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-5900 | Preorder Induced by Preorder Category | Let $\struct {S, \precsim}$ be a preordered set.
Let $\mathbf S$ be its associated preorder category.
Let $\precsim'$ be the preorder induced by $\mathbf S$ as on Category Induces Preorder.
Then $\precsim'$ is the same as $\precsim$. | Suppose that for some $a, b \in S$, we have:
:$a \precsim' b$
By Category Induces Preorder, this happens {{iff}} there exists an $\mathbf S$-morphism $f: a \to b$.
By definition of $\mathbf S$, this $f: a \to b$ exists {{iff}}:
:$a \precsim b$
Hence the result.
{{qed}}
Category:Preorder Categories
3vb5btihlg5h22v8p3fq6... | Let $\struct {S, \precsim}$ be a [[Definition:Preordered Set|preordered set]].
Let $\mathbf S$ be its associated [[Definition:Preorder Category|preorder category]].
Let $\precsim'$ be the [[Definition:Preordering|preorder]] induced by $\mathbf S$ as on [[Category Induces Preorder]].
Then $\precsim'$ is the same as ... | Suppose that for some $a, b \in S$, we have:
:$a \precsim' b$
By [[Category Induces Preorder]], this happens {{iff}} there exists an $\mathbf S$-[[Definition:Morphism (Category Theory)|morphism]] $f: a \to b$.
By definition of $\mathbf S$, this $f: a \to b$ exists {{iff}}:
:$a \precsim b$
Hence the result.
{{qed... | Preorder Induced by Preorder Category | https://proofwiki.org/wiki/Preorder_Induced_by_Preorder_Category | https://proofwiki.org/wiki/Preorder_Induced_by_Preorder_Category | [
"Preorder Categories"
] | [
"Definition:Preordering/Preordered Set",
"Definition:Preorder Category",
"Definition:Preordering",
"Category Induces Preorder"
] | [
"Category Induces Preorder",
"Definition:Morphism",
"Category:Preorder Categories"
] |
proofwiki-5901 | Functor between Order Categories | Let $\struct {S, \preceq}$ and $\struct {T, \preceq'}$ be ordered sets.
Let $\mathbf S$ and $\mathbf T$ be their associated order categories, respectively.
Let $F: \mathbf S \to \mathbf T$ be a functor.
Then its object functor $F: S \to T$ is a monotone mapping. | Suppose that for some $a, b \in S$, we have:
:$a \preceq b$
Then there is a morphism $a \to b$ in $\mathbf S$.
As $F$ is a functor, it follows that there is a morphism:
:$F a \to F b$
in $\mathbf T$ as well, that is:
:$F a \preceq' F b$
Hence the result.
{{qed}} | Let $\struct {S, \preceq}$ and $\struct {T, \preceq'}$ be [[Definition:Ordered Set|ordered sets]].
Let $\mathbf S$ and $\mathbf T$ be their associated [[Definition:Order Category|order categories]], respectively.
Let $F: \mathbf S \to \mathbf T$ be a [[Definition:Covariant Functor|functor]].
Then its [[Definition:O... | Suppose that for some $a, b \in S$, we have:
:$a \preceq b$
Then there is a [[Definition:Morphism (Category Theory)|morphism]] $a \to b$ in $\mathbf S$.
As $F$ is a [[Definition:Functor|functor]], it follows that there is a [[Definition:Morphism (Category Theory)|morphism]]:
:$F a \to F b$
in $\mathbf T$ as well,... | Functor between Order Categories | https://proofwiki.org/wiki/Functor_between_Order_Categories | https://proofwiki.org/wiki/Functor_between_Order_Categories | [
"Order Categories"
] | [
"Definition:Ordered Set",
"Definition:Order Category",
"Definition:Functor/Covariant",
"Definition:Object Functor",
"Definition:Monotone (Order Theory)/Mapping"
] | [
"Definition:Morphism",
"Definition:Functor",
"Definition:Morphism"
] |
proofwiki-5902 | Discrete Category on Set is Discrete Category | Let $S$ be a set.
Let $\map {\mathbf {Dis} } S$ be the discrete category on $S$.
Then $\map {\mathbf {Dis} } S$ determines a unique (up to isomorphism) discrete category $\map {\mathbf {Dis} } S$ whose objects precisely comprise $S$.
{{explain|What is unique up to isomorphism in this context? This is the definition of ... | {{finish|Trivial.}}
{{qed}} | Let $S$ be a [[Definition:Set|set]].
Let $\map {\mathbf {Dis} } S$ be the [[Definition:Discrete Category on Set|discrete category]] on $S$.
Then $\map {\mathbf {Dis} } S$ determines a unique (up to [[Definition:Isomorphism of Categories|isomorphism]]) [[Definition:Discrete Category|discrete category]] $\map {\mathbf... | {{finish|Trivial.}}
{{qed}} | Discrete Category on Set is Discrete Category | https://proofwiki.org/wiki/Discrete_Category_on_Set_is_Discrete_Category | https://proofwiki.org/wiki/Discrete_Category_on_Set_is_Discrete_Category | [
"Discrete Categories"
] | [
"Definition:Set",
"Definition:Discrete Category on Set",
"Definition:Isomorphism of Categories",
"Definition:Discrete Category",
"Definition:Object (Category Theory)",
"Definition:Isomorphism of Categories",
"Definition:Discrete Category on Set"
] | [] |
proofwiki-5903 | Discrete Category is Order Category | Let $\map {\mathbf{Dis} } S$ be a discrete category.
Then $\map {\mathbf{Dis} } S$ is also an order category. | We have, for any morphism $a \to b$ in $\map {\mathbf{Dis} } S$ that $a = b$.
Thus we see that $\map {\mathbf{Dis} } S$ will be an order category {{iff}}:
:$\forall a, b \in S: a \preceq b \iff a = b$
holds for some ordering $\preceq$ on $S$.
The trivial ordering on $S$ accomplishes this.
Hence the result.
{{qed}} | Let $\map {\mathbf{Dis} } S$ be a [[Definition:Discrete Category|discrete category]].
Then $\map {\mathbf{Dis} } S$ is also an [[Definition:Order Category|order category]]. | We have, for any [[Definition:Morphism (Category Theory)|morphism]] $a \to b$ in $\map {\mathbf{Dis} } S$ that $a = b$.
Thus we see that $\map {\mathbf{Dis} } S$ will be an [[Definition:Order Category|order category]] {{iff}}:
:$\forall a, b \in S: a \preceq b \iff a = b$
holds for some [[Definition:Ordering|orderi... | Discrete Category is Order Category | https://proofwiki.org/wiki/Discrete_Category_is_Order_Category | https://proofwiki.org/wiki/Discrete_Category_is_Order_Category | [
"Discrete Categories",
"Order Categories"
] | [
"Definition:Discrete Category",
"Definition:Order Category"
] | [
"Definition:Morphism",
"Definition:Order Category",
"Definition:Ordering",
"Definition:Trivial Ordering"
] |
proofwiki-5904 | Image of Small Class under Mapping is Small | Let $A$ be a mapping.
Let $a$ be a small class.
Then, the image of $a$ under $A$ is small.
{{improve|$A$ being a class variable, I strongly suggest a different letter to denote the mapping}} | {{NotZFC}}
Since $A$ is a mapping:
:$\forall y: \exists x: \forall z: \left({ y A z \implies z = x }\right)$
This satisfies the antecedent of the axiom of replacement. Therefore:
:$\forall w: \exists x: \forall y: \left({ y \in w \implies \forall z: \left({ y A z \implies z \in x }\right) }\right)$
Universal Instantia... | Let $A$ be a [[Definition:Mapping|mapping]].
Let $a$ be a [[Definition:Small Class|small class]].
Then, the [[Definition:Image of Class under Mapping|image]] of $a$ under $A$ is [[Definition:Small Class|small]].
{{improve|$A$ being a class variable, I strongly suggest a different letter to denote the mapping}} | {{NotZFC}}
Since $A$ is a [[Definition:Mapping|mapping]]:
:$\forall y: \exists x: \forall z: \left({ y A z \implies z = x }\right)$
This satisfies the [[Definition:Antecedent|antecedent]] of the [[Axiom:Axiom of Replacement|axiom of replacement]]. Therefore:
:$\forall w: \exists x: \forall y: \left({ y \in w \imp... | Image of Small Class under Mapping is Small | https://proofwiki.org/wiki/Image_of_Small_Class_under_Mapping_is_Small | https://proofwiki.org/wiki/Image_of_Small_Class_under_Mapping_is_Small | [
"Class Theory",
"Class Mappings"
] | [
"Definition:Mapping",
"Definition:Small Class",
"Definition:Image of Class under Mapping",
"Definition:Small Class"
] | [
"Definition:Mapping",
"Definition:Conditional/Antecedent",
"Axiom:Axiom of Replacement",
"Universal Instantiation",
"Definition:Restricted Universal Quantifier",
"Definition:Image of Class under Mapping",
"Axiom of Subsets Equivalents",
"Definition:Image of Class under Mapping",
"Definition:Small Cl... |
proofwiki-5905 | Inverse of Small Relation is Small | Let $a$ be a small class.
Let $a$ also be a relation.
Then the inverse relation of $a$ is small. | Let $A$ equal:
:$\set {\tuple {\tuple {x, y}, \tuple {y, x} } : \tuple {x, y} \in a}$
Then $A$ maps $a$ to its inverse.
{{link wanted|$A$ in general has a name but I can't find it atm}}
Thus, the inverse of $a$ is the image of $a$ under $A$.
By Image of Small Class under Mapping is Small, the inverse of $a$ is small.
{... | Let $a$ be a [[Definition:Small Class|small class]].
Let $a$ also be a [[Definition:Relation|relation]].
Then the [[Definition:Inverse Relation|inverse relation]] of $a$ is [[Definition:Small Class|small]]. | Let $A$ equal:
:$\set {\tuple {\tuple {x, y}, \tuple {y, x} } : \tuple {x, y} \in a}$
Then $A$ [[Definition:Mapping|maps]] $a$ to its [[Definition:Inverse Relation|inverse]].
{{link wanted|$A$ in general has a name but I can't find it atm}}
Thus, the inverse of $a$ is the image of $a$ under $A$.
By [[Image of Sma... | Inverse of Small Relation is Small | https://proofwiki.org/wiki/Inverse_of_Small_Relation_is_Small | https://proofwiki.org/wiki/Inverse_of_Small_Relation_is_Small | [
"Zermelo-Fraenkel Class Theory"
] | [
"Definition:Small Class",
"Definition:Relation",
"Definition:Inverse Relation",
"Definition:Small Class"
] | [
"Definition:Mapping",
"Definition:Inverse Relation",
"Image of Small Class under Mapping is Small",
"Definition:Inverse Relation"
] |
proofwiki-5906 | Functor between Monoid Categories | Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be monoids.
Let $\mathbf S$ and $\mathbf T$ be the associated monoid categories.
Let $F: \mathbf S \to \mathbf T$ be a functor.
Then the morphism functor $F_1$ of $F$ is a monoid homomorphism. | By definition of monoid category, we have:
:$\mathbf S_1 = S, \mathbf T_1 = T$
which in particular are sets.
Hence the morphism functor $F_1: \mathbf S_1 \to \mathbf T_1$ is a mapping $F_1: S \to T$.
For $F_1$ to be a monoid homomorphism, it suffices to show that the following hold:
:$F_1 \left({e_S}\right) = e_T$
:$\f... | Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be [[Definition:Monoid|monoids]].
Let $\mathbf S$ and $\mathbf T$ be the associated [[Definition:Monoid Category|monoid categories]].
Let $F: \mathbf S \to \mathbf T$ be a [[Definition:Covariant Functor|functor]].
Then the [[Definition:Morphism Functor|morphis... | By definition of [[Definition:Monoid Category|monoid category]], we have:
:$\mathbf S_1 = S, \mathbf T_1 = T$
which in particular are [[Definition:Set|sets]].
Hence the [[Definition:Morphism Functor|morphism functor]] $F_1: \mathbf S_1 \to \mathbf T_1$ is a [[Definition:Mapping|mapping]] $F_1: S \to T$.
For $F_1$ ... | Functor between Monoid Categories | https://proofwiki.org/wiki/Functor_between_Monoid_Categories | https://proofwiki.org/wiki/Functor_between_Monoid_Categories | [
"Monoid Categories",
"Monoid Homomorphisms"
] | [
"Definition:Monoid",
"Definition:Monoid Category",
"Definition:Functor/Covariant",
"Definition:Morphism Functor",
"Definition:Monoid Homomorphism"
] | [
"Definition:Monoid Category",
"Definition:Set",
"Definition:Morphism Functor",
"Definition:Mapping",
"Definition:Monoid Homomorphism",
"Definition:Composable Morphisms",
"Definition:Functor"
] |
proofwiki-5907 | Category of Monoids is Category | Let $\mathbf{Mon}$ be the category of monoids.
Then $\mathbf{Mon}$ is a metacategory. | Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory.
We have Composite of Homomorphisms on Algebraic Structure is Homomorphism, verifying $(C1)$.
We have Identity Mapping is Automorphism providing $\operatorname{id}_S$ for every monoid $\left({S, \circ}\right)$.
Now, $(C2)$ follows from Identity Mapping is ... | Let $\mathbf{Mon}$ be the [[Definition:Category of Monoids|category of monoids]].
Then $\mathbf{Mon}$ is a [[Definition:Metacategory|metacategory]]. | Let us verify the axioms $(C1)$ up to $(C3)$ for a [[Definition:Metacategory|metacategory]].
We have [[Composite of Homomorphisms on Algebraic Structure is Homomorphism]], verifying $(C1)$.
We have [[Identity Mapping is Automorphism]] providing $\operatorname{id}_S$ for every [[Definition:Monoid|monoid]] $\left({S,... | Category of Monoids is Category | https://proofwiki.org/wiki/Category_of_Monoids_is_Category | https://proofwiki.org/wiki/Category_of_Monoids_is_Category | [
"Category of Monoids"
] | [
"Definition:Category of Monoids",
"Definition:Metacategory"
] | [
"Definition:Metacategory",
"Composite of Homomorphisms is Homomorphism/Algebraic Structure",
"Identity Mapping is Automorphism",
"Definition:Monoid",
"Identity Mapping is Left Identity",
"Identity Mapping is Right Identity",
"Composition of Mappings is Associative",
"Definition:Metacategory",
"Categ... |
proofwiki-5908 | Cayley's Representation Theorem/General Case | Let $\struct {G, \cdot}$ be a group.
Then there exists a permutation group $P$ on some set $S$ such that:
:$G \cong P$
That is, such that $G$ is isomorphic to $P$. | Let $G$ be a group and let $a \in G$.
Consider the left regular representation $\lambda_a: G \to G$ defined as:
:$\map {\lambda_a} x = a \cdot x$
From Regular Representations in Group are Permutations we have that $\lambda_a$ is a permutation.
Now let $b \in G$ and consider $\lambda_b: G \to G$ defined as:
:$\map {\lam... | Let $\struct {G, \cdot}$ be a [[Definition:Group|group]].
Then there exists a [[Definition:Permutation Group|permutation group]] $P$ on some [[Definition:Set|set]] $S$ such that:
:$G \cong P$
That is, such that $G$ is [[Definition:Group Isomorphism|isomorphic]] to $P$. | Let $G$ be a [[Definition:Group|group]] and let $a \in G$.
Consider the [[Definition:Left Regular Representation|left regular representation]] $\lambda_a: G \to G$ defined as:
:$\map {\lambda_a} x = a \cdot x$
From [[Regular Representations in Group are Permutations]] we have that $\lambda_a$ is a [[Definition:Permu... | Cayley's Representation Theorem/General Case/Proof 1 | https://proofwiki.org/wiki/Cayley's_Representation_Theorem/General_Case | https://proofwiki.org/wiki/Cayley's_Representation_Theorem/General_Case/Proof_1 | [
"Cayley's Representation Theorem",
"Group Theory"
] | [
"Definition:Group",
"Definition:Permutation Group",
"Definition:Set",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Definition:Group",
"Definition:Regular Representations/Left Regular Representation",
"Regular Representations in Group are Permutations",
"Definition:Permutation",
"Cancellation Laws",
"Definition:Bijection",
"Composition of Regular Representations",
"Definition:Composition of Mappings",
"Definitio... |
proofwiki-5909 | Cayley's Representation Theorem/General Case | Let $\struct {G, \cdot}$ be a group.
Then there exists a permutation group $P$ on some set $S$ such that:
:$G \cong P$
That is, such that $G$ is isomorphic to $P$. | We interpret $\struct {G,\cdot}$ as an group category:
{{DefineCategory
| ob = Only one, say $*$
| mor = $a: * \to *$ for all $a \in G$
| comp = $a \cdot b: * \to *$
| id = $e: * \to *$
}}
where $e$ is the identity element of $G$.
In particular, $G$ is a small category.
By Cayley's Theorem, $G$ is isomorphic to a sub... | Let $\struct {G, \cdot}$ be a [[Definition:Group|group]].
Then there exists a [[Definition:Permutation Group|permutation group]] $P$ on some [[Definition:Set|set]] $S$ such that:
:$G \cong P$
That is, such that $G$ is [[Definition:Group Isomorphism|isomorphic]] to $P$. | We interpret $\struct {G,\cdot}$ as an [[Definition:Group Category|group category]]:
{{DefineCategory
| ob = Only one, say $*$
| mor = $a: * \to *$ for all $a \in G$
| comp = $a \cdot b: * \to *$
| id = $e: * \to *$
}}
where $e$ is the [[Definition:Identity Element|identity element]] of $G$.
In particular, $G$ is a ... | Cayley's Representation Theorem/General Case/Proof 2 | https://proofwiki.org/wiki/Cayley's_Representation_Theorem/General_Case | https://proofwiki.org/wiki/Cayley's_Representation_Theorem/General_Case/Proof_2 | [
"Cayley's Representation Theorem",
"Group Theory"
] | [
"Definition:Group",
"Definition:Permutation Group",
"Definition:Set",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Definition:Group Category",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Small Category",
"Cayley's Theorem (Category Theory)",
"Definition:Isomorphism of Categories/Isomorphic Categories",
"Definition:Subcategory",
"Definition:Functor/Covariant",
"Definition:Identity Func... |
proofwiki-5910 | Permutation of Cosets/Corollary 1 | Let $G$ be a group.
Let $H \le G$ such that $\index G H = n$ where $n \in \Z$.
Then:
:$\exists N \lhd G: N \lhd H: n \divides \index G H \divides n!$ | Apply Permutation of Cosets to $H$ and let $N = \map \ker \theta$.
Then:
:$N \lhd G$ and $N \lhd H$
so from the Correspondence Theorem:
:$H / N \le G / N$
such that:
:$\index {G / N} {H / N} = n$
Thus:
:$n \divides \index G N$
Also by Permutation of Cosets:
:$\exists K \in S_n: G / N \cong K$
Thus:
:$\index G N \divide... | Let $G$ be a [[Definition:Group|group]].
Let $H \le G$ such that $\index G H = n$ where $n \in \Z$.
Then:
:$\exists N \lhd G: N \lhd H: n \divides \index G H \divides n!$ | Apply [[Permutation of Cosets]] to $H$ and let $N = \map \ker \theta$.
Then:
:$N \lhd G$ and $N \lhd H$
so from the [[Correspondence Theorem (Group Theory)|Correspondence Theorem]]:
:$H / N \le G / N$
such that:
:$\index {G / N} {H / N} = n$
Thus:
:$n \divides \index G N$
Also by [[Permutation of Cosets]]:
:$\exists... | Permutation of Cosets/Corollary 1 | https://proofwiki.org/wiki/Permutation_of_Cosets/Corollary_1 | https://proofwiki.org/wiki/Permutation_of_Cosets/Corollary_1 | [
"Symmetric Groups",
"Cosets"
] | [
"Definition:Group"
] | [
"Permutation of Cosets",
"Correspondence Theorem (Group Theory)",
"Permutation of Cosets"
] |
proofwiki-5911 | Cayley's Theorem (Category Theory) | Let $\mathbf C$ be a small category.
Denote with $\mathbf{Set}$ the category of sets.
Then there exists a category $\mathbf D$, subject to:
:$(1): \quad $ The objects of $\mathbf D$ are sets.
:$(2): \quad $ The morphisms of $\mathbf D$ are mappings.
:$(3): \quad \mathbf C \cong \mathbf D$, i.e. $\mathbf C$ and $\mathbf... | Define a functor $H: \mathbf C \to \mathbf{Set}$ by:
:$H C := \set {f \in \operatorname{mor} \mathbf C: \operatorname{cod} f = C}$
:$H f: H A \to H B, g \mapsto f \circ g$
for $f: A \to B$ a morphism of $\mathbf C$.
It is immediate by the definition of identity morphism that:
:$\map H {\operatorname{id}_A} = \operatorn... | Let $\mathbf C$ be a [[Definition:Small Category|small category]].
Denote with $\mathbf{Set}$ the [[Definition:Category of Sets|category of sets]].
Then there exists a [[Definition:Category|category]] $\mathbf D$, subject to:
:$(1): \quad $ The objects of $\mathbf D$ are [[Definition:Set|sets]].
:$(2): \quad $ The ... | Define a [[Definition:Covariant Functor|functor]] $H: \mathbf C \to \mathbf{Set}$ by:
:$H C := \set {f \in \operatorname{mor} \mathbf C: \operatorname{cod} f = C}$
:$H f: H A \to H B, g \mapsto f \circ g$
for $f: A \to B$ a [[Definition:Morphism (Category Theory)|morphism]] of $\mathbf C$.
It is immediate by the de... | Cayley's Theorem (Category Theory) | https://proofwiki.org/wiki/Cayley's_Theorem_(Category_Theory) | https://proofwiki.org/wiki/Cayley's_Theorem_(Category_Theory) | [
"Category Theory"
] | [
"Definition:Small Category",
"Definition:Category of Sets",
"Definition:Category",
"Definition:Set",
"Definition:Mapping",
"Definition:Isomorphism of Categories/Isomorphic Categories",
"Definition:Isomorphism of Categories/Isomorphic Categories",
"Definition:Subcategory"
] | [
"Definition:Functor/Covariant",
"Definition:Morphism",
"Definition:Identity Morphism",
"Composition of Mappings is Associative",
"Equality of Mappings",
"Definition:Functor/Covariant",
"Definition:Injective on Objects",
"Definition:Faithful Functor",
"Definition:Morphism",
"Equality of Mappings",
... |
proofwiki-5912 | Domain of Small Relation is Small | Let $a$ be a small class.
Let $a$ also be a relation.
Then the domain of $a$ is small. | Let $A$ equal:
:$\set {\tuple {\tuple {x, y}, x}: \tuple {x, y} \in a}$
Then, $A$ maps $a$ to its domain.
Thus, the domain of $a$ is the image of $a$ under $A$.
By Image of Small Class under Mapping is Small, the domain of $a$ is small.
{{explain|And exactly why is $A$ small, so that this result applies? Same qn for Ra... | Let $a$ be a [[Definition:Small Class|small class]].
Let $a$ also be a [[Definition:Relation|relation]].
Then the [[Definition:Domain of Relation|domain]] of $a$ is [[Definition:Small Class|small]]. | Let $A$ equal:
:$\set {\tuple {\tuple {x, y}, x}: \tuple {x, y} \in a}$
Then, $A$ [[Definition:Mapping|maps]] $a$ to its [[Definition:Domain of Relation|domain]].
Thus, the domain of $a$ is the [[Definition:Image of Element under Mapping|image]] of $a$ under $A$.
By [[Image of Small Class under Mapping is Small]],... | Domain of Small Relation is Small | https://proofwiki.org/wiki/Domain_of_Small_Relation_is_Small | https://proofwiki.org/wiki/Domain_of_Small_Relation_is_Small | [
"Zermelo-Fraenkel Class Theory"
] | [
"Definition:Small Class",
"Definition:Relation",
"Definition:Domain (Set Theory)/Relation",
"Definition:Small Class"
] | [
"Definition:Mapping",
"Definition:Domain (Set Theory)/Relation",
"Definition:Image (Set Theory)/Mapping/Element",
"Image of Small Class under Mapping is Small",
"Definition:Domain (Set Theory)/Relation",
"Range of Small Relation is Small"
] |
proofwiki-5913 | Range of Small Relation is Small | Let $a$ be a small class.
Let $a$ also be a relation.
Then the range of $a$ is small. | Let $A$ equal:
:$\set {\tuple {\tuple {x, y}, y}: \tuple {x, y} \in a}$
Then, $A$ maps $a$ to its range.
Thus, the range of $a$ is the image of $A$.
By Image of Small Class under Mapping is Small, the range of $a$ is small.
{{explain|Why is $A$ small, so that Image of Small Class under Mapping is Small of applies?}}
{{... | Let $a$ be a [[Definition:Small Class|small class]].
Let $a$ also be a [[Definition:Relation|relation]].
Then the [[Definition:Range of Relation|range]] of $a$ is [[Definition:Small Class|small]]. | Let $A$ equal:
:$\set {\tuple {\tuple {x, y}, y}: \tuple {x, y} \in a}$
Then, $A$ [[Definition:Mapping|maps]] $a$ to its [[Definition:Range of Relation|range]].
Thus, the [[Definition:Range of Relation|range]] of $a$ is the [[Definition:Image of Mapping|image]] of $A$.
By [[Image of Small Class under Mapping is Sm... | Range of Small Relation is Small | https://proofwiki.org/wiki/Range_of_Small_Relation_is_Small | https://proofwiki.org/wiki/Range_of_Small_Relation_is_Small | [
"Zermelo-Fraenkel Class Theory"
] | [
"Definition:Small Class",
"Definition:Relation",
"Definition:Range of Relation",
"Definition:Small Class"
] | [
"Definition:Mapping",
"Definition:Range of Relation",
"Definition:Range of Relation",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Image of Small Class under Mapping is Small",
"Definition:Range of Relation",
"Image of Small Class under Mapping is Small"
] |
proofwiki-5914 | Center of Opposite Group | Let $\struct {G, \circ}$ be a group.
Let $\struct {G, *}$ be the opposite group to $G$.
Let $\map Z {G, \circ}$ and $\map Z {G, *}$ be the centers of $\struct {G, \circ}$ and $\struct {G, *}$, respectively.
Then:
:$\map Z {G, \circ} = \map Z {G, *}$ | We have, for $g \in G$:
{{begin-eqn}}
{{eqn | l = g \in \map Z {G, \circ}
| o = \leadstoandfrom
| r = \forall h \in G: g \circ h = h \circ g
| c = {{Defof|Center of Group}}
}}
{{eqn | o = \leadstoandfrom
| r = \forall h \in G: h * g = g * h
| c = {{Defof|Opposite Group}}
}}
{{eqn | o = \le... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $\struct {G, *}$ be the [[Definition:Opposite Group|opposite group]] to $G$.
Let $\map Z {G, \circ}$ and $\map Z {G, *}$ be the [[Definition:Center of Group|centers]] of $\struct {G, \circ}$ and $\struct {G, *}$, respectively.
Then:
:$\map Z {G, \circ} ... | We have, for $g \in G$:
{{begin-eqn}}
{{eqn | l = g \in \map Z {G, \circ}
| o = \leadstoandfrom
| r = \forall h \in G: g \circ h = h \circ g
| c = {{Defof|Center of Group}}
}}
{{eqn | o = \leadstoandfrom
| r = \forall h \in G: h * g = g * h
| c = {{Defof|Opposite Group}}
}}
{{eqn | o = \l... | Center of Opposite Group | https://proofwiki.org/wiki/Center_of_Opposite_Group | https://proofwiki.org/wiki/Center_of_Opposite_Group | [
"Opposite Groups",
"Centers of Groups"
] | [
"Definition:Group",
"Definition:Opposite Group",
"Definition:Center (Abstract Algebra)/Group"
] | [
"Definition:Set Equality",
"Category:Opposite Groups",
"Category:Centers of Groups"
] |
proofwiki-5915 | Opposite Group of Opposite Group | Let $\struct {G, \circ}$ be a group.
Let $\struct {G, *}$ be the opposite group to $\struct {G, \circ}$.
Let $\struct {G, \circ'}$ be the opposite group to $\struct {G, *}$.
Then:
:$\struct {G, \circ} = \struct {G, \circ'}$ | We have, for all $a, b \in G$:
:$a \circ b = b * a = a \circ' b$
by definition of opposite group.
Hence the result.
{{qed}}
Category:Opposite Groups
hbswvtiqyapehbhxyoz37xhcjs0c5n4 | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $\struct {G, *}$ be the [[Definition:Opposite Group|opposite group]] to $\struct {G, \circ}$.
Let $\struct {G, \circ'}$ be the [[Definition:Opposite Group|opposite group]] to $\struct {G, *}$.
Then:
:$\struct {G, \circ} = \struct {G, \circ'}$ | We have, for all $a, b \in G$:
:$a \circ b = b * a = a \circ' b$
by definition of [[Definition:Opposite Group|opposite group]].
Hence the result.
{{qed}}
[[Category:Opposite Groups]]
hbswvtiqyapehbhxyoz37xhcjs0c5n4 | Opposite Group of Opposite Group | https://proofwiki.org/wiki/Opposite_Group_of_Opposite_Group | https://proofwiki.org/wiki/Opposite_Group_of_Opposite_Group | [
"Opposite Groups"
] | [
"Definition:Group",
"Definition:Opposite Group",
"Definition:Opposite Group"
] | [
"Definition:Opposite Group",
"Category:Opposite Groups"
] |
proofwiki-5916 | Slice Category is Category | Let $\mathbf C$ be a metacategory.
Let $C \in \mathbf C_0$ be a object of $\mathbf C$.
Let $\mathbf C / C$ be the slice category of $\mathbf C$ over $C$.
Then $\mathbf C / C$ is a metacategory. | Let us verify the axioms $(\text C 1)$ up to $(\text C 3)$ for a metacategory.
Suppose that $a: f \to g$ and $b: g \to h$ are morphisms in $\mathbf C / C$.
To show that $b \circ a: f \to h$ is a morphism as well, compute:
{{begin-eqn}}
{{eqn|l = \paren {b \circ a} \circ f
|r = b \circ \paren {a \circ f}
|c = ... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $C \in \mathbf C_0$ be a [[Definition:Object|object]] of $\mathbf C$.
Let $\mathbf C / C$ be the [[Definition:Slice Category|slice category]] of $\mathbf C$ over $C$.
Then $\mathbf C / C$ is a [[Definition:Metacategory|metacategory]]. | Let us verify the axioms $(\text C 1)$ up to $(\text C 3)$ for a [[Definition:Metacategory|metacategory]].
Suppose that $a: f \to g$ and $b: g \to h$ are [[Definition:Morphism (Category Theory)|morphisms]] in $\mathbf C / C$.
To show that $b \circ a: f \to h$ is a [[Definition:Morphism (Category Theory)|morphism]] a... | Slice Category is Category | https://proofwiki.org/wiki/Slice_Category_is_Category | https://proofwiki.org/wiki/Slice_Category_is_Category | [
"Slice Categories"
] | [
"Definition:Metacategory",
"Definition:Object",
"Definition:Slice Category",
"Definition:Metacategory"
] | [
"Definition:Metacategory",
"Definition:Morphism",
"Definition:Morphism",
"Definition:Composition of Morphisms",
"Definition:Associative Operation",
"Definition:Morphism",
"Definition:Morphism",
"Definition:Morphism",
"Definition:Object (Category Theory)",
"Definition:Metacategory",
"Definition:C... |
proofwiki-5917 | Monoid Category is Category | Let $\struct {S, \circ}$ be a monoid with identity $e_S$.
Let $\mathbf S$ be the associated monoid category.
Then $\mathbf S$ is a category. | Let us verify the axioms $(\text C 1)$ up to $(\text C 3)$ for a metacategory.
Suppose that $a, b \in \mathbf S_1$ are morphisms.
From {{MonoidAxiom|0}}, $a \circ b$ is also a morphism.
Axiom $(\text C 2)$ follows directly from {{MonoidAxiom|2}}.
Axiom $(\text C 3)$ follows directly from {{MonoidAxiom|1}}.
Hence $\math... | Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]] with [[Definition:Identity Element|identity]] $e_S$.
Let $\mathbf S$ be the associated [[Definition:Monoid Category|monoid category]].
Then $\mathbf S$ is a [[Definition:Category|category]]. | Let us verify the axioms $(\text C 1)$ up to $(\text C 3)$ for a [[Definition:Metacategory|metacategory]].
Suppose that $a, b \in \mathbf S_1$ are [[Definition:Morphism (Category Theory)|morphisms]].
From {{MonoidAxiom|0}}, $a \circ b$ is also a [[Definition:Morphism (Category Theory)|morphism]].
Axiom $(\text C 2... | Monoid Category is Category | https://proofwiki.org/wiki/Monoid_Category_is_Category | https://proofwiki.org/wiki/Monoid_Category_is_Category | [
"Monoid Categories"
] | [
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Monoid Category",
"Definition:Category"
] | [
"Definition:Metacategory",
"Definition:Morphism",
"Definition:Morphism",
"Definition:Metacategory",
"Definition:Object",
"Definition:Category",
"Category:Monoid Categories"
] |
proofwiki-5918 | Dual Category is Category | Let $\mathbf C$ be a metacategory.
Let $\mathbf C^{\text{op} }$ be its dual category.
Then $\mathbf C^{\text{op} }$ is also a metacategory. | Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory.
Let $f^{\text{op} }: C^{\text{op} } \to D^{\text{op} }$ and $g^{\text{op} }: D^{\text{op} } \to E^{\text{op} }$ be morphisms in $\mathbf C^{\text{op} }$.
Then $f: D \to C$ and $g: E \to D$ are morphisms in $\mathbf C$, and so is $f \circ g: E \to C$.
Ther... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $\mathbf C^{\text{op} }$ be its [[Definition:Dual Category|dual category]].
Then $\mathbf C^{\text{op} }$ is also a [[Definition:Metacategory|metacategory]]. | Let us verify the axioms $(C1)$ up to $(C3)$ for a [[Definition:Metacategory|metacategory]].
Let $f^{\text{op} }: C^{\text{op} } \to D^{\text{op} }$ and $g^{\text{op} }: D^{\text{op} } \to E^{\text{op} }$ be [[Definition:Morphism (Category Theory)|morphisms]] in $\mathbf C^{\text{op} }$.
Then $f: D \to C$ and $g: E ... | Dual Category is Category | https://proofwiki.org/wiki/Dual_Category_is_Category | https://proofwiki.org/wiki/Dual_Category_is_Category | [
"Dual Categories"
] | [
"Definition:Metacategory",
"Definition:Dual Category",
"Definition:Metacategory"
] | [
"Definition:Metacategory",
"Definition:Morphism",
"Definition:Morphism",
"Definition:Morphism",
"Definition:Metacategory",
"Definition:Metacategory",
"Definition:Metacategory",
"Definition:Metacategory",
"Category:Dual Categories"
] |
proofwiki-5919 | Signum Function on Integers is Extension of Signum on Natural Numbers | Let $\sgn_\Z: \Z \to \set {-1, 0, 1}$ be the signum function on the integers.
Let $\sgn_\N: \N \to \set {0, 1}$ be the signum function on the natural numbers.
Then $\sgn_\Z: \Z \to \Z$ is an extension of $\sgn_\N: \N \to \N$. | Let $n \in \Z: n \ge 0$.
Then by definition of the signum function:
:$\map {\sgn_\Z} n = \begin {cases} 0 & : n = 0 \\ 1 & : n > 0 \end {cases}$
So by definition of the signum function on the natural numbers:
:$\forall n \in \N: \map {\sgn_\Z} n = \map {\sgn_\N} n$
Hence the result, by definition of extension.
{{qed}}
... | Let $\sgn_\Z: \Z \to \set {-1, 0, 1}$ be the [[Definition:Signum Function|signum function on the integers]].
Let $\sgn_\N: \N \to \set {0, 1}$ be the [[Definition:Signum Function/Natural Numbers|signum function on the natural numbers]].
Then $\sgn_\Z: \Z \to \Z$ is an [[Definition:Extension of Mapping|extension]] of... | Let $n \in \Z: n \ge 0$.
Then by definition of the [[Definition:Signum Function|signum function]]:
:$\map {\sgn_\Z} n = \begin {cases} 0 & : n = 0 \\ 1 & : n > 0 \end {cases}$
So by definition of the [[Definition:Signum Function/Natural Numbers|signum function on the natural numbers]]:
:$\forall n \in \N: \map {\sgn_... | Signum Function on Integers is Extension of Signum on Natural Numbers | https://proofwiki.org/wiki/Signum_Function_on_Integers_is_Extension_of_Signum_on_Natural_Numbers | https://proofwiki.org/wiki/Signum_Function_on_Integers_is_Extension_of_Signum_on_Natural_Numbers | [
"Number Theory"
] | [
"Definition:Signum Function",
"Definition:Signum Function/Natural Numbers",
"Definition:Extension of Mapping"
] | [
"Definition:Signum Function",
"Definition:Signum Function/Natural Numbers",
"Definition:Extension of Mapping",
"Category:Number Theory"
] |
proofwiki-5920 | Cartesian Product is Small iff Inverse is Small | Let $A$ and $B$ be classes.
Then the Cartesian product $A \times B$ is a small class {{iff}} $B \times A$ is small. | {{begin-eqn}}
{{eqn | l = A \times B
| r = \set {\tuple {x, y} : x \in A \land y \in B}
| c = {{Defof|Cartesian Product}}
}}
{{eqn | r = \set {\tuple {y, x} : x \in A \land y \in B}^{-1}
| c = {{Defof|Inverse Relation}}
}}
{{eqn | r = \paren {B \times A}^{-1}
| c = {{Defof|Cartesian Product}}
}}... | Let $A$ and $B$ be [[Definition:Class (Class Theory)|classes]].
Then the [[Definition:Cartesian Product|Cartesian product]] $A \times B$ is a [[Definition:Small Class|small class]] {{iff}} $B \times A$ is [[Definition:Small Class|small]]. | {{begin-eqn}}
{{eqn | l = A \times B
| r = \set {\tuple {x, y} : x \in A \land y \in B}
| c = {{Defof|Cartesian Product}}
}}
{{eqn | r = \set {\tuple {y, x} : x \in A \land y \in B}^{-1}
| c = {{Defof|Inverse Relation}}
}}
{{eqn | r = \paren {B \times A}^{-1}
| c = {{Defof|Cartesian Product}}
}}... | Cartesian Product is Small iff Inverse is Small | https://proofwiki.org/wiki/Cartesian_Product_is_Small_iff_Inverse_is_Small | https://proofwiki.org/wiki/Cartesian_Product_is_Small_iff_Inverse_is_Small | [
"Zermelo-Fraenkel Class Theory"
] | [
"Definition:Class (Class Theory)",
"Definition:Cartesian Product",
"Definition:Small Class",
"Definition:Small Class"
] | [
"Definition:Small Class",
"Inverse of Small Relation is Small",
"Inverse of Small Relation is Small"
] |
proofwiki-5921 | Cartesian Product with Proper Class is Proper Class | Let $A$ be a proper class.
Let $B$ be a class which is not empty.
Then the Cartesian product $\paren {A \times B}$ is a proper class. | {{NotZFC}}
{{AimForCont}} that $\paren {A \times B}$ is small.
By Domain of Small Relation is Small, the domain of $\paren {A \times B}$ is small.
Since $B \ne \O$, Nonempty Class has Members shows that $\exists y: y \in B$.
The domain of $\paren {A \times B}$ is the collection of all $x \in A$ such that $\exists y: y ... | Let $A$ be a [[Definition:Proper Class|proper class]].
Let $B$ be a [[Definition:Class (Class Theory)|class]] which is not [[Definition:Empty Set|empty]].
Then the [[Definition:Cartesian Product|Cartesian product]] $\paren {A \times B}$ is a [[Definition:Proper Class|proper class]]. | {{NotZFC}}
{{AimForCont}} that $\paren {A \times B}$ is [[Definition:Small Class|small]].
By [[Domain of Small Relation is Small]], the [[Definition:Domain of Relation|domain]] of $\paren {A \times B}$ is [[Definition:Small Class|small]].
Since $B \ne \O$, [[Nonempty Class has Members]] shows that $\exists y: y \in... | Cartesian Product with Proper Class is Proper Class | https://proofwiki.org/wiki/Cartesian_Product_with_Proper_Class_is_Proper_Class | https://proofwiki.org/wiki/Cartesian_Product_with_Proper_Class_is_Proper_Class | [
"Class Theory"
] | [
"Definition:Class (Class Theory)/Proper Class",
"Definition:Class (Class Theory)",
"Definition:Empty Set",
"Definition:Cartesian Product",
"Definition:Class (Class Theory)/Proper Class"
] | [
"Definition:Small Class",
"Domain of Small Relation is Small",
"Definition:Domain (Set Theory)/Relation",
"Definition:Small Class",
"Nonempty Class has Members",
"Definition:Domain (Set Theory)/Relation",
"Definition:Domain (Set Theory)/Relation",
"Definition:Small Class",
"Definition:Class (Class T... |
proofwiki-5922 | Uniqueness Condition for Relation Value | Let $\RR$ be a relation.
Let $\tuple {x, y} \in \RR$.
Let:
:$\exists ! y: \tuple {x, y} \in \RR$
Then:
:$\map \RR x = y$
where $\map \RR x$ denotes the image of $\RR$ at $x$.
If $y$ is not unique, then:
:$\map \RR x = \O$ | {{begin-eqn}}
{{eqn | n = 1
| l = z \in \map \RR x
| o = \implies
| r = \exists y: \paren {z \in y \land \tuple {x, y} \in \RR} \land \exists ! y: \tuple {x, y} \in \RR
| c = {{Defof|Image (Set Theory)/Relation/Element/Singleton|Image of Element under Relation}}
}}
{{eqn | o = \implies
| r... | Let $\RR$ be a [[Definition:Relation|relation]].
Let $\tuple {x, y} \in \RR$.
Let:
:$\exists ! y: \tuple {x, y} \in \RR$
Then:
:$\map \RR x = y$
where $\map \RR x$ denotes the [[Definition:Image (Set Theory)/Relation/Element/Singleton|image of $\RR$ at $x$]].
If $y$ is not unique, then:
:$\map \RR x = \O$ | {{begin-eqn}}
{{eqn | n = 1
| l = z \in \map \RR x
| o = \implies
| r = \exists y: \paren {z \in y \land \tuple {x, y} \in \RR} \land \exists ! y: \tuple {x, y} \in \RR
| c = {{Defof|Image (Set Theory)/Relation/Element/Singleton|Image of Element under Relation}}
}}
{{eqn | o = \implies
| r... | Uniqueness Condition for Relation Value | https://proofwiki.org/wiki/Uniqueness_Condition_for_Relation_Value | https://proofwiki.org/wiki/Uniqueness_Condition_for_Relation_Value | [
"Relation Theory"
] | [
"Definition:Relation",
"Definition:Image (Set Theory)/Relation/Element/Singleton"
] | [
"Existential Instantiation",
"Substitutivity of Equality",
"Definition:Class Equality"
] |
proofwiki-5923 | Value of Relation is Small | The value of a relation is always a small class. | Let $\RR$ be an arbitrary relation.
Let $s$ be any set.
The value of a relation is either equal to some set $y$ or $\O$ by Uniqueness Condition for Relation Value.
If it is equal to some set $y$, then the value of $s$ under $\RR$ is a small class by the definition of small class.
If it is equal to $\O$, then the result... | The [[Definition:Image (Set Theory)/Relation/Element/Singleton|value of a relation]] is always a [[Definition:Small Class|small class]]. | Let $\RR$ be an [[Definition:Arbitrary|arbitrary]] [[Definition:Relation|relation]].
Let $s$ be any [[Definition:Set|set]].
The [[Definition:Image (Set Theory)/Relation/Element/Singleton|value of a relation]] is either equal to some set $y$ or $\O$ by [[Uniqueness Condition for Relation Value]].
If it is equal to so... | Value of Relation is Small | https://proofwiki.org/wiki/Value_of_Relation_is_Small | https://proofwiki.org/wiki/Value_of_Relation_is_Small | [
"Zermelo-Fraenkel Class Theory"
] | [
"Definition:Image (Set Theory)/Relation/Element/Singleton",
"Definition:Small Class"
] | [
"Definition:Arbitrary",
"Definition:Relation",
"Definition:Set",
"Definition:Image (Set Theory)/Relation/Element/Singleton",
"Uniqueness Condition for Relation Value",
"Definition:Small Class",
"Definition:Small Class",
"Empty Set is Small"
] |
proofwiki-5924 | Mapping whose Domain is Small Class is Small | Let $F$ be a mapping.
Let the domain of $F$ be a small class.
Then, $F$ is a small class. | Let $A$ denote the domain of $F$.
Let $B$ denote the image of $F$.
Since $F$ is a mapping, $F$ is also a relation.
Therefore:
:$F \subseteq A \times B$
where $A \times B$ denotes the Cartesian product of $A$ and $B$.
$B$ is the image of $A$ under $F$ and is therefore a small class by Image of Small Class under Mapping... | Let $F$ be a [[Definition:Mapping|mapping]].
Let the [[Definition:Domain of Mapping|domain]] of $F$ be a [[Definition:Small Class|small class]].
Then, $F$ is a [[Definition:Small Class|small class]]. | Let $A$ denote the [[Definition:Domain of Mapping|domain]] of $F$.
Let $B$ denote the [[Definition:Image of Mapping|image]] of $F$.
Since $F$ is a [[Definition:Mapping|mapping]], $F$ is also a [[Definition:Relation|relation]].
Therefore:
:$F \subseteq A \times B$
where $A \times B$ denotes the [[Definition:Carte... | Mapping whose Domain is Small Class is Small | https://proofwiki.org/wiki/Mapping_whose_Domain_is_Small_Class_is_Small | https://proofwiki.org/wiki/Mapping_whose_Domain_is_Small_Class_is_Small | [
"Class Theory"
] | [
"Definition:Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Small Class",
"Definition:Small Class"
] | [
"Definition:Domain (Set Theory)/Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Mapping",
"Definition:Relation",
"Definition:Cartesian Product",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Small Class",
"Image of Small Class under Mapping is Small",
"Definition... |
proofwiki-5925 | Restriction of Mapping to Small Class is Small | Let $F$ be a mapping.
Let $A$ be a small class.
Then the restriction $F {\restriction_A}$ is a small class. | The domain of $F {\restriction_A}$ is a subset of $A$.
By Axiom of Subsets Equivalents, the domain is a small class.
By Mapping whose Domain is Small Class is Small, it follows that $F {\restriction_A}$ is a small class.
{{qed}} | Let $F$ be a [[Definition:Mapping|mapping]].
Let $A$ be a [[Definition:Small Class|small class]].
Then the [[Definition:Restriction of Mapping|restriction]] $F {\restriction_A}$ is a [[Definition:Small Class|small class]]. | The [[Definition:Domain of Mapping|domain]] of $F {\restriction_A}$ is a [[Definition:Subset|subset]] of $A$.
By [[Axiom of Subsets Equivalents]], the [[Definition:Domain of Mapping|domain]] is a [[Definition:Small Class|small class]].
By [[Mapping whose Domain is Small Class is Small]], it follows that $F {\restrict... | Restriction of Mapping to Small Class is Small | https://proofwiki.org/wiki/Restriction_of_Mapping_to_Small_Class_is_Small | https://proofwiki.org/wiki/Restriction_of_Mapping_to_Small_Class_is_Small | [
"Zermelo-Fraenkel Class Theory"
] | [
"Definition:Mapping",
"Definition:Small Class",
"Definition:Restriction/Mapping",
"Definition:Small Class"
] | [
"Definition:Domain (Set Theory)/Mapping",
"Definition:Subset",
"Axiom of Subsets Equivalents",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Small Class",
"Mapping whose Domain is Small Class is Small",
"Definition:Small Class"
] |
proofwiki-5926 | Preimage of Singleton | Let $\RR$ be a relation.
Let $\map {\RR^{-1} } t$ denote the preimage of $t$ under $\RR$.
Let $\RR^{-1} \sqbrk {\set t}$ denote the preimage of $\set t$ under $\RR$.
Then:
:$\RR^{-1} \sqbrk {\set t} = \map {\RR^{-1} } t$ | {{begin-eqn}}
{{eqn | l = \map {\RR^{-1} } t
| r = \set {s: \paren {s, t} \in \RR}
| c = {{Defof|Preimage of Element under Relation}}
}}
{{eqn | r = \set {s: \exists y: \paren {y = t \land \tuple {s, y} \in \RR} }
| c = Equality implies Substitution
}}
{{eqn | r = \set {s: \exists y \in \set t: \tuple... | Let $\RR$ be a [[Definition:Relation|relation]].
Let $\map {\RR^{-1} } t$ denote the [[Definition:Preimage of Element under Relation|preimage]] of $t$ under $\RR$.
Let $\RR^{-1} \sqbrk {\set t}$ denote the [[Definition:Preimage of Subset under Relation|preimage]] of $\set t$ under $\RR$.
Then:
:$\RR^{-1} \sqbrk {\s... | {{begin-eqn}}
{{eqn | l = \map {\RR^{-1} } t
| r = \set {s: \paren {s, t} \in \RR}
| c = {{Defof|Preimage of Element under Relation}}
}}
{{eqn | r = \set {s: \exists y: \paren {y = t \land \tuple {s, y} \in \RR} }
| c = [[Equality implies Substitution]]
}}
{{eqn | r = \set {s: \exists y \in \set t: \t... | Preimage of Singleton | https://proofwiki.org/wiki/Preimage_of_Singleton | https://proofwiki.org/wiki/Preimage_of_Singleton | [
"Relation Theory",
"Singletons"
] | [
"Definition:Relation",
"Definition:Preimage/Relation/Element",
"Definition:Preimage/Relation/Subset"
] | [
"Equality implies Substitution"
] |
proofwiki-5927 | Order Isomorphism Preserves Strictly Minimal Elements | Let $A_1$ and $A_2$ be classes.
Let $\prec_1$ and $\prec_2$ be relations.
Let $\phi : A_1 \to A_2$ create an order isomorphism between $\struct {A_1, \prec_1}$ and $\struct {A_1, \prec_2}$.
Let $B \subseteq A_1$.
Then $\phi$ maps the strictly minimal elements under $\prec_1$ of $B$ to the strictly minimal elements unde... | Suppose $x$ is a strictly minimal element of $B$. That is:
:$\neg \exists y \in B: x \prec_1 y$
This is equivalent to the statement:
:$\neg \exists z \in \phi \sqbrk B: \map \phi x \prec_2 z$ since $z$ is of the form $\map \phi y$ by the definition of order isomorphism.
{{handwaving|Too much corners cut}}
Thus, this s... | Let $A_1$ and $A_2$ be [[Definition:Class (Class Theory)|classes]].
Let $\prec_1$ and $\prec_2$ be [[Definition:Relation|relations]].
Let $\phi : A_1 \to A_2$ create an [[Definition:Order Isomorphism|order isomorphism]] between $\struct {A_1, \prec_1}$ and $\struct {A_1, \prec_2}$.
Let $B \subseteq A_1$.
Then $\ph... | Suppose $x$ is a [[Definition:Strictly Minimal Element|strictly minimal element]] of $B$. That is:
:$\neg \exists y \in B: x \prec_1 y$
This is equivalent to the statement:
:$\neg \exists z \in \phi \sqbrk B: \map \phi x \prec_2 z$ since $z$ is of the form $\map \phi y$ by the definition of [[Definition:Order Isomo... | Order Isomorphism Preserves Strictly Minimal Elements | https://proofwiki.org/wiki/Order_Isomorphism_Preserves_Strictly_Minimal_Elements | https://proofwiki.org/wiki/Order_Isomorphism_Preserves_Strictly_Minimal_Elements | [
"Order Isomorphisms"
] | [
"Definition:Class (Class Theory)",
"Definition:Relation",
"Definition:Order Isomorphism",
"Definition:Strictly Minimal Element",
"Definition:Strictly Minimal Element"
] | [
"Definition:Strictly Minimal Element",
"Definition:Order Isomorphism",
"Definition:Strictly Minimal Element"
] |
proofwiki-5928 | Order Isomorphism Preserves Initial Segments | Let $A_1$ and $A_2$ be classes.
Let $\prec_1$ and $\prec_2$ be strict orderings.
Let $\phi: A_1 \to A_2$ create an order isomorphism between $\struct {A_1, \prec_1}$ and $\struct {A_2, \prec_2}$.
Suppose $x \in A_1$.
Then $\phi$ maps the $\prec_1$-initial segment of $x$ to the $\prec_2$-initial segment of $\map \phi x$... | $\phi$ maps the $\prec_1$-initial segment of $x$ to:
{{begin-eqn}}
{{eqn | l = \phi \sqbrk {\set {y \in A: y \prec_1 x} }
| r = \phi \sqbrk {\set {y \in A: \map \phi y \prec_2 \map \phi x} }
| c = {{Defof|Initial Segment}} and {{Defof|Order Isomorphism}}
}}
{{eqn | r = \set {\map \phi y \in \phi \sqbrk A: \... | Let $A_1$ and $A_2$ be [[Definition:Class (Class Theory)|classes]].
Let $\prec_1$ and $\prec_2$ be [[Definition:Strict Ordering|strict orderings]].
Let $\phi: A_1 \to A_2$ create an [[Definition:Order Isomorphism|order isomorphism]] between $\struct {A_1, \prec_1}$ and $\struct {A_2, \prec_2}$.
Suppose $x \in A_1$.
... | $\phi$ maps the $\prec_1$-[[Definition:Initial Segment|initial segment]] of $x$ to:
{{begin-eqn}}
{{eqn | l = \phi \sqbrk {\set {y \in A: y \prec_1 x} }
| r = \phi \sqbrk {\set {y \in A: \map \phi y \prec_2 \map \phi x} }
| c = {{Defof|Initial Segment}} and {{Defof|Order Isomorphism}}
}}
{{eqn | r = \set {... | Order Isomorphism Preserves Initial Segments | https://proofwiki.org/wiki/Order_Isomorphism_Preserves_Initial_Segments | https://proofwiki.org/wiki/Order_Isomorphism_Preserves_Initial_Segments | [
"Order Isomorphisms"
] | [
"Definition:Class (Class Theory)",
"Definition:Strict Ordering",
"Definition:Order Isomorphism",
"Definition:Initial Segment",
"Definition:Initial Segment"
] | [
"Definition:Initial Segment",
"Definition:Initial Segment"
] |
proofwiki-5929 | Order Isomorphism on Strictly Well-Founded Relation preserves Strictly Well-Founded Structure | Let $A_1$ and $A_2$ be classes.
Let $\prec_1$ and $\prec_2$ be relations.
Let $\phi: \struct {A_1, \prec_1} \to \struct {A_2, \prec_2}$ be an order isomorphism.
Then $\struct {A_1, \prec_1}$ is a strictly well-founded structure {{iff}} $\struct {A_2, \prec_2}$ is also a strictly well-founded structure. | Take any nonempty subset $B \subseteq A_1$.
From Order Isomorphism Preserves Strictly Minimal Elements:
:$x$ is a strictly minimal element of $B$ {{iff}} $\map \phi x$ is a strictly minimal element of $\phi \sqbrk B$.
By the definition of strictly well-founded relation, $\struct {A_1, \prec_1}$ is strictly well-founded... | Let $A_1$ and $A_2$ be [[Definition:Class (Class Theory)|classes]].
Let $\prec_1$ and $\prec_2$ be [[Definition:Relation|relations]].
Let $\phi: \struct {A_1, \prec_1} \to \struct {A_2, \prec_2}$ be an [[Definition:Order Isomorphism|order isomorphism]].
Then $\struct {A_1, \prec_1}$ is a [[Definition:Strictly Well-... | Take any [[Definition:Non-Empty Set|nonempty]] [[Definition:Subset|subset]] $B \subseteq A_1$.
From [[Order Isomorphism Preserves Strictly Minimal Elements]]:
:$x$ is a [[Definition:Strictly Minimal Element|strictly minimal element]] of $B$ {{iff}} $\map \phi x$ is a [[Definition:Strictly Minimal Element|strictly mini... | Order Isomorphism on Strictly Well-Founded Relation preserves Strictly Well-Founded Structure | https://proofwiki.org/wiki/Order_Isomorphism_on_Strictly_Well-Founded_Relation_preserves_Strictly_Well-Founded_Structure | https://proofwiki.org/wiki/Order_Isomorphism_on_Strictly_Well-Founded_Relation_preserves_Strictly_Well-Founded_Structure | [
"Order Isomorphisms"
] | [
"Definition:Class (Class Theory)",
"Definition:Relation",
"Definition:Order Isomorphism",
"Definition:Strictly Well-Founded Relation",
"Definition:Strictly Well-Founded Relation"
] | [
"Definition:Non-Empty Set",
"Definition:Subset",
"Order Isomorphism Preserves Strictly Minimal Elements",
"Definition:Strictly Minimal Element",
"Definition:Strictly Minimal Element",
"Definition:Strictly Well-Founded Relation",
"Definition:Strictly Well-Founded Relation",
"Definition:Strictly Well-Fo... |
proofwiki-5930 | Alternating Group is Set of Even Permutations | Let $S_n$ denote the symmetric group on $n$ letters.
Let $A_n$ be the alternating group on $n$ letters.
Then $A_n$ consists of the set of even permutations of $S_n$. | Let $\sgn$ denote the sign of permutation on n Letters.
We have that $\map \sgn {S_n}$ is onto $C_2$.
Thus from the First Isomorphism Theorem, $A_n$ consists of the set of even permutations of $S_n$.
{{Qed}} | Let $S_n$ denote the [[Definition:Symmetric Group|symmetric group on $n$ letters]].
Let $A_n$ be the [[Definition:Alternating Group|alternating group on $n$ letters]].
Then $A_n$ consists of the set of [[Definition:Even Permutation|even permutations]] of $S_n$. | Let $\sgn$ denote the [[Definition:Sign of Permutation on n Letters|sign of permutation on n Letters]].
We have that $\map \sgn {S_n}$ is [[Definition:Surjection|onto]] $C_2$.
Thus from the [[First Isomorphism Theorem for Groups|First Isomorphism Theorem]], $A_n$ consists of the [[Definition:Set|set]] of [[Definition... | Alternating Group is Set of Even Permutations | https://proofwiki.org/wiki/Alternating_Group_is_Set_of_Even_Permutations | https://proofwiki.org/wiki/Alternating_Group_is_Set_of_Even_Permutations | [
"Symmetric Groups",
"Alternating Groups"
] | [
"Definition:Symmetric Group",
"Definition:Alternating Group",
"Definition:Even Permutation"
] | [
"Definition:Sign of Permutation on n Letters",
"Definition:Surjection",
"First Isomorphism Theorem/Groups",
"Definition:Set",
"Definition:Even Permutation"
] |
proofwiki-5931 | Induced Relation Generates Order Isomorphism | Let $\struct {A_1, \preceq_1}$ be an ordered set.
Let $\phi: A_1 \to A_2$ be a bijection.
Let:
:$\preceq_2 \mathop{:=} \set {\tuple {\map \phi x, \map \phi y}: x \in A_1 \land y \in A_1 \land x \mathop{\preceq_1} y}$
Then $\phi: \struct {A_1, \preceq_1} \to \struct {A_2, \preceq_2}$ is an order isomorphism. | Take any $x, y \in A_1$ such that $x \preceq_1 y$.
Since $x, y \in A_1$, it follows by the definition of a mapping that:
:$\map \phi x, \map \phi y \in A_2$
So $x \in A_1$ and $y \in A_1$ and $x \preceq_1 y$.
It follows from the definition of $\preceq_2$ that:
:$\map \phi x \preceq_2 \map \phi y$
Conversely, suppose th... | Let $\struct {A_1, \preceq_1}$ be an [[Definition:Ordered Set|ordered set]].
Let $\phi: A_1 \to A_2$ be a [[Definition:Bijection|bijection]].
Let:
:$\preceq_2 \mathop{:=} \set {\tuple {\map \phi x, \map \phi y}: x \in A_1 \land y \in A_1 \land x \mathop{\preceq_1} y}$
Then $\phi: \struct {A_1, \preceq_1} \to \struc... | Take any $x, y \in A_1$ such that $x \preceq_1 y$.
Since $x, y \in A_1$, it follows by the definition of a [[Definition:Mapping|mapping]] that:
:$\map \phi x, \map \phi y \in A_2$
So $x \in A_1$ and $y \in A_1$ and $x \preceq_1 y$.
It follows from the definition of $\preceq_2$ that:
:$\map \phi x \preceq_2 \map \ph... | Induced Relation Generates Order Isomorphism | https://proofwiki.org/wiki/Induced_Relation_Generates_Order_Isomorphism | https://proofwiki.org/wiki/Induced_Relation_Generates_Order_Isomorphism | [
"Order Isomorphisms"
] | [
"Definition:Ordered Set",
"Definition:Bijection",
"Definition:Order Isomorphism"
] | [
"Definition:Mapping",
"Definition:Order Isomorphism"
] |
proofwiki-5932 | Ordinal Multiplication is Left Distributive | Let $x$, $y$, and $z$ be ordinals.
Let $\times$ denote ordinal multiplication.
Let $+$ denote ordinal addition.
Then:
:$x \times \paren {y + z} = \paren {x \times y} + \paren {x \times z}$ | The proof shall proceed by Transfinite Induction, as follows: | Let $x$, $y$, and $z$ be [[Definition:Ordinal|ordinals]].
Let $\times$ denote [[Definition:Ordinal Multiplication|ordinal multiplication]].
Let $+$ denote [[Definition:Ordinal Addition|ordinal addition]].
Then:
:$x \times \paren {y + z} = \paren {x \times y} + \paren {x \times z}$ | The proof shall proceed by [[Transfinite Induction/Schema 2|Transfinite Induction]], as follows: | Ordinal Multiplication is Left Distributive | https://proofwiki.org/wiki/Ordinal_Multiplication_is_Left_Distributive | https://proofwiki.org/wiki/Ordinal_Multiplication_is_Left_Distributive | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Ordinal Multiplication",
"Definition:Ordinal Addition"
] | [
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5933 | Ordinal Multiplication is Associative | Let $x, y, z$ be ordinals.
Let $\times$ denote ordinal multiplication.
Then:
:$x \times \paren {y \times z} = \paren {x \times y} \times z$ | The proof shall proceed by Transfinite Induction on $z$: | Let $x, y, z$ be [[Definition:Ordinal|ordinals]].
Let $\times$ denote [[Definition:Ordinal Multiplication|ordinal multiplication]].
Then:
:$x \times \paren {y \times z} = \paren {x \times y} \times z$ | The proof shall proceed by [[Transfinite Induction/Schema 2|Transfinite Induction]] on $z$: | Ordinal Multiplication is Associative | https://proofwiki.org/wiki/Ordinal_Multiplication_is_Associative | https://proofwiki.org/wiki/Ordinal_Multiplication_is_Associative | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Ordinal Multiplication"
] | [
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5934 | Division Theorem for Ordinals | Let $x$ and $y$ be ordinals.
Let $0$ denote the zero ordinal.
Suppose $y \ne 0$.
Then there exist unique ordinals $z$ and $w$ such that:
:$x = \paren {y \times z} + w$ and $w < y$. | === Existence of $z$ and $w$ ===
First, it must be proven that such $z$ and $w$ exist.
By Relation between Two Ordinals, it follows that:
:$x < y$ or $y \le x$
The proof shall proceed by cases. | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Let $0$ denote the [[Definition:Zero (Ordinal)|zero ordinal]].
Suppose $y \ne 0$.
Then there exist [[Definition:Unique|unique]] ordinals $z$ and $w$ such that:
:$x = \paren {y \times z} + w$ and $w < y$. | === Existence of $z$ and $w$ ===
First, it must be proven that such $z$ and $w$ exist.
By [[Relation between Two Ordinals]], it follows that:
:$x < y$ or $y \le x$
The proof shall proceed by [[Proof by Cases|cases]]. | Division Theorem for Ordinals | https://proofwiki.org/wiki/Division_Theorem_for_Ordinals | https://proofwiki.org/wiki/Division_Theorem_for_Ordinals | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Zero (Ordinal)",
"Definition:Unique"
] | [
"Relation between Two Ordinals",
"Proof by Cases"
] |
proofwiki-5935 | Finite Ordinal Times Ordinal | Let $m$ and $n$ be finite ordinals.
Let $m \ne 0$, where $0$ is the zero ordinal.
Let $x$ be a limit ordinal.
Then:
:$m \times \paren {x + n} = x + \paren {m \times n}$
{{expand|Via Cantor normal form, all ordinals are of the form $x + n$}} | === Lemma ===
{{:Finite Ordinal Times Ordinal/Lemma}}{{qed|lemma}}
By Ordinal Multiplication is Left Distributive:
:$m \times \paren {x + n} = \paren {m \times x} + \paren {m \times n}$
It remains to prove that $x = m \times x$.
Since $x$ is a limit ordinal, it follows that:
{{begin-eqn}}
{{eqn | q = \exists y \in \On
... | Let $m$ and $n$ be [[Definition:Finite Ordinal|finite ordinals]].
Let $m \ne 0$, where $0$ is the [[Definition:Zero (Ordinal)|zero ordinal]].
Let $x$ be a [[Definition:Limit Ordinal|limit ordinal]].
Then:
:$m \times \paren {x + n} = x + \paren {m \times n}$
{{expand|Via [[Definition:Cantor Normal Form|Cantor norma... | === [[Finite Ordinal Times Ordinal/Lemma|Lemma]] ===
{{:Finite Ordinal Times Ordinal/Lemma}}{{qed|lemma}}
By [[Ordinal Multiplication is Left Distributive]]:
:$m \times \paren {x + n} = \paren {m \times x} + \paren {m \times n}$
It remains to prove that $x = m \times x$.
Since $x$ is a [[Definition:Limit Ordinal... | Finite Ordinal Times Ordinal | https://proofwiki.org/wiki/Finite_Ordinal_Times_Ordinal | https://proofwiki.org/wiki/Finite_Ordinal_Times_Ordinal | [
"Ordinal Arithmetic",
"Transfinite Arithmetic",
"Finite Ordinals"
] | [
"Definition:Finite Ordinal",
"Definition:Zero (Ordinal)",
"Definition:Limit Ordinal",
"Definition:Cantor Normal Form"
] | [
"Finite Ordinal Times Ordinal/Lemma",
"Ordinal Multiplication is Left Distributive",
"Definition:Limit Ordinal",
"Factorization of Limit Ordinals",
"Substitutivity of Class Equality",
"Ordinal Multiplication is Associative",
"Finite Ordinal Times Ordinal/Lemma"
] |
proofwiki-5936 | Factorization of Limit Ordinals | Let $x$ be a limit ordinal.
Then:
:$x = \paren {\omega \times y}$ for some $y \in \On$
where $\omega$ is the minimally inductive set. | By the Division Theorem for Ordinals:
:$x = \paren {\omega \times y} + z$
for some unique $y$ and $z \in \omega$.
{{AimForCont}} $z \ne 0$.
Because $z \in \omega$, $z$ is not a limit ordinal.
Therefore, by the definition of limit ordinal:
:$z = w^+$
for some $w \in \omega$.
But this means that:
{{begin-eqn}}
{{eqn | l ... | Let $x$ be a [[Definition:Limit Ordinal|limit ordinal]].
Then:
:$x = \paren {\omega \times y}$ for some $y \in \On$
where $\omega$ is the [[Definition:Minimally Inductive Set|minimally inductive set]]. | By the [[Division Theorem for Ordinals]]:
:$x = \paren {\omega \times y} + z$
for some [[Definition:Unique|unique]] $y$ and $z \in \omega$.
{{AimForCont}} $z \ne 0$.
Because $z \in \omega$, $z$ is not a [[Definition:Limit Ordinal|limit ordinal]].
Therefore, by the definition of [[Definition:Limit Ordinal|limit ord... | Factorization of Limit Ordinals | https://proofwiki.org/wiki/Factorization_of_Limit_Ordinals | https://proofwiki.org/wiki/Factorization_of_Limit_Ordinals | [
"Ordinal Arithmetic",
"Limit Ordinals"
] | [
"Definition:Limit Ordinal",
"Definition:Minimally Inductive Set"
] | [
"Division Theorem for Ordinals",
"Definition:Unique",
"Definition:Limit Ordinal",
"Definition:Limit Ordinal",
"Division Theorem for Ordinals",
"Definition:Ordinal",
"Definition:Limit Ordinal",
"Definition:Contradiction",
"Definition:Limit Ordinal",
"Division Theorem for Ordinals",
"Ordinal Addit... |
proofwiki-5937 | Ordinal Exponentiation is Closed | Let $x$ and $y$ be ordinals.
Then:
:$x^y \in \On$
That is, ordinal exponentiation is closed. | Let $x = 0$ and $y = 0$.
Then by the definition of ordinal exponentiation:
:$x^y = 1$
Let $x = 0$ and $y \ne 0$.
Then by the definition of ordinal exponentiation:
:$x^y = 0$
In either case, $x^y$ is an ordinal.
Now suppose that $x \ne 0$.
The proof shall proceed by Transfinite Induction on $y$. | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Then:
:$x^y \in \On$
That is, [[Definition:Ordinal Exponentiation|ordinal exponentiation]] is [[Definition:Closed Algebraic Structure|closed]]. | Let $x = 0$ and $y = 0$.
Then by the definition of [[Definition:Ordinal Exponentiation|ordinal exponentiation]]:
:$x^y = 1$
Let $x = 0$ and $y \ne 0$.
Then by the definition of [[Definition:Ordinal Exponentiation|ordinal exponentiation]]:
:$x^y = 0$
In either case, $x^y$ is an [[Definition:Ordinal|ordinal]].
Now ... | Ordinal Exponentiation is Closed | https://proofwiki.org/wiki/Ordinal_Exponentiation_is_Closed | https://proofwiki.org/wiki/Ordinal_Exponentiation_is_Closed | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Ordinal Exponentiation",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] | [
"Definition:Ordinal Exponentiation",
"Definition:Ordinal Exponentiation",
"Definition:Ordinal",
"Transfinite Induction/Schema 2",
"Definition:Ordinal Exponentiation",
"Definition:Ordinal",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5938 | Exponent Base of One | Let $x$ be an ordinal.
Then:
:$1^x = 1$ | The proof shall proceed by Transfinite Induction on $x$. | Let $x$ be an [[Definition:Ordinal|ordinal]].
Then:
:$1^x = 1$ | The proof shall proceed by [[Transfinite Induction/Schema 2|Transfinite Induction]] on $x$. | Exponent Base of One | https://proofwiki.org/wiki/Exponent_Base_of_One | https://proofwiki.org/wiki/Exponent_Base_of_One | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal"
] | [
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5939 | Exponent Not Equal to Zero | Let $x$ and $y$ be ordinals.
Let $x \ne 0$.
Then:
:$x^y \ne 0$ | The proof shall proceed by Transfinite Induction on $y$. | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Let $x \ne 0$.
Then:
:$x^y \ne 0$ | The proof shall proceed by [[Transfinite Induction/Schema 2|Transfinite Induction]] on $y$. | Exponent Not Equal to Zero | https://proofwiki.org/wiki/Exponent_Not_Equal_to_Zero | https://proofwiki.org/wiki/Exponent_Not_Equal_to_Zero | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal"
] | [
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5940 | Membership is Left Compatible with Ordinal Exponentiation | Let $x$, $y$, and $z$ be ordinals.
Suppose $1 < z$.
Then:
:$x < y \iff z^x < z^y$ | === Sufficient Condition ===
The proof shall proceed by Transfinite Induction on $y$. | Let $x$, $y$, and $z$ be [[Definition:Ordinal|ordinals]].
Suppose $1 < z$.
Then:
:$x < y \iff z^x < z^y$ | === Sufficient Condition ===
The proof shall proceed by [[Transfinite Induction/Schema 2|Transfinite Induction]] on $y$. | Membership is Left Compatible with Ordinal Exponentiation | https://proofwiki.org/wiki/Membership_is_Left_Compatible_with_Ordinal_Exponentiation | https://proofwiki.org/wiki/Membership_is_Left_Compatible_with_Ordinal_Exponentiation | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal"
] | [
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5941 | Conjugacy Action on Group Elements is Group Action | Let $\struct {G, \circ}$ be a group whose identity is $e$.
The conjugacy action on $G$:
: $\forall g, h \in G: g * h = g \circ h \circ g^{-1}$
is a group action on itself. | We have that:
:$e * x = e \circ x \circ e^{-1} = x$
and so {{GroupActionAxiom|2}} is fulfilled.
{{GroupActionAxiom|1}} is shown to be fulfilled thus:
{{begin-eqn}}
{{eqn | l = \paren {g_1 \circ g_2} * x
| r = \paren {g_1 \circ g_2} \circ x \circ \paren {g_1 \circ g_2}^{-1}
| c = Definition of $*$
}}
{{eqn |... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
The [[Definition:Conjugacy Action|conjugacy action]] on $G$:
: $\forall g, h \in G: g * h = g \circ h \circ g^{-1}$
is a [[Definition:Group Action|group action]] on itself. | We have that:
:$e * x = e \circ x \circ e^{-1} = x$
and so {{GroupActionAxiom|2}} is fulfilled.
{{GroupActionAxiom|1}} is shown to be fulfilled thus:
{{begin-eqn}}
{{eqn | l = \paren {g_1 \circ g_2} * x
| r = \paren {g_1 \circ g_2} \circ x \circ \paren {g_1 \circ g_2}^{-1}
| c = Definition of $*$
}}
{{eq... | Conjugacy Action on Group Elements is Group Action | https://proofwiki.org/wiki/Conjugacy_Action_on_Group_Elements_is_Group_Action | https://proofwiki.org/wiki/Conjugacy_Action_on_Group_Elements_is_Group_Action | [
"Conjugacy Action on Group Elements is Group Action",
"Conjugacy Action"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Conjugacy Action",
"Definition:Group Action"
] | [
"Inverse of Group Product"
] |
proofwiki-5942 | Conjugacy Action on Group Elements is Group Action | Let $\struct {G, \circ}$ be a group whose identity is $e$.
The conjugacy action on $G$:
: $\forall g, h \in G: g * h = g \circ h \circ g^{-1}$
is a group action on itself. | Let $X$ be the set of all subgroups of $G$.
By definition, the (left) conjugacy action on subgroups is the group action $*_X : G \times X \to X$ defined as:
:$g *_X X = g \circ X \circ g^{-1}$
By Conjugacy Action on Subgroups is Group Action, the (left) conjugacy action on subgroups $*_X$ is a group action.
By Subset P... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
The [[Definition:Conjugacy Action|conjugacy action]] on $G$:
: $\forall g, h \in G: g * h = g \circ h \circ g^{-1}$
is a [[Definition:Group Action|group action]] on itself. | Let $X$ be the set of all [[Definition:Subgroup|subgroups]] of $G$.
By definition, the [[Definition:Conjugacy Action on Subgroups|(left) conjugacy action on subgroups]] is the [[Definition:Group Action|group action]] $*_X : G \times X \to X$ defined as:
:$g *_X X = g \circ X \circ g^{-1}$
By [[Conjugacy Action on Sub... | Conjugacy Action on Group Elements is Group Action/Proof 2 | https://proofwiki.org/wiki/Conjugacy_Action_on_Group_Elements_is_Group_Action | https://proofwiki.org/wiki/Conjugacy_Action_on_Group_Elements_is_Group_Action/Proof_2 | [
"Conjugacy Action on Group Elements is Group Action",
"Conjugacy Action"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Conjugacy Action",
"Definition:Group Action"
] | [
"Definition:Subgroup",
"Definition:Conjugacy Action/Subgroups",
"Definition:Group Action",
"Conjugacy Action on Subgroups is Group Action",
"Definition:Conjugacy Action/Subgroups",
"Definition:Group Action",
"Subset Product Action is Group Action",
"Subset Product Action is Group Action",
"Definitio... |
proofwiki-5943 | Conjugacy Action on Subgroups is Group Action | Let $X$ be the set of all subgroups of $G$.
For any $H \le G$ and for any $g \in G$, the conjugacy action:
:$g * H := g \circ H \circ g^{-1}$
is a group action. | Clearly {{GroupActionAxiom|1}} is fulfilled as $e * H = H$.
{{GroupActionAxiom|2}} is shown to be fulfilled thus:
{{begin-eqn}}
{{eqn | l = \paren {g_1 \circ g_2} * H
| r = \paren {g_1 \circ g_2} \circ H \circ \paren {g_1 \circ g_2}^{-1}
| c = Definition of $*$
}}
{{eqn | r = g_1 \circ g_2 \circ H \circ g_2... | Let $X$ be the set of all [[Definition:Subgroup|subgroups]] of $G$.
For any $H \le G$ and for any $g \in G$, the [[Definition:Conjugacy Action on Subgroups|conjugacy action]]:
:$g * H := g \circ H \circ g^{-1}$
is a [[Definition:Group Action|group action]]. | Clearly {{GroupActionAxiom|1}} is fulfilled as $e * H = H$.
{{GroupActionAxiom|2}} is shown to be fulfilled thus:
{{begin-eqn}}
{{eqn | l = \paren {g_1 \circ g_2} * H
| r = \paren {g_1 \circ g_2} \circ H \circ \paren {g_1 \circ g_2}^{-1}
| c = Definition of $*$
}}
{{eqn | r = g_1 \circ g_2 \circ H \circ g... | Conjugacy Action on Subgroups is Group Action | https://proofwiki.org/wiki/Conjugacy_Action_on_Subgroups_is_Group_Action | https://proofwiki.org/wiki/Conjugacy_Action_on_Subgroups_is_Group_Action | [
"Conjugacy Action"
] | [
"Definition:Subgroup",
"Definition:Conjugacy Action/Subgroups",
"Definition:Group Action"
] | [
"Inverse of Group Product"
] |
proofwiki-5944 | Slice Category of Order Category | Let $\mathbf P$ be a order category, and denote its ordering by $\preceq$.
Let $p \in \mathbf P_0$ be an object of $\mathbf P$.
Then:
:$\mathbf P \mathbin / p \cong p^\preceq$
where:
:$\mathbf P \mathbin / p$ is the slice of $\mathbf P$ over $p$
:$p^\preceq$ is the order category defined by the weak lower closure of $p... | The objects of $\mathbf P \mathbin / p$ are morphisms $q \to p$ of $\mathbf P$.
The morphisms are $q \to r$ fitting into a commutative diagram:
$\quad\quad \begin{xy}\xymatrix@C=1em{
q
\ar[rr]
\ar[rd]
& &
r
\ar[ld]
\\ &
p
}\end{xy}$
Define a functor $U: \mathbf P \mathbin / p \to \mathbf P$ by:
:$U \left({q \to p}\righ... | Let $\mathbf P$ be a [[Definition:Order Category|order category]], and denote its [[Definition:Ordering|ordering]] by $\preceq$.
Let $p \in \mathbf P_0$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf P$.
Then:
:$\mathbf P \mathbin / p \cong p^\preceq$
where:
:$\mathbf P \mathbin / p$ is the [[De... | The [[Definition:Object (Category Theory)|objects]] of $\mathbf P \mathbin / p$ are [[Definition:Morphism (Category Theory)|morphisms]] $q \to p$ of $\mathbf P$.
The [[Definition:Morphism (Category Theory)|morphisms]] are $q \to r$ fitting into a [[Definition:Commutative Diagram|commutative diagram]]:
$\quad\quad \be... | Slice Category of Order Category | https://proofwiki.org/wiki/Slice_Category_of_Order_Category | https://proofwiki.org/wiki/Slice_Category_of_Order_Category | [
"Slice Categories",
"Order Categories"
] | [
"Definition:Order Category",
"Definition:Ordering",
"Definition:Object (Category Theory)",
"Definition:Slice Category",
"Definition:Order Category",
"Definition:Lower Closure/Element"
] | [
"Definition:Object (Category Theory)",
"Definition:Morphism",
"Definition:Morphism",
"Definition:Commutative Diagram",
"Definition:Functor/Covariant",
"Definition:Morphism",
"Definition:Injective on Objects",
"Definition:Faithful Functor",
"Functor is Embedding iff Faithful and Injective on Objects"... |
proofwiki-5945 | Category of Pointed Sets as Coslice Category | Let $\mathbf{Set}_*$ be the category of pointed sets.
Let $\mathbf{Set}$ be the category of sets.
Let $1 := \left\{{*}\right\}$ be any singleton.
Then:
:$\mathbf{Set}_* \cong 1 \mathbin / \mathbf{Set}$
where $1 \mathbin / \mathbf{Set}$ denotes the coslice of $\mathbf{Set}$ under $1$ and $\cong$ signifies isomorphic cat... | Define the functor $F: \mathbf{Set}_* \to 1 \mathbin / \mathbf{Set}$ by:
:$\map F {C, c} := \bar c: 1 \to C$
:$F f := f$
where $\bar c: 1 \to C$ is defined by $\bar c (*) = c$.
Further, define $G: 1 \mathbin / \mathbf{Set} \to \mathbf{Set}_*$ by:
:$\map G {x: 1 \to C} := \struct {C, x \paren *}$
:$G f := f$ | Let $\mathbf{Set}_*$ be the [[Definition:Category of Pointed Sets|category of pointed sets]].
Let $\mathbf{Set}$ be the [[Definition:Category of Sets|category of sets]].
Let $1 := \left\{{*}\right\}$ be any [[Definition:Singleton|singleton]].
Then:
:$\mathbf{Set}_* \cong 1 \mathbin / \mathbf{Set}$
where $1 \mathb... | Define the [[Definition:Covariant Functor|functor]] $F: \mathbf{Set}_* \to 1 \mathbin / \mathbf{Set}$ by:
:$\map F {C, c} := \bar c: 1 \to C$
:$F f := f$
where $\bar c: 1 \to C$ is defined by $\bar c (*) = c$.
Further, define $G: 1 \mathbin / \mathbf{Set} \to \mathbf{Set}_*$ by:
:$\map G {x: 1 \to C} := \struct {C,... | Category of Pointed Sets as Coslice Category | https://proofwiki.org/wiki/Category_of_Pointed_Sets_as_Coslice_Category | https://proofwiki.org/wiki/Category_of_Pointed_Sets_as_Coslice_Category | [
"Category of Pointed Sets",
"Coslice Categories"
] | [
"Definition:Category of Pointed Sets",
"Definition:Category of Sets",
"Definition:Singleton",
"Definition:Coslice Category",
"Definition:Isomorphism of Categories/Isomorphic Categories"
] | [
"Definition:Functor/Covariant",
"Definition:Functor/Covariant",
"Definition:Functor/Covariant"
] |
proofwiki-5946 | Subset is Right Compatible with Ordinal Exponentiation | Let $x, y, z$ be ordinals.
Then:
:$x \le y \implies x^z \le y^z$ | The proof shall proceed by Transfinite Induction on $z$. | Let $x, y, z$ be [[Definition:Ordinal|ordinals]].
Then:
:$x \le y \implies x^z \le y^z$ | The proof shall proceed by [[Transfinite Induction/Schema 2|Transfinite Induction]] on $z$. | Subset is Right Compatible with Ordinal Exponentiation | https://proofwiki.org/wiki/Subset_is_Right_Compatible_with_Ordinal_Exponentiation | https://proofwiki.org/wiki/Subset_is_Right_Compatible_with_Ordinal_Exponentiation | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal"
] | [
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5947 | Condition for Membership is Right Compatible with Ordinal Exponentiation | Let $x, y, z$ be ordinals.
Let $z$ be the successor of some ordinal $w$.
Then:
:$x < y \iff x^z < y^z$ | === Sufficient Condition ===
Suppose $x < y$.
By Subset is Right Compatible with Ordinal Exponentiation:
:$x^w \le y^w$
Then:
{{begin-eqn}}
{{eqn | l = x^z
| r = x^w \times x
| c = {{Defof|Ordinal Exponentiation}}
}}
{{eqn | l = y^z
| r = y^w \times y
| c = {{Defof|Ordinal Exponentiation}}
}}
{{... | Let $x, y, z$ be [[Definition:Ordinal|ordinals]].
Let $z$ be the [[Definition:Successor Set|successor]] of some [[Definition:Ordinal|ordinal]] $w$.
Then:
:$x < y \iff x^z < y^z$ | === Sufficient Condition ===
Suppose $x < y$.
By [[Subset is Right Compatible with Ordinal Exponentiation]]:
:$x^w \le y^w$
Then:
{{begin-eqn}}
{{eqn | l = x^z
| r = x^w \times x
| c = {{Defof|Ordinal Exponentiation}}
}}
{{eqn | l = y^z
| r = y^w \times y
| c = {{Defof|Ordinal Exponentiation}... | Condition for Membership is Right Compatible with Ordinal Exponentiation | https://proofwiki.org/wiki/Condition_for_Membership_is_Right_Compatible_with_Ordinal_Exponentiation | https://proofwiki.org/wiki/Condition_for_Membership_is_Right_Compatible_with_Ordinal_Exponentiation | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set",
"Definition:Ordinal"
] | [
"Subset is Right Compatible with Ordinal Exponentiation",
"Subset is Right Compatible with Ordinal Exponentiation",
"Membership is Left Compatible with Ordinal Exponentiation",
"Definition:Conditional/Sufficient Condition",
"Subset is Right Compatible with Ordinal Exponentiation"
] |
proofwiki-5948 | Lower Bound for Ordinal Exponentiation | Let $x$ and $y$ be ordinals.
Let $x$ be greater than $1$, where $1$ denotes the successor of the zero ordinal.
Then:
:$y \le x^y$ | The proof shall proceed by Transfinite Induction on $y$. | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Let $x$ be greater than $1$, where $1$ denotes the [[Definition:Successor Set|successor]] of the [[Definition:Zero (Ordinal)|zero ordinal]].
Then:
:$y \le x^y$ | The proof shall proceed by [[Transfinite Induction/Schema 2|Transfinite Induction]] on $y$. | Lower Bound for Ordinal Exponentiation | https://proofwiki.org/wiki/Lower_Bound_for_Ordinal_Exponentiation | https://proofwiki.org/wiki/Lower_Bound_for_Ordinal_Exponentiation | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set",
"Definition:Zero (Ordinal)"
] | [
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5949 | Unique Ordinal Exponentiation Inequality | Let $x$ and $y$ be ordinals.
Let $x > 1$ and $y > 0$.
Then there exists a unique ordinal $z$ such that:
:$x^z \le y$ and $y < x^{z^+}$ | === Existence of $z$ ===
By Lower Bound for Ordinal Exponentiation, $y \le x^y$ so $y < x^{y^+}$.
Therefore, $y$ is bounded above by $x^w$ for some $w$, so there is a smallest $w$ such that:
:$y < x^w$ by Subset of Ordinals has Minimal Element.
Moreover, if $w$ is a limit ordinal then there is some $v ∈ w$ such that $y... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Let $x > 1$ and $y > 0$.
Then there exists a [[Definition:Unique|unique]] [[Definition:Ordinal|ordinal]] $z$ such that:
:$x^z \le y$ and $y < x^{z^+}$ | === Existence of $z$ ===
By [[Lower Bound for Ordinal Exponentiation]], $y \le x^y$ so $y < x^{y^+}$.
Therefore, $y$ is bounded above by $x^w$ for some $w$, so there is a smallest $w$ such that:
:$y < x^w$ by [[Subset of Ordinals has Minimal Element]].
Moreover, if $w$ is a [[Definition:Limit Ordinal|limit ordinal... | Unique Ordinal Exponentiation Inequality | https://proofwiki.org/wiki/Unique_Ordinal_Exponentiation_Inequality | https://proofwiki.org/wiki/Unique_Ordinal_Exponentiation_Inequality | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Unique",
"Definition:Ordinal"
] | [
"Lower Bound for Ordinal Exponentiation",
"Subset of Ordinals has Minimal Element",
"Definition:Limit Ordinal",
"Definition:Limit Ordinal",
"Definition:Successor Mapping/Successor Set",
"Definition:Ordinal"
] |
proofwiki-5950 | Limit Ordinals Closed under Ordinal Exponentiation | Let $x$ and $y$ be ordinals.
Let $y$ be a limit ordinal.
Let $x^y$ denote ordinal exponentiation.
Then:
:If $x > 1$, then $x^y$ is a limit ordinal.
:If $x \ne \O$, then $y^x$ is a limit ordinal. | Suppose $x > 1$.
{{AimForCont}} also that $x^y$ is the successor of some ordinal $w$.
By definition of ordinal exponentiation:
:$\ds x^y = \bigcup_{z \mathop \in y} x^z$
Then:
{{begin-eqn}}
{{eqn | l = w
| o = \in
| r = x^y
| c = Ordinal is Less than Successor
}}
{{eqn | ll= \leadsto
| q = \exis... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Let $y$ be a [[Definition:Limit Ordinal|limit ordinal]].
Let $x^y$ denote [[Definition:Ordinal Exponentiation|ordinal exponentiation]].
Then:
:If $x > 1$, then $x^y$ is a [[Definition:Limit Ordinal|limit ordinal]].
:If $x \ne \O$, then $y^x$ is a [[Definition:Li... | Suppose $x > 1$.
{{AimForCont}} also that $x^y$ is the [[Definition:Successor Set|successor]] of some [[Definition:Ordinal|ordinal]] $w$.
By definition of [[Definition:Ordinal Exponentiation|ordinal exponentiation]]:
:$\ds x^y = \bigcup_{z \mathop \in y} x^z$
Then:
{{begin-eqn}}
{{eqn | l = w
| o = \in
|... | Limit Ordinals Closed under Ordinal Exponentiation | https://proofwiki.org/wiki/Limit_Ordinals_Closed_under_Ordinal_Exponentiation | https://proofwiki.org/wiki/Limit_Ordinals_Closed_under_Ordinal_Exponentiation | [
"Ordinal Arithmetic",
"Limit Ordinals"
] | [
"Definition:Ordinal",
"Definition:Limit Ordinal",
"Definition:Ordinal Exponentiation",
"Definition:Limit Ordinal",
"Definition:Limit Ordinal"
] | [
"Definition:Successor Mapping/Successor Set",
"Definition:Ordinal",
"Definition:Ordinal Exponentiation",
"Ordinal is Less than Successor",
"Successor of Element of Ordinal is Subset",
"Membership is Left Compatible with Ordinal Exponentiation",
"Successor of Ordinal Smaller than Limit Ordinal is also Sm... |
proofwiki-5951 | Successor of Element of Ordinal is Subset | Let $x$ and $y$ be ordinals.
Then:
:$x \in y \iff x^+ \subseteq y$ | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = y
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x^+
| o = \in
| r = y^+
| c = Successor is Less than Successor
}}
{{eqn | ll= \leadstoandfrom
| l = x^+
| o = \in
| r = y
| c = {{Defof|Successor Set}}
}}
{{eqn | lo... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Then:
:$x \in y \iff x^+ \subseteq y$ | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = y
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x^+
| o = \in
| r = y^+
| c = [[Successor is Less than Successor]]
}}
{{eqn | ll= \leadstoandfrom
| l = x^+
| o = \in
| r = y
| c = {{Defof|Successor Set}}
}}
{{eqn ... | Successor of Element of Ordinal is Subset | https://proofwiki.org/wiki/Successor_of_Element_of_Ordinal_is_Subset | https://proofwiki.org/wiki/Successor_of_Element_of_Ordinal_is_Subset | [
"Ordinals"
] | [
"Definition:Ordinal"
] | [
"Successor is Less than Successor",
"Transitive Set is Proper Subset of Ordinal iff Element of Ordinal",
"Category:Ordinals"
] |
proofwiki-5952 | Ordinal is Less than Ordinal to Limit Power | Let $x$, $y$, and $z$ be ordinals.
Let $z$ be a limit ordinal.
Then:
:$x < y^z \iff \exists w \in z: x < y^w$ | {{begin-eqn}}
{{eqn | o =
| r = x < y^z
| c =
}}
{{eqn | ll= \leadstoandfrom
| o =
| r = x < \bigcup_{w \mathop \in z} y^w
| c = {{Defof|Ordinal Exponentiation}}
}}
{{eqn | ll= \leadstoandfrom
| o =
| r = \exists w \in z: x < y^w
| c = {{Defof|Union of Family}}
}}
{{... | Let $x$, $y$, and $z$ be [[Definition:Ordinal|ordinals]].
Let $z$ be a [[Definition:Limit Ordinal|limit ordinal]].
Then:
:$x < y^z \iff \exists w \in z: x < y^w$ | {{begin-eqn}}
{{eqn | o =
| r = x < y^z
| c =
}}
{{eqn | ll= \leadstoandfrom
| o =
| r = x < \bigcup_{w \mathop \in z} y^w
| c = {{Defof|Ordinal Exponentiation}}
}}
{{eqn | ll= \leadstoandfrom
| o =
| r = \exists w \in z: x < y^w
| c = {{Defof|Union of Family}}
}}
{{... | Ordinal is Less than Ordinal to Limit Power | https://proofwiki.org/wiki/Ordinal_is_Less_than_Ordinal_to_Limit_Power | https://proofwiki.org/wiki/Ordinal_is_Less_than_Ordinal_to_Limit_Power | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Limit Ordinal"
] | [] |
proofwiki-5953 | Ordinal Sum of Powers | Let $x$, $y$, and $z$ be ordinals.
Then:
:$x^y \times x^z = x^{y + z}$ | The proof shall proceed by Transfinite Induction on $z$. | Let $x$, $y$, and $z$ be [[Definition:Ordinal|ordinals]].
Then:
:$x^y \times x^z = x^{y + z}$ | The proof shall proceed by [[Transfinite Induction/Schema 2|Transfinite Induction]] on $z$. | Ordinal Sum of Powers | https://proofwiki.org/wiki/Ordinal_Sum_of_Powers | https://proofwiki.org/wiki/Ordinal_Sum_of_Powers | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal"
] | [
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5954 | Kleene Closure is Monoid | Let $S$ be a set, and let $S^*$ be its Kleene closure.
Let $*$ denote concatenation of ordered tuples.
Then $\struct {S^*, *}$ is a monoid. | First, to prove that $\struct {S^*, *}$ is a semigroup.
That is, to prove $*$ is associative.
Let $s, s', s'' \in S^*$ be sequences of lengths $n, n', n''$, respectively.
Then:
{{begin-eqn}}
{{eqn | l = \map {s * \paren {s' * s''} } i
| r = \begin {cases} \map s i & \text {if } 1 \le i \le n \\ \map {s' * s''} {i... | Let $S$ be a [[Definition:Set|set]], and let $S^*$ be its [[Definition:Kleene Closure|Kleene closure]].
Let $*$ denote [[Definition:Concatenation of Ordered Tuples|concatenation of ordered tuples]].
Then $\struct {S^*, *}$ is a [[Definition:Monoid|monoid]]. | First, to prove that $\struct {S^*, *}$ is a [[Definition:Semigroup|semigroup]].
That is, to prove $*$ is [[Definition:Associative Operation|associative]].
Let $s, s', s'' \in S^*$ be [[Definition:Finite Sequence|sequences]] of [[Definition:Length of Sequence|lengths]] $n, n', n''$, respectively.
Then:
{{begin-eqn}... | Kleene Closure is Monoid | https://proofwiki.org/wiki/Kleene_Closure_is_Monoid | https://proofwiki.org/wiki/Kleene_Closure_is_Monoid | [
"Sequences"
] | [
"Definition:Set",
"Definition:Kleene Closure",
"Definition:Concatenation of Ordered Tuples",
"Definition:Monoid"
] | [
"Definition:Semigroup",
"Definition:Associative Operation",
"Definition:Finite Sequence",
"Definition:Length of Sequence",
"Equality of Mappings",
"Definition:Associative Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Length of Sequence",
"Definition:Concatenati... |
proofwiki-5955 | Kleene Closure is Free Monoid | Let $S$ be a set.
Let $S^*$ be its Kleene closure, and let $i: S \to S^*$ be the insertion of generators.
Then $\struct {S^*, i}$ is a free monoid over $S$. | By Kleene Closure is Monoid, $S^*$ is a monoid.
It remains to verify the universal mapping property:{{Property|Free Monoid UMP}} | Let $S$ be a [[Definition:Set|set]].
Let $S^*$ be its [[Definition:Kleene Closure|Kleene closure]], and let $i: S \to S^*$ be the [[Definition:Insertion of Generators|insertion of generators]].
Then $\struct {S^*, i}$ is a [[Definition:Free Monoid|free monoid]] over $S$. | By [[Kleene Closure is Monoid]], $S^*$ is a [[Definition:Monoid|monoid]].
It remains to verify the [[Definition:Free Monoid UMP|universal mapping property]]:{{Property|Free Monoid UMP}} | Kleene Closure is Free Monoid | https://proofwiki.org/wiki/Kleene_Closure_is_Free_Monoid | https://proofwiki.org/wiki/Kleene_Closure_is_Free_Monoid | [
"Abstract Algebra"
] | [
"Definition:Set",
"Definition:Kleene Closure",
"Definition:Insertion of Generators",
"Definition:Free Monoid"
] | [
"Kleene Closure is Monoid",
"Definition:Monoid",
"Definition:Free Monoid UMP",
"Definition:Monoid",
"Definition:Free Monoid"
] |
proofwiki-5956 | Free Monoid is Unique | Let $S$ be a set.
Let $\struct {M, i}$ and $\struct {N, j}$ be free monoids over $S$.
Then there is a unique monoid isomorphism $f: M \to N$ such that:
:$\size f \circ i = j$
:$\size {f^{-1} } \circ j = i$
where $\size {\, \cdot \,}$ denotes the underlying set functor on $\mathbf{Mon}$. | By the (categorial) definition of free monoid, we have the following commutative diagram:
$\quad\quad \begin{xy}
<-4em,4em>*{\mathbf{Mon} :},
<-4em,1em>*{\mathbf{Set} :},
<0em,4em>*+{N} = "N",
<4em,4em>*+{M} = "M",
<8em,4em>*+{N} = "N2",
"N";"M" **@{.} ?>*@{>} ?*!/_1em/{\bar i},
"M";"N2" **@{.} ?>*@{>} ?*!/_1em/{\bar j... | Let $S$ be a [[Definition:Set|set]].
Let $\struct {M, i}$ and $\struct {N, j}$ be [[Definition:Free Monoid|free monoids]] over $S$.
Then there is a unique [[Definition:Monoid Isomorphism|monoid isomorphism]] $f: M \to N$ such that:
:$\size f \circ i = j$
:$\size {f^{-1} } \circ j = i$
where $\size {\, \cdot \,}$ d... | By the (categorial) definition of [[Definition:Free Monoid|free monoid]], we have the following [[Definition:Commutative Diagram|commutative diagram]]:
$\quad\quad \begin{xy}
<-4em,4em>*{\mathbf{Mon} :},
<-4em,1em>*{\mathbf{Set} :},
<0em,4em>*+{N} = "N",
<4em,4em>*+{M} = "M",
<8em,4em>*+{N} = "N2",
"N";"M" **@{.} ?>*@... | Free Monoid is Unique | https://proofwiki.org/wiki/Free_Monoid_is_Unique | https://proofwiki.org/wiki/Free_Monoid_is_Unique | [
"Free Monoids",
"Category of Monoids"
] | [
"Definition:Set",
"Definition:Free Monoid",
"Definition:Isomorphism (Abstract Algebra)/Monoid Isomorphism",
"Definition:Underlying Set Functor/Category of Monoids"
] | [
"Definition:Free Monoid",
"Definition:Commutative Diagram",
"Definition:Isomorphism (Abstract Algebra)/Monoid Isomorphism",
"Definition:Free Monoid"
] |
proofwiki-5957 | Ordinal Power of Power | Let $x$, $y$, and $z$ be ordinals.
Then:
:$\paren {x^y}^z = x^{y \mathop \times z}$ | The proof shall proceed by Transfinite Induction on $z$. | Let $x$, $y$, and $z$ be [[Definition:Ordinal|ordinals]].
Then:
:$\paren {x^y}^z = x^{y \mathop \times z}$ | The proof shall proceed by [[Transfinite Induction/Schema 2|Transfinite Induction]] on $z$. | Ordinal Power of Power | https://proofwiki.org/wiki/Ordinal_Power_of_Power | https://proofwiki.org/wiki/Ordinal_Power_of_Power | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal"
] | [
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5958 | Upper Bound of Ordinal Sum | Let $x$ and $y$ be ordinals.
Suppose $x > 1$.
Let $\sequence {a_n}$ be a finite sequence of ordinals such that:
:$a_n < x$ for all $n$
Let $\sequence {b_n}$ be a strictly decreasing finite sequence of ordinals such that:
:$b_n < y$ for all $n$
Then:
:$\ds \sum_{i \mathop = 1}^n x^{b_i} a_i < x^y$ | The proof shall proceed by finite induction on $n$:
For all $n \in \N_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n x^{b_i} a_i < x^y$ | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Suppose $x > 1$.
Let $\sequence {a_n}$ be a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Ordinal|ordinals]] such that:
:$a_n < x$ for all $n$
Let $\sequence {b_n}$ be a [[Definition:Strictly Decreasing Sequence|strictly decreasing]] [[Definition:... | The proof shall proceed by [[Principle of Mathematical Induction|finite induction]] on $n$:
For all $n \in \N_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \sum_{i \mathop = 1}^n x^{b_i} a_i < x^y$ | Upper Bound of Ordinal Sum | https://proofwiki.org/wiki/Upper_Bound_of_Ordinal_Sum | https://proofwiki.org/wiki/Upper_Bound_of_Ordinal_Sum | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Finite Sequence",
"Definition:Ordinal",
"Definition:Strictly Decreasing/Sequence",
"Definition:Finite Sequence",
"Definition:Ordinal"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-5959 | General Associative Law for Ordinal Sum | Let $x$ be a finite ordinal.
Let $\sequence {a_i}$ be a sequence of ordinals.
Then:
:$\ds \sum_{i \mathop = 1}^{x + 1} a_i = a_1 + \sum_{i \mathop = 1}^x a_{i + 1}$ | The proof shall proceed by induction on $x$.
=== Basis for the Induction ===
If $x = 0$, then:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^{0 + 1} a_i
| r = \sum_{i \mathop = 1}^0 a_i + a_1
| c = {{Defof|Ordinal Sum}}
}}
{{eqn | r = a_1
| c = Ordinal Addition by Zero
}}
{{eqn | r = a_1 + \sum_{i \m... | Let $x$ be a [[Definition:Finite Ordinal|finite ordinal]].
Let $\sequence {a_i}$ be a [[Definition:Sequence|sequence]] of [[Definition:Ordinal|ordinals]].
Then:
:$\ds \sum_{i \mathop = 1}^{x + 1} a_i = a_1 + \sum_{i \mathop = 1}^x a_{i + 1}$ | The proof shall proceed by [[Principle of Mathematical Induction for Minimally Inductive Set|induction]] on $x$.
=== Basis for the Induction ===
If $x = 0$, then:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^{0 + 1} a_i
| r = \sum_{i \mathop = 1}^0 a_i + a_1
| c = {{Defof|Ordinal Sum}}
}}
{{eqn | r = ... | General Associative Law for Ordinal Sum/Proof 1 | https://proofwiki.org/wiki/General_Associative_Law_for_Ordinal_Sum | https://proofwiki.org/wiki/General_Associative_Law_for_Ordinal_Sum/Proof_1 | [
"Ordinal Arithmetic",
"General Associative Law for Ordinal Sum"
] | [
"Definition:Finite Ordinal",
"Definition:Sequence",
"Definition:Ordinal"
] | [
"Principle of Mathematical Induction for Minimally Inductive Set",
"Ordinal Addition by Zero",
"Ordinal Addition by Zero",
"Definition:Basis for the Induction",
"Ordinal Addition is Associative",
"Definition:Induction Step"
] |
proofwiki-5960 | General Associative Law for Ordinal Sum | Let $x$ be a finite ordinal.
Let $\sequence {a_i}$ be a sequence of ordinals.
Then:
:$\ds \sum_{i \mathop = 1}^{x + 1} a_i = a_1 + \sum_{i \mathop = 1}^x a_{i + 1}$ | From Ordinal Addition is Associative we have that:
:$\forall a, b, c \in \On: a + \paren {b + c} = \paren {a + b} + c$
The result follows directly from the General Associativity Theorem.
{{qed}} | Let $x$ be a [[Definition:Finite Ordinal|finite ordinal]].
Let $\sequence {a_i}$ be a [[Definition:Sequence|sequence]] of [[Definition:Ordinal|ordinals]].
Then:
:$\ds \sum_{i \mathop = 1}^{x + 1} a_i = a_1 + \sum_{i \mathop = 1}^x a_{i + 1}$ | From [[Ordinal Addition is Associative]] we have that:
:$\forall a, b, c \in \On: a + \paren {b + c} = \paren {a + b} + c$
The result follows directly from the [[General Associativity Theorem]].
{{qed}} | General Associative Law for Ordinal Sum/Proof 2 | https://proofwiki.org/wiki/General_Associative_Law_for_Ordinal_Sum | https://proofwiki.org/wiki/General_Associative_Law_for_Ordinal_Sum/Proof_2 | [
"Ordinal Arithmetic",
"General Associative Law for Ordinal Sum"
] | [
"Definition:Finite Ordinal",
"Definition:Sequence",
"Definition:Ordinal"
] | [
"Ordinal Addition is Associative",
"General Associativity Theorem"
] |
proofwiki-5961 | Basis Representation Theorem for Ordinals | Let $x$ and $y$ be ordinals.
Let $x > 1$ and $y > 0$.
Then there exist unique finite sequences of ordinals:
:$\sequence {a_i}, \sequence {b_i}$
both of unique length $n$ such that:
:$(1): \quad \sequence {a_i}$ is a strictly decreasing sequence for $1 \le i \le n$
:$(2): \quad 0 < b_i < x$ for all $1 \le i \le n$
:$(3)... | The proof shall proceed by Transfinite Induction (Strong Induction) on $y$.
The inductive hypothesis states that for all $v < y$, there exist unique finite sequences of ordinals:
:$\sequence {c_i}, \sequence {d_i}$
both of unique length $n$ such that:
:$(1): \quad \sequence {c_i}$ is strictly decreasing sequence for $1... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Let $x > 1$ and $y > 0$.
Then there exist [[Definition:Unique|unique]] [[Definition:Finite Sequence|finite sequences]] of [[Definition:Ordinal|ordinals]]:
:$\sequence {a_i}, \sequence {b_i}$
both of [[Definition:Unique|unique]] [[Definition:Length of Sequence|lengt... | The proof shall proceed by [[Transfinite Induction/Schema 1|Transfinite Induction (Strong Induction)]] on $y$.
The inductive hypothesis states that for all $v < y$, there exist [[Definition:Unique|unique]] [[Definition:Finite Sequence|finite sequences]] of [[Definition:Ordinal|ordinals]]:
:$\sequence {c_i}, \sequence... | Basis Representation Theorem for Ordinals | https://proofwiki.org/wiki/Basis_Representation_Theorem_for_Ordinals | https://proofwiki.org/wiki/Basis_Representation_Theorem_for_Ordinals | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Unique",
"Definition:Finite Sequence",
"Definition:Ordinal",
"Definition:Unique",
"Definition:Length of Sequence",
"Definition:Strictly Decreasing/Sequence"
] | [
"Transfinite Induction/Schema 1",
"Definition:Unique",
"Definition:Finite Sequence",
"Definition:Ordinal",
"Definition:Unique",
"Definition:Length of Sequence",
"Definition:Strictly Decreasing/Sequence",
"Unique Ordinal Exponentiation Inequality",
"Division Theorem for Ordinals",
"Division Theorem... |
proofwiki-5962 | Ordinal Multiplication via Cantor Normal Form/Infinite Exponent | Let $x$ and $y$ be ordinals.
Let $x > 1$.
Let $y \ge \omega$ where $\omega$ denotes the minimally inductive set.
Let $\sequence {a_i}$ be a sequence of ordinals that is strictly decreasing on $1 \le i \le n$.
Let $\sequence {b_i}$ be a sequence of ordinals such that $0 < b_i < x$ for all $1 \le i \le n$.
Then:
:$\ds \s... | It follows that:
{{begin-eqn}}
{{eqn | l = x^{a_1}
| o = \le
| r = \sum_{i \mathop = 1}^n \paren {x^{a_i} b_i}
| c = Ordinal is Less than Sum
}}
{{eqn | o = <
| r = x^{a_1 + 1}
| c = Upper Bound of Ordinal Sum
}}
{{end-eqn}}
By multiplying the inequalities by $x^y$ on the left:
{{begin-eqn... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Let $x > 1$.
Let $y \ge \omega$ where $\omega$ denotes the [[Definition:Minimally Inductive Set|minimally inductive set]].
Let $\sequence {a_i}$ be a [[Definition:Sequence|sequence]] of [[Definition:Ordinal|ordinals]] that is [[Definition:Strictly Decreasing Sequen... | It follows that:
{{begin-eqn}}
{{eqn | l = x^{a_1}
| o = \le
| r = \sum_{i \mathop = 1}^n \paren {x^{a_i} b_i}
| c = [[Ordinal is Less than Sum]]
}}
{{eqn | o = <
| r = x^{a_1 + 1}
| c = [[Upper Bound of Ordinal Sum]]
}}
{{end-eqn}}
By multiplying the inequalities by $x^y$ on the left:
... | Ordinal Multiplication via Cantor Normal Form/Infinite Exponent | https://proofwiki.org/wiki/Ordinal_Multiplication_via_Cantor_Normal_Form/Infinite_Exponent | https://proofwiki.org/wiki/Ordinal_Multiplication_via_Cantor_Normal_Form/Infinite_Exponent | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Minimally Inductive Set",
"Definition:Sequence",
"Definition:Ordinal",
"Definition:Strictly Decreasing/Sequence",
"Definition:Sequence",
"Definition:Ordinal"
] | [
"Ordinal is Less than Sum",
"Upper Bound of Ordinal Sum",
"Subset is Right Compatible with Ordinal Multiplication",
"Ordinal Sum of Powers",
"Ordinal Sum of Powers",
"Ordinal Addition is Associative",
"Finite Ordinal Plus Transfinite Ordinal",
"Substitutivity of Class Equality",
"Substitutivity of C... |
proofwiki-5963 | Ordinal Multiplication via Cantor Normal Form/Limit Base | Let $x$ and $y$ be ordinals.
Let $x$ be a limit ordinal.
Let $y > 0$.
Let $\sequence {a_i}$ be a sequence of ordinals that is strictly decreasing on $1 \le i \le n$.
Let $\sequence {b_i}$ be a sequence of finite ordinals.
Then:
:$\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} b_i} \times x^y = x^{a_1 \mathop + y}$ | The proof shall proceed by finite induction on $n$:
For all $n \in \N_{\le 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} b_i} \times x^y = x^{a_1 \mathop + y}$
Since $x$ is a limit ordinal, it follows that $x^y$ is a limit ordinal by Limit Ordinals Closed under Ordinal Exponentiat... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Let $x$ be a [[Definition:Limit Ordinal|limit ordinal]].
Let $y > 0$.
Let $\sequence {a_i}$ be a [[Definition:Sequence|sequence]] of [[Definition:Ordinal|ordinals]] that is [[Definition:Strictly Decreasing Sequence|strictly decreasing]] on $1 \le i \le n$.
Let $\s... | The proof shall proceed by [[Principle of Mathematical Induction|finite induction]] on $n$:
For all $n \in \N_{\le 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} b_i} \times x^y = x^{a_1 \mathop + y}$
Since $x$ is a [[Definition:Limit Ordinal|limit ord... | Ordinal Multiplication via Cantor Normal Form/Limit Base | https://proofwiki.org/wiki/Ordinal_Multiplication_via_Cantor_Normal_Form/Limit_Base | https://proofwiki.org/wiki/Ordinal_Multiplication_via_Cantor_Normal_Form/Limit_Base | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Limit Ordinal",
"Definition:Sequence",
"Definition:Ordinal",
"Definition:Strictly Decreasing/Sequence",
"Definition:Sequence",
"Definition:Finite Ordinal"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Limit Ordinal",
"Definition:Limit Ordinal",
"Limit Ordinals Closed under Ordinal Exponentiation"
] |
proofwiki-5964 | Ordinal Exponentiation of Terms | Let $x, y, z$ be ordinals.
Let $n$ be a finite ordinal.
Let $x$ be a limit ordinal.
Let $y, z, n$ all be greater than $0$.
Then:
:$\paren {x^y \times n}^z = x^{y \mathop \times z} \times n$ if $z$ is not a limit ordinal
:$\paren {x^y \times n}^z = x^{y \mathop \times z}$ if $z$ is a limit ordinal. | The proof shall proceed by Transfinite Induction on $z$. | Let $x, y, z$ be [[Definition:Ordinal|ordinals]].
Let $n$ be a [[Definition:Finite Ordinal|finite ordinal]].
Let $x$ be a [[Definition:Limit Ordinal|limit ordinal]].
Let $y, z, n$ all be greater than $0$.
Then:
:$\paren {x^y \times n}^z = x^{y \mathop \times z} \times n$ if $z$ is not a [[Definition:Limit Ordinal... | The proof shall proceed by [[Transfinite Induction/Schema 2|Transfinite Induction]] on $z$. | Ordinal Exponentiation of Terms | https://proofwiki.org/wiki/Ordinal_Exponentiation_of_Terms | https://proofwiki.org/wiki/Ordinal_Exponentiation_of_Terms | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Finite Ordinal",
"Definition:Limit Ordinal",
"Definition:Limit Ordinal",
"Definition:Limit Ordinal"
] | [
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5965 | Unique Sequence of Consecutive Odd Numbers which are Prime | Let $n \in \Z$ be an integer such that $n > 3$.
Then $n$, $n + 2$, $n + 4$ cannot all be prime.
That is, the only set of $3$ consecutive odd numbers all of which are prime is $\set {3, 5, 7}$. | Let $n \in \Z_{>0}$.
For $n \le 3$ the cases can be examined in turn.
:$\set {1, 3, 5}$ are not all prime, as $1$ is not classified as prime.
:$\set {2, 4, 6}$ are not all prime, as $4$ and $6$ both have $2$ as a divisor.
:$\set {3, 5, 7}$ is the set of three consecutive odd primes which is asserted as being the unique... | Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n > 3$.
Then $n$, $n + 2$, $n + 4$ cannot all be [[Definition:Prime Number|prime]].
That is, the only [[Definition:Set|set]] of $3$ consecutive [[Definition:Odd Number|odd numbers]] all of which are [[Definition:Prime Number|prime]] is $\set {3, 5, 7}$. | Let $n \in \Z_{>0}$.
For $n \le 3$ the cases can be examined in turn.
:$\set {1, 3, 5}$ are not all [[Definition:Prime Number|prime]], as $1$ is not classified as [[Definition:Prime Number|prime]].
:$\set {2, 4, 6}$ are not all [[Definition:Prime Number|prime]], as $4$ and $6$ both have $2$ as a [[Definition:Divisor... | Unique Sequence of Consecutive Odd Numbers which are Prime | https://proofwiki.org/wiki/Unique_Sequence_of_Consecutive_Odd_Numbers_which_are_Prime | https://proofwiki.org/wiki/Unique_Sequence_of_Consecutive_Odd_Numbers_which_are_Prime | [
"Prime Numbers"
] | [
"Definition:Integer",
"Definition:Prime Number",
"Definition:Set",
"Definition:Odd Integer",
"Definition:Prime Number"
] | [
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Odd Prime",
"Definition:Unique",
"Definition:Set",
"Definition:Odd Integer",
"Definition:Prime Number",
"Definition:Even Integer",
"Definition:Divisor (Algebra)/In... |
proofwiki-5966 | Inequality for Ordinal Exponentiation | Let $x$ and $y$ be ordinals.
Let $x$ be a limit ordinal and let $y > 0$.
Let $\sequence {a_i}$ be a sequence of ordinals that is strictly decreasing on $1 \le i \le n$.
Let $\sequence {b_i}$ be a sequence of natural numbers.
Then:
:$\ds \paren {\sum_{i \mathop = 1}^n x^{a_i} \times b_i}^y \le x^{a_1 \mathop \times y} \... | By Upper Bound of Ordinal Sum:
:$\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} \times b_i} \le x^{a_1} \times \paren {b_1 + 1}$
So:
{{begin-eqn}}
{{eqn | l = \paren {\sum_{i \mathop = 1}^n x^{a_i} \times b_i}^y
| o = \le
| r = \paren {x^{a_1} \times \paren {b_1 + 1} }^y
| c = Subset is Right Compatible w... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Let $x$ be a [[Definition:Limit Ordinal|limit ordinal]] and let $y > 0$.
Let $\sequence {a_i}$ be a [[Definition:Sequence|sequence]] of [[Definition:Ordinal|ordinals]] that is [[Definition:Strictly Decreasing Sequence|strictly decreasing]] on $1 \le i \le n$.
Let $... | By [[Upper Bound of Ordinal Sum]]:
:$\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} \times b_i} \le x^{a_1} \times \paren {b_1 + 1}$
So:
{{begin-eqn}}
{{eqn | l = \paren {\sum_{i \mathop = 1}^n x^{a_i} \times b_i}^y
| o = \le
| r = \paren {x^{a_1} \times \paren {b_1 + 1} }^y
| c = [[Subset is Right Com... | Inequality for Ordinal Exponentiation | https://proofwiki.org/wiki/Inequality_for_Ordinal_Exponentiation | https://proofwiki.org/wiki/Inequality_for_Ordinal_Exponentiation | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Limit Ordinal",
"Definition:Sequence",
"Definition:Ordinal",
"Definition:Strictly Decreasing/Sequence",
"Definition:Sequence",
"Definition:Natural Numbers"
] | [
"Upper Bound of Ordinal Sum",
"Subset is Right Compatible with Ordinal Exponentiation",
"Ordinal Multiplication via Cantor Normal Form/Limit Base",
"Ordinal Power of Power"
] |
proofwiki-5967 | Ordinal Exponentiation via Cantor Normal Form/Limit Exponents | Let $x$ and $y$ be ordinals.
Let $x$ and $y$ be limit ordinals.
Let $\sequence {a_i}$ be a sequence of ordinals that is strictly decreasing on $1 \le i \le n$.
Let $\sequence {b_i}$ be a sequence of natural numbers.
Then:
:$\ds \paren {\sum_{i \mathop = 1}^n x^{a_i} \times b_i}^y = x^{a_1 \mathop \times y}$ | By Upper Bound of Ordinal Sum:
:$\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} \times b_i} \le x^{a_1} \times \paren {b_1 + 1}$
Furthermore:
:$\ds x^{a_1} \le \sum_{i \mathop = 1}^n \paren {x^{a_i} \times b_i}$
It follows that:
{{begin-eqn}}
{{eqn | l = \paren {x^{a_1} }^y
| o = \le
| r = \paren {\sum_{i \math... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Let $x$ and $y$ be [[Definition:Limit Ordinal|limit ordinals]].
Let $\sequence {a_i}$ be a [[Definition:Sequence|sequence]] of [[Definition:Ordinal|ordinals]] that is [[Definition:Strictly Decreasing Sequence|strictly decreasing]] on $1 \le i \le n$.
Let $\sequence... | By [[Upper Bound of Ordinal Sum]]:
:$\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} \times b_i} \le x^{a_1} \times \paren {b_1 + 1}$
Furthermore:
:$\ds x^{a_1} \le \sum_{i \mathop = 1}^n \paren {x^{a_i} \times b_i}$
It follows that:
{{begin-eqn}}
{{eqn | l = \paren {x^{a_1} }^y
| o = \le
| r = \paren {\sum... | Ordinal Exponentiation via Cantor Normal Form/Limit Exponents | https://proofwiki.org/wiki/Ordinal_Exponentiation_via_Cantor_Normal_Form/Limit_Exponents | https://proofwiki.org/wiki/Ordinal_Exponentiation_via_Cantor_Normal_Form/Limit_Exponents | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Limit Ordinal",
"Definition:Sequence",
"Definition:Ordinal",
"Definition:Strictly Decreasing/Sequence",
"Definition:Sequence",
"Definition:Natural Numbers"
] | [
"Upper Bound of Ordinal Sum",
"Subset is Right Compatible with Ordinal Exponentiation",
"Subset is Right Compatible with Ordinal Exponentiation",
"Ordinal Exponentiation of Terms",
"Ordinal Power of Power"
] |
proofwiki-5968 | Ordinal Exponentiation via Cantor Normal Form/Corollary | Let $x$ and $y$ be ordinals.
Let $x$ be a limit ordinal and let $y > 0$.
Let $\sequence {a_i}$ be a sequence of ordinals that is strictly decreasing on $1 \le i \le n$.
Let $\sequence {b_i}$ be a sequence of natural numbers.
Then:
:$\ds \paren {\sum_{i \mathop = 1}^n x^{a_i} \times b_i}^{x^y} = x^{a_1 \mathop \times x^... | By the hypothesis, $x^y$ is a limit ordinal by Limit Ordinals Closed under Ordinal Exponentiation.
The result follows from Ordinal Exponentiation via Cantor Normal Form/Limit Exponents.
{{qed}} | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Let $x$ be a [[Definition:Limit Ordinal|limit ordinal]] and let $y > 0$.
Let $\sequence {a_i}$ be a [[Definition:Sequence|sequence]] of [[Definition:Ordinal|ordinals]] that is [[Definition:Strictly Decreasing Sequence|strictly decreasing]] on $1 \le i \le n$.
Let $... | By the hypothesis, $x^y$ is a [[Definition:Limit Ordinal|limit ordinal]] by [[Limit Ordinals Closed under Ordinal Exponentiation]].
The result follows from [[Ordinal Exponentiation via Cantor Normal Form/Limit Exponents]].
{{qed}} | Ordinal Exponentiation via Cantor Normal Form/Corollary | https://proofwiki.org/wiki/Ordinal_Exponentiation_via_Cantor_Normal_Form/Corollary | https://proofwiki.org/wiki/Ordinal_Exponentiation_via_Cantor_Normal_Form/Corollary | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Limit Ordinal",
"Definition:Sequence",
"Definition:Ordinal",
"Definition:Strictly Decreasing/Sequence",
"Definition:Sequence",
"Definition:Natural Numbers"
] | [
"Definition:Limit Ordinal",
"Limit Ordinals Closed under Ordinal Exponentiation",
"Ordinal Exponentiation via Cantor Normal Form/Limit Exponents"
] |
proofwiki-5969 | Inverse Relation Functor is Contravariant Functor | Let $\mathbf{Rel}$ be the category of relations.
Let $C: \mathbf{Rel} \to \mathbf{Rel}$ be the inverse relation functor.
Then $C$ is a contravariant functor. | For any set $X$. we have:
{{begin-eqn}}
{{eqn|l = C \operatorname{id}_X
|r = \operatorname{id}_X^{-1}
|c = {{Defof|Inverse Relation Functor}}
}}
{{eqn|r = \set {\tuple {x, x}: \tuple {x, x} \in \operatorname{id}_X}
|c = {{Defof|Inverse Relation}}
}}
{{eqn|r = \operatorname{id}_X
}}
{{end-eqn}}
For any tw... | Let $\mathbf{Rel}$ be the [[Definition:Category of Relations|category of relations]].
Let $C: \mathbf{Rel} \to \mathbf{Rel}$ be the [[Definition:Inverse Relation Functor|inverse relation functor]].
Then $C$ is a [[Definition:Contravariant Functor|contravariant functor]]. | For any [[Definition:Set|set]] $X$. we have:
{{begin-eqn}}
{{eqn|l = C \operatorname{id}_X
|r = \operatorname{id}_X^{-1}
|c = {{Defof|Inverse Relation Functor}}
}}
{{eqn|r = \set {\tuple {x, x}: \tuple {x, x} \in \operatorname{id}_X}
|c = {{Defof|Inverse Relation}}
}}
{{eqn|r = \operatorname{id}_X
}}
{{... | Inverse Relation Functor is Contravariant Functor | https://proofwiki.org/wiki/Inverse_Relation_Functor_is_Contravariant_Functor | https://proofwiki.org/wiki/Inverse_Relation_Functor_is_Contravariant_Functor | [
"Category of Relations",
"Functors"
] | [
"Definition:Category of Relations",
"Definition:Inverse Relation Functor",
"Definition:Functor/Contravariant"
] | [
"Definition:Set",
"Definition:Relation",
"Inverse of Composite Relation",
"Definition:Functor/Contravariant"
] |
proofwiki-5970 | Injection iff Monomorphism in Category of Sets | Let $\mathbf{Set}$ be the category of sets.
Let $f: X \to Y$ be a morphism in $\mathbf{Set}$, that is, a mapping.
Then $f$ is an injection {{iff}} it is a monomorphism. | === Necessary Condition ===
Suppose that $f$ is injective.
Suppose further that we have mappings $g, h: Z \to X$ such that $g \ne h$.
Then necessarily there exists some $z \in Z$ such that $\map g z \ne \map h z$ by Equality of Mappings.
As $f$ is injective, it follows that:
:$\map f {\map g z} \ne \map f {\map h z}$
w... | Let $\mathbf{Set}$ be the [[Definition:Category of Sets|category of sets]].
Let $f: X \to Y$ be a [[Definition:Morphism (Category Theory)|morphism]] in $\mathbf{Set}$, that is, a [[Definition:Mapping|mapping]].
Then $f$ is an [[Definition:Injection|injection]] {{iff}} it is a [[Definition:Monomorphism (Category Theo... | === Necessary Condition ===
Suppose that $f$ is [[Definition:Injection|injective]].
Suppose further that we have [[Definition:Mapping|mappings]] $g, h: Z \to X$ such that $g \ne h$.
Then necessarily there exists some $z \in Z$ such that $\map g z \ne \map h z$ by [[Equality of Mappings]].
As $f$ is [[Definition:In... | Injection iff Monomorphism in Category of Sets | https://proofwiki.org/wiki/Injection_iff_Monomorphism_in_Category_of_Sets | https://proofwiki.org/wiki/Injection_iff_Monomorphism_in_Category_of_Sets | [
"Injections",
"Category of Sets"
] | [
"Definition:Category of Sets",
"Definition:Morphism",
"Definition:Mapping",
"Definition:Injection",
"Definition:Monomorphism (Category Theory)"
] | [
"Definition:Injection",
"Definition:Mapping",
"Equality of Mappings",
"Definition:Injection",
"Equality of Mappings",
"Definition:Monomorphism (Category Theory)",
"Rule of Transposition",
"Definition:Monomorphism (Category Theory)",
"Definition:Injection",
"Definition:Monomorphism (Category Theory... |
proofwiki-5971 | Transitive Closure of Set Always Exists | Let $S$ be a set.
Let $G$ be a mapping such that $\map G x = x \cup \bigcup x$.
{{explain|Domain and range of $G$ needed}}
Let $F$ be defined using the Principle of Recursive Definition:
:$\map F 0 = S$
:$\map F {n^+} = \map G {\map F n}$
Let $\ds T = \bigcup_{n \mathop \in \omega} \map F n$.
Then:
:$T$ is a set and is... | $\omega$ is a set by the {{Axiom-link|Infinity}}.
Thus by the {{Axiom-link|Replacement}}, the image of $\omega$ under $F$ is also a set.
Since $T$ is the union of $\map F \omega$, it is thus a set by the {{Axiom-link|Unions|Set Theory}}.
Furthermore:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = T
| c =... | Let $S$ be a [[Definition:Set|set]].
Let $G$ be a [[Definition:Mapping|mapping]] such that $\map G x = x \cup \bigcup x$.
{{explain|Domain and range of $G$ needed}}
Let $F$ be defined using the [[Principle of Recursive Definition]]:
:$\map F 0 = S$
:$\map F {n^+} = \map G {\map F n}$
Let $\ds T = \bigcup_{n \mat... | $\omega$ is a [[Definition:Set|set]] by the {{Axiom-link|Infinity}}.
Thus by the {{Axiom-link|Replacement}}, the [[Definition:Image of Subset under Mapping|image]] of $\omega$ under $F$ is also a [[Definition:Set|set]].
Since $T$ is the [[Definition:Union of Set of Sets|union]] of $\map F \omega$, it is thus a [[Defi... | Transitive Closure of Set Always Exists | https://proofwiki.org/wiki/Transitive_Closure_of_Set_Always_Exists | https://proofwiki.org/wiki/Transitive_Closure_of_Set_Always_Exists | [
"Relational Closures"
] | [
"Definition:Set",
"Definition:Mapping",
"Principle of Recursive Definition",
"Definition:Set",
"Definition:Transitive Class",
"Definition:Transitive Class",
"Definition:Transitive Closure of Set"
] | [
"Definition:Set",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Set",
"Definition:Set Union/Set of Sets",
"Definition:Set",
"Set is Subset of Union/Family of Sets",
"Definition:Transitive Class",
"Set is Subset of Union/Family of Sets",
"Definition:Transitive Class",
"Principle of Ma... |
proofwiki-5972 | Morphism in Preorder Category is Monic | Let $\mathbf P$ be a preorder category.
Let $f \in \mathbf P_1$ be a morphism.
Then $f$ is monic. | Suppose that $g, h \in \mathbf P_1$ are morphisms such that:
:$f \circ g = f \circ h$
In particular then, $g$ and $h$ have equal domain and codomain.
Since $\mathbf P$ is a preorder category, there is at most one morphism between any two objects.
Thus necessarily $g = h$, and hence $f$ is monic.
{{qed}} | Let $\mathbf P$ be a [[Definition:Preorder Category|preorder category]].
Let $f \in \mathbf P_1$ be a [[Definition:Morphism (Category Theory)|morphism]].
Then $f$ is [[Definition:Monomorphism (Category Theory)|monic]]. | Suppose that $g, h \in \mathbf P_1$ are [[Definition:Morphism (Category Theory)|morphisms]] such that:
:$f \circ g = f \circ h$
In particular then, $g$ and $h$ have equal [[Definition:Domain (Category Theory)|domain]] and [[Definition:Codomain (Category Theory)|codomain]].
Since $\mathbf P$ is a [[Definition:Preord... | Morphism in Preorder Category is Monic | https://proofwiki.org/wiki/Morphism_in_Preorder_Category_is_Monic | https://proofwiki.org/wiki/Morphism_in_Preorder_Category_is_Monic | [
"Preorder Categories"
] | [
"Definition:Preorder Category",
"Definition:Morphism",
"Definition:Monomorphism (Category Theory)"
] | [
"Definition:Morphism",
"Definition:Domain (Category Theory)",
"Definition:Codomain (Category Theory)",
"Definition:Preorder Category",
"Definition:Morphism",
"Definition:Object (Category Theory)",
"Definition:Monomorphism (Category Theory)"
] |
proofwiki-5973 | Morphism in Preorder Category is Epic | Let $\mathbf P$ be a preorder category.
Let $f \in \mathbf P_1$ be a morphism.
Then $f$ is epic. | Suppose that $g, h \in \mathbf P_1$ are morphisms such that:
:$g \circ f = h \circ f$
In particular then, $g$ and $h$ have equal domain and codomain.
Since $\mathbf P$ is a preorder category, there is at most one morphism between any two objects.
Thus necessarily $g = h$, and hence $f$ is epic.
{{qed}} | Let $\mathbf P$ be a [[Definition:Preorder Category|preorder category]].
Let $f \in \mathbf P_1$ be a [[Definition:Morphism (Category Theory)|morphism]].
Then $f$ is [[Definition:Epimorphism (Category Theory)|epic]]. | Suppose that $g, h \in \mathbf P_1$ are [[Definition:Morphism (Category Theory)|morphisms]] such that:
:$g \circ f = h \circ f$
In particular then, $g$ and $h$ have equal [[Definition:Domain (Category Theory)|domain]] and [[Definition:Codomain (Category Theory)|codomain]].
Since $\mathbf P$ is a [[Definition:Preord... | Morphism in Preorder Category is Epic | https://proofwiki.org/wiki/Morphism_in_Preorder_Category_is_Epic | https://proofwiki.org/wiki/Morphism_in_Preorder_Category_is_Epic | [
"Preorder Categories"
] | [
"Definition:Preorder Category",
"Definition:Morphism",
"Definition:Epimorphism (Category Theory)"
] | [
"Definition:Morphism",
"Definition:Domain (Category Theory)",
"Definition:Codomain (Category Theory)",
"Definition:Preorder Category",
"Definition:Morphism",
"Definition:Object",
"Definition:Epimorphism (Category Theory)"
] |
proofwiki-5974 | Surjection iff Epimorphism in Category of Sets | Let $\mathbf{Set}$ be the category of sets.
Let $f: X \to Y$ be a morphism in $\mathbf{Set}$, i.e. a mapping.
Then:
:$f$ is a surjection {{iff}} it is an epimorphism. | === Necessary Condition (proof 1 - contrapositive) ===
Let $f$ be surjective.
{{:Surjection iff Epimorphism in Category of Sets/Necessary Condition Proof 1}}{{qed|lemma}} | Let $\mathbf{Set}$ be the [[Definition:Category of Sets|category of sets]].
Let $f: X \to Y$ be a [[Definition:Morphism (Category Theory)|morphism]] in $\mathbf{Set}$, i.e. a [[Definition:Mapping|mapping]].
Then:
:$f$ is a [[Definition:Surjection|surjection]] {{iff}} it is an [[Definition:Epimorphism (Category Theor... | === [[Surjection iff Epimorphism in Category of Sets/Necessary Condition Proof 1|Necessary Condition (proof 1 - contrapositive)]] ===
Let $f$ be [[Definition:Surjection|surjective]].
{{:Surjection iff Epimorphism in Category of Sets/Necessary Condition Proof 1}}{{qed|lemma}} | Surjection iff Epimorphism in Category of Sets | https://proofwiki.org/wiki/Surjection_iff_Epimorphism_in_Category_of_Sets | https://proofwiki.org/wiki/Surjection_iff_Epimorphism_in_Category_of_Sets | [
"Surjection iff Epimorphism in Category of Sets",
"Surjections",
"Category of Sets"
] | [
"Definition:Category of Sets",
"Definition:Morphism",
"Definition:Mapping",
"Definition:Surjection",
"Definition:Epimorphism (Category Theory)"
] | [
"Surjection iff Epimorphism in Category of Sets/Necessary Condition Proof 1",
"Definition:Surjection",
"Definition:Surjection"
] |
proofwiki-5975 | Surjection iff Epimorphism in Category of Sets | Let $\mathbf{Set}$ be the category of sets.
Let $f: X \to Y$ be a morphism in $\mathbf{Set}$, i.e. a mapping.
Then:
:$f$ is a surjection {{iff}} it is an epimorphism. | By definition of epimorphism:
:$f$ is an epimorphism
{{iff}}
:for all mappings $g, h: Y \to Z : g \circ f = h \circ f \implies g = h$
In the following we prove the contrapositive statement:
:for all mappings $g, h: Y \to Z : g \ne h \implies g \circ f \ne h \circ f$
Let $g, h: Y \to Z$ be mappings such that $g \ne h$.
... | Let $\mathbf{Set}$ be the [[Definition:Category of Sets|category of sets]].
Let $f: X \to Y$ be a [[Definition:Morphism (Category Theory)|morphism]] in $\mathbf{Set}$, i.e. a [[Definition:Mapping|mapping]].
Then:
:$f$ is a [[Definition:Surjection|surjection]] {{iff}} it is an [[Definition:Epimorphism (Category Theor... | By definition of [[Definition:Epimorphism (Category Theory)|epimorphism]]:
:$f$ is an [[Definition:Epimorphism (Category Theory)|epimorphism]]
{{iff}}
:for all [[Definition:Mapping|mappings]] $g, h: Y \to Z : g \circ f = h \circ f \implies g = h$
In the following we prove the [[Definition:Contrapositive Statement|con... | Surjection iff Epimorphism in Category of Sets/Necessary Condition Proof 1 | https://proofwiki.org/wiki/Surjection_iff_Epimorphism_in_Category_of_Sets | https://proofwiki.org/wiki/Surjection_iff_Epimorphism_in_Category_of_Sets/Necessary_Condition_Proof_1 | [
"Surjection iff Epimorphism in Category of Sets",
"Surjections",
"Category of Sets"
] | [
"Definition:Category of Sets",
"Definition:Morphism",
"Definition:Mapping",
"Definition:Surjection",
"Definition:Epimorphism (Category Theory)"
] | [
"Definition:Epimorphism (Category Theory)",
"Definition:Epimorphism (Category Theory)",
"Definition:Mapping",
"Definition:Contrapositive Statement",
"Definition:Mapping",
"Definition:Mapping",
"Equality of Mappings",
"Definition:Surjection",
"Equality of Mappings",
"Rule of Transposition",
"Defi... |
proofwiki-5976 | Surjection iff Epimorphism in Category of Sets | Let $\mathbf{Set}$ be the category of sets.
Let $f: X \to Y$ be a morphism in $\mathbf{Set}$, i.e. a mapping.
Then:
:$f$ is a surjection {{iff}} it is an epimorphism. | Let $g, h: Y \to Z$ be mappings such that $g \circ f = h \circ f$.
Let $y \in Y$ be an arbitrary element of $Y$.
By definition of surjection:
:$(1) \quad \exists x \in X : \map f x = y$.
We have:
{{begin-eqn}}
{{eqn | l = \map g y
| r = \map g {\map f x}
| c = By $(1)$
}}
{{eqn | r = \map {\paren{g \circ f}... | Let $\mathbf{Set}$ be the [[Definition:Category of Sets|category of sets]].
Let $f: X \to Y$ be a [[Definition:Morphism (Category Theory)|morphism]] in $\mathbf{Set}$, i.e. a [[Definition:Mapping|mapping]].
Then:
:$f$ is a [[Definition:Surjection|surjection]] {{iff}} it is an [[Definition:Epimorphism (Category Theor... | Let $g, h: Y \to Z$ be [[Definition:Mapping|mappings]] such that $g \circ f = h \circ f$.
Let $y \in Y$ be an arbitrary [[Definition:Element|element]] of $Y$.
By definition of [[Definition:Surjection|surjection]]:
:$(1) \quad \exists x \in X : \map f x = y$.
We have:
{{begin-eqn}}
{{eqn | l = \map g y
| r = \... | Surjection iff Epimorphism in Category of Sets/Necessary Condition Proof 2 | https://proofwiki.org/wiki/Surjection_iff_Epimorphism_in_Category_of_Sets | https://proofwiki.org/wiki/Surjection_iff_Epimorphism_in_Category_of_Sets/Necessary_Condition_Proof_2 | [
"Surjection iff Epimorphism in Category of Sets",
"Surjections",
"Category of Sets"
] | [
"Definition:Category of Sets",
"Definition:Morphism",
"Definition:Mapping",
"Definition:Surjection",
"Definition:Epimorphism (Category Theory)"
] | [
"Definition:Mapping",
"Definition:Element",
"Definition:Surjection",
"Equality of Mappings",
"Definition:Epimorphism (Category Theory)"
] |
proofwiki-5977 | Inclusion of Natural Numbers in Integers is Epimorphism | Let $\mathbf{Mon}$ be the category of monoids.
Let $\struct {\N, +}$ denote the monoid of natural numbers as on Natural Numbers under Addition form Commutative Monoid.
Let $\struct {\Z, +}$ denote the additive group of integers.
Denote with $\iota: \N \to \Z$ the inclusion mapping.
Then $\iota: \N \to \Z$ is an epimorp... | Let $\struct {M, \circ}$ be a monoid with identity $e$.
Let $f, g: \Z \to M$ be monoid homomorphisms such that:
:$f \circ \iota = g \circ \iota$
that is, by definition of inclusion, such that:
:$\forall n \in \N: \map f n = \map g n$
Now $\iota$ will be epic if we can show that $f = g$.
The morphism property of $f$ and... | Let $\mathbf{Mon}$ be the [[Definition:Category of Monoids|category of monoids]].
Let $\struct {\N, +}$ denote the [[Definition:Monoid|monoid]] of [[Definition:Natural Numbers|natural numbers]] as on [[Natural Numbers under Addition form Commutative Monoid]].
Let $\struct {\Z, +}$ denote the [[Definition:Additive Gro... | Let $\struct {M, \circ}$ be a [[Definition:Monoid|monoid]] with [[Definition:Identity Element|identity]] $e$.
Let $f, g: \Z \to M$ be [[Definition:Monoid Homomorphism|monoid homomorphisms]] such that:
:$f \circ \iota = g \circ \iota$
that is, by definition of [[Definition:Inclusion Mapping|inclusion]], such that:
:... | Inclusion of Natural Numbers in Integers is Epimorphism | https://proofwiki.org/wiki/Inclusion_of_Natural_Numbers_in_Integers_is_Epimorphism | https://proofwiki.org/wiki/Inclusion_of_Natural_Numbers_in_Integers_is_Epimorphism | [
"Category of Monoids",
"Natural Numbers",
"Integers",
"Inclusion Mappings"
] | [
"Definition:Category of Monoids",
"Definition:Monoid",
"Definition:Natural Numbers",
"Natural Numbers under Addition form Commutative Monoid",
"Definition:Additive Group of Integers",
"Definition:Inclusion Mapping",
"Definition:Epimorphism (Category Theory)"
] | [
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Monoid Homomorphism",
"Definition:Inclusion Mapping",
"Definition:Epimorphism (Category Theory)",
"Definition:Morphism Property"
] |
proofwiki-5978 | Isomorphism (Category Theory) is Monic | Let $\mathbf C$ be a metacategory.
Let $C$ and $D$ be objects of $\mathbf C$.
Let $f: C \to D$ be an isomorphism.
Then $f: C \rightarrowtail D$ is monic. | Since $f$ is an isomorphism, it is a fortiori a split monomorphism.
The result follows from Split Monomorphism is Monic.
{{qed}} | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $C$ and $D$ be [[Definition:Object (Category Theory)|objects]] of $\mathbf C$.
Let $f: C \to D$ be an [[Definition:Isomorphism (Category Theory)|isomorphism]].
Then $f: C \rightarrowtail D$ is [[Definition:Monomorphism (Category Theory)|monic]]. | Since $f$ is an [[Definition:Isomorphism (Category Theory)|isomorphism]], it is [[Definition:A Fortiori|a fortiori]] a [[Definition:Split Monomorphism|split monomorphism]].
The result follows from [[Split Monomorphism is Monic]].
{{qed}} | Isomorphism (Category Theory) is Monic | https://proofwiki.org/wiki/Isomorphism_(Category_Theory)_is_Monic | https://proofwiki.org/wiki/Isomorphism_(Category_Theory)_is_Monic | [
"Category Theory"
] | [
"Definition:Metacategory",
"Definition:Object (Category Theory)",
"Definition:Isomorphism (Category Theory)",
"Definition:Monomorphism (Category Theory)"
] | [
"Definition:Isomorphism (Category Theory)",
"Definition:A Fortiori",
"Definition:Split Monomorphism",
"Split Monomorphism is Monic"
] |
proofwiki-5979 | Stabilizer of Polynomial | Let $n \in \Z: n > 0$.
Let $\map f {x_1, x_2, \ldots, x_n}$ be a polynomial in $n$ variables $x_1, x_2, \ldots, x_n$.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi, \rho \in S_n$.
Let the group action $\pi * f$ be defined as the permutation on the polynomial $f$ by $\pi$.
Then the stabilizer of $f$ is t... | Follows directly from the definition of the stabilizer of $f$.
{{qed}} | Let $n \in \Z: n > 0$.
Let $\map f {x_1, x_2, \ldots, x_n}$ be a [[Definition:Polynomial (Abstract Algebra)|polynomial]] in $n$ variables $x_1, x_2, \ldots, x_n$.
Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
Let $\pi, \rho \in S_n$.
Let the [[Definition:Group Act... | Follows directly from the definition of the [[Definition:Stabilizer|stabilizer]] of $f$.
{{qed}} | Stabilizer of Polynomial | https://proofwiki.org/wiki/Stabilizer_of_Polynomial | https://proofwiki.org/wiki/Stabilizer_of_Polynomial | [
"Polynomial Theory",
"Stabilizers"
] | [
"Definition:Polynomial over Ring",
"Definition:Symmetric Group/n Letters",
"Definition:Group Action",
"Definition:Permutation on Polynomial",
"Definition:Stabilizer",
"Definition:Permutation on n Letters",
"Definition:Fixed Element under Permutation"
] | [
"Definition:Stabilizer"
] |
proofwiki-5980 | Stabilizer of Element of Group Acting on Itself is Trivial | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $*$ be the group action of $\struct {G, \circ}$ on itself by the rule:
:$\forall g, h \in G: g * h = g \circ h$
Then the stabilizer of an element $x \in G$ is given by:
:$\Stab x = \set e$ | Let $g \in \Stab x$.
Then:
{{begin-eqn}}
{{eqn | l = g * x
| r = x
| c = {{Defof|Stabilizer}}
}}
{{eqn | ll= \leadsto
| l = g \circ x
| r = x
| c = {{Defof|Group Action}} (this particular one)
}}
{{eqn | ll= \leadsto
| l = g
| r = e
| c = {{Defof|Identity Element}}
}}
{{e... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $*$ be the [[Definition:Group Action|group action]] of $\struct {G, \circ}$ on itself by the rule:
:$\forall g, h \in G: g * h = g \circ h$
Then the [[Definition:Stabilizer|stabilizer]] of an element ... | Let $g \in \Stab x$.
Then:
{{begin-eqn}}
{{eqn | l = g * x
| r = x
| c = {{Defof|Stabilizer}}
}}
{{eqn | ll= \leadsto
| l = g \circ x
| r = x
| c = {{Defof|Group Action}} (this particular one)
}}
{{eqn | ll= \leadsto
| l = g
| r = e
| c = {{Defof|Identity Element}}
}}
{{... | Stabilizer of Element of Group Acting on Itself is Trivial | https://proofwiki.org/wiki/Stabilizer_of_Element_of_Group_Acting_on_Itself_is_Trivial | https://proofwiki.org/wiki/Stabilizer_of_Element_of_Group_Acting_on_Itself_is_Trivial | [
"Stabilizers"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group Action",
"Definition:Stabilizer"
] | [
"Definition:Trivial Subgroup"
] |
proofwiki-5981 | Stabilizer of Element under Conjugacy Action is Centralizer | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $*$ be the conjugacy action on $G$ defined by the rule:
:$\forall g, h \in G: g * h = g \circ h \circ g^{-1}$
Let $x \in G$.
Then the stabilizer of $x$ under this conjugacy action is:
:$\Stab x = \map {C_G} x$
where $\map {C_G} x$ is the centralizer of $x$ ... | From the definition of centralizer:
:$\map {C_G} x = \set {g \in G: g \circ x = x \circ g}$
Then:
{{begin-eqn}}
{{eqn | l = z
| o = \in
| r = \Stab x
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = z
| o = \in
| r = \set {g \in G: g \circ x \circ g^{-1} = x}
| c =
}}
{{eqn | ll= \... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $*$ be the [[Definition:Conjugacy Action|conjugacy action]] on $G$ defined by the rule:
:$\forall g, h \in G: g * h = g \circ h \circ g^{-1}$
Let $x \in G$.
Then the [[Definition:Stabilizer|stabilizer... | From the definition of [[Definition:Centralizer of Group Element|centralizer]]:
:$\map {C_G} x = \set {g \in G: g \circ x = x \circ g}$
Then:
{{begin-eqn}}
{{eqn | l = z
| o = \in
| r = \Stab x
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = z
| o = \in
| r = \set {g \in G: g \circ x \c... | Stabilizer of Element under Conjugacy Action is Centralizer | https://proofwiki.org/wiki/Stabilizer_of_Element_under_Conjugacy_Action_is_Centralizer | https://proofwiki.org/wiki/Stabilizer_of_Element_under_Conjugacy_Action_is_Centralizer | [
"Conjugacy Action",
"Centralizers",
"Stabilizers"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Conjugacy Action",
"Definition:Stabilizer",
"Definition:Conjugacy Action",
"Definition:Centralizer/Group Element"
] | [
"Definition:Centralizer/Group Element"
] |
proofwiki-5982 | Stabilizer of Subset Product Action on Power Set | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\powerset G$ be the power set of $\struct {G, \circ}$.
Let $*: G \times \powerset G \to \powerset G$ be the subset product action on $\powerset G$ defined as:
:$\forall g \in G: \forall S \in \powerset G: g * S = g \circ S$
where $g \circ S$ is the subset ... | From the definition of stabilizer:
:$\Stab S = \set {g \in G: g * S = S}$
The result follows from the definition of the group action $*$ given.
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\powerset G$ be the [[Definition:Power Set|power set]] of $\struct {G, \circ}$.
Let $*: G \times \powerset G \to \powerset G$ be the [[Definition:Subset Product Action|subset product action]] on $\pow... | From the definition of [[Definition:Stabilizer|stabilizer]]:
:$\Stab S = \set {g \in G: g * S = S}$
The result follows from the definition of the [[Definition:Group Action|group action]] $*$ given.
{{qed}} | Stabilizer of Subset Product Action on Power Set | https://proofwiki.org/wiki/Stabilizer_of_Subset_Product_Action_on_Power_Set | https://proofwiki.org/wiki/Stabilizer_of_Subset_Product_Action_on_Power_Set | [
"Subset Product Action",
"Stabilizers"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Power Set",
"Definition:Subset Product Action",
"Definition:Subset Product",
"Definition:Stabilizer",
"Definition:Set"
] | [
"Definition:Stabilizer",
"Definition:Group Action"
] |
proofwiki-5983 | Stabilizer of Coset Action on Set of Subgroups | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\powerset G$ denote the power set of $G$.
Let $\HH \subseteq \powerset G$ denote the set of subgroups of $G$.
Let $*$ be the subset product action on $\HH \subseteq \powerset G$ defined as:
:$\forall g \in G: \forall H \in \HH: g * H = g \circ H$
where $g ... | From the definition of Stabilizer of Subset Product Action on Power Set:
:$\Stab H = H = \set {g \in G: g * H = H}$
The result follows from Left Coset Equals Subgroup iff Element in Subgroup.
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\powerset G$ denote the [[Definition:Power Set|power set]] of $G$.
Let $\HH \subseteq \powerset G$ denote the [[Definition:Set|set]] of [[Definition:Subgroup|subgroups]] of $G$.
Let $*$ be the [[Defi... | From the definition of [[Stabilizer of Subset Product Action on Power Set]]:
:$\Stab H = H = \set {g \in G: g * H = H}$
The result follows from [[Left Coset Equals Subgroup iff Element in Subgroup]].
{{qed}} | Stabilizer of Coset Action on Set of Subgroups | https://proofwiki.org/wiki/Stabilizer_of_Coset_Action_on_Set_of_Subgroups | https://proofwiki.org/wiki/Stabilizer_of_Coset_Action_on_Set_of_Subgroups | [
"Subset Product Action",
"Stabilizers"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Power Set",
"Definition:Set",
"Definition:Subgroup",
"Definition:Subset Product Action",
"Definition:Coset/Left Coset",
"Definition:Stabilizer"
] | [
"Stabilizer of Subset Product Action on Power Set",
"Left Coset Equals Subgroup iff Element in Subgroup"
] |
proofwiki-5984 | Isomorphism (Category Theory) is Epic | Let $\mathbf C$ be a metacategory.
Let $C$ and $D$ be objects of $\mathbf C$.
Let $f: C \to D$ be an isomorphism.
Then $f: C \twoheadrightarrow D$ is epic. | Since $f$ is an isomorphism, it is a fortiori a split epimorphism.
The result follows from Split Epimorphism is Epic.
{{qed}} | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $C$ and $D$ be [[Definition:Object (Category Theory)|objects]] of $\mathbf C$.
Let $f: C \to D$ be an [[Definition:Isomorphism (Category Theory)|isomorphism]].
Then $f: C \twoheadrightarrow D$ is [[Definition:Epimorphism (Category Theory)|epic]]. | Since $f$ is an [[Definition:Isomorphism (Category Theory)|isomorphism]], it is [[Definition:A Fortiori|a fortiori]] a [[Definition:Split Epimorphism|split epimorphism]].
The result follows from [[Split Epimorphism is Epic]].
{{qed}} | Isomorphism (Category Theory) is Epic | https://proofwiki.org/wiki/Isomorphism_(Category_Theory)_is_Epic | https://proofwiki.org/wiki/Isomorphism_(Category_Theory)_is_Epic | [
"Category Theory"
] | [
"Definition:Metacategory",
"Definition:Object (Category Theory)",
"Definition:Isomorphism (Category Theory)",
"Definition:Epimorphism (Category Theory)"
] | [
"Definition:Isomorphism (Category Theory)",
"Definition:A Fortiori",
"Definition:Split Epimorphism",
"Split Epimorphism is Epic"
] |
proofwiki-5985 | Split Epimorphism is Epic | Let $\mathbf C$ be a metacategory.
Let $C$ and $D$ be objects of $\mathbf C$.
Let $f: C \to D$ be a split epimorphism.
Then $f: C \twoheadrightarrow D$ is epic. | Let $g: D \to C$ be a morphism such that:
:$f \circ g = \operatorname{id}_D$
which is guaranteed to exist by definition of a split epimorphism.
Suppose $x, y: D \to E$ are morphisms, such that:
:$x \circ f = y \circ f$
Then:
:$x \circ f \circ g = y \circ f \circ g$
Hence:
:$f \circ g = \operatorname{id}_D$
It follows t... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $C$ and $D$ be [[Definition:Object (Category Theory)|objects]] of $\mathbf C$.
Let $f: C \to D$ be a [[Definition:Split Epimorphism|split epimorphism]].
Then $f: C \twoheadrightarrow D$ is [[Definition:Epimorphism (Category Theory)|epic]]. | Let $g: D \to C$ be a [[Definition:Morphism (Category Theory)|morphism]] such that:
:$f \circ g = \operatorname{id}_D$
which is guaranteed to exist by definition of a [[Definition:Split Epimorphism|split epimorphism]].
Suppose $x, y: D \to E$ are [[Definition:Morphism (Category Theory)|morphisms]], such that:
:$x ... | Split Epimorphism is Epic | https://proofwiki.org/wiki/Split_Epimorphism_is_Epic | https://proofwiki.org/wiki/Split_Epimorphism_is_Epic | [
"Category Theory"
] | [
"Definition:Metacategory",
"Definition:Object (Category Theory)",
"Definition:Split Epimorphism",
"Definition:Epimorphism (Category Theory)"
] | [
"Definition:Morphism",
"Definition:Split Epimorphism",
"Definition:Morphism",
"Definition:Identity Morphism",
"Definition:Commutative Diagram"
] |
proofwiki-5986 | Split Monomorphism is Monic | Let $\mathbf C$ be a metacategory.
Let $C$ and $D$ be objects of $\mathbf C$.
Let $f: C \to D$ be a split monomorphism.
Then $f: C \rightarrowtail D$ is monic. | Let $g: D \to C$ be a morphism such that:
:$g \circ f = \operatorname{id}_C$
which is guaranteed to exist by definition of split monomorphism.
Suppose that $x, y: B \to C$ are morphisms such that:
:$f \circ x = f \circ y$
Then necessarily also:
:$g \circ \paren {f \circ x} = g \circ \paren {f \circ y}$
and by Compositi... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $C$ and $D$ be [[Definition:Object (Category Theory)|objects]] of $\mathbf C$.
Let $f: C \to D$ be a [[Definition:Split Monomorphism|split monomorphism]].
Then $f: C \rightarrowtail D$ is [[Definition:Monomorphism (Category Theory)|monic]]. | Let $g: D \to C$ be a [[Definition:Morphism (Category Theory)|morphism]] such that:
:$g \circ f = \operatorname{id}_C$
which is guaranteed to exist by definition of [[Definition:Split Monomorphism|split monomorphism]].
Suppose that $x, y: B \to C$ are [[Definition:Morphism (Category Theory)|morphisms]] such that:
... | Split Monomorphism is Monic | https://proofwiki.org/wiki/Split_Monomorphism_is_Monic | https://proofwiki.org/wiki/Split_Monomorphism_is_Monic | [
"Category Theory"
] | [
"Definition:Metacategory",
"Definition:Object (Category Theory)",
"Definition:Split Monomorphism",
"Definition:Monomorphism (Category Theory)"
] | [
"Definition:Morphism",
"Definition:Split Monomorphism",
"Definition:Morphism",
"Composition of Morphisms is Associative",
"Definition:Given",
"Definition:Identity Morphism",
"Definition:Morphism",
"Definition:Object (Category Theory)",
"Definition:Metacategory",
"Definition:Morphism",
"Definitio... |
proofwiki-5987 | Epimorphism into Projective Object Splits | Let $\mathbf C$ be a metacategory.
Let $P \in \mathbf C_0$ be a projective object of $\mathbf C$.
Let $e: E \twoheadrightarrow P$ be an epimorphism.
Then $e$ is a split epimorphism, i.e. it admits a retraction $f: P \to E$. | Consider the identity morphism $\operatorname{id}_P: P \to P$.
By definition of projective object, we obtain the following commutative diagram:
$\quad\quad \begin{xy}
<0em,0em>*+{P} = "P",
<4em,0em>*+{P} = "P2",
<4em,4em>*+{E} = "E",
"P";"E" **@{-} ?>*@{>} ?*!/_.6em/{f},
"P";"P2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $P \in \mathbf C_0$ be a [[Definition:Projective Object|projective object]] of $\mathbf C$.
Let $e: E \twoheadrightarrow P$ be an [[Definition:Epimorphism (Category Theory)|epimorphism]].
Then $e$ is a [[Definition:Split Epimorphism|split epimorphis... | Consider the [[Definition:Identity Morphism|identity morphism]] $\operatorname{id}_P: P \to P$.
By definition of [[Definition:Projective Object|projective object]], we obtain the following [[Definition:Commutative Diagram|commutative diagram]]:
$\quad\quad \begin{xy}
<0em,0em>*+{P} = "P",
<4em,0em>*+{P} = "P2",
<4em,... | Epimorphism into Projective Object Splits | https://proofwiki.org/wiki/Epimorphism_into_Projective_Object_Splits | https://proofwiki.org/wiki/Epimorphism_into_Projective_Object_Splits | [
"Epimorphisms"
] | [
"Definition:Metacategory",
"Definition:Projective Object",
"Definition:Epimorphism (Category Theory)",
"Definition:Split Epimorphism",
"Definition:Retraction (Category Theory)"
] | [
"Definition:Identity Morphism",
"Definition:Projective Object",
"Definition:Commutative Diagram",
"Definition:Split Epimorphism"
] |
proofwiki-5988 | Relational Closure Exists for Set-Like Relation | Let $A$ be a class.
Let $\prec$ be a relation on $A$.
Furthermore, let $\map {\prec^{-1} } a$ be a small class for each $a \in A$.
Let $S$ be a small class that is a subset of the class $A$.
Let $G$ be a mapping such that:
:$\map G x = A \cap \paren {\map {\prec^{-1} } x}$
Let $F$ be defined using the Principle of Recu... | === Proof of $(1)$ ===
Let $x \in A$ and $y \in T$ such that $x \prec y$.
Then by definition:
:$x \in \map {\prec^{-1} } y$
Moreover, since $y \in T$:
:$\exists n \in \omega: y \in \map F n$
Therefore:
:$x \in \map F {n + 1}$
and so:
:$x \in T$
{{qed|lemma}} | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $\prec$ be a [[Definition:Endorelation|relation]] on $A$.
Furthermore, let $\map {\prec^{-1} } a$ be a [[Definition:Small Class|small class]] for each $a \in A$.
Let $S$ be a [[Definition:Small Class|small class]] that is a subset of the [[Definition:Class... | === Proof of $(1)$ ===
Let $x \in A$ and $y \in T$ such that $x \prec y$.
Then by definition:
:$x \in \map {\prec^{-1} } y$
Moreover, since $y \in T$:
:$\exists n \in \omega: y \in \map F n$
Therefore:
:$x \in \map F {n + 1}$
and so:
:$x \in T$
{{qed|lemma}} | Relational Closure Exists for Set-Like Relation | https://proofwiki.org/wiki/Relational_Closure_Exists_for_Set-Like_Relation | https://proofwiki.org/wiki/Relational_Closure_Exists_for_Set-Like_Relation | [
"Relational Closures"
] | [
"Definition:Class (Class Theory)",
"Definition:Endorelation",
"Definition:Small Class",
"Definition:Small Class",
"Definition:Class (Class Theory)",
"Definition:Mapping",
"Principle of Recursive Definition",
"Definition:Set",
"Definition:Transitive with Respect to a Relation",
"Definition:Transiti... | [] |
proofwiki-5989 | Well-Founded Proper Relational Structure Determines Minimal Elements | Let $A$ and $B$ be classes.
Let $\struct {A, \prec}$ be a proper relational structure.
Let $\prec$ be a strictly well-founded relation.
Suppose $B \subset A$ and $B \ne \O$.
Then $B$ has a strictly minimal element under $\prec$. | $B$ is not empty.
So $B$ has at least one element $x$.
By Singleton of Element is Subset:
:$\set x \subseteq B$
By Relational Closure Exists for Set-Like Relation:
:$\set x$ has a $\prec$-relational closure.
{{explain|Set-like relation: it needs to be shown that $\prec$ is one of those.}}
This $\prec$-relational closur... | Let $A$ and $B$ be [[Definition:Class (Class Theory)|classes]].
Let $\struct {A, \prec}$ be a [[Definition:Proper Relational Structure|proper relational structure]].
Let $\prec$ be a [[Definition:Strictly Well-Founded Relation|strictly well-founded relation]].
Suppose $B \subset A$ and $B \ne \O$.
Then $B$ has a ... | $B$ is not [[Definition:Empty Set|empty]].
So $B$ has at least one [[Definition:Element|element]] $x$.
By [[Singleton of Element is Subset]]:
:$\set x \subseteq B$
By [[Relational Closure Exists for Set-Like Relation]]:
:$\set x$ has a [[Definition:Relational Closure|$\prec$-relational closure]].
{{explain|Set-like... | Well-Founded Proper Relational Structure Determines Minimal Elements | https://proofwiki.org/wiki/Well-Founded_Proper_Relational_Structure_Determines_Minimal_Elements | https://proofwiki.org/wiki/Well-Founded_Proper_Relational_Structure_Determines_Minimal_Elements | [
"Relational Closures"
] | [
"Definition:Class (Class Theory)",
"Definition:Proper Relational Structure",
"Definition:Strictly Well-Founded Relation",
"Definition:Strictly Minimal Element"
] | [
"Definition:Empty Set",
"Definition:Element",
"Singleton of Element is Subset",
"Relational Closure Exists for Set-Like Relation",
"Definition:Relational Closure",
"Definition:Relational Closure",
"Intersection is Subset",
"Definition:Strictly Well-Founded Relation",
"Definition:Set Intersection",
... |
proofwiki-5990 | Initial Object is Unique | Let $\mathbf C$ be a metacategory.
Let $0$ and $0'$ be two initial objects of $\mathbf C$.
Then there is a unique isomorphism $u: 0 \to 0'$.
Hence, initial objects are unique up to unique isomorphism. | Consider the following commutative diagram:
$\quad\quad \begin{xy}
<-4em,0em>*+{0} = "M",
<0em,0em> *+{0'}= "N",
<0em,-4em>*+{0} = "M2",
<4em,-4em>*+{0'}= "N2",
"M";"N" **@{-} ?>*@{>} ?*!/_.6em/{u},
"M";"M2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_0},
"N";"M2" **@{-} ?>*@{>} ?*!/_.6em/{v},
"N";"N2" **@{-} ?>*@{>} ... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $0$ and $0'$ be two [[Definition:Initial Object|initial objects]] of $\mathbf C$.
Then there is a [[Definition:Unique|unique]] [[Definition:Isomorphism (Category Theory)|isomorphism]] $u: 0 \to 0'$.
Hence, [[Definition:Initial Object|initial objects... | Consider the following [[Definition:Commutative Diagram|commutative diagram]]:
$\quad\quad \begin{xy}
<-4em,0em>*+{0} = "M",
<0em,0em> *+{0'}= "N",
<0em,-4em>*+{0} = "M2",
<4em,-4em>*+{0'}= "N2",
"M";"N" **@{-} ?>*@{>} ?*!/_.6em/{u},
"M";"M2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_0},
"N";"M2" **@{-} ?>*@{>} ?*!... | Initial Object is Unique | https://proofwiki.org/wiki/Initial_Object_is_Unique | https://proofwiki.org/wiki/Initial_Object_is_Unique | [
"Objects (Category Theory)"
] | [
"Definition:Metacategory",
"Definition:Initial Object",
"Definition:Unique",
"Definition:Isomorphism (Category Theory)",
"Definition:Initial Object",
"Definition:Unique up to Unique Isomorphism"
] | [
"Definition:Commutative Diagram",
"Definition:Morphism",
"Definition:Initial Object",
"Definition:Inverse Morphism",
"Definition:Isomorphism (Category Theory)"
] |
proofwiki-5991 | Terminal Object is Unique | Let $\mathbf C$ be a metacategory.
Let $1$ and $1'$ be two terminal objects of $\mathbf C$.
Then there is a unique isomorphism $u: 1 \to 1'$.
Hence, terminal objects are unique up to unique isomorphism. | Consider the following commutative diagram:
$\quad\quad \begin{xy}
<-4em,0em>*+{1} = "M",
<0em,0em> *+{1'}= "N",
<0em,-4em>*+{1} = "M2",
<4em,-4em>*+{1'}= "N2",
"M";"N" **@{-} ?>*@{>} ?*!/_.6em/{u},
"M";"M2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_1},
"N";"M2" **@{-} ?>*@{>} ?*!/_.6em/{v},
"N";"N2" **@{-} ?>*@{>} ... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $1$ and $1'$ be two [[Definition:Terminal Object|terminal objects]] of $\mathbf C$.
Then there is a [[Definition:Unique|unique]] [[Definition:Isomorphism (Category Theory)|isomorphism]] $u: 1 \to 1'$.
Hence, [[Definition:Terminal Object|terminal obj... | Consider the following [[Definition:Commutative Diagram|commutative diagram]]:
$\quad\quad \begin{xy}
<-4em,0em>*+{1} = "M",
<0em,0em> *+{1'}= "N",
<0em,-4em>*+{1} = "M2",
<4em,-4em>*+{1'}= "N2",
"M";"N" **@{-} ?>*@{>} ?*!/_.6em/{u},
"M";"M2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_1},
"N";"M2" **@{-} ?>*@{>} ?*!... | Terminal Object is Unique | https://proofwiki.org/wiki/Terminal_Object_is_Unique | https://proofwiki.org/wiki/Terminal_Object_is_Unique | [
"Objects (Category Theory)"
] | [
"Definition:Metacategory",
"Definition:Terminal Object",
"Definition:Unique",
"Definition:Isomorphism (Category Theory)",
"Definition:Terminal Object",
"Definition:Unique up to Unique Isomorphism"
] | [
"Definition:Commutative Diagram",
"Definition:Morphism",
"Definition:Terminal Object",
"Definition:Inverse Morphism",
"Definition:Isomorphism (Category Theory)"
] |
proofwiki-5992 | Empty Set is Initial Object | Let $\mathbf{Set}$ be the category of sets.
Then the empty set $\O$ is an initial object of $\mathbf{Set}$. | Follows from Empty Mapping is Unique and the definition of initial object.
{{qed}} | Let $\mathbf{Set}$ be the [[Definition:Category of Sets|category of sets]].
Then the [[Definition:Empty Set|empty set]] $\O$ is an [[Definition:Initial Object|initial object]] of $\mathbf{Set}$. | Follows from [[Empty Mapping is Unique]] and the definition of [[Definition:Initial Object|initial object]].
{{qed}} | Empty Set is Initial Object | https://proofwiki.org/wiki/Empty_Set_is_Initial_Object | https://proofwiki.org/wiki/Empty_Set_is_Initial_Object | [
"Category of Sets",
"Empty Set"
] | [
"Definition:Category of Sets",
"Definition:Empty Set",
"Definition:Initial Object"
] | [
"Empty Mapping is Unique",
"Definition:Initial Object"
] |
proofwiki-5993 | Singleton is Terminal Object of Category of Sets | Let $\mathbf {Set}$ be the category of sets.
Let $S = \set x$ be any singleton set.
Then $S$ is a terminal object of $\mathbf {Set}$. | Let $T$ be a set, and let $f: T \to S$ be a mapping.
Then since for all $t \in T$, we have $\map f t \in S$, it follows that:
:$\forall t \in T: \map f t = x$
By Equality of Mappings, there is precisely one such mapping $f: T \to S$.
Hence the result, by definition of terminal object.
{{qed}} | Let $\mathbf {Set}$ be the [[Definition:Category of Sets|category of sets]].
Let $S = \set x$ be any [[Definition:Singleton|singleton set]].
Then $S$ is a [[Definition:Terminal Object|terminal object]] of $\mathbf {Set}$. | Let $T$ be a [[Definition:Set|set]], and let $f: T \to S$ be a [[Definition:Mapping|mapping]].
Then since for all $t \in T$, we have $\map f t \in S$, it follows that:
:$\forall t \in T: \map f t = x$
By [[Equality of Mappings]], there is precisely one such [[Definition:Mapping|mapping]] $f: T \to S$.
Hence the re... | Singleton is Terminal Object of Category of Sets | https://proofwiki.org/wiki/Singleton_is_Terminal_Object_of_Category_of_Sets | https://proofwiki.org/wiki/Singleton_is_Terminal_Object_of_Category_of_Sets | [
"Category Theory",
"Singletons"
] | [
"Definition:Category of Sets",
"Definition:Singleton",
"Definition:Terminal Object"
] | [
"Definition:Set",
"Definition:Mapping",
"Equality of Mappings",
"Definition:Mapping",
"Definition:Terminal Object"
] |
proofwiki-5994 | Zero (Category) is Initial Object | Let $\mathbf{Cat}$ be the category of categories.
Let $\mathbf 0$ be the zero category.
Then $\mathbf 0$ is an initial object of $\mathbf{Cat}$. | Let $\mathbf C$ be an object of $\mathbf{Cat}$, i.e. a small category.
By Empty Mapping is Unique, there are unique mappings:
:$F_0: \mathbf 0_0 = \O \to \mathbf C_0$
:$F_1: \mathbf 0_1 = \O \to \mathbf C_1$
making $F: \mathbf 0 \to \mathbf C$ a functor by vacuous truth.
That $F_0$ and $F_1$ are actually mappings follo... | Let $\mathbf{Cat}$ be the [[Definition:Category of Categories|category of categories]].
Let $\mathbf 0$ be the [[Definition:Zero (Category)|zero category]].
Then $\mathbf 0$ is an [[Definition:Initial Object|initial object]] of $\mathbf{Cat}$. | Let $\mathbf C$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf{Cat}$, i.e. a [[Definition:Small Category|small category]].
By [[Empty Mapping is Unique]], there are unique [[Definition:Mapping|mappings]]:
:$F_0: \mathbf 0_0 = \O \to \mathbf C_0$
:$F_1: \mathbf 0_1 = \O \to \mathbf C_1$
making $F: ... | Zero (Category) is Initial Object | https://proofwiki.org/wiki/Zero_(Category)_is_Initial_Object | https://proofwiki.org/wiki/Zero_(Category)_is_Initial_Object | [
"Category of Categories"
] | [
"Definition:Category of Categories",
"Definition:Zero (Category)",
"Definition:Initial Object"
] | [
"Definition:Object (Category Theory)",
"Definition:Small Category",
"Empty Mapping is Unique",
"Definition:Mapping",
"Definition:Functor/Covariant",
"Definition:Vacuous Truth",
"Definition:Mapping",
"Definition:Small Category",
"Definition:Initial Object"
] |
proofwiki-5995 | One (Category) is Terminal Object | Let $\mathbf{Cat}$ be the category of categories.
Let $\mathbf 1$ be the category one.
Then $\mathbf 1$ is a terminal object of $\mathbf{Cat}$. | Let $\mathbf C$ be an object of $\mathbf{Cat}$, i.e. a small category.
From Singleton is Terminal Object of Category of Sets, there exist unique mappings:
:$F_0: \mathbf C_0 \to \mathbf 1_0 = \left\{{*}\right\}$
:$F_1: \mathbf C_1 \to \mathbf 1_1 = \left\{{\operatorname{id}_*}\right\}$
since the latter sets are singlet... | Let $\mathbf{Cat}$ be the [[Definition:Category of Categories|category of categories]].
Let $\mathbf 1$ be the [[Definition:One (Category)|category one]].
Then $\mathbf 1$ is a [[Definition:Terminal Object|terminal object]] of $\mathbf{Cat}$. | Let $\mathbf C$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf{Cat}$, i.e. a [[Definition:Small Category|small category]].
From [[Singleton is Terminal Object of Category of Sets]], there exist unique [[Definition:Mapping|mappings]]:
:$F_0: \mathbf C_0 \to \mathbf 1_0 = \left\{{*}\right\}$
:$F_1: \... | One (Category) is Terminal Object | https://proofwiki.org/wiki/One_(Category)_is_Terminal_Object | https://proofwiki.org/wiki/One_(Category)_is_Terminal_Object | [
"Category of Categories"
] | [
"Definition:Category of Categories",
"Definition:One (Category)",
"Definition:Terminal Object"
] | [
"Definition:Object (Category Theory)",
"Definition:Small Category",
"Singleton is Terminal Object of Category of Sets",
"Definition:Mapping",
"Definition:Singleton",
"Definition:Functor/Covariant",
"Definition:Identity Morphism",
"Definition:Morphism Property",
"Definition:Functor/Covariant",
"Def... |
proofwiki-5996 | Empty Mapping is Mapping | Let $T$ be a set.
Let $e: \O \to T$ be the empty mapping.
Then $e$ is indeed a mapping. | Let $e: \O \to T$ be the empty mapping.
Vacuously:
:$\forall x \in \O: \exists y \in T: \tuple {x, y} \in e$
thus showing that $e$ is left-total.
Also vacuously:
:$\forall x \in \O: \forall y_1, y_2 \in T: \tuple {x, y_1} \in e \land \tuple {x, y_2} \in e \implies y_1 = y_2$
thus showing that $e$ is many-to-one.
Hence ... | Let $T$ be a [[Definition:Set|set]].
Let $e: \O \to T$ be the [[Definition:Empty Mapping|empty mapping]].
Then $e$ is indeed a [[Definition:Mapping|mapping]]. | Let $e: \O \to T$ be the [[Definition:Empty Mapping|empty mapping]].
[[Definition:Vacuous Truth|Vacuously]]:
:$\forall x \in \O: \exists y \in T: \tuple {x, y} \in e$
thus showing that $e$ is [[Definition:Left-Total Relation|left-total]].
Also [[Definition:Vacuous Truth|vacuously]]:
:$\forall x \in \O: \forall y_1, ... | Empty Mapping is Mapping | https://proofwiki.org/wiki/Empty_Mapping_is_Mapping | https://proofwiki.org/wiki/Empty_Mapping_is_Mapping | [
"Empty Mapping"
] | [
"Definition:Set",
"Definition:Empty Mapping",
"Definition:Mapping"
] | [
"Definition:Empty Mapping",
"Definition:Vacuous Truth",
"Definition:Left-Total Relation",
"Definition:Vacuous Truth",
"Definition:Many-to-One Relation",
"Definition:Mapping"
] |
proofwiki-5997 | Empty Mapping is Unique | For each set $T$ there exists exactly one empty mapping $e: \O \to T$ whose domain is the empty set. | By definition a mapping from $\O$ to $T$ is a subset of the cartesian product $\O \times T$:
:$e: \O \to T \subseteq \O \times T$
From Empty Mapping is Mapping, we have that the empty mapping from $\O$ to $T$ exists.
From Cartesian Product is Empty iff Factor is Empty it follows that the empty mapping equals the empty ... | For each [[Definition:Set|set]] $T$ there exists exactly one [[Definition:Empty Mapping|empty mapping]] $e: \O \to T$ whose [[Definition:Domain of Mapping|domain]] is the [[Definition:Empty Set|empty set]]. | By definition a [[Definition:Mapping|mapping]] from $\O$ to $T$ is a [[Definition:Subset|subset]] of the [[Definition:Cartesian Product|cartesian product]] $\O \times T$:
:$e: \O \to T \subseteq \O \times T$
From [[Empty Mapping is Mapping]], we have that the [[Definition:Empty Mapping|empty mapping]] from $\O$ to $T$... | Empty Mapping is Unique | https://proofwiki.org/wiki/Empty_Mapping_is_Unique | https://proofwiki.org/wiki/Empty_Mapping_is_Unique | [
"Empty Mapping"
] | [
"Definition:Set",
"Definition:Empty Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Empty Set"
] | [
"Definition:Mapping",
"Definition:Subset",
"Definition:Cartesian Product",
"Empty Mapping is Mapping",
"Definition:Empty Mapping",
"Cartesian Product is Empty iff Factor is Empty",
"Definition:Empty Mapping",
"Definition:Empty Set",
"Empty Set is Unique"
] |
proofwiki-5998 | Null Relation is Mapping iff Domain is Empty Set | Let $S$ and $T$ be sets.
The null relation $\RR = \O \subseteq S \times T$ is a mapping {{iff}} $S = \O$. | === Sufficient Condition ===
Let $S = \O$.
Then the null relation $\RR = \O \subseteq S \times T$ is a mapping from Empty Mapping is Mapping.
{{qed|lemma}} | Let $S$ and $T$ be [[Definition:Set|sets]].
The [[Definition:Null Relation|null relation]] $\RR = \O \subseteq S \times T$ is a [[Definition:Mapping|mapping]] {{iff}} $S = \O$. | === Sufficient Condition ===
Let $S = \O$.
Then the [[Definition:Null Relation|null relation]] $\RR = \O \subseteq S \times T$ is a [[Definition:Mapping|mapping]] from [[Empty Mapping is Mapping]].
{{qed|lemma}} | Null Relation is Mapping iff Domain is Empty Set | https://proofwiki.org/wiki/Null_Relation_is_Mapping_iff_Domain_is_Empty_Set | https://proofwiki.org/wiki/Null_Relation_is_Mapping_iff_Domain_is_Empty_Set | [
"Null Relation",
"Empty Mapping",
"Empty Set"
] | [
"Definition:Set",
"Definition:Null Relation",
"Definition:Mapping"
] | [
"Definition:Null Relation",
"Definition:Mapping",
"Empty Mapping is Mapping"
] |
proofwiki-5999 | Orbit of Element of Group Acting on Itself is Group | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $*$ be the group action of $\struct {G, \circ}$ on itself by the rule:
:$\forall g, h \in G: g * h = g \circ h$
Then the orbit of an element $x \in G$ is given by:
:$\Orb x = G$ | Let $y \in G$.
Then:
{{begin-eqn}}
{{eqn | q = \exists g \in G
| l = y
| r = g \circ x
| c = Group has Latin Square Property
}}
{{eqn | ll= \leadsto
| l = y
| o = \in
| r = \Orb x
| c = {{Defof|Orbit (Group Theory)|Orbit}}
}}
{{end-eqn}}
Hence the result.
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $*$ be the [[Definition:Group Action|group action]] of $\struct {G, \circ}$ on itself by the rule:
:$\forall g, h \in G: g * h = g \circ h$
Then the [[Definition:Orbit (Group Theory)|orbit]] of an ele... | Let $y \in G$.
Then:
{{begin-eqn}}
{{eqn | q = \exists g \in G
| l = y
| r = g \circ x
| c = [[Group has Latin Square Property]]
}}
{{eqn | ll= \leadsto
| l = y
| o = \in
| r = \Orb x
| c = {{Defof|Orbit (Group Theory)|Orbit}}
}}
{{end-eqn}}
Hence the result.
{{qed}} | Orbit of Element of Group Acting on Itself is Group | https://proofwiki.org/wiki/Orbit_of_Element_of_Group_Acting_on_Itself_is_Group | https://proofwiki.org/wiki/Orbit_of_Element_of_Group_Acting_on_Itself_is_Group | [
"Group Actions"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group Action",
"Definition:Orbit (Group Theory)"
] | [
"Group has Latin Square Property"
] |
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