \to H$ be a [[Definition:Group Epimorphism|group epimorphism]].
Let $A$ be a [[Definition:Generator of Group|generator]] for $\struct {G, \circ}$.
Then $\phi \sqbrk A$ is a [[Definition:Generator of Group|generator]] for $\st... | By definition of [[Definition:Generator of Group|generator]]:
:$A$ is the [[Definition:Set Intersection|intersection]] of all [[Definition:Subgroup|subgroups]] of $G$ containing $A$.
{{ProofWanted}} | Group Epimorphism Preserves Generator | https://proofwiki.org/wiki/Group_Epimorphism_Preserves_Generator | https://proofwiki.org/wiki/Group_Epimorphism_Preserves_Generator | [
"Generated Subgroups",
"Group Epimorphisms"
] | [
"Definition:Group",
"Definition:Group Epimorphism",
"Definition:Generator of Group",
"Definition:Generator of Group"
] | [
"Definition:Generator of Group",
"Definition:Set Intersection",
"Definition:Subgroup"
] |
proofwiki-19244 | Lattice of Subgroups forming Totally Ordered Set is Indecomposable | Let $\struct {G, \circ}$ be a Group.
Let $\mathbb G$ be the set of subgroups of $G$.
Let $\struct {\mathbb G, \subseteq}$ be the complete lattice formed by the subset ordering on $\mathbb G$.
Let $\struct {\mathbb G, \subseteq}$ be totally ordered.
Then $\struct {G, \circ}$ is an indecomposable group. | First we note that from Set of Subgroups forms Complete Lattice, $\struct {\mathbb G, \subseteq}$ is indeed a complete lattice.
As postulated, let $\struct {\mathbb G, \subseteq}$ be totally ordered.
{{AimForCont}} there exists a decomposition of $\struct {G, \circ}$.
Then:
:$\exists H, K \in \mathbb G: H \cap K = \set... | Let $\struct {G, \circ}$ be a [[Definition:Group|Group]].
Let $\mathbb G$ be the [[Definition:Set|set]] of [[Definition:Subgroup|subgroups]] of $G$.
Let $\struct {\mathbb G, \subseteq}$ be the [[Definition:Complete Lattice|complete lattice]] formed by the [[Definition:Subset Ordering|subset ordering]] on $\mathbb G$.... | First we note that from [[Set of Subgroups forms Complete Lattice]], $\struct {\mathbb G, \subseteq}$ is indeed a [[Definition:Complete Lattice|complete lattice]].
As postulated, let $\struct {\mathbb G, \subseteq}$ be [[Definition:Totally Ordered Set|totally ordered]].
{{AimForCont}} there exists a [[Definition:Grou... | Lattice of Subgroups forming Totally Ordered Set is Indecomposable | https://proofwiki.org/wiki/Lattice_of_Subgroups_forming_Totally_Ordered_Set_is_Indecomposable | https://proofwiki.org/wiki/Lattice_of_Subgroups_forming_Totally_Ordered_Set_is_Indecomposable | [
"Decomposable Groups"
] | [
"Definition:Group",
"Definition:Set",
"Definition:Subgroup",
"Definition:Complete Lattice",
"Definition:Set Ordered by Subset Relation",
"Definition:Totally Ordered Set",
"Definition:Decomposable Group/Indecomposable"
] | [
"Set of Subgroups forms Complete Lattice",
"Definition:Complete Lattice",
"Definition:Totally Ordered Set",
"Definition:Internal Group Direct Product/Decomposition",
"Definition:Non-Trivial Group",
"Definition:Normal Subgroup",
"Definition:Totally Ordered Set",
"Proof by Contradiction"
] |
proofwiki-19245 | Indecomposable Lattice of Subgroups does not necessarily form Totally Ordered Set | Let $\struct {G, \circ}$ be a Group.
Let $\mathbb G$ be the set of subgroups of $G$.
Let $\struct {\mathbb G, \subseteq}$ be the complete lattice formed by the subset ordering on $\mathbb G$.
Let $\struct {G, \circ}$ be an indecomposable group.
Then it is not necessarily the case that $\struct {\mathbb G, \subseteq}$ i... | First we note that from Set of Subgroups forms Complete Lattice, $\struct {\mathbb G, \subseteq}$ is indeed a complete lattice.
;Proof by Counterexample
Let $D_4$ denote the dihedral group of order $8$, also known as the symmetry group of the square.
From Internal Group Direct Product Examples: $D_4$, it is seen that $... | Let $\struct {G, \circ}$ be a [[Definition:Group|Group]].
Let $\mathbb G$ be the [[Definition:Set|set]] of [[Definition:Subgroup|subgroups]] of $G$.
Let $\struct {\mathbb G, \subseteq}$ be the [[Definition:Complete Lattice|complete lattice]] formed by the [[Definition:Subset Ordering|subset ordering]] on $\mathbb G$.... | First we note that from [[Set of Subgroups forms Complete Lattice]], $\struct {\mathbb G, \subseteq}$ is indeed a [[Definition:Complete Lattice|complete lattice]].
;[[Proof by Counterexample]]
Let $D_4$ denote the [[Definition:Dihedral Group|dihedral group of order $8$]], also known as the [[Definition:Symmetry Gro... | Indecomposable Lattice of Subgroups does not necessarily form Totally Ordered Set | https://proofwiki.org/wiki/Indecomposable_Lattice_of_Subgroups_does_not_necessarily_form_Totally_Ordered_Set | https://proofwiki.org/wiki/Indecomposable_Lattice_of_Subgroups_does_not_necessarily_form_Totally_Ordered_Set | [
"Decomposable Groups"
] | [
"Definition:Group",
"Definition:Set",
"Definition:Subgroup",
"Definition:Complete Lattice",
"Definition:Set Ordered by Subset Relation",
"Definition:Decomposable Group/Indecomposable",
"Definition:Totally Ordered Set"
] | [
"Set of Subgroups forms Complete Lattice",
"Definition:Complete Lattice",
"Proof by Counterexample",
"Definition:Dihedral Group",
"Definition:Symmetry Group of Square",
"Internal Group Direct Product/Examples/D4",
"Definition:Decomposable Group/Indecomposable",
"Hasse Diagram/Examples/Subgroups of Sym... |
proofwiki-19246 | Conditions for Relation to be Well-Ordering | Let $\struct {S, \RR}$ be a relational structure.
$\RR$ is a well-ordering {{iff}}:
:$(1): \quad$ For all $x, y \in S$, either $x \mathrel \RR y$ or $y \mathrel \RR x$
:$(2): \quad$ For every non-empty subset $T$ of $S$, there exists $z \in T$ such that:
::::$\forall x \in T: \paren {x \mathrel \RR z \iff x = z}$ | === Sufficient Condition ===
Let $\RR$ be a well-ordering.
Then $\RR$ is {{afortiori}} a total ordering, and so by definition:
:$\forall x, y \in S: x \mathrel \RR y \text { or } y \mathrel \RR x$
Thus condition $(1)$ is fulfilled.
Let $T$ be a non-empty subset of $S$.
Then by definition of well-ordering:
:$\exists z \... | Let $\struct {S, \RR}$ be a [[Definition:Relational Structure|relational structure]].
$\RR$ is a [[Definition:Well-Ordering|well-ordering]] {{iff}}:
:$(1): \quad$ For all $x, y \in S$, either $x \mathrel \RR y$ or $y \mathrel \RR x$
:$(2): \quad$ For every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|s... | === Sufficient Condition ===
Let $\RR$ be a [[Definition:Well-Ordering|well-ordering]].
Then $\RR$ is {{afortiori}} a [[Definition:Total Ordering|total ordering]], and so by definition:
:$\forall x, y \in S: x \mathrel \RR y \text { or } y \mathrel \RR x$
Thus condition $(1)$ is fulfilled.
Let $T$ be a [[Definiti... | Conditions for Relation to be Well-Ordering | https://proofwiki.org/wiki/Conditions_for_Relation_to_be_Well-Ordering | https://proofwiki.org/wiki/Conditions_for_Relation_to_be_Well-Ordering | [
"Well-Orderings"
] | [
"Definition:Relational Structure",
"Definition:Well-Ordering",
"Definition:Non-Empty Set",
"Definition:Subset"
] | [
"Definition:Well-Ordering",
"Definition:Total Ordering",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Well-Ordering",
"Definition:Well-Ordering",
"Definition:Ordering",
"Definition:Asymmetric Relation",
"Definition:Ordering",
"Definition:Reflexive Relation",
"Definition:Non-Empty... |
proofwiki-19247 | Simple Order Product of Pair of Ordered Sets is Lattice iff Factors are Lattices | Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ denote the '''simple (order) product''' of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$.
Then $\struct {S_1, \preccurlyeq_1} \otimes... | Let $\struct {T, \preccurlyeq_s} := \struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$.
First we note that from Simple Order Product of Pair of Ordered Sets is Ordered Set, $\struct {T, \preccurlyeq_s}$ is an ordered set. | Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ denote the '''[[Definition:Simple Order Product|simple (order) product]]''' of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_... | Let $\struct {T, \preccurlyeq_s} := \struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$.
First we note that from [[Simple Order Product of Pair of Ordered Sets is Ordered Set]], $\struct {T, \preccurlyeq_s}$ is an [[Definition:Ordered Set|ordered set]]. | Simple Order Product of Pair of Ordered Sets is Lattice iff Factors are Lattices | https://proofwiki.org/wiki/Simple_Order_Product_of_Pair_of_Ordered_Sets_is_Lattice_iff_Factors_are_Lattices | https://proofwiki.org/wiki/Simple_Order_Product_of_Pair_of_Ordered_Sets_is_Lattice_iff_Factors_are_Lattices | [
"Simple Order Product",
"Lattice Theory"
] | [
"Definition:Ordered Set",
"Definition:Simple Order Product",
"Definition:Lattice (Ordered Set)",
"Definition:Lattice (Ordered Set)"
] | [
"Simple Order Product of Pair of Ordered Sets is Ordered Set",
"Definition:Ordered Set"
] |
proofwiki-19248 | Simple Order Product of Pair of Totally Ordered Sets is Total iff One Factor is Singleton | Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ denote the '''simple (order) product''' of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$.
Then $\struct {S_1, \preccurlyeq_1} \otimes... | Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\struct {T, \preccurlyeq_s} := \struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$.
First we note that from Simple Order Product of Pair of Ordered Sets is Ordered Set, $\struct {T, \preccurlyeq_s}$ is an or... | Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ denote the '''[[Definition:Simple Order Product|simple (order) product]]''' of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_... | Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $\struct {T, \preccurlyeq_s} := \struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$.
First we note that from [[Simple Order Product of Pair of Ordered Sets is Ordered Set]], $\str... | Simple Order Product of Pair of Totally Ordered Sets is Total iff One Factor is Singleton | https://proofwiki.org/wiki/Simple_Order_Product_of_Pair_of_Totally_Ordered_Sets_is_Total_iff_One_Factor_is_Singleton | https://proofwiki.org/wiki/Simple_Order_Product_of_Pair_of_Totally_Ordered_Sets_is_Total_iff_One_Factor_is_Singleton | [
"Simple Order Product",
"Total Orderings"
] | [
"Definition:Ordered Set",
"Definition:Simple Order Product",
"Definition:Totally Ordered Set",
"Definition:Singleton",
"Definition:Total Ordering"
] | [
"Definition:Ordered Set",
"Simple Order Product of Pair of Ordered Sets is Ordered Set",
"Definition:Ordered Set",
"Ordering on Singleton is Total Ordering",
"Definition:Singleton",
"Definition:Totally Ordered Set",
"Definition:Totally Ordered Set",
"Definition:Singleton",
"Definition:Ordered Set",
... |
proofwiki-19249 | Ordering on Singleton is Total Ordering | Let $S = \set s$ be a singleton.
Let $\RR$ be an ordering on $S$.
Then $\RR$ is a total ordering on $S$. | By definition of ordering, $\RR$ is {{afortiori}} a reflexive relation.
Hence from Reflexive Relation on Singleton is Well-Ordering:
:$\struct {S, \RR}$ is a well-ordered set.
Hence by definition of well-ordered set:
:$\RR$ is a total ordering on $S$.
Hence the result.
{{qed}}
Category:Singletons
Category:Total Orderin... | Let $S = \set s$ be a [[Definition:Singleton|singleton]].
Let $\RR$ be an [[Definition:Ordering|ordering]] on $S$.
Then $\RR$ is a [[Definition:Total Ordering|total ordering]] on $S$. | By definition of [[Definition:Ordering|ordering]], $\RR$ is {{afortiori}} a [[Definition:Reflexive Relation|reflexive relation]].
Hence from [[Reflexive Relation on Singleton is Well-Ordering]]:
:$\struct {S, \RR}$ is a [[Definition:Well-Ordered Set|well-ordered set]].
Hence by definition of [[Definition:Well-Ordered... | Ordering on Singleton is Total Ordering | https://proofwiki.org/wiki/Ordering_on_Singleton_is_Total_Ordering | https://proofwiki.org/wiki/Ordering_on_Singleton_is_Total_Ordering | [
"Singletons",
"Total Orderings"
] | [
"Definition:Singleton",
"Definition:Ordering",
"Definition:Total Ordering"
] | [
"Definition:Ordering",
"Definition:Reflexive Relation",
"Reflexive Relation on Singleton is Well-Ordering",
"Definition:Well-Ordered Set",
"Definition:Well-Ordered Set",
"Definition:Total Ordering",
"Category:Singletons",
"Category:Total Orderings"
] |
proofwiki-19250 | Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice | Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\preccurlyeq_l$ denote the lexicographic order on $S_1 \times S_2$''':
:$\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2} \iff \tuple {x_1 \prec_1 y_1} \lor \tuple {x_1 = y_1 \land x_2 \preccurlyeq_2 y_2}$
Then:
:$\struct {S... | Let $\struct {T, \preccurlyeq_l} := \struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$.
From Lexicographic Order is Ordering we have that $\struct {T, \preccurlyeq_l}$ is an ordered set.
Recall the definition of lexicographic order:
{{:Definition:Lexicographic Order}} | Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $\preccurlyeq_l$ denote the [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]''':
:$\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2} \iff \tuple {x_1 \prec_1 y_1} \lor \tup... | Let $\struct {T, \preccurlyeq_l} := \struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$.
From [[Lexicographic Order is Ordering]] we have that $\struct {T, \preccurlyeq_l}$ is an [[Definition:Ordered Set|ordered set]].
Recall the definition of [[Definition:Lexicographic Order|lexicographic order]]... | Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice | https://proofwiki.org/wiki/Conditions_for_Lexicographic_Order_on_Pair_of_Ordered_Sets_to_be_Lattice | https://proofwiki.org/wiki/Conditions_for_Lexicographic_Order_on_Pair_of_Ordered_Sets_to_be_Lattice | [
"Lexicographic Order",
"Lattice Theory",
"Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice"
] | [
"Definition:Ordered Set",
"Definition:Lexicographic Order",
"Definition:Lattice (Ordered Set)",
"Definition:Lattice (Ordered Set)",
"Definition:Total Ordering",
"Definition:Greatest Element",
"Definition:Smallest Element",
"Definition:Doubleton",
"Definition:Subset",
"Definition:Bounded Above Set/... | [
"Lexicographic Order is Ordering",
"Definition:Ordered Set",
"Definition:Lexicographic Order"
] |
proofwiki-19251 | Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice/Lemma 1 | Let $\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2}$.
Then:
:$x_1 \preccurlyeq_1 y_1$ | Recall the definition of lexicographic order:
{{:Definition:Lexicographic Order}}
Then:
{{begin-eqn}}
{{eqn | l = \tuple {x_1, x_2}
| o = \preccurlyeq_l
| r = \tuple {y_1, y_2}
| c = {{Defof|Upper Bound of Set}}
}}
{{eqn | ll= \leadsto
| l = x_1
| o = \prec_1
| r = y_1
| c = {{... | Let $\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2}$.
Then:
:$x_1 \preccurlyeq_1 y_1$ | Recall the definition of [[Definition:Lexicographic Order|lexicographic order]]:
{{:Definition:Lexicographic Order}}
Then:
{{begin-eqn}}
{{eqn | l = \tuple {x_1, x_2}
| o = \preccurlyeq_l
| r = \tuple {y_1, y_2}
| c = {{Defof|Upper Bound of Set}}
}}
{{eqn | ll= \leadsto
| l = x_1
| o = ... | Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice/Lemma 1 | https://proofwiki.org/wiki/Conditions_for_Lexicographic_Order_on_Pair_of_Ordered_Sets_to_be_Lattice/Lemma_1 | https://proofwiki.org/wiki/Conditions_for_Lexicographic_Order_on_Pair_of_Ordered_Sets_to_be_Lattice/Lemma_1 | [
"Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice"
] | [] | [
"Definition:Lexicographic Order",
"Category:Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice"
] |
proofwiki-19252 | Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice/Lemma 2 | Let $x_1$ and $y_1$ be non-comparable elements in $S_1$:
:$\lnot \paren {x_1 \preccurlyeq_1 y_1}$ and $\lnot \paren {y_1 \preccurlyeq_1 x_1}$
Then $\tuple {x_1, x_2}$ and $\tuple {y_1, y_2}$ are non-comparable elements in $S_1 \times S_2$:
:$\lnot \paren {\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2} }$ and $\lnot... | Suppose $x_1$ and $y_1$ are non-comparable elements in $S_1$:
:$\lnot \paren {x_1 \preccurlyeq_1 y_1}$ and $\lnot \paren {y_1 \preccurlyeq_1 x_1}$
{{AimForCont}} that:
:$\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2}$ or $\tuple {y_1, y_2} \preccurlyeq_l \tuple {x_1, x_2}$
Then from Lemma $1$:
:$x_1 \preccurlyeq_1 ... | Let $x_1$ and $y_1$ be [[Definition:Non-Comparable Elements|non-comparable elements]] in $S_1$:
:$\lnot \paren {x_1 \preccurlyeq_1 y_1}$ and $\lnot \paren {y_1 \preccurlyeq_1 x_1}$
Then $\tuple {x_1, x_2}$ and $\tuple {y_1, y_2}$ are [[Definition:Non-Comparable Elements|non-comparable elements]] in $S_1 \times S_2$:
:... | Suppose $x_1$ and $y_1$ are [[Definition:Non-Comparable Elements|non-comparable elements]] in $S_1$:
:$\lnot \paren {x_1 \preccurlyeq_1 y_1}$ and $\lnot \paren {y_1 \preccurlyeq_1 x_1}$
{{AimForCont}} that:
:$\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2}$ or $\tuple {y_1, y_2} \preccurlyeq_l \tuple {x_1, x_2}$
... | Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice/Lemma 2 | https://proofwiki.org/wiki/Conditions_for_Lexicographic_Order_on_Pair_of_Ordered_Sets_to_be_Lattice/Lemma_2 | https://proofwiki.org/wiki/Conditions_for_Lexicographic_Order_on_Pair_of_Ordered_Sets_to_be_Lattice/Lemma_2 | [
"Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice"
] | [
"Definition:Non-Comparable Elements",
"Definition:Non-Comparable Elements"
] | [
"Definition:Non-Comparable Elements",
"Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice/Lemma 1",
"Definition:Contradiction",
"Proof by Contradiction",
"Category:Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice"
] |
proofwiki-19253 | Ordering is Directed iff Composite with Inverse is Trivial Ordering | Let $\struct {S, \RR}$ be an ordered set.
Then $\RR$ is a '''directed ordering''' {{iff}}:
:$\RR^{-1} \circ \RR = S \times S$
where:
:$\circ$ denotes composite relation
:$\RR^{-1}$ denotes inverse relation
:$S \times S$ denotes the trivial relation, that is, the Cartesian product of $S$ with itself. | We are given that $\RR$ is an ordering. | Let $\struct {S, \RR}$ be an [[Definition:Ordered Set|ordered set]].
Then $\RR$ is a '''[[Definition:Directed Ordering|directed ordering]]''' {{iff}}:
:$\RR^{-1} \circ \RR = S \times S$
where:
:$\circ$ denotes [[Definition:Composite Relation|composite relation]]
:$\RR^{-1}$ denotes [[Definition:Inverse Relation|inve... | We are given that $\RR$ is an [[Definition:Ordering|ordering]]. | Ordering is Directed iff Composite with Inverse is Trivial Ordering | https://proofwiki.org/wiki/Ordering_is_Directed_iff_Composite_with_Inverse_is_Trivial_Ordering | https://proofwiki.org/wiki/Ordering_is_Directed_iff_Composite_with_Inverse_is_Trivial_Ordering | [
"Directed Orderings",
"Composite Relations"
] | [
"Definition:Ordered Set",
"Definition:Directed Ordering",
"Definition:Composition of Relations",
"Definition:Inverse Relation",
"Definition:Trivial Relation",
"Definition:Cartesian Product"
] | [
"Definition:Ordering"
] |
proofwiki-19254 | Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice/Corollary | Let $\struct {S_2, \preccurlyeq_2}$ have neither a greatest element nor a smallest element.
Then:
:$\preccurlyeq_l$ is a lattice ordering
{{iff}}:
:$\preccurlyeq_1$ is a total ordering
and:
:$\preccurlyeq_2$ is a lattice ordering. | === Sufficient Condition ===
Let $\preccurlyeq_l$ be a lattice ordering.
We are given that $\struct {S_2, \preccurlyeq_2}$ has neither a greatest element nor a smallest element.
From Condition $(2)$ of Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice, it follows that $\preccurlyeq_1$ is a total ... | Let $\struct {S_2, \preccurlyeq_2}$ have neither a [[Definition:Greatest Element|greatest element]] nor a [[Definition:Smallest Element|smallest element]].
Then:
:$\preccurlyeq_l$ is a [[Definition:Lattice Ordering|lattice ordering]]
{{iff}}:
:$\preccurlyeq_1$ is a [[Definition:Total Ordering|total ordering]]
and:
:$... | === Sufficient Condition ===
Let $\preccurlyeq_l$ be a [[Definition:Lattice Ordering|lattice ordering]].
We are given that $\struct {S_2, \preccurlyeq_2}$ has neither a [[Definition:Greatest Element|greatest element]] nor a [[Definition:Smallest Element|smallest element]].
From Condition $(2)$ of [[Conditions for Le... | Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice/Corollary | https://proofwiki.org/wiki/Conditions_for_Lexicographic_Order_on_Pair_of_Ordered_Sets_to_be_Lattice/Corollary | https://proofwiki.org/wiki/Conditions_for_Lexicographic_Order_on_Pair_of_Ordered_Sets_to_be_Lattice/Corollary | [
"Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice"
] | [
"Definition:Greatest Element",
"Definition:Smallest Element",
"Definition:Lattice Ordering",
"Definition:Total Ordering",
"Definition:Lattice Ordering"
] | [
"Definition:Lattice Ordering",
"Definition:Greatest Element",
"Definition:Smallest Element",
"Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice",
"Definition:Total Ordering",
"Definition:Smallest Element",
"Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice",... |
proofwiki-19255 | Ordered Set of All Mappings is Lattice iff Codomain is Lattice or Domain is Empty | Let $S$ be a set.
Let $\struct {T, \preccurlyeq}$ be an ordered set.
Let $\struct {T^S, \preccurlyeq}$ denote the ordered set of all mappings from $S$ to $T$.
Then:
:$\struct {T^S, \preccurlyeq}$ is a lattice
{{iff}} either:
:$\struct {T, \preccurlyeq}$ is a lattice
or:
:$S = \O$ | Recall the definition of lattice:
{{:Definition:Lattice (Ordered Set)}}
First a lemma to take care of the case where $S = \O$: | Let $S$ be a [[Definition:Set|set]].
Let $\struct {T, \preccurlyeq}$ be an [[Definition:Ordered Set|ordered set]].
Let $\struct {T^S, \preccurlyeq}$ denote the [[Definition:Ordered Set of All Mappings|ordered set of all mappings]] from $S$ to $T$.
Then:
:$\struct {T^S, \preccurlyeq}$ is a [[Definition:Lattice (Orde... | Recall the definition of [[Definition:Lattice (Ordered Set)|lattice]]:
{{:Definition:Lattice (Ordered Set)}}
First a [[Ordered Set of All Mappings is Lattice iff Codomain is Lattice or Domain is Empty/Lemma|lemma]] to take care of the case where $S = \O$: | Ordered Set of All Mappings is Lattice iff Codomain is Lattice or Domain is Empty | https://proofwiki.org/wiki/Ordered_Set_of_All_Mappings_is_Lattice_iff_Codomain_is_Lattice_or_Domain_is_Empty | https://proofwiki.org/wiki/Ordered_Set_of_All_Mappings_is_Lattice_iff_Codomain_is_Lattice_or_Domain_is_Empty | [
"Ordered Sets",
"Lattice Theory",
"Ordered Set of All Mappings is Lattice iff Codomain is Lattice or Domain is Empty"
] | [
"Definition:Set",
"Definition:Ordered Set",
"Definition:Ordered Set of All Mappings",
"Definition:Lattice (Ordered Set)",
"Definition:Lattice (Ordered Set)"
] | [
"Definition:Lattice (Ordered Set)",
"Ordered Set of All Mappings is Lattice iff Codomain is Lattice or Domain is Empty/Lemma",
"Ordered Set of All Mappings is Lattice iff Codomain is Lattice or Domain is Empty/Lemma",
"Definition:Lattice (Ordered Set)",
"Definition:Lattice (Ordered Set)",
"Definition:Latt... |
proofwiki-19256 | Condition for Ordered Set of All Mappings to be Total Ordering | Let $S$ be a set.
Let $\struct {T, \preccurlyeq}$ be an ordered set.
Let $\struct {T^S, \preccurlyeq}$ denote the ordered set of all mappings from $S$ to $T$.
Then:
:$\struct {T^S, \preccurlyeq}$ is a totally ordered set
{{iff}}:
:$\card S \le 1$
or:
:$\card T \le 1$ | === Sufficient Condition ===
Let $\struct {T^S, \preccurlyeq}$ be a totally ordered set.
Let $f, g \in T^S$ be arbitrary.
Let $S$ and $T$ be such that:
{{begin-eqn}}
{{eqn | q = \exists a, b \in S
| l = a
| o = \ne
| r = b
}}
{{eqn | q = \exists x, y \in T
| l = x
| o = \ne
| r = y
... | Let $S$ be a [[Definition:Set|set]].
Let $\struct {T, \preccurlyeq}$ be an [[Definition:Ordered Set|ordered set]].
Let $\struct {T^S, \preccurlyeq}$ denote the [[Definition:Ordered Set of All Mappings|ordered set of all mappings]] from $S$ to $T$.
Then:
:$\struct {T^S, \preccurlyeq}$ is a [[Definition:Totally Order... | === Sufficient Condition ===
Let $\struct {T^S, \preccurlyeq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $f, g \in T^S$ be arbitrary.
Let $S$ and $T$ be such that:
{{begin-eqn}}
{{eqn | q = \exists a, b \in S
| l = a
| o = \ne
| r = b
}}
{{eqn | q = \exists x, y \in T
... | Condition for Ordered Set of All Mappings to be Total Ordering | https://proofwiki.org/wiki/Condition_for_Ordered_Set_of_All_Mappings_to_be_Total_Ordering | https://proofwiki.org/wiki/Condition_for_Ordered_Set_of_All_Mappings_to_be_Total_Ordering | [
"Ordered Sets",
"Total Orderings"
] | [
"Definition:Set",
"Definition:Ordered Set",
"Definition:Ordered Set of All Mappings",
"Definition:Totally Ordered Set"
] | [
"Definition:Totally Ordered Set",
"Definition:Totally Ordered Set",
"Definition:Antisymmetric Relation",
"Definition:Contradiction",
"Definition:Ordered Set",
"Definition:Totally Ordered Set",
"Rule of Transposition"
] |
proofwiki-19257 | Natural Numbers with Divisor Operation is Isomorphic to Subgroups of Integer Multiples under Inclusion | Let $\N_{>0}$ denote the set of strictly positive natural numbers.
For $n \in \N_{>0}$, let $n \Z$ denote the set of integer multiples of $n$.
Let $\struct {\Z, +}$ denote the additive group of integers.
Let $\mathscr G$ be the set of all subgroups of $\struct {\Z, +}$.
Consider the algebraic structure $\struct {\N_{>0... | We note that from Subgroups of Additive Group of Integers, the subgroups of $\struct {\Z, +}$ are precisely the sets of integer multiples $n \Z$, for $n \in \N_{>0}$.
For each $n \in \N_{>0}$, there is a unique $n \Z \in \mathscr G$.
Hence $\phi$ is a bijection.
It remains to be demonstrated that $\phi$ is order-preser... | Let $\N_{>0}$ denote the [[Definition:Set|set]] of [[Definition:Strictly Positive Integer|strictly positive]] [[Definition:Natural Numbers|natural numbers]].
For $n \in \N_{>0}$, let $n \Z$ denote the [[Definition:Set of Integer Multiples|set of integer multiples]] of $n$.
Let $\struct {\Z, +}$ denote the [[Definitio... | We note that from [[Subgroups of Additive Group of Integers]], the [[Definition:Subgroup|subgroups]] of $\struct {\Z, +}$ are precisely the [[Definition:Set of Integer Multiples|sets of integer multiples]] $n \Z$, for $n \in \N_{>0}$.
For each $n \in \N_{>0}$, there is a [[Definition:Unique|unique]] $n \Z \in \mathscr... | Natural Numbers with Divisor Operation is Isomorphic to Subgroups of Integer Multiples under Inclusion | https://proofwiki.org/wiki/Natural_Numbers_with_Divisor_Operation_is_Isomorphic_to_Subgroups_of_Integer_Multiples_under_Inclusion | https://proofwiki.org/wiki/Natural_Numbers_with_Divisor_Operation_is_Isomorphic_to_Subgroups_of_Integer_Multiples_under_Inclusion | [
"Natural Numbers",
"Additive Groups of Integer Multiples",
"Divisors",
"Examples of Order Isomorphisms"
] | [
"Definition:Set",
"Definition:Strictly Positive/Integer",
"Definition:Natural Numbers",
"Definition:Set of Integer Multiples",
"Definition:Additive Group of Integers",
"Definition:Set",
"Definition:Subgroup",
"Definition:Algebraic Structure",
"Definition:Divisor (Algebra)/Integer",
"Definition:Div... | [
"Subgroups of Additive Group of Integers",
"Definition:Subgroup",
"Definition:Set of Integer Multiples",
"Definition:Unique",
"Definition:Bijection",
"Definition:Increasing/Mapping"
] |
proofwiki-19258 | Subset of Join Semilattice on Total Ordering is Closed | Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $\struct {S, \vee, \preccurlyeq}$ be the join semilattice with respect to $\preccurlyeq$.
Let $T \subseteq S$ be an arbitrary subset of $S$.
Then $\struct {T, \vee, \preccurlyeq}$ is closed under $\vee$. | Let $x \in T$.
Then from Supremum of Singleton:
:$\sup \set x = x$
That is:
:$x \vee x = x$
and so:
:$x \vee x \in T$
Let $x, y \in T$.
From Supremum of Doubleton in Totally Ordered Set:
:$\sup \set {x, y} \in \set {x, y}$
That is:
:$x \vee y \in \set {x, y}$
But:
:$\set {x, y} \subseteq T$
Hence by definition of subse... | Let $\struct {S, \preccurlyeq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $\struct {S, \vee, \preccurlyeq}$ be the [[Definition:Join Semilattice|join semilattice]] with respect to $\preccurlyeq$.
Let $T \subseteq S$ be an arbitrary [[Definition:Subset|subset]] of $S$.
Then $\struct {T, \vee, ... | Let $x \in T$.
Then from [[Supremum of Singleton]]:
:$\sup \set x = x$
That is:
:$x \vee x = x$
and so:
:$x \vee x \in T$
Let $x, y \in T$.
From [[Supremum of Doubleton in Totally Ordered Set]]:
:$\sup \set {x, y} \in \set {x, y}$
That is:
:$x \vee y \in \set {x, y}$
But:
:$\set {x, y} \subseteq T$
Hence by def... | Subset of Join Semilattice on Total Ordering is Closed | https://proofwiki.org/wiki/Subset_of_Join_Semilattice_on_Total_Ordering_is_Closed | https://proofwiki.org/wiki/Subset_of_Join_Semilattice_on_Total_Ordering_is_Closed | [
"Join Semilattices",
"Total Orderings"
] | [
"Definition:Totally Ordered Set",
"Definition:Join Semilattice",
"Definition:Subset",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] | [
"Supremum of Singleton",
"Supremum of Doubleton in Totally Ordered Set",
"Definition:Subset",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] |
proofwiki-19259 | Supremum of Doubleton in Totally Ordered Set | Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $D = \set {a, b} \subseteq S$ be an arbitrary doubleton subset of $S$.
Then:
:$\map \sup D \in D$
where $\sup D$ denotes the supremum of $D$. | Let $D = \set {a, b} \subseteq S$ as asserted.
As $S$ is a totally ordered set, either $a \preccurlyeq b$ or $b \preccurlyeq a$.
{{WLOG}} let $a \preccurlyeq b$.
We have by definition of supremum that:
:$\forall x \in D: x \preccurlyeq \map \sup D$
:$\forall y \in S:$ if $y$ is an upper bound of $D$, then $\map \sup D ... | Let $\struct {S, \preccurlyeq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $D = \set {a, b} \subseteq S$ be an arbitrary [[Definition:Doubleton|doubleton]] [[Definition:Subset|subset]] of $S$.
Then:
:$\map \sup D \in D$
where $\sup D$ denotes the [[Definition:Supremum of Set|supremum]] of $D$. | Let $D = \set {a, b} \subseteq S$ as asserted.
As $S$ is a [[Definition:Totally Ordered Set|totally ordered set]], either $a \preccurlyeq b$ or $b \preccurlyeq a$.
{{WLOG}} let $a \preccurlyeq b$.
We have by definition of [[Definition:Supremum of Set|supremum]] that:
:$\forall x \in D: x \preccurlyeq \map \sup D$
:$... | Supremum of Doubleton in Totally Ordered Set | https://proofwiki.org/wiki/Supremum_of_Doubleton_in_Totally_Ordered_Set | https://proofwiki.org/wiki/Supremum_of_Doubleton_in_Totally_Ordered_Set | [
"Suprema",
"Doubletons"
] | [
"Definition:Totally Ordered Set",
"Definition:Doubleton",
"Definition:Subset",
"Definition:Supremum of Set"
] | [
"Definition:Totally Ordered Set",
"Definition:Supremum of Set",
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set",
"Category:Suprema",
"Category:Doubletons"
] |
proofwiki-19260 | Total Semilattice has Unique Total Ordering | Let $\struct {S, \vee}$ be a total semilattice.
There exists a unique total ordering $\preccurlyeq$ on $S$ such that:
:$x \vee y = \max \set {x, y}$
where:
:$\max \set {x, y} := \begin {cases} b & : a \preccurlyeq b \\ a & : b \preccurlyeq a \end {cases}$ | Recall the definition of total semilattice:
A '''total semilattice''' $\struct {S, \odot}$ is a '''semilattice''' which has the property that every subset of $\struct {S, \odot}$ is a '''subsemilattice'''.
That is, such that every subset of $\struct {S, \odot}$ is closed under $\odot$.
Let us define a relation $\RR$ on... | Let $\struct {S, \vee}$ be a [[Definition:Total Semilattice|total semilattice]].
There exists a [[Definition:Unique|unique]] [[Definition:Total Ordering|total ordering]] $\preccurlyeq$ on $S$ such that:
:$x \vee y = \max \set {x, y}$
where:
:$\max \set {x, y} := \begin {cases} b & : a \preccurlyeq b \\ a & : b \precc... | Recall the definition of [[Definition:Total Semilattice|total semilattice]]:
A '''[[Definition:Total Semilattice|total semilattice]]''' $\struct {S, \odot}$ is a '''[[Definition:Semilattice|semilattice]]''' which has the property that every [[Definition:Subset|subset]] of $\struct {S, \odot}$ is a '''[[Definition:Subs... | Total Semilattice has Unique Total Ordering | https://proofwiki.org/wiki/Total_Semilattice_has_Unique_Total_Ordering | https://proofwiki.org/wiki/Total_Semilattice_has_Unique_Total_Ordering | [
"Total Semilattices",
"Total Orderings"
] | [
"Definition:Total Semilattice",
"Definition:Unique",
"Definition:Total Ordering"
] | [
"Definition:Total Semilattice",
"Definition:Total Semilattice",
"Definition:Semilattice",
"Definition:Subset",
"Definition:Subsemilattice",
"Definition:Subset",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Relation",
"Definition:Total Ordering",
"Semilattice Induces Ord... |
proofwiki-19261 | Semilattice has Unique Ordering such that Operation is Supremum | Let $\struct {S, \vee}$ be a semilattice.
Then there exists a unique ordering $\preccurlyeq$ on $S$ such that:
:$x \vee y = \sup \set {x, y}$
where $\sup \set {x, y}$ is the supremum of $\set {x, y}$ {{WRT}} $\preccurlyeq$. | Let us define a relation $\RR$ on $S$ as:
:$\forall x, y \in S: x \mathrel \RR y \iff x \vee y = y$
From Semilattice Induces Ordering, we have that $\RR$ is an ordering.
Let us denote this ordering by $\preccurlyeq$, and recall its definition:
:$x \vee y = y \iff x \preccurlyeq y$ | Let $\struct {S, \vee}$ be a [[Definition:Semilattice|semilattice]].
Then there exists a [[Definition:Unique|unique]] [[Definition:Ordering|ordering]] $\preccurlyeq$ on $S$ such that:
:$x \vee y = \sup \set {x, y}$
where $\sup \set {x, y}$ is the [[Definition:Supremum of Set|supremum]] of $\set {x, y}$ {{WRT}} $\prec... | Let us define a [[Definition:Relation|relation]] $\RR$ on $S$ as:
:$\forall x, y \in S: x \mathrel \RR y \iff x \vee y = y$
From [[Semilattice Induces Ordering]], we have that $\RR$ is an [[Definition:Ordering|ordering]].
Let us denote this [[Definition:Ordering|ordering]] by $\preccurlyeq$, and recall its definiti... | Semilattice has Unique Ordering such that Operation is Supremum | https://proofwiki.org/wiki/Semilattice_has_Unique_Ordering_such_that_Operation_is_Supremum | https://proofwiki.org/wiki/Semilattice_has_Unique_Ordering_such_that_Operation_is_Supremum | [
"Semilattices",
"Suprema",
"Orderings"
] | [
"Definition:Semilattice",
"Definition:Unique",
"Definition:Ordering",
"Definition:Supremum of Set"
] | [
"Definition:Relation",
"Semilattice Induces Ordering",
"Definition:Ordering",
"Definition:Ordering",
"Definition:Ordering",
"Definition:Ordering"
] |
proofwiki-19262 | Max Operation is Idempotent | The max operation operation is idempotent:
:$\map \max {x, x} = x$ | Follows immediately from the definition of max operation:
:$\map \max {a, b} = \begin {cases} b & : a \le b \\ a & : b \le a \end {cases}$
Setting $x = a = b$ returns the result.
{{Qed}} | The [[Definition:Max Operation|max operation]] operation is [[Definition:Idempotent Operation|idempotent]]:
:$\map \max {x, x} = x$ | Follows immediately from the definition of [[Definition:Max Operation|max operation]]:
:$\map \max {a, b} = \begin {cases} b & : a \le b \\ a & : b \le a \end {cases}$
Setting $x = a = b$ returns the result.
{{Qed}} | Max Operation is Idempotent | https://proofwiki.org/wiki/Max_Operation_is_Idempotent | https://proofwiki.org/wiki/Max_Operation_is_Idempotent | [
"Max Operation",
"Examples of Idempotence"
] | [
"Definition:Max Operation",
"Definition:Idempotence/Operation"
] | [
"Definition:Max Operation"
] |
proofwiki-19263 | Natural Numbers under Min Operation forms Total Semilattice | Let $\struct {\N, \wedge}$ denote the set of natural numbers $\N$ under the min operation:
:$\forall a, b \in \N: a \wedge b := \min \set {a, b}$
Then $\struct {\N, \wedge}$ forms a total semilattice. | Taking the semilattice axioms in turn: | Let $\struct {\N, \wedge}$ denote the [[Definition:Natural Numbers|set of natural numbers]] $\N$ under the [[Definition:Min Operation|min operation]]:
:$\forall a, b \in \N: a \wedge b := \min \set {a, b}$
Then $\struct {\N, \wedge}$ forms a [[Definition:Total Semilattice|total semilattice]]. | Taking the [[Axiom:Semilattice Axioms|semilattice axioms]] in turn: | Natural Numbers under Min Operation forms Total Semilattice | https://proofwiki.org/wiki/Natural_Numbers_under_Min_Operation_forms_Total_Semilattice | https://proofwiki.org/wiki/Natural_Numbers_under_Min_Operation_forms_Total_Semilattice | [
"Total Semilattices",
"Natural Numbers",
"Min Operation"
] | [
"Definition:Natural Numbers",
"Definition:Min Operation",
"Definition:Total Semilattice"
] | [
"Axiom:Semilattice Axioms",
"Axiom:Semilattice Axioms"
] |
proofwiki-19264 | Structure with Commutative Idempotent Associative Operations satisfying Absorption Laws is Lattice | Let $S$ be a set.
Let $\vee$ and $\wedge$ be binary operations which, when applied to $S$, are both:
:closed operations
:commutative operations
:idempotent operations
:associative operations.
Furthermore, let $\vee$ and $\wedge$ satisfy the absorption laws:
:$\forall a, b \in S: a \vee \paren {a \wedge b} = a = a \wedg... | We have {{hypothesis}} that:
:$\struct {S, \vee}$ is a commutative idempotent semigroup
:$\struct {S, \wedge}$ is a commutative idempotent semigroup.
That is, $\struct {S, \vee}$ and $\struct {S, \wedge}$ are semilattices.
We are also given that $\vee$ and $\wedge$ satisfy the absorption laws:
:$\forall a, b \in S: a ... | Let $S$ be a [[Definition:Set|set]].
Let $\vee$ and $\wedge$ be [[Definition:Binary Operation|binary operations]] which, when applied to $S$, are both:
:[[Definition:Closed Operation|closed operations]]
:[[Definition:Commutative Operation|commutative operations]]
:[[Definition:Idempotent Operation|idempotent operatio... | We have {{hypothesis}} that:
:$\struct {S, \vee}$ is a [[Definition:Commutative Semigroup|commutative]] [[Definition:Idempotent Semigroup|idempotent semigroup]]
:$\struct {S, \wedge}$ is a [[Definition:Commutative Semigroup|commutative]] [[Definition:Idempotent Semigroup|idempotent semigroup]].
That is, $\struct {S... | Structure with Commutative Idempotent Associative Operations satisfying Absorption Laws is Lattice | https://proofwiki.org/wiki/Structure_with_Commutative_Idempotent_Associative_Operations_satisfying_Absorption_Laws_is_Lattice | https://proofwiki.org/wiki/Structure_with_Commutative_Idempotent_Associative_Operations_satisfying_Absorption_Laws_is_Lattice | [
"Lattice Theory"
] | [
"Definition:Set",
"Definition:Operation/Binary Operation",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Commutative/Operation",
"Definition:Idempotence/Operation",
"Definition:Associative Operation",
"Definition:Absorption Law",
"Definition:Unique",
"Definition:Lattice Or... | [
"Definition:Commutative Semigroup",
"Definition:Idempotent Semigroup",
"Definition:Commutative Semigroup",
"Definition:Idempotent Semigroup",
"Definition:Semilattice",
"Definition:Absorption Law",
"Semilattice has Unique Ordering such that Operation is Supremum",
"Definition:Unique",
"Definition:Ord... |
proofwiki-19265 | Join Operation on Ordered Set such that Every Doubleton admits Supremum is Entropic | Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $\struct {S, \preccurlyeq}$ be such that every doubleton subset of $S$ admits a supremum.
Let $\vee$ be the join operation on $S$, defined as:
:$\forall a, b \in S: a \vee b = \sup_\preccurlyeq \set {a, b}$
Then $\vee$ is an entropic operation. | By definition, $\struct {S, \vee, \preccurlyeq}$ is a join semilattice.
From Join Semilattice is Semilattice, $\struct {S, \vee, \preccurlyeq}$ is indeed a semilattice.
Then:
{{begin-eqn}}
{{eqn | q = \forall a, b, c, d, \in S
| l = \paren {a \vee b} \vee \paren {c \vee d}
| r = a \vee \paren {\paren {b \v... | Let $\struct {S, \preccurlyeq}$ be an [[Definition:Ordered Set|ordered set]].
Let $\struct {S, \preccurlyeq}$ be such that every [[Definition:Doubleton|doubleton]] [[Definition:Subset|subset]] of $S$ admits a [[Definition:Supremum of Set|supremum]].
Let $\vee$ be the [[Definition:Join (Order Theory)|join operation]]... | By definition, $\struct {S, \vee, \preccurlyeq}$ is a [[Definition:Join Semilattice|join semilattice]].
From [[Join Semilattice is Semilattice]], $\struct {S, \vee, \preccurlyeq}$ is indeed a [[Definition:Semilattice|semilattice]].
Then:
{{begin-eqn}}
{{eqn | q = \forall a, b, c, d, \in S
| l = \paren {a \ve... | Join Operation on Ordered Set such that Every Doubleton admits Supremum is Entropic | https://proofwiki.org/wiki/Join_Operation_on_Ordered_Set_such_that_Every_Doubleton_admits_Supremum_is_Entropic | https://proofwiki.org/wiki/Join_Operation_on_Ordered_Set_such_that_Every_Doubleton_admits_Supremum_is_Entropic | [
"Entropic Operations",
"Ordered Structures",
"Join Operation"
] | [
"Definition:Ordered Set",
"Definition:Doubleton",
"Definition:Subset",
"Definition:Supremum of Set",
"Definition:Join (Order Theory)",
"Definition:Entropic Operation"
] | [
"Definition:Join Semilattice",
"Join Semilattice is Semilattice",
"Definition:Semilattice"
] |
proofwiki-19266 | Set of Closed Elements wrt Closure Operator under Subset Operation is Complete Lattice | Let $S$ be a set.
Let $\cl$ be a closure operator on the power set $\powerset S$ of $S$.
Let $\mathscr C$ be the set of all subsets $T$ of $S$ such that:
:$\map \cl T = T$
Then the algebraic structure $\struct {\mathscr C, \subseteq}$ forms a complete lattice. | Recall the closure axioms:
{{:Axiom:Closure Axioms for Power Set}}
First we note that from {{Closure-axiom-powerset|1}} we have that:
:$\map \cl S = S$
and so:
:$S \in \mathscr C$
Let $\AA \subseteq \mathscr C$.
Thus $\AA$ is a set of subsets $T$ of $S$ for all of which $\map \cl T = T$.
From Intersection of Closed Set... | Let $S$ be a [[Definition:Set|set]].
Let $\cl$ be a [[Definition:Closure Operator|closure operator]] on the [[Definition:Power Set|power set]] $\powerset S$ of $S$.
Let $\mathscr C$ be the [[Definition:Set of Sets|set]] of all [[Definition:Subset|subsets]] $T$ of $S$ such that:
:$\map \cl T = T$
Then the [[Definitio... | Recall the [[Axiom:Closure Axioms for Power Set|closure axioms]]:
{{:Axiom:Closure Axioms for Power Set}}
First we note that from {{Closure-axiom-powerset|1}} we have that:
:$\map \cl S = S$
and so:
:$S \in \mathscr C$
Let $\AA \subseteq \mathscr C$.
Thus $\AA$ is a [[Definition:Set of Sets|set]] of [[Definition:Sub... | Set of Closed Elements wrt Closure Operator under Subset Operation is Complete Lattice | https://proofwiki.org/wiki/Set_of_Closed_Elements_wrt_Closure_Operator_under_Subset_Operation_is_Complete_Lattice | https://proofwiki.org/wiki/Set_of_Closed_Elements_wrt_Closure_Operator_under_Subset_Operation_is_Complete_Lattice | [
"Examples of Complete Lattices",
"Closure Operators"
] | [
"Definition:Set",
"Definition:Closure Operator",
"Definition:Power Set",
"Definition:Set of Sets",
"Definition:Subset",
"Definition:Algebraic Structure",
"Definition:Complete Lattice"
] | [
"Axiom:Closure Axioms/Power Set",
"Definition:Set of Sets",
"Definition:Subset",
"Intersection of Closed Sets is Closed/Closure Operator",
"Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice"
] |
proofwiki-19267 | Inverse Relational Structures of Isomorphic Structures are Isomorphic | Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures.
Let ${\RR_1}^{-1}$ and ${\RR_2}^{-1}$ be the inverses of $\RR_1$ and $\RR_2$ respectively.
Let $f: \struct {S, \RR_1} \to \struct {T, \RR_2}$ be a relation isomorphism.
Then $f: \struct {S, {\RR_1}^{-1} } \to \struct {T, {\RR_2}^{-1} } $ is als... | {{begin-eqn}}
{{eqn | q =\forall x, y \in S
| l = x
| o = {\RR_1}^{-1}
| r = y
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = y
| o = \RR_1
| r = x
| c = {{Defof|Inverse Relation}}
}}
{{eqn | ll= \leadstoandfrom
| l = \map f y
| o = \RR_2
| r = \map f x
... | Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be [[Definition:Relational Structure|relational structures]].
Let ${\RR_1}^{-1}$ and ${\RR_2}^{-1}$ be the [[Definition:Inverse Relation|inverses]] of $\RR_1$ and $\RR_2$ respectively.
Let $f: \struct {S, \RR_1} \to \struct {T, \RR_2}$ be a [[Definition:Relation Isom... | {{begin-eqn}}
{{eqn | q =\forall x, y \in S
| l = x
| o = {\RR_1}^{-1}
| r = y
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = y
| o = \RR_1
| r = x
| c = {{Defof|Inverse Relation}}
}}
{{eqn | ll= \leadstoandfrom
| l = \map f y
| o = \RR_2
| r = \map f x
... | Inverse Relational Structures of Isomorphic Structures are Isomorphic | https://proofwiki.org/wiki/Inverse_Relational_Structures_of_Isomorphic_Structures_are_Isomorphic | https://proofwiki.org/wiki/Inverse_Relational_Structures_of_Isomorphic_Structures_are_Isomorphic | [
"Relation Isomorphisms",
"Inverse Relations"
] | [
"Definition:Relational Structure",
"Definition:Inverse Relation",
"Definition:Relation Isomorphism",
"Definition:Relation Isomorphism"
] | [] |
proofwiki-19268 | Characterisation of Terminal P-adic Expansion/Necessary Condition | Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $x \in \Q_p$.
Let the $p$-adic expansion of $x$ terminate.
Then:
:$\exists a \in \N : \exists k \in \Z : x = \dfrac a {p^k}$ | Let the $p$-adic expansion of $x$ be:
:$x = \ds \sum_{n \mathop = m}^\infty d_n p^n$
where:
:$m \in \Z_{\le 0}$
:$\forall n \in \Z_{\ge m}: d_n$ is a $p$-adic digit
:$m < 0 \implies d_m \ne 0$
By the definition of terminates:
:$\exists n_0 \in \N : n_0 \ge m : \forall n \ge n_0 : d_n = 0$
We have:
{{begin-eqn}}
{{eqn ... | Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the [[Definition:Valued Field of P-adic Numbers|$p$-adic numbers]] for some [[Definition:Prime Number|prime]] $p$.
Let $x \in \Q_p$.
Let the [[Definition:P-adic Expansion|$p$-adic expansion]] of $x$ [[Definition:Terminal P-adic Expansion|terminate]].
Then:
:$\exists a \i... | Let the [[Definition:P-adic Expansion|$p$-adic expansion]] of $x$ be:
:$x = \ds \sum_{n \mathop = m}^\infty d_n p^n$
where:
:$m \in \Z_{\le 0}$
:$\forall n \in \Z_{\ge m}: d_n$ is a [[Definition:P-adic Digit|$p$-adic digit]]
:$m < 0 \implies d_m \ne 0$
By the definition of [[Definition:Terminal P-adic Expansion|term... | Characterisation of Terminal P-adic Expansion/Necessary Condition | https://proofwiki.org/wiki/Characterisation_of_Terminal_P-adic_Expansion/Necessary_Condition | https://proofwiki.org/wiki/Characterisation_of_Terminal_P-adic_Expansion/Necessary_Condition | [
"Characterisation of Terminal P-adic Expansion"
] | [
"Definition:Valued Field of P-adic Numbers",
"Definition:Prime Number",
"Definition:P-adic Expansion",
"Definition:Terminal P-adic Expansion"
] | [
"Definition:P-adic Expansion",
"Definition:P-adic Digit",
"Definition:Terminal P-adic Expansion"
] |
proofwiki-19269 | Characterisation of Terminal P-adic Expansion/Sufficient Condition | Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $a \in \N$.
Let $k \in \Z$.
Let $x = \dfrac a {p^k}$.
Then:
:the $p$-adic expansion of $x$ terminates | From Basis Representation Theorem, $a$ can be expressed uniquely in the form:
:$\ds a = \sum_{j \mathop = 0}^n d_j p^j$
where:
:$n$ is such that $p^n \le a < p^{n + 1}$
:all the $d_j$ are such that $0 \le d_j < p$.
We have:
{{begin-eqn}}
{{eqn | l = x
| r = \dfrac a {p^k}
| c = Hypothesis
}}
{{eqn | r = \df... | Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the [[Definition:Valued Field of P-adic Numbers|$p$-adic numbers]] for some [[Definition:Prime Number|prime]] $p$.
Let $a \in \N$.
Let $k \in \Z$.
Let $x = \dfrac a {p^k}$.
Then:
:the [[Definition:P-adic Expansion|$p$-adic expansion]] of $x$ [[Definition:Terminal P-adic... | From [[Basis Representation Theorem]], $a$ can be expressed uniquely in the form:
:$\ds a = \sum_{j \mathop = 0}^n d_j p^j$
where:
:$n$ is such that $p^n \le a < p^{n + 1}$
:all the $d_j$ are such that $0 \le d_j < p$.
We have:
{{begin-eqn}}
{{eqn | l = x
| r = \dfrac a {p^k}
| c = Hypothesis
}}
{{eqn |... | Characterisation of Terminal P-adic Expansion/Sufficient Condition | https://proofwiki.org/wiki/Characterisation_of_Terminal_P-adic_Expansion/Sufficient_Condition | https://proofwiki.org/wiki/Characterisation_of_Terminal_P-adic_Expansion/Sufficient_Condition | [
"Characterisation of Terminal P-adic Expansion"
] | [
"Definition:Valued Field of P-adic Numbers",
"Definition:Prime Number",
"Definition:P-adic Expansion",
"Definition:Terminal P-adic Expansion"
] | [
"Basis Representation Theorem",
"Definition:Number Base",
"Definition:Terminal P-adic Expansion",
"P-adic Expansion Representative of P-adic Number is Unique",
"Definition:P-adic Expansion"
] |
proofwiki-19270 | Intersection of Antisymmetric Relations is Antisymmetric | The intersection of two antisymmetric relations is also an antisymmetric relation. | Let $\RR_1$ and $\RR_2$ be antisymmetric relations on a set $S$.
Let $\RR_3 = \RR_1 \cap \RR_2$.
Hence we have:
{{begin-eqn}}
{{eqn | q = \forall \tuple {x, y} \in \RR_3
| l = \tuple {y, x}
| o = \in
| r = \RR_3
| c =
}}
{{eqn | q = \forall \tuple {x, y} \in \RR_3
| l = \tuple {x, y}
... | The [[Definition:Set Intersection|intersection]] of two [[Definition:Antisymmetric Relation|antisymmetric relations]] is also an [[Definition:Antisymmetric Relation|antisymmetric relation]]. | Let $\RR_1$ and $\RR_2$ be [[Definition:Antisymmetric Relation|antisymmetric relations]] on a set $S$.
Let $\RR_3 = \RR_1 \cap \RR_2$.
Hence we have:
{{begin-eqn}}
{{eqn | q = \forall \tuple {x, y} \in \RR_3
| l = \tuple {y, x}
| o = \in
| r = \RR_3
| c =
}}
{{eqn | q = \forall \tuple {x, y} ... | Intersection of Antisymmetric Relations is Antisymmetric | https://proofwiki.org/wiki/Intersection_of_Antisymmetric_Relations_is_Antisymmetric | https://proofwiki.org/wiki/Intersection_of_Antisymmetric_Relations_is_Antisymmetric | [
"Antisymmetric Relations",
"Set Intersection"
] | [
"Definition:Set Intersection",
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation"
] | [
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation",
"Category:Antisymmetric Relations",
"Category:Set Intersection"
] |
proofwiki-19271 | Intersection of Orderings is Ordering | Let $A$ be a set.
Let $\RR$ and $\SS$ be orderings on $A$.
Then $\RR \cap \SS$ is an ordering on $A$. | By definition of ordering:
:$\RR$ and $\SS$ are reflexive
:$\RR$ and $\SS$ are transitive
:$\RR$ and $\SS$ are antisymmetric.
We have:
:Intersection of Reflexive Relations is Reflexive
:Intersection of Transitive Relations is Transitive
:Intersection of Antisymmetric Relations is Antisymmetric
and the result follows.
{... | Let $A$ be a [[Definition:Set|set]].
Let $\RR$ and $\SS$ be [[Definition:Ordering|orderings]] on $A$.
Then $\RR \cap \SS$ is an [[Definition:Ordering|ordering]] on $A$. | By definition of [[Definition:Ordering|ordering]]:
:$\RR$ and $\SS$ are [[Definition:Reflexive Relation|reflexive]]
:$\RR$ and $\SS$ are [[Definition:Transitive Relation|transitive]]
:$\RR$ and $\SS$ are [[Definition:Antisymmetric Relation|antisymmetric]].
We have:
:[[Intersection of Reflexive Relations is Reflexi... | Intersection of Orderings is Ordering | https://proofwiki.org/wiki/Intersection_of_Orderings_is_Ordering | https://proofwiki.org/wiki/Intersection_of_Orderings_is_Ordering | [
"Orderings",
"Set Intersection"
] | [
"Definition:Set",
"Definition:Ordering",
"Definition:Ordering"
] | [
"Definition:Ordering",
"Definition:Reflexive Relation",
"Definition:Transitive Relation",
"Definition:Antisymmetric Relation",
"Intersection of Reflexive Relations is Reflexive",
"Intersection of Transitive Relations is Transitive",
"Intersection of Antisymmetric Relations is Antisymmetric"
] |
proofwiki-19272 | Inverse of Diagonal Relation | Let $S$ be a set.
Let $\Delta_S$ denote the diagonal relation on $S$.
Let ${\Delta_S}^{-1}$ denote the inverse of $\Delta_S$.
Then:
:${\Delta_S}^{-1} = \Delta_S$ | By definition of diagonal relation:
:$\Delta_S = \set {\tuple {x, x} \in S \times S: x \in S}$
By definition of inverse relation:
:${\Delta_S}^{-1} = \set {\tuple {x, x} \in S \times S: x \in S}$
Hence it follows that:
:$\tuple {x, x} \in \Delta_S \iff \tuple {x, x} \in {\Delta_S}^{-1}$
{{qed}}
Category:Diagonal Relati... | Let $S$ be a [[Definition:Set|set]].
Let $\Delta_S$ denote the [[Definition:Diagonal Relation|diagonal relation]] on $S$.
Let ${\Delta_S}^{-1}$ denote the [[Definition:Inverse Relation|inverse]] of $\Delta_S$.
Then:
:${\Delta_S}^{-1} = \Delta_S$ | By definition of [[Definition:Diagonal Relation|diagonal relation]]:
:$\Delta_S = \set {\tuple {x, x} \in S \times S: x \in S}$
By definition of [[Definition:Inverse Relation|inverse relation]]:
:${\Delta_S}^{-1} = \set {\tuple {x, x} \in S \times S: x \in S}$
Hence it follows that:
:$\tuple {x, x} \in \Delta_S \iff ... | Inverse of Diagonal Relation | https://proofwiki.org/wiki/Inverse_of_Diagonal_Relation | https://proofwiki.org/wiki/Inverse_of_Diagonal_Relation | [
"Diagonal Relation",
"Inverse Relations"
] | [
"Definition:Set",
"Definition:Diagonal Relation",
"Definition:Inverse Relation"
] | [
"Definition:Diagonal Relation",
"Definition:Inverse Relation",
"Category:Diagonal Relation",
"Category:Inverse Relations"
] |
proofwiki-19273 | Union of Antisymmetric Relation with Inverse is Antisymmetric iff Diagonal | Let $\RR$ be an antisymmetric relation on a set $S$.
Then:
:$\RR \cup \RR^{-1}$ is antisymmetric
{{iff}}
:$\RR = \Delta_S$
where:
:$\RR^{-1}$ denotes the inverse of $\RR$
:$\Delta_S$ denotes the diagonal relation
:$\cup$ denotes set union. | As asserted, let $\RR$ be an antisymmetric relation. | Let $\RR$ be an [[Definition:Antisymmetric Relation|antisymmetric relation]] on a [[Definition:Set|set]] $S$.
Then:
:$\RR \cup \RR^{-1}$ is [[Definition:Antisymmetric Relation|antisymmetric]]
{{iff}}
:$\RR = \Delta_S$
where:
:$\RR^{-1}$ denotes the [[Definition:Inverse Relation|inverse]] of $\RR$
:$\Delta_S$ denotes ... | As asserted, let $\RR$ be an [[Definition:Antisymmetric Relation|antisymmetric relation]]. | Union of Antisymmetric Relation with Inverse is Antisymmetric iff Diagonal | https://proofwiki.org/wiki/Union_of_Antisymmetric_Relation_with_Inverse_is_Antisymmetric_iff_Diagonal | https://proofwiki.org/wiki/Union_of_Antisymmetric_Relation_with_Inverse_is_Antisymmetric_iff_Diagonal | [
"Diagonal Relation",
"Antisymmetric Relations",
"Set Union"
] | [
"Definition:Antisymmetric Relation",
"Definition:Set",
"Definition:Antisymmetric Relation",
"Definition:Inverse Relation",
"Definition:Diagonal Relation",
"Definition:Set Union"
] | [
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation"
] |
proofwiki-19274 | Union of Orderings is not necessarily Ordering | Let $A$ be a set.
Let $\RR$ and $\SS$ be orderings on $A$.
Then $\RR \cup \SS$ is not necessarily an ordering on $A$. | Let $\RR$ be an ordering as asserted.
Let $\SS$ be the dual ordering of $\RR$.
From Dual Ordering is Ordering, $\SS$ is an ordering.
By definition, the dual of $\RR$ is the inverse of $\RR$:
:$\SS = \RR^{-1}$
From Union of Antisymmetric Relation with Inverse is Antisymmetric iff Diagonal, it is generally not the case t... | Let $A$ be a [[Definition:Set|set]].
Let $\RR$ and $\SS$ be [[Definition:Ordering|orderings]] on $A$.
Then $\RR \cup \SS$ is not necessarily an [[Definition:Ordering|ordering]] on $A$. | Let $\RR$ be an [[Definition:Ordering|ordering]] as asserted.
Let $\SS$ be the [[Definition:Dual Ordering|dual ordering]] of $\RR$.
From [[Dual Ordering is Ordering]], $\SS$ is an [[Definition:Ordering|ordering]].
By definition, the [[Definition:Dual Ordering|dual]] of $\RR$ is the [[Definition:Inverse Relation|inve... | Union of Orderings is not necessarily Ordering | https://proofwiki.org/wiki/Union_of_Orderings_is_not_necessarily_Ordering | https://proofwiki.org/wiki/Union_of_Orderings_is_not_necessarily_Ordering | [
"Orderings",
"Set Union"
] | [
"Definition:Set",
"Definition:Ordering",
"Definition:Ordering"
] | [
"Definition:Ordering",
"Definition:Dual Ordering",
"Dual Ordering is Ordering",
"Definition:Ordering",
"Definition:Dual Ordering",
"Definition:Inverse Relation",
"Union of Antisymmetric Relation with Inverse is Antisymmetric iff Diagonal",
"Definition:Antisymmetric Relation",
"Definition:Ordering"
] |
proofwiki-19275 | Composite of Reflexive Relations is Reflexive | Let $A$ be a set.
Let $\RR$ and $\SS$ be reflexive relations on $A$.
Then their composite $\RR \circ \SS$ is also reflexive. | Recall the definition of composition of relations:
{{:Definition:Composition of Relations}}
Hence in this particular context:
:$\RR \circ \SS := \set {\tuple {x, z} \in A \times A: \exists y \in A: \tuple {x, y} \in \SS \land \tuple {y, z} \in \RR}$
Let $x \in A$ be arbitrary.
By definition of reflexive relation:
:$\tu... | Let $A$ be a [[Definition:Set|set]].
Let $\RR$ and $\SS$ be [[Definition:Reflexive Relation|reflexive relations]] on $A$.
Then their [[Definition:Composition of Relations|composite]] $\RR \circ \SS$ is also [[Definition:Reflexive Relation|reflexive]]. | Recall the definition of [[Definition:Composition of Relations|composition of relations]]:
{{:Definition:Composition of Relations}}
Hence in this particular context:
:$\RR \circ \SS := \set {\tuple {x, z} \in A \times A: \exists y \in A: \tuple {x, y} \in \SS \land \tuple {y, z} \in \RR}$
Let $x \in A$ be arbitrary... | Composite of Reflexive Relations is Reflexive | https://proofwiki.org/wiki/Composite_of_Reflexive_Relations_is_Reflexive | https://proofwiki.org/wiki/Composite_of_Reflexive_Relations_is_Reflexive | [
"Composite Relations",
"Reflexive Relations"
] | [
"Definition:Set",
"Definition:Reflexive Relation",
"Definition:Composition of Relations",
"Definition:Reflexive Relation"
] | [
"Definition:Composition of Relations",
"Definition:Reflexive Relation",
"Category:Composite Relations",
"Category:Reflexive Relations"
] |
proofwiki-19276 | Composite of Symmetric Relations is not necessarily Symmetric | Let $A$ be a set.
Let $\RR$ and $\SS$ be symmetric relations on $A$.
Then their composite $\RR \circ \SS$ is not necessarily symmetric. | Proof by Counterexample:
Let:
{{begin-eqn}}
{{eqn | l = A
| r = \set {1, 2, 3}
}}
{{eqn | l = \RR
| r = \set {\tuple {1, 2}, \tuple {2, 1} }
}}
{{eqn | l = \SS
| r = \set {\tuple {2, 3}, \tuple {3, 2} }
}}
{{end-eqn}}
We note that both $\RR$ and $\SS$ are symmetric relations on $A$.
We have by definit... | Let $A$ be a [[Definition:Set|set]].
Let $\RR$ and $\SS$ be [[Definition:Symmetric Relation|symmetric relations]] on $A$.
Then their [[Definition:Composition of Relations|composite]] $\RR \circ \SS$ is not necessarily [[Definition:Symmetric Relation|symmetric]]. | [[Proof by Counterexample]]:
Let:
{{begin-eqn}}
{{eqn | l = A
| r = \set {1, 2, 3}
}}
{{eqn | l = \RR
| r = \set {\tuple {1, 2}, \tuple {2, 1} }
}}
{{eqn | l = \SS
| r = \set {\tuple {2, 3}, \tuple {3, 2} }
}}
{{end-eqn}}
We note that both $\RR$ and $\SS$ are [[Definition:Symmetric Relation|symmetri... | Composite of Symmetric Relations is not necessarily Symmetric | https://proofwiki.org/wiki/Composite_of_Symmetric_Relations_is_not_necessarily_Symmetric | https://proofwiki.org/wiki/Composite_of_Symmetric_Relations_is_not_necessarily_Symmetric | [
"Composite Relations",
"Symmetric Relations"
] | [
"Definition:Set",
"Definition:Symmetric Relation",
"Definition:Composition of Relations",
"Definition:Symmetric Relation"
] | [
"Proof by Counterexample",
"Definition:Symmetric Relation",
"Definition:Composition of Relations",
"Definition:Symmetric Relation",
"Category:Composite Relations",
"Category:Symmetric Relations"
] |
proofwiki-19277 | Composite of Antisymmetric Relations is not necessarily Antisymmetric | Let $A$ be a set.
Let $\RR$ and $\SS$ be antisymmetric relations on $A$.
Then their composite $\RR \circ \SS$ is not necessarily also antisymmetric. | ;Proof by Counterexample:
Consider the ordering $\le$ on the natural numbers $\N$.
Consider its dual ordering $\ge$ also on $\N$.
Note that Dual Ordering is Ordering.
Both $\le$ and $\ge$ are {{afortiori}} antisymmetric relations.
We have:
:$1 \le 3$
:$3 \ge 2$
and similarly:
:$2 \le 3$
:$3 \ge 1$
Hence it follows that... | Let $A$ be a [[Definition:Set|set]].
Let $\RR$ and $\SS$ be [[Definition:Antisymmetric Relation|antisymmetric relations]] on $A$.
Then their [[Definition:Composition of Relations|composite]] $\RR \circ \SS$ is not necessarily also [[Definition:Antisymmetric Relation|antisymmetric]]. | ;[[Proof by Counterexample]]:
Consider the [[Definition:Ordering|ordering]] $\le$ on the [[Definition:Natural Numbers|natural numbers]] $\N$.
Consider its [[Definition:Dual Ordering|dual ordering]] $\ge$ also on $\N$.
Note that [[Dual Ordering is Ordering]].
Both $\le$ and $\ge$ are {{afortiori}} [[Definition:Antis... | Composite of Antisymmetric Relations is not necessarily Antisymmetric | https://proofwiki.org/wiki/Composite_of_Antisymmetric_Relations_is_not_necessarily_Antisymmetric | https://proofwiki.org/wiki/Composite_of_Antisymmetric_Relations_is_not_necessarily_Antisymmetric | [
"Antisymmetric Relations",
"Composite Relations"
] | [
"Definition:Set",
"Definition:Antisymmetric Relation",
"Definition:Composition of Relations",
"Definition:Antisymmetric Relation"
] | [
"Proof by Counterexample",
"Definition:Ordering",
"Definition:Natural Numbers",
"Definition:Dual Ordering",
"Dual Ordering is Ordering",
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation",
"Definition:Composition of Relations",
"Category:Antisymmetric Relations",
"Category:Co... |
proofwiki-19278 | Composite of Orderings is not necessarily Ordering | Let $A$ be a set.
Let $\RR$ and $\SS$ be orderings on $A$.
Then their composite $\RR \circ \SS$ is not necessarily also an ordering on $A$. | Let $\RR$ and $\SS$ be orderings as asserted.
Both $\RR$ and $\SS$ are {{afortiori}} both antisymmetric and transitive.
But we have:
:From Composite of Antisymmetric Relations is not necessarily Antisymmetric, it is not necessarily the case that $\RR \circ \SS$ is itself antisymmetric.
:From Composite of Transitive Rel... | Let $A$ be a [[Definition:Set|set]].
Let $\RR$ and $\SS$ be [[Definition:Ordering|orderings]] on $A$.
Then their [[Definition:Composition of Relations|composite]] $\RR \circ \SS$ is not necessarily also an [[Definition:Ordering|ordering]] on $A$. | Let $\RR$ and $\SS$ be [[Definition:Ordering|orderings]] as asserted.
Both $\RR$ and $\SS$ are {{afortiori}} both [[Definition:Antisymmetric Relation|antisymmetric]] and [[Definition:Transitive Relation|transitive]].
But we have:
:From [[Composite of Antisymmetric Relations is not necessarily Antisymmetric]], it is... | Composite of Orderings is not necessarily Ordering | https://proofwiki.org/wiki/Composite_of_Orderings_is_not_necessarily_Ordering | https://proofwiki.org/wiki/Composite_of_Orderings_is_not_necessarily_Ordering | [
"Orderings",
"Composite Relations"
] | [
"Definition:Set",
"Definition:Ordering",
"Definition:Composition of Relations",
"Definition:Ordering"
] | [
"Definition:Ordering",
"Definition:Antisymmetric Relation",
"Definition:Transitive Relation",
"Composite of Antisymmetric Relations is not necessarily Antisymmetric",
"Definition:Antisymmetric Relation",
"Composite of Transitive Relations is not necessarily Transitive",
"Definition:Transitive Relation",... |
proofwiki-19279 | Composite of Transitive Relations is not necessarily Transitive | Let $A$ be a set.
Let $\RR$ and $\SS$ be transitive relations on $A$.
Then their composite $\RR \circ \SS$ is not necessarily also transitive. | Recall the definition of composition of relations:
{{:Definition:Composition of Relations}}
;Proof by Counterexample
:240pxthumbright
Let $A = \set {a, b, c}$.
Let $\RR$ be defined as:
:$\RR = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {b, c} }$
Let $\SS$ be defined as:
:$\SS = \set {\tuple {a, a}, \tupl... | Let $A$ be a [[Definition:Set|set]].
Let $\RR$ and $\SS$ be [[Definition:Transitive Relation|transitive relations]] on $A$.
Then their [[Definition:Composition of Relations|composite]] $\RR \circ \SS$ is not necessarily also [[Definition:Transitive Relation|transitive]]. | Recall the definition of [[Definition:Composition of Relations|composition of relations]]:
{{:Definition:Composition of Relations}}
;[[Proof by Counterexample]]
:[[File:Non-Transitive-Composite.png|240px|thumb|right]]
Let $A = \set {a, b, c}$.
Let $\RR$ be defined as:
:$\RR = \set {\tuple {a, a}, \tuple {b, b}, \tup... | Composite of Transitive Relations is not necessarily Transitive | https://proofwiki.org/wiki/Composite_of_Transitive_Relations_is_not_necessarily_Transitive | https://proofwiki.org/wiki/Composite_of_Transitive_Relations_is_not_necessarily_Transitive | [
"Composite Relations",
"Transitive Relations"
] | [
"Definition:Set",
"Definition:Transitive Relation",
"Definition:Composition of Relations",
"Definition:Transitive Relation"
] | [
"Definition:Composition of Relations",
"Proof by Counterexample",
"File:Non-Transitive-Composite.png",
"Definition:Transitive Relation",
"Definition:Transitive Relation",
"Category:Composite Relations",
"Category:Transitive Relations"
] |
proofwiki-19280 | Identity Mapping is Automorphism/Ordered Semigroups | Let $\struct {S, \circ, \preccurlyeq}$ be an ordered semigroup.
Then $I_S: \struct {S, \circ, \preccurlyeq} \to \struct {S, \circ, \preccurlyeq}$ is a ordered semigroup automorphism. | From Identity Mapping is Semigroup Automorphism:
:$I_S: \struct {S, \circ} \to \struct {S, \circ}$ is a semigroup automorphism.
From Identity Mapping is Order Isomorphism:
:$I_S: \struct {S, \preccurlyeq} \to \struct {S, \preccurlyeq}$ is an order isomorphism.
{{qed}} | Let $\struct {S, \circ, \preccurlyeq}$ be an [[Definition:Ordered Semigroup|ordered semigroup]].
Then $I_S: \struct {S, \circ, \preccurlyeq} \to \struct {S, \circ, \preccurlyeq}$ is a [[Definition:Ordered Semigroup Automorphism|ordered semigroup automorphism]]. | From [[Identity Mapping is Semigroup Automorphism]]:
:$I_S: \struct {S, \circ} \to \struct {S, \circ}$ is a [[Definition:Semigroup Automorphism|semigroup automorphism]].
From [[Identity Mapping is Order Isomorphism]]:
:$I_S: \struct {S, \preccurlyeq} \to \struct {S, \preccurlyeq}$ is an [[Definition:Order Isomorphism|... | Identity Mapping is Automorphism/Ordered Semigroups | https://proofwiki.org/wiki/Identity_Mapping_is_Automorphism/Ordered_Semigroups | https://proofwiki.org/wiki/Identity_Mapping_is_Automorphism/Ordered_Semigroups | [
"Semigroup Automorphisms",
"Order Isomorphisms",
"Ordered Semigroups",
"Identity Mappings"
] | [
"Definition:Ordered Semigroup",
"Definition:Ordered Semigroup Automorphism"
] | [
"Identity Mapping is Automorphism/Semigroups",
"Definition:Semigroup Automorphism",
"Identity Mapping is Order Isomorphism",
"Definition:Order Isomorphism"
] |
proofwiki-19281 | Inverse of Ordered Semigroup Isomorphism is Isomorphism | Let $\struct {S, \circ, \preceq}$ and $\struct {T, *, \preccurlyeq}$ be ordered semigroups.
Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be a mapping.
Then:
:$\phi$ is an ordered semigroup isomorphism
{{iff}}:
:$\phi^{-1}: \struct {T, *, \preccurlyeq} \to \struct {S, \circ, \preceq}$ is also... | === Sufficient Condition ===
Let $\phi$ be an ordered semigroup isomorphism.
Then by definition $\phi$ is a bijection.
Thus $\exists \phi^{-1}$ such that $\phi^{-1}$ is also a bijection from Bijection iff Inverse is Bijection.
That is:
:$\exists \phi^{-1}: \struct {T, *, \preccurlyeq} \to \struct {S, \circ, \preceq}$
F... | Let $\struct {S, \circ, \preceq}$ and $\struct {T, *, \preccurlyeq}$ be [[Definition:Ordered Semigroup|ordered semigroups]].
Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be a [[Definition:Mapping|mapping]].
Then:
:$\phi$ is an [[Definition:Ordered Semigroup Isomorphism|ordered semigroup i... | === Sufficient Condition ===
Let $\phi$ be an [[Definition:Ordered Semigroup Isomorphism|ordered semigroup isomorphism]].
Then by definition $\phi$ is a [[Definition:Bijection|bijection]].
Thus $\exists \phi^{-1}$ such that $\phi^{-1}$ is also a [[Definition:Bijection|bijection]] from [[Bijection iff Inverse is Bije... | Inverse of Ordered Semigroup Isomorphism is Isomorphism | https://proofwiki.org/wiki/Inverse_of_Ordered_Semigroup_Isomorphism_is_Isomorphism | https://proofwiki.org/wiki/Inverse_of_Ordered_Semigroup_Isomorphism_is_Isomorphism | [
"Ordered Semigroup Isomorphisms",
"Inverse Mappings"
] | [
"Definition:Ordered Semigroup",
"Definition:Mapping",
"Definition:Ordered Semigroup Isomorphism",
"Definition:Ordered Semigroup Isomorphism"
] | [
"Definition:Ordered Semigroup Isomorphism",
"Definition:Bijection",
"Definition:Bijection",
"Inverse of Bijection is Bijection",
"Inverse of Algebraic Structure Isomorphism is Isomorphism",
"Definition:Isomorphism (Abstract Algebra)",
"Isomorphism Preserves Semigroups",
"Definition:Isomorphism (Abstra... |
proofwiki-19282 | Composite of Ordered Semigroup Isomorphisms is Isomorphism | Let $\struct {S_1, \circ_1, \preccurlyeq_1}, \struct {S_2, \circ_2, \preccurlyeq_2}, \struct {S_3, \circ_3, \preccurlyeq_3}$ be ordered semigroups.
Let $\phi: S_1 \to S_2$ and $\psi: S_2 \to S_3$ be ordered semigroup isomorphisms.
Then the composite mapping $\psi \circ \phi: S_1 \to S_3$ is also an ordered semigroup is... | From Composite of Isomorphisms in Algebraic Structure is Isomorphism, $\psi \circ \phi$ is an algebraic structure isomorphism.
From Isomorphism Preserves Semigroups, it follows that $\psi \circ \phi$ is a semigroup isomorphism from $\struct {S_1, \circ_1}$ to $\struct {S_3, \circ_3}$.
From Composite of Order Isomorphis... | Let $\struct {S_1, \circ_1, \preccurlyeq_1}, \struct {S_2, \circ_2, \preccurlyeq_2}, \struct {S_3, \circ_3, \preccurlyeq_3}$ be [[Definition:Ordered Semigroup|ordered semigroups]].
Let $\phi: S_1 \to S_2$ and $\psi: S_2 \to S_3$ be [[Definition:Ordered Semigroup Isomorphism|ordered semigroup isomorphisms]].
Then the... | From [[Composite of Isomorphisms in Algebraic Structure is Isomorphism]], $\psi \circ \phi$ is an [[Definition:Isomorphism (Abstract Algebra)|algebraic structure isomorphism]].
From [[Isomorphism Preserves Semigroups]], it follows that $\psi \circ \phi$ is a [[Definition:Semigroup Isomorphism|semigroup isomorphism]] f... | Composite of Ordered Semigroup Isomorphisms is Isomorphism | https://proofwiki.org/wiki/Composite_of_Ordered_Semigroup_Isomorphisms_is_Isomorphism | https://proofwiki.org/wiki/Composite_of_Ordered_Semigroup_Isomorphisms_is_Isomorphism | [
"Ordered Semigroup Isomorphisms",
"Composite Mappings"
] | [
"Definition:Ordered Semigroup",
"Definition:Ordered Semigroup Isomorphism",
"Definition:Composition of Mappings",
"Definition:Ordered Semigroup Isomorphism"
] | [
"Composite of Isomorphisms is Isomorphism/Algebraic Structure",
"Definition:Isomorphism (Abstract Algebra)",
"Isomorphism Preserves Semigroups",
"Definition:Isomorphism (Abstract Algebra)/Semigroup Isomorphism",
"Composite of Order Isomorphisms is Order Isomorphism",
"Definition:Order Isomorphism"
] |
proofwiki-19283 | Integers Modulo m under Max Operation form Ordered Semigroup | Let $\Z_m$ denote the set of integers modulo $m$:
:$\Z_m = \set {0, 1, \ldots, m - 1}$
Let $\vee_m$ be the operation on $\Z_m$ defined as:
:$\forall a, b \in \Z_m: a \vee_m b = \max \set {a, b}$
Then the ordered algebraic structure $\struct {\Z_m, \vee_m, \le}$ is an ordered semigroup. | Taking the semigroup axioms in turn: | Let $\Z_m$ denote the [[Definition:Set|set]] of [[Definition:Integers Modulo m|integers modulo $m$]]:
:$\Z_m = \set {0, 1, \ldots, m - 1}$
Let $\vee_m$ be the [[Definition:Binary Operation|operation]] on $\Z_m$ defined as:
:$\forall a, b \in \Z_m: a \vee_m b = \max \set {a, b}$
Then the [[Definition:Ordered Structur... | Taking the [[Axiom:Semigroup Axioms|semigroup axioms]] in turn: | Integers Modulo m under Max Operation form Ordered Semigroup | https://proofwiki.org/wiki/Integers_Modulo_m_under_Max_Operation_form_Ordered_Semigroup | https://proofwiki.org/wiki/Integers_Modulo_m_under_Max_Operation_form_Ordered_Semigroup | [
"Examples of Ordered Semigroups"
] | [
"Definition:Set",
"Definition:Integers Modulo m",
"Definition:Operation/Binary Operation",
"Definition:Ordered Structure",
"Definition:Ordered Semigroup"
] | [
"Axiom:Semigroup Axioms",
"Axiom:Semigroup Axioms"
] |
proofwiki-19284 | Composite of Connected Relation is not necessarily Connected | Let $A$ be a set.
Let $\RR$ and $\SS$ be connected relations on $A$.
Then their composite $\RR \circ \SS$ is not necessarily also connected. | Recall the definition of composition of relations:
{{:Definition:Composition of Relations}}
;Proof by Counterexample
:240pxthumbright
Let $A = \set {a, b, c}$.
Let $\RR$ be defined as:
:$\RR = \set {\tuple {a, b}, \tuple {b, c}, \tuple {c, a} }$
Let $\SS$ be defined as $\RR^{-1}$, that is, the inverse of $\RR$:
:$\SS =... | Let $A$ be a [[Definition:Set|set]].
Let $\RR$ and $\SS$ be [[Definition:Connected Relation|connected relations]] on $A$.
Then their [[Definition:Composition of Relations|composite]] $\RR \circ \SS$ is not necessarily also [[Definition:Connected Relation|connected]]. | Recall the definition of [[Definition:Composition of Relations|composition of relations]]:
{{:Definition:Composition of Relations}}
;[[Proof by Counterexample]]
:[[File:Non-Connected-Composite.png|240px|thumb|right]]
Let $A = \set {a, b, c}$.
Let $\RR$ be defined as:
:$\RR = \set {\tuple {a, b}, \tuple {b, c}, \tupl... | Composite of Connected Relation is not necessarily Connected | https://proofwiki.org/wiki/Composite_of_Connected_Relation_is_not_necessarily_Connected | https://proofwiki.org/wiki/Composite_of_Connected_Relation_is_not_necessarily_Connected | [
"Composite Relations",
"Connected Relations"
] | [
"Definition:Set",
"Definition:Connected Relation",
"Definition:Composition of Relations",
"Definition:Connected Relation"
] | [
"Definition:Composition of Relations",
"Proof by Counterexample",
"File:Non-Connected-Composite.png",
"Definition:Inverse Relation",
"Definition:Connected Relation",
"Definition:Connected Relation",
"Category:Composite Relations",
"Category:Connected Relations"
] |
proofwiki-19285 | Composite of Total Relations is Total | Let $A$ be a set.
Let $\RR$ and $\SS$ be total relations on $A$.
Then their composite $\RR \circ \SS$ is also total. | Recall the definition of composition of relations:
{{:Definition:Composition of Relations}}
Hence in this particular context:
:$\RR \circ \SS := \set {\tuple {x, z} \in A \times A: \exists y \in A: \tuple {x, y} \in \SS \land \tuple {y, z} \in \RR}$
Let $x$ and $y$ in $A$ be arbitrary.
Because $\SS$ is total, either $x... | Let $A$ be a [[Definition:Set|set]].
Let $\RR$ and $\SS$ be [[Definition:Total Relation|total relations]] on $A$.
Then their [[Definition:Composition of Relations|composite]] $\RR \circ \SS$ is also [[Definition:Total Relation|total]]. | Recall the definition of [[Definition:Composition of Relations|composition of relations]]:
{{:Definition:Composition of Relations}}
Hence in this particular context:
:$\RR \circ \SS := \set {\tuple {x, z} \in A \times A: \exists y \in A: \tuple {x, y} \in \SS \land \tuple {y, z} \in \RR}$
Let $x$ and $y$ in $A$ be a... | Composite of Total Relations is Total | https://proofwiki.org/wiki/Composite_of_Total_Relations_is_Total | https://proofwiki.org/wiki/Composite_of_Total_Relations_is_Total | [
"Composite Relations",
"Total Relations"
] | [
"Definition:Set",
"Definition:Total Relation",
"Definition:Composition of Relations",
"Definition:Total Relation"
] | [
"Definition:Composition of Relations",
"Definition:Total Relation",
"Relation is Connected and Reflexive iff Total",
"Definition:Reflexive Relation",
"Definition:Composition of Relations",
"Definition:Total Relation",
"Category:Composite Relations",
"Category:Total Relations"
] |
proofwiki-19286 | Dual of Ordered Semigroup is Ordered Semigroup | Let $\struct {S, \circ, \preccurlyeq}$ be an ordered semigroup.
Then its dual $\struct {S, \circ, \succcurlyeq}$ is also an ordered semigroup. | From Dual Ordering is Ordering, we have that $\struct {S, \succcurlyeq}$ is an ordered set.
We also note from the definition that $\struct {S, \circ}$ is a semigroup.
It remains to be demonstrated that $\succcurlyeq$ is compatible with $\circ$.
Recall that $\struct {S, \circ, \preccurlyeq}$ is an ordered semigroup.
Hen... | Let $\struct {S, \circ, \preccurlyeq}$ be an [[Definition:Ordered Semigroup|ordered semigroup]].
Then its [[Definition:Dual Ordered Set|dual]] $\struct {S, \circ, \succcurlyeq}$ is also an [[Definition:Ordered Semigroup|ordered semigroup]]. | From [[Dual Ordering is Ordering]], we have that $\struct {S, \succcurlyeq}$ is an [[Definition:Ordered Set|ordered set]].
We also note from the definition that $\struct {S, \circ}$ is a [[Definition:Semigroup|semigroup]].
It remains to be demonstrated that $\succcurlyeq$ is [[Definition:Relation Compatible with Ope... | Dual of Ordered Semigroup is Ordered Semigroup | https://proofwiki.org/wiki/Dual_of_Ordered_Semigroup_is_Ordered_Semigroup | https://proofwiki.org/wiki/Dual_of_Ordered_Semigroup_is_Ordered_Semigroup | [
"Ordered Semigroups",
"Dual Orderings"
] | [
"Definition:Ordered Semigroup",
"Definition:Dual Ordering/Dual Ordered Set",
"Definition:Ordered Semigroup"
] | [
"Dual Ordering is Ordering",
"Definition:Ordered Set",
"Definition:Semigroup",
"Definition:Relation Compatible with Operation",
"Definition:Ordered Semigroup",
"Definition:Relation Compatible with Operation"
] |
proofwiki-19287 | Inversion Mapping is Isomorphism from Ordered Abelian Group to its Dual | Let $\struct {G, \circ, \preccurlyeq}$ be an ordered Abelian group.
Let $\struct {G, \circ, \succcurlyeq}$ be the dual of $\struct {G, \circ, \preccurlyeq}$.
Let $\phi: \struct {G, \circ, \preccurlyeq} \to \struct {G, \circ, \succcurlyeq}$ be the inversion mapping from $\struct {G, \circ, \preccurlyeq}$ to $\struct {G,... | $\struct {G, \circ, \preccurlyeq}$ is {{afortiori}} an ordered semigroup.
Hence from Dual of Ordered Semigroup is Ordered Semigroup, $\struct {G, \circ, \succcurlyeq}$ is also an ordered semigroup.
That is, $\succcurlyeq$ is compatible with $\circ$.
From Inversion Mapping is Permutation, {{afortiori}} the inversion map... | Let $\struct {G, \circ, \preccurlyeq}$ be an [[Definition:Ordered Group|ordered]] [[Definition:Abelian Group|Abelian group]].
Let $\struct {G, \circ, \succcurlyeq}$ be the [[Definition:Dual Ordered Set|dual]] of $\struct {G, \circ, \preccurlyeq}$.
Let $\phi: \struct {G, \circ, \preccurlyeq} \to \struct {G, \circ, \s... | $\struct {G, \circ, \preccurlyeq}$ is {{afortiori}} an [[Definition:Ordered Semigroup|ordered semigroup]].
Hence from [[Dual of Ordered Semigroup is Ordered Semigroup]], $\struct {G, \circ, \succcurlyeq}$ is also an [[Definition:Ordered Semigroup|ordered semigroup]].
That is, $\succcurlyeq$ is [[Definition:Relation C... | Inversion Mapping is Isomorphism from Ordered Abelian Group to its Dual | https://proofwiki.org/wiki/Inversion_Mapping_is_Isomorphism_from_Ordered_Abelian_Group_to_its_Dual | https://proofwiki.org/wiki/Inversion_Mapping_is_Isomorphism_from_Ordered_Abelian_Group_to_its_Dual | [
"Ordered Groups",
"Abelian Groups",
"Dual Orderings",
"Inversion Mappings",
"Group Isomorphisms",
"Order Isomorphisms"
] | [
"Definition:Ordered Group",
"Definition:Abelian Group",
"Definition:Dual Ordering/Dual Ordered Set",
"Definition:Inversion Mapping",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Ordered Group Isomorphism"
] | [
"Definition:Ordered Semigroup",
"Dual of Ordered Semigroup is Ordered Semigroup",
"Definition:Ordered Semigroup",
"Definition:Relation Compatible with Operation",
"Inversion Mapping is Permutation",
"Definition:Inversion Mapping",
"Definition:Bijection",
"Inversion Mapping is Automorphism iff Group is... |
proofwiki-19288 | Subsets Greater Than and Less Than Identity of Ordered Abelian Group are Isomorphic Ordered Semigroups | Let $\struct {G, \circ, \preccurlyeq}$ be an ordered abelian group whose identity element is $e$.
Let $G^+$ and $G^-$ denote the subsets of $G$ defined as:
:$G^+ = \set {x \in G: e \preccurlyeq x}$
:$G^- = \set {x \in G: x \preccurlyeq e}$
Then $\struct {G^+, \circ, \preccurlyeq}$ and $\struct {G^-, \circ, \succcurlyeq... | Let $x, y \in G^+$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = e
| o = \preccurlyeq
| r = x
| c = Definition of $G^+$
}}
{{eqn | l = e
| o = \preccurlyeq
| r = y
| c = Definition of $G^+$
}}
{{eqn | ll= \leadsto
| l = e \circ y
| o = \preccurlyeq
| r = x \circ y
... | Let $\struct {G, \circ, \preccurlyeq}$ be an [[Definition:Ordered Group|ordered]] [[Definition:Abelian Group|abelian group]] whose [[Definition:Identity Element|identity element]] is $e$.
Let $G^+$ and $G^-$ denote the [[Definition:Subset|subsets]] of $G$ defined as:
:$G^+ = \set {x \in G: e \preccurlyeq x}$
:$G^- = \... | Let $x, y \in G^+$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = e
| o = \preccurlyeq
| r = x
| c = Definition of $G^+$
}}
{{eqn | l = e
| o = \preccurlyeq
| r = y
| c = Definition of $G^+$
}}
{{eqn | ll= \leadsto
| l = e \circ y
| o = \preccurlyeq
| r = x \circ y... | Subsets Greater Than and Less Than Identity of Ordered Abelian Group are Isomorphic Ordered Semigroups | https://proofwiki.org/wiki/Subsets_Greater_Than_and_Less_Than_Identity_of_Ordered_Abelian_Group_are_Isomorphic_Ordered_Semigroups | https://proofwiki.org/wiki/Subsets_Greater_Than_and_Less_Than_Identity_of_Ordered_Abelian_Group_are_Isomorphic_Ordered_Semigroups | [
"Ordered Groups",
"Abelian Groups",
"Inversion Mappings",
"Ordered Semigroup Isomorphisms"
] | [
"Definition:Ordered Group",
"Definition:Abelian Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subset",
"Definition:Subsemigroup",
"Definition:Inversion Mapping",
"Definition:Ordered Semigroup Isomorphism",
"Definition:Ordered Semigroup Isomorphism"
] | [
"Definition:Relation Compatible with Operation",
"Definition:Ordering",
"Definition:Transitive Relation",
"Subsemigroup Closure Test",
"Definition:Subsemigroup",
"Definition:Relation Compatible with Operation",
"Definition:Ordering",
"Definition:Transitive Relation",
"Subsemigroup Closure Test",
"... |
proofwiki-19289 | Hensel's Lemma/P-adic Integers/Lemma 1 | :There exists a unique $p$-adic expansion $\ds \sum_{n \mathop = 0}^\infty d_n p^n$:
::$\ds \forall k : a_k = \sum_{n \mathop = 0}^k d_n p^n$ satisfies:
{{begin-eqn}}
{{eqn | n = 1
| l = \map F {a_k}
| o = \equiv
| r = 0
| rr= \pmod {p^{k+1}\Z_p}
}}
{{eqn | n = 2
| l = a_k
| o = \equ... | The Second Principle of Recursive Definition is used to construct the sequence
$\sequence {d_n}$.
Let $T$ be the set of $p$-adic digits.
For each $k \in \N_{>0}$, let:
:$\ds S_k = \set{\tuple{b_0, b_1, \ldots, b_{k-1}} \subseteq T^k : \map F {\sum_{n \mathop = 0}^{k-1} b_n p^n } \equiv 0 \pmod{p^k\Z_p} \quad \text{and}... | :There exists a [[Definition:Unique|unique]] [[Definition:P-adic Expansion|$p$-adic expansion]] $\ds \sum_{n \mathop = 0}^\infty d_n p^n$:
::$\ds \forall k : a_k = \sum_{n \mathop = 0}^k d_n p^n$ satisfies:
{{begin-eqn}}
{{eqn | n = 1
| l = \map F {a_k}
| o = \equiv
| r = 0
| rr= \pmod {p^{k+1}\... | The [[Second Principle of Recursive Definition]] is used to construct the [[Definition:Sequence|sequence]]
$\sequence {d_n}$.
Let $T$ be the [[Definition:Set|set]] of [[Definition:P-adic Digit|$p$-adic digits]].
For each $k \in \N_{>0}$, let:
:$\ds S_k = \set{\tuple{b_0, b_1, \ldots, b_{k-1}} \subseteq T^k : \map F {... | Hensel's Lemma/P-adic Integers/Lemma 1 | https://proofwiki.org/wiki/Hensel's_Lemma/P-adic_Integers/Lemma_1 | https://proofwiki.org/wiki/Hensel's_Lemma/P-adic_Integers/Lemma_1 | [
"Hensel's Lemma"
] | [
"Definition:Unique",
"Definition:P-adic Expansion"
] | [
"Second Principle of Recursive Definition",
"Definition:Sequence",
"Definition:Set",
"Definition:P-adic Digit",
"Definition:P-adic Digit",
"Definition:Canonical P-adic Expansion",
"Second Principle of Recursive Definition"
] |
proofwiki-19290 | Simple Order Product of Pair of Ordered Semigroups is Ordered Semigroup | Let $\struct {S_1, \circ_1, \preccurlyeq_1}$ and $\struct {S_2, \circ_2, \preccurlyeq_2}$ be ordered semigroups.
Let $\struct {S_1 \times S_2, \odot} := \struct {S_1, \circ_1} \times \struct {S_2, \circ_2}$ denote the '''external direct product''' of $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$.
Let $\struct {... | From Simple Order Product of Pair of Ordered Sets is Ordered Set, $\struct {S_1 \times S_2, \otimes^s}$ is an ordered set.
From External Direct Product of Semigroups, $\struct {S_1 \times S_2, \odot}$ is a semigroup.
It remains to be shown that $\preccurlyeq_s$ is compatible with $\odot$.
Let $\tuple {x_1, x_2}, \tuple... | Let $\struct {S_1, \circ_1, \preccurlyeq_1}$ and $\struct {S_2, \circ_2, \preccurlyeq_2}$ be [[Definition:Ordered Semigroup|ordered semigroups]].
Let $\struct {S_1 \times S_2, \odot} := \struct {S_1, \circ_1} \times \struct {S_2, \circ_2}$ denote the '''[[Definition:External Direct Product|external direct product]]''... | From [[Simple Order Product of Pair of Ordered Sets is Ordered Set]], $\struct {S_1 \times S_2, \otimes^s}$ is an [[Definition:Ordered Set|ordered set]].
From [[External Direct Product of Semigroups]], $\struct {S_1 \times S_2, \odot}$ is a [[Definition:Semigroup|semigroup]].
It remains to be shown that $\preccurlye... | Simple Order Product of Pair of Ordered Semigroups is Ordered Semigroup | https://proofwiki.org/wiki/Simple_Order_Product_of_Pair_of_Ordered_Semigroups_is_Ordered_Semigroup | https://proofwiki.org/wiki/Simple_Order_Product_of_Pair_of_Ordered_Semigroups_is_Ordered_Semigroup | [
"Simple Order Product",
"Ordered Semigroups"
] | [
"Definition:Ordered Semigroup",
"Definition:External Direct Product",
"Definition:Simple Order Product",
"Definition:Ordered Semigroup"
] | [
"Simple Order Product of Pair of Ordered Sets is Ordered Set",
"Definition:Ordered Set",
"External Direct Product of Semigroups",
"Definition:Semigroup",
"Definition:Relation Compatible with Operation"
] |
proofwiki-19291 | Convergent Series can be Added Term by Term | Let $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ be two convergent series converging to $A$ and $B$ respectively.
Then:
:$\ds \sum_{n \mathop = 1}^\infty \paren {a_n + b_n} = A + B$ | {{begin-eqn}}
{{eqn | l = A + B
| r = \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n
}}
{{eqn | r = \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N + \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N b_n
| c = {{Defof|Convergent Series of Numbers}}
}}
{{eqn | r = \lim_{N \mathop \to \... | Let $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ be two [[Definition:Convergent Series of Numbers|convergent series]] converging to $A$ and $B$ respectively.
Then:
:$\ds \sum_{n \mathop = 1}^\infty \paren {a_n + b_n} = A + B$ | {{begin-eqn}}
{{eqn | l = A + B
| r = \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n
}}
{{eqn | r = \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N + \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N b_n
| c = {{Defof|Convergent Series of Numbers}}
}}
{{eqn | r = \lim_{N \mathop \to \... | Convergent Series can be Added Term by Term | https://proofwiki.org/wiki/Convergent_Series_can_be_Added_Term_by_Term | https://proofwiki.org/wiki/Convergent_Series_can_be_Added_Term_by_Term | [
"Series",
"Convergence"
] | [
"Definition:Convergent Series/Number Field"
] | [
"Combination Theorem for Sequences/Real/Sum Rule",
"Definition:Finite Sum",
"Category:Series",
"Category:Convergence"
] |
proofwiki-19292 | Lexicographically Ordered Pair of Ordered Semigroups with Cancellable Elements | Let $\struct {S_1, \circ_1, \preccurlyeq_1}$ and $\struct {S_2, \circ_2, \preccurlyeq_2}$ be ordered semigroups.
Let $\struct {S_1 \times S_2, \odot} := \struct {S_1, \circ_1} \times \struct {S_2, \circ_2}$ denote the '''external direct product''' of $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$.
Let $\struct {... | From Lexicographic Order is Ordering we have that $\struct {S_1 \times S_2, \preccurlyeq_l}$ is an ordered set.
From External Direct Product of Semigroups, $\struct {S_1 \times S_2, \odot}$ is a semigroup.
It remains to be shown that $\preccurlyeq_l$ is compatible with $\odot$.
Let $\tuple {x_1, x_2}, \tuple {y_1, y_2}... | Let $\struct {S_1, \circ_1, \preccurlyeq_1}$ and $\struct {S_2, \circ_2, \preccurlyeq_2}$ be [[Definition:Ordered Semigroup|ordered semigroups]].
Let $\struct {S_1 \times S_2, \odot} := \struct {S_1, \circ_1} \times \struct {S_2, \circ_2}$ denote the '''[[Definition:External Direct Product|external direct product]]''... | From [[Lexicographic Order is Ordering]] we have that $\struct {S_1 \times S_2, \preccurlyeq_l}$ is an [[Definition:Ordered Set|ordered set]].
From [[External Direct Product of Semigroups]], $\struct {S_1 \times S_2, \odot}$ is a [[Definition:Semigroup|semigroup]].
It remains to be shown that $\preccurlyeq_l$ is [[De... | Lexicographically Ordered Pair of Ordered Semigroups with Cancellable Elements | https://proofwiki.org/wiki/Lexicographically_Ordered_Pair_of_Ordered_Semigroups_with_Cancellable_Elements | https://proofwiki.org/wiki/Lexicographically_Ordered_Pair_of_Ordered_Semigroups_with_Cancellable_Elements | [
"Lexicographic Order",
"Ordered Semigroups",
"Cancellability",
"Lexicographically Ordered Pair of Ordered Semigroups with Cancellable Elements"
] | [
"Definition:Ordered Semigroup",
"Definition:External Direct Product",
"Definition:Lexicographic Order",
"Definition:Element",
"Definition:Cancellable Element",
"Definition:Ordered Semigroup"
] | [
"Lexicographic Order is Ordering",
"Definition:Ordered Set",
"External Direct Product of Semigroups",
"Definition:Semigroup",
"Definition:Relation Compatible with Operation",
"Reflexive Reduction of Relation Compatible with Cancellable Operation is Compatible"
] |
proofwiki-19293 | Reflexive Reduction of Relation Compatible with Cancellable Operation is Compatible | Let $\struct {S, \circ}$ be an algebraic structure such that $\circ$ is a cancellable operation.
Let $\RR$ be a relation on $S$ which is compatible with $\circ$.
Let $\RR^\ne$ be the reflexive reduction of $\RR$.
Then $\RR^\ne$ is compatible with $\circ$. | {{AimForCont}} $\RR^\ne$ is not compatible with $\circ$.
Let $x, y \in S$ such that:
:$x \mathrel \RR y$
but:
:$x \mathrel {\RR^\ne} y$
Then by definition of reflexive reduction:
:$x \ne y$
Then as $\RR^\ne$ is not compatible with $\circ$:
:$\exists z \in S: \lnot \paren {z \circ x \mathrel {\RR^\ne} z \circ y}$
But as... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]] such that $\circ$ is a [[Definition:Cancellable Operation|cancellable operation]].
Let $\RR$ be a [[Definition:Endorelation|relation]] on $S$ which is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Let $\... | {{AimForCont}} $\RR^\ne$ is not [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Let $x, y \in S$ such that:
:$x \mathrel \RR y$
but:
:$x \mathrel {\RR^\ne} y$
Then by definition of [[Definition:Reflexive Reduction|reflexive reduction]]:
:$x \ne y$
Then as $\RR^\ne$ is not [[Definition:Rel... | Reflexive Reduction of Relation Compatible with Cancellable Operation is Compatible | https://proofwiki.org/wiki/Reflexive_Reduction_of_Relation_Compatible_with_Cancellable_Operation_is_Compatible | https://proofwiki.org/wiki/Reflexive_Reduction_of_Relation_Compatible_with_Cancellable_Operation_is_Compatible | [
"Cancellability",
"Compatible Relations",
"Reflexive Reductions"
] | [
"Definition:Algebraic Structure",
"Definition:Cancellable Operation",
"Definition:Endorelation",
"Definition:Relation Compatible with Operation",
"Definition:Reflexive Reduction",
"Definition:Relation Compatible with Operation"
] | [
"Definition:Relation Compatible with Operation",
"Definition:Reflexive Reduction",
"Definition:Relation Compatible with Operation",
"Definition:Relation Compatible with Operation",
"Definition:Cancellable Operation",
"Definition:Contradiction",
"Definition:Relation Compatible with Operation",
"Categor... |
proofwiki-19294 | Lexicographically Ordered Pair of Ordered Semigroups with Cancellable Elements/Converse | If $\circ_1$ is not a cancellable operation, then it may not necessarily be the case that $\preccurlyeq_l$ is compatible with $\odot$. | Recall from Lexicographically Ordered Pair of Ordered Semigroups with Cancellable Elements that if $\circ_1$ is cancellable, then $\preccurlyeq_l$ is compatible with $\odot$.
Let $\Z_2$ denote the set of integers modulo $2$:
:$\Z_2 = \set {0, 1}$
Let $\vee_2$ be the operation on $\Z_2$ defined as:
:$\forall a, b \in \Z... | If $\circ_1$ is not a [[Definition:Cancellable Operation|cancellable operation]], then it may not necessarily be the case that $\preccurlyeq_l$ is [[Definition:Relation Compatible with Operation|compatible]] with $\odot$. | Recall from [[Lexicographically Ordered Pair of Ordered Semigroups with Cancellable Elements]] that if $\circ_1$ is [[Definition:Cancellable Operation|cancellable]], then $\preccurlyeq_l$ is [[Definition:Relation Compatible with Operation|compatible]] with $\odot$.
Let $\Z_2$ denote the [[Definition:Set|set]] of [[Def... | Lexicographically Ordered Pair of Ordered Semigroups with Cancellable Elements/Converse | https://proofwiki.org/wiki/Lexicographically_Ordered_Pair_of_Ordered_Semigroups_with_Cancellable_Elements/Converse | https://proofwiki.org/wiki/Lexicographically_Ordered_Pair_of_Ordered_Semigroups_with_Cancellable_Elements/Converse | [
"Lexicographically Ordered Pair of Ordered Semigroups with Cancellable Elements"
] | [
"Definition:Cancellable Operation",
"Definition:Relation Compatible with Operation"
] | [
"Lexicographically Ordered Pair of Ordered Semigroups with Cancellable Elements",
"Definition:Cancellable Operation",
"Definition:Relation Compatible with Operation",
"Definition:Set",
"Definition:Integers Modulo m",
"Definition:Operation/Binary Operation",
"Integers Modulo m under Max Operation form Or... |
proofwiki-19295 | Structure Induced by Semigroup Operation is Semigroup | Let $\struct {T, \circ}$ be a semigroup.
Let $S$ be a set.
Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$.
Then $\struct {T^S, \oplus}$ is a semigroup. | Taking the semigroup axioms in turn: | Let $\struct {T, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $S$ be a [[Definition:Set|set]].
Let $\struct {T^S, \oplus}$ be the [[Definition:Induced Structure|structure on $T^S$ induced]] by $\circ$.
Then $\struct {T^S, \oplus}$ is a [[Definition:Semigroup|semigroup]]. | Taking the [[Axiom:Semigroup Axioms|semigroup axioms]] in turn: | Structure Induced by Semigroup Operation is Semigroup | https://proofwiki.org/wiki/Structure_Induced_by_Semigroup_Operation_is_Semigroup | https://proofwiki.org/wiki/Structure_Induced_by_Semigroup_Operation_is_Semigroup | [
"Semigroups",
"Pointwise Operations"
] | [
"Definition:Semigroup",
"Definition:Set",
"Definition:Pointwise Operation/Induced Structure",
"Definition:Semigroup"
] | [
"Axiom:Semigroup Axioms",
"Axiom:Semigroup Axioms"
] |
proofwiki-19296 | Structure Induced on Set of All Mappings to Ordered Semigroup is Ordered Semigroup | Let $S$ be a set.
Let $\struct {T, \odot, \preccurlyeq}$ be an ordered semigroup.
Let $\struct {T^S, \otimes, \preccurlyeq}$ denote the algebraic structure on $T^S$ induced by $\odot$.
Then $\struct {T^S, \otimes, \preccurlyeq}$ is an ordered semigroup. | From Structure Induced by Semigroup Operation is Semigroup, $\struct {T^S, \otimes}$ is a semigroup
From Ordered Set of All Mappings is Ordered Set, $\struct {T^S, \preccurlyeq}$ is an ordered set.
It remains to be demonstrated that $\preccurlyeq$ is compatible with $\odot$.
Let $f, g \in T^S$ such that $f \preccurlyeq... | Let $S$ be a [[Definition:Set|set]].
Let $\struct {T, \odot, \preccurlyeq}$ be an [[Definition:Ordered Semigroup|ordered semigroup]].
Let $\struct {T^S, \otimes, \preccurlyeq}$ denote the [[Definition:Induced Structure|algebraic structure on $T^S$ induced by $\odot$]].
Then $\struct {T^S, \otimes, \preccurlyeq}$ is... | From [[Structure Induced by Semigroup Operation is Semigroup]], $\struct {T^S, \otimes}$ is a [[Definition:Semigroup|semigroup]]
From [[Ordered Set of All Mappings is Ordered Set]], $\struct {T^S, \preccurlyeq}$ is an [[Definition:Ordered Set|ordered set]].
It remains to be demonstrated that $\preccurlyeq$ is [[Defi... | Structure Induced on Set of All Mappings to Ordered Semigroup is Ordered Semigroup | https://proofwiki.org/wiki/Structure_Induced_on_Set_of_All_Mappings_to_Ordered_Semigroup_is_Ordered_Semigroup | https://proofwiki.org/wiki/Structure_Induced_on_Set_of_All_Mappings_to_Ordered_Semigroup_is_Ordered_Semigroup | [
"Examples of Ordered Semigroups",
"Pointwise Operations"
] | [
"Definition:Set",
"Definition:Ordered Semigroup",
"Definition:Pointwise Operation/Induced Structure",
"Definition:Ordered Semigroup"
] | [
"Structure Induced by Semigroup Operation is Semigroup",
"Definition:Semigroup",
"Ordered Set of All Mappings is Ordered Set",
"Definition:Ordered Set",
"Definition:Relation Compatible with Operation",
"Definition:Ordered Semigroup",
"Definition:Relation Compatible with Operation"
] |
proofwiki-19297 | Power Set with Intersection and Subset Relation is Ordered Semigroup | Let $S$ be a set and let $\powerset S$ be its power set.
Let $\struct {\powerset S, \cap, \subseteq}$ be the ordered structure formed from the set intersection operation and subset relation.
Then $\struct {\powerset S, \cap, \subseteq}$ is an ordered semigroup. | From Power Set with Intersection is Commutative Monoid, $\struct {\powerset S, \cap}$ is {{afortiori}} a semigroup.
From Subset Relation is Ordering, $\struct {\powerset S, \subseteq}$ is an ordered set.
It remains to be shown that $\subseteq$ is compatible with $\cap$.
Let $A, B \subseteq S$ be arbitrary such that $A ... | Let $S$ be a [[Definition:Set|set]] and let $\powerset S$ be its [[Definition:Power Set|power set]].
Let $\struct {\powerset S, \cap, \subseteq}$ be the [[Definition:Ordered Structure|ordered structure]] formed from the [[Definition:Set Intersection|set intersection operation]] and [[Definition:Subset|subset relation]... | From [[Power Set with Intersection is Commutative Monoid]], $\struct {\powerset S, \cap}$ is {{afortiori}} a [[Definition:Semigroup|semigroup]].
From [[Subset Relation is Ordering]], $\struct {\powerset S, \subseteq}$ is an [[Definition:Ordered Set|ordered set]].
It remains to be shown that $\subseteq$ is [[Definiti... | Power Set with Intersection and Subset Relation is Ordered Semigroup | https://proofwiki.org/wiki/Power_Set_with_Intersection_and_Subset_Relation_is_Ordered_Semigroup | https://proofwiki.org/wiki/Power_Set_with_Intersection_and_Subset_Relation_is_Ordered_Semigroup | [
"Examples of Ordered Semigroups",
"Power Set",
"Set Intersection",
"Subsets"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Ordered Structure",
"Definition:Set Intersection",
"Definition:Subset",
"Definition:Ordered Semigroup"
] | [
"Power Set with Intersection is Commutative Monoid",
"Definition:Semigroup",
"Subset Relation is Ordering",
"Definition:Ordered Set",
"Definition:Relation Compatible with Operation",
"Set Intersection Preserves Subsets/Corollary",
"Intersection is Commutative"
] |
proofwiki-19298 | Power Set with Intersection and Superset Relation is Ordered Semigroup | Let $S$ be a set and let $\powerset S$ be its power set.
Let $\struct {\powerset S, \cap, \supseteq}$ be the ordered structure formed from the set intersection operation and superset relation.
Then $\struct {\powerset S, \cap, \supseteq}$ is an ordered semigroup. | From Power Set with Intersection is Commutative Monoid, $\struct {\powerset S, \cap}$ is {{afortiori}} a semigroup.
From Subset Relation is Ordering, $\struct {\powerset S, \subseteq}$ is an ordered set.
We have that $\supseteq$ is the dual to $\subseteq$.
Hence $\struct {\powerset S, \supseteq}$ is an ordered set.
It ... | Let $S$ be a [[Definition:Set|set]] and let $\powerset S$ be its [[Definition:Power Set|power set]].
Let $\struct {\powerset S, \cap, \supseteq}$ be the [[Definition:Ordered Structure|ordered structure]] formed from the [[Definition:Set Intersection|set intersection operation]] and [[Definition:Superset|superset relat... | From [[Power Set with Intersection is Commutative Monoid]], $\struct {\powerset S, \cap}$ is {{afortiori}} a [[Definition:Semigroup|semigroup]].
From [[Subset Relation is Ordering]], $\struct {\powerset S, \subseteq}$ is an [[Definition:Ordered Set|ordered set]].
We have that $\supseteq$ is the [[Definition:Dual Orde... | Power Set with Intersection and Superset Relation is Ordered Semigroup | https://proofwiki.org/wiki/Power_Set_with_Intersection_and_Superset_Relation_is_Ordered_Semigroup | https://proofwiki.org/wiki/Power_Set_with_Intersection_and_Superset_Relation_is_Ordered_Semigroup | [
"Examples of Ordered Semigroups",
"Power Set",
"Set Intersection",
"Subsets"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Ordered Structure",
"Definition:Set Intersection",
"Definition:Subset/Superset",
"Definition:Ordered Semigroup"
] | [
"Power Set with Intersection is Commutative Monoid",
"Definition:Semigroup",
"Subset Relation is Ordering",
"Definition:Ordered Set",
"Definition:Dual Ordering",
"Definition:Ordered Set",
"Definition:Relation Compatible with Operation",
"Set Intersection Preserves Subsets/Corollary",
"Intersection i... |
proofwiki-19299 | Power Set with Union and Subset Relation is Ordered Semigroup | Let $S$ be a set and let $\powerset S$ be its power set.
Let $\struct {\powerset S, \cup, \subseteq}$ be the ordered structure formed from the set union operation and subset relation.
Then $\struct {\powerset S, \cup, \subseteq}$ is an ordered semigroup. | From Power Set with Union is Commutative Monoid, $\struct {\powerset S, \cup}$ is {{afortiori}} a semigroup.
From Subset Relation is Ordering, $\struct {\powerset S, \subseteq}$ is an ordered set.
It remains to be shown that $\subseteq$ is compatible with $\cup$.
Let $A, B \subseteq S$ be arbitrary such that $A \subset... | Let $S$ be a [[Definition:Set|set]] and let $\powerset S$ be its [[Definition:Power Set|power set]].
Let $\struct {\powerset S, \cup, \subseteq}$ be the [[Definition:Ordered Structure|ordered structure]] formed from the [[Definition:Set Union|set union operation]] and [[Definition:Subset|subset relation]].
Then $\st... | From [[Power Set with Union is Commutative Monoid]], $\struct {\powerset S, \cup}$ is {{afortiori}} a [[Definition:Semigroup|semigroup]].
From [[Subset Relation is Ordering]], $\struct {\powerset S, \subseteq}$ is an [[Definition:Ordered Set|ordered set]].
It remains to be shown that $\subseteq$ is [[Definition:Rela... | Power Set with Union and Subset Relation is Ordered Semigroup | https://proofwiki.org/wiki/Power_Set_with_Union_and_Subset_Relation_is_Ordered_Semigroup | https://proofwiki.org/wiki/Power_Set_with_Union_and_Subset_Relation_is_Ordered_Semigroup | [
"Examples of Ordered Semigroups",
"Power Set",
"Set Union",
"Subsets"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Ordered Structure",
"Definition:Set Union",
"Definition:Subset",
"Definition:Ordered Semigroup"
] | [
"Power Set with Union is Commutative Monoid",
"Definition:Semigroup",
"Subset Relation is Ordering",
"Definition:Ordered Set",
"Definition:Relation Compatible with Operation",
"Set Union Preserves Subsets/Corollary",
"Union is Commutative"
] |
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