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-300 | Symmetric Difference is Associative | Symmetric difference is associative:
:$R \symdif \paren {S \symdif T} = \paren {R \symdif S} \symdif T$ | Expanding the {{RHS}}:
{{begin-eqn}}
{{eqn | o =
| r = \paren {R \symdif S} \symdif T
| c =
}}
{{eqn | r = \paren {\paren {\paren {R \cup S} \cap \paren {\overline R \cup \overline S} } \cup T} \cap \paren {\overline {R \symdif S} \cup \overline T}
| c = {{Defof|Symmetric Difference|index = 4}}
}}
{... | [[Definition:Symmetric Difference|Symmetric difference]] is [[Definition:Associative Operation|associative]]:
:$R \symdif \paren {S \symdif T} = \paren {R \symdif S} \symdif T$ | Expanding the {{RHS}}:
{{begin-eqn}}
{{eqn | o =
| r = \paren {R \symdif S} \symdif T
| c =
}}
{{eqn | r = \paren {\paren {\paren {R \cup S} \cap \paren {\overline R \cup \overline S} } \cup T} \cap \paren {\overline {R \symdif S} \cup \overline T}
| c = {{Defof|Symmetric Difference|index = 4}}
}}
... | Symmetric Difference is Associative/Proof 2 | https://proofwiki.org/wiki/Symmetric_Difference_is_Associative | https://proofwiki.org/wiki/Symmetric_Difference_is_Associative/Proof_2 | [
"Symmetric Difference is Associative",
"Symmetric Difference",
"Associative Laws of Set Theory",
"Examples of Associative Operations"
] | [
"Definition:Symmetric Difference",
"Definition:Associative Operation"
] | [
"Union Distributes over Intersection",
"De Morgan's Laws (Set Theory)/Set Complement/Complement of Union",
"De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection",
"Union Distributes over Intersection",
"Union Distributes over Intersection",
"De Morgan's Laws (Set Theory)/Set Complement/... |
proofwiki-301 | Cartesian Product is Empty iff Factor is Empty | :$S \times T = \O \iff S = \O \lor T = \O$
Thus:
:$S \times \O = \O = \O \times T$ | {{begin-eqn}}
{{eqn | l = S \times T
| o = \ne
| r = \O
}}
{{eqn | ll= \leadstoandfrom
| l = \exists \tuple {s, t}
| o = \in
| r = S \times T
| c = {{Defof|Empty Set}}
}}
{{eqn | ll= \leadstoandfrom
| l = \exists s \in S
| o = \land
| r = \exists t \in T
| c =... | :$S \times T = \O \iff S = \O \lor T = \O$
Thus:
:$S \times \O = \O = \O \times T$ | {{begin-eqn}}
{{eqn | l = S \times T
| o = \ne
| r = \O
}}
{{eqn | ll= \leadstoandfrom
| l = \exists \tuple {s, t}
| o = \in
| r = S \times T
| c = {{Defof|Empty Set}}
}}
{{eqn | ll= \leadstoandfrom
| l = \exists s \in S
| o = \land
| r = \exists t \in T
| c =... | Cartesian Product is Empty iff Factor is Empty | https://proofwiki.org/wiki/Cartesian_Product_is_Empty_iff_Factor_is_Empty | https://proofwiki.org/wiki/Cartesian_Product_is_Empty_iff_Factor_is_Empty | [
"Cartesian Product",
"Empty Set"
] | [] | [
"De Morgan's Laws (Logic)/Conjunction of Negations",
"Rule of Transposition"
] |
proofwiki-302 | Cartesian Product is Anticommutative | Let $S, T \ne \O$.
Then:
:$S \times T = T \times S \iff S = T$ | Suppose $S \times T = T \times S$.
Then:
{{begin-eqn}}
{{eqn | l = x \in S
| o = \land
| r = y \in T
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \tuple {x, y}
| o = \in
| r = S \times T
| c = {{Defof|Cartesian Product}}
}}
{{eqn | ll= \leadstoandfrom
| l = \tuple {x, y}
... | Let $S, T \ne \O$.
Then:
:$S \times T = T \times S \iff S = T$ | Suppose $S \times T = T \times S$.
Then:
{{begin-eqn}}
{{eqn | l = x \in S
| o = \land
| r = y \in T
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \tuple {x, y}
| o = \in
| r = S \times T
| c = {{Defof|Cartesian Product}}
}}
{{eqn | ll= \leadstoandfrom
| l = \tuple {x, y}... | Cartesian Product is Anticommutative | https://proofwiki.org/wiki/Cartesian_Product_is_Anticommutative | https://proofwiki.org/wiki/Cartesian_Product_is_Anticommutative | [
"Cartesian Product"
] | [] | [
"Definition:Set Equality",
"Cartesian Product is Empty iff Factor is Empty"
] |
proofwiki-303 | Cartesian Product of Subsets | Let $A, B, S, T$ be sets such that $A \subseteq B$ and $S \subseteq T$.
Then:
:$A \times S \subseteq B \times T$
In addition, if $A, S \ne \O$, then:
:$A \times S \subseteq B \times T \iff A \subseteq B \land S \subseteq T$ | First we show that $A \subseteq B \land S \subseteq T \implies A \times S \subseteq B \times T$.
First, let $A = \O$ or $S = \O$.
Then from Cartesian Product is Empty iff Factor is Empty:
:$A \times S = \O \subseteq B \times T$
so the result holds.
Next, let $A, S \ne \O$.
Then from Cartesian Product is Empty iff Facto... | Let $A, B, S, T$ be [[Definition:Set|sets]] such that $A \subseteq B$ and $S \subseteq T$.
Then:
:$A \times S \subseteq B \times T$
In addition, if $A, S \ne \O$, then:
:$A \times S \subseteq B \times T \iff A \subseteq B \land S \subseteq T$ | First we show that $A \subseteq B \land S \subseteq T \implies A \times S \subseteq B \times T$.
First, let $A = \O$ or $S = \O$.
Then from [[Cartesian Product is Empty iff Factor is Empty]]:
:$A \times S = \O \subseteq B \times T$
so the result holds.
Next, let $A, S \ne \O$.
Then from [[Cartesian Product is Empt... | Cartesian Product of Subsets | https://proofwiki.org/wiki/Cartesian_Product_of_Subsets | https://proofwiki.org/wiki/Cartesian_Product_of_Subsets | [
"Cartesian Product",
"Subsets",
"Cartesian Product of Subsets"
] | [
"Definition:Set"
] | [
"Cartesian Product is Empty iff Factor is Empty",
"Cartesian Product is Empty iff Factor is Empty"
] |
proofwiki-304 | Cartesian Product of Intersections | :$\paren {S_1 \cap S_2} \times \paren {T_1 \cap T_2} = \paren {S_1 \times T_1} \cap \paren {S_2 \times T_2}$
where $S_1, S_2, T_1, T_2$ are sets. | {{begin-eqn}}
{{eqn | o =
| r = \tuple {x, y} \in \paren {S_1 \cap S_2} \times \paren {T_1 \cap T_2}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in S_1 \land x \in S_2} \land \paren {y \in T_1 \land y \in T_2}
| c = {{Defof|Cartesian Product}} and {{Defof|Set Intersection}}
}}
{{eqn | o = \leadst... | :$\paren {S_1 \cap S_2} \times \paren {T_1 \cap T_2} = \paren {S_1 \times T_1} \cap \paren {S_2 \times T_2}$
where $S_1, S_2, T_1, T_2$ are [[Definition:Set|sets]]. | {{begin-eqn}}
{{eqn | o =
| r = \tuple {x, y} \in \paren {S_1 \cap S_2} \times \paren {T_1 \cap T_2}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in S_1 \land x \in S_2} \land \paren {y \in T_1 \land y \in T_2}
| c = {{Defof|Cartesian Product}} and {{Defof|Set Intersection}}
}}
{{eqn | o = \leadst... | Cartesian Product of Intersections | https://proofwiki.org/wiki/Cartesian_Product_of_Intersections | https://proofwiki.org/wiki/Cartesian_Product_of_Intersections | [
"Cartesian Product",
"Set Intersection",
"Cartesian Product of Intersections"
] | [
"Definition:Set"
] | [
"Rule of Commutation",
"Rule of Association"
] |
proofwiki-305 | Cartesian Product Distributes over Union | Cartesian product is distributive over union:
:$A \times \paren {B \cup C} = \paren {A \times B} \cup \paren {A \times C}$
:$\paren {B \cup C} \times A = \paren {B \times A} \cup \paren {C \times A}$ | Take the result Cartesian Product of Unions:
:$\paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2} = \paren {S_1 \times T_1} \cup \paren {S_2 \times T_2} \cup \paren {S_1 \times T_2} \cup \paren {S_2 \times T_1}$
Put $S_1 = S_2 = A, T_1 = B, T_2 = C$:
{{begin-eqn}}
{{eqn | o =
| r = A \times \paren {B \cup C}
... | [[Definition:Cartesian Product|Cartesian product]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Set Union|union]]:
:$A \times \paren {B \cup C} = \paren {A \times B} \cup \paren {A \times C}$
:$\paren {B \cup C} \times A = \paren {B \times A} \cup \paren {C \times A}$ | Take the result [[Cartesian Product of Unions]]:
:$\paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2} = \paren {S_1 \times T_1} \cup \paren {S_2 \times T_2} \cup \paren {S_1 \times T_2} \cup \paren {S_2 \times T_1}$
Put $S_1 = S_2 = A, T_1 = B, T_2 = C$:
{{begin-eqn}}
{{eqn | o =
| r = A \times \paren {B \cup... | Cartesian Product Distributes over Union | https://proofwiki.org/wiki/Cartesian_Product_Distributes_over_Union | https://proofwiki.org/wiki/Cartesian_Product_Distributes_over_Union | [
"Cartesian Product of Unions",
"Cartesian Product",
"Set Union",
"Examples of Distributive Operations"
] | [
"Definition:Cartesian Product",
"Definition:Distributive Operation",
"Definition:Set Union"
] | [
"Cartesian Product of Unions",
"Set Union is Idempotent",
"Cartesian Product of Unions",
"Set Union is Idempotent"
] |
proofwiki-306 | Cartesian Product of Unions | :$\paren {A \cup B} \times \paren {C \cup D} = \paren {A \times C} \cup \paren {B \times D} \cup \paren {A \times D} \cup \paren {B \times C}$ | {{begin-eqn}}
{{eqn | o =
| r = \tuple {x, y} \in \paren {A \cup B} \times \paren {C \cup D}
}}
{{eqn | ll= \leadstoandfrom
| o =
| r = \paren {x \in A \lor x \in B}
}}
{{eqn | o = \land
| r = \paren {y \in C \lor y \in D}
| c = {{Defof|Cartesian Product}} and {{Defof|Set Union}}
}}
{{eq... | :$\paren {A \cup B} \times \paren {C \cup D} = \paren {A \times C} \cup \paren {B \times D} \cup \paren {A \times D} \cup \paren {B \times C}$ | {{begin-eqn}}
{{eqn | o =
| r = \tuple {x, y} \in \paren {A \cup B} \times \paren {C \cup D}
}}
{{eqn | ll= \leadstoandfrom
| o =
| r = \paren {x \in A \lor x \in B}
}}
{{eqn | o = \land
| r = \paren {y \in C \lor y \in D}
| c = {{Defof|Cartesian Product}} and {{Defof|Set Union}}
}}
{{eq... | Cartesian Product of Unions | https://proofwiki.org/wiki/Cartesian_Product_of_Unions | https://proofwiki.org/wiki/Cartesian_Product_of_Unions | [
"Cartesian Product",
"Set Union",
"Cartesian Product of Unions"
] | [] | [
"Rule of Distribution",
"Rule of Distribution"
] |
proofwiki-307 | Cartesian Product Distributes over Set Difference | Cartesian product is distributive over set difference:
:$(1): \quad S \times \paren {T_1 \setminus T_2} = \paren {S \times T_1} \setminus \paren {S \times T_2}$
:$(2): \quad \paren {T_1 \setminus T_2} \times S = \paren {T_1 \times S} \setminus \paren {T_2 \times S}$ | {{begin-eqn}}
{{eqn | o =
| r = \tuple {x, y} \in S \times \paren {T_1 \setminus T_2}
| c =
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in S} \land \paren {y \in \paren {T_1 \setminus T_2} }
| c = {{Defof|Cartesian Product}}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in S} \land... | [[Definition:Cartesian Product|Cartesian product]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Set Difference|set difference]]:
:$(1): \quad S \times \paren {T_1 \setminus T_2} = \paren {S \times T_1} \setminus \paren {S \times T_2}$
:$(2): \quad \paren {T_1 \setminus T_2} \times S = \paren... | {{begin-eqn}}
{{eqn | o =
| r = \tuple {x, y} \in S \times \paren {T_1 \setminus T_2}
| c =
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in S} \land \paren {y \in \paren {T_1 \setminus T_2} }
| c = {{Defof|Cartesian Product}}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in S} \land... | Cartesian Product Distributes over Set Difference | https://proofwiki.org/wiki/Cartesian_Product_Distributes_over_Set_Difference | https://proofwiki.org/wiki/Cartesian_Product_Distributes_over_Set_Difference | [
"Cartesian Product",
"Set Difference",
"Examples of Distributive Operations"
] | [
"Definition:Cartesian Product",
"Definition:Distributive Operation",
"Definition:Set Difference"
] | [] |
proofwiki-308 | Set Difference of Cartesian Products | :$\paren {S_1 \times S_2} \setminus \paren {T_1 \times T_2} = \paren {S_1 \times \paren {S_2 \setminus T_2} } \cup \paren {\paren {S_1 \setminus T_1} \times S_2}$ | Let $\tuple {x, y} \in \paren {S_1 \times S_2} \setminus \paren {T_1 \times T_2}$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {x, y} \in S_1 \times S_2
| o = \land
| r = \tuple {x, y} \notin T_1 \times T_2
| c = {{Defof|Set Difference}}
}}
{{eqn | ll= \leadstoandfrom
| l = x \in S_1 \land y \in S_2
... | :$\paren {S_1 \times S_2} \setminus \paren {T_1 \times T_2} = \paren {S_1 \times \paren {S_2 \setminus T_2} } \cup \paren {\paren {S_1 \setminus T_1} \times S_2}$ | Let $\tuple {x, y} \in \paren {S_1 \times S_2} \setminus \paren {T_1 \times T_2}$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {x, y} \in S_1 \times S_2
| o = \land
| r = \tuple {x, y} \notin T_1 \times T_2
| c = {{Defof|Set Difference}}
}}
{{eqn | ll= \leadstoandfrom
| l = x \in S_1 \land y \in S_... | Set Difference of Cartesian Products | https://proofwiki.org/wiki/Set_Difference_of_Cartesian_Products | https://proofwiki.org/wiki/Set_Difference_of_Cartesian_Products | [
"Cartesian Product",
"Set Difference"
] | [] | [
"De Morgan's Laws (Logic)/Disjunction of Negations",
"Rule of Distribution",
"Definition:Subset",
"Definition:Set Equality"
] |
proofwiki-309 | Inverse of Inverse Relation | The inverse of an inverse relation is the relation itself:
:$\paren {\RR^{-1} }^{-1} = \RR$ | {{begin-eqn}}
{{eqn | l = \tuple {s, t}
| o = \in
| r = \RR
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \tuple {t, s}
| o = \in
| r = \RR^{-1}
| c = {{Defof|Inverse Relation}}
}}
{{eqn | ll= \leadstoandfrom
| l = \tuple {s, t}
| o = \in
| r = \paren {\RR^{-1} ... | The inverse of an [[Definition:Inverse Relation|inverse relation]] is the [[Definition:Relation|relation]] itself:
:$\paren {\RR^{-1} }^{-1} = \RR$ | {{begin-eqn}}
{{eqn | l = \tuple {s, t}
| o = \in
| r = \RR
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \tuple {t, s}
| o = \in
| r = \RR^{-1}
| c = {{Defof|Inverse Relation}}
}}
{{eqn | ll= \leadstoandfrom
| l = \tuple {s, t}
| o = \in
| r = \paren {\RR^{-1} ... | Inverse of Inverse Relation | https://proofwiki.org/wiki/Inverse_of_Inverse_Relation | https://proofwiki.org/wiki/Inverse_of_Inverse_Relation | [
"Inverse Relations"
] | [
"Definition:Inverse Relation",
"Definition:Relation"
] | [] |
proofwiki-310 | Diagonal Relation is Equivalence | The diagonal relation $\Delta_S$ on a set $S$ is always an equivalence in $S$. | Checking in turn each of the criteria for equivalence: | The [[Definition:Diagonal Relation|diagonal relation]] $\Delta_S$ on a [[Definition:Set|set]] $S$ is always an [[Definition:Equivalence Relation|equivalence]] in $S$. | Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Diagonal Relation is Equivalence | https://proofwiki.org/wiki/Diagonal_Relation_is_Equivalence | https://proofwiki.org/wiki/Diagonal_Relation_is_Equivalence | [
"Diagonal Relation is Equivalence",
"Diagonal Relation",
"Examples of Equivalence Relations",
"Equality"
] | [
"Definition:Diagonal Relation",
"Definition:Set",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation"
] |
proofwiki-311 | Trivial Relation is Equivalence | The trivial relation on $S$:
:$\RR = S \times S$
is always an equivalence in $S$. | Let us verify the conditions for an equivalence in turn. | The [[Definition:Trivial Relation|trivial relation]] on $S$:
:$\RR = S \times S$
is always an [[Definition:Equivalence Relation|equivalence]] in $S$. | Let us verify the conditions for an [[Definition:Equivalence Relation|equivalence]] in turn. | Trivial Relation is Equivalence | https://proofwiki.org/wiki/Trivial_Relation_is_Equivalence | https://proofwiki.org/wiki/Trivial_Relation_is_Equivalence | [
"Examples of Equivalence Relations",
"Trivial Relation"
] | [
"Definition:Trivial Relation",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-312 | Equality of Relations | Let $\RR_1$ and $\RR_2$ be relations on $S_1 \times T_1$ and $S_2 \times T_2$ respectively.
Then $\RR_1$ and $\RR_2$ are '''equal''' {{iff}}:
:$S_1 = S_2$
:$T_1 = T_2$
:$\tuple {s, t} \in \RR_1 \iff \tuple {s, t} \in \RR_2$
It is worth labouring the point that for two relations to be '''equal''', not only must their do... | This follows from set equality and Equality of Ordered Pairs.
A '''relation''' on $S \times T$ is an ordered triple, thus:
{{begin-eqn}}
{{eqn | l = \RR_1
| r = \tuple {S_1, T_1, R_1}
| c = where $R_1 \subseteq S_1 \times T_1$
}}
{{eqn | l = \RR_2
| r = \tuple {S_2, T_2, R_2}
| c = where $R_2 \s... | Let $\RR_1$ and $\RR_2$ be [[Definition:Relation|relations]] on $S_1 \times T_1$ and $S_2 \times T_2$ respectively.
Then $\RR_1$ and $\RR_2$ are '''[[Definition:Equality|equal]]''' {{iff}}:
:$S_1 = S_2$
:$T_1 = T_2$
:$\tuple {s, t} \in \RR_1 \iff \tuple {s, t} \in \RR_2$
It is worth labouring the point that for two... | This follows from [[Definition:Set Equality|set equality]] and [[Equality of Ordered Pairs]].
A '''relation''' on $S \times T$ is an [[Definition:Ordered Triple|ordered triple]], thus:
{{begin-eqn}}
{{eqn | l = \RR_1
| r = \tuple {S_1, T_1, R_1}
| c = where $R_1 \subseteq S_1 \times T_1$
}}
{{eqn | l = \R... | Equality of Relations | https://proofwiki.org/wiki/Equality_of_Relations | https://proofwiki.org/wiki/Equality_of_Relations | [
"Relation Theory",
"Equality"
] | [
"Definition:Relation",
"Definition:Equals",
"Definition:Domain (Set Theory)/Relation",
"Definition:Codomain (Set Theory)/Relation"
] | [
"Definition:Set Equality",
"Equality of Ordered Pairs",
"Definition:Ordered Tuple as Ordered Set/Ordered Triple",
"Equality of Ordered Tuples",
"Definition:Set",
"Definition:Ordered Pair",
"Definition:Set Equality"
] |
proofwiki-313 | Image of Singleton under Relation | Let $\RR \subseteq S \times T$ be a relation.
Then the image of an element of $S$ is equal to the image of a singleton containing that element, the singleton being a subset of $S$:
:$\forall s \in S: \map \RR s = \RR \sqbrk {\set s}$ | We have the definitions:
{{begin-eqn}}
{{eqn | l = \map \RR s
| r = \set {t \in T: \tuple {s, t} \in \RR}
| c = {{Defof|Image of Element under Relation}}
}}
{{eqn | l = \RR \sqbrk {\set s}
| r = \set {t \in T: \exists s \in \set s: \tuple {s, t} \in \RR}
| c = {{Defof|Image of Subset under Relat... | Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Then the [[Definition:Image of Element under Relation|image]] of an [[Definition:Element|element]] of $S$ is equal to the [[Definition:Image of Subset under Relation|image]] of a [[Definition:Singleton|singleton]] containing that element, the single... | We have the definitions:
{{begin-eqn}}
{{eqn | l = \map \RR s
| r = \set {t \in T: \tuple {s, t} \in \RR}
| c = {{Defof|Image of Element under Relation}}
}}
{{eqn | l = \RR \sqbrk {\set s}
| r = \set {t \in T: \exists s \in \set s: \tuple {s, t} \in \RR}
| c = {{Defof|Image of Subset under Rela... | Image of Singleton under Relation | https://proofwiki.org/wiki/Image_of_Singleton_under_Relation | https://proofwiki.org/wiki/Image_of_Singleton_under_Relation | [
"Relation Theory",
"Singletons"
] | [
"Definition:Relation",
"Definition:Image (Set Theory)/Relation/Element",
"Definition:Element",
"Definition:Image (Set Theory)/Relation/Subset",
"Definition:Singleton",
"Definition:Subset"
] | [
"Singleton Equality",
"Definition:Set Equality",
"Category:Relation Theory",
"Category:Singletons"
] |
proofwiki-314 | Image of Subset under Relation is Subset of Image | Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation from $S$ to $T$.
Let $A, B \subseteq S$ such that $A \subseteq B$.
Then the image of $A$ is a subset of the image of $B$:
:$A \subseteq B \implies \RR \sqbrk A \subseteq \RR \sqbrk B$
In the notation of direct image mappings, this can be written:
:$A... | Suppose $\RR \sqbrk A \nsubseteq \RR \sqbrk B$.
{{begin-eqn}}
{{eqn | l = \RR \sqbrk A
| o = \nsubseteq
| r = \RR \sqbrk B
}}
{{eqn | ll= \leadsto
| q = \exists t \in \RR \sqbrk A: \exists \tuple {s, t} \in \RR
| l = s
| o = \notin
| r = B
| c = {{Defof|Image of Subset under Re... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] from $S$ to $T$.
Let $A, B \subseteq S$ such that $A \subseteq B$.
Then the [[Definition:Image of Subset under Relation|image]] of $A$ is a [[Definition:Subset|subset]] of the [[Definition:Image of Subse... | Suppose $\RR \sqbrk A \nsubseteq \RR \sqbrk B$.
{{begin-eqn}}
{{eqn | l = \RR \sqbrk A
| o = \nsubseteq
| r = \RR \sqbrk B
}}
{{eqn | ll= \leadsto
| q = \exists t \in \RR \sqbrk A: \exists \tuple {s, t} \in \RR
| l = s
| o = \notin
| r = B
| c = {{Defof|Image of Subset under R... | Image of Subset under Relation is Subset of Image | https://proofwiki.org/wiki/Image_of_Subset_under_Relation_is_Subset_of_Image | https://proofwiki.org/wiki/Image_of_Subset_under_Relation_is_Subset_of_Image | [
"Subsets",
"Relation Theory"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Image (Set Theory)/Relation/Subset",
"Definition:Subset",
"Definition:Image (Set Theory)/Relation/Subset",
"Definition:Direct Image Mapping/Relation"
] | [
"Rule of Transposition"
] |
proofwiki-315 | Image of Element is Subset | Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $A \subseteq S$.
Then:
:$s \in A \implies \map \RR s \subseteq \RR \sqbrk A$ | From Image of Singleton under Relation:
:$\map \RR s = \RR \sqbrk {\set s}$
From Singleton of Element is Subset:
:$s \in A \implies \set s \subseteq A$
The result follows from Image of Subset is Subset of Image.
{{qed}}
Category:Relation Theory
81ns2wlo73dzga5xv99qsw98k988h3w | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Let $A \subseteq S$.
Then:
:$s \in A \implies \map \RR s \subseteq \RR \sqbrk A$ | From [[Image of Singleton under Relation]]:
:$\map \RR s = \RR \sqbrk {\set s}$
From [[Singleton of Element is Subset]]:
:$s \in A \implies \set s \subseteq A$
The result follows from [[Image of Subset is Subset of Image]].
{{qed}}
[[Category:Relation Theory]]
81ns2wlo73dzga5xv99qsw98k988h3w | Image of Element is Subset | https://proofwiki.org/wiki/Image_of_Element_is_Subset | https://proofwiki.org/wiki/Image_of_Element_is_Subset | [
"Relation Theory"
] | [
"Definition:Set",
"Definition:Relation"
] | [
"Image of Singleton under Relation",
"Singleton of Element is Subset",
"Image of Subset under Relation is Subset of Image",
"Category:Relation Theory"
] |
proofwiki-316 | Image is Subset of Codomain | Let $\RR = S \times T$ be a relation.
For all subsets $A$ of the domain of $\RR$, the image of $A$ is a subset of the codomain of $\RR$:
:$\forall A \subseteq \Dom \RR: \RR \sqbrk A \subseteq T$
In the notation of direct image mappings, this can be written as:
:$\forall A \in \powerset S: \map {\RR^\to} A \in \powerset... | {{begin-eqn}}
{{eqn | l = A
| o = \subseteq
| r = \Dom \RR
| c =
}}
{{eqn | ll= \leadsto
| l = \RR \sqbrk A
| o = \subseteq
| r = \Img \RR
| c = Image of Subset is Subset of Image
}}
{{eqn | o = \subseteq
| r = T
| c =
}}
{{end-eqn}}
{{qed}}
Category:Relation Theo... | Let $\RR = S \times T$ be a [[Definition:Relation|relation]].
For all [[Definition:Subset|subsets]] $A$ of the [[Definition:Domain of Relation|domain]] of $\RR$, the [[Definition:Image of Subset under Relation|image]] of $A$ is a subset of the [[Definition:Codomain of Relation|codomain]] of $\RR$:
:$\forall A \subse... | {{begin-eqn}}
{{eqn | l = A
| o = \subseteq
| r = \Dom \RR
| c =
}}
{{eqn | ll= \leadsto
| l = \RR \sqbrk A
| o = \subseteq
| r = \Img \RR
| c = [[Image of Subset is Subset of Image]]
}}
{{eqn | o = \subseteq
| r = T
| c =
}}
{{end-eqn}}
{{qed}}
[[Category:Relati... | Image is Subset of Codomain | https://proofwiki.org/wiki/Image_is_Subset_of_Codomain | https://proofwiki.org/wiki/Image_is_Subset_of_Codomain | [
"Relation Theory"
] | [
"Definition:Relation",
"Definition:Subset",
"Definition:Domain (Set Theory)/Relation",
"Definition:Image (Set Theory)/Relation/Subset",
"Definition:Codomain (Set Theory)/Relation",
"Definition:Direct Image Mapping/Relation"
] | [
"Image of Subset under Relation is Subset of Image",
"Category:Relation Theory"
] |
proofwiki-317 | Image of Empty Set is Empty Set | Let $\RR \subseteq S \times T$ be a relation.
The image of the empty set is the empty set:
:$\RR \sqbrk \O = \O$ | {{begin-eqn}}
{{eqn | l = \RR \sqbrk \O
| r = \set {t \in \Rng \RR: \exists s \in \O: \tuple {s, t} \in \RR}
| c = {{Defof|Image of Subset under Relation}}
}}
{{eqn | l = \neg \exists s
| o = \in
| r = \O
| c = {{Defof|Empty Set}}
}}
{{eqn | ll= \leadsto
| l = \neg \exists t
| ... | Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
The [[Definition:Image of Subset under Relation|image]] of the [[Definition:Empty Set|empty set]] is the [[Definition:Empty Set|empty set]]:
:$\RR \sqbrk \O = \O$ | {{begin-eqn}}
{{eqn | l = \RR \sqbrk \O
| r = \set {t \in \Rng \RR: \exists s \in \O: \tuple {s, t} \in \RR}
| c = {{Defof|Image of Subset under Relation}}
}}
{{eqn | l = \neg \exists s
| o = \in
| r = \O
| c = {{Defof|Empty Set}}
}}
{{eqn | ll= \leadsto
| l = \neg \exists t
| ... | Image of Empty Set is Empty Set | https://proofwiki.org/wiki/Image_of_Empty_Set_is_Empty_Set | https://proofwiki.org/wiki/Image_of_Empty_Set_is_Empty_Set | [
"Image of Empty Set is Empty Set",
"Empty Set",
"Images",
"Relation Theory"
] | [
"Definition:Relation",
"Definition:Image (Set Theory)/Relation/Subset",
"Definition:Empty Set",
"Definition:Empty Set"
] | [
"Category:Image of Empty Set is Empty Set",
"Category:Empty Set",
"Category:Images",
"Category:Relation Theory"
] |
proofwiki-318 | Domain of Composite Relation | Let $\RR_2 \circ \RR_1$ be a composite relation.
Then the domain of $\RR_2 \circ \RR_1$ is the domain of $\RR_1$:
{{Mistake|All that can be said is that the domain of the composition is a subset of the domain of the first relation}}
:$\Dom {\RR_2 \circ \RR_1} = \Dom {\RR_1}$ | Let $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$.
The domain of $\RR_1$ is $S_1$.
The composite of $\RR_1$ and $\RR_2$ is defined as:
:$\RR_2 \circ \RR_1 = \set {\tuple {x, z}: x \in S_1, z \in S_3: \exists y \in S_2: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}$
From this definition:... | Let $\RR_2 \circ \RR_1$ be a [[Definition:Composition of Relations|composite relation]].
Then the [[Definition:Domain of Relation|domain]] of $\RR_2 \circ \RR_1$ is the [[Definition:Domain of Relation|domain]] of $\RR_1$:
{{Mistake|All that can be said is that the domain of the composition is a subset of the domain ... | Let $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$.
The [[Definition:Domain of Relation|domain]] of $\RR_1$ is $S_1$.
The composite of $\RR_1$ and $\RR_2$ is defined as:
:$\RR_2 \circ \RR_1 = \set {\tuple {x, z}: x \in S_1, z \in S_3: \exists y \in S_2: \tuple {x, y} \in \RR_1 \land \tuple {y... | Domain of Composite Relation | https://proofwiki.org/wiki/Domain_of_Composite_Relation | https://proofwiki.org/wiki/Domain_of_Composite_Relation | [
"Composite Relations"
] | [
"Definition:Composition of Relations",
"Definition:Domain (Set Theory)/Relation",
"Definition:Domain (Set Theory)/Relation"
] | [
"Definition:Domain (Set Theory)/Relation",
"Definition:Domain (Set Theory)/Relation",
"Category:Composite Relations"
] |
proofwiki-319 | Codomain of Composite Relation | Let $\RR_2 \circ \RR_1$ be a composite relation.
Then the codomain of $\RR_2 \circ \RR_1$ is the codomain of $\RR_2$:
:$\Cdm {\RR_2 \circ \RR_1} = \Cdm {\RR_2}$ | Let $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$.
The codomain of $\RR_2$ is $S_3$.
The composite of $\RR_1$ and $\RR_2$ is defined as:
:$\RR_2 \circ \RR_1 = \set {\tuple {x, z}: x \in S_1, z \in S_3: \exists y \in S_2: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}$
From this definitio... | Let $\RR_2 \circ \RR_1$ be a [[Definition:Composition of Relations|composite relation]].
Then the [[Definition:Codomain of Relation|codomain]] of $\RR_2 \circ \RR_1$ is the [[Definition:Codomain of Relation|codomain]] of $\RR_2$:
:$\Cdm {\RR_2 \circ \RR_1} = \Cdm {\RR_2}$ | Let $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$.
The [[Definition:Codomain of Relation|codomain]] of $\RR_2$ is $S_3$.
The composite of $\RR_1$ and $\RR_2$ is defined as:
:$\RR_2 \circ \RR_1 = \set {\tuple {x, z}: x \in S_1, z \in S_3: \exists y \in S_2: \tuple {x, y} \in \RR_1 \land \tupl... | Codomain of Composite Relation | https://proofwiki.org/wiki/Codomain_of_Composite_Relation | https://proofwiki.org/wiki/Codomain_of_Composite_Relation | [
"Composite Relations"
] | [
"Definition:Composition of Relations",
"Definition:Codomain (Set Theory)/Relation",
"Definition:Codomain (Set Theory)/Relation"
] | [
"Definition:Codomain (Set Theory)/Relation",
"Definition:Codomain (Set Theory)/Relation",
"Category:Composite Relations"
] |
proofwiki-320 | Preimage of Mapping equals Domain | The preimage of a mapping is the same set as its domain:
:$\Preimg f = \Dom f$ | Let $f \subseteq S \times T$ be a mapping.
Then:
{{begin-eqn}}
{{eqn | o =
| r = \forall x \in S: \exists y \in T: \tuple {x, y} \in f
| c = {{Defof|Mapping}}
}}
{{eqn | o = \leadsto
| r = \forall x \in S: x \in \Preimg f
| c = {{Defof|Preimage of Mapping}}
}}
{{eqn | o = \leadsto
| r = S... | The [[Definition:Preimage of Mapping|preimage of a mapping]] is the same [[Definition:Set|set]] as its [[Definition:Domain of Mapping|domain]]:
:$\Preimg f = \Dom f$ | Let $f \subseteq S \times T$ be a [[Definition:Mapping|mapping]].
Then:
{{begin-eqn}}
{{eqn | o =
| r = \forall x \in S: \exists y \in T: \tuple {x, y} \in f
| c = {{Defof|Mapping}}
}}
{{eqn | o = \leadsto
| r = \forall x \in S: x \in \Preimg f
| c = {{Defof|Preimage of Mapping}}
}}
{{eqn | ... | Preimage of Mapping equals Domain | https://proofwiki.org/wiki/Preimage_of_Mapping_equals_Domain | https://proofwiki.org/wiki/Preimage_of_Mapping_equals_Domain | [
"Preimages under Mappings",
"Domains of Mappings"
] | [
"Definition:Preimage/Mapping/Mapping",
"Definition:Set",
"Definition:Domain (Set Theory)/Mapping"
] | [
"Definition:Mapping",
"Preimage of Relation is Subset of Domain",
"Definition:Set Equality",
"Category:Preimages under Mappings",
"Category:Domains of Mappings"
] |
proofwiki-321 | Preimage of Relation is Subset of Domain | Let $\RR \subseteq S \times T$ be a relation.
Then the preimage of $\RR$ is a subset of its domain:
:$\Preimg \RR \subseteq S$ | The preimage of $\RR$ is defined as:
:$\Preimg \RR = \set {s \in \Dom \RR: \exists t \in \Rng \RR: \tuple {s, t} \in \RR}$
Hence:
{{begin-eqn}}
{{eqn | l = s
| o = \in
| r = \Preimg \RR
}}
{{eqn | ll= \leadsto
| l = s
| o = \in
| r = \Dom \RR
| c = {{Defof|Preimage of Relation}}
}}
{... | Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Then the [[Definition:Preimage of Relation|preimage]] of $\RR$ is a [[Definition:Subset|subset]] of its [[Definition:Domain of Relation|domain]]:
:$\Preimg \RR \subseteq S$ | The [[Definition:Preimage of Relation|preimage]] of $\RR$ is defined as:
:$\Preimg \RR = \set {s \in \Dom \RR: \exists t \in \Rng \RR: \tuple {s, t} \in \RR}$
Hence:
{{begin-eqn}}
{{eqn | l = s
| o = \in
| r = \Preimg \RR
}}
{{eqn | ll= \leadsto
| l = s
| o = \in
| r = \Dom \RR
|... | Preimage of Relation is Subset of Domain | https://proofwiki.org/wiki/Preimage_of_Relation_is_Subset_of_Domain | https://proofwiki.org/wiki/Preimage_of_Relation_is_Subset_of_Domain | [
"Relation Theory"
] | [
"Definition:Relation",
"Definition:Preimage/Relation/Relation",
"Definition:Subset",
"Definition:Domain (Set Theory)/Relation"
] | [
"Definition:Preimage/Relation/Relation",
"Subset of Set with Propositional Function",
"Category:Relation Theory"
] |
proofwiki-322 | Preimage of Image under Left-Total Relation is Superset | Let $\RR \subseteq S \times T$ be a left-total relation.
Then:
:$A \subseteq S \implies A \subseteq \paren {\RR^{-1} \circ \RR} \sqbrk A$
where:
:$\RR \sqbrk A$ denotes the image of $A$ under $\RR$
:$\RR^{-1} \sqbrk A$ denotes the preimage of $A$ under $\RR$
:$\RR^{-1} \circ \RR$ denotes composition of $\RR^{-1}$ and $... | Suppose $A \subseteq S$.
We have:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = A
}}
{{eqn | ll= \leadsto
| l = \set x
| o = \subseteq
| r = A
| c = Singleton of Element is Subset
}}
{{eqn | ll= \leadsto
| l = \RR \sqbrk {\set x}
| o = \subseteq
| r = \RR \sqbrk A
... | Let $\RR \subseteq S \times T$ be a [[Definition:Left-Total Relation|left-total relation]].
Then:
:$A \subseteq S \implies A \subseteq \paren {\RR^{-1} \circ \RR} \sqbrk A$
where:
:$\RR \sqbrk A$ denotes the [[Definition:Image of Subset under Relation|image of $A$ under $\RR$]]
:$\RR^{-1} \sqbrk A$ denotes the [[Defi... | Suppose $A \subseteq S$.
We have:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = A
}}
{{eqn | ll= \leadsto
| l = \set x
| o = \subseteq
| r = A
| c = [[Singleton of Element is Subset]]
}}
{{eqn | ll= \leadsto
| l = \RR \sqbrk {\set x}
| o = \subseteq
| r = \RR \sqbrk... | Preimage of Image under Left-Total Relation is Superset | https://proofwiki.org/wiki/Preimage_of_Image_under_Left-Total_Relation_is_Superset | https://proofwiki.org/wiki/Preimage_of_Image_under_Left-Total_Relation_is_Superset | [
"Preimages under Relations",
"Composite Relations",
"Left-Total Relations"
] | [
"Definition:Left-Total Relation",
"Definition:Image (Set Theory)/Relation/Subset",
"Definition:Preimage/Relation/Subset",
"Definition:Composition of Relations",
"Definition:Direct Image Mapping/Relation",
"Definition:Inverse Image Mapping/Relation"
] | [
"Singleton of Element is Subset",
"Image of Subset under Relation is Subset of Image",
"Definition:Left-Total Relation",
"Definition:Inverse Relation",
"Definition:Subset"
] |
proofwiki-323 | Inverse of Composite Relation | Let $\RR_2 \circ \RR_1 \subseteq S_1 \times S_3$ be the composite of the two relations $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$.
Then:
:$\paren {\RR_2 \circ \RR_1}^{-1} = \RR_1^{-1} \circ \RR_2^{-1}$ | Let $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$ be relations.
We assume that:
:$\Dom {\RR_2} = \Cdm {\RR_1}$
where $\Dom \RR$ denotes domain and $\Cdm \RR$ denotes codomain of a relation $\RR$.
This is necessary for $\RR_2 \circ \RR_1$ to exist.
From the definition of an inverse relation, we h... | Let $\RR_2 \circ \RR_1 \subseteq S_1 \times S_3$ be the [[Definition:Composition of Relations|composite]] of the two [[Definition:Relation|relations]] $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$.
Then:
:$\paren {\RR_2 \circ \RR_1}^{-1} = \RR_1^{-1} \circ \RR_2^{-1}$ | Let $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$ be [[Definition:Relation|relations]].
We assume that:
:$\Dom {\RR_2} = \Cdm {\RR_1}$
where $\Dom \RR$ denotes [[Definition:Domain of Relation|domain]] and $\Cdm \RR$ denotes [[Definition:Codomain of Relation|codomain]] of a [[Definition:Relation... | Inverse of Composite Relation | https://proofwiki.org/wiki/Inverse_of_Composite_Relation | https://proofwiki.org/wiki/Inverse_of_Composite_Relation | [
"Inverse Relations",
"Composite Relations"
] | [
"Definition:Composition of Relations",
"Definition:Relation"
] | [
"Definition:Relation",
"Definition:Domain (Set Theory)/Relation",
"Definition:Codomain (Set Theory)/Relation",
"Definition:Relation",
"Definition:Inverse Relation"
] |
proofwiki-324 | Image of Union under Relation | Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $S_1$ and $S_2$ be subsets of $S$.
Then:
:$\RR \sqbrk {S_1 \cup S_2} = \RR \sqbrk {S_1} \cup \RR \sqbrk {S_2}$
That is, the image of the union of subsets of $S$ is equal to the union of their images. | {{begin-eqn}}
{{eqn | l = t
| o = \in
| r = \RR \sqbrk {S_1 \cup S_2}
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \exists s \in S_1 \cup S_2
| l = t
| o = \in
| r = \RR \sqbrk s
| c = {{Defof|Image of Subset under Relation}}
}}
{{eqn | ll= \leadstoandfrom
| q = \exis... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Let $S_1$ and $S_2$ be [[Definition:Subset|subsets]] of $S$.
Then:
:$\RR \sqbrk {S_1 \cup S_2} = \RR \sqbrk {S_1} \cup \RR \sqbrk {S_2}$
That is, the [[Definition:Image of Subset under Relation|image]... | {{begin-eqn}}
{{eqn | l = t
| o = \in
| r = \RR \sqbrk {S_1 \cup S_2}
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \exists s \in S_1 \cup S_2
| l = t
| o = \in
| r = \RR \sqbrk s
| c = {{Defof|Image of Subset under Relation}}
}}
{{eqn | ll= \leadstoandfrom
| q = \exis... | Image of Union under Relation | https://proofwiki.org/wiki/Image_of_Union_under_Relation | https://proofwiki.org/wiki/Image_of_Union_under_Relation | [
"Relation Theory",
"Set Union",
"Image of Union under Relation"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Subset",
"Definition:Image (Set Theory)/Relation/Subset",
"Definition:Set Union",
"Definition:Subset",
"Definition:Set Union",
"Definition:Image (Set Theory)/Relation/Subset"
] | [] |
proofwiki-325 | Image of Intersection under Relation | Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $S_1$ and $S_2$ be subsets of $S$.
Then:
:$\RR \sqbrk {S_1 \cap S_2} \subseteq \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$
That is, the image of the intersection of subsets of $S$ is a subset of the intersection of their images. | {{begin-eqn}}
{{eqn | l = S_1 \cap S_2
| o = \subseteq
| r = S_1
| c = Intersection is Subset
}}
{{eqn | ll= \leadsto
| l = \RR \sqbrk {S_1 \cap S_2}
| o = \subseteq
| r = \RR \sqbrk {S_1}
| c = Image of Subset is Subset of Image
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = S_1 \ca... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Let $S_1$ and $S_2$ be [[Definition:Subset|subsets]] of $S$.
Then:
:$\RR \sqbrk {S_1 \cap S_2} \subseteq \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$
That is, the [[Definition:Image of Subset under Relatio... | {{begin-eqn}}
{{eqn | l = S_1 \cap S_2
| o = \subseteq
| r = S_1
| c = [[Intersection is Subset]]
}}
{{eqn | ll= \leadsto
| l = \RR \sqbrk {S_1 \cap S_2}
| o = \subseteq
| r = \RR \sqbrk {S_1}
| c = [[Image of Subset is Subset of Image]]
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l... | Image of Intersection under Relation | https://proofwiki.org/wiki/Image_of_Intersection_under_Relation | https://proofwiki.org/wiki/Image_of_Intersection_under_Relation | [
"Relation Theory",
"Set Intersection",
"Image of Intersection under Relation"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Subset",
"Definition:Image (Set Theory)/Relation/Subset",
"Definition:Set Intersection",
"Definition:Subset",
"Definition:Subset",
"Definition:Set Intersection",
"Definition:Image (Set Theory)/Relation/Subset"
] | [
"Intersection is Subset",
"Image of Subset under Relation is Subset of Image",
"Intersection is Subset",
"Image of Subset under Relation is Subset of Image",
"Intersection is Largest Subset"
] |
proofwiki-326 | Image of Set Difference under Relation | Let $\RR \subseteq S \times T$ be a relation.
Let $A$ and $B$ be subsets of $S$.
Then:
:$\RR \sqbrk A \setminus \RR \sqbrk B \subseteq \RR \sqbrk {A \setminus B}$
where:
:$\setminus$ denotes set difference
:$\RR \sqbrk A$ denotes image of $A$ under $\RR$. | {{begin-eqn}}
{{eqn | l = y
| o = \in
| r = \RR \sqbrk A \setminus \RR \sqbrk B
| c =
}}
{{eqn | ll= \leadsto
| q = \exists x \in A: x \notin B
| l = \tuple {x, y}
| o = \in
| r = \RR
| c = {{Defof|Image of Subset under Relation}}
}}
{{eqn | ll= \leadsto
| q = \exi... | Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Let $A$ and $B$ be [[Definition:Subset|subsets]] of $S$.
Then:
:$\RR \sqbrk A \setminus \RR \sqbrk B \subseteq \RR \sqbrk {A \setminus B}$
where:
:$\setminus$ denotes [[Definition:Set Difference|set difference]]
:$\RR \sqbrk A$ denotes [[Definitio... | {{begin-eqn}}
{{eqn | l = y
| o = \in
| r = \RR \sqbrk A \setminus \RR \sqbrk B
| c =
}}
{{eqn | ll= \leadsto
| q = \exists x \in A: x \notin B
| l = \tuple {x, y}
| o = \in
| r = \RR
| c = {{Defof|Image of Subset under Relation}}
}}
{{eqn | ll= \leadsto
| q = \exi... | Image of Set Difference under Relation | https://proofwiki.org/wiki/Image_of_Set_Difference_under_Relation | https://proofwiki.org/wiki/Image_of_Set_Difference_under_Relation | [
"Relation Theory",
"Set Difference",
"Image of Set Difference under Relation"
] | [
"Definition:Relation",
"Definition:Subset",
"Definition:Set Difference",
"Definition:Image (Set Theory)/Relation/Subset"
] | [] |
proofwiki-327 | Preimage of Union under Relation | Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $T_1$ and $T_2$ be subsets of $T$.
Then:
:$\RR^{-1} \sqbrk {T_1 \cup T_2} = \RR^{-1} \sqbrk {T_1} \cup \RR^{-1} \sqbrk {T_2}$ | We have that $\RR^{-1}$ is a relation.
The result follows from Image of Union under Relation.
{{qed}} | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Let $T_1$ and $T_2$ be [[Definition:Subset|subsets]] of $T$.
Then:
:$\RR^{-1} \sqbrk {T_1 \cup T_2} = \RR^{-1} \sqbrk {T_1} \cup \RR^{-1} \sqbrk {T_2}$ | We have that $\RR^{-1}$ is a [[Definition:Relation|relation]].
The result follows from [[Image of Union under Relation]].
{{qed}} | Preimage of Union under Relation | https://proofwiki.org/wiki/Preimage_of_Union_under_Relation | https://proofwiki.org/wiki/Preimage_of_Union_under_Relation | [
"Preimages under Relations",
"Set Union",
"Preimage of Union under Relation"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Subset"
] | [
"Definition:Relation",
"Image of Union under Relation"
] |
proofwiki-328 | Preimage of Intersection under Relation | Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $C$ and $D$ be subsets of $T$.
Then:
:$\RR^{-1} \sqbrk {C \cap D} \subseteq \RR^{-1} \sqbrk C \cap \RR^{-1} \sqbrk D$ | This follows from Image of Intersection under Relation, and the fact that $\RR^{-1}$ is itself a relation, and therefore obeys the same rules.
{{qed}}
Category:Relation Theory
Category:Set Intersection
osjfreff1fjhbf3bsq2mmphacmym5ec | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Let $C$ and $D$ be [[Definition:Subset|subsets]] of $T$.
Then:
:$\RR^{-1} \sqbrk {C \cap D} \subseteq \RR^{-1} \sqbrk C \cap \RR^{-1} \sqbrk D$ | This follows from [[Image of Intersection under Relation]], and the fact that $\RR^{-1}$ is itself a [[Definition:Relation|relation]], and therefore obeys the same rules.
{{qed}}
[[Category:Relation Theory]]
[[Category:Set Intersection]]
osjfreff1fjhbf3bsq2mmphacmym5ec | Preimage of Intersection under Relation | https://proofwiki.org/wiki/Preimage_of_Intersection_under_Relation | https://proofwiki.org/wiki/Preimage_of_Intersection_under_Relation | [
"Relation Theory",
"Set Intersection"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Subset"
] | [
"Image of Intersection under Relation",
"Definition:Relation",
"Category:Relation Theory",
"Category:Set Intersection"
] |
proofwiki-329 | Preimage of Set Difference under Relation | Let $\RR \subseteq S \times T$ be a relation.
Let $C$ and $D$ be subsets of $T$.
Then:
:$\RR^{-1} \sqbrk C \setminus \RR^{-1} \sqbrk D \subseteq \RR^{-1} \sqbrk {C \setminus D}$
where:
:$\setminus$ denotes set difference
:$\RR^{-1} \sqbrk C$ denotes the preimage of $C$ under $\RR$. | We have that $\RR^{-1}$ is itself a relation
The result then follows from Image of Set Difference under Relation.
{{qed}}
Category:Preimages under Relations
Category:Set Difference
jg0wummvseuge8ns642yx293yqmwlw0 | Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Let $C$ and $D$ be [[Definition:Subset|subsets]] of $T$.
Then:
:$\RR^{-1} \sqbrk C \setminus \RR^{-1} \sqbrk D \subseteq \RR^{-1} \sqbrk {C \setminus D}$
where:
:$\setminus$ denotes [[Definition:Set Difference|set difference]]
:$\RR^{-1} \sqbrk C$... | We have that $\RR^{-1}$ is itself a [[Definition:Relation|relation]]
The result then follows from [[Image of Set Difference under Relation]].
{{qed}}
[[Category:Preimages under Relations]]
[[Category:Set Difference]]
jg0wummvseuge8ns642yx293yqmwlw0 | Preimage of Set Difference under Relation | https://proofwiki.org/wiki/Preimage_of_Set_Difference_under_Relation | https://proofwiki.org/wiki/Preimage_of_Set_Difference_under_Relation | [
"Preimages under Relations",
"Set Difference"
] | [
"Definition:Relation",
"Definition:Subset",
"Definition:Set Difference",
"Definition:Preimage/Relation/Subset"
] | [
"Definition:Relation",
"Image of Set Difference under Relation",
"Category:Preimages under Relations",
"Category:Set Difference"
] |
proofwiki-330 | Restriction is Subset of Relation | Let $\RR \subseteq S \times T$ be a relation.
Let $X \subseteq S$.
Then the restriction of $\RR$ to $X$ is a subset of $\RR$. | From the definition of restriction:
:$\forall x \in X: \map {\RR \restriction_X} x = \map \RR x$
Thus:
:$\forall x \in X: \exists t \in T: \tuple {x, t} \in \RR \restriction_X$
But $\tuple {x, t}$ is also (by definition) in $\RR$.
It follows that:
:$\RR \restriction_X \subseteq \RR$
{{qed}}
Category:Restrictions
73kf7l... | Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Let $X \subseteq S$.
Then the [[Definition:Restriction of Relation|restriction]] of $\RR$ to $X$ is a [[Definition:Subset|subset]] of $\RR$. | From the definition of [[Definition:Restriction of Relation|restriction]]:
:$\forall x \in X: \map {\RR \restriction_X} x = \map \RR x$
Thus:
:$\forall x \in X: \exists t \in T: \tuple {x, t} \in \RR \restriction_X$
But $\tuple {x, t}$ is also (by definition) in $\RR$.
It follows that:
:$\RR \restriction_X \subset... | Restriction is Subset of Relation | https://proofwiki.org/wiki/Restriction_is_Subset_of_Relation | https://proofwiki.org/wiki/Restriction_is_Subset_of_Relation | [
"Restrictions"
] | [
"Definition:Relation",
"Definition:Restriction/Relation",
"Definition:Subset"
] | [
"Definition:Restriction/Relation",
"Category:Restrictions"
] |
proofwiki-331 | Null Relation is Antireflexive, Symmetric and Transitive | Let $S$ be a set which is non-empty.
Let $\RR \subseteq S \times S$ be the null relation.
Then $\RR$ is antireflexive, symmetric and transitive.
If $S = \O$ then Relation on Empty Set is Equivalence applies. | From the definition of null relation:
:$\RR = \O$ | Let $S$ be a [[Definition:Set|set]] which is [[Definition:Non-Empty Set|non-empty]].
Let $\RR \subseteq S \times S$ be the [[Definition:Null Relation|null relation]].
Then $\RR$ is [[Definition:Antireflexive Relation|antireflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|tr... | From the definition of [[Definition:Null Relation|null relation]]:
:$\RR = \O$ | Null Relation is Antireflexive, Symmetric and Transitive | https://proofwiki.org/wiki/Null_Relation_is_Antireflexive,_Symmetric_and_Transitive | https://proofwiki.org/wiki/Null_Relation_is_Antireflexive,_Symmetric_and_Transitive | [
"Examples of Antireflexive Relations",
"Examples of Symmetric Relations",
"Examples of Transitive Relations",
"Null Relation"
] | [
"Definition:Set",
"Definition:Non-Empty Set",
"Definition:Null Relation",
"Definition:Antireflexive Relation",
"Definition:Symmetric Relation",
"Definition:Transitive Relation",
"Relation on Empty Set is Equivalence"
] | [
"Definition:Null Relation"
] |
proofwiki-332 | Symmetry of Relations is Symmetric | Let $\RR$ be a relation on $S$ which is symmetric. Then:
:$\tuple {x, y} \in \RR \iff \tuple {y, x} \in \RR$. | Let $\RR$ be symmetric.
{{begin-eqn}}
{{eqn | l = \tuple {x, y} \in \RR
| o = \implies
| r = \tuple {y, x} \in \RR
| c = {{Defof|Symmetric Relation}}
}}
{{eqn | l = \tuple {y, x} \in \RR
| o = \implies
| r = \tuple {x, y} \in \RR
| c = {{Defof|Symmetric Relation}}
}}
{{eqn | ll= \lea... | Let $\RR$ be a [[Definition:Relation|relation]] on $S$ which is [[Definition:Symmetric Relation|symmetric]]. Then:
:$\tuple {x, y} \in \RR \iff \tuple {y, x} \in \RR$. | Let $\RR$ be [[Definition:Symmetric Relation|symmetric]].
{{begin-eqn}}
{{eqn | l = \tuple {x, y} \in \RR
| o = \implies
| r = \tuple {y, x} \in \RR
| c = {{Defof|Symmetric Relation}}
}}
{{eqn | l = \tuple {y, x} \in \RR
| o = \implies
| r = \tuple {x, y} \in \RR
| c = {{Defof|Symme... | Symmetry of Relations is Symmetric | https://proofwiki.org/wiki/Symmetry_of_Relations_is_Symmetric | https://proofwiki.org/wiki/Symmetry_of_Relations_is_Symmetric | [
"Symmetric Relations"
] | [
"Definition:Relation",
"Definition:Symmetric Relation"
] | [
"Definition:Symmetric Relation",
"Category:Symmetric Relations"
] |
proofwiki-333 | Relation is Symmetric and Antisymmetric iff Coreflexive | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation in $S$.
Then:
:$\RR$ is both symmetric and antisymmetric
{{iff}}:
:$\RR$ is coreflexive. | === Necessary Condition ===
Let $\RR$ be both symmetric and antisymmetric
Suppose $\RR \nsubseteq \Delta_S$.
Then:
:$\exists \tuple {x, y} \in \RR: x \ne y$
But then as $\RR$ is symmetric:
:$\tuple {y, x} \in \RR$
So we have:
:$\tuple {x, y} \in \RR$
and:
:$\tuple {y, x} \in \RR$
where $x \ne y$.
Thus $\RR$ is not anti... | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be a [[Definition:Relation|relation]] in $S$.
Then:
:$\RR$ is both [[Definition:Symmetric Relation|symmetric]] and [[Definition:Antisymmetric Relation|antisymmetric]]
{{iff}}:
:$\RR$ is [[Definition:Coreflexive Relation|coreflexive]]. | === Necessary Condition ===
Let $\RR$ be both [[Definition:Symmetric Relation|symmetric]] and [[Definition:Antisymmetric Relation|antisymmetric]]
Suppose $\RR \nsubseteq \Delta_S$.
Then:
:$\exists \tuple {x, y} \in \RR: x \ne y$
But then as $\RR$ is [[Definition:Symmetric Relation|symmetric]]:
:$\tuple {y, x} \in \... | Relation is Symmetric and Antisymmetric iff Coreflexive | https://proofwiki.org/wiki/Relation_is_Symmetric_and_Antisymmetric_iff_Coreflexive | https://proofwiki.org/wiki/Relation_is_Symmetric_and_Antisymmetric_iff_Coreflexive | [
"Symmetric Relations",
"Antisymmetric Relations",
"Coreflexive Relations"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Symmetric Relation",
"Definition:Antisymmetric Relation",
"Definition:Coreflexive Relation"
] | [
"Definition:Symmetric Relation",
"Definition:Antisymmetric Relation",
"Definition:Symmetric Relation",
"Definition:Antisymmetric Relation",
"Definition:Coreflexive Relation/Definition 2",
"Definition:Coreflexive Relation",
"Definition:Symmetric Relation",
"Definition:Antisymmetric Relation",
"Defini... |
proofwiki-334 | Relation both Symmetric and Asymmetric is Null | Let $\RR$ be a relation in $S$ which is both symmetric and asymmetric.
Then:
:$\RR = \O$ | Let $\RR \ne \O$.
Then:
:$\exists \tuple {x, y} \in \RR$
Now, either $\tuple {y, x} \in \RR$ or $\tuple {y, x} \notin \RR$.
If $\tuple {y, x} \in \RR$, then $\RR$ is not asymmetric.
If $\tuple {y, x} \notin \RR$, then $\RR$ is not symmetric.
Therefore, if $\RR \ne \O$, $\RR$ cannot be both symmetric and asymmetric.
{{q... | Let $\RR$ be a [[Definition:Relation|relation]] in $S$ which is both [[Definition:Symmetric Relation|symmetric]] and [[Definition:Asymmetric Relation|asymmetric]].
Then:
:$\RR = \O$ | Let $\RR \ne \O$.
Then:
:$\exists \tuple {x, y} \in \RR$
Now, either $\tuple {y, x} \in \RR$ or $\tuple {y, x} \notin \RR$.
If $\tuple {y, x} \in \RR$, then $\RR$ is not [[Definition:Asymmetric Relation|asymmetric]].
If $\tuple {y, x} \notin \RR$, then $\RR$ is not [[Definition:Symmetric Relation|symmetric]].
Ther... | Relation both Symmetric and Asymmetric is Null | https://proofwiki.org/wiki/Relation_both_Symmetric_and_Asymmetric_is_Null | https://proofwiki.org/wiki/Relation_both_Symmetric_and_Asymmetric_is_Null | [
"Symmetric Relations",
"Asymmetric Relations",
"Null Relation"
] | [
"Definition:Relation",
"Definition:Symmetric Relation",
"Definition:Asymmetric Relation"
] | [
"Definition:Asymmetric Relation",
"Definition:Symmetric Relation",
"Definition:Symmetric Relation",
"Definition:Asymmetric Relation",
"Biconditional as Disjunction of Conjunctions",
"Definition:Symmetric Relation",
"False Statement implies Every Statement",
"Definition:Asymmetric Relation",
"Definit... |
proofwiki-335 | Equivalent Characterizations of Abelian Group | Let $G$ be a group.
{{TFAE}}
{{begin-axiom}}
{{axiom | n = 1
| t = $G$ is abelian
}}
{{axiom | n = 2
| t = $\forall a, b \in G: \paren {a b}^{-1} = a^{-1} b^{-1}$
}}
{{axiom | n = 3
| t = {{Defof|Cross Cancellation Property}}: $\forall a, b, c \in G: a b = c a \implies b = c$
}}
{{axiom | n = 4
... | === $(1)$ iff $(2)$ ===
See Inversion Mapping is Automorphism iff Group is Abelian.
{{qed|lemma}}
{{explain|There's a better link that proves this one, but IIRC it's difficult to find. I've been tidying up this recently, I'll see if I can find it in a bit.}} | Let $G$ be a [[Definition:Group|group]].
{{TFAE}}
{{begin-axiom}}
{{axiom | n = 1
| t = $G$ is [[Definition:Abelian Group|abelian]]
}}
{{axiom | n = 2
| t = $\forall a, b \in G: \paren {a b}^{-1} = a^{-1} b^{-1}$
}}
{{axiom | n = 3
| t = {{Defof|Cross Cancellation Property}}: $\forall a, b, c... | === $(1)$ iff $(2)$ ===
See [[Inversion Mapping is Automorphism iff Group is Abelian]].
{{qed|lemma}}
{{explain|There's a better link that proves this one, but IIRC it's difficult to find. I've been tidying up this recently, I'll see if I can find it in a bit.}} | Equivalent Characterizations of Abelian Group | https://proofwiki.org/wiki/Equivalent_Characterizations_of_Abelian_Group | https://proofwiki.org/wiki/Equivalent_Characterizations_of_Abelian_Group | [
"Abelian Groups"
] | [
"Definition:Group",
"Definition:Abelian Group",
"Definition:Opposite Group"
] | [
"Inversion Mapping is Automorphism iff Group is Abelian"
] |
proofwiki-336 | Relation Symmetry | Every non-null relation has exactly one of these properties: it is either:
:symmetric,
:asymmetric or
:non-symmetric. | Let $\RR \subseteq S \times T$ denote an arbitrary relation on $S \times T$, for $S$ and $T$ arbitrary sets.
First we deal with the case where $\RR = \O$.
From Relation both Symmetric and Asymmetric is Null, the null relation ''is'' both symmetric and asymmetric.
So, let us assume that $\RR \ne \O$.
Let $\RR$ be symmet... | Every non-[[Definition:Null Relation|null]] [[Definition:Relation|relation]] has exactly one of these properties: it is either:
:[[Definition:Symmetric Relation|symmetric]],
:[[Definition:Asymmetric Relation|asymmetric]] or
:[[Definition:Non-Symmetric Relation|non-symmetric]]. | Let $\RR \subseteq S \times T$ denote an [[Definition:Arbitrary|arbitrary]] [[Definition:Relation|relation]] on $S \times T$, for $S$ and $T$ [[Definition:Arbitrary|arbitrary]] [[Definition:Set|sets]].
First we deal with the case where $\RR = \O$.
From [[Relation both Symmetric and Asymmetric is Null]], the [[Defini... | Relation Symmetry | https://proofwiki.org/wiki/Relation_Symmetry | https://proofwiki.org/wiki/Relation_Symmetry | [
"Symmetric Relations"
] | [
"Definition:Null Relation",
"Definition:Relation",
"Definition:Symmetric Relation",
"Definition:Asymmetric Relation",
"Definition:Non-Symmetric Relation"
] | [
"Definition:Arbitrary",
"Definition:Relation",
"Definition:Arbitrary",
"Definition:Set",
"Relation both Symmetric and Asymmetric is Null",
"Definition:Null Relation",
"Definition:Symmetric Relation",
"Definition:Asymmetric Relation",
"Definition:Symmetric Relation",
"Definition:Asymmetric Relation... |
proofwiki-337 | Asymmetric Relation is Antisymmetric | Let $\RR$ be an asymmetric relation.
Then $\RR$ is also antisymmetric. | Let $\RR$ be asymmetric.
Then from the definition of asymmetric:
:$\tuple {x, y} \in \RR \implies \tuple {y, x} \notin \RR$
Thus:
:$\neg \exists \tuple {x, y} \in \RR: \tuple {y, x} \in \RR$
Thus:
:$\set {\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR} = \O$
Thus:
:$\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR ... | Let $\RR$ be an [[Definition:Asymmetric Relation|asymmetric relation]].
Then $\RR$ is also [[Definition:Antisymmetric Relation|antisymmetric]]. | Let $\RR$ be [[Definition:Asymmetric Relation|asymmetric]].
Then from the definition of [[Definition:Asymmetric Relation|asymmetric]]:
:$\tuple {x, y} \in \RR \implies \tuple {y, x} \notin \RR$
Thus:
:$\neg \exists \tuple {x, y} \in \RR: \tuple {y, x} \in \RR$
Thus:
:$\set {\tuple {x, y} \in \RR \land \tuple {y, x} ... | Asymmetric Relation is Antisymmetric | https://proofwiki.org/wiki/Asymmetric_Relation_is_Antisymmetric | https://proofwiki.org/wiki/Asymmetric_Relation_is_Antisymmetric | [
"Asymmetric Relations",
"Antisymmetric Relations"
] | [
"Definition:Asymmetric Relation",
"Definition:Antisymmetric Relation"
] | [
"Definition:Asymmetric Relation",
"Definition:Asymmetric Relation",
"Definition:Vacuous Truth"
] |
proofwiki-338 | Asymmetric Relation is Antireflexive | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation on $S$.
Let $\RR$ be asymmetric.
Then $\RR$ is also antireflexive. | Let $\RR$ be asymmetric.
Then, by definition:
:$\tuple {x, y} \in \RR \implies \tuple {y, x} \notin \RR$
{{AimForCont}} $\tuple {x, x} \in \RR$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {x, x} \in \RR
| o = \implies
| r = \tuple {x, x} \notin \RR
| c = {{Defof|Asymmetric Relation}}
}}
{{eqn | ll= \leads... | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be a [[Definition:Endorelation|relation]] on $S$.
Let $\RR$ be [[Definition:Asymmetric Relation|asymmetric]].
Then $\RR$ is also [[Definition:Antireflexive Relation|antireflexive]]. | Let $\RR$ be [[Definition:Asymmetric Relation|asymmetric]].
Then, by definition:
:$\tuple {x, y} \in \RR \implies \tuple {y, x} \notin \RR$
{{AimForCont}} $\tuple {x, x} \in \RR$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {x, x} \in \RR
| o = \implies
| r = \tuple {x, x} \notin \RR
| c = {{Defof|Asy... | Asymmetric Relation is Antireflexive | https://proofwiki.org/wiki/Asymmetric_Relation_is_Antireflexive | https://proofwiki.org/wiki/Asymmetric_Relation_is_Antireflexive | [
"Asymmetric Relations",
"Antireflexive Relations"
] | [
"Definition:Set",
"Definition:Endorelation",
"Definition:Asymmetric Relation",
"Definition:Antireflexive Relation"
] | [
"Definition:Asymmetric Relation",
"Proof by Contradiction",
"Definition:Antireflexive Relation"
] |
proofwiki-339 | Antireflexive and Transitive Relation is Asymmetric | Let $\RR \subseteq S \times S$ be a relation which is not null.
Let $\RR$ be antireflexive and transitive.
Then $\RR$ is also asymmetric. | Let $\RR \subseteq S \times S$ be antireflexive and transitive.
That is:
{{begin-eqn}}
{{eqn | l = \forall x \in S:
| o =
| r = \tuple {x, x} \notin \RR
| c = {{Defof|Antireflexive Relation}}
}}
{{eqn | l = \tuple {x, y} \in \RR \land \tuple {y, z} \in \RR
| o = \implies
| r = \tuple {x,... | Let $\RR \subseteq S \times S$ be a [[Definition:Relation|relation]] which is not [[Definition:Null Relation|null]].
Let $\RR$ be [[Definition:Antireflexive Relation|antireflexive]] and [[Definition:Transitive Relation|transitive]].
Then $\RR$ is also [[Definition:Asymmetric Relation|asymmetric]]. | Let $\RR \subseteq S \times S$ be [[Definition:Antireflexive Relation|antireflexive]] and [[Definition:Transitive Relation|transitive]].
That is:
{{begin-eqn}}
{{eqn | l = \forall x \in S:
| o =
| r = \tuple {x, x} \notin \RR
| c = {{Defof|Antireflexive Relation}}
}}
{{eqn | l = \tuple {x, y} \in... | Antireflexive and Transitive Relation is Asymmetric | https://proofwiki.org/wiki/Antireflexive_and_Transitive_Relation_is_Asymmetric | https://proofwiki.org/wiki/Antireflexive_and_Transitive_Relation_is_Asymmetric | [
"Antireflexive Relations",
"Asymmetric Relations",
"Transitive Relations"
] | [
"Definition:Relation",
"Definition:Null Relation",
"Definition:Antireflexive Relation",
"Definition:Transitive Relation",
"Definition:Asymmetric Relation"
] | [
"Definition:Antireflexive Relation",
"Definition:Transitive Relation",
"Definition:Null Relation",
"Definition:Asymmetric Relation",
"Definition:Transitive Relation",
"Definition:Antireflexive Relation",
"Definition:Asymmetric Relation"
] |
proofwiki-340 | Antitransitive Relation is Antireflexive | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation on $S$.
Let $\RR$ be antitransitive.
Then $\RR$ is also antireflexive. | Suppose $\RR \subseteq S \times S$ is not antireflexive.
Then $\exists x \in S: \tuple {x, x} \in \RR$. (In the case of $\RR$ being reflexive, the property holds for ''all'' $x \in S$.)
Thus $\RR$ is not antitransitive:
{{begin-eqn}}
{{eqn | o =
| r = \tuple {x, x} \in \RR \land \tuple {x, x} \in \RR
| c =
... | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be a [[Definition:Endorelation|relation]] on $S$.
Let $\RR$ be [[Definition:Antitransitive Relation|antitransitive]].
Then $\RR$ is also [[Definition:Antireflexive Relation|antireflexive]]. | Suppose $\RR \subseteq S \times S$ is not [[Definition:Antireflexive Relation|antireflexive]].
Then $\exists x \in S: \tuple {x, x} \in \RR$. (In the case of $\RR$ being [[Definition:Reflexive Relation|reflexive]], the property holds for ''all'' $x \in S$.)
Thus $\RR$ is not [[Definition:Antitransitive Relation|antit... | Antitransitive Relation is Antireflexive | https://proofwiki.org/wiki/Antitransitive_Relation_is_Antireflexive | https://proofwiki.org/wiki/Antitransitive_Relation_is_Antireflexive | [
"Antireflexive Relations",
"Antitransitive Relations"
] | [
"Definition:Set",
"Definition:Endorelation",
"Definition:Antitransitive Relation",
"Definition:Antireflexive Relation"
] | [
"Definition:Antireflexive Relation",
"Definition:Reflexive Relation",
"Definition:Antitransitive Relation",
"Rule of Idempotence",
"Definition:Antitransitive Relation",
"Definition:Antireflexive Relation"
] |
proofwiki-341 | Symmetric Transitive and Serial Relation is Reflexive | Let $\RR$ be a relation which is:
:symmetric
:transitive
:serial.
Then $\RR$ is reflexive.
Thus such a relation is an equivalence. | Let $S$ be a set on which $\RR$ is a relation which is symmetric, transitive and serial.
As $\RR$ is symmetric:
:$x \mathrel \RR y \implies y \mathrel \RR x$
As $\RR$ is transitive:
:$x \mathrel \RR y \land y \mathrel \RR x \implies x \mathrel \RR x$
As $\RR$ is serial:
:$\forall x \in S: \exists y \in S: x \mathrel \... | Let $\RR$ be a [[Definition:Relation|relation]] which is:
:[[Definition:Symmetric Relation|symmetric]]
:[[Definition:Transitive Relation|transitive]]
:[[Definition:Serial Relation|serial]].
Then $\RR$ is [[Definition:Reflexive Relation|reflexive]].
Thus such a [[Definition:Relation|relation]] is an [[Definition:Equ... | Let $S$ be a [[Definition:Set|set]] on which $\RR$ is a [[Definition:Relation|relation]] which is [[Definition:Symmetric Relation|symmetric]], [[Definition:Transitive Relation|transitive]] and [[Definition:Serial Relation|serial]].
As $\RR$ is [[Definition:Symmetric Relation|symmetric]]:
:$x \mathrel \RR y \implies ... | Symmetric Transitive and Serial Relation is Reflexive | https://proofwiki.org/wiki/Symmetric_Transitive_and_Serial_Relation_is_Reflexive | https://proofwiki.org/wiki/Symmetric_Transitive_and_Serial_Relation_is_Reflexive | [
"Reflexive Relations",
"Symmetric Relations",
"Transitive Relations",
"Serial Relations",
"Equivalence Relations"
] | [
"Definition:Relation",
"Definition:Symmetric Relation",
"Definition:Transitive Relation",
"Definition:Serial Relation",
"Definition:Reflexive Relation",
"Definition:Relation",
"Definition:Equivalence Relation"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Symmetric Relation",
"Definition:Transitive Relation",
"Definition:Serial Relation",
"Definition:Symmetric Relation",
"Definition:Transitive Relation",
"Definition:Serial Relation",
"Definition:Serial Relation",
"Definition:Symmetric Relation",
... |
proofwiki-342 | Inverse Relation Properties | Let $\RR$ be a relation on a set $S$.
If $\RR$ has any of the properties:
:Reflexive
:Antireflexive
:Non-reflexive
:Symmetric
:Asymmetric
:Antisymmetric
:Non-symmetric
:Transitive
:Antitransitive
:Non-transitive
then its inverse $\RR^{-1}$ has the same properties. | === Reflexivity ===
{{:Inverse of Reflexive Relation is Reflexive}} | Let $\RR$ be a [[Definition:Relation|relation]] on a set $S$.
If $\RR$ has any of the properties:
:[[Definition:Reflexive Relation|Reflexive]]
:[[Definition:Antireflexive Relation|Antireflexive]]
:[[Definition:Non-Reflexive Relation|Non-reflexive]]
:[[Definition:Symmetric Relation|Symmetric]]
:[[Definition:Asymmetric... | === [[Inverse of Reflexive Relation is Reflexive|Reflexivity]] ===
{{:Inverse of Reflexive Relation is Reflexive}} | Inverse Relation Properties | https://proofwiki.org/wiki/Inverse_Relation_Properties | https://proofwiki.org/wiki/Inverse_Relation_Properties | [
"Inverse Relations"
] | [
"Definition:Relation",
"Definition:Reflexive Relation",
"Definition:Antireflexive Relation",
"Definition:Non-Reflexive Relation",
"Definition:Symmetric Relation",
"Definition:Asymmetric Relation",
"Definition:Antisymmetric Relation",
"Definition:Non-Symmetric Relation",
"Definition:Transitive Relati... | [
"Inverse of Reflexive Relation is Reflexive"
] |
proofwiki-343 | Properties of Restriction of Relation | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
If $\RR$ on $S$ has any of the properties:
* Reflexive
* Antireflexive
* Symmetric
* Antisymmetric
* Asymmetric
* Transitive
... | === Reflexivity === | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be a [[Definition:Relation|relation]] on $S$.
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the [[Definition:Restriction of Relation|restriction]] of $\RR$ to $T$.
If $\RR$ on $S$... | === Reflexivity === | Properties of Restriction of Relation | https://proofwiki.org/wiki/Properties_of_Restriction_of_Relation | https://proofwiki.org/wiki/Properties_of_Restriction_of_Relation | [
"Restrictions"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Subset",
"Definition:Restriction/Relation",
"Definition:Reflexive Relation",
"Definition:Antireflexive Relation",
"Definition:Symmetric Relation",
"Definition:Antisymmetric Relation",
"Definition:Asymmetric Relation",
"Definition:Transitive Rela... | [] |
proofwiki-344 | Relation is Antireflexive iff Disjoint from Diagonal Relation | Let $\RR \subseteq S \times S$ be a relation on a set $S$.
Then:
:$\RR$ is antireflexive
{{iff}}
:$\Delta_S \cap \RR = \O$
where $\Delta_S$ is the diagonal relation. | === Necessary Condition ===
Let $\RR$ be an antireflexive relation.
Let $\tuple {x, y} \in \Delta_S \cap \RR$.
By definition:
:$\tuple {x, y} \in \Delta_S \implies x = y$
Likewise, by definition:
:$\tuple {x, y} \in \RR \implies x \ne y$.
Thus:
:$\Delta_S \cap \RR = \set {\tuple {x, y}: x = y \land x \ne y}$
and so:
:$... | Let $\RR \subseteq S \times S$ be a [[Definition:Relation|relation]] on a [[Definition:Set|set]] $S$.
Then:
:$\RR$ is [[Definition:Antireflexive Relation|antireflexive]]
{{iff}}
:$\Delta_S \cap \RR = \O$
where $\Delta_S$ is the [[Definition:Diagonal Relation|diagonal relation]]. | === Necessary Condition ===
Let $\RR$ be an [[Definition:Antireflexive Relation|antireflexive relation]].
Let $\tuple {x, y} \in \Delta_S \cap \RR$.
By definition:
:$\tuple {x, y} \in \Delta_S \implies x = y$
Likewise, by definition:
:$\tuple {x, y} \in \RR \implies x \ne y$.
Thus:
:$\Delta_S \cap \RR = \set {\tup... | Relation is Antireflexive iff Disjoint from Diagonal Relation | https://proofwiki.org/wiki/Relation_is_Antireflexive_iff_Disjoint_from_Diagonal_Relation | https://proofwiki.org/wiki/Relation_is_Antireflexive_iff_Disjoint_from_Diagonal_Relation | [
"Antireflexive Relations",
"Diagonal Relation"
] | [
"Definition:Relation",
"Definition:Set",
"Definition:Antireflexive Relation",
"Definition:Diagonal Relation"
] | [
"Definition:Antireflexive Relation",
"Definition:Antireflexive Relation"
] |
proofwiki-345 | Inverse Relation Equal iff Subset | Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation.
If $\RR$ is a subset or superset of its inverse, then it equals its inverse.
That is, the following are equivalent:
{{begin-eqn}}
{{eqn | n = 1
| l = \RR
| o = \subseteq
| r = \RR^{-1}
}}
{{eqn | n = 2
| l = \RR^{-1}
| ... | === $(1)$ implies $(2)$ ===
Suppose $\RR \subseteq \RR^{-1}$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {a, b}
| o = \in
| r = \RR^{-1}
| c =
}}
{{eqn | ll= \leadsto
| l = \tuple {b, a}
| o = \in
| r = \RR
| c = {{Defof|Inverse Relation}}
}}
{{eqn | ll= \leadsto
| l = \tuple... | Let $S$ and $T$ be [[Definition:Set|sets]]. Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
If $\RR$ is a [[Definition:Subset|subset]] or [[Definition:Superset|superset]] of its [[Definition:Inverse Relation|inverse]], then it equals its inverse.
That is, the following are [[Definition:Logical Eq... | === $(1)$ implies $(2)$ ===
Suppose $\RR \subseteq \RR^{-1}$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {a, b}
| o = \in
| r = \RR^{-1}
| c =
}}
{{eqn | ll= \leadsto
| l = \tuple {b, a}
| o = \in
| r = \RR
| c = {{Defof|Inverse Relation}}
}}
{{eqn | ll= \leadsto
| l = \tu... | Inverse Relation Equal iff Subset | https://proofwiki.org/wiki/Inverse_Relation_Equal_iff_Subset | https://proofwiki.org/wiki/Inverse_Relation_Equal_iff_Subset | [
"Inverse Relations"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Subset",
"Definition:Subset/Superset",
"Definition:Inverse Relation",
"Definition:Logical Equivalence"
] | [] |
proofwiki-346 | Relation is Symmetric iff Inverse is Symmetric | A relation $\RR$ is symmetric {{iff}} its inverse $\RR^{-1}$ is also symmetric. | Let $\RR$ be symmetric.
Then from Relation equals Inverse iff Symmetric:
:$\RR = \RR^{-1}$
The result follows.
{{qed}} | A [[Definition:Relation|relation]] $\RR$ is [[Definition:Symmetric Relation|symmetric]] {{iff}} its [[Definition:Inverse Relation|inverse]] $\RR^{-1}$ is also [[Definition:Symmetric Relation|symmetric]]. | Let $\RR$ be [[Definition:Symmetric Relation|symmetric]].
Then from [[Relation equals Inverse iff Symmetric]]:
:$\RR = \RR^{-1}$
The result follows.
{{qed}} | Relation is Symmetric iff Inverse is Symmetric | https://proofwiki.org/wiki/Relation_is_Symmetric_iff_Inverse_is_Symmetric | https://proofwiki.org/wiki/Relation_is_Symmetric_iff_Inverse_is_Symmetric | [
"Symmetric Relations"
] | [
"Definition:Relation",
"Definition:Symmetric Relation",
"Definition:Inverse Relation",
"Definition:Symmetric Relation"
] | [
"Definition:Symmetric Relation",
"Equivalence of Definitions of Symmetric Relation"
] |
proofwiki-347 | Composition of Relation with Inverse is Symmetric | Let $\RR \subseteq S \times T$ be a relation.
Then the composition of $\RR$ with its inverse $\RR^{-1}$ is symmetric:
:$(1): \quad \RR^{-1} \circ \RR$ is a symmetric relation on $S$
:$(2): \quad \RR \circ \RR^{-1}$ is a symmetric relation on $T$. | {{Rewrite|In light of the more relaxed definition of composite of relation}}
Note that this result holds for any $\RR \subseteq S \times T$, and does not require that $\struct {S, \RR}$ necessarily be a relational structure.
{{begin-eqn}}
{{eqn | l = \tuple {a, b}
| o = \in
| r =\RR^{-1} \circ \RR
| c... | Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Then the [[Definition:Composition of Relations|composition]] of $\RR$ with its [[Definition:Inverse Relation|inverse]] $\RR^{-1}$ is [[Definition:Symmetric Relation|symmetric]]:
:$(1): \quad \RR^{-1} \circ \RR$ is a [[Definition:Symmetric Relation|... | {{Rewrite|In light of the more relaxed definition of composite of relation}}
Note that this result holds for any $\RR \subseteq S \times T$, and does not require that $\struct {S, \RR}$ necessarily be a [[Definition:Relational Structure|relational structure]].
{{begin-eqn}}
{{eqn | l = \tuple {a, b}
| o = \in
... | Composition of Relation with Inverse is Symmetric | https://proofwiki.org/wiki/Composition_of_Relation_with_Inverse_is_Symmetric | https://proofwiki.org/wiki/Composition_of_Relation_with_Inverse_is_Symmetric | [
"Symmetric Relations",
"Composite Relations"
] | [
"Definition:Relation",
"Definition:Composition of Relations",
"Definition:Inverse Relation",
"Definition:Symmetric Relation",
"Definition:Symmetric Relation",
"Definition:Symmetric Relation"
] | [
"Definition:Relational Structure",
"Definition:Symmetric Relation",
"Inverse of Inverse Relation",
"Definition:Symmetric Relation",
"Domain of Composite Relation",
"Codomain of Composite Relation",
"Definition:Inverse Relation",
"Definition:Codomain (Set Theory)/Relation",
"Definition:Domain (Set Th... |
proofwiki-348 | Inverse of Many-to-One Relation is One-to-Many | The inverse of a many-to-one relation is a one-to-many relation, and vice versa. | From the definition, a many-to-one relation $\RR \subseteq S \times T$ is such that:
:$\RR \subseteq S \times T: \forall x \in \Dom \RR: \tuple {x, y_1} \in \RR \land \tuple {x, y_2} \in \RR \implies y_1 = y_2$
Also from the definition, a one-to-many relation $\RR \subseteq S \times T$ is such that:
:$\RR \subseteq S \... | The [[Definition:Inverse Relation|inverse]] of a [[Definition:Many-to-One Relation|many-to-one relation]] is a [[Definition:One-to-Many Relation|one-to-many relation]], and vice versa. | From the definition, a [[Definition:Many-to-One Relation|many-to-one relation]] $\RR \subseteq S \times T$ is such that:
:$\RR \subseteq S \times T: \forall x \in \Dom \RR: \tuple {x, y_1} \in \RR \land \tuple {x, y_2} \in \RR \implies y_1 = y_2$
Also from the definition, a [[Definition:One-to-Many Relation|one-to-m... | Inverse of Many-to-One Relation is One-to-Many | https://proofwiki.org/wiki/Inverse_of_Many-to-One_Relation_is_One-to-Many | https://proofwiki.org/wiki/Inverse_of_Many-to-One_Relation_is_One-to-Many | [
"Inverse Relations"
] | [
"Definition:Inverse Relation",
"Definition:Many-to-One Relation",
"Definition:One-to-Many Relation"
] | [
"Definition:Many-to-One Relation",
"Definition:One-to-Many Relation",
"Definition:Inverse Relation",
"Definition:Relation",
"Definition:Inverse Relation",
"Definition:Many-to-One Relation",
"Definition:One-to-Many Relation",
"Definition:One-to-Many Relation",
"Definition:Many-to-One Relation",
"Ca... |
proofwiki-349 | Many-to-One Relation Composite with Inverse is Transitive | Let $\RR \subseteq S \times T$ be a relation which is many-to-one.
Then the composites (both ways) of $\RR$ and its inverse $\RR^{-1}$, that is, both $\RR^{-1} \circ \RR$ and $\RR \circ \RR^{-1}$, are transitive. | Let $\RR \subseteq S \times T$ be many-to-one.
Then, from the definition of many-to-one:
:$\RR \subseteq S \times T: \forall x \in S: \tuple {x, y_1} \in \RR \land \tuple {x, y_2} \in \RR \implies y_1 = y_2$
Also, note that from Inverse of Many-to-One Relation is One-to-Many, $\RR^{-1}$ is one-to-many.
Let $\tuple {a, ... | Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] which is [[Definition:Many-to-One Relation|many-to-one]].
Then the [[Definition:Composition of Relations|composites]] (both ways) of $\RR$ and its [[Definition:Inverse Relation|inverse]] $\RR^{-1}$, that is, both $\RR^{-1} \circ \RR$ and $\RR \circ \... | Let $\RR \subseteq S \times T$ be [[Definition:Many-to-One Relation|many-to-one]].
Then, from the definition of [[Definition:Many-to-One Relation|many-to-one]]:
:$\RR \subseteq S \times T: \forall x \in S: \tuple {x, y_1} \in \RR \land \tuple {x, y_2} \in \RR \implies y_1 = y_2$
Also, note that from [[Inverse of Man... | Many-to-One Relation Composite with Inverse is Transitive | https://proofwiki.org/wiki/Many-to-One_Relation_Composite_with_Inverse_is_Transitive | https://proofwiki.org/wiki/Many-to-One_Relation_Composite_with_Inverse_is_Transitive | [
"Composite Relations",
"Transitive Relations"
] | [
"Definition:Relation",
"Definition:Many-to-One Relation",
"Definition:Composition of Relations",
"Definition:Inverse Relation",
"Definition:Transitive Relation"
] | [
"Definition:Many-to-One Relation",
"Definition:Many-to-One Relation",
"Inverse of Many-to-One Relation is One-to-Many",
"Definition:One-to-Many Relation",
"Definition:Many-to-One Relation",
"Definition:Transitive Relation",
"Definition:Many-to-One Relation",
"Definition:Many-to-One Relation",
"Defin... |
proofwiki-350 | One-to-Many Relation Composite with Inverse is Coreflexive | Let $\RR \subseteq S \times S$ be a relation which is one-to-many.
Then the composite of $\RR$ with its inverse is a coreflexive relation:
:$\RR^{-1} \circ \RR \subseteq \Delta_X$
{{refactor|This is a separate result.|level = basic}}
That is, by Relation is Symmetric and Antisymmetric iff Coreflexive, $\RR^{-1} \circ \... | As $\RR$ is one-to-many, it follows from Inverse of Many-to-One Relation is One-to-Many that $\RR^{-1}$ is many-to-one.
Let $\tuple {x, z} \in \RR^{-1} \circ \RR$.
Then:
:$\exists y \in S: \tuple {x, y} \in \RR, \tuple {y, z} \in \RR^{-1}$
As $\tuple {x, y} \in \RR$, from the definition of inverse relation:
:$\tuple {y... | Let $\RR \subseteq S \times S$ be a [[Definition:Relation|relation]] which is [[Definition:One-to-Many Relation|one-to-many]].
Then the [[Definition:Composition of Relations|composite]] of $\RR$ with its [[Definition:Inverse Relation|inverse]] is a [[Definition:Coreflexive Relation|coreflexive relation]]:
:$\RR^{-1} \... | As $\RR$ is [[Definition:One-to-Many Relation|one-to-many]], it follows from [[Inverse of Many-to-One Relation is One-to-Many]] that $\RR^{-1}$ is [[Definition:Many-to-One Relation|many-to-one]].
Let $\tuple {x, z} \in \RR^{-1} \circ \RR$.
Then:
:$\exists y \in S: \tuple {x, y} \in \RR, \tuple {y, z} \in \RR^{-1}$
A... | One-to-Many Relation Composite with Inverse is Coreflexive | https://proofwiki.org/wiki/One-to-Many_Relation_Composite_with_Inverse_is_Coreflexive | https://proofwiki.org/wiki/One-to-Many_Relation_Composite_with_Inverse_is_Coreflexive | [
"Coreflexive Relations"
] | [
"Definition:Relation",
"Definition:One-to-Many Relation",
"Definition:Composition of Relations",
"Definition:Inverse Relation",
"Definition:Coreflexive Relation",
"Relation is Symmetric and Antisymmetric iff Coreflexive",
"Definition:Symmetric Relation",
"Definition:Antisymmetric Relation"
] | [
"Definition:One-to-Many Relation",
"Inverse of Many-to-One Relation is One-to-Many",
"Definition:Many-to-One Relation",
"Definition:Inverse Relation",
"Definition:Many-to-One Relation",
"Definition:Coreflexive Relation",
"Category:Coreflexive Relations"
] |
proofwiki-351 | Element in its own Equivalence Class | Let $\RR$ be an equivalence relation on a set $S$.
Then every element of $S$ is in its own $\RR$-class:
:$\forall x \in S: x \in \eqclass x \RR$ | {{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \tuple {x, x}
| o = \in
| r = \RR
| c = {{Defof|Equivalence Relation}}: $\RR$ is Reflexive
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \eqclass x \RR
| c = {{Defof|Equivalence Class}}
}}
{{end-eqn}}
{{qed}} | Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on a [[Definition:Set|set]] $S$.
Then every [[Definition:Element|element]] of $S$ is in its own [[Definition:Equivalence Class|$\RR$-class]]:
:$\forall x \in S: x \in \eqclass x \RR$ | {{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \tuple {x, x}
| o = \in
| r = \RR
| c = {{Defof|Equivalence Relation}}: $\RR$ is [[Definition:Reflexive Relation|Reflexive]]
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \eqclass x \RR
| c = {{Defof|Equivalence Class}}
}}... | Element in its own Equivalence Class | https://proofwiki.org/wiki/Element_in_its_own_Equivalence_Class | https://proofwiki.org/wiki/Element_in_its_own_Equivalence_Class | [
"Equivalence Classes",
"Fundamental Theorem on Equivalence Relations"
] | [
"Definition:Equivalence Relation",
"Definition:Set",
"Definition:Element",
"Definition:Equivalence Class"
] | [
"Definition:Reflexive Relation"
] |
proofwiki-352 | Equivalence Class of Element is Subset | Let $\RR$ be an equivalence relation on a set $S$.
The $\RR$-class of every element of $S$ is a subset of the set the element is in:
:$\forall x \in S: \eqclass x \RR \subseteq S$ | {{begin-eqn}}
{{eqn | l = y
| o = \in
| r = \eqclass x \RR
| c =
}}
{{eqn | ll= \leadsto
| l = \tuple {x, y}
| o = \in
| r = \RR
| c = {{Defof|Equivalence Class}}
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = S \land y \in S
| c = {{Defof|Relation}}
}... | Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on a [[Definition:Set|set]] $S$.
The [[Definition:Equivalence Class|$\RR$-class]] of every [[Definition:Element|element]] of $S$ is a [[Definition:Subset|subset]] of the [[Definition:Set|set]] the [[Definition:Element|element]] is in:
:$\forall x... | {{begin-eqn}}
{{eqn | l = y
| o = \in
| r = \eqclass x \RR
| c =
}}
{{eqn | ll= \leadsto
| l = \tuple {x, y}
| o = \in
| r = \RR
| c = {{Defof|Equivalence Class}}
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = S \land y \in S
| c = {{Defof|Relation}}
}... | Equivalence Class of Element is Subset | https://proofwiki.org/wiki/Equivalence_Class_of_Element_is_Subset | https://proofwiki.org/wiki/Equivalence_Class_of_Element_is_Subset | [
"Equivalence Classes"
] | [
"Definition:Equivalence Relation",
"Definition:Set",
"Definition:Equivalence Class",
"Definition:Element",
"Definition:Subset",
"Definition:Set",
"Definition:Element"
] | [
"Category:Equivalence Classes"
] |
proofwiki-353 | Equivalence Class is not Empty | Let $\RR$ be an equivalence relation on a set $S$.
Then no $\RR$-class is empty. | {{begin-eqn}}
{{eqn | q = \forall \eqclass x \RR \subseteq S: \exists x \in S
| l = x
| o = \in
| r = \eqclass x \RR
| c = {{Defof|Equivalence Class}}
}}
{{eqn | ll= \leadsto
| l = \eqclass x \RR
| o = \ne
| r = \O
| c = {{Defof|Empty Set}}
}}
{{end-eqn}} | Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on a [[Definition:Set|set]] $S$.
Then no [[Definition:Equivalence Class|$\RR$-class]] is [[Definition:Empty Set|empty]]. | {{begin-eqn}}
{{eqn | q = \forall \eqclass x \RR \subseteq S: \exists x \in S
| l = x
| o = \in
| r = \eqclass x \RR
| c = {{Defof|Equivalence Class}}
}}
{{eqn | ll= \leadsto
| l = \eqclass x \RR
| o = \ne
| r = \O
| c = {{Defof|Empty Set}}
}}
{{end-eqn}} | Equivalence Class is not Empty | https://proofwiki.org/wiki/Equivalence_Class_is_not_Empty | https://proofwiki.org/wiki/Equivalence_Class_is_not_Empty | [
"Equivalence Classes",
"Empty Set",
"Fundamental Theorem on Equivalence Relations"
] | [
"Definition:Equivalence Relation",
"Definition:Set",
"Definition:Equivalence Class",
"Definition:Empty Set"
] | [] |
proofwiki-354 | Handshake Lemma | Let $G$ be a $\tuple {p, q}$-undirected graph, which may be a multigraph or a loop-graph, or both.
Let $V = \set {v_1, v_2, \ldots, v_p}$ be the vertex set of $G$.
Then:
:$\ds \sum_{i \mathop = 1}^p \map {\deg_G} {v_i} = 2 q$
where $\map {\deg_G} {v_i}$ is the degree of vertex $v_i$.
That is, the sum of all the degrees... | In the notation $\tuple {p, q}$-graph, $p$ is its order and $q$ its size.
That is, $p$ is the number of vertices in $G$, and $q$ is the number of edges in $G$.
Each edge is incident to exactly two vertices.
The degree of each vertex is defined as the number of edges to which it is incident.
So when we add up the degree... | Let $G$ be a $\tuple {p, q}$-[[Definition:Undirected Graph|undirected graph]], which may be a [[Definition:Multigraph|multigraph]] or a [[Definition:Loop-Graph|loop-graph]], or both.
Let $V = \set {v_1, v_2, \ldots, v_p}$ be the [[Definition:Vertex Set|vertex set]] of $G$.
Then:
:$\ds \sum_{i \mathop = 1}^p \map {\d... | In the [[Definition:Graph Notation|notation]] $\tuple {p, q}$-[[Definition:Graph (Graph Theory)|graph]], $p$ is its [[Definition:Order of Graph|order]] and $q$ its [[Definition:Size of Graph|size]].
That is, $p$ is the number of [[Definition:Vertex of Graph|vertices]] in $G$, and $q$ is the number of [[Definition:Edge... | Handshake Lemma | https://proofwiki.org/wiki/Handshake_Lemma | https://proofwiki.org/wiki/Handshake_Lemma | [
"Graph Theory",
"Vertices of Graphs",
"Direct Proofs",
"Named Theorems",
"Handshake Lemma"
] | [
"Definition:Undirected Graph",
"Definition:Multigraph",
"Definition:Loop-Graph",
"Definition:Vertex Set",
"Definition:Degree of Vertex",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Addition/Integers",
"Definition:Degree of Vertex",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Un... | [
"Definition:Graph (Graph Theory)/Notation",
"Definition:Graph (Graph Theory)",
"Definition:Graph (Graph Theory)/Order",
"Definition:Graph (Graph Theory)/Size",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Graph (Graph Theory)/Edge",
"Definition:Incident... |
proofwiki-355 | Equivalence Class holds Equivalent Elements | Let $\RR$ be an equivalence relation on a set $S$.
Then:
:$\tuple {x, y} \in \RR \iff \eqclass x \RR = \eqclass y \RR$ | === Necessary Condition ===
First we prove that $\tuple {x, y} \in \RR \implies \eqclass x \RR = \eqclass y \RR$.
Suppose:
: $\tuple {x, y} \in \RR: x, y \in S$
Then:
{{begin-eqn}}
{{eqn | l = z
| o = \in
| r = \eqclass x \RR
| c =
}}
{{eqn | ll= \leadsto
| l = \tuple {x, z}
| o = \in
... | Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on a [[Definition:Set|set]] $S$.
Then:
:$\tuple {x, y} \in \RR \iff \eqclass x \RR = \eqclass y \RR$ | === Necessary Condition ===
First we prove that $\tuple {x, y} \in \RR \implies \eqclass x \RR = \eqclass y \RR$.
Suppose:
: $\tuple {x, y} \in \RR: x, y \in S$
Then:
{{begin-eqn}}
{{eqn | l = z
| o = \in
| r = \eqclass x \RR
| c =
}}
{{eqn | ll= \leadsto
| l = \tuple {x, z}
| o = \in
... | Equivalence Class holds Equivalent Elements | https://proofwiki.org/wiki/Equivalence_Class_holds_Equivalent_Elements | https://proofwiki.org/wiki/Equivalence_Class_holds_Equivalent_Elements | [
"Equivalence Classes"
] | [
"Definition:Equivalence Relation",
"Definition:Set"
] | [
"Definition:Symmetric Relation",
"Definition:Transitive Relation",
"Definition:Symmetric Relation",
"Definition:Symmetric Relation",
"Definition:Set Equality",
"Definition:Equivalence Class",
"Definition:Equivalence Relation",
"Definition:Symmetric Relation",
"Definition:Biconditional"
] |
proofwiki-356 | Equivalence Classes are Disjoint | Let $\RR$ be an equivalence relation on a set $S$.
Then all $\RR$-classes are pairwise disjoint:
:$\tuple {x, y} \notin \RR \iff \eqclass x \RR \cap \eqclass y \RR = \O$ | First we show that:
:$\tuple {x, y} \notin \RR \implies \eqclass x \RR \cap \eqclass y \RR = \O$
Suppose two $\RR$-classes are ''not'' disjoint:
{{begin-eqn}}
{{eqn | l = \eqclass x \RR \cap \eqclass y \RR \ne \O
| o = \leadsto
| r = \exists z: z \in \eqclass x \RR \cap \eqclass y \RR
| c = {{Defof|Em... | Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on a [[Definition:Set|set]] $S$.
Then all [[Definition:Equivalence Class|$\RR$-classes]] are [[Definition:Pairwise Disjoint|pairwise disjoint]]:
:$\tuple {x, y} \notin \RR \iff \eqclass x \RR \cap \eqclass y \RR = \O$ | First we show that:
:$\tuple {x, y} \notin \RR \implies \eqclass x \RR \cap \eqclass y \RR = \O$
Suppose two $\RR$-classes are ''not'' [[Definition:Disjoint Sets|disjoint]]:
{{begin-eqn}}
{{eqn | l = \eqclass x \RR \cap \eqclass y \RR \ne \O
| o = \leadsto
| r = \exists z: z \in \eqclass x \RR \cap \eqcl... | Equivalence Classes are Disjoint/Proof 1 | https://proofwiki.org/wiki/Equivalence_Classes_are_Disjoint | https://proofwiki.org/wiki/Equivalence_Classes_are_Disjoint/Proof_1 | [
"Equivalence Classes",
"Equivalence Classes are Disjoint",
"Fundamental Theorem on Equivalence Relations"
] | [
"Definition:Equivalence Relation",
"Definition:Set",
"Definition:Equivalence Class",
"Definition:Pairwise Disjoint"
] | [
"Definition:Disjoint Sets",
"Rule of Transposition",
"Definition:By Hypothesis",
"Rule of Conjunction",
"Rule of Transposition",
"Biconditional Introduction"
] |
proofwiki-357 | Equivalence Classes are Disjoint | Let $\RR$ be an equivalence relation on a set $S$.
Then all $\RR$-classes are pairwise disjoint:
:$\tuple {x, y} \notin \RR \iff \eqclass x \RR \cap \eqclass y \RR = \O$ | Suppose that for $x, y \in S$:
:$\eqclass x \RR \cap \eqclass y \RR \ne \O$
Let:
:$z \in \eqclass x \RR$
:$z \in \eqclass y \RR$
Then by definition of equivalence class:
:$\tuple {x, z} \in \RR$
:$\tuple {y, z} \in \RR$
Let $c \in \eqclass x \RR$.
That is:
:$\tuple {x, c} \in \RR$
By definition of equivalence relation,... | Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on a [[Definition:Set|set]] $S$.
Then all [[Definition:Equivalence Class|$\RR$-classes]] are [[Definition:Pairwise Disjoint|pairwise disjoint]]:
:$\tuple {x, y} \notin \RR \iff \eqclass x \RR \cap \eqclass y \RR = \O$ | Suppose that for $x, y \in S$:
:$\eqclass x \RR \cap \eqclass y \RR \ne \O$
Let:
:$z \in \eqclass x \RR$
:$z \in \eqclass y \RR$
Then by definition of [[Definition:Equivalence Class|equivalence class]]:
:$\tuple {x, z} \in \RR$
:$\tuple {y, z} \in \RR$
Let $c \in \eqclass x \RR$.
That is:
:$\tuple {x, c} \in \RR$
... | Equivalence Classes are Disjoint/Proof 2 | https://proofwiki.org/wiki/Equivalence_Classes_are_Disjoint | https://proofwiki.org/wiki/Equivalence_Classes_are_Disjoint/Proof_2 | [
"Equivalence Classes",
"Equivalence Classes are Disjoint",
"Fundamental Theorem on Equivalence Relations"
] | [
"Definition:Equivalence Relation",
"Definition:Set",
"Definition:Equivalence Class",
"Definition:Pairwise Disjoint"
] | [
"Definition:Equivalence Class",
"Definition:Equivalence Relation",
"Definition:Symmetric Relation",
"Definition:Equivalence Relation",
"Definition:Transitive Relation",
"Definition:Subset",
"Definition:Equivalence Relation",
"Definition:Symmetric Relation",
"Definition:Equivalence Relation",
"Defi... |
proofwiki-358 | Fundamental Theorem on Equivalence Relations | Let $\RR \subseteq S \times S$ be an equivalence on a set $S$.
Then the quotient set $S / \RR$ of $S$ by $\RR$ forms a partition of $S$. | To prove that $S / \RR$ is a partition of $S$, we have to prove:
:$(1): \quad \ds \bigcup {S / \RR} = S$
:$(2): \quad \eqclass x \RR \ne \eqclass y \RR \iff \eqclass x \RR \cap \eqclass y \RR = \O$
:$(3): \quad \forall \eqclass x \RR \in S / \RR: \eqclass x \RR \ne \O$
Taking each proposition in turn: | Let $\RR \subseteq S \times S$ be an [[Definition:Equivalence Relation|equivalence]] on a [[Definition:Set|set]] $S$.
Then the [[Definition:Quotient Set|quotient set]] $S / \RR$ of $S$ by $\RR$ forms a [[Definition:Set Partition|partition]] of $S$. | To prove that $S / \RR$ is a [[Definition:Set Partition|partition]] of $S$, we have to prove:
:$(1): \quad \ds \bigcup {S / \RR} = S$
:$(2): \quad \eqclass x \RR \ne \eqclass y \RR \iff \eqclass x \RR \cap \eqclass y \RR = \O$
:$(3): \quad \forall \eqclass x \RR \in S / \RR: \eqclass x \RR \ne \O$
Taking each [[De... | Fundamental Theorem on Equivalence Relations | https://proofwiki.org/wiki/Fundamental_Theorem_on_Equivalence_Relations | https://proofwiki.org/wiki/Fundamental_Theorem_on_Equivalence_Relations | [
"Fundamental Theorem on Equivalence Relations",
"Equivalence Relations",
"Quotient Sets",
"Fundamental Theorems"
] | [
"Definition:Equivalence Relation",
"Definition:Set",
"Definition:Quotient Set",
"Definition:Set Partition"
] | [
"Definition:Set Partition",
"Definition:Proposition",
"Definition:Set Partition"
] |
proofwiki-359 | Equivalence Class is Unique | Let $\RR$ be an equivalence relation on $S$.
For each $x \in S$, the one and only one $\RR$-class to which $x$ belongs is $\eqclass x \RR$. | This follows directly from the Fundamental Theorem on Equivalence Relations: the set of $\RR$-classes forms a partition of $S$.
{{qed}} | Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on $S$.
For each $x \in S$, the [[Definition:Exactly One|one and only one]] [[Definition:Equivalence Class|$\RR$-class]] to which $x$ belongs is $\eqclass x \RR$. | This follows directly from the [[Fundamental Theorem on Equivalence Relations]]: the [[Definition:Quotient Set|set]] of $\RR$-classes forms a [[Definition:Set Partition|partition]] of $S$.
{{qed}} | Equivalence Class is Unique | https://proofwiki.org/wiki/Equivalence_Class_is_Unique | https://proofwiki.org/wiki/Equivalence_Class_is_Unique | [
"Equivalence Classes"
] | [
"Definition:Equivalence Relation",
"Definition:Unique",
"Definition:Equivalence Class"
] | [
"Fundamental Theorem on Equivalence Relations",
"Definition:Quotient Set",
"Definition:Set Partition"
] |
proofwiki-360 | Equivalence Class Equivalent Statements | Let $\RR$ be an equivalence relation on $S$.
Let $x, y \in S$.
{{TFAE}}
:$(1): \quad x$ and $y$ are in the same $\RR$-class
:$(2): \quad \eqclass x \RR = \eqclass y \RR$
:$(3): \quad x \mathrel \RR y$
:$(4): \quad x \in \eqclass y \RR$
:$(5): \quad y \in \eqclass x \RR$
:$(6): \quad \eqclass x \RR \cap \eqclass y \RR \... | === $(1)$ Equivalent to $(2)$ ===
{{:Equivalence Class Equivalent Statements/1 iff 2}} | Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on $S$.
Let $x, y \in S$.
{{TFAE}}
:$(1): \quad x$ and $y$ are in the same [[Definition:Equivalence Class|$\RR$-class]]
:$(2): \quad \eqclass x \RR = \eqclass y \RR$
:$(3): \quad x \mathrel \RR y$
:$(4): \quad x \in \eqclass y \RR$
:$(5): \quad... | === [[Equivalence Class Equivalent Statements/1 iff 2|$(1)$ Equivalent to $(2)$]] ===
{{:Equivalence Class Equivalent Statements/1 iff 2}} | Equivalence Class Equivalent Statements | https://proofwiki.org/wiki/Equivalence_Class_Equivalent_Statements | https://proofwiki.org/wiki/Equivalence_Class_Equivalent_Statements | [
"Equivalence Classes",
"Equivalence Class Equivalent Statements"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Class"
] | [
"Equivalence Class Equivalent Statements/1 iff 2"
] |
proofwiki-361 | Relation Induced by Partition is Equivalence | Let $\mathbb S$ be a partition of a set $S$.
Let $\RR$ be the relation induced by $\mathbb S$.
Then:
:$(1): \quad \RR$ is unique
:$(2): \quad \RR$ is an equivalence relation on $S$.
Hence $\mathbb S$ is the quotient set of $S$ by $\RR$, that is:
:$\mathbb S = S / \RR$ | From the definition of Relation Induced by Partition, we define the relation $\RR$ on $S$ by:
:$\RR = \set {\tuple {x, y}: \paren {\exists T \in \mathbb S: x \in T \land y \in T} }$ | Let $\mathbb S$ be a [[Definition:Set Partition|partition]] of a [[Definition:Set|set]] $S$.
Let $\RR$ be the [[Definition:Equivalence Relation Induced by Partition|relation induced by $\mathbb S$]].
Then:
:$(1): \quad \RR$ is [[Definition:Unique|unique]]
:$(2): \quad \RR$ is an [[Definition:Equivalence Relation|equ... | From the definition of [[Definition:Equivalence Relation Induced by Partition|Relation Induced by Partition]], we define the [[Definition:Relation|relation]] $\RR$ on $S$ by:
:$\RR = \set {\tuple {x, y}: \paren {\exists T \in \mathbb S: x \in T \land y \in T} }$ | Relation Induced by Partition is Equivalence | https://proofwiki.org/wiki/Relation_Induced_by_Partition_is_Equivalence | https://proofwiki.org/wiki/Relation_Induced_by_Partition_is_Equivalence | [
"Quotient Sets",
"Equivalence Relations"
] | [
"Definition:Set Partition",
"Definition:Set",
"Definition:Equivalence Relation Induced by Partition",
"Definition:Unique",
"Definition:Equivalence Relation",
"Definition:Quotient Set"
] | [
"Definition:Equivalence Relation Induced by Partition",
"Definition:Relation",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Relation"
] |
proofwiki-362 | Relation Partitions Set iff Equivalence | Let $\RR$ be a relation on a set $S$.
Then $S$ can be partitioned into subsets by $\RR$ {{iff}} $\RR$ is an equivalence relation on $S$.
The partition of $S$ defined by $\RR$ is the quotient set $S / \RR$. | Let $\RR$ be an equivalence relation on $S$.
From the Fundamental Theorem on Equivalence Relations, we have that the equivalence classes of $\RR$ form a partition.
{{qed|lemma}}
Let $S$ be partitioned into subsets by a relation $\RR$.
From Relation Induced by Partition is Equivalence, $\RR$ is an equivalence relation.
... | Let $\RR$ be a [[Definition:Relation|relation]] on a [[Definition:Set|set]] $S$.
Then $S$ can be [[Definition:Set Partition|partitioned]] into [[Definition:Subset|subsets]] by $\RR$ {{iff}} $\RR$ is an [[Definition:Equivalence Relation|equivalence relation]] on $S$.
The [[Definition:Set Partition|partition]] of $S$... | Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on $S$.
From the [[Fundamental Theorem on Equivalence Relations]], we have that the [[Definition:Equivalence Class|equivalence classes]] of $\RR$ form a [[Definition:Set Partition|partition]].
{{qed|lemma}}
Let $S$ be [[Definition:Set Partition... | Relation Partitions Set iff Equivalence/Proof | https://proofwiki.org/wiki/Relation_Partitions_Set_iff_Equivalence | https://proofwiki.org/wiki/Relation_Partitions_Set_iff_Equivalence/Proof | [
"Equivalence Relations",
"Quotient Sets",
"Set Partitions",
"Relation Partitions Set iff Equivalence"
] | [
"Definition:Relation",
"Definition:Set",
"Definition:Set Partition",
"Definition:Subset",
"Definition:Equivalence Relation",
"Definition:Set Partition",
"Definition:Quotient Set"
] | [
"Definition:Equivalence Relation",
"Fundamental Theorem on Equivalence Relations",
"Definition:Equivalence Class",
"Definition:Set Partition",
"Definition:Set Partition",
"Definition:Subset",
"Definition:Relation",
"Relation Induced by Partition is Equivalence",
"Definition:Equivalence Relation"
] |
proofwiki-363 | Intersection of Equivalences | The intersection of two equivalence relations is itself an equivalence relation. | Let $\RR_1$ and $\RR_2$ be equivalence relations on $S$.
Let $\RR_3 = \RR_1 \cap \RR_2$.
Checking in turn each of the criteria for equivalence: | The [[Definition:Set Intersection|intersection]] of two [[Definition:Equivalence Relation|equivalence relations]] is itself an [[Definition:Equivalence Relation|equivalence relation]]. | Let $\RR_1$ and $\RR_2$ be [[Definition:Equivalence Relation|equivalence relations]] on $S$.
Let $\RR_3 = \RR_1 \cap \RR_2$.
Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Intersection of Equivalences | https://proofwiki.org/wiki/Intersection_of_Equivalences | https://proofwiki.org/wiki/Intersection_of_Equivalences | [
"Equivalence Relations",
"Set Intersection"
] | [
"Definition:Set Intersection",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-364 | Union of Equivalences | The union of two equivalence relations is '''not''' necessarily an equivalence relation itself. | This can be shown by giving an example.
Let $S = \set {a, b, c}$, and let $\RR_1$ and $\RR_2$ be equivalences on $S$ such that:
: $\eqclass a {\RR_1} = \eqclass b {\RR_1} = \set {a, b}$
: $\eqclass c {\RR_1} = \set c$
: $\eqclass a {\RR_2} = \set a$
: $\eqclass b {\RR_2} = \eqclass c {\RR_2} = \set {b, c}$
Let $\RR_3 =... | The [[Definition:Set Union|union]] of two [[Definition:Equivalence Relation|equivalence relations]] is '''not''' necessarily an [[Definition:Equivalence Relation|equivalence relation]] itself. | This can be shown by giving an example.
Let $S = \set {a, b, c}$, and let $\RR_1$ and $\RR_2$ be [[Definition:Equivalence Relation|equivalences]] on $S$ such that:
: $\eqclass a {\RR_1} = \eqclass b {\RR_1} = \set {a, b}$
: $\eqclass c {\RR_1} = \set c$
: $\eqclass a {\RR_2} = \set a$
: $\eqclass b {\RR_2} = \eqclass... | Union of Equivalences/Proof 1 | https://proofwiki.org/wiki/Union_of_Equivalences | https://proofwiki.org/wiki/Union_of_Equivalences/Proof_1 | [
"Union",
"Set Union",
"Equivalence Relations",
"Set Union",
"Union of Equivalences"
] | [
"Definition:Set Union",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Transitive Relation",
"Definition:Equivalence Relation"
] |
proofwiki-365 | Union of Equivalences | The union of two equivalence relations is '''not''' necessarily an equivalence relation itself. | We have that the Union of Reflexive Relations is Reflexive.
We also have that the Union of Symmetric Relations is Symmetric.
However, we also have that the Union of Transitive Relations Not Always Transitive.
Hence the union of two equivalence relations is not always an equivalence relation.
{{qed}} | The [[Definition:Set Union|union]] of two [[Definition:Equivalence Relation|equivalence relations]] is '''not''' necessarily an [[Definition:Equivalence Relation|equivalence relation]] itself. | We have that the [[Union of Reflexive Relations is Reflexive]].
We also have that the [[Union of Symmetric Relations is Symmetric]].
However, we also have that the [[Union of Transitive Relations Not Always Transitive]].
Hence the union of two [[Definition:Equivalence Relation|equivalence relations]] is not always a... | Union of Equivalences/Proof 2 | https://proofwiki.org/wiki/Union_of_Equivalences | https://proofwiki.org/wiki/Union_of_Equivalences/Proof_2 | [
"Union",
"Set Union",
"Equivalence Relations",
"Set Union",
"Union of Equivalences"
] | [
"Definition:Set Union",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] | [
"Union of Reflexive Relations is Reflexive",
"Union of Symmetric Relations is Symmetric",
"Union of Transitive Relations Not Always Transitive",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-366 | Equivalence of Definitions of Equivalence Relation | Let $\RR$ be a relation on a set $S$.
{{TFAE|def = Equivalence Relation}} | === Definition 1 implies Definition 2 ===
Let $\RR$ be an equivalence relation by definition 1.
By definition, $\RR$ is reflexive, symmetric and transitive.
From Relation Contains Diagonal Relation iff Reflexive:
:$\Delta_S \subseteq \RR$
From Relation equals Inverse iff Symmetric:
:$\RR^{-1} = \RR$
and so by definitio... | Let $\RR$ be a [[Definition:Endorelation|relation]] on a [[Definition:Set|set]] $S$.
{{TFAE|def = Equivalence Relation}} | === Definition 1 implies Definition 2 ===
Let $\RR$ be an [[Definition:Equivalence Relation/Definition 1|equivalence relation by definition 1]].
By definition, $\RR$ is [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
From [[... | Equivalence of Definitions of Equivalence Relation | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Equivalence_Relation | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Equivalence_Relation | [
"Equivalence Relations"
] | [
"Definition:Endorelation",
"Definition:Set"
] | [
"Definition:Equivalence Relation/Definition 1",
"Definition:Reflexive Relation",
"Definition:Symmetric Relation",
"Definition:Transitive Relation",
"Equivalence of Definitions of Reflexive Relation",
"Equivalence of Definitions of Symmetric Relation",
"Definition:Set Equality/Definition 2",
"Equivalen... |
proofwiki-367 | Equivalence iff Diagonal and Inverse Composite | Let $\RR$ be a relation on $S$.
Then $\RR$ is an equivalence relation on $S$ {{iff}}:
:$\Delta_S \subseteq \RR$
and:
:$\RR = \RR \circ \RR^{-1}$ | === Necessary Condition ===
Let $\RR$ be an equivalence relation.
Then by definition, it is reflexive, symmetric and transitive.
As $\RR$ is reflexive, we have $\Delta_S \subseteq \RR$ from Relation Contains Diagonal Relation iff Reflexive.
As $\RR$ is transitive:
:$\RR \circ \RR \subseteq \RR$
from Relation contains C... | Let $\RR$ be a [[Definition:Relation|relation]] on $S$.
Then $\RR$ is an [[Definition:Equivalence Relation|equivalence relation]] on $S$ {{iff}}:
:$\Delta_S \subseteq \RR$
and:
:$\RR = \RR \circ \RR^{-1}$ | === Necessary Condition ===
Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]].
Then by definition, it is [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
As $\RR$ is [[Definition:Reflexive Relation|refl... | Equivalence iff Diagonal and Inverse Composite | https://proofwiki.org/wiki/Equivalence_iff_Diagonal_and_Inverse_Composite | https://proofwiki.org/wiki/Equivalence_iff_Diagonal_and_Inverse_Composite | [
"Equivalence Relations"
] | [
"Definition:Relation",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Reflexive Relation",
"Definition:Symmetric Relation",
"Definition:Transitive Relation",
"Definition:Reflexive Relation",
"Equivalence of Definitions of Reflexive Relation",
"Definition:Transitive Relation",
"Equivalence of Definitions of Transitive Relati... |
proofwiki-368 | Equality of Elements in Range of Mapping | Let $f: S \to T$ be a mapping.
Then:
:$\exists y \in \Rng f: \tuple {x_1, y} \in f \land \tuple {x_2, y} \in f \iff \map f {x_1} = \map f {x_2}$ | === Necessary Condition ===
Let:
:$\exists y \in \Rng f: \tuple {x_1, y} \in f \land \tuple {x_2, y} \in f$
Then:
{{begin-eqn}}
{{eqn | o =
| r = \tuple {x_1, y} \in f \land \tuple {x_2, y} \in f
| c =
}}
{{eqn | o = \leadsto
| r = \map f {x_1} = y \land \map f {x_2} = y
| c = {{Defof|Mapping}}
... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$\exists y \in \Rng f: \tuple {x_1, y} \in f \land \tuple {x_2, y} \in f \iff \map f {x_1} = \map f {x_2}$ | === Necessary Condition ===
Let:
:$\exists y \in \Rng f: \tuple {x_1, y} \in f \land \tuple {x_2, y} \in f$
Then:
{{begin-eqn}}
{{eqn | o =
| r = \tuple {x_1, y} \in f \land \tuple {x_2, y} \in f
| c =
}}
{{eqn | o = \leadsto
| r = \map f {x_1} = y \land \map f {x_2} = y
| c = {{Defof|Mapping... | Equality of Elements in Range of Mapping | https://proofwiki.org/wiki/Equality_of_Elements_in_Range_of_Mapping | https://proofwiki.org/wiki/Equality_of_Elements_in_Range_of_Mapping | [
"Mapping Theory"
] | [
"Definition:Mapping"
] | [
"Diagonal Relation is Equivalence"
] |
proofwiki-369 | Inverse of Mapping is One-to-Many Relation | Let $f$ be a mapping.
Then its inverse $f^{-1}$ is a one-to-many relation.
Hence $f^{-1}$ is not necessarily a mapping itself. | We have that $f$ is a mapping.
Hence $f$ is {{afortiori}} a many-to-one relation.
Then from Inverse of Many-to-One Relation is One-to-Many, $f^{-1}$ is one-to-many.
{{Qed}} | Let $f$ be a [[Definition:Mapping|mapping]].
Then its [[Definition:Inverse of Mapping|inverse]] $f^{-1}$ is a [[Definition:One-to-Many Relation|one-to-many relation]].
Hence $f^{-1}$ is not necessarily a [[Definition:Mapping|mapping]] itself. | We have that $f$ is a [[Definition:Mapping|mapping]].
Hence $f$ is {{afortiori}} a [[Definition:Many-to-One Relation|many-to-one relation]].
Then from [[Inverse of Many-to-One Relation is One-to-Many]], $f^{-1}$ is [[Definition:One-to-Many Relation|one-to-many]].
{{Qed}} | Inverse of Mapping is One-to-Many Relation | https://proofwiki.org/wiki/Inverse_of_Mapping_is_One-to-Many_Relation | https://proofwiki.org/wiki/Inverse_of_Mapping_is_One-to-Many_Relation | [
"Inverse Relations",
"Inverse Mappings"
] | [
"Definition:Mapping",
"Definition:Inverse of Mapping",
"Definition:One-to-Many Relation",
"Definition:Mapping"
] | [
"Definition:Mapping",
"Definition:Many-to-One Relation",
"Inverse of Many-to-One Relation is One-to-Many",
"Definition:One-to-Many Relation"
] |
proofwiki-370 | Equality of Mappings | Two mappings $f_1: S_1 \to T_1, f_2: S_2 \to T_2$ are equal {{iff}}:
:$(1): \quad S_1 = S_2$
:$(2): \quad T_1 = T_2$
:$(3): \quad \forall x \in S_1: \map {f_1} x = \map {f_2} x$ | This follows directly from Equality of Relations.
{{qed}} | Two [[Definition:Mapping|mappings]] $f_1: S_1 \to T_1, f_2: S_2 \to T_2$ are equal {{iff}}:
:$(1): \quad S_1 = S_2$
:$(2): \quad T_1 = T_2$
:$(3): \quad \forall x \in S_1: \map {f_1} x = \map {f_2} x$ | This follows directly from [[Equality of Relations]].
{{qed}} | Equality of Mappings | https://proofwiki.org/wiki/Equality_of_Mappings | https://proofwiki.org/wiki/Equality_of_Mappings | [
"Mapping Theory",
"Equality of Mappings"
] | [
"Definition:Mapping"
] | [
"Equality of Relations"
] |
proofwiki-371 | Mapping is Constant iff Image is Singleton | A mapping is a constant mapping {{iff}} its image is a singleton. | === Necessary Condition ===
Let $f_c: S \to T$ be a constant mapping.
Then from the definition of the image of an element:
:$\forall x \in S: \map {f_c} x = c \implies \Img S = \set c$
Thus the image of $f_c: S \to T$ is a singleton.
{{qed|lemma}} | A [[Definition:Mapping|mapping]] is a [[Definition:Constant Mapping|constant mapping]] {{iff}} its [[Definition:Image of Mapping|image]] is a [[Definition:Singleton|singleton]]. | === Necessary Condition ===
Let $f_c: S \to T$ be a [[Definition:Constant Mapping|constant mapping]].
Then from the definition of the [[Definition:Image of Element under Mapping|image of an element]]:
:$\forall x \in S: \map {f_c} x = c \implies \Img S = \set c$
Thus the [[Definition:Image of Mapping|image]] of $f_... | Mapping is Constant iff Image is Singleton | https://proofwiki.org/wiki/Mapping_is_Constant_iff_Image_is_Singleton | https://proofwiki.org/wiki/Mapping_is_Constant_iff_Image_is_Singleton | [
"Constant Mappings",
"Singletons"
] | [
"Definition:Mapping",
"Definition:Constant Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Singleton"
] | [
"Definition:Constant Mapping",
"Definition:Image (Set Theory)/Mapping/Element",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Singleton",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Singleton",
"Definition:Constant Mapping"
] |
proofwiki-372 | Diagonal Relation is Right Identity | Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Then:
:$\RR \circ \Delta_S = \RR$
where $\Delta_S$ is the diagonal relation on $S$, and $\circ$ signifies composition of relations. | {{Rewrite|In light of the more relaxed definition of composite of relation}}
We use the definition of relation equality, as follows: | Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] on $S \times T$.
Then:
:$\RR \circ \Delta_S = \RR$
where $\Delta_S$ is the [[Definition:Diagonal Relation|diagonal relation]] on $S$, and $\circ$ signifies [[Definition:Composition of Relations|composition of relations]]. | {{Rewrite|In light of the more relaxed definition of composite of relation}}
We use the [[Equality of Relations|definition of relation equality]], as follows: | Diagonal Relation is Right Identity | https://proofwiki.org/wiki/Diagonal_Relation_is_Right_Identity | https://proofwiki.org/wiki/Diagonal_Relation_is_Right_Identity | [
"Diagonal Relation",
"Identity Mappings"
] | [
"Definition:Relation",
"Definition:Diagonal Relation",
"Definition:Composition of Relations"
] | [
"Equality of Relations"
] |
proofwiki-373 | Identity Mapping is Left Identity | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
:$I_T \circ f = f$
where $I_T$ is the identity mapping on $T$, and $\circ$ signifies composition of mappings. | === Equality of Domains ===
The domains of $f$ and $I_T \circ f$ are equal from Domain of Composite Relation:
:$\Dom {I_T \circ f} = \Dom f$
{{qed|lemma}}
=== Equality of Codomains ===
The codomains of $f$ and $f \circ I_S$ are also easily shown to be equal.
From Codomain of Composite Relation, the codomains of $I_T \c... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$I_T \circ f = f$
where $I_T$ is the [[Definition:Identity Mapping|identity mapping]] on $T$, and $\circ$ signifies [[Definition:Composition of Mappings|composition of mappings]]. | === Equality of Domains ===
The [[Definition:Domain of Mapping|domains]] of $f$ and $I_T \circ f$ are equal from [[Domain of Composite Relation]]:
:$\Dom {I_T \circ f} = \Dom f$
{{qed|lemma}}
=== Equality of Codomains ===
The [[Definition:Codomain of Mapping|codomains]] of $f$ and $f \circ I_S$ are also easily sho... | Identity Mapping is Left Identity/Proof 1 | https://proofwiki.org/wiki/Identity_Mapping_is_Left_Identity | https://proofwiki.org/wiki/Identity_Mapping_is_Left_Identity/Proof_1 | [
"Identity Mappings",
"Identity Mapping is Left Identity"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Identity Mapping",
"Definition:Composition of Mappings"
] | [
"Definition:Domain (Set Theory)/Mapping",
"Domain of Composite Relation",
"Definition:Codomain (Set Theory)/Mapping",
"Codomain of Composite Relation",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Identity Mapping",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Composition of Mapp... |
proofwiki-374 | Identity Mapping is Left Identity | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
:$I_T \circ f = f$
where $I_T$ is the identity mapping on $T$, and $\circ$ signifies composition of mappings. | By definition, a mapping is also a relation.
Also by definition, the identity mapping is the same as the diagonal relation.
Thus Diagonal Relation is Left Identity can be applied directly.
{{qed}} | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$I_T \circ f = f$
where $I_T$ is the [[Definition:Identity Mapping|identity mapping]] on $T$, and $\circ$ signifies [[Definition:Composition of Mappings|composition of mappings]]. | By definition, a [[Definition:Mapping|mapping]] is also a [[Definition:Relation|relation]].
Also by definition, the [[Definition:Identity Mapping|identity mapping]] is the same as the [[Definition:Diagonal Relation|diagonal relation]].
Thus [[Diagonal Relation is Left Identity]] can be applied directly.
{{qed}} | Identity Mapping is Left Identity/Proof 2 | https://proofwiki.org/wiki/Identity_Mapping_is_Left_Identity | https://proofwiki.org/wiki/Identity_Mapping_is_Left_Identity/Proof_2 | [
"Identity Mappings",
"Identity Mapping is Left Identity"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Identity Mapping",
"Definition:Composition of Mappings"
] | [
"Definition:Mapping",
"Definition:Relation",
"Definition:Identity Mapping",
"Definition:Diagonal Relation",
"Diagonal Relation is Left Identity"
] |
proofwiki-375 | Image of Element under Inverse Mapping | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping such that its inverse $f^{-1}: T \to S$ is also a mapping.
Then:
:$\forall x \in S, y \in T: \map f x = y \iff \map {f^{-1} } y = x$ | === Sufficient Condition ===
Let $f: S \to T$ be a mapping.
From the definition of inverse mapping:
:$f^{-1} = \set {\tuple {y, x}: \tuple {x, y} \in f}$
Let $y = \map f x$.
From the definition of the preimage of an element:
:$\map {f^{-1} } y = \set {s \in S: \tuple {y, x} \in f}$
Thus:
:$x \in \map {f^{-1} } y$
Howev... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]] such that its [[Definition:Inverse of Mapping|inverse]] $f^{-1}: T \to S$ is also a [[Definition:Mapping|mapping]].
Then:
:$\forall x \in S, y \in T: \map f x = y \iff \map {f^{-1} } y = x$ | === Sufficient Condition ===
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
From the definition of [[Definition:Inverse of Mapping|inverse mapping]]:
:$f^{-1} = \set {\tuple {y, x}: \tuple {x, y} \in f}$
Let $y = \map f x$.
From the definition of the [[Definition:Preimage of Element under Mapping|preimage o... | Image of Element under Inverse Mapping | https://proofwiki.org/wiki/Image_of_Element_under_Inverse_Mapping | https://proofwiki.org/wiki/Image_of_Element_under_Inverse_Mapping | [
"Image of Element under Inverse Mapping",
"Inverse Mappings"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Inverse of Mapping",
"Definition:Mapping"
] | [
"Definition:Mapping",
"Definition:Inverse of Mapping",
"Definition:Preimage/Mapping/Element",
"Definition:Mapping"
] |
proofwiki-376 | Injection iff Left Cancellable | A mapping $f$ is an injection {{iff}} $f$ is left cancellable. | From the definition: a mapping $f: Y \to Z$ is left cancellable {{iff}}:
:$\forall X: \forall g_1: X \to Y, g_2: X \to Y: f \circ g_1 = f \circ g_2 \implies g_1 = g_2$ | A [[Definition:Mapping|mapping]] $f$ is an [[Definition:Injection|injection]] {{iff}} $f$ is [[Definition:Left Cancellable Mapping|left cancellable]]. | From the definition: a [[Definition:Mapping|mapping]] $f: Y \to Z$ is [[Definition:Left Cancellable Mapping|left cancellable]] {{iff}}:
:$\forall X: \forall g_1: X \to Y, g_2: X \to Y: f \circ g_1 = f \circ g_2 \implies g_1 = g_2$ | Injection iff Left Cancellable | https://proofwiki.org/wiki/Injection_iff_Left_Cancellable | https://proofwiki.org/wiki/Injection_iff_Left_Cancellable | [
"Injections",
"Cancellability",
"Injection iff Left Cancellable"
] | [
"Definition:Mapping",
"Definition:Injection",
"Definition:Left Cancellable Mapping"
] | [
"Definition:Mapping",
"Definition:Left Cancellable Mapping"
] |
proofwiki-377 | Identity Mapping is Injection | On any set $S$, the identity mapping $I_S: S \to S$ is an injection. | From the definition of the identity mapping:
:$\forall x \in S: \map {I_S} x = x$
So:
:$\map {I_S} x = \map {I_S} y \implies x = y$
So by definition $I_S$ is an injection.
{{Qed}} | On any [[Definition:Set|set]] $S$, the [[Definition:Identity Mapping|identity mapping]] $I_S: S \to S$ is an [[Definition:Injection|injection]]. | From the definition of the [[Definition:Identity Mapping|identity mapping]]:
:$\forall x \in S: \map {I_S} x = x$
So:
:$\map {I_S} x = \map {I_S} y \implies x = y$
So by definition $I_S$ is an [[Definition:Injection|injection]].
{{Qed}} | Identity Mapping is Injection | https://proofwiki.org/wiki/Identity_Mapping_is_Injection | https://proofwiki.org/wiki/Identity_Mapping_is_Injection | [
"Injections",
"Identity Mappings"
] | [
"Definition:Set",
"Definition:Identity Mapping",
"Definition:Injection"
] | [
"Definition:Identity Mapping",
"Definition:Injection"
] |
proofwiki-378 | Composite of Injections is Injection | A composite of injections is an injection.
That is:
:If $f$ and $g$ are injections, then so is $f \circ g$. | Let $f$ and $g$ be injections.
Then:
{{begin-eqn}}
{{eqn | l = \map {f \circ g} x
| r = \map {f \circ g} y
| c =
}}
{{eqn | ll= \leadsto
| l = \map f {\map g x}
| r = \map f {\map g y}
| c = {{Defof|Composition of Mappings}}
}}
{{eqn | ll= \leadsto
| l = \map g x
| r = \map g ... | A [[Definition:Composition of Mappings|composite]] of [[Definition:Injection|injections]] is an [[Definition:Injection|injection]].
That is:
:If $f$ and $g$ are [[Definition:Injection|injections]], then so is $f \circ g$. | Let $f$ and $g$ be [[Definition:Injection|injections]].
Then:
{{begin-eqn}}
{{eqn | l = \map {f \circ g} x
| r = \map {f \circ g} y
| c =
}}
{{eqn | ll= \leadsto
| l = \map f {\map g x}
| r = \map f {\map g y}
| c = {{Defof|Composition of Mappings}}
}}
{{eqn | ll= \leadsto
| l = \... | Composite of Injections is Injection | https://proofwiki.org/wiki/Composite_of_Injections_is_Injection | https://proofwiki.org/wiki/Composite_of_Injections_is_Injection | [
"Injections",
"Composite Mappings"
] | [
"Definition:Composition of Mappings",
"Definition:Injection",
"Definition:Injection",
"Definition:Injection"
] | [
"Definition:Injection",
"Definition:Injection",
"Definition:Injection"
] |
proofwiki-379 | Injection if Composite is Injection | Let $f$ and $g$ be mappings such that their composite $g \circ f$ is an injection.
Then $f$ is an injection. | Let $g \circ f$ be injective.
We need to show that $\map f a = \map f b \implies a = b$.
So suppose $\map f a = \map f b$.
Then:
{{begin-eqn}}
{{eqn | l = \map {g \circ f} a
| r = \map g {\map f a}
| c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \map g {\map f b}
| c = by hypothesis
}}
{{eqn | ... | Let $f$ and $g$ be [[Definition:Mapping|mappings]] such that their [[Definition:Composition of Mappings|composite]] $g \circ f$ is an [[Definition:Injection|injection]].
Then $f$ is an [[Definition:Injection|injection]]. | Let $g \circ f$ be [[Definition:Injection|injective]].
We need to show that $\map f a = \map f b \implies a = b$.
So suppose $\map f a = \map f b$.
Then:
{{begin-eqn}}
{{eqn | l = \map {g \circ f} a
| r = \map g {\map f a}
| c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \map g {\map f b}
| ... | Injection if Composite is Injection | https://proofwiki.org/wiki/Injection_if_Composite_is_Injection | https://proofwiki.org/wiki/Injection_if_Composite_is_Injection | [
"Injections",
"Composite Mappings"
] | [
"Definition:Mapping",
"Definition:Composition of Mappings",
"Definition:Injection",
"Definition:Injection"
] | [
"Definition:Injection",
"Definition:By Hypothesis",
"Definition:Injection",
"Definition:Injection"
] |
proofwiki-380 | Injection iff Left Inverse | A mapping $f: S \to T, S \ne \O$ is an injection {{iff}}:
:$\exists g: T \to S: g \circ f = I_S$
where $g$ is a mapping.
That is, {{iff}} $f$ has a left inverse. | Let:
:$\exists g: T \to S: g \circ f = I_S$
From Identity Mapping is Injection, $I_S$ is injective, so $g \circ f$ is injective.
So from Injection if Composite is Injection, $f$ is an injection.
Note that the existence of such a $g$ ''requires'' that $S \ne \O$.
{{qed|lemma}}
Now, assume $f$ is an injection.
We now def... | A [[Definition:Mapping|mapping]] $f: S \to T, S \ne \O$ is an [[Definition:Injection|injection]] {{iff}}:
:$\exists g: T \to S: g \circ f = I_S$
where $g$ is a [[Definition:Mapping|mapping]].
That is, {{iff}} $f$ has a [[Definition:Left Inverse Mapping|left inverse]]. | Let:
:$\exists g: T \to S: g \circ f = I_S$
From [[Identity Mapping is Injection]], $I_S$ is [[Definition:Injection|injective]], so $g \circ f$ is [[Definition:Injection|injective]].
So from [[Injection if Composite is Injection]], $f$ is an [[Definition:Injection|injection]].
Note that the existence of such a $g$ '... | Injection iff Left Inverse/Proof 1 | https://proofwiki.org/wiki/Injection_iff_Left_Inverse | https://proofwiki.org/wiki/Injection_iff_Left_Inverse/Proof_1 | [
"Injections",
"Left Inverse Mappings",
"Injection iff Left Inverse"
] | [
"Definition:Mapping",
"Definition:Injection",
"Definition:Mapping",
"Definition:Left Inverse Mapping"
] | [
"Identity Mapping is Injection",
"Definition:Injection",
"Definition:Injection",
"Injection if Composite is Injection",
"Definition:Injection",
"Definition:Injection",
"Definition:Mapping",
"Definition:Injection/Definition 3",
"Definition:Mapping",
"File:InjectionIffLeftInverse.png",
"Definition... |
proofwiki-381 | Injection iff Left Inverse | A mapping $f: S \to T, S \ne \O$ is an injection {{iff}}:
:$\exists g: T \to S: g \circ f = I_S$
where $g$ is a mapping.
That is, {{iff}} $f$ has a left inverse. | Take the result Condition for Composite Mapping on Left:
Let $A, B, C$ be sets.
Let $f: A \to B$ and $g: A \to C$ be mappings.
Then:
:$\forall x, y \in A: \map f x = \map f y \implies \map g x = \map g y$
{{iff}}:
:$\exists h: B \to C$ such that $h$ is a mapping and $h \circ f = g$.
Let $C = A = S$, let $B = T$ and let... | A [[Definition:Mapping|mapping]] $f: S \to T, S \ne \O$ is an [[Definition:Injection|injection]] {{iff}}:
:$\exists g: T \to S: g \circ f = I_S$
where $g$ is a [[Definition:Mapping|mapping]].
That is, {{iff}} $f$ has a [[Definition:Left Inverse Mapping|left inverse]]. | Take the result [[Condition for Composite Mapping on Left]]:
Let $A, B, C$ be [[Definition:Set|sets]].
Let $f: A \to B$ and $g: A \to C$ be [[Definition:Mapping|mappings]].
Then:
:$\forall x, y \in A: \map f x = \map f y \implies \map g x = \map g y$
{{iff}}:
:$\exists h: B \to C$ such that $h$ is a [[Definition:Map... | Injection iff Left Inverse/Proof 2 | https://proofwiki.org/wiki/Injection_iff_Left_Inverse | https://proofwiki.org/wiki/Injection_iff_Left_Inverse/Proof_2 | [
"Injections",
"Left Inverse Mappings",
"Injection iff Left Inverse"
] | [
"Definition:Mapping",
"Definition:Injection",
"Definition:Mapping",
"Definition:Left Inverse Mapping"
] | [
"Condition for Composite Mapping on Left",
"Definition:Set",
"Definition:Mapping",
"Definition:Mapping",
"Definition:Mapping",
"Definition:Identity Mapping",
"Definition:Mapping"
] |
proofwiki-382 | Injection iff Left Inverse | A mapping $f: S \to T, S \ne \O$ is an injection {{iff}}:
:$\exists g: T \to S: g \circ f = I_S$
where $g$ is a mapping.
That is, {{iff}} $f$ has a left inverse. | Let $f: S \to T$ be an injection.
Then $f$ is a one-to-many relation.
By Inverse of Many-to-One Relation is One-to-Many, $f^{-1}: T \to S$ is many-to-one.
By Many-to-One Relation Extends to Mapping, there is a mapping $g: T \to S$ such that:
:$f^{-1} \subseteq g$
Let $\tuple {x, y} \in g \circ f$.
Then:
:$\exists z \in... | A [[Definition:Mapping|mapping]] $f: S \to T, S \ne \O$ is an [[Definition:Injection|injection]] {{iff}}:
:$\exists g: T \to S: g \circ f = I_S$
where $g$ is a [[Definition:Mapping|mapping]].
That is, {{iff}} $f$ has a [[Definition:Left Inverse Mapping|left inverse]]. | Let $f: S \to T$ be an [[Definition:Injection|injection]].
Then $f$ is a [[Definition:One-to-Many Relation|one-to-many relation]].
By [[Inverse of Many-to-One Relation is One-to-Many]], $f^{-1}: T \to S$ is [[Definition:Many-to-One Relation|many-to-one]].
By [[Many-to-One Relation Extends to Mapping]], there is a [[... | Injection iff Left Inverse/Proof 3 | https://proofwiki.org/wiki/Injection_iff_Left_Inverse | https://proofwiki.org/wiki/Injection_iff_Left_Inverse/Proof_3 | [
"Injections",
"Left Inverse Mappings",
"Injection iff Left Inverse"
] | [
"Definition:Mapping",
"Definition:Injection",
"Definition:Mapping",
"Definition:Left Inverse Mapping"
] | [
"Definition:Injection",
"Definition:One-to-Many Relation",
"Inverse of Many-to-One Relation is One-to-Many",
"Definition:Many-to-One Relation",
"Many-to-One Relation Extends to Mapping",
"Definition:Mapping",
"Definition:Mapping",
"Definition:Set Equality/Definition 2"
] |
proofwiki-383 | Inclusion Mapping is Injection | Let $S, T$ be sets such that $S$ is a subset of $T$.
Then the inclusion mapping $i_S: S \to T$ defined as:
:$\forall x \in S: \map {i_S} x = x$
is an injection.
For this reason the inclusion mapping can be known as the '''canonical injection of $S$ to $T$'''. | Suppose $\map {i_S} {s_1} = \map {i_S} {s_2}$.
{{begin-eqn}}
{{eqn | l = \map {i_S} {s_1}
| r = s_1
| c = {{Defof|Inclusion Mapping}}
}}
{{eqn | l = \map {i_S} {s_2}
| r = s_2
| c = {{Defof|Inclusion Mapping}}
}}
{{eqn | l = \map {i_S} {s_1}
| r = \map {i_S} {s_2}
| c = by definition... | Let $S, T$ be [[Definition:Set|sets]] such that $S$ is a [[Definition:Subset|subset]] of $T$.
Then the [[Definition:Inclusion Mapping|inclusion mapping]] $i_S: S \to T$ defined as:
:$\forall x \in S: \map {i_S} x = x$
is an [[Definition:Injection|injection]].
For this reason the [[Definition:Inclusion Mapping|inclu... | Suppose $\map {i_S} {s_1} = \map {i_S} {s_2}$.
{{begin-eqn}}
{{eqn | l = \map {i_S} {s_1}
| r = s_1
| c = {{Defof|Inclusion Mapping}}
}}
{{eqn | l = \map {i_S} {s_2}
| r = s_2
| c = {{Defof|Inclusion Mapping}}
}}
{{eqn | l = \map {i_S} {s_1}
| r = \map {i_S} {s_2}
| c = by definitio... | Inclusion Mapping is Injection | https://proofwiki.org/wiki/Inclusion_Mapping_is_Injection | https://proofwiki.org/wiki/Inclusion_Mapping_is_Injection | [
"Injections",
"Inclusion Mappings"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Inclusion Mapping",
"Definition:Injection",
"Definition:Inclusion Mapping"
] | [
"Definition:Injection"
] |
proofwiki-384 | Surjection iff Right Cancellable | Let $f$ be a mapping.
Then $f$ is a surjection {{iff}} $f$ is right cancellable. | Let $f: X \to Y$ be surjective.
Let $h_1: Y \to Z, h_2: Y \to Z: h_1 \circ f = h_2 \circ f$.
As $f$ is a surjection:
:$\Img f = Y$
by definition.
But in order for $h_1 \circ f$ to be defined, it is necessary that $Y = \Dom {h_1}$.
Similarly, for $h_2 \circ f$ to be defined, it is necessary that $Y = \Dom {h_2}$.
So it ... | Let $f$ be a [[Definition:Mapping|mapping]].
Then $f$ is a [[Definition:Surjection|surjection]] {{iff}} $f$ is [[Definition:Right Cancellable Mapping|right cancellable]]. | Let $f: X \to Y$ be [[Definition:Surjection|surjective]].
Let $h_1: Y \to Z, h_2: Y \to Z: h_1 \circ f = h_2 \circ f$.
As $f$ is a [[Definition:Surjection|surjection]]:
:$\Img f = Y$
by definition.
But in order for $h_1 \circ f$ to be [[Definition:Composition of Mappings|defined]], it is necessary that $Y = \Dom {h... | Surjection iff Right Cancellable/Necessary Condition/Proof 1 | https://proofwiki.org/wiki/Surjection_iff_Right_Cancellable | https://proofwiki.org/wiki/Surjection_iff_Right_Cancellable/Necessary_Condition/Proof_1 | [
"Surjection iff Right Cancellable",
"Surjections",
"Cancellability"
] | [
"Definition:Mapping",
"Definition:Surjection",
"Definition:Right Cancellable Mapping"
] | [
"Definition:Surjection",
"Definition:Surjection",
"Definition:Composition of Mappings",
"Definition:Composition of Mappings",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Codomain (Set Theory)/Mapping",
... |
proofwiki-385 | Surjection iff Right Cancellable | Let $f$ be a mapping.
Then $f$ is a surjection {{iff}} $f$ is right cancellable. | Let $f: X \to Y$ be surjective.
Then from Surjection iff Right Inverse:
:$\exists g: Y \to X: f \circ g = I_Y$
Suppose $h \circ f = k \circ f$ for two mappings $h: Y \to Z$ and $k: Y \to Z$.
Then:
{{begin-eqn}}
{{eqn | l = h
| r = h \circ I_Y
| c =
}}
{{eqn | r = h \circ \paren {f \circ g}
| c =
}}
... | Let $f$ be a [[Definition:Mapping|mapping]].
Then $f$ is a [[Definition:Surjection|surjection]] {{iff}} $f$ is [[Definition:Right Cancellable Mapping|right cancellable]]. | Let $f: X \to Y$ be [[Definition:Surjection|surjective]].
Then from [[Surjection iff Right Inverse]]:
:$\exists g: Y \to X: f \circ g = I_Y$
Suppose $h \circ f = k \circ f$ for two mappings $h: Y \to Z$ and $k: Y \to Z$.
Then:
{{begin-eqn}}
{{eqn | l = h
| r = h \circ I_Y
| c =
}}
{{eqn | r = h \circ \... | Surjection iff Right Cancellable/Necessary Condition/Proof 2 | https://proofwiki.org/wiki/Surjection_iff_Right_Cancellable | https://proofwiki.org/wiki/Surjection_iff_Right_Cancellable/Necessary_Condition/Proof_2 | [
"Surjection iff Right Cancellable",
"Surjections",
"Cancellability"
] | [
"Definition:Mapping",
"Definition:Surjection",
"Definition:Right Cancellable Mapping"
] | [
"Definition:Surjection",
"Surjection iff Right Inverse",
"Composition of Mappings is Associative",
"Composition of Mappings is Associative",
"Definition:Right Cancellable Mapping",
"Definition:Surjection",
"Definition:Right Cancellable Mapping"
] |
proofwiki-386 | Surjection iff Right Cancellable | Let $f$ be a mapping.
Then $f$ is a surjection {{iff}} $f$ is right cancellable. | Suppose $f$ is a mapping which is not surjective.
Then:
:$\exists y_1 \in Y: \neg \exists x \in X: \map f x = y_1$
Let $Z = \set {a, b}$.
Let $h_1$ and $h_2$ be defined as follows.
:$\map {h_1} y = a: y \in Y$
:<nowiki>$\map {h_2} y = \begin {cases}
a & : y \ne y_1 \\
b & : y = y_1
\end {cases}$</nowiki>
Thus we have ... | Let $f$ be a [[Definition:Mapping|mapping]].
Then $f$ is a [[Definition:Surjection|surjection]] {{iff}} $f$ is [[Definition:Right Cancellable Mapping|right cancellable]]. | Suppose $f$ is a [[Definition:Mapping|mapping]] which is not [[Definition:Surjection|surjective]].
Then:
:$\exists y_1 \in Y: \neg \exists x \in X: \map f x = y_1$
Let $Z = \set {a, b}$.
Let $h_1$ and $h_2$ be defined as follows.
:$\map {h_1} y = a: y \in Y$
:<nowiki>$\map {h_2} y = \begin {cases}
a & : y \ne y_... | Surjection iff Right Cancellable/Sufficient Condition/Proof 1 | https://proofwiki.org/wiki/Surjection_iff_Right_Cancellable | https://proofwiki.org/wiki/Surjection_iff_Right_Cancellable/Sufficient_Condition/Proof_1 | [
"Surjection iff Right Cancellable",
"Surjections",
"Cancellability"
] | [
"Definition:Mapping",
"Definition:Surjection",
"Definition:Right Cancellable Mapping"
] | [
"Definition:Mapping",
"Definition:Surjection",
"Definition:Right Cancellable Mapping",
"Rule of Transposition",
"Definition:Right Cancellable Mapping",
"Definition:Surjection"
] |
proofwiki-387 | Surjection iff Right Cancellable | Let $f$ be a mapping.
Then $f$ is a surjection {{iff}} $f$ is right cancellable. | Let $f: X \to Y$ be a right cancellable mapping.
Let $Y$ contain exactly one element.
Then by definition $Y$ is a singleton.
From Mapping to Singleton is Surjection it follows that $f$ is a surjection.
So let $Y$ contain at least two elements.
Call those two elements $a$ and $b$, and we note that $a \ne b$.
We define t... | Let $f$ be a [[Definition:Mapping|mapping]].
Then $f$ is a [[Definition:Surjection|surjection]] {{iff}} $f$ is [[Definition:Right Cancellable Mapping|right cancellable]]. | Let $f: X \to Y$ be a [[Definition:Right Cancellable Mapping|right cancellable mapping]].
Let $Y$ contain exactly one element.
Then by definition $Y$ is a [[Definition:Singleton|singleton]].
From [[Mapping to Singleton is Surjection]] it follows that $f$ is a [[Definition:Surjection|surjection]].
So let $Y$ contai... | Surjection iff Right Cancellable/Sufficient Condition/Proof 2 | https://proofwiki.org/wiki/Surjection_iff_Right_Cancellable | https://proofwiki.org/wiki/Surjection_iff_Right_Cancellable/Sufficient_Condition/Proof_2 | [
"Surjection iff Right Cancellable",
"Surjections",
"Cancellability"
] | [
"Definition:Mapping",
"Definition:Surjection",
"Definition:Right Cancellable Mapping"
] | [
"Definition:Right Cancellable Mapping",
"Definition:Singleton",
"Mapping to Singleton is Surjection",
"Definition:Surjection",
"Definition:Right Cancellable Mapping",
"Definition:Surjection"
] |
proofwiki-388 | Identity Mapping is Surjection | On any set $S$, the identity mapping $I_S: S \to S$ is a surjection. | The identity mapping is defined as:
:$\forall y \in S: \map {I_S} y = y$
Then we have:
{{begin-eqn}}
{{eqn | q = \forall y \in S: \exists x \in S
| l = x
| r = y
| c = that is, $y$ itself
}}
{{eqn | ll= \leadsto
| q = \forall y \in S: \exists x \in S
| l = \map {I_S} x
| r = y
... | On any set $S$, the [[Definition:Identity Mapping|identity mapping]] $I_S: S \to S$ is a [[Definition:Surjection|surjection]]. | The [[Definition:Identity Mapping|identity mapping]] is defined as:
:$\forall y \in S: \map {I_S} y = y$
Then we have:
{{begin-eqn}}
{{eqn | q = \forall y \in S: \exists x \in S
| l = x
| r = y
| c = that is, $y$ itself
}}
{{eqn | ll= \leadsto
| q = \forall y \in S: \exists x \in S
| l =... | Identity Mapping is Surjection | https://proofwiki.org/wiki/Identity_Mapping_is_Surjection | https://proofwiki.org/wiki/Identity_Mapping_is_Surjection | [
"Surjections",
"Identity Mappings"
] | [
"Definition:Identity Mapping",
"Definition:Surjection"
] | [
"Definition:Identity Mapping"
] |
proofwiki-389 | Restriction of Mapping to Image is Surjection | Let $f: S \to T$ be a mapping.
Let $g: S \to \Img f$ be the restriction of $f$ to $S \times \Img f$.
Then $g$ is a surjective restriction of $f$. | From Image is Subset of Codomain:
:$\Img g \subseteq T$
Furthermore, by definition of image, we have:
:$\forall s \in S: g \paren s \in \Img g$
Therefore, $g$ may be viewed as a mapping $g: S \to \Img g$.
Thus, by definition, $g$ is a surjection.
{{qed}} | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $g: S \to \Img f$ be the [[Definition:Restriction of Mapping|restriction]] of $f$ to $S \times \Img f$.
Then $g$ is a [[Definition:Surjective Restriction|surjective restriction]] of $f$. | From [[Image is Subset of Codomain]]:
:$\Img g \subseteq T$
Furthermore, by definition of [[Definition:Image of Mapping|image]], we have:
:$\forall s \in S: g \paren s \in \Img g$
Therefore, $g$ may be viewed as a [[Definition:Mapping|mapping]] $g: S \to \Img g$.
Thus, by definition, $g$ is a [[Definition:Surjecti... | Restriction of Mapping to Image is Surjection | https://proofwiki.org/wiki/Restriction_of_Mapping_to_Image_is_Surjection | https://proofwiki.org/wiki/Restriction_of_Mapping_to_Image_is_Surjection | [
"Restrictions",
"Surjections"
] | [
"Definition:Mapping",
"Definition:Restriction/Mapping",
"Definition:Surjective Restriction"
] | [
"Image is Subset of Codomain",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Mapping",
"Definition:Surjection/Definition 2"
] |
proofwiki-390 | Composite of Surjections is Surjection | A composite of surjections is a surjection.
That is:
:If $g$ and $f$ are surjections, then so is $g \circ f$. | Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be surjections.
Then:
{{begin-eqn}}
{{eqn | q = \forall z \in S_3: \exists y \in S_2
| l = \map g y
| r = z
| c = {{Defof|Surjection}}
}}
{{eqn | ll= \leadsto
| q = \exists x \in S_1
| l = \map f x
| r = y
| c = {{Defof|Surjection}}
}}
... | A [[Definition:Composition of Mappings|composite]] of [[Definition:Surjection|surjections]] is a [[Definition:Surjection|surjection]].
That is:
:If $g$ and $f$ are [[Definition:Surjection|surjections]], then so is $g \circ f$. | Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be [[Definition:Surjection|surjections]].
Then:
{{begin-eqn}}
{{eqn | q = \forall z \in S_3: \exists y \in S_2
| l = \map g y
| r = z
| c = {{Defof|Surjection}}
}}
{{eqn | ll= \leadsto
| q = \exists x \in S_1
| l = \map f x
| r = y
| ... | Composite of Surjections is Surjection | https://proofwiki.org/wiki/Composite_of_Surjections_is_Surjection | https://proofwiki.org/wiki/Composite_of_Surjections_is_Surjection | [
"Surjections",
"Composite Mappings"
] | [
"Definition:Composition of Mappings",
"Definition:Surjection",
"Definition:Surjection",
"Definition:Surjection"
] | [
"Definition:Surjection",
"Definition:Composition of Mappings",
"Definition:Surjection"
] |
proofwiki-391 | Surjection if Composite is Surjection | Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be mappings such that $g \circ f$ is a surjection.
Then $g$ is a surjection. | Let $g \circ f$ be surjective.
Fix $z \in S_3$.
Now find an $x \in S_1: \map {g \circ f} x = z$.
The surjectivity of $g \circ f$ guarantees this can be done.
Now find an $y \in S_2: f \paren x = y$.
$f$ is a mapping and therefore a left-total relation; which guarantees this too can be done.
It follows that:
{{begin-eq... | Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be [[Definition:Mapping|mappings]] such that $g \circ f$ is a [[Definition:Surjection|surjection]].
Then $g$ is a [[Definition:Surjection|surjection]]. | Let $g \circ f$ be [[Definition:Surjection|surjective]].
Fix $z \in S_3$.
Now find an $x \in S_1: \map {g \circ f} x = z$.
The [[Definition:Surjection|surjectivity]] of $g \circ f$ guarantees this can be done.
Now find an $y \in S_2: f \paren x = y$.
$f$ is a [[Definition:Mapping|mapping]] and therefore a [[Defin... | Surjection if Composite is Surjection | https://proofwiki.org/wiki/Surjection_if_Composite_is_Surjection | https://proofwiki.org/wiki/Surjection_if_Composite_is_Surjection | [
"Surjections",
"Composite Mappings"
] | [
"Definition:Mapping",
"Definition:Surjection",
"Definition:Surjection"
] | [
"Definition:Surjection",
"Definition:Surjection",
"Definition:Mapping",
"Definition:Left-Total Relation"
] |
proofwiki-392 | Surjection iff Right Inverse | A mapping $f: S \to T, S \ne \O$ is a surjection {{iff}}:
:$\exists g: T \to S: f \circ g = I_T$
where:
:$g$ is a mapping
:$I_T$ is the identity mapping on $T$.
That is, {{iff}} $f$ has a right inverse. | Assume $\exists g: T \to S: f \circ g = I_T$.
From Identity Mapping is Surjection, $I_T$ is surjective, so $f \circ g$ is surjective.
So from Surjection if Composite is Surjection, $f$ is a surjection.
Note that the existence of such a $g$ ''requires'' that $S$ is non-empty.
Now, assume $f$ is a surjection.
Consider th... | A [[Definition:Mapping|mapping]] $f: S \to T, S \ne \O$ is a [[Definition:Surjection|surjection]] {{iff}}:
:$\exists g: T \to S: f \circ g = I_T$
where:
:$g$ is a [[Definition:Mapping|mapping]]
:$I_T$ is the [[Definition:Identity Mapping|identity mapping]] on $T$.
That is, {{iff}} $f$ has a [[Definition:Right Inverse... | Assume $\exists g: T \to S: f \circ g = I_T$.
From [[Identity Mapping is Surjection]], $I_T$ is [[Definition:Surjection|surjective]], so $f \circ g$ is [[Definition:Surjection|surjective]].
So from [[Surjection if Composite is Surjection]], $f$ is a [[Definition:Surjection|surjection]].
Note that the existence of su... | Surjection iff Right Inverse/Proof 1 | https://proofwiki.org/wiki/Surjection_iff_Right_Inverse | https://proofwiki.org/wiki/Surjection_iff_Right_Inverse/Proof_1 | [
"Surjections",
"Right Inverse Mappings",
"Surjection iff Right Inverse"
] | [
"Definition:Mapping",
"Definition:Surjection",
"Definition:Mapping",
"Definition:Identity Mapping",
"Definition:Right Inverse Mapping"
] | [
"Identity Mapping is Surjection",
"Definition:Surjection",
"Definition:Surjection",
"Surjection if Composite is Surjection",
"Definition:Surjection",
"Definition:Non-Empty Set",
"Definition:Surjection",
"Definition:Indexing Set/Family of Sets",
"Definition:Non-Empty Set",
"Definition:Set",
"Defi... |
proofwiki-393 | Surjection iff Right Inverse | A mapping $f: S \to T, S \ne \O$ is a surjection {{iff}}:
:$\exists g: T \to S: f \circ g = I_T$
where:
:$g$ is a mapping
:$I_T$ is the identity mapping on $T$.
That is, {{iff}} $f$ has a right inverse. | Take the result Condition for Composite Mapping on Right:
Let $A, B, C$ be sets.
Let $f: B \to A$ and $g: C \to A$ be mappings.
Then:
:$\Img g \subseteq \Img f$
{{iff}}:
:$\exists h: C \to B$ such that $h$ is a mapping and $f \circ h = g$.
Let $C = A = T$, let $B = S$ and let $g = I_T$.
Then the above translates into:
... | A [[Definition:Mapping|mapping]] $f: S \to T, S \ne \O$ is a [[Definition:Surjection|surjection]] {{iff}}:
:$\exists g: T \to S: f \circ g = I_T$
where:
:$g$ is a [[Definition:Mapping|mapping]]
:$I_T$ is the [[Definition:Identity Mapping|identity mapping]] on $T$.
That is, {{iff}} $f$ has a [[Definition:Right Inverse... | Take the result [[Condition for Composite Mapping on Right]]:
Let $A, B, C$ be [[Definition:Set|sets]].
Let $f: B \to A$ and $g: C \to A$ be [[Definition:Mapping|mappings]].
Then:
:$\Img g \subseteq \Img f$
{{iff}}:
:$\exists h: C \to B$ such that $h$ is a [[Definition:Mapping|mapping]] and $f \circ h = g$.
Let $C... | Surjection iff Right Inverse/Proof 2 | https://proofwiki.org/wiki/Surjection_iff_Right_Inverse | https://proofwiki.org/wiki/Surjection_iff_Right_Inverse/Proof_2 | [
"Surjections",
"Right Inverse Mappings",
"Surjection iff Right Inverse"
] | [
"Definition:Mapping",
"Definition:Surjection",
"Definition:Mapping",
"Definition:Identity Mapping",
"Definition:Right Inverse Mapping"
] | [
"Condition for Composite Mapping on Right",
"Definition:Set",
"Definition:Mapping",
"Definition:Mapping",
"Definition:Mapping",
"Definition:Set Equality"
] |
proofwiki-394 | Identity Mapping is Bijection | The identity mapping $I_S: S \to S$ on the set $S$ is a bijection. | The identity mapping is:
:an injection, from Identity Mapping is Injection
:a surjection, from Identity Mapping is Surjection
and therefore a bijection.
{{qed}} | The [[Definition:Identity Mapping|identity mapping]] $I_S: S \to S$ on the [[Definition:Set|set]] $S$ is a [[Definition:Bijection|bijection]]. | The [[Definition:Identity Mapping|identity mapping]] is:
:an [[Definition:Injection|injection]], from [[Identity Mapping is Injection]]
:a [[Definition:Surjection|surjection]], from [[Identity Mapping is Surjection]]
and therefore a [[Definition:Bijection|bijection]].
{{qed}} | Identity Mapping is Bijection | https://proofwiki.org/wiki/Identity_Mapping_is_Bijection | https://proofwiki.org/wiki/Identity_Mapping_is_Bijection | [
"Bijections",
"Identity Mappings"
] | [
"Definition:Identity Mapping",
"Definition:Set",
"Definition:Bijection"
] | [
"Definition:Identity Mapping",
"Definition:Injection",
"Identity Mapping is Injection",
"Definition:Surjection",
"Identity Mapping is Surjection",
"Definition:Bijection"
] |
proofwiki-395 | Identity Mapping is Permutation | The identity mapping $I_S: S \to S$ on the set $S$ is a permutation. | The identity mapping $I_S$ is a bijection from $S$ to itself, by Identity Mapping is Bijection.
{{qed}} | The [[Definition:Identity Mapping|identity mapping]] $I_S: S \to S$ on the [[Definition:Set|set]] $S$ is a [[Definition:Permutation|permutation]]. | The identity mapping $I_S$ is a [[Definition:Bijection|bijection]] from $S$ to itself, by [[Identity Mapping is Bijection]].
{{qed}} | Identity Mapping is Permutation | https://proofwiki.org/wiki/Identity_Mapping_is_Permutation | https://proofwiki.org/wiki/Identity_Mapping_is_Permutation | [
"Identity Mappings"
] | [
"Definition:Identity Mapping",
"Definition:Set",
"Definition:Permutation"
] | [
"Definition:Bijection",
"Identity Mapping is Bijection"
] |
proofwiki-396 | Bijection iff Left and Right Inverse | Let $f: S \to T$ be a mapping.
$f$ is a bijection {{iff}}:
{{begin-eqn}}
{{eqn | n = 1
| q = \exists g_1: T \to S
| l = g_1 \circ f
| r = I_S
}}
{{eqn | n = 2
| q = \exists g_2: T \to S
| l = f \circ g_2
| r = I_T
}}
{{end-eqn}}
where:
:$g_1$ and $g_2$ are mappings
:$\circ$ denotes c... | === Necessary Condition ===
Let $f: S \to T$ be a mapping.
Let $f$ be such that:
{{begin-eqn}}
{{eqn | q = \exists g_1: T \to S
| l = g_1 \circ f
| r = I_S
}}
{{eqn | q = \exists g_2: T \to S
| l = f \circ g_2
| r = I_T
}}
{{end-eqn}}
where both $g_1$ and $g_2$ are mappings.
Then from Left and R... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
$f$ is a [[Definition:Bijection|bijection]] {{iff}}:
{{begin-eqn}}
{{eqn | n = 1
| q = \exists g_1: T \to S
| l = g_1 \circ f
| r = I_S
}}
{{eqn | n = 2
| q = \exists g_2: T \to S
| l = f \circ g_2
| r = I_T
}}
{{end-eqn}}
wher... | === Necessary Condition ===
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $f$ be such that:
{{begin-eqn}}
{{eqn | q = \exists g_1: T \to S
| l = g_1 \circ f
| r = I_S
}}
{{eqn | q = \exists g_2: T \to S
| l = f \circ g_2
| r = I_T
}}
{{end-eqn}}
where both $g_1$ and $g_2$ are [[D... | Bijection iff Left and Right Inverse | https://proofwiki.org/wiki/Bijection_iff_Left_and_Right_Inverse | https://proofwiki.org/wiki/Bijection_iff_Left_and_Right_Inverse | [
"Bijections",
"Inverse Mappings"
] | [
"Definition:Mapping",
"Definition:Bijection",
"Definition:Mapping",
"Definition:Composition of Mappings"
] | [
"Definition:Mapping",
"Definition:Mapping",
"Left and Right Inverse Mappings Implies Bijection",
"Definition:Bijection",
"Left and Right Inverses of Mapping are Inverse Mapping",
"Definition:Inverse Mapping",
"Definition:Bijection",
"Definition:Inverse Mapping"
] |
proofwiki-397 | Inverse of Bijection is Bijection | Let $f: S \to T$ be a bijection in the sense that:
:$(1): \quad f$ is an injection
:$(2): \quad f$ is a surjection.
Then the inverse $f^{-1}$ of $f$ is itself a bijection by the same definition. | Let $f$ be both an injection and a surjection.
From Mapping is Injection and Surjection iff Inverse is Mapping it follows that its inverse $f^{-1}$ is a mapping.
From Inverse of Inverse Relation:
:$\paren {f^{-1} }^{-1} = f$
Thus the inverse of $f^{-1}$ is $f$.
But then $f$, being a bijection, is by definition a mappin... | Let $f: S \to T$ be a [[Definition:Bijection/Definition 1|bijection]] in the sense that:
:$(1): \quad f$ is an [[Definition:Injection|injection]]
:$(2): \quad f$ is a [[Definition:Surjection|surjection]].
Then the [[Definition:Inverse of Mapping|inverse]] $f^{-1}$ of $f$ is itself a [[Definition:Bijection/Definition ... | Let $f$ be both an [[Definition:Injection|injection]] and a [[Definition:Surjection|surjection]].
From [[Mapping is Injection and Surjection iff Inverse is Mapping]] it follows that its [[Definition:Inverse of Mapping|inverse $f^{-1}$]] is a [[Definition:Mapping|mapping]].
From [[Inverse of Inverse Relation]]:
:$\pa... | Inverse of Bijection is Bijection | https://proofwiki.org/wiki/Inverse_of_Bijection_is_Bijection | https://proofwiki.org/wiki/Inverse_of_Bijection_is_Bijection | [
"Bijections",
"Inverse Mappings"
] | [
"Definition:Bijection/Definition 1",
"Definition:Injection",
"Definition:Surjection",
"Definition:Inverse of Mapping",
"Definition:Bijection/Definition 1"
] | [
"Definition:Injection",
"Definition:Surjection",
"Mapping is Injection and Surjection iff Inverse is Mapping",
"Definition:Inverse of Mapping",
"Definition:Mapping",
"Inverse of Inverse Relation",
"Definition:Inverse of Mapping",
"Definition:Bijection",
"Definition:Mapping",
"Mapping is Injection ... |
proofwiki-398 | Inverse Element of Bijection | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a bijection.
Then:
:$\map {f^{-1} } y = x \iff \map f x = y$
where $f^{-1}$ is the inverse mapping of $f$. | Suppose $f$ is a bijection.
Because $f^{-1}$ is a bijection from Bijection iff Inverse is Bijection, it is by definition a mapping.
The result follows directly from Image of Element under Inverse Mapping.
{{qed}} | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Bijection|bijection]].
Then:
:$\map {f^{-1} } y = x \iff \map f x = y$
where $f^{-1}$ is the [[Definition:Inverse Mapping|inverse mapping]] of $f$. | Suppose $f$ is a [[Definition:Bijection|bijection]].
Because $f^{-1}$ is a [[Definition:Bijection|bijection]] from [[Bijection iff Inverse is Bijection]], it is by definition a [[Definition:Mapping|mapping]].
The result follows directly from [[Image of Element under Inverse Mapping]].
{{qed}} | Inverse Element of Bijection | https://proofwiki.org/wiki/Inverse_Element_of_Bijection | https://proofwiki.org/wiki/Inverse_Element_of_Bijection | [
"Bijections",
"Inverse Mappings"
] | [
"Definition:Set",
"Definition:Bijection",
"Definition:Inverse Mapping"
] | [
"Definition:Bijection",
"Definition:Bijection",
"Inverse of Bijection is Bijection",
"Definition:Mapping",
"Image of Element under Inverse Mapping"
] |
proofwiki-399 | Composite of Bijection with Inverse is Identity Mapping | Let $f: S \to T$ be a bijection.
Then:
{{begin-eqn}}
{{eqn | l = f^{-1} \circ f
| r = I_S
}}
{{eqn | l = f \circ f^{-1}
| r = I_T
}}
{{end-eqn}}
where $I_S$ and $I_T$ are the identity mappings on $S$ and $T$ respectively. | Let $f: S \to T$ be a bijection.
From Inverse of Bijection is Bijection, $f^{-1}$ is also a bijection.
Let $x \in S$.
From Inverse Element of Bijection:
:$\exists y \in T: y = \map f x \implies x = \map {f^{-1} } y$
Then:
{{begin-eqn}}
{{eqn | l = \map {f^{-1} \circ f} x
| r = \map {f^{-1} } {\map f x}
| c ... | Let $f: S \to T$ be a [[Definition:Bijection|bijection]].
Then:
{{begin-eqn}}
{{eqn | l = f^{-1} \circ f
| r = I_S
}}
{{eqn | l = f \circ f^{-1}
| r = I_T
}}
{{end-eqn}}
where $I_S$ and $I_T$ are the [[Definition:Identity Mapping|identity mappings]] on $S$ and $T$ respectively. | Let $f: S \to T$ be a [[Definition:Bijection|bijection]].
From [[Inverse of Bijection is Bijection]], $f^{-1}$ is also a [[Definition:Bijection|bijection]].
Let $x \in S$.
From [[Inverse Element of Bijection]]:
:$\exists y \in T: y = \map f x \implies x = \map {f^{-1} } y$
Then:
{{begin-eqn}}
{{eqn | l = \map {... | Composite of Bijection with Inverse is Identity Mapping | https://proofwiki.org/wiki/Composite_of_Bijection_with_Inverse_is_Identity_Mapping | https://proofwiki.org/wiki/Composite_of_Bijection_with_Inverse_is_Identity_Mapping | [
"Bijections",
"Inverse Mappings",
"Identity Mappings"
] | [
"Definition:Bijection",
"Definition:Identity Mapping"
] | [
"Definition:Bijection",
"Inverse of Bijection is Bijection",
"Definition:Bijection",
"Inverse Element of Bijection",
"Domain of Composite Relation",
"Codomain of Composite Relation",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Identity Mapping",
... |
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