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-400 | Inverse of Inverse of Bijection | Let $f: S \to T$ be a bijection.
Then:
:$\paren {f^{-1} }^{-1} = f$
where $f^{-1}$ is the inverse of $f$. | Let $f: S \to T$ be a bijection.
From Composite of Bijection with Inverse is Identity Mapping we have:
:$f^{-1} \circ f = I_S$
:$f \circ f^{-1} = I_T$
where $I_S$ and $I_T$ are the identity mappings on $S$ and $T$ respectively.
The result follows from Left and Right Inverses of Mapping are Inverse Mapping.
{{qed}} | Let $f: S \to T$ be a [[Definition:Bijection|bijection]].
Then:
:$\paren {f^{-1} }^{-1} = f$
where $f^{-1}$ is the [[Definition:Inverse of Mapping|inverse of $f$]]. | Let $f: S \to T$ be a [[Definition:Bijection|bijection]].
From [[Composite of Bijection with Inverse is Identity Mapping]] we have:
:$f^{-1} \circ f = I_S$
:$f \circ f^{-1} = I_T$
where $I_S$ and $I_T$ are the [[Definition:Identity Mapping|identity mappings]] on $S$ and $T$ respectively.
The result follows from [[Le... | Inverse of Inverse of Bijection/Proof 1 | https://proofwiki.org/wiki/Inverse_of_Inverse_of_Bijection | https://proofwiki.org/wiki/Inverse_of_Inverse_of_Bijection/Proof_1 | [
"Bijections",
"Inverse Mappings",
"Inverse of Inverse of Bijection"
] | [
"Definition:Bijection",
"Definition:Inverse of Mapping"
] | [
"Definition:Bijection",
"Composite of Bijection with Inverse is Identity Mapping",
"Definition:Identity Mapping",
"Left and Right Inverses of Mapping are Inverse Mapping"
] |
proofwiki-401 | Inverse of Inverse of Bijection | Let $f: S \to T$ be a bijection.
Then:
:$\paren {f^{-1} }^{-1} = f$
where $f^{-1}$ is the inverse of $f$. | A mapping is a relation.
Thus it follows that Inverse of Inverse Relation can be applied directly.
{{qed}} | Let $f: S \to T$ be a [[Definition:Bijection|bijection]].
Then:
:$\paren {f^{-1} }^{-1} = f$
where $f^{-1}$ is the [[Definition:Inverse of Mapping|inverse of $f$]]. | A [[Definition:Mapping|mapping]] is a [[Definition:Relation|relation]].
Thus it follows that [[Inverse of Inverse Relation]] can be applied directly.
{{qed}} | Inverse of Inverse of Bijection/Proof 2 | https://proofwiki.org/wiki/Inverse_of_Inverse_of_Bijection | https://proofwiki.org/wiki/Inverse_of_Inverse_of_Bijection/Proof_2 | [
"Bijections",
"Inverse Mappings",
"Inverse of Inverse of Bijection"
] | [
"Definition:Bijection",
"Definition:Inverse of Mapping"
] | [
"Definition:Mapping",
"Definition:Relation",
"Inverse of Inverse Relation"
] |
proofwiki-402 | Composite of Bijections is Bijection | Let $f$ and $g$ be mappings such that $\Dom f = \Cdm g$.
Then:
:If $f$ and $g$ are both bijections, then so is $f \circ g$
where $f \circ g$ is the composite mapping of $f$ with $g$. | As every bijection is also by definition an injection, a composite of bijections is also a composite of injections.
Every composite of injections is also an injection by Composite of Injections is Injection.
As every bijection is also by definition a surjection, a composite of bijections is also a composite of surjecti... | Let $f$ and $g$ be [[Definition:Mapping|mappings]] such that $\Dom f = \Cdm g$.
Then:
:If $f$ and $g$ are both [[Definition:Bijection|bijections]], then so is $f \circ g$
where $f \circ g$ is the [[Definition:Composition of Mappings|composite mapping]] of $f$ with $g$. | As every [[Definition:Bijection|bijection]] is also by definition an [[Definition:Injection|injection]], a [[Definition:Composition of Mappings|composite]] of [[Definition:Bijection|bijections]] is also a [[Definition:Composition of Mappings|composite]] of [[Definition:Injection|injections]].
Every [[Definition:Compos... | Composite of Bijections is Bijection/Proof 1 | https://proofwiki.org/wiki/Composite_of_Bijections_is_Bijection | https://proofwiki.org/wiki/Composite_of_Bijections_is_Bijection/Proof_1 | [
"Bijections",
"Composite Mappings",
"Composite of Bijections is Bijection"
] | [
"Definition:Mapping",
"Definition:Bijection",
"Definition:Composition of Mappings"
] | [
"Definition:Bijection",
"Definition:Injection",
"Definition:Composition of Mappings",
"Definition:Bijection",
"Definition:Composition of Mappings",
"Definition:Injection",
"Definition:Composition of Mappings",
"Definition:Injection",
"Definition:Injection",
"Composite of Injections is Injection",
... |
proofwiki-403 | Composite of Bijections is Bijection | Let $f$ and $g$ be mappings such that $\Dom f = \Cdm g$.
Then:
:If $f$ and $g$ are both bijections, then so is $f \circ g$
where $f \circ g$ is the composite mapping of $f$ with $g$. | Let $g: X \to Y$ and $f: Y \to Z$ be bijections.
Then from Bijection iff Inverse is Bijection, both $f^{-1}$ and $g^{-1}$ are bijections.
From Inverse of Composite Relation we have that $g^{-1} \circ f^{-1}$ is the inverse of $f \circ g$.
Then:
:$\paren {f \circ g} \circ \paren {g^{-1} \circ f^{-1} } = I_Z$
:$\paren {g... | Let $f$ and $g$ be [[Definition:Mapping|mappings]] such that $\Dom f = \Cdm g$.
Then:
:If $f$ and $g$ are both [[Definition:Bijection|bijections]], then so is $f \circ g$
where $f \circ g$ is the [[Definition:Composition of Mappings|composite mapping]] of $f$ with $g$. | Let $g: X \to Y$ and $f: Y \to Z$ be [[Definition:Bijection|bijections]].
Then from [[Bijection iff Inverse is Bijection]], both $f^{-1}$ and $g^{-1}$ are [[Definition:Bijection|bijections]].
From [[Inverse of Composite Relation]] we have that $g^{-1} \circ f^{-1}$ is the [[Definition:Inverse Mapping|inverse]] of $f ... | Composite of Bijections is Bijection/Proof 2 | https://proofwiki.org/wiki/Composite_of_Bijections_is_Bijection | https://proofwiki.org/wiki/Composite_of_Bijections_is_Bijection/Proof_2 | [
"Bijections",
"Composite Mappings",
"Composite of Bijections is Bijection"
] | [
"Definition:Mapping",
"Definition:Bijection",
"Definition:Composition of Mappings"
] | [
"Definition:Bijection",
"Inverse of Bijection is Bijection",
"Definition:Bijection",
"Inverse of Composite Relation",
"Definition:Inverse Mapping"
] |
proofwiki-404 | Inverse of Composite Bijection | Let $f$ and $g$ be bijections such that $\Dom g = \Cdm f$.
Then:
:$\paren {g \circ f}^{-1} = f^{-1} \circ g^{-1}$
and $f^{-1} \circ g^{-1}$ is itself a bijection. | $\left({g \circ f}\right)^{-1} = f^{-1} \circ g^{-1}$ is a specific example of Inverse of Composite Relation.
As $f$ and $g$ are bijections then by Bijection iff Inverse is Bijection, so are both $f^{-1}$ and $g^{-1}$.
By Composite of Bijections is Bijection, it follows that $f^{-1} \circ g^{-1}$ is a bijection.
{{qed}... | Let $f$ and $g$ be [[Definition:Bijection|bijections]] such that $\Dom g = \Cdm f$.
Then:
:$\paren {g \circ f}^{-1} = f^{-1} \circ g^{-1}$
and $f^{-1} \circ g^{-1}$ is itself a [[Definition:Bijection|bijection]]. | $\left({g \circ f}\right)^{-1} = f^{-1} \circ g^{-1}$ is a specific example of [[Inverse of Composite Relation]].
As $f$ and $g$ are [[Definition:Bijection|bijections]] then by [[Bijection iff Inverse is Bijection]], so are both $f^{-1}$ and $g^{-1}$.
By [[Composite of Bijections is Bijection]], it follows that $f^{-... | Inverse of Composite Bijection/Proof 1 | https://proofwiki.org/wiki/Inverse_of_Composite_Bijection | https://proofwiki.org/wiki/Inverse_of_Composite_Bijection/Proof_1 | [
"Bijections",
"Inverse Mappings",
"Composite Mappings",
"Inverse of Composite Bijection"
] | [
"Definition:Bijection",
"Definition:Bijection"
] | [
"Inverse of Composite Relation",
"Definition:Bijection",
"Inverse of Bijection is Bijection",
"Composite of Bijections is Bijection",
"Definition:Bijection"
] |
proofwiki-405 | Inverse of Composite Bijection | Let $f$ and $g$ be bijections such that $\Dom g = \Cdm f$.
Then:
:$\paren {g \circ f}^{-1} = f^{-1} \circ g^{-1}$
and $f^{-1} \circ g^{-1}$ is itself a bijection. | Let $f: X \to Y$ and $g: Y \to Z$ be bijections.
Then:
{{begin-eqn}}
{{eqn | l = \paren {g \circ f} \circ \paren {f^{-1} \circ g^{-1} }
| r = g \circ \paren {\paren {f \circ f^{-1} } \circ g^{-1} }
| c = Composition of Mappings is Associative
}}
{{eqn | r = g \circ \paren {I_Y \circ g^{-1} }
| c = Com... | Let $f$ and $g$ be [[Definition:Bijection|bijections]] such that $\Dom g = \Cdm f$.
Then:
:$\paren {g \circ f}^{-1} = f^{-1} \circ g^{-1}$
and $f^{-1} \circ g^{-1}$ is itself a [[Definition:Bijection|bijection]]. | Let $f: X \to Y$ and $g: Y \to Z$ be [[Definition:Bijection|bijections]].
Then:
{{begin-eqn}}
{{eqn | l = \paren {g \circ f} \circ \paren {f^{-1} \circ g^{-1} }
| r = g \circ \paren {\paren {f \circ f^{-1} } \circ g^{-1} }
| c = [[Composition of Mappings is Associative]]
}}
{{eqn | r = g \circ \paren {I_Y... | Inverse of Composite Bijection/Proof 2 | https://proofwiki.org/wiki/Inverse_of_Composite_Bijection | https://proofwiki.org/wiki/Inverse_of_Composite_Bijection/Proof_2 | [
"Bijections",
"Inverse Mappings",
"Composite Mappings",
"Inverse of Composite Bijection"
] | [
"Definition:Bijection",
"Definition:Bijection"
] | [
"Definition:Bijection",
"Composition of Mappings is Associative",
"Composite of Bijection with Inverse is Identity Mapping",
"Identity Mapping is Left Identity",
"Composite of Bijection with Inverse is Identity Mapping",
"Composition of Mappings is Associative",
"Composite of Bijection with Inverse is I... |
proofwiki-406 | Composite of Permutations is Permutation | Let $f, g$ are permutations of a set $S$.
Then their composite $g \circ f$ is also a permutation of $S$. | This follows from the fact that a permutation is a bijection.
The domain and codomain are coincident.
The result follows from Composite of Bijections is Bijection.
{{Qed}} | Let $f, g$ are [[Definition:Permutation|permutations]] of a [[Definition:Set|set]] $S$.
Then their [[Definition:Composition of Mappings|composite]] $g \circ f$ is also a [[Definition:Permutation|permutation]] of $S$. | This follows from the fact that a [[Definition:Permutation|permutation]] is a [[Definition:Bijection|bijection]].
The [[Definition:Domain of Mapping|domain]] and [[Definition:Codomain of Mapping|codomain]] are coincident.
The result follows from [[Composite of Bijections is Bijection]].
{{Qed}} | Composite of Permutations is Permutation | https://proofwiki.org/wiki/Composite_of_Permutations_is_Permutation | https://proofwiki.org/wiki/Composite_of_Permutations_is_Permutation | [
"Permutations",
"Composite Mappings"
] | [
"Definition:Permutation",
"Definition:Set",
"Definition:Composition of Mappings",
"Definition:Permutation"
] | [
"Definition:Permutation",
"Definition:Bijection",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Codomain (Set Theory)/Mapping",
"Composite of Bijections is Bijection"
] |
proofwiki-407 | Inverse of Permutation is Permutation | If $f$ is a permutation of $S$, then so is its inverse $f^{-1}$. | Let $f: S \to S$ is a permutation of $S$.
By definition, a permutation is a bijection such that the domain and codomain are the same set.
From Bijection iff Inverse is Bijection, it follows $f^{-1}$ is a bijection.
From the definition of inverse relation, the domain of a relation is the codomain of its inverse and vice... | If $f$ is a [[Definition:Permutation|permutation]] of $S$, then so is its [[Definition:Inverse of Mapping|inverse]] $f^{-1}$. | Let $f: S \to S$ is a [[Definition:Permutation|permutation]] of $S$.
By definition, a [[Definition:Permutation|permutation]] is a [[Definition:Bijection|bijection]] such that the [[Definition:Domain of Mapping|domain]] and [[Definition:Codomain of Mapping|codomain]] are the same [[Definition:Set|set]].
From [[Bijecti... | Inverse of Permutation is Permutation | https://proofwiki.org/wiki/Inverse_of_Permutation_is_Permutation | https://proofwiki.org/wiki/Inverse_of_Permutation_is_Permutation | [
"Permutations",
"Inverse Mappings"
] | [
"Definition:Permutation",
"Definition:Inverse of Mapping"
] | [
"Definition:Permutation",
"Definition:Permutation",
"Definition:Bijection",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Set",
"Inverse of Bijection is Bijection",
"Definition:Bijection",
"Definition:Inverse Relation",
"Definition:Domain (Set Th... |
proofwiki-408 | Left Inverse Mapping is Surjection | Let $f: S \to T$ be an injection.
Let $g: T \to S$ be a left inverse of $f$.
Then $g$ is a surjection. | Let $f: S \to T$ be an injection.
Then from Injection iff Left Inverse there exists at least one left inverse $g: T \to S$ of $f$ such that $g \circ f = I_S$.
$I_S$ is a surjection.
Thus $g \circ f$ is a surjection.
By Surjection if Composite is Surjection it follows that $g$ is also a surjection.
{{Qed}} | Let $f: S \to T$ be an [[Definition:Injection|injection]].
Let $g: T \to S$ be a [[Definition:Left Inverse Mapping|left inverse]] of $f$.
Then $g$ is a [[Definition:Surjection|surjection]]. | Let $f: S \to T$ be an [[Definition:Injection|injection]].
Then from [[Injection iff Left Inverse]] there exists at least one [[Definition:Left Inverse Mapping|left inverse]] $g: T \to S$ of $f$ such that $g \circ f = I_S$.
$I_S$ is a [[Identity Mapping is Surjection|surjection]].
Thus $g \circ f$ is a [[Definition:... | Left Inverse Mapping is Surjection | https://proofwiki.org/wiki/Left_Inverse_Mapping_is_Surjection | https://proofwiki.org/wiki/Left_Inverse_Mapping_is_Surjection | [
"Left Inverse Mappings",
"Surjections"
] | [
"Definition:Injection",
"Definition:Left Inverse Mapping",
"Definition:Surjection"
] | [
"Definition:Injection",
"Injection iff Left Inverse",
"Definition:Left Inverse Mapping",
"Identity Mapping is Surjection",
"Definition:Surjection",
"Surjection if Composite is Surjection",
"Definition:Surjection"
] |
proofwiki-409 | Right Inverse Mapping is Injection | Let $f: S \to T$ be a mapping.
Let $g: T \to S$ be a right inverse of $f$.
Then $g$ is an injection. | By the definition of right inverse:
:$f \circ g = I_T$
where $I_T$ is the identity mapping on $T$.
By Identity Mapping is Injection, $I_T$ is an injection.
By Injection if Composite is Injection, it follows that $g$ is an injection.
{{qed}} | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $g: T \to S$ be a [[Definition:Right Inverse Mapping|right inverse]] of $f$.
Then $g$ is an [[Definition:Injection|injection]]. | By the definition of [[Definition:Right Inverse Mapping|right inverse]]:
:$f \circ g = I_T$
where $I_T$ is the [[Definition:Identity Mapping|identity mapping]] on $T$.
By [[Identity Mapping is Injection]], $I_T$ is an [[Definition:Injection|injection]].
By [[Injection if Composite is Injection]], it follows that $... | Right Inverse Mapping is Injection | https://proofwiki.org/wiki/Right_Inverse_Mapping_is_Injection | https://proofwiki.org/wiki/Right_Inverse_Mapping_is_Injection | [
"Injections",
"Right Inverse Mappings"
] | [
"Definition:Mapping",
"Definition:Right Inverse Mapping",
"Definition:Injection"
] | [
"Definition:Right Inverse Mapping",
"Definition:Identity Mapping",
"Identity Mapping is Injection",
"Definition:Injection",
"Injection if Composite is Injection",
"Definition:Injection"
] |
proofwiki-410 | Set Equivalence behaves like Equivalence Relation | Set equivalence behaves like an equivalence relation.
That is:
{{begin-axiom}}
{{axiom | q = \forall S
| t = Reflexivity
| m = S \sim S
}}
{{axiom | q = \forall S, T
| t = Symmetry
| m = S \sim T \implies T \sim S
}}
{{axiom | q = \forall S_1, S_2, S_3
| t = Transitivity
... | For two sets to be equivalent, there needs to exist a bijection between them.
In the following, let $\phi$, $\phi_1$ and $\phi_2$ be understood to be bijections. | [[Definition:Set Equivalence|Set equivalence]] behaves like an [[Definition:Equivalence Relation|equivalence relation]].
That is:
{{begin-axiom}}
{{axiom | q = \forall S
| t = [[Definition:Reflexive Relation|Reflexivity]]
| m = S \sim S
}}
{{axiom | q = \forall S, T
| t = [[Definition:Symmetr... | For two [[Definition:Set|sets]] to be [[Definition:Set Equivalence|equivalent]], there needs to exist a [[Definition:Bijection|bijection]] between them.
In the following, let $\phi$, $\phi_1$ and $\phi_2$ be understood to be [[Definition:Bijection|bijections]]. | Set Equivalence behaves like Equivalence Relation | https://proofwiki.org/wiki/Set_Equivalence_behaves_like_Equivalence_Relation | https://proofwiki.org/wiki/Set_Equivalence_behaves_like_Equivalence_Relation | [
"Examples of Equivalence Relations",
"Set Equivalence",
"Set Equivalence behaves like Equivalence Relation"
] | [
"Definition:Set Equivalence",
"Definition:Equivalence Relation",
"Definition:Reflexive Relation",
"Definition:Symmetric Relation",
"Definition:Transitive Relation",
"Definition:Set"
] | [
"Definition:Set",
"Definition:Set Equivalence",
"Definition:Bijection",
"Definition:Bijection"
] |
proofwiki-411 | Set Equivalence Less One Element | Let $S$ and $T$ be sets which are equivalent:
:$S \sim T$
Let $a \in S$ and $b \in T$.
Then:
:$S \setminus \set a \sim T \setminus \set b$
where $\setminus$ denotes set difference. | As $S \sim T$, there exists a bijection $f: S \to T$.
Let $g: \paren {S \setminus \set a} \to \paren {T \setminus \set b}$ be the mapping defined as follows:
$\forall x \in S \setminus \set a: \map g x =
\begin{cases}
\map f x: \map f x \ne b \\
\map f a: \map f x = b
\end{cases}$
First it is shown that $g$ is an injec... | Let $S$ and $T$ be [[Definition:Set|sets]] which are [[Definition:Set Equivalence|equivalent]]:
:$S \sim T$
Let $a \in S$ and $b \in T$.
Then:
:$S \setminus \set a \sim T \setminus \set b$
where $\setminus$ denotes [[Definition:Set Difference|set difference]]. | As $S \sim T$, there exists a [[Definition:Bijection|bijection]] $f: S \to T$.
Let $g: \paren {S \setminus \set a} \to \paren {T \setminus \set b}$ be the [[Definition:Mapping|mapping]] defined as follows:
$\forall x \in S \setminus \set a: \map g x =
\begin{cases}
\map f x: \map f x \ne b \\
\map f a: \map f x = b
\... | Set Equivalence Less One Element | https://proofwiki.org/wiki/Set_Equivalence_Less_One_Element | https://proofwiki.org/wiki/Set_Equivalence_Less_One_Element | [
"Equivalence Relations"
] | [
"Definition:Set",
"Definition:Set Equivalence",
"Definition:Set Difference"
] | [
"Definition:Bijection",
"Definition:Mapping",
"Definition:Injection",
"Definition:Injection",
"Definition:Set Difference",
"Definition:Injection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Surjection",
"Definition:By Hypothesis",
"Rule of Transposition",
"Rule of Transpositi... |
proofwiki-412 | Image of Union under Mapping | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $A$ and $B$ be subsets of $S$.
Then:
:$f \sqbrk {A \cup B} = f \sqbrk A \cup f \sqbrk B$
This can be expressed in the language and notation of direct image mappings as:
:$\forall A, B \in \powerset S: \map {f^\to} {A \cup B} = \map {f^\to} A \cup \map {f^\to} ... | First we have:
{{begin-eqn}}
{{eqn | l = A
| o = \subseteq
| r = A \cup B
| c = Set is Subset of Union
}}
{{eqn | ll= \leadsto
| l = f \sqbrk A
| o = \subseteq
| r = f \sqbrk {A \cup B}
| c = Image of Subset under Mapping is Subset of Image
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | ... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $A$ and $B$ be [[Definition:Subset|subsets]] of $S$.
Then:
:$f \sqbrk {A \cup B} = f \sqbrk A \cup f \sqbrk B$
This can be expressed in the language and notation of [[Definition:Direct Image Mapping|direct image ... | First we have:
{{begin-eqn}}
{{eqn | l = A
| o = \subseteq
| r = A \cup B
| c = [[Set is Subset of Union]]
}}
{{eqn | ll= \leadsto
| l = f \sqbrk A
| o = \subseteq
| r = f \sqbrk {A \cup B}
| c = [[Image of Subset under Mapping is Subset of Image]]
}}
{{end-eqn}}
{{begin-eqn... | Image of Union under Mapping/Proof 1 | https://proofwiki.org/wiki/Image_of_Union_under_Mapping | https://proofwiki.org/wiki/Image_of_Union_under_Mapping/Proof_1 | [
"Images",
"Set Union",
"Image of Union under Mapping"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Subset",
"Definition:Direct Image Mapping"
] | [
"Set is Subset of Union",
"Image of Subset under Mapping is Subset of Image",
"Set is Subset of Union",
"Image of Subset under Mapping is Subset of Image",
"Union is Smallest Superset"
] |
proofwiki-413 | Image of Union under Mapping | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $A$ and $B$ be subsets of $S$.
Then:
:$f \sqbrk {A \cup B} = f \sqbrk A \cup f \sqbrk B$
This can be expressed in the language and notation of direct image mappings as:
:$\forall A, B \in \powerset S: \map {f^\to} {A \cup B} = \map {f^\to} A \cup \map {f^\to} ... | As $f$, being a mapping, is also a relation, we can apply Image of Union under Relation:
:$\RR \sqbrk {A \cup B} = \RR \sqbrk A \cup \RR \sqbrk B$
{{qed}} | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $A$ and $B$ be [[Definition:Subset|subsets]] of $S$.
Then:
:$f \sqbrk {A \cup B} = f \sqbrk A \cup f \sqbrk B$
This can be expressed in the language and notation of [[Definition:Direct Image Mapping|direct image ... | As $f$, being a [[Definition:Mapping|mapping]], is also a [[Definition:Relation|relation]], we can apply [[Image of Union under Relation]]:
:$\RR \sqbrk {A \cup B} = \RR \sqbrk A \cup \RR \sqbrk B$
{{qed}} | Image of Union under Mapping/Proof 2 | https://proofwiki.org/wiki/Image_of_Union_under_Mapping | https://proofwiki.org/wiki/Image_of_Union_under_Mapping/Proof_2 | [
"Images",
"Set Union",
"Image of Union under Mapping"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Subset",
"Definition:Direct Image Mapping"
] | [
"Definition:Mapping",
"Definition:Relation",
"Image of Union under Relation"
] |
proofwiki-414 | Image of Intersection under Mapping | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $S_1$ and $S_2$ be subsets of $S$.
Then:
:$f \sqbrk {S_1 \cap S_2} \subseteq f \sqbrk {S_1} \cap f \sqbrk {S_2}$
This can be expressed in the language and notation of direct image mappings as:
:$\forall S_1, S_2 \in \powerset S: \map {f^\to} {S_1 \cap S_2} \su... | {{begin-eqn}}
{{eqn | l = S_1 \cap S_2
| o = \subseteq
| r = S_1
| c = Intersection is Subset
}}
{{eqn | ll= \leadsto
| l = f \sqbrk {S_1 \cap S_2}
| o = \subseteq
| r = f \sqbrk {S_1}
| c = Image of Subset under Mapping is Subset of Image
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $S_1$ and $S_2$ be [[Definition:Subset|subsets]] of $S$.
Then:
:$f \sqbrk {S_1 \cap S_2} \subseteq f \sqbrk {S_1} \cap f \sqbrk {S_2}$
This can be expressed in the language and notation of [[Definition:Direct Im... | {{begin-eqn}}
{{eqn | l = S_1 \cap S_2
| o = \subseteq
| r = S_1
| c = [[Intersection is Subset]]
}}
{{eqn | ll= \leadsto
| l = f \sqbrk {S_1 \cap S_2}
| o = \subseteq
| r = f \sqbrk {S_1}
| c = [[Image of Subset under Mapping is Subset of Image]]
}}
{{end-eqn}}
{{begin-eqn}}... | Image of Intersection under Mapping/Proof 1 | https://proofwiki.org/wiki/Image_of_Intersection_under_Mapping | https://proofwiki.org/wiki/Image_of_Intersection_under_Mapping/Proof_1 | [
"Images",
"Set Intersection",
"Image of Intersection under Mapping"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Subset",
"Definition:Direct Image Mapping"
] | [
"Intersection is Subset",
"Image of Subset under Mapping is Subset of Image",
"Intersection is Subset",
"Image of Subset under Mapping is Subset of Image",
"Intersection is Largest Subset"
] |
proofwiki-415 | Image of Intersection under Mapping | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $S_1$ and $S_2$ be subsets of $S$.
Then:
:$f \sqbrk {S_1 \cap S_2} \subseteq f \sqbrk {S_1} \cap f \sqbrk {S_2}$
This can be expressed in the language and notation of direct image mappings as:
:$\forall S_1, S_2 \in \powerset S: \map {f^\to} {S_1 \cap S_2} \su... | As $f$, being a mapping, is also a relation, we can apply Image of Intersection under Relation:
:$\RR \sqbrk {S_1 \cap S_2} \subseteq \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$
{{qed}} | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $S_1$ and $S_2$ be [[Definition:Subset|subsets]] of $S$.
Then:
:$f \sqbrk {S_1 \cap S_2} \subseteq f \sqbrk {S_1} \cap f \sqbrk {S_2}$
This can be expressed in the language and notation of [[Definition:Direct Im... | As $f$, being a [[Definition:Mapping|mapping]], is also a [[Definition:Relation|relation]], we can apply [[Image of Intersection under Relation]]:
:$\RR \sqbrk {S_1 \cap S_2} \subseteq \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$
{{qed}} | Image of Intersection under Mapping/Proof 2 | https://proofwiki.org/wiki/Image_of_Intersection_under_Mapping | https://proofwiki.org/wiki/Image_of_Intersection_under_Mapping/Proof_2 | [
"Images",
"Set Intersection",
"Image of Intersection under Mapping"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Subset",
"Definition:Direct Image Mapping"
] | [
"Definition:Mapping",
"Definition:Relation",
"Image of Intersection under Relation"
] |
proofwiki-416 | Image of Intersection under Injection | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
:$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$
{{iff}} $f$ is an injection.
This can be expressed in the language and notation of direct image mappings as:
:$\forall A, B \subseteq S: \map {f^\to} {A \cap B} = \map {f^\to} A \c... | An injection is a type of one-to-one relation, and therefore also a one-to-many relation.
Therefore Image of Intersection under One-to-Many Relation applies:
:$\forall A, B \subseteq S: \RR \sqbrk {A \cap B} = \RR \sqbrk A \cap \RR \sqbrk B$
{{iff}} $\RR$ is a one-to-many relation.
We have that $f$ is a mapping and the... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$
{{iff}} $f$ is an [[Definition:Injection|injection]].
This can be expressed in the language and notation of [[Definition:Direct Image... | An [[Definition:Injection|injection]] is a type of [[Definition:One-to-One Relation|one-to-one relation]], and therefore also a [[Definition:One-to-Many Relation|one-to-many relation]].
Therefore [[Image of Intersection under One-to-Many Relation]] applies:
:$\forall A, B \subseteq S: \RR \sqbrk {A \cap B} = \RR \sq... | Image of Intersection under Injection/Proof 1 | https://proofwiki.org/wiki/Image_of_Intersection_under_Injection | https://proofwiki.org/wiki/Image_of_Intersection_under_Injection/Proof_1 | [
"Injections",
"Set Intersection",
"Image of Intersection under Injection"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Injection",
"Definition:Direct Image Mapping"
] | [
"Definition:Injection",
"Definition:One-to-One Relation",
"Definition:One-to-Many Relation",
"Image of Intersection under One-to-Many Relation",
"Definition:One-to-Many Relation",
"Definition:Mapping",
"Definition:Many-to-One Relation",
"Definition:One-to-Many Relation",
"Definition:Injection",
"D... |
proofwiki-417 | Image of Intersection under Injection | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
:$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$
{{iff}} $f$ is an injection.
This can be expressed in the language and notation of direct image mappings as:
:$\forall A, B \subseteq S: \map {f^\to} {A \cap B} = \map {f^\to} A \c... | From Image of Intersection under Mapping:
:$f \sqbrk {A \cap B} \subseteq f \sqbrk A \cap f \sqbrk B$
which holds for all mappings.
It remains to be shown that:
:$f \sqbrk A \cap f \sqbrk B \subseteq f \sqbrk {A \cap B}$
{{iff}} $f$ is an injection.
=== Sufficient Condition ===
Suppose that:
:$\forall A, B \subseteq S:... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$
{{iff}} $f$ is an [[Definition:Injection|injection]].
This can be expressed in the language and notation of [[Definition:Direct Image... | From [[Image of Intersection under Mapping]]:
:$f \sqbrk {A \cap B} \subseteq f \sqbrk A \cap f \sqbrk B$
which holds for all [[Definition:Mapping|mappings]].
It remains to be shown that:
:$f \sqbrk A \cap f \sqbrk B \subseteq f \sqbrk {A \cap B}$
{{iff}} $f$ is an [[Definition:Injection|injection]].
=== Sufficient ... | Image of Intersection under Injection/Proof 2 | https://proofwiki.org/wiki/Image_of_Intersection_under_Injection | https://proofwiki.org/wiki/Image_of_Intersection_under_Injection/Proof_2 | [
"Injections",
"Set Intersection",
"Image of Intersection under Injection"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Injection",
"Definition:Direct Image Mapping"
] | [
"Image of Intersection under Mapping",
"Definition:Mapping",
"Definition:Injection",
"Definition:Singleton",
"Definition:A Fortiori",
"Definition:Injection",
"Mapping from Singleton is Injection",
"Definition:Singleton",
"Definition:Injection",
"Definition:Set Intersection",
"Definition:Contradi... |
proofwiki-418 | Image of Intersection under One-to-Many Relation | Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Then:
:$\forall S_1, S_2 \subseteq S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$
{{iff}} $\RR$ is one-to-many. | === Sufficient Condition ===
Suppose that:
:$\forall S_1, S_2 \subseteq S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$
If $S$ is singleton, the result follows immediately as $\RR$ would ''have'' to be one-to-many.
So, assume $S$ is not singleton.
Suppose $\RR$ is specifically ''not'' one-to-many... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Then:
:$\forall S_1, S_2 \subseteq S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$
{{iff}} $\RR$ is [[Definition:One-to-Many Relation|one-to-many]]. | === Sufficient Condition ===
Suppose that:
:$\forall S_1, S_2 \subseteq S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$
If $S$ is [[Definition:Singleton|singleton]], the result follows immediately as $\RR$ would ''have'' to be [[Definition:One-to-Many Relation|one-to-many]].
So, assume $S$ is ... | Image of Intersection under One-to-Many Relation | https://proofwiki.org/wiki/Image_of_Intersection_under_One-to-Many_Relation | https://proofwiki.org/wiki/Image_of_Intersection_under_One-to-Many_Relation | [
"One-to-Many Relations",
"Set Intersection",
"Image of Intersection under One-to-Many Relation"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:One-to-Many Relation"
] | [
"Definition:Singleton",
"Definition:One-to-Many Relation",
"Definition:Singleton",
"Definition:One-to-Many Relation",
"Definition:Set Intersection",
"Image of Empty Set is Empty Set",
"Definition:One-to-Many Relation",
"Definition:One-to-Many Relation",
"Definition:One-to-Many Relation"
] |
proofwiki-419 | One-to-Many Image of Set Difference | Let $\RR \subseteq S \times T$ be a relation.
Let $A$ and $B$ be subsets of $S$.
Then:
:$(1): \quad \RR \sqbrk A \setminus \RR \sqbrk B = \RR \sqbrk {A \setminus B}$
{{iff}} $\RR$ is one-to-many. | === Sufficient Condition ===
First, to show that $(1)$ holds if $\RR$ is one-to-many.
From Image of Set Difference under Relation, we already have:
:$\RR \sqbrk A \setminus \RR \sqbrk B \subseteq \RR \sqbrk {A \setminus B}$
So we just need to show:
:$\RR \sqbrk {A \setminus B} \subseteq \RR \sqbrk A \setminus \RR \sqbr... | Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Let $A$ and $B$ be [[Definition:Subset|subsets]] of $S$.
Then:
:$(1): \quad \RR \sqbrk A \setminus \RR \sqbrk B = \RR \sqbrk {A \setminus B}$
{{iff}} $\RR$ is [[Definition:One-to-Many Relation|one-to-many]]. | === Sufficient Condition ===
First, to show that $(1)$ holds if $\RR$ is [[Definition:One-to-Many Relation|one-to-many]].
From [[Image of Set Difference under Relation]], we already have:
:$\RR \sqbrk A \setminus \RR \sqbrk B \subseteq \RR \sqbrk {A \setminus B}$
So we just need to show:
:$\RR \sqbrk {A \setminu... | One-to-Many Image of Set Difference | https://proofwiki.org/wiki/One-to-Many_Image_of_Set_Difference | https://proofwiki.org/wiki/One-to-Many_Image_of_Set_Difference | [
"Set Difference",
"One-to-Many Relations",
"One-to-Many Image of Set Difference"
] | [
"Definition:Relation",
"Definition:Subset",
"Definition:One-to-Many Relation"
] | [
"Definition:One-to-Many Relation",
"Image of Set Difference under Relation",
"De Morgan's Laws (Logic)",
"Definition:Relation",
"Image of Subset under Relation is Subset of Image",
"Definition:Subset",
"Rule of Transposition",
"Definition:One-to-Many Relation",
"Proof by Cases",
"Set Complement in... |
proofwiki-420 | Preimage of Union under Mapping | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $T_1$ and $T_2$ be subsets of $T$.
Then:
:$f^{-1} \sqbrk {T_1 \cup T_2} = f^{-1} \sqbrk {T_1} \cup f^{-1} \sqbrk {T_2}$
This can be expressed in the language and notation of inverse image mappings as:
:$\forall T_1, T_2 \in \powerset T: \map {f^\gets} {T_1 \cu... | As $f$, being a mapping, is also a relation, we can apply Preimage of Union under Relation:
:$\RR^{-1} \sqbrk {T_1 \cup T_2} = \RR^{-1} \sqbrk {T_1} \cup \RR^{-1} \sqbrk {T_2}$
{{qed}} | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $T_1$ and $T_2$ be [[Definition:Subset|subsets]] of $T$.
Then:
:$f^{-1} \sqbrk {T_1 \cup T_2} = f^{-1} \sqbrk {T_1} \cup f^{-1} \sqbrk {T_2}$
This can be expressed in the language and notation of [[Definition:Inv... | As $f$, being a [[Definition:Mapping|mapping]], is also a [[Definition:Relation|relation]], we can apply [[Preimage of Union under Relation]]:
:$\RR^{-1} \sqbrk {T_1 \cup T_2} = \RR^{-1} \sqbrk {T_1} \cup \RR^{-1} \sqbrk {T_2}$
{{qed}} | Preimage of Union under Mapping | https://proofwiki.org/wiki/Preimage_of_Union_under_Mapping | https://proofwiki.org/wiki/Preimage_of_Union_under_Mapping | [
"Preimage of Union under Mapping",
"Preimages under Mappings",
"Set Union"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Subset",
"Definition:Inverse Image Mapping"
] | [
"Definition:Mapping",
"Definition:Relation",
"Preimage of Union under Relation"
] |
proofwiki-421 | Preimage of Union under Mapping | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $T_1$ and $T_2$ be subsets of $T$.
Then:
:$f^{-1} \sqbrk {T_1 \cup T_2} = f^{-1} \sqbrk {T_1} \cup f^{-1} \sqbrk {T_2}$
This can be expressed in the language and notation of inverse image mappings as:
:$\forall T_1, T_2 \in \powerset T: \map {f^\gets} {T_1 \cu... | As $f$, being a mapping, is also a relation, we can apply Preimage of Union under Relation: Family of Sets:
:$\ds \RR^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \RR^{-1} \sqbrk {T_i}$
where $\RR^{-1} \sqbrk {T_i}$ denotes the preimage of $T_i$ under $\RR^{-1}$.
{{qed}} | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $T_1$ and $T_2$ be [[Definition:Subset|subsets]] of $T$.
Then:
:$f^{-1} \sqbrk {T_1 \cup T_2} = f^{-1} \sqbrk {T_1} \cup f^{-1} \sqbrk {T_2}$
This can be expressed in the language and notation of [[Definition:Inv... | As $f$, being a [[Definition:Mapping|mapping]], is also a [[Definition:Relation|relation]], we can apply [[Preimage of Union under Relation/Family of Sets|Preimage of Union under Relation: Family of Sets]]:
:$\ds \RR^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \RR^{-1} \sqbrk {T_i}$
where $... | Preimage of Union under Mapping/Family of Sets/Proof 1 | https://proofwiki.org/wiki/Preimage_of_Union_under_Mapping | https://proofwiki.org/wiki/Preimage_of_Union_under_Mapping/Family_of_Sets/Proof_1 | [
"Preimage of Union under Mapping",
"Preimages under Mappings",
"Set Union"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Subset",
"Definition:Inverse Image Mapping"
] | [
"Definition:Mapping",
"Definition:Relation",
"Preimage of Union under Relation/Family of Sets",
"Definition:Preimage/Relation/Subset"
] |
proofwiki-422 | Preimage of Union under Mapping | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $T_1$ and $T_2$ be subsets of $T$.
Then:
:$f^{-1} \sqbrk {T_1 \cup T_2} = f^{-1} \sqbrk {T_1} \cup f^{-1} \sqbrk {T_2}$
This can be expressed in the language and notation of inverse image mappings as:
:$\forall T_1, T_2 \in \powerset T: \map {f^\gets} {T_1 \cu... | We have that $f$ is a mapping, and so also a relation.
Thus its inverse $f^{-1}$ is also a relation.
Hence we can apply Image of Union under Relation: Family of Sets:
:$\ds \RR \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \RR \sqbrk {T_i}$
where $\RR \sqbrk {T_i}$ denotes the image of $T_i$ under ... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $T_1$ and $T_2$ be [[Definition:Subset|subsets]] of $T$.
Then:
:$f^{-1} \sqbrk {T_1 \cup T_2} = f^{-1} \sqbrk {T_1} \cup f^{-1} \sqbrk {T_2}$
This can be expressed in the language and notation of [[Definition:Inv... | We have that $f$ is a [[Definition:Mapping|mapping]], and so also a [[Definition:Relation|relation]].
Thus its [[Definition:Inverse of Mapping|inverse]] $f^{-1}$ is also a [[Definition:Relation|relation]].
Hence we can apply [[Image of Union under Relation/Family of Sets|Image of Union under Relation: Family of Sets]... | Preimage of Union under Mapping/Family of Sets/Proof 2 | https://proofwiki.org/wiki/Preimage_of_Union_under_Mapping | https://proofwiki.org/wiki/Preimage_of_Union_under_Mapping/Family_of_Sets/Proof_2 | [
"Preimage of Union under Mapping",
"Preimages under Mappings",
"Set Union"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Subset",
"Definition:Inverse Image Mapping"
] | [
"Definition:Mapping",
"Definition:Relation",
"Definition:Inverse of Mapping",
"Definition:Relation",
"Image of Union under Relation/Family of Sets",
"Definition:Image (Set Theory)/Relation/Subset"
] |
proofwiki-423 | Preimage of Intersection under Mapping | Let $f: S \to T$ be a mapping.
Let $T_1$ and $T_2$ be subsets of $T$.
Then:
:$f^{-1} \sqbrk {T_1 \cap T_2} = f^{-1} \sqbrk {T_1} \cap f^{-1} \sqbrk {T_2}$
This can be expressed in the language and notation of inverse image mappings as:
:$\forall T_1, T_2 \in \powerset T: \map {f^\gets} {T_1 \cap T_2} = \map {f^\gets} {... | As $f$, being a mapping, is also a many-to-one relation, it follows from Inverse of Many-to-One Relation is One-to-Many that its inverse $f^{-1}$ is a one-to-many relation.
Thus Image of Intersection under One-to-Many Relation applies:
:$\RR \sqbrk {T_1 \cap T_2} = \RR \sqbrk {T_1} \cap \RR \sqbrk {T_2}$
where here $\R... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $T_1$ and $T_2$ be [[Definition:Subset|subsets]] of $T$.
Then:
:$f^{-1} \sqbrk {T_1 \cap T_2} = f^{-1} \sqbrk {T_1} \cap f^{-1} \sqbrk {T_2}$
This can be expressed in the language and notation of [[Definition:Inverse Image Mapping|inverse image mappings]] a... | As $f$, being a [[Definition:Mapping|mapping]], is also a [[Definition:Many-to-One Relation|many-to-one relation]], it follows from [[Inverse of Many-to-One Relation is One-to-Many]] that its [[Definition:Inverse of Mapping|inverse]] $f^{-1}$ is a [[Definition:One-to-Many Relation|one-to-many relation]].
Thus [[Image... | Preimage of Intersection under Mapping | https://proofwiki.org/wiki/Preimage_of_Intersection_under_Mapping | https://proofwiki.org/wiki/Preimage_of_Intersection_under_Mapping | [
"Preimage of Intersection under Mapping",
"Preimages under Mappings",
"Set Intersection"
] | [
"Definition:Mapping",
"Definition:Subset",
"Definition:Inverse Image Mapping"
] | [
"Definition:Mapping",
"Definition:Many-to-One Relation",
"Inverse of Many-to-One Relation is One-to-Many",
"Definition:Inverse of Mapping",
"Definition:One-to-Many Relation",
"Image of Intersection under One-to-Many Relation"
] |
proofwiki-424 | Preimage of Intersection under Mapping | Let $f: S \to T$ be a mapping.
Let $T_1$ and $T_2$ be subsets of $T$.
Then:
:$f^{-1} \sqbrk {T_1 \cap T_2} = f^{-1} \sqbrk {T_1} \cap f^{-1} \sqbrk {T_2}$
This can be expressed in the language and notation of inverse image mappings as:
:$\forall T_1, T_2 \in \powerset T: \map {f^\gets} {T_1 \cap T_2} = \map {f^\gets} {... | As $f$ is a mapping, it is by definition also a many-to-one relation.
It follows from Inverse of Many-to-One Relation is One-to-Many that its inverse $f^{-1}$ is a one-to-many relation.
Thus Image of Intersection under One-to-Many Relation: Family of Sets can be applied for $\RR = f^{-1}$:
:$\ds \RR \sqbrk {\bigcap_{i ... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $T_1$ and $T_2$ be [[Definition:Subset|subsets]] of $T$.
Then:
:$f^{-1} \sqbrk {T_1 \cap T_2} = f^{-1} \sqbrk {T_1} \cap f^{-1} \sqbrk {T_2}$
This can be expressed in the language and notation of [[Definition:Inverse Image Mapping|inverse image mappings]] a... | As $f$ is a [[Definition:Mapping|mapping]], it is by definition also a [[Definition:Many-to-One Relation|many-to-one relation]].
It follows from [[Inverse of Many-to-One Relation is One-to-Many]] that its [[Definition:Inverse of Mapping|inverse]] $f^{-1}$ is a [[Definition:One-to-Many Relation|one-to-many relation]].
... | Preimage of Intersection under Mapping/Family of Sets/Proof 1 | https://proofwiki.org/wiki/Preimage_of_Intersection_under_Mapping | https://proofwiki.org/wiki/Preimage_of_Intersection_under_Mapping/Family_of_Sets/Proof_1 | [
"Preimage of Intersection under Mapping",
"Preimages under Mappings",
"Set Intersection"
] | [
"Definition:Mapping",
"Definition:Subset",
"Definition:Inverse Image Mapping"
] | [
"Definition:Mapping",
"Definition:Many-to-One Relation",
"Inverse of Many-to-One Relation is One-to-Many",
"Definition:Inverse of Mapping",
"Definition:One-to-Many Relation",
"Image of Intersection under One-to-Many Relation/Family of Sets",
"Definition:Image (Set Theory)/Relation/Subset"
] |
proofwiki-425 | Preimage of Intersection under Mapping | Let $f: S \to T$ be a mapping.
Let $T_1$ and $T_2$ be subsets of $T$.
Then:
:$f^{-1} \sqbrk {T_1 \cap T_2} = f^{-1} \sqbrk {T_1} \cap f^{-1} \sqbrk {T_2}$
This can be expressed in the language and notation of inverse image mappings as:
:$\forall T_1, T_2 \in \powerset T: \map {f^\gets} {T_1 \cap T_2} = \map {f^\gets} {... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = f^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \map f x
| o = \in
| r = \bigcap_{i \mathop \in I} T_i
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \forall i \in I
| l = \map f x
... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $T_1$ and $T_2$ be [[Definition:Subset|subsets]] of $T$.
Then:
:$f^{-1} \sqbrk {T_1 \cap T_2} = f^{-1} \sqbrk {T_1} \cap f^{-1} \sqbrk {T_2}$
This can be expressed in the language and notation of [[Definition:Inverse Image Mapping|inverse image mappings]] a... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = f^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \map f x
| o = \in
| r = \bigcap_{i \mathop \in I} T_i
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \forall i \in I
| l = \map f x
... | Preimage of Intersection under Mapping/Family of Sets/Proof 2 | https://proofwiki.org/wiki/Preimage_of_Intersection_under_Mapping | https://proofwiki.org/wiki/Preimage_of_Intersection_under_Mapping/Family_of_Sets/Proof_2 | [
"Preimage of Intersection under Mapping",
"Preimages under Mappings",
"Set Intersection"
] | [
"Definition:Mapping",
"Definition:Subset",
"Definition:Inverse Image Mapping"
] | [] |
proofwiki-426 | Preimage of Set Difference under Mapping | Let $f: S \to T$ be a mapping.
Let $T_1$ and $T_2$ be subsets of $T$.
Then:
:$f^{-1} \sqbrk {T_1 \setminus T_2} = f^{-1} \sqbrk {T_1} \setminus f^{-1} \sqbrk {T_2}$
where:
:$\setminus$ denotes set difference
:$f^{-1} \sqbrk {T_1}$ denotes preimage. | From Inverse of Mapping is One-to-Many Relation, we have that $f^{-1}: T \to S$ is one-to-many.
Thus we can apply One-to-Many Image of Set Difference:
:$\RR \sqbrk {T_1 \setminus T_2} = \RR \sqbrk {T_1} \setminus \RR \sqbrk {T_2}$
where in this context $\RR = f^{-1}$.
{{qed}} | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $T_1$ and $T_2$ be [[Definition:Subset|subsets]] of $T$.
Then:
:$f^{-1} \sqbrk {T_1 \setminus T_2} = f^{-1} \sqbrk {T_1} \setminus f^{-1} \sqbrk {T_2}$
where:
:$\setminus$ denotes [[Definition:Set Difference|set difference]]
:$f^{-1} \sqbrk {T_1}$ denotes [[D... | From [[Inverse of Mapping is One-to-Many Relation]], we have that $f^{-1}: T \to S$ is [[Definition:One-to-Many Relation|one-to-many]].
Thus we can apply [[One-to-Many Image of Set Difference]]:
:$\RR \sqbrk {T_1 \setminus T_2} = \RR \sqbrk {T_1} \setminus \RR \sqbrk {T_2}$
where in this context $\RR = f^{-1}$.
{{qed... | Preimage of Set Difference under Mapping | https://proofwiki.org/wiki/Preimage_of_Set_Difference_under_Mapping | https://proofwiki.org/wiki/Preimage_of_Set_Difference_under_Mapping | [
"Preimages under Mappings",
"Set Difference"
] | [
"Definition:Mapping",
"Definition:Subset",
"Definition:Set Difference",
"Definition:Preimage/Mapping/Subset"
] | [
"Inverse of Mapping is One-to-Many Relation",
"Definition:One-to-Many Relation",
"One-to-Many Image of Set Difference"
] |
proofwiki-427 | Projection is Surjection | Let $S$ and $T$ be non-empty sets.
Let $S \times T$ be the Cartesian product of $S$ and $T$.
Let $\pr_1: S \times T \to S$ and $\pr_2: S \times T \to T$ be the first projection and second projection respectively on $S \times T$.
Then $\pr_1$ and $\pr_2$ are both surjections. | Let $S$ and $T$ be sets such that neither is empty.
Let $\pr_1: S \times T \to S$ be the first projection on $S \times T$.
Then by definition of first projection:
:$\forall x \in S: \exists \tuple {x, t} \in S \times T: \map {\pr_1} {x, t} = x$
Similarly, let $\pr_2: S \times T \to T$ be the second projection on $S \ti... | Let $S$ and $T$ be [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|sets]].
Let $S \times T$ be the [[Definition:Cartesian Product|Cartesian product]] of $S$ and $T$.
Let $\pr_1: S \times T \to S$ and $\pr_2: S \times T \to T$ be the [[Definition:First Projection|first projection]] and [[Definition:Second Proj... | Let $S$ and $T$ be [[Definition:Set|sets]] such that neither is [[Definition:Empty Set|empty]].
Let $\pr_1: S \times T \to S$ be the [[Definition:First Projection|first projection]] on $S \times T$.
Then by definition of [[Definition:First Projection|first projection]]:
:$\forall x \in S: \exists \tuple {x, t} \in S... | Projection is Surjection | https://proofwiki.org/wiki/Projection_is_Surjection | https://proofwiki.org/wiki/Projection_is_Surjection | [
"Surjections",
"Projections"
] | [
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Cartesian Product",
"Definition:Projection (Mapping Theory)/First Projection",
"Definition:Projection (Mapping Theory)/Second Projection",
"Definition:Surjection"
] | [
"Definition:Set",
"Definition:Empty Set",
"Definition:Projection (Mapping Theory)/First Projection",
"Definition:Projection (Mapping Theory)/First Projection",
"Definition:Projection (Mapping Theory)/Second Projection",
"Definition:Projection (Mapping Theory)/Second Projection"
] |
proofwiki-428 | Sum Rule for Derivatives | Let $\map f x, \map j x, \map k x$ be real functions defined on the open interval $I$.
Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are differentiable.
Let $\map f x = \map j x + \map k x$.
Then $f$ is differentiable at $\xi$ and:
:$\map {f'} \xi = \map {j'} \xi + \map {k'} \xi$
It follows from the defin... | {{begin-eqn}}
{{eqn | l = \map {f'} \xi
| r = \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi} h
| c = {{Defof|Derivative}}
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {\paren {\map j {\xi + h} + \map k {\xi + h} } - \paren {\map j \xi + \map k \xi } } h
| c =
}}
{{eqn | r = \lim_{h \math... | Let $\map f x, \map j x, \map k x$ be [[Definition:Real Function|real functions]] defined on the [[Definition:Open Real Interval|open interval]] $I$.
Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are [[Definition:Differentiable Real Function at Point|differentiable]].
Let $\map f x = \map j x + \map k ... | {{begin-eqn}}
{{eqn | l = \map {f'} \xi
| r = \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi} h
| c = {{Defof|Derivative}}
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {\paren {\map j {\xi + h} + \map k {\xi + h} } - \paren {\map j \xi + \map k \xi } } h
| c =
}}
{{eqn | r = \lim_{h \math... | Sum Rule for Derivatives/Proof 1 | https://proofwiki.org/wiki/Sum_Rule_for_Derivatives | https://proofwiki.org/wiki/Sum_Rule_for_Derivatives/Proof_1 | [
"Sum Rule for Derivatives",
"Derivatives",
"Differential Calculus"
] | [
"Definition:Real Function",
"Definition:Real Interval/Open",
"Definition:Differentiable Mapping/Real Function/Point",
"Definition:Differentiable Mapping/Real Function/Point",
"Definition:Derivative/Real Function/Derivative on Interval",
"Definition:Differentiable Mapping/Real Function/Interval"
] | [] |
proofwiki-429 | Sum Rule for Derivatives | Let $\map f x, \map j x, \map k x$ be real functions defined on the open interval $I$.
Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are differentiable.
Let $\map f x = \map j x + \map k x$.
Then $f$ is differentiable at $\xi$ and:
:$\map {f'} \xi = \map {j'} \xi + \map {k'} \xi$
It follows from the defin... | It can be observed that this is an example of a Linear Combination of Derivatives with $\lambda = \mu = 1$.
{{qed}} | Let $\map f x, \map j x, \map k x$ be [[Definition:Real Function|real functions]] defined on the [[Definition:Open Real Interval|open interval]] $I$.
Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are [[Definition:Differentiable Real Function at Point|differentiable]].
Let $\map f x = \map j x + \map k ... | It can be observed that this is an example of a [[Linear Combination of Derivatives]] with $\lambda = \mu = 1$.
{{qed}} | Sum Rule for Derivatives/Proof 2 | https://proofwiki.org/wiki/Sum_Rule_for_Derivatives | https://proofwiki.org/wiki/Sum_Rule_for_Derivatives/Proof_2 | [
"Sum Rule for Derivatives",
"Derivatives",
"Differential Calculus"
] | [
"Definition:Real Function",
"Definition:Real Interval/Open",
"Definition:Differentiable Mapping/Real Function/Point",
"Definition:Differentiable Mapping/Real Function/Point",
"Definition:Derivative/Real Function/Derivative on Interval",
"Definition:Differentiable Mapping/Real Function/Interval"
] | [
"Linear Combination of Derivatives"
] |
proofwiki-430 | Quotient Mapping is Surjection | Let $S$ be a set.
Let $\RR$ be an equivalence relation on $S$.
Then the quotient mapping $q_\RR: S \to S / \RR$ is a surjection. | Suppose $S$ is empty.
<!-- The following sentence is annoying, normally we talk about the equclass of an *element* (under some relation).
Intuitively, it's really should be "the set of equclass is empty" and more or less directly go to "$S / \RR$ is empty", but it feel as if a step would be missed. -->
Then, vacuousl... | Let $S$ be a [[Definition:Set|set]].
Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on $S$.
Then the [[Definition:Quotient Mapping|quotient mapping]] $q_\RR: S \to S / \RR$ is a [[Definition:Surjection|surjection]]. | Suppose $S$ is [[Definition:Empty Set| empty]].
<!-- The following sentence is annoying, normally we talk about the equclass of an *element* (under some relation).
Intuitively, it's really should be "the set of equclass is empty" and more or less directly go to "$S / \RR$ is empty", but it feel as if a step would be ... | Quotient Mapping is Surjection | https://proofwiki.org/wiki/Quotient_Mapping_is_Surjection | https://proofwiki.org/wiki/Quotient_Mapping_is_Surjection | [
"Quotient Mappings",
"Surjections"
] | [
"Definition:Set",
"Definition:Equivalence Relation",
"Definition:Quotient Mapping",
"Definition:Surjection"
] | [
"Definition:Empty Set",
"Definition:Vacuous Truth",
"Definition:Equivalence Class",
"Definition:Set",
"Definition:Equivalence Class",
"Definition:Empty Set",
"Definition:Codomain",
"Definition:Empty Set",
"Definition:Vacuous Truth",
"Definition:Surjection",
"Definition:Non-Empty Set",
"Equival... |
proofwiki-431 | Trivial Quotient is a Bijection | Let $\Delta_S$ be the diagonal relation on a set $S$.
Let $q_{\Delta_S}: S \to S / \Delta_S$ be the trivial quotient of $S$.
Then $q_{\Delta_S}: S \to S / \Delta_S$ is a bijection. | The diagonal relation is defined as:
:$\Delta_S = \set {\tuple {x, x}: x \in S} \subseteq S \times S$
From the fact that $q_{\Delta_S}$ is a quotient mapping, we know that it is a surjection.
It relates each $x \in S$ to the singleton $\set x$.
Thus:
:$\set x = \eqclass x {\Delta_S} \subseteq S$
So $\map {q_{\Delta_S} ... | Let $\Delta_S$ be the [[Definition:Diagonal Relation|diagonal relation]] on a [[Definition:Set|set]] $S$.
Let $q_{\Delta_S}: S \to S / \Delta_S$ be the [[Definition:Trivial Quotient|trivial quotient of $S$]].
Then $q_{\Delta_S}: S \to S / \Delta_S$ is a [[Definition:Bijection|bijection]]. | The [[Definition:Diagonal Relation|diagonal relation]] is defined as:
:$\Delta_S = \set {\tuple {x, x}: x \in S} \subseteq S \times S$
From the fact that $q_{\Delta_S}$ is a [[Definition:Quotient Mapping|quotient mapping]], we know that it is a [[Quotient Mapping is Surjection|surjection]].
It relates each $x \in S... | Trivial Quotient is a Bijection | https://proofwiki.org/wiki/Trivial_Quotient_is_a_Bijection | https://proofwiki.org/wiki/Trivial_Quotient_is_a_Bijection | [
"Quotient Mappings",
"Bijections"
] | [
"Definition:Diagonal Relation",
"Definition:Set",
"Definition:Trivial Quotient",
"Definition:Bijection"
] | [
"Definition:Diagonal Relation",
"Definition:Quotient Mapping",
"Quotient Mapping is Surjection",
"Definition:Singleton",
"Definition:Injection",
"Definition:Bijection"
] |
proofwiki-432 | Composite of Quotient Mappings | Let $S$ be a set.
Let $\RR_1$ be an equivalence on $S$, and $\RR_2$ be an equivalence on the quotient set $S / \RR_1$.
We can find an equivalence $\RR_3$ on $S$ such that $\paren {S / \RR_1} / \RR_2$ is in one-to-one correspondence with $S / \RR_3$ under the mapping:
:$\phi: \paren {S / \RR_1} / \RR_2 \to S / \RR_3: \e... | Define $\RR_3$ to be the equivalence induced by:
:$x \mapsto \eqclass {\eqclass x {\RR_1} } {\RR_2}$
By definition of $\RR_3$:
:$\eqclass {\eqclass x {\RR_1} } {\RR_2} = \eqclass {\eqclass y {\RR_1} } {\RR_2} \implies \eqclass x {\RR_3} = \eqclass y {\RR_3}$
Therefore, $\phi$ is well-defined.
Again by definition of $\R... | Let $S$ be a [[Definition:Set|set]].
Let $\RR_1$ be an [[Definition:Equivalence Relation|equivalence]] on $S$, and $\RR_2$ be an [[Definition:Equivalence Relation|equivalence]] on the [[Definition:Quotient Set|quotient set]] $S / \RR_1$.
We can find an [[Definition:Equivalence Relation|equivalence]] $\RR_3$ on $S$ s... | Define $\RR_3$ to be the [[Definition:Equivalence Relation Induced by Mapping|equivalence induced by]]:
:$x \mapsto \eqclass {\eqclass x {\RR_1} } {\RR_2}$
By definition of $\RR_3$:
:$\eqclass {\eqclass x {\RR_1} } {\RR_2} = \eqclass {\eqclass y {\RR_1} } {\RR_2} \implies \eqclass x {\RR_3} = \eqclass y {\RR_3}$
Th... | Composite of Quotient Mappings | https://proofwiki.org/wiki/Composite_of_Quotient_Mappings | https://proofwiki.org/wiki/Composite_of_Quotient_Mappings | [
"Quotient Mappings",
"Composite Mappings"
] | [
"Definition:Set",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Quotient Set",
"Definition:Equivalence Relation",
"Definition:Bijection",
"Definition:Mapping"
] | [
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Well-Defined/Mapping",
"Definition:Injection",
"Definition:Surjection",
"Definition:Bijection"
] |
proofwiki-433 | Renaming Mapping is Well-Defined | Let $f: S \to T$ be a mapping.
Let $r: S / \RR_f \to \Img f$ be the renaming mapping, defined as:
:$r: S / \RR_f \to \Img f: \map r {\eqclass x {\RR_f} } = \map f x$
where:
:$\RR_f$ is the equivalence induced by the mapping $f$
:$S / \RR_f$ is the quotient set of $S$ determined by $\RR_f$
:$\eqclass x {\RR_f}$ is the ... | By Relation Induced by Mapping is Equivalence Relation, we have that $\RR_f$ is an equivalence relation.
To determine whether $r$ is well-defined, we have to determine whether $r: S / \RR_f \to \Img f$ actually defines a mapping at all.
Consider a typical element $\eqclass x {\RR_f}$ of $S / \RR_f$.
Suppose we were to ... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $r: S / \RR_f \to \Img f$ be the [[Definition:Renaming Mapping|renaming mapping]], defined as:
:$r: S / \RR_f \to \Img f: \map r {\eqclass x {\RR_f} } = \map f x$
where:
:$\RR_f$ is the [[Definition:Equivalence Relation Induced by Mapping|equivalence induced... | By [[Relation Induced by Mapping is Equivalence Relation]], we have that $\RR_f$ is an [[Definition:Equivalence Relation|equivalence relation]].
To determine whether $r$ is [[Definition:Well-Defined Mapping|well-defined]], we have to determine whether $r: S / \RR_f \to \Img f$ actually defines a [[Definition:Mapping|m... | Renaming Mapping is Well-Defined/Proof 1 | https://proofwiki.org/wiki/Renaming_Mapping_is_Well-Defined | https://proofwiki.org/wiki/Renaming_Mapping_is_Well-Defined/Proof_1 | [
"Mapping Theory",
"Quotient Sets",
"Equivalence Relations",
"Renaming Mapping is Well-Defined"
] | [
"Definition:Mapping",
"Definition:Renaming Mapping",
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Mapping",
"Definition:Quotient Set",
"Definition:Equivalence Class",
"Definition:Well-Defined/Mapping"
] | [
"Relation Induced by Mapping is Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Well-Defined/Mapping",
"Definition:Mapping",
"Definition:Singleton"
] |
proofwiki-434 | Renaming Mapping is Well-Defined | Let $f: S \to T$ be a mapping.
Let $r: S / \RR_f \to \Img f$ be the renaming mapping, defined as:
:$r: S / \RR_f \to \Img f: \map r {\eqclass x {\RR_f} } = \map f x$
where:
:$\RR_f$ is the equivalence induced by the mapping $f$
:$S / \RR_f$ is the quotient set of $S$ determined by $\RR_f$
:$\eqclass x {\RR_f}$ is the ... | From Condition for Mapping from Quotient Set to be Well-Defined:
:there exists a mapping $\phi: S / \RR \to T$ such that $\phi \circ q_\RR = f$
{{iff}}:
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \map f x = \map f y$
But by definition of the equivalence induced by the mapping $f$:
:$\forall x, y \in S: \tuple... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $r: S / \RR_f \to \Img f$ be the [[Definition:Renaming Mapping|renaming mapping]], defined as:
:$r: S / \RR_f \to \Img f: \map r {\eqclass x {\RR_f} } = \map f x$
where:
:$\RR_f$ is the [[Definition:Equivalence Relation Induced by Mapping|equivalence induced... | From [[Condition for Mapping from Quotient Set to be Well-Defined]]:
:there exists a [[Definition:Mapping|mapping]] $\phi: S / \RR \to T$ such that $\phi \circ q_\RR = f$
{{iff}}:
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \map f x = \map f y$
But by definition of the [[Definition:Equivalence Relation Induc... | Renaming Mapping is Well-Defined/Proof 2 | https://proofwiki.org/wiki/Renaming_Mapping_is_Well-Defined | https://proofwiki.org/wiki/Renaming_Mapping_is_Well-Defined/Proof_2 | [
"Mapping Theory",
"Quotient Sets",
"Equivalence Relations",
"Renaming Mapping is Well-Defined"
] | [
"Definition:Mapping",
"Definition:Renaming Mapping",
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Mapping",
"Definition:Quotient Set",
"Definition:Equivalence Class",
"Definition:Well-Defined/Mapping"
] | [
"Condition for Mapping from Quotient Set to be Well-Defined",
"Definition:Mapping",
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Mapping"
] |
proofwiki-435 | Renaming Mapping is Bijection | Let $f: S \to T$ be a mapping.
Let $r: S / \RR_f \to \Img f$ be the renaming mapping, defined as:
:$r: S / \RR_f \to \Img f: \map r {\eqclass x {\RR_f} } = \map f x$
where:
:$\RR_f$ is the equivalence induced by the mapping $f$
:$S / \RR_f$ is the quotient set of $S$ determined by $\RR_f$
:$\eqclass x {\RR_f}$ is the e... | === Proof of Injectivity ===
To show that $r: S / \RR_f \to \Img f$ is an injection:
{{begin-eqn}}
{{eqn | l = \map r {\eqclass x {\RR_f} }
| r = \map r {\eqclass y {\RR_f} }
| c =
}}
{{eqn | ll= \leadsto
| l = \map f x
| r = \map f y
| c = {{Defof|Renaming Mapping}}
}}
{{eqn | ll= \leads... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $r: S / \RR_f \to \Img f$ be the [[Definition:Renaming Mapping|renaming mapping]], defined as:
:$r: S / \RR_f \to \Img f: \map r {\eqclass x {\RR_f} } = \map f x$
where:
:$\RR_f$ is the [[Definition:Equivalence Relation Induced by Mapping|equivalence induced]... | === Proof of Injectivity ===
To show that $r: S / \RR_f \to \Img f$ is an [[Definition:Injection|injection]]:
{{begin-eqn}}
{{eqn | l = \map r {\eqclass x {\RR_f} }
| r = \map r {\eqclass y {\RR_f} }
| c =
}}
{{eqn | ll= \leadsto
| l = \map f x
| r = \map f y
| c = {{Defof|Renaming Mapp... | Renaming Mapping is Bijection/Proof 1 | https://proofwiki.org/wiki/Renaming_Mapping_is_Bijection | https://proofwiki.org/wiki/Renaming_Mapping_is_Bijection/Proof_1 | [
"Equivalence Relations",
"Bijections",
"Quotient Sets",
"Renaming Mapping is Bijection"
] | [
"Definition:Mapping",
"Definition:Renaming Mapping",
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Mapping",
"Definition:Quotient Set",
"Definition:Equivalence Class",
"Definition:Renaming Mapping",
"Definition:Bijection"
] | [
"Definition:Injection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Mapping",
"Definition:Surjection",
"Restriction of Mapping to Image is Surjection",
"Equivalence Class is not Empty",
"Definition:Surjection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Bijecti... |
proofwiki-436 | Renaming Mapping is Bijection | Let $f: S \to T$ be a mapping.
Let $r: S / \RR_f \to \Img f$ be the renaming mapping, defined as:
:$r: S / \RR_f \to \Img f: \map r {\eqclass x {\RR_f} } = \map f x$
where:
:$\RR_f$ is the equivalence induced by the mapping $f$
:$S / \RR_f$ is the quotient set of $S$ determined by $\RR_f$
:$\eqclass x {\RR_f}$ is the e... | From Renaming Mapping is Well-Defined, $r: S / \RR_f \to \Img f$ is a well-defined mapping.
By definition, $\RR_f$ is the equivalence relation induced by the mapping $f$.
Hence by definition:
:$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$
From Condition for Mapping from Quotient Set to be Injection, th... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $r: S / \RR_f \to \Img f$ be the [[Definition:Renaming Mapping|renaming mapping]], defined as:
:$r: S / \RR_f \to \Img f: \map r {\eqclass x {\RR_f} } = \map f x$
where:
:$\RR_f$ is the [[Definition:Equivalence Relation Induced by Mapping|equivalence induced]... | From [[Renaming Mapping is Well-Defined]], $r: S / \RR_f \to \Img f$ is a [[Definition:Well-Defined Mapping|well-defined mapping]].
By definition, $\RR_f$ is the [[Definition:Equivalence Relation Induced by Mapping|equivalence relation induced]] by the [[Definition:Mapping|mapping]] $f$.
Hence by definition:
:$\tuple... | Renaming Mapping is Bijection/Proof 2 | https://proofwiki.org/wiki/Renaming_Mapping_is_Bijection | https://proofwiki.org/wiki/Renaming_Mapping_is_Bijection/Proof_2 | [
"Equivalence Relations",
"Bijections",
"Quotient Sets",
"Renaming Mapping is Bijection"
] | [
"Definition:Mapping",
"Definition:Renaming Mapping",
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Mapping",
"Definition:Quotient Set",
"Definition:Equivalence Class",
"Definition:Renaming Mapping",
"Definition:Bijection"
] | [
"Renaming Mapping is Well-Defined",
"Definition:Well-Defined/Mapping",
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Mapping",
"Condition for Mapping from Quotient Set to be Injection",
"Definition:Injection",
"Definition:Codomain (Set Theory)/Mapping",
"Restriction of Mapping to I... |
proofwiki-437 | Composite of Mapping with Inverse | Let $f: S \to T$ be a mapping.
Then:
:$\forall x \in S: \map {f^{-1} \circ f} x = \eqclass x {\RR_f}$
where:
:$\RR_f$ is the equivalence induced by $f$
:$\eqclass x {\RR_f}$ is the $\RR_f$-equivalence class of $x$. | Let $y = \map f x$.
Then by the definition of induced equivalence:
:$x \in \eqclass x {\RR_f}$
By the definition of the inverse of a mapping:
:$f^{-1} = \set {\tuple {y, x}: \tuple {x, y} \in f}$
Thus:
:$\eqclass x {\RR_f} = \set {s \in \Dom f: \map f s = \map f x}$
By definition:
:$\map {f^{-1} } y = \eqclass x {\RR_f... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$\forall x \in S: \map {f^{-1} \circ f} x = \eqclass x {\RR_f}$
where:
:$\RR_f$ is the [[Definition:Equivalence Relation Induced by Mapping|equivalence induced by $f$]]
:$\eqclass x {\RR_f}$ is the [[Definition:Equivalence Class|$\RR_f$-equivalence class of... | Let $y = \map f x$.
Then by the definition of [[Definition:Equivalence Relation Induced by Mapping|induced equivalence]]:
:$x \in \eqclass x {\RR_f}$
By the definition of the [[Definition:Inverse of Mapping|inverse of a mapping]]:
:$f^{-1} = \set {\tuple {y, x}: \tuple {x, y} \in f}$
Thus:
:$\eqclass x {\RR_f} = \se... | Composite of Mapping with Inverse | https://proofwiki.org/wiki/Composite_of_Mapping_with_Inverse | https://proofwiki.org/wiki/Composite_of_Mapping_with_Inverse | [
"Equivalence Relations",
"Inverse Mappings",
"Composite Mappings"
] | [
"Definition:Mapping",
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Equivalence Class"
] | [
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Inverse of Mapping"
] |
proofwiki-438 | Factoring Mapping into Surjection and Inclusion | Every mapping $f:S \to T$ can be uniquely '''factored into''' a surjection $g$ followed by the inclusion mapping $i_T$.
That is, $f = i_T \circ g$ where:
:$g: S \to \Img f: \map g x = \map f x$
:$i_T: \Img f \to T: \map {i_T} x = x$
This can be illustrated using a commutative diagram as follows:
$\quad\quad \begin{xy}\... | From Surjection by Restriction of Codomain, $f: S \to \Img f$ is a surjection.
The mapping $g: S \to \Img f$ where $\map g x = \map f x$ is therefore also clearly a surjection.
The mapping $g: S \to \Img f: \map g x = \map f x$ is clearly unique, by Equality of Mappings.
From Inclusion Mapping is Injection, $i_T: \Img ... | Every [[Definition:Mapping|mapping]] $f:S \to T$ can be uniquely '''factored into''' a [[Definition:Surjection|surjection]] $g$ followed by the [[Definition:Inclusion Mapping|inclusion mapping]] $i_T$.
That is, $f = i_T \circ g$ where:
:$g: S \to \Img f: \map g x = \map f x$
:$i_T: \Img f \to T: \map {i_T} x = x$
... | From [[Surjection by Restriction of Codomain]], $f: S \to \Img f$ is a [[Definition:Surjection|surjection]].
The [[Definition:Mapping|mapping]] $g: S \to \Img f$ where $\map g x = \map f x$ is therefore also clearly a [[Definition:Surjection|surjection]].
The [[Definition:Mapping|mapping]] $g: S \to \Img f: \map g x ... | Factoring Mapping into Surjection and Inclusion | https://proofwiki.org/wiki/Factoring_Mapping_into_Surjection_and_Inclusion | https://proofwiki.org/wiki/Factoring_Mapping_into_Surjection_and_Inclusion | [
"Surjections",
"Inclusion Mappings"
] | [
"Definition:Mapping",
"Definition:Surjection",
"Definition:Inclusion Mapping",
"Definition:Commutative Diagram"
] | [
"Restriction of Mapping to Image is Surjection",
"Definition:Surjection",
"Definition:Mapping",
"Definition:Surjection",
"Definition:Mapping",
"Equality of Mappings",
"Inclusion Mapping is Injection",
"Definition:Injection"
] |
proofwiki-439 | Quotient Theorem for Surjections | Let $f: S \to T$ be a surjection.
Then there is one and only one bijection $r: S / \RR_f \to T$ such that:
:$r \circ q_{\RR_f} = f$
where:
:$\RR_f$ is the equivalence induced by $f$
:$r: S / \RR_f \to T$ is the renaming mapping
:$q_{\RR_f}: S \to S / \RR_f$ is the quotient mapping induced by $\RR_f$.
This can be illust... | From the definition of Induced Equivalence, the mapping $f: S \to T$ induces an equivalence $\RR_f$ on its domain.
As $f: S \to T$ is a surjection, $T = \Img f$ by definition.
From Renaming Mapping is Bijection, the renaming mapping $r: S / \RR_f \to T$ is a bijection, where $S / \RR_f$ is the quotient set of $S$ by $\... | Let $f: S \to T$ be a [[Definition:Surjection|surjection]].
Then there is one and only one [[Definition:Bijection|bijection]] $r: S / \RR_f \to T$ such that:
:$r \circ q_{\RR_f} = f$
where:
:$\RR_f$ is the [[Definition:Equivalence Relation Induced by Mapping|equivalence induced by $f$]]
:$r: S / \RR_f \to T$ is the... | From the definition of [[Definition:Equivalence Relation Induced by Mapping|Induced Equivalence]], the [[Definition:Mapping|mapping]] $f: S \to T$ induces an [[Definition:Equivalence Relation|equivalence]] $\RR_f$ on its [[Definition:Domain of Mapping|domain]].
As $f: S \to T$ is a [[Definition:Surjection|surjection]]... | Quotient Theorem for Surjections | https://proofwiki.org/wiki/Quotient_Theorem_for_Surjections | https://proofwiki.org/wiki/Quotient_Theorem_for_Surjections | [
"Surjections",
"Quotient Sets",
"Quotient Mappings",
"Quotient Theorems",
"Named Theorems"
] | [
"Definition:Surjection",
"Definition:Bijection",
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Renaming Mapping",
"Definition:Quotient Mapping",
"Definition:Commutative Diagram"
] | [
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Mapping",
"Definition:Equivalence Relation",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Surjection",
"Renaming Mapping is Bijection",
"Definition:Renaming Mapping",
"Definition:Bijection",
"Definition:Quotient Set"
] |
proofwiki-440 | Quotient Theorem for Sets | A mapping $f: S \to T$ can be uniquely '''factored into''' a surjection, followed by a bijection, followed by an injection.
Thus:
:$f = i \circ r \circ q_{\RR_f}$
where:
{{begin-eqn}}
{{eqn | ll= q_{\RR_f}:
| lo= S \to S / \RR_f:
| l = \map {q_{\RR_f} } s
| r = \eqclass s {\RR_f}
| c = Quotient ... | From Factoring Mapping into Surjection and Inclusion, $f$ can be factored uniquely into:
:A surjection $g: S \to \Img f$, followed by:
:The inclusion mapping $i: \Img f \to T$ (an injection).
$\quad\quad \begin{xy}\xymatrix@L+2mu@+1em {
S \ar[drdr]_*{g}
\ar@{-->}[rr]^*{f = i_T \circ g}
& & T
\\ \\
& & \Img f \ar[uu]_*{... | A [[Definition:Mapping|mapping]] $f: S \to T$ can be [[Definition:Unique|uniquely]] '''factored into''' a [[Definition:Surjection|surjection]], followed by a [[Definition:Bijection|bijection]], followed by an [[Definition:Injection|injection]].
Thus:
:$f = i \circ r \circ q_{\RR_f}$
where:
{{begin-eqn}}
{{eqn | ll= q... | From [[Factoring Mapping into Surjection and Inclusion]], $f$ can be factored uniquely into:
:A [[Definition:Surjection|surjection]] $g: S \to \Img f$, followed by:
:The [[Definition:Inclusion Mapping|inclusion mapping]] $i: \Img f \to T$ (an [[Definition:Injection|injection]]).
$\quad\quad \begin{xy}\xymatrix@L+2mu... | Quotient Theorem for Sets/Proof | https://proofwiki.org/wiki/Quotient_Theorem_for_Sets | https://proofwiki.org/wiki/Quotient_Theorem_for_Sets/Proof | [
"Mapping Theory",
"Quotient Mappings",
"Quotient Sets",
"Named Theorems",
"Quotient Theorems",
"Quotient Theorem for Sets"
] | [
"Definition:Mapping",
"Definition:Unique",
"Definition:Surjection",
"Definition:Bijection",
"Definition:Injection",
"Definition:Quotient Mapping",
"Definition:Renaming Mapping",
"Definition:Inclusion Mapping",
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Quotient Set",
"Defi... | [
"Factoring Mapping into Surjection and Inclusion",
"Definition:Surjection",
"Definition:Inclusion Mapping",
"Definition:Injection",
"Quotient Theorem for Surjections",
"Definition:Surjection",
"Definition:Quotient Mapping",
"Definition:Surjection",
"Definition:Renaming Mapping",
"Definition:Biject... |
proofwiki-441 | Empty Set is Element of Power Set | The empty set is an element of all power sets:
:$\forall S: \O \in \powerset S$ | {{begin-eqn}}
{{eqn | q = \forall S
| l = \O
| o = \subseteq
| r = S
| c = Empty Set is Subset of All Sets
}}
{{eqn | ll= \leadsto
| q = \forall S
| l = \O
| o = \in
| r = \powerset S
| c = {{Defof|Power Set}}
}}
{{end-eqn}}
{{qed}} | The [[Definition:Empty Set|empty set]] is an [[Definition:Element|element]] of all [[Definition:Power Set|power sets]]:
:$\forall S: \O \in \powerset S$ | {{begin-eqn}}
{{eqn | q = \forall S
| l = \O
| o = \subseteq
| r = S
| c = [[Empty Set is Subset of All Sets]]
}}
{{eqn | ll= \leadsto
| q = \forall S
| l = \O
| o = \in
| r = \powerset S
| c = {{Defof|Power Set}}
}}
{{end-eqn}}
{{qed}} | Empty Set is Element of Power Set | https://proofwiki.org/wiki/Empty_Set_is_Element_of_Power_Set | https://proofwiki.org/wiki/Empty_Set_is_Element_of_Power_Set | [
"Empty Set",
"Power Set"
] | [
"Definition:Empty Set",
"Definition:Element",
"Definition:Power Set"
] | [
"Empty Set is Subset of All Sets"
] |
proofwiki-442 | Direct Image Mapping of Relation is Mapping | Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Let $\RR^\to: \powerset S \to \powerset T$ be the direct image mapping of $\RR$:
:$\forall X \in \powerset S: \map {\RR^\to} X = \set {t \in T: \exists s \in X: \tuple {s, t} \in \RR}$
Then $\RR^\to$ is indeed a mapping. | Take the general relation $\RR \subseteq S \times T$.
Let $X \subseteq S$, that is $X \in \powerset S$.
Suppose $X = \O$.
Then from Image of Empty Set is Empty Set:
:$\map {\RR^\to} X = \O \subseteq T$
Suppose $X = S$.
Then from Image is Subset of Codomain: Corollary 1:
:$\map {\RR^\to} X = \Img \RR \subseteq T$
Finall... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] on $S \times T$.
Let $\RR^\to: \powerset S \to \powerset T$ be the [[Definition:Direct Image Mapping of Relation|direct image mapping]] of $\RR$:
:$\forall X \in \powerset S: \map {\RR^\to} X = \set {t \i... | Take the general relation $\RR \subseteq S \times T$.
Let $X \subseteq S$, that is $X \in \powerset S$.
Suppose $X = \O$.
Then from [[Image of Empty Set is Empty Set]]:
:$\map {\RR^\to} X = \O \subseteq T$
Suppose $X = S$.
Then from [[Image is Subset of Codomain/Corollary 1|Image is Subset of Codomain: Corollary... | Direct Image Mapping of Relation is Mapping | https://proofwiki.org/wiki/Direct_Image_Mapping_of_Relation_is_Mapping | https://proofwiki.org/wiki/Direct_Image_Mapping_of_Relation_is_Mapping | [
"Direct Image Mappings"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Direct Image Mapping/Relation",
"Definition:Mapping"
] | [
"Image of Empty Set is Empty Set",
"Image is Subset of Codomain/Corollary 1",
"Image is Subset of Codomain",
"Definition:Power Set",
"Definition:Many-to-One Relation",
"Definition:Mapping"
] |
proofwiki-443 | Inverse Image Mapping of Relation is Mapping | Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Let $\RR^\gets$ be the inverse image mapping of $\RR$:
:$\RR^\gets: \powerset T \to \powerset S: \map {\RR^\gets} Y = \RR^{-1} \sqbrk Y$
Then $\RR^\gets$ is indeed a mapping. | $\RR^{-1}$, being a relation, obeys the same laws as $\RR$.
So Direct Image Mapping of Relation is Mapping applies directly.
{{qed}} | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] on $S \times T$.
Let $\RR^\gets$ be the [[Definition:Inverse Image Mapping of Relation|inverse image mapping]] of $\RR$:
:$\RR^\gets: \powerset T \to \powerset S: \map {\RR^\gets} Y = \RR^{-1} \sqbrk Y$
... | $\RR^{-1}$, being a [[Definition:Relation|relation]], obeys the same laws as $\RR$.
So [[Direct Image Mapping of Relation is Mapping]] applies directly.
{{qed}} | Inverse Image Mapping of Relation is Mapping | https://proofwiki.org/wiki/Inverse_Image_Mapping_of_Relation_is_Mapping | https://proofwiki.org/wiki/Inverse_Image_Mapping_of_Relation_is_Mapping | [
"Inverse Image Mappings"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Inverse Image Mapping/Relation",
"Definition:Mapping"
] | [
"Definition:Relation",
"Direct Image Mapping of Relation is Mapping"
] |
proofwiki-444 | Direct Image Mapping of Surjection is Surjection | Let $f: S \to T$ be a surjection.
Then the direct image mapping of $f$:
:$f^\to: \powerset S \to \powerset T$
is a surjection. | Let $f: S \to T$ be a surjection.
Then:
:$\forall y \in T: \exists x \in S: f \paren x = y$
From the Quotient Theorem for Surjections, there is one and only one bijection $r: S / \RR_f \to T$ such that $r \circ q_{\RR_f} = f$.
Each element of $S / \RR_f$ is a subset of $S$ and therefore an element of $\powerset S$.
Thu... | Let $f: S \to T$ be a [[Definition:Surjection|surjection]].
Then the [[Definition:Direct Image Mapping of Mapping|direct image mapping]] of $f$:
:$f^\to: \powerset S \to \powerset T$
is a [[Definition:Surjection|surjection]]. | Let $f: S \to T$ be a [[Definition:Surjection|surjection]].
Then:
:$\forall y \in T: \exists x \in S: f \paren x = y$
From the [[Quotient Theorem for Surjections]], there is [[Definition:Unique|one and only one]] [[Definition:Bijection|bijection]] $r: S / \RR_f \to T$ such that $r \circ q_{\RR_f} = f$.
Each [[Defini... | Direct Image Mapping of Surjection is Surjection/Proof 1 | https://proofwiki.org/wiki/Direct_Image_Mapping_of_Surjection_is_Surjection | https://proofwiki.org/wiki/Direct_Image_Mapping_of_Surjection_is_Surjection/Proof_1 | [
"Surjections",
"Direct Image Mappings",
"Power Set",
"Direct Image Mapping of Surjection is Surjection"
] | [
"Definition:Surjection",
"Definition:Direct Image Mapping/Mapping",
"Definition:Surjection"
] | [
"Definition:Surjection",
"Quotient Theorem for Surjections",
"Definition:Unique",
"Definition:Bijection",
"Definition:Element",
"Definition:Subset",
"Definition:Element",
"Definition:Bijection",
"Definition:A Fortiori",
"Definition:Injection",
"Definition:Surjection",
"Definition:Unique",
"D... |
proofwiki-445 | Direct Image Mapping of Surjection is Surjection | Let $f: S \to T$ be a surjection.
Then the direct image mapping of $f$:
:$f^\to: \powerset S \to \powerset T$
is a surjection. | Let $f: S \to T$ be a surjection.
By definition, $f^\to$ is defined by sending subsets of $S$ to their image under $f$.
That is:
:$\forall X \subseteq S: \map {f^\to} X = \set{\map f x: x \in X} \subseteq T$
To prove that $f^\to$ is a surjection, we need to show that every subset of $T$ is the image under $f^\to$ of so... | Let $f: S \to T$ be a [[Definition:Surjection|surjection]].
Then the [[Definition:Direct Image Mapping of Mapping|direct image mapping]] of $f$:
:$f^\to: \powerset S \to \powerset T$
is a [[Definition:Surjection|surjection]]. | Let $f: S \to T$ be a [[Definition:Surjection|surjection]].
By definition, $f^\to$ is defined by sending [[Definition:Subset|subsets]] of $S$ to their [[Definition:Image of Subset under Mapping|image]] under $f$.
That is:
:$\forall X \subseteq S: \map {f^\to} X = \set{\map f x: x \in X} \subseteq T$
To prove that ... | Direct Image Mapping of Surjection is Surjection/Proof 2 | https://proofwiki.org/wiki/Direct_Image_Mapping_of_Surjection_is_Surjection | https://proofwiki.org/wiki/Direct_Image_Mapping_of_Surjection_is_Surjection/Proof_2 | [
"Surjections",
"Direct Image Mappings",
"Power Set",
"Direct Image Mapping of Surjection is Surjection"
] | [
"Definition:Surjection",
"Definition:Direct Image Mapping/Mapping",
"Definition:Surjection"
] | [
"Definition:Surjection",
"Definition:Subset",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Surjection",
"Definition:Subset",
"Definition:Image (Set Theory)/Mapping/Element",
"Definition:Subset",
"Definition:Surjection",
"Definition:Surjection"
] |
proofwiki-446 | Direct Image Mapping of Surjection is Surjection | Let $f: S \to T$ be a surjection.
Then the direct image mapping of $f$:
:$f^\to: \powerset S \to \powerset T$
is a surjection. | Let $f^\gets$ be the inverse image mapping of $f$.
Let $Y \in \powerset T$.
Let $X = \map {f^\gets} Y$.
By Subset equals Image of Preimage iff Mapping is Surjection:
:$\map {f^\to} X = Y$
As such an $X$ exists for each $Y \in \powerset S$, $f^\to$ is surjective.
{{qed}} | Let $f: S \to T$ be a [[Definition:Surjection|surjection]].
Then the [[Definition:Direct Image Mapping of Mapping|direct image mapping]] of $f$:
:$f^\to: \powerset S \to \powerset T$
is a [[Definition:Surjection|surjection]]. | Let $f^\gets$ be the [[Definition:Inverse Image Mapping of Mapping|inverse image mapping]] of $f$.
Let $Y \in \powerset T$.
Let $X = \map {f^\gets} Y$.
By [[Subset equals Image of Preimage iff Mapping is Surjection]]:
:$\map {f^\to} X = Y$
As such an $X$ exists for each $Y \in \powerset S$, $f^\to$ is [[Definition:... | Direct Image Mapping of Surjection is Surjection/Proof 3 | https://proofwiki.org/wiki/Direct_Image_Mapping_of_Surjection_is_Surjection | https://proofwiki.org/wiki/Direct_Image_Mapping_of_Surjection_is_Surjection/Proof_3 | [
"Surjections",
"Direct Image Mappings",
"Power Set",
"Direct Image Mapping of Surjection is Surjection"
] | [
"Definition:Surjection",
"Definition:Direct Image Mapping/Mapping",
"Definition:Surjection"
] | [
"Definition:Inverse Image Mapping/Mapping",
"Subset equals Image of Preimage iff Mapping is Surjection",
"Definition:Surjection"
] |
proofwiki-447 | Cantor's Theorem | There is no surjection from a set $S$ to its power set for any set $S$.
That is, $S$ is strictly smaller than its power set. | We temporarily introduce the notation:
:<nowiki>$a^n = \begin{cases}
a : & n = 0 \\
\set {a^{n - 1} } : & n > 0
\end{cases}$</nowiki>
where $a \in S$.
Thus $a^n$ consists of a single element $a^{n - 1} \in \map {\PP^{n - 1} } S$.
Let there be a bijection $f: S \to \QQ^n$ where $\QQ^n \subseteq \map {\PP^n} S$.
Then def... | There is no [[Definition:Surjection|surjection]] from a [[Definition:Set|set]] $S$ to its [[Definition:Power Set|power set]] for any set $S$.
That is, $S$ is [[Definition:Strictly Smaller Set|strictly smaller]] than its [[Definition:Power Set|power set]]. | We temporarily introduce the notation:
:<nowiki>$a^n = \begin{cases}
a : & n = 0 \\
\set {a^{n - 1} } : & n > 0
\end{cases}$</nowiki>
where $a \in S$.
Thus $a^n$ consists of a single element $a^{n - 1} \in \map {\PP^{n - 1} } S$.
Let there be a [[Definition:Bijection|bijection]] $f: S \to \QQ^n$ where $\QQ^n \subsete... | Cantor's Theorem (Strong Version)/Proof 1 | https://proofwiki.org/wiki/Cantor's_Theorem | https://proofwiki.org/wiki/Cantor's_Theorem_(Strong_Version)/Proof_1 | [
"Cantor's Theorem",
"Surjections",
"Power Set"
] | [
"Definition:Surjection",
"Definition:Set",
"Definition:Power Set",
"Definition:Strictly Smaller Set",
"Definition:Power Set"
] | [
"Definition:Bijection",
"Definition:Image (Set Theory)/Mapping/Element",
"Definition:Element",
"Definition:Proper Subset"
] |
proofwiki-448 | Cantor's Theorem | There is no surjection from a set $S$ to its power set for any set $S$.
That is, $S$ is strictly smaller than its power set. | The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:There is no surjection from $S$ onto $\map {\PP^n} S$.
=== Basis for the Induction ===
$\map P 1$ is Cantor's Theorem.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ ... | There is no [[Definition:Surjection|surjection]] from a [[Definition:Set|set]] $S$ to its [[Definition:Power Set|power set]] for any set $S$.
That is, $S$ is [[Definition:Strictly Smaller Set|strictly smaller]] than its [[Definition:Power Set|power set]]. | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:There is no [[Definition:Surjection|surjection]] from $S$ onto $\map {\PP^n} S$.
=== Basis for the Induction ===
$\map P 1$ is [[Cantor's Theorem]].
This... | Cantor's Theorem (Strong Version)/Proof 2 | https://proofwiki.org/wiki/Cantor's_Theorem | https://proofwiki.org/wiki/Cantor's_Theorem_(Strong_Version)/Proof_2 | [
"Cantor's Theorem",
"Surjections",
"Power Set"
] | [
"Definition:Surjection",
"Definition:Set",
"Definition:Power Set",
"Definition:Strictly Smaller Set",
"Definition:Power Set"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Surjection",
"Cantor's Theorem",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Surjection",
"Definition:Surjection",
"Cantor's Theorem (Strong Version)/Proof 2/Induction Step"
] |
proofwiki-449 | Cantor's Theorem | There is no surjection from a set $S$ to its power set for any set $S$.
That is, $S$ is strictly smaller than its power set. | By Power Set is Nonempty, $\powerset {\map {\PP^k} S}$ is non-empty.
By definition:
:$\map {\PP^{k + 1} } S = \powerset {\map {\PP^k} S}$
Then:
:$\map {\PP^{k + 1} } S \ne \O$
By Law of Excluded Middle, there are two choices:
:$S = \O$
or:
:$S \ne \O$
Suppose that $S = \O$.
By {{Corollary|Image of Empty Set is Empty S... | There is no [[Definition:Surjection|surjection]] from a [[Definition:Set|set]] $S$ to its [[Definition:Power Set|power set]] for any set $S$.
That is, $S$ is [[Definition:Strictly Smaller Set|strictly smaller]] than its [[Definition:Power Set|power set]]. | By [[Power Set is Nonempty]], $\powerset {\map {\PP^k} S}$ is [[Definition:Non-Empty Set|non-empty]].
By definition:
:$\map {\PP^{k + 1} } S = \powerset {\map {\PP^k} S}$
Then:
:$\map {\PP^{k + 1} } S \ne \O$
By [[Law of Excluded Middle]], there are two choices:
:$S = \O$
or:
:$S \ne \O$
Suppose that $S = \O$.
... | Cantor's Theorem (Strong Version)/Proof 2/Induction Step/Proof 2 | https://proofwiki.org/wiki/Cantor's_Theorem | https://proofwiki.org/wiki/Cantor's_Theorem_(Strong_Version)/Proof_2/Induction_Step/Proof_2 | [
"Cantor's Theorem",
"Surjections",
"Power Set"
] | [
"Definition:Surjection",
"Definition:Set",
"Definition:Power Set",
"Definition:Strictly Smaller Set",
"Definition:Power Set"
] | [
"Power Set is Nonempty",
"Definition:Non-Empty Set",
"Law of Excluded Middle",
"Definition:Surjection",
"Cantor's Theorem (Strong Version)/Proof 2",
"Definition:Surjection",
"Injection from Set to Power Set",
"Definition:Injection",
"Injection has Surjective Left Inverse Mapping",
"Definition:Surj... |
proofwiki-450 | Cantor's Theorem | There is no surjection from a set $S$ to its power set for any set $S$.
That is, $S$ is strictly smaller than its power set. | {{AimForCont}} $S$ is a set with a surjection $f: S \to \powerset S$.
Then:
{{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map f x
| o = \in
| r = \powerset S
| c = {{Hypothesis}}
}}
{{eqn | ll= \leadsto
| q = \forall x \in S
| l = \map f x
| o = \subseteq
| r = S
... | There is no [[Definition:Surjection|surjection]] from a [[Definition:Set|set]] $S$ to its [[Definition:Power Set|power set]] for any set $S$.
That is, $S$ is [[Definition:Strictly Smaller Set|strictly smaller]] than its [[Definition:Power Set|power set]]. | {{AimForCont}} $S$ is a [[Definition:Set|set]] with a [[Definition:Surjection|surjection]] $f: S \to \powerset S$.
Then:
{{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map f x
| o = \in
| r = \powerset S
| c = {{Hypothesis}}
}}
{{eqn | ll= \leadsto
| q = \forall x \in S
| l = \ma... | Cantor's Theorem/Proof 1 | https://proofwiki.org/wiki/Cantor's_Theorem | https://proofwiki.org/wiki/Cantor's_Theorem/Proof_1 | [
"Cantor's Theorem",
"Surjections",
"Power Set"
] | [
"Definition:Surjection",
"Definition:Set",
"Definition:Power Set",
"Definition:Strictly Smaller Set",
"Definition:Power Set"
] | [
"Definition:Set",
"Definition:Surjection",
"Law of Excluded Middle",
"Definition:Surjection",
"Proof by Contradiction",
"Definition:Surjection"
] |
proofwiki-451 | Cantor's Theorem | There is no surjection from a set $S$ to its power set for any set $S$.
That is, $S$ is strictly smaller than its power set. | Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $f: S \to \powerset S$ be a mapping.
Let $T = \set {x \in S: \neg \paren {x \in \map f x} }$.
Then $T \subseteq S$, so $T \in \powerset S$ by the definition of power set.
We will show that $T$ is not in the image of $f$ and therefore $f$ is not surjective... | There is no [[Definition:Surjection|surjection]] from a [[Definition:Set|set]] $S$ to its [[Definition:Power Set|power set]] for any set $S$.
That is, $S$ is [[Definition:Strictly Smaller Set|strictly smaller]] than its [[Definition:Power Set|power set]]. | Let $S$ be a [[Definition:Set|set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $f: S \to \powerset S$ be a [[Definition:Mapping|mapping]].
Let $T = \set {x \in S: \neg \paren {x \in \map f x} }$.
Then $T \subseteq S$, so $T \in \powerset S$ by the definition of [[Definition:Power Set|p... | Cantor's Theorem/Proof 2 | https://proofwiki.org/wiki/Cantor's_Theorem | https://proofwiki.org/wiki/Cantor's_Theorem/Proof_2 | [
"Cantor's Theorem",
"Surjections",
"Power Set"
] | [
"Definition:Surjection",
"Definition:Set",
"Definition:Power Set",
"Definition:Strictly Smaller Set",
"Definition:Power Set"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Mapping",
"Definition:Power Set",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Surjection",
"Rule of Implication",
"Rule of Implication",
"Non-Equivalence of Proposition and Negation/Formulation 1",
"Definition:Contradiction",
... |
proofwiki-452 | Cantor's Theorem | There is no surjection from a set $S$ to its power set for any set $S$.
That is, $S$ is strictly smaller than its power set. | Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $f: S \to \powerset S$ be a mapping.
Let $T = \set {x \in S: \neg \paren {x \in \map f x} }$.
The set $T$ exists by the {{axiom-link|Specification}}.
Then:
:$T \subseteq S$
so $T \in \powerset S$ by the definition of power set.
We will show that $T$ is no... | There is no [[Definition:Surjection|surjection]] from a [[Definition:Set|set]] $S$ to its [[Definition:Power Set|power set]] for any set $S$.
That is, $S$ is [[Definition:Strictly Smaller Set|strictly smaller]] than its [[Definition:Power Set|power set]]. | Let $S$ be a [[Definition:Set|set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $f: S \to \powerset S$ be a [[Definition:Mapping|mapping]].
Let $T = \set {x \in S: \neg \paren {x \in \map f x} }$.
The set $T$ exists by the {{axiom-link|Specification}}.
Then:
:$T \subseteq S$
so $T \in ... | Cantor's Theorem/Proof 3 | https://proofwiki.org/wiki/Cantor's_Theorem | https://proofwiki.org/wiki/Cantor's_Theorem/Proof_3 | [
"Cantor's Theorem",
"Surjections",
"Power Set"
] | [
"Definition:Surjection",
"Definition:Set",
"Definition:Power Set",
"Definition:Strictly Smaller Set",
"Definition:Power Set"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Mapping",
"Definition:Power Set",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Surjection",
"Definition:Contradiction",
"Proof by Contradiction",
"Principle of Non-Contradiction",
"Definition:Contradiction",
"Proof by Contradi... |
proofwiki-453 | Surjection Induced by Powerset is Induced by Surjection | Let $\RR \subseteq S \times T$ be a relation.
Let $\RR^\to: \powerset S \to \powerset T$ be the direct image mapping $\RR$.
Let $\RR^\to$ be a surjection.
Let $X = \Preimg \RR$, that is, the preimage of $\RR$.
Then $\RR {\restriction_X} \subseteq X \times T$, that is, the restriction of $\RR$ to $X$, is a surjection. | Let $X$ be the preimage of $\RR$.
Suppose $\RR {\restriction_X} \subseteq X \times T$ is a mapping, but not a surjection.
Then:
:$\exists y \in T: \neg \exists x \in S: \tuple {x, y} \in \RR$
Because no element of $S$ relates to $y$, no subset of $S$ contains any element of $S$ that relates to the subset $\set y \subse... | Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Let $\RR^\to: \powerset S \to \powerset T$ be the [[Definition:Direct Image Mapping of Relation|direct image mapping]] $\RR$.
Let $\RR^\to$ be a [[Definition:Surjection|surjection]].
Let $X = \Preimg \RR$, that is, the [[Definition:Preimage of Rel... | Let $X$ be the [[Definition:Preimage of Relation|preimage]] of $\RR$.
Suppose $\RR {\restriction_X} \subseteq X \times T$ is a [[Definition:Mapping|mapping]], but not a [[Definition:Surjection|surjection]].
Then:
:$\exists y \in T: \neg \exists x \in S: \tuple {x, y} \in \RR$
Because no [[Definition:Element|element... | Surjection Induced by Powerset is Induced by Surjection | https://proofwiki.org/wiki/Surjection_Induced_by_Powerset_is_Induced_by_Surjection | https://proofwiki.org/wiki/Surjection_Induced_by_Powerset_is_Induced_by_Surjection | [
"Surjections",
"Direct Image Mappings"
] | [
"Definition:Relation",
"Definition:Direct Image Mapping/Relation",
"Definition:Surjection",
"Definition:Preimage/Relation/Relation",
"Definition:Restriction/Relation",
"Definition:Surjection"
] | [
"Definition:Preimage/Relation/Relation",
"Definition:Mapping",
"Definition:Surjection",
"Definition:Element",
"Definition:Subset",
"Definition:Element",
"Definition:Subset",
"Definition:Surjection",
"Definition:Mapping",
"Definition:Mapping",
"Definition:Preimage/Relation/Relation",
"Definitio... |
proofwiki-454 | Mapping is Injection iff Direct Image Mapping is Injection | Let $f: S \to T$ be a mapping.
Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$.
Then:
:$f^\to$ is an injection
{{iff}}
:$f: S \to T$ is also an injection. | === Necessary Condition ===
Suppose $f: S \to T$ is a mapping, but not injective.
Then:
:$\exists x_1 \ne x_2 \in S: \map f {x_1} = \map f {x_2} = y$
Let:
:$X_1 = \set {x_1}$
:$X_2 = \set {x_2}$
:$Y = \set y$
Then it follows that:
:$\map {f^\to} {X_1} = \map {f^\to} {X_2} = Y$
Thus $f^\to: \powerset S \to \powerset T$ ... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $f^\to: \powerset S \to \powerset T$ be the [[Definition:Direct Image Mapping of Mapping|direct image mapping]] of $f$.
Then:
:$f^\to$ is an [[Definition:Injection|injection]]
{{iff}}
:$f: S \to T$ is also an [[Definition:Injection|injection]]. | === Necessary Condition ===
Suppose $f: S \to T$ is a [[Definition:Mapping|mapping]], but not [[Definition:Injection|injective]].
Then:
:$\exists x_1 \ne x_2 \in S: \map f {x_1} = \map f {x_2} = y$
Let:
:$X_1 = \set {x_1}$
:$X_2 = \set {x_2}$
:$Y = \set y$
Then it follows that:
:$\map {f^\to} {X_1} = \map {f^\to} {... | Mapping is Injection iff Direct Image Mapping is Injection | https://proofwiki.org/wiki/Mapping_is_Injection_iff_Direct_Image_Mapping_is_Injection | https://proofwiki.org/wiki/Mapping_is_Injection_iff_Direct_Image_Mapping_is_Injection | [
"Injections",
"Direct Image Mappings"
] | [
"Definition:Mapping",
"Definition:Direct Image Mapping/Mapping",
"Definition:Injection",
"Definition:Injection"
] | [
"Definition:Mapping",
"Definition:Injection",
"Definition:Injection",
"Rule of Transposition",
"Definition:Injection",
"Definition:Mapping",
"Definition:Injection",
"Definition:Injection",
"Definition:Injection",
"Rule of Transposition"
] |
proofwiki-455 | Direct Image Mapping is Bijection iff Mapping is Bijection | Let $\RR \subseteq S \times T$ be a relation.
Let $\RR^\to: \powerset S \to \powerset T$ be the direct image mapping of $\RR$.
Then $\RR \subseteq S \times T$ is a bijection {{iff}} $\RR^\to: \powerset S \to \powerset T$ is a bijection. | {{ProofWanted}}
Category:Bijections
Category:Direct Image Mappings
jhcdarz7hd2e162xhx0dlkuai318jn8 | Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Let $\RR^\to: \powerset S \to \powerset T$ be the [[Definition:Direct Image Mapping of Relation|direct image mapping]] of $\RR$.
Then $\RR \subseteq S \times T$ is a [[Definition:Bijection|bijection]] {{iff}} $\RR^\to: \powerset S \to \powerset T$... | {{ProofWanted}}
[[Category:Bijections]]
[[Category:Direct Image Mappings]]
jhcdarz7hd2e162xhx0dlkuai318jn8 | Direct Image Mapping is Bijection iff Mapping is Bijection | https://proofwiki.org/wiki/Direct_Image_Mapping_is_Bijection_iff_Mapping_is_Bijection | https://proofwiki.org/wiki/Direct_Image_Mapping_is_Bijection_iff_Mapping_is_Bijection | [
"Bijections",
"Direct Image Mappings"
] | [
"Definition:Relation",
"Definition:Direct Image Mapping/Relation",
"Definition:Bijection",
"Definition:Bijection"
] | [
"Category:Bijections",
"Category:Direct Image Mappings"
] |
proofwiki-456 | No Bijection from Set to its Power Set | Let $S$ be a set.
Let $\powerset S$ denote the power set of $S$.
There is no bijection $f: S \to \powerset S$. | A bijection is by its definition also a surjection.
By Cantor's Theorem there is no surjection from $S$ to $\powerset S$.
Hence the result.
{{Qed}} | Let $S$ be a [[Definition:Set|set]].
Let $\powerset S$ denote the [[Definition:Power Set|power set]] of $S$.
There is no [[Definition:Bijection|bijection]] $f: S \to \powerset S$. | A [[Definition:Bijection|bijection]] is by its definition also a [[Definition:Surjection|surjection]].
By [[Cantor's Theorem]] there is no [[Definition:Surjection|surjection]] from $S$ to $\powerset S$.
Hence the result.
{{Qed}} | No Bijection from Set to its Power Set | https://proofwiki.org/wiki/No_Bijection_from_Set_to_its_Power_Set | https://proofwiki.org/wiki/No_Bijection_from_Set_to_its_Power_Set | [
"Power Set",
"Bijections"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Bijection"
] | [
"Definition:Bijection",
"Definition:Surjection",
"Cantor's Theorem",
"Definition:Surjection"
] |
proofwiki-457 | Set is Equivalent to Proper Subset of Power Set | Every set is equivalent to a proper subset of its power set:
:$\forall S: \exists T \subset \powerset S: S \sim T$ | To show equivalence between two sets, we need to demonstrate that a bijection exists between them.
We will now define such a bijection.
Let $T = \set {\set x: x \in S}$.
{{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \set x
| o = \subseteq
| r = S
| c = {{Defof|Subset}}
}}
{{eqn | ll= \leadsto
... | Every [[Definition:Set|set]] is [[Definition:Set Equivalence|equivalent]] to a [[Definition:Proper Subset|proper subset]] of its [[Definition:Power Set|power set]]:
:$\forall S: \exists T \subset \powerset S: S \sim T$ | To show [[Definition:Set Equivalence|equivalence]] between two [[Definition:Set|sets]], we need to demonstrate that a [[Definition:Bijection|bijection]] exists between them.
We will now define such a [[Definition:Bijection|bijection]].
Let $T = \set {\set x: x \in S}$.
{{begin-eqn}}
{{eqn | q = \forall x \in S
... | Set is Equivalent to Proper Subset of Power Set | https://proofwiki.org/wiki/Set_is_Equivalent_to_Proper_Subset_of_Power_Set | https://proofwiki.org/wiki/Set_is_Equivalent_to_Proper_Subset_of_Power_Set | [
"Power Set",
"Proper Subsets"
] | [
"Definition:Set",
"Definition:Set Equivalence",
"Definition:Proper Subset",
"Definition:Power Set"
] | [
"Definition:Set Equivalence",
"Definition:Set",
"Definition:Bijection",
"Definition:Bijection",
"Definition:Injection",
"Definition:Set Equality",
"Definition:Surjection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Bijection",
"Definition:Proper Subset",
"Empty Set is Element... |
proofwiki-458 | Union of Power Sets | The union of the power sets of two sets $S$ and $T$ is a subset of the power set of their union:
:$\powerset S \cup \powerset T \subseteq \powerset {S \cup T}$ | {{begin-eqn}}
{{eqn | l = X
| o = \in
| r = \powerset S \cup \powerset T
| c =
}}
{{eqn | ll= \leadsto
| l = X
| o = \subseteq
| r = S \lor X \subseteq T
| c = {{Defof|Set Union}} and {{Defof|Power Set}}
}}
{{eqn | ll= \leadsto
| l = X
| o = \subseteq
| r = S... | The [[Definition:Set Union|union]] of the [[Definition:Power Set|power sets]] of two [[Definition:Set|sets]] $S$ and $T$ is a [[Definition:Subset|subset]] of the [[Definition:Power Set|power set]] of their [[Definition:Set Union|union]]:
:$\powerset S \cup \powerset T \subseteq \powerset {S \cup T}$ | {{begin-eqn}}
{{eqn | l = X
| o = \in
| r = \powerset S \cup \powerset T
| c =
}}
{{eqn | ll= \leadsto
| l = X
| o = \subseteq
| r = S \lor X \subseteq T
| c = {{Defof|Set Union}} and {{Defof|Power Set}}
}}
{{eqn | ll= \leadsto
| l = X
| o = \subseteq
| r = S... | Union of Power Sets | https://proofwiki.org/wiki/Union_of_Power_Sets | https://proofwiki.org/wiki/Union_of_Power_Sets | [
"Power Set",
"Set Union",
"Subsets"
] | [
"Definition:Set Union",
"Definition:Power Set",
"Definition:Set",
"Definition:Subset",
"Definition:Power Set",
"Definition:Set Union"
] | [] |
proofwiki-459 | Intersection of Power Sets | The intersection of the power sets of two sets $S$ and $T$ is equal to the power set of their intersection:
:$\powerset S \cap \powerset T = \powerset {S \cap T}$ | {{begin-eqn}}
{{eqn | l = X
| o = \in
| r = \powerset {S \cap T}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = X
| o = \subseteq
| r = S \cap T
| c = {{Defof|Power Set}}
}}
{{eqn | ll= \leadstoandfrom
| l = X
| o = \subseteq
| r = S \land X \subseteq T
| c... | The [[Definition:Set Intersection|intersection]] of the [[Definition:Power Set|power sets]] of two [[Definition:Set|sets]] $S$ and $T$ is equal to the [[Definition:Power Set|power set]] of their [[Definition:Set Intersection|intersection]]:
:$\powerset S \cap \powerset T = \powerset {S \cap T}$ | {{begin-eqn}}
{{eqn | l = X
| o = \in
| r = \powerset {S \cap T}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = X
| o = \subseteq
| r = S \cap T
| c = {{Defof|Power Set}}
}}
{{eqn | ll= \leadstoandfrom
| l = X
| o = \subseteq
| r = S \land X \subseteq T
| c... | Intersection of Power Sets | https://proofwiki.org/wiki/Intersection_of_Power_Sets | https://proofwiki.org/wiki/Intersection_of_Power_Sets | [
"Power Set",
"Set Intersection"
] | [
"Definition:Set Intersection",
"Definition:Power Set",
"Definition:Set",
"Definition:Power Set",
"Definition:Set Intersection"
] | [] |
proofwiki-460 | Cantor-Bernstein-Schröder Theorem | If a subset of one set is equivalent to the other, and a subset of the other is equivalent to the first, then the two sets are themselves equivalent:
:$\forall S, T: T \sim S_1 \subseteq S \land S \sim T_1 \subseteq T \implies S \sim T$ | Recursively define a sequence $\langle C_n \rangle$ in the power set of $S$ as follows:
:$C_0 = S \setminus T$, the difference between $S$ and $T$.
:$C_{n + 1} = f \sqbrk {C_n}$, the image of $C_n$ under $f$.
Let $\ds C := \bigcup_{n \mathop \in \N} C_n$.
Define a mapping $h: S \to T$ as follows:
:$\map h x = \begin {c... | If a [[Definition:Subset|subset]] of one [[Definition:Set|set]] is [[Definition:Set Equivalence|equivalent]] to the other, and a [[Definition:Subset|subset]] of the other is [[Definition:Set Equivalence|equivalent]] to the first, then the two [[Definition:Set|sets]] are themselves [[Definition:Set Equivalence|equivalen... | [[Principle of Recursive Definition|Recursively]] define a [[Definition:Sequence|sequence]] $\langle C_n \rangle$ in the [[Definition:Power Set|power set]] of $S$ as follows:
:$C_0 = S \setminus T$, the [[Definition:Set Difference|difference]] between $S$ and $T$.
:$C_{n + 1} = f \sqbrk {C_n}$, the [[Definition:Image ... | Cantor-Bernstein-Schröder Theorem/Lemma/Proof 1 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Lemma/Proof_1 | [
"Cantor-Bernstein-Schröder Theorem",
"Set Theory",
"Injections",
"Bijections",
"Cardinals"
] | [
"Definition:Subset",
"Definition:Set",
"Definition:Set Equivalence",
"Definition:Subset",
"Definition:Set Equivalence",
"Definition:Set",
"Definition:Set Equivalence"
] | [
"Principle of Recursive Definition",
"Definition:Sequence",
"Definition:Power Set",
"Definition:Set Difference",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Mapping",
"Definition:Mapping",
"Law of Excluded Middle",
"Definition:Mapping",
"Set is Subset of Union/Family of Sets",
"... |
proofwiki-461 | Cantor-Bernstein-Schröder Theorem | If a subset of one set is equivalent to the other, and a subset of the other is equivalent to the first, then the two sets are themselves equivalent:
:$\forall S, T: T \sim S_1 \subseteq S \land S \sim T_1 \subseteq T \implies S \sim T$ | Define a mapping $E: \powerset S \to \powerset S$ as:
:$\map E K = S \setminus \paren {T \setminus f \sqbrk K}$
where $f \sqbrk K$ is the image of $K$ under $f$.
Then:
:$\map E K = \paren {S \setminus T} \cup f \sqbrk K$
By Image of Subset under Relation is Subset of Image and {{Corollary|Set Union Preserves Subsets}},... | If a [[Definition:Subset|subset]] of one [[Definition:Set|set]] is [[Definition:Set Equivalence|equivalent]] to the other, and a [[Definition:Subset|subset]] of the other is [[Definition:Set Equivalence|equivalent]] to the first, then the two [[Definition:Set|sets]] are themselves [[Definition:Set Equivalence|equivalen... | Define a mapping $E: \powerset S \to \powerset S$ as:
:$\map E K = S \setminus \paren {T \setminus f \sqbrk K}$
where $f \sqbrk K$ is the [[Definition:Image of Subset under Mapping|image of $K$ under $f$]].
Then:
:$\map E K = \paren {S \setminus T} \cup f \sqbrk K$
By [[Image of Subset under Relation is Subset of I... | Cantor-Bernstein-Schröder Theorem/Lemma/Proof 2 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Lemma/Proof_2 | [
"Cantor-Bernstein-Schröder Theorem",
"Set Theory",
"Injections",
"Bijections",
"Cardinals"
] | [
"Definition:Subset",
"Definition:Set",
"Definition:Set Equivalence",
"Definition:Subset",
"Definition:Set Equivalence",
"Definition:Set",
"Definition:Set Equivalence"
] | [
"Definition:Image (Set Theory)/Mapping/Subset",
"Image of Subset under Relation is Subset of Image",
"Definition:Increasing/Mapping",
"Knaster-Tarski Lemma/Power Set",
"Definition:Fixed Point",
"Definition:Fixed Point",
"Definition:Set Difference",
"Definition:Restriction/Mapping",
"Injection to Ima... |
proofwiki-462 | Cantor-Bernstein-Schröder Theorem | If a subset of one set is equivalent to the other, and a subset of the other is equivalent to the first, then the two sets are themselves equivalent:
:$\forall S, T: T \sim S_1 \subseteq S \land S \sim T_1 \subseteq T \implies S \sim T$ | Define $C = \{f^k(x) | k \in \N, x \in S \setminus T \}$.
Clearly, $C = C_0 \cup C_1$, where:
$C_0 = S \setminus T$, the difference between $S$ and $T$,
$C_1 = \{f^k(x) | k \in \N_{> 0}, x \in S \setminus T \}$.
Note, that $S \setminus C_0 = S \setminus (S \setminus T) = S \cap T = T$ (use this theorem and $T \subsete... | If a [[Definition:Subset|subset]] of one [[Definition:Set|set]] is [[Definition:Set Equivalence|equivalent]] to the other, and a [[Definition:Subset|subset]] of the other is [[Definition:Set Equivalence|equivalent]] to the first, then the two [[Definition:Set|sets]] are themselves [[Definition:Set Equivalence|equivalen... | Define $C = \{f^k(x) | k \in \N, x \in S \setminus T \}$.
Clearly, $C = C_0 \cup C_1$, where:
$C_0 = S \setminus T$, the [[Definition:Set Difference|difference]] between $S$ and $T$,
$C_1 = \{f^k(x) | k \in \N_{> 0}, x \in S \setminus T \}$.
Note, that $S \setminus C_0 = S \setminus (S \setminus T) = S \cap T = T$... | Cantor-Bernstein-Schröder Theorem/Lemma/Proof 3 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Lemma/Proof_3 | [
"Cantor-Bernstein-Schröder Theorem",
"Set Theory",
"Injections",
"Bijections",
"Cardinals"
] | [
"Definition:Subset",
"Definition:Set",
"Definition:Set Equivalence",
"Definition:Subset",
"Definition:Set Equivalence",
"Definition:Set",
"Definition:Set Equivalence"
] | [
"Definition:Set Difference",
"Set Difference with Set Difference",
"Definition:Identity Mapping",
"Restriction of Injection is Injection"
] |
proofwiki-463 | Cantor-Bernstein-Schröder Theorem | If a subset of one set is equivalent to the other, and a subset of the other is equivalent to the first, then the two sets are themselves equivalent:
:$\forall S, T: T \sim S_1 \subseteq S \land S \sim T_1 \subseteq T \implies S \sim T$ | From the facts that $T \sim S_1$ and $S \sim T_1$, we can set up the two bijections:
:$f: S \to T_1$
:$g: T \to S_1$
Let:
:$S_2 = g \sqbrk {f \sqbrk S} = g \sqbrk {T_1} \subseteq S_1$
and:
:$T_2 = f \sqbrk {g \sqbrk T} = f \sqbrk {S_1} \subseteq T_1$
So $S_2 \subseteq S_1$ and $S_2 \sim S$, while $T_2 \subseteq T_1$, a... | If a [[Definition:Subset|subset]] of one [[Definition:Set|set]] is [[Definition:Set Equivalence|equivalent]] to the other, and a [[Definition:Subset|subset]] of the other is [[Definition:Set Equivalence|equivalent]] to the first, then the two [[Definition:Set|sets]] are themselves [[Definition:Set Equivalence|equivalen... | From the facts that $T \sim S_1$ and $S \sim T_1$, we can set up the two [[Definition:Bijection|bijections]]:
:$f: S \to T_1$
:$g: T \to S_1$
Let:
:$S_2 = g \sqbrk {f \sqbrk S} = g \sqbrk {T_1} \subseteq S_1$
and:
:$T_2 = f \sqbrk {g \sqbrk T} = f \sqbrk {S_1} \subseteq T_1$
So $S_2 \subseteq S_1$ and $S_2 \sim S... | Cantor-Bernstein-Schröder Theorem/Proof 1 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Proof_1 | [
"Cantor-Bernstein-Schröder Theorem",
"Set Theory",
"Injections",
"Bijections",
"Cardinals"
] | [
"Definition:Subset",
"Definition:Set",
"Definition:Set Equivalence",
"Definition:Subset",
"Definition:Set Equivalence",
"Definition:Set",
"Definition:Set Equivalence"
] | [
"Definition:Bijection",
"Definition:Natural Numbers",
"Definition:Set Difference",
"Definition:Bijection"
] |
proofwiki-464 | Cantor-Bernstein-Schröder Theorem | If a subset of one set is equivalent to the other, and a subset of the other is equivalent to the first, then the two sets are themselves equivalent:
:$\forall S, T: T \sim S_1 \subseteq S \land S \sim T_1 \subseteq T \implies S \sim T$ | Suppose $S \preccurlyeq T$ and $T \preccurlyeq S$.
By definition, we have that there exist injections $f: S \to T$ and $g: T \to S$.
We are going to try to build a sequence $t_1, s_1, t_2, s_2, t_3 \ldots$ as follows.
Consider any $t_1 \in T$.
By Law of Excluded Middle there are two choices:
{{begin-eqn}}
{{eqn | q = \... | If a [[Definition:Subset|subset]] of one [[Definition:Set|set]] is [[Definition:Set Equivalence|equivalent]] to the other, and a [[Definition:Subset|subset]] of the other is [[Definition:Set Equivalence|equivalent]] to the first, then the two [[Definition:Set|sets]] are themselves [[Definition:Set Equivalence|equivalen... | Suppose $S \preccurlyeq T$ and $T \preccurlyeq S$.
By [[Definition:Dominate (Set Theory)|definition]], we have that there exist [[Definition:Injection|injections]] $f: S \to T$ and $g: T \to S$.
We are going to try to build a [[Definition:Sequence|sequence]] $t_1, s_1, t_2, s_2, t_3 \ldots$ as follows.
Consider any... | Cantor-Bernstein-Schröder Theorem/Proof 2 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Proof_2 | [
"Cantor-Bernstein-Schröder Theorem",
"Set Theory",
"Injections",
"Bijections",
"Cardinals"
] | [
"Definition:Subset",
"Definition:Set",
"Definition:Set Equivalence",
"Definition:Subset",
"Definition:Set Equivalence",
"Definition:Set",
"Definition:Set Equivalence"
] | [
"Definition:Dominate (Set Theory)",
"Definition:Injection",
"Definition:Sequence",
"Law of Excluded Middle",
"Definition:Injection",
"Law of Excluded Middle",
"Definition:Injection",
"Definition:Surjection",
"Definition:Surjection",
"Definition:Set Partition",
"Definition:Subset",
"Definition:... |
proofwiki-465 | Cantor-Bernstein-Schröder Theorem | If a subset of one set is equivalent to the other, and a subset of the other is equivalent to the first, then the two sets are themselves equivalent:
:$\forall S, T: T \sim S_1 \subseteq S \land S \sim T_1 \subseteq T \implies S \sim T$ | Let $S, T$ be sets.
Let $\powerset S, \powerset T$ denote their power sets.
Let $f: S \to T$ and $g: T \to S$ be injections that we know to exist between $S$ and $T$.
Consider the relative complements of elements of $\powerset S$ and $\powerset T$ as mappings:
:$\complement_S: \powerset S \to \powerset S: \forall X \in... | If a [[Definition:Subset|subset]] of one [[Definition:Set|set]] is [[Definition:Set Equivalence|equivalent]] to the other, and a [[Definition:Subset|subset]] of the other is [[Definition:Set Equivalence|equivalent]] to the first, then the two [[Definition:Set|sets]] are themselves [[Definition:Set Equivalence|equivalen... | Let $S, T$ be [[Definition:Set|sets]].
Let $\powerset S, \powerset T$ denote their [[Definition:Power Set|power sets]].
Let $f: S \to T$ and $g: T \to S$ be [[Definition:Injection|injections]] that we know to exist between $S$ and $T$.
Consider the [[Definition:Relative Complement|relative complements]] of elements... | Cantor-Bernstein-Schröder Theorem/Proof 3 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Proof_3 | [
"Cantor-Bernstein-Schröder Theorem",
"Set Theory",
"Injections",
"Bijections",
"Cardinals"
] | [
"Definition:Subset",
"Definition:Set",
"Definition:Set Equivalence",
"Definition:Subset",
"Definition:Set Equivalence",
"Definition:Set",
"Definition:Set Equivalence"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Injection",
"Definition:Relative Complement",
"Definition:Mapping",
"Definition:Relative Complement",
"Definition:Direct Image Mapping/Mapping",
"Definition:Mapping",
"Definition:Composition of Mappings",
"Image of Subset under Mapping is Subse... |
proofwiki-466 | Cantor-Bernstein-Schröder Theorem | If a subset of one set is equivalent to the other, and a subset of the other is equivalent to the first, then the two sets are themselves equivalent:
:$\forall S, T: T \sim S_1 \subseteq S \land S \sim T_1 \subseteq T \implies S \sim T$ | From the facts that $T \sim S_1$ and $S \sim T_1$, we can set up the two injections:
:$f \sqbrk S = T_1 \subseteq T$
:$g \sqbrk T = S_1 \subseteq S$
Using $f$ and $g$, $S$ and $T$ are divided into disjoint subsets such that there exists a bijection between the subsets of $S$ and those of $T$, as follows.
Let $a \in S$.... | If a [[Definition:Subset|subset]] of one [[Definition:Set|set]] is [[Definition:Set Equivalence|equivalent]] to the other, and a [[Definition:Subset|subset]] of the other is [[Definition:Set Equivalence|equivalent]] to the first, then the two [[Definition:Set|sets]] are themselves [[Definition:Set Equivalence|equivalen... | From the facts that $T \sim S_1$ and $S \sim T_1$, we can set up the two [[Definition:Injection|injections]]:
:$f \sqbrk S = T_1 \subseteq T$
:$g \sqbrk T = S_1 \subseteq S$
Using $f$ and $g$, $S$ and $T$ are divided into [[Definition:Disjoint Sets|disjoint subsets]] such that there exists a [[Definition:Bijection|b... | Cantor-Bernstein-Schröder Theorem/Proof 4 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Proof_4 | [
"Cantor-Bernstein-Schröder Theorem",
"Set Theory",
"Injections",
"Bijections",
"Cardinals"
] | [
"Definition:Subset",
"Definition:Set",
"Definition:Set Equivalence",
"Definition:Subset",
"Definition:Set Equivalence",
"Definition:Set",
"Definition:Set Equivalence"
] | [
"Definition:Injection",
"Definition:Disjoint Sets",
"Definition:Bijection",
"Definition:Subset",
"Definition:Element",
"Definition:Element",
"Principle of Recursive Definition",
"Definition:Negative/Integer",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Definition:Element... |
proofwiki-467 | Cantor-Bernstein-Schröder Theorem | If a subset of one set is equivalent to the other, and a subset of the other is equivalent to the first, then the two sets are themselves equivalent:
:$\forall S, T: T \sim S_1 \subseteq S \land S \sim T_1 \subseteq T \implies S \sim T$ | By Injection to Image is Bijection:
:$g: T \to g \sqbrk T$ is a bijection.
Thus $T$ is equivalent to $g \sqbrk T$.
By Composite of Injections is Injection $g \circ f: S \to g \sqbrk T \subset S$ is also an injection (to a subset of the domain of $g \circ f$).
Then by Cantor-Bernstein-Schröder Theorem: Lemma:
:There exi... | If a [[Definition:Subset|subset]] of one [[Definition:Set|set]] is [[Definition:Set Equivalence|equivalent]] to the other, and a [[Definition:Subset|subset]] of the other is [[Definition:Set Equivalence|equivalent]] to the first, then the two [[Definition:Set|sets]] are themselves [[Definition:Set Equivalence|equivalen... | By [[Injection to Image is Bijection]]:
:$g: T \to g \sqbrk T$ is a [[Definition:Bijection|bijection]].
Thus $T$ is [[Definition:Set Equivalence|equivalent]] to $g \sqbrk T$.
By [[Composite of Injections is Injection]] $g \circ f: S \to g \sqbrk T \subset S$ is also an [[Definition:Injection|injection]] (to a [[Defi... | Cantor-Bernstein-Schröder Theorem/Proof 5 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Proof_5 | [
"Cantor-Bernstein-Schröder Theorem",
"Set Theory",
"Injections",
"Bijections",
"Cardinals"
] | [
"Definition:Subset",
"Definition:Set",
"Definition:Set Equivalence",
"Definition:Subset",
"Definition:Set Equivalence",
"Definition:Set",
"Definition:Set Equivalence"
] | [
"Injection to Image is Bijection",
"Definition:Bijection",
"Definition:Set Equivalence",
"Composite of Injections is Injection",
"Definition:Injection",
"Definition:Subset",
"Definition:Domain (Set Theory)/Mapping",
"Cantor-Bernstein-Schröder Theorem/Lemma",
"Definition:Bijection",
"Definition:Set... |
proofwiki-468 | Cantor-Bernstein-Schröder Theorem | If a subset of one set is equivalent to the other, and a subset of the other is equivalent to the first, then the two sets are themselves equivalent:
:$\forall S, T: T \sim S_1 \subseteq S \land S \sim T_1 \subseteq T \implies S \sim T$ | Let $\powerset A$ be the power set of $A$.
Define a mapping $E: \powerset A \to \powerset A$ thus:
:$\map E S = A \setminus g \sqbrk {B \setminus f \sqbrk S}$
=== $E$ is increasing ===
Let $S, T \in \powerset A$ such that $S \subseteq T$.
Then:
{{begin-eqn}}
{{eqn | l = f \sqbrk S
| o = \subseteq
| r = f \s... | If a [[Definition:Subset|subset]] of one [[Definition:Set|set]] is [[Definition:Set Equivalence|equivalent]] to the other, and a [[Definition:Subset|subset]] of the other is [[Definition:Set Equivalence|equivalent]] to the first, then the two [[Definition:Set|sets]] are themselves [[Definition:Set Equivalence|equivalen... | Let $\powerset A$ be the [[Definition:Power Set|power set]] of $A$.
Define a mapping $E: \powerset A \to \powerset A$ thus:
:$\map E S = A \setminus g \sqbrk {B \setminus f \sqbrk S}$
=== $E$ is increasing ===
Let $S, T \in \powerset A$ such that $S \subseteq T$.
Then:
{{begin-eqn}}
{{eqn | l = f \sqbrk S
... | Cantor-Bernstein-Schröder Theorem/Proof 6 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Proof_6 | [
"Cantor-Bernstein-Schröder Theorem",
"Set Theory",
"Injections",
"Bijections",
"Cardinals"
] | [
"Definition:Subset",
"Definition:Set",
"Definition:Set Equivalence",
"Definition:Subset",
"Definition:Set Equivalence",
"Definition:Set",
"Definition:Set Equivalence"
] | [
"Definition:Power Set",
"Image of Subset under Relation is Subset of Image",
"Set Difference with Subset is Superset of Set Difference",
"Image of Subset under Relation is Subset of Image",
"Set Difference with Subset is Superset of Set Difference",
"Knaster-Tarski Lemma/Power Set",
"Definition:Fixed Po... |
proofwiki-469 | Trivial Ordering is Universally Compatible | Let $S$ be a set.
Let $\RR$ be the trivial ordering on $S$.
Then $\RR$ is universally compatible. | To prove that the trivial ordering is in fact an ordering, we need to checking each of the criteria for an ordering: | Let $S$ be a [[Definition:Set|set]].
Let $\RR$ be the [[Definition:Trivial Ordering|trivial ordering]] on $S$.
Then $\RR$ is [[Definition:Universally Compatible Relation|universally compatible]]. | To prove that the trivial ordering is in fact an ordering, we need to checking each of the criteria for an [[Definition:Ordering|ordering]]: | Trivial Ordering is Universally Compatible | https://proofwiki.org/wiki/Trivial_Ordering_is_Universally_Compatible | https://proofwiki.org/wiki/Trivial_Ordering_is_Universally_Compatible | [
"Order Theory"
] | [
"Definition:Set",
"Definition:Trivial Ordering",
"Definition:Universally Compatible Relation"
] | [
"Definition:Ordering"
] |
proofwiki-470 | Identity Mapping is Order Isomorphism | Let $\struct {S, \preceq}$ be an ordered set.
The identity mapping $I_S$ is an order isomorphism from $\struct {S, \preceq}$ to itself. | By definition:
:$\forall x \in S: \map {I_S} x = x$
So:
:$x \preceq y \implies \map {I_S} x \preceq \map {I_S} y$
As $I_S$ is a bijection, we also have:
:$\map {I_S^{-1} } x = x$
So:
:$x \preceq y \implies \map {I_S^{-1} } x \preceq \map {I_S^{-1} } y$
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
The [[Definition:Identity Mapping|identity mapping]] $I_S$ is an [[Definition:Order Isomorphism|order isomorphism]] from $\struct {S, \preceq}$ to itself. | By definition:
:$\forall x \in S: \map {I_S} x = x$
So:
:$x \preceq y \implies \map {I_S} x \preceq \map {I_S} y$
As $I_S$ is a [[Identity Mapping is Bijection|bijection]], we also have:
:$\map {I_S^{-1} } x = x$
So:
:$x \preceq y \implies \map {I_S^{-1} } x \preceq \map {I_S^{-1} } y$
{{qed}} | Identity Mapping is Order Isomorphism/Proof 1 | https://proofwiki.org/wiki/Identity_Mapping_is_Order_Isomorphism | https://proofwiki.org/wiki/Identity_Mapping_is_Order_Isomorphism/Proof_1 | [
"Examples of Order Isomorphisms",
"Identity Mappings",
"Identity Mapping is Order Isomorphism"
] | [
"Definition:Ordered Set",
"Definition:Identity Mapping",
"Definition:Order Isomorphism"
] | [
"Identity Mapping is Bijection"
] |
proofwiki-471 | Identity Mapping is Order Isomorphism | Let $\struct {S, \preceq}$ be an ordered set.
The identity mapping $I_S$ is an order isomorphism from $\struct {S, \preceq}$ to itself. | An ordered set is a relational structure where order isomorphism is a special case of relation isomorphism.
The result follows directly from Identity Mapping is Relation Isomorphism.
{{Qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
The [[Definition:Identity Mapping|identity mapping]] $I_S$ is an [[Definition:Order Isomorphism|order isomorphism]] from $\struct {S, \preceq}$ to itself. | An [[Definition:Ordered Set|ordered set]] is a [[Definition:Relational Structure|relational structure]] where [[Definition:Order Isomorphism|order isomorphism]] is a special case of [[Definition:Relation Isomorphism|relation isomorphism]].
The result follows directly from [[Identity Mapping is Relation Isomorphism]].
... | Identity Mapping is Order Isomorphism/Proof 2 | https://proofwiki.org/wiki/Identity_Mapping_is_Order_Isomorphism | https://proofwiki.org/wiki/Identity_Mapping_is_Order_Isomorphism/Proof_2 | [
"Examples of Order Isomorphisms",
"Identity Mappings",
"Identity Mapping is Order Isomorphism"
] | [
"Definition:Ordered Set",
"Definition:Identity Mapping",
"Definition:Order Isomorphism"
] | [
"Definition:Ordered Set",
"Definition:Relational Structure",
"Definition:Order Isomorphism",
"Definition:Relation Isomorphism",
"Identity Mapping is Relation Isomorphism"
] |
proofwiki-472 | Inverse of Order Isomorphism is Order Isomorphism | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $\phi$ be a bijection from $\struct {S, \preceq_1}$ to $\struct {T, \preceq_2}$.
Then:
:$\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$
is an order isomorphism {{iff}}:
:$\phi^{-1}: \struct {T, \preceq_2} \to \struct {S, \preceq_1}... | Follows directly from the definition of order isomorphism.
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be an order isomorphism.
Then:
:$\forall x, y, \in S: x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$
and:
:$\forall \map \phi x, \map \phi y \in T: \map \phi x \preceq_2 \map \phi y \impli... | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $\phi$ be a [[Definition:Bijection|bijection]] from $\struct {S, \preceq_1}$ to $\struct {T, \preceq_2}$.
Then:
:$\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$
is an [[Definition:Order Isomorphism|ord... | Follows directly from the definition of [[Definition:Order Isomorphism|order isomorphism]].
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be an [[Definition:Order Isomorphism|order isomorphism]].
Then:
:$\forall x, y, \in S: x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$
and:
:$\forall \ma... | Inverse of Order Isomorphism is Order Isomorphism | https://proofwiki.org/wiki/Inverse_of_Order_Isomorphism_is_Order_Isomorphism | https://proofwiki.org/wiki/Inverse_of_Order_Isomorphism_is_Order_Isomorphism | [
"Order Isomorphisms"
] | [
"Definition:Ordered Set",
"Definition:Bijection",
"Definition:Order Isomorphism",
"Definition:Order Isomorphism"
] | [
"Definition:Order Isomorphism",
"Definition:Order Isomorphism"
] |
proofwiki-473 | Composite of Order Isomorphisms is Order Isomorphism | Let $\struct {S_1, \preceq_1}$, $\struct {S_2, \preceq_2}$ and $\struct {S_3, \preceq_3}$ be ordered sets.
Let:
:$\phi: \struct {S_1, \preceq_1} \to \struct {S_2, \preceq_2}$
and:
:$\psi: \struct {S_2, \preceq_2} \to \struct {S_3, \preceq_3}$
be order isomorphisms.
Then $\psi \circ \phi: \struct {S_1, \preceq_1} \to \s... | From Composite of Bijections is Bijection, $\psi \circ \phi$ is a bijection, as, by definition, an order isomorphism is also a bijection.
From Inverse of Composite Bijection, the inverse of $\psi \circ \phi$ is given by:
:$\paren {\psi \circ \phi}^{-1} = \phi^{-1} \circ \psi^{-1}$
By definition of composition of mappin... | Let $\struct {S_1, \preceq_1}$, $\struct {S_2, \preceq_2}$ and $\struct {S_3, \preceq_3}$ be [[Definition:Ordered Set|ordered sets]].
Let:
:$\phi: \struct {S_1, \preceq_1} \to \struct {S_2, \preceq_2}$
and:
:$\psi: \struct {S_2, \preceq_2} \to \struct {S_3, \preceq_3}$
be [[Definition:Order Isomorphism|order isomorphi... | From [[Composite of Bijections is Bijection]], $\psi \circ \phi$ is a [[Definition:Bijection|bijection]], as, by definition, an [[Definition:Order Isomorphism|order isomorphism]] is also a [[Definition:Bijection|bijection]].
From [[Inverse of Composite Bijection]], the [[Definition:Inverse Mapping|inverse]] of $\psi \... | Composite of Order Isomorphisms is Order Isomorphism | https://proofwiki.org/wiki/Composite_of_Order_Isomorphisms_is_Order_Isomorphism | https://proofwiki.org/wiki/Composite_of_Order_Isomorphisms_is_Order_Isomorphism | [
"Order Isomorphisms",
"Composite Mappings"
] | [
"Definition:Ordered Set",
"Definition:Order Isomorphism",
"Definition:Order Isomorphism"
] | [
"Composite of Bijections is Bijection",
"Definition:Bijection",
"Definition:Order Isomorphism",
"Definition:Bijection",
"Inverse of Composite Bijection",
"Definition:Inverse Mapping",
"Definition:Composition of Mappings",
"Definition:Order Isomorphism",
"Definition:Increasing/Mapping",
"Definition... |
proofwiki-474 | Order Isomorphism is Equivalence Relation | Order isomorphism between ordered sets is an equivalence relation.
So any given family of ordered sets can be partitioned into disjoint classes of isomorphic sets. | Let $\struct {S_1, \preccurlyeq_1} \cong \struct {S_2, \preccurlyeq_2}$ denote that $\struct {S_1, \preccurlyeq_1}$ is isomorphic to $\struct {S_2, \preccurlyeq_2}$.
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
{{:Order Isomorphism is Reflexive}}
Thus $\cong$ is seen to be reflexive.
{{qed... | [[Definition:Order Isomorphism|Order isomorphism]] between [[Definition:Ordered Set|ordered sets]] is an [[Definition:Equivalence Relation|equivalence relation]].
So any given family of [[Definition:Ordered Set|ordered sets]] can be [[Definition:Partition (Set Theory)|partitioned]] into [[Definition:Disjoint Sets|disj... | Let $\struct {S_1, \preccurlyeq_1} \cong \struct {S_2, \preccurlyeq_2}$ denote that $\struct {S_1, \preccurlyeq_1}$ is [[Definition:Isomorphic Ordered Sets|isomorphic]] to $\struct {S_2, \preccurlyeq_2}$.
Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]:
=== [[Order Isomorph... | Order Isomorphism is Equivalence Relation/Proof 1 | https://proofwiki.org/wiki/Order_Isomorphism_is_Equivalence_Relation | https://proofwiki.org/wiki/Order_Isomorphism_is_Equivalence_Relation/Proof_1 | [
"Order Isomorphisms",
"Examples of Equivalence Relations",
"Order Isomorphism is Equivalence Relation"
] | [
"Definition:Order Isomorphism",
"Definition:Ordered Set",
"Definition:Equivalence Relation",
"Definition:Ordered Set",
"Definition:Set Partition",
"Definition:Disjoint Sets",
"Definition:Equivalence Class",
"Definition:Set"
] | [
"Definition:Order Isomorphism/Isomorphic Sets",
"Definition:Equivalence Relation",
"Order Isomorphism is Reflexive",
"Definition:Reflexive Relation",
"Order Isomorphism is Symmetric",
"Definition:Symmetric Relation",
"Order Isomorphism is Transitive",
"Definition:Transitive Relation",
"Definition:Re... |
proofwiki-475 | Order Isomorphism is Equivalence Relation | Order isomorphism between ordered sets is an equivalence relation.
So any given family of ordered sets can be partitioned into disjoint classes of isomorphic sets. | An ordered set is a relational structure where order isomorphism is a special case of relation isomorphism.
The result follows directly from Relation Isomorphism is Equivalence Relation.
{{qed}} | [[Definition:Order Isomorphism|Order isomorphism]] between [[Definition:Ordered Set|ordered sets]] is an [[Definition:Equivalence Relation|equivalence relation]].
So any given family of [[Definition:Ordered Set|ordered sets]] can be [[Definition:Partition (Set Theory)|partitioned]] into [[Definition:Disjoint Sets|disj... | An [[Definition:Ordered Set|ordered set]] is a [[Definition:Relational Structure|relational structure]] where [[Definition:Order Isomorphism|order isomorphism]] is a special case of [[Definition:Relation Isomorphism|relation isomorphism]].
The result follows directly from [[Relation Isomorphism is Equivalence Relation... | Order Isomorphism is Equivalence Relation/Proof 2 | https://proofwiki.org/wiki/Order_Isomorphism_is_Equivalence_Relation | https://proofwiki.org/wiki/Order_Isomorphism_is_Equivalence_Relation/Proof_2 | [
"Order Isomorphisms",
"Examples of Equivalence Relations",
"Order Isomorphism is Equivalence Relation"
] | [
"Definition:Order Isomorphism",
"Definition:Ordered Set",
"Definition:Equivalence Relation",
"Definition:Ordered Set",
"Definition:Set Partition",
"Definition:Disjoint Sets",
"Definition:Equivalence Class",
"Definition:Set"
] | [
"Definition:Ordered Set",
"Definition:Relational Structure",
"Definition:Order Isomorphism",
"Definition:Relation Isomorphism",
"Relation Isomorphism is Equivalence Relation"
] |
proofwiki-476 | Dual Ordering is Ordering | Let $\struct {S, \preceq}$ be an ordered set.
Let $\succeq$ denote the dual ordering of $\preceq$.
Then $\succeq$ is an ordering on $S$. | By definition of ordering, $\preceq$ is reflexive, transitive and antisymmetric.
By definition, $\succeq$ is the inverse relation to $\preceq$.
By Inverse of Reflexive Relation is Reflexive, $\succeq$ is reflexive.
By Inverse of Antisymmetric Relation is Antisymmetric, $\succeq$ is antisymmetric.
By Inverse of Transiti... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $\succeq$ denote the [[Definition:Dual Ordering|dual ordering]] of $\preceq$.
Then $\succeq$ is an [[Definition:Ordering|ordering]] on $S$. | By definition of [[Definition:Ordering|ordering]], $\preceq$ is [[Definition:Reflexive Relation|reflexive]], [[Definition:Transitive Relation|transitive]] and [[Definition:Antisymmetric Relation|antisymmetric]].
By definition, $\succeq$ is the [[Definition:Inverse Relation|inverse relation]] to $\preceq$.
By [[Invers... | Dual Ordering is Ordering | https://proofwiki.org/wiki/Dual_Ordering_is_Ordering | https://proofwiki.org/wiki/Dual_Ordering_is_Ordering | [
"Dual Orderings"
] | [
"Definition:Ordered Set",
"Definition:Dual Ordering",
"Definition:Ordering"
] | [
"Definition:Ordering",
"Definition:Reflexive Relation",
"Definition:Transitive Relation",
"Definition:Antisymmetric Relation",
"Definition:Inverse Relation",
"Inverse of Reflexive Relation is Reflexive",
"Definition:Reflexive Relation",
"Inverse of Antisymmetric Relation is Antisymmetric",
"Definiti... |
proofwiki-477 | Power Set is Complete Lattice | Let $S$ be a set.
Let $\struct {\powerset S, \subseteq}$ be the relational structure defined on the power set $\powerset S$ of $S$ by the relation $\subseteq$.
Then:
:$\struct {\powerset S, \subseteq}$ is a complete lattice
where for every subset $\mathbb S$ of $\powerset S$:
:the infimum of $\mathbb S$ necessarily adm... | From Set is Subset of Itself:
:$S \in \powerset S$
Let $\mathbb S$ be a non-empty subset of $\powerset S$.
From Intersection is Subset:
:$\bigcap \mathbb S \in \powerset S$
Hence, from Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice:
:$\struct {\powerset S, \subseteq}$ is a complete la... | Let $S$ be a [[Definition:Set|set]].
Let $\struct {\powerset S, \subseteq}$ be the [[Definition:Relational Structure|relational structure]] defined on the [[Definition:Power Set|power set]] $\powerset S$ of $S$ by the [[Definition:Relation|relation]] $\subseteq$.
Then:
:$\struct {\powerset S, \subseteq}$ is a [[Defi... | From [[Set is Subset of Itself]]:
:$S \in \powerset S$
Let $\mathbb S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $\powerset S$.
From [[Intersection is Subset]]:
:$\bigcap \mathbb S \in \powerset S$
Hence, from [[Set of Subsets which contains Set and Intersection of Subsets is Comple... | Power Set is Complete Lattice/Proof 2 | https://proofwiki.org/wiki/Power_Set_is_Complete_Lattice | https://proofwiki.org/wiki/Power_Set_is_Complete_Lattice/Proof_2 | [
"Examples of Complete Lattices",
"Power Set",
"Power Set is Complete Lattice"
] | [
"Definition:Set",
"Definition:Relational Structure",
"Definition:Power Set",
"Definition:Relation",
"Definition:Complete Lattice",
"Definition:Subset",
"Definition:Infimum of Set"
] | [
"Set is Subset of Itself",
"Definition:Non-Empty Set",
"Definition:Subset",
"Intersection is Subset",
"Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice",
"Definition:Complete Lattice",
"Definition:Infimum of Set"
] |
proofwiki-478 | Totally Ordered Set is Lattice | Every totally ordered set is a lattice. | Let $\struct {S, \preceq}$ be a totally ordered set.
Then we have:
:$\forall x, y \in S: x \preceq y \lor y \preceq x$
:$\forall x, y \in S: x \preceq y \implies \sup \set {x, y} = y \land \inf \set {x, y} = x$
:$\forall x, y \in S: y \preceq x \implies \sup \set {x, y} = x \land \inf \set {x, y} = y$
Thus the conditio... | Every [[Definition:Totally Ordered Set|totally ordered set]] is a [[Definition:Lattice (Ordered Set)|lattice]]. | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Then we have:
:$\forall x, y \in S: x \preceq y \lor y \preceq x$
:$\forall x, y \in S: x \preceq y \implies \sup \set {x, y} = y \land \inf \set {x, y} = x$
:$\forall x, y \in S: y \preceq x \implies \sup \set {x, y} = x \land \i... | Totally Ordered Set is Lattice | https://proofwiki.org/wiki/Totally_Ordered_Set_is_Lattice | https://proofwiki.org/wiki/Totally_Ordered_Set_is_Lattice | [
"Total Orderings",
"Lattice Theory"
] | [
"Definition:Totally Ordered Set",
"Definition:Lattice (Ordered Set)"
] | [
"Definition:Totally Ordered Set",
"Definition:Lattice (Ordered Set)"
] |
proofwiki-479 | Trichotomy Law (Ordering) | Let $\struct {S, \preceq}$ be an ordered set.
Then $\preceq$ is a total ordering {{iff}}:
:$\forall a, b \in S: \paren {a \prec b} \lor \paren {a = b} \lor \paren {a \succ b}$
That is, every element either strictly precedes, is the same as, or strictly succeeds, every other element.
In other words, {{iff}} $\prec$ is a... | {{begin-eqn}}
{{eqn | q = \forall a, b \in S
| o =
| r = a \preceq b \lor b \preceq a
| c = {{Defof|Total Ordering}}
}}
{{eqn | ll= \leadstoandfrom
| q = \forall a, b \in S
| o =
| r = a \preceq b \lor a \succeq b
| c = {{Defof|Dual Ordering}}
}}
{{eqn | ll= \leadstoandfrom
... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Then $\preceq$ is a [[Definition:Total Ordering|total ordering]] {{iff}}:
:$\forall a, b \in S: \paren {a \prec b} \lor \paren {a = b} \lor \paren {a \succ b}$
That is, every element either [[Definition:Strictly Precede|strictly precedes]], is ... | {{begin-eqn}}
{{eqn | q = \forall a, b \in S
| o =
| r = a \preceq b \lor b \preceq a
| c = {{Defof|Total Ordering}}
}}
{{eqn | ll= \leadstoandfrom
| q = \forall a, b \in S
| o =
| r = a \preceq b \lor a \succeq b
| c = {{Defof|Dual Ordering}}
}}
{{eqn | ll= \leadstoandfrom
... | Trichotomy Law (Ordering) | https://proofwiki.org/wiki/Trichotomy_Law_(Ordering) | https://proofwiki.org/wiki/Trichotomy_Law_(Ordering) | [
"Trichotomies",
"Trichotomy Law",
"Total Orderings",
"Named Theorems"
] | [
"Definition:Ordered Set",
"Definition:Total Ordering",
"Definition:Strictly Precede",
"Definition:Equals",
"Definition:Strictly Succeed",
"Definition:Trichotomy"
] | [
"Strictly Precedes is Strict Ordering",
"Rule of Commutation",
"Rule of Association",
"Rule of Idempotence"
] |
proofwiki-480 | Complement of Reflexive Relation | Let $\RR \subseteq S \times S$ be a relation.
Then $\RR$ is reflexive {{iff}} its complement $\relcomp {S \times S} \RR \subseteq S \times S$ is antireflexive.
Likewise, $\RR$ is antireflexive {{iff}} its complement $\relcomp {S \times S} \RR \subseteq S \times S$ is reflexive. | Let $\RR \subseteq S \times T$ be reflexive.
Then:
:$\forall x \in S: \tuple {x, x} \in \RR$
By the definition of complement of $\RR$:
:$\tuple {x, y} \in \RR \implies \tuple {x, y} \notin \relcomp {S \times S} \RR$
The same applies to $\tuple {x, x}$, and thus:
:$\forall x \in S: \tuple {x, x} \notin \relcomp {S \time... | Let $\RR \subseteq S \times S$ be a [[Definition:Relation|relation]].
Then $\RR$ is [[Definition:Reflexive Relation|reflexive]] {{iff}} its [[Definition:Complement of Relation|complement]] $\relcomp {S \times S} \RR \subseteq S \times S$ is [[Definition:Antireflexive Relation|antireflexive]].
Likewise, $\RR$ is [[Def... | Let $\RR \subseteq S \times T$ be [[Definition:Reflexive Relation|reflexive]].
Then:
:$\forall x \in S: \tuple {x, x} \in \RR$
By the definition of [[Definition:Complement of Relation|complement]] of $\RR$:
:$\tuple {x, y} \in \RR \implies \tuple {x, y} \notin \relcomp {S \times S} \RR$
The same applies to $\tuple {... | Complement of Reflexive Relation | https://proofwiki.org/wiki/Complement_of_Reflexive_Relation | https://proofwiki.org/wiki/Complement_of_Reflexive_Relation | [
"Reflexive Relations",
"Relative Complement"
] | [
"Definition:Relation",
"Definition:Reflexive Relation",
"Definition:Complement of Relation",
"Definition:Antireflexive Relation",
"Definition:Antireflexive Relation",
"Definition:Complement of Relation",
"Definition:Reflexive Relation"
] | [
"Definition:Reflexive Relation",
"Definition:Complement of Relation",
"Definition:Antireflexive Relation",
"Double Negation",
"Definition:Converse Statement",
"Relative Complement of Relative Complement",
"Category:Reflexive Relations",
"Category:Relative Complement"
] |
proofwiki-481 | Complement of Symmetric Relation | Let $\RR \subseteq S \times S$ be a relation.
Then $\RR$ is symmetric {{iff}} its complement $\relcomp {S \times S} \RR \subseteq S \times S$ is also symmetric. | Let $\RR \subseteq S \times S$ be symmetric.
Then from Symmetry of Relations is Symmetric:
:$\tuple {x, y} \in \RR \iff \tuple {y, x} \in \RR$
{{AimForCont}} $\relcomp {S \times S} \RR \subseteq S \times S$ is not symmetric.
Then:
:$\exists \tuple {x, y} \in \relcomp {S \times S} \RR: \tuple {y, x} \notin \relcomp {S \... | Let $\RR \subseteq S \times S$ be a [[Definition:Relation|relation]].
Then $\RR$ is [[Definition:Symmetric Relation|symmetric]] {{iff}} its [[Definition:Complement of Relation|complement]] $\relcomp {S \times S} \RR \subseteq S \times S$ is also [[Definition:Symmetric Relation|symmetric]]. | Let $\RR \subseteq S \times S$ be [[Definition:Symmetric Relation|symmetric]].
Then from [[Symmetry of Relations is Symmetric]]:
:$\tuple {x, y} \in \RR \iff \tuple {y, x} \in \RR$
{{AimForCont}} $\relcomp {S \times S} \RR \subseteq S \times S$ is not [[Definition:Symmetric Relation|symmetric]].
Then:
:$\exists \tu... | Complement of Symmetric Relation | https://proofwiki.org/wiki/Complement_of_Symmetric_Relation | https://proofwiki.org/wiki/Complement_of_Symmetric_Relation | [
"Symmetric Relations",
"Relative Complement"
] | [
"Definition:Relation",
"Definition:Symmetric Relation",
"Definition:Complement of Relation",
"Definition:Symmetric Relation"
] | [
"Definition:Symmetric Relation",
"Symmetry of Relations is Symmetric",
"Definition:Symmetric Relation",
"Definition:Complement of Relation",
"Definition:Symmetric Relation",
"Definition:Contradiction",
"Proof by Contradiction",
"Definition:Symmetric Relation",
"Category:Symmetric Relations",
"Cate... |
proofwiki-482 | Strictly Increasing Mapping is Increasing | A mapping that is strictly increasing is an increasing mapping. | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be strictly increasing.
From Strictly Precedes is Strict Ordering:
:$x \preceq_1 y \implies x = y \lor x \prec_1 y$
So:
{{begin-eqn}}
{{eqn | l = x
| r = y
| c =
}}
{{eqn... | A [[Definition:Mapping|mapping]] that is [[Definition:Strictly Increasing Mapping|strictly increasing]] is an [[Definition:Increasing Mapping|increasing]] mapping. | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be [[Definition:Strictly Increasing Mapping|strictly increasing]].
From [[Strictly Precedes is Strict Ordering]]:
:$x \preceq_1 y \implies x = y \lor x \... | Strictly Increasing Mapping is Increasing | https://proofwiki.org/wiki/Strictly_Increasing_Mapping_is_Increasing | https://proofwiki.org/wiki/Strictly_Increasing_Mapping_is_Increasing | [
"Increasing Mappings"
] | [
"Definition:Mapping",
"Definition:Strictly Increasing/Mapping",
"Definition:Increasing/Mapping"
] | [
"Definition:Ordered Set",
"Definition:Strictly Increasing/Mapping",
"Strictly Precedes is Strict Ordering",
"Definition:Ordering",
"Definition:Reflexive Relation",
"Strictly Precedes is Strict Ordering"
] |
proofwiki-483 | Strictly Decreasing Mapping is Decreasing | A mapping that is strictly decreasing is a decreasing mapping. | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be strictly decreasing.
From Strictly Precedes is Strict Ordering:
:$x \preceq_1 y \iff x = y \lor x \prec_1 y$
So:
{{begin-eqn}}
{{eqn | l = x
| r = y
| c =
}}
{{eqn | l... | A [[Definition:Mapping|mapping]] that is [[Definition:Strictly Decreasing Mapping|strictly decreasing]] is a [[Definition:Decreasing Mapping|decreasing]] mapping. | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be [[Definition:Strictly Decreasing Mapping|strictly decreasing]].
From [[Strictly Precedes is Strict Ordering]]:
:$x \preceq_1 y \iff x = y \lor x \prec... | Strictly Decreasing Mapping is Decreasing | https://proofwiki.org/wiki/Strictly_Decreasing_Mapping_is_Decreasing | https://proofwiki.org/wiki/Strictly_Decreasing_Mapping_is_Decreasing | [
"Decreasing Mappings"
] | [
"Definition:Mapping",
"Definition:Strictly Decreasing/Mapping",
"Definition:Decreasing/Mapping"
] | [
"Definition:Ordered Set",
"Definition:Strictly Decreasing/Mapping",
"Strictly Precedes is Strict Ordering",
"Definition:Ordering",
"Definition:Reflexive Relation",
"Strictly Precedes is Strict Ordering"
] |
proofwiki-484 | Strictly Monotone Mapping is Monotone | A mapping that is strictly monotone is a monotone mapping. | If $\phi$ is strictly monotone, then it is either strictly increasing or strictly decreasing.
If $\phi$ is strictly increasing, then by Strictly Increasing Mapping is Increasing, $\phi$ is increasing.
If $\phi$ is strictly decreasing, then by Strictly Decreasing Mapping is Decreasing, $\phi$ is decreasing.
Thus $\phi$ ... | A [[Definition:Mapping|mapping]] that is [[Definition:Strictly Monotone Mapping|strictly monotone]] is a [[Definition:Monotone Mapping|monotone mapping]]. | If $\phi$ is strictly monotone, then it is either [[Definition:Strictly Increasing Mapping|strictly increasing]] or [[Definition:Strictly Decreasing Mapping|strictly decreasing]].
If $\phi$ is strictly increasing, then by [[Strictly Increasing Mapping is Increasing]], $\phi$ is [[Definition:Increasing Mapping|increasi... | Strictly Monotone Mapping is Monotone | https://proofwiki.org/wiki/Strictly_Monotone_Mapping_is_Monotone | https://proofwiki.org/wiki/Strictly_Monotone_Mapping_is_Monotone | [
"Strictly Monotone Mappings",
"Monotone Mappings"
] | [
"Definition:Mapping",
"Definition:Strictly Monotone/Mapping",
"Definition:Monotone (Order Theory)/Mapping"
] | [
"Definition:Strictly Increasing/Mapping",
"Definition:Strictly Decreasing/Mapping",
"Strictly Increasing Mapping is Increasing",
"Definition:Increasing/Mapping",
"Strictly Decreasing Mapping is Decreasing",
"Definition:Decreasing/Mapping",
"Definition:Monotone (Order Theory)/Mapping"
] |
proofwiki-485 | Order Embedding into Image is Isomorphism | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $S'$ be the image of a mapping $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$.
Then:
:$\phi$ is an order embedding from $\struct {S, \preceq_1}$ into $\struct {T, \preceq_2}$
{{iff}}:
:$\phi$ is an order isomorphism from $\struct ... | Let $\phi$ be an order embedding from $\struct {S, \preceq_1}$ into $\struct {T, \preceq_2}$.
Then $\phi$ is an injection into $\struct {T, \preceq_2}$ by definition.
From Restriction of Mapping to Image is Surjection, a mapping from a set to the image of that mapping is a surjection.
Thus the surjective restriction of... | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $S'$ be the [[Definition:Image of Mapping|image]] of a [[Definition:Mapping|mapping]] $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$.
Then:
:$\phi$ is an [[Definition:Order Embedding|order embedding]]... | Let $\phi$ be an [[Definition:Order Embedding|order embedding]] from $\struct {S, \preceq_1}$ into $\struct {T, \preceq_2}$.
Then $\phi$ is an [[Definition:Injection|injection]] into $\struct {T, \preceq_2}$ by definition.
From [[Restriction of Mapping to Image is Surjection]], a [[Definition:Mapping|mapping]] from a... | Order Embedding into Image is Isomorphism | https://proofwiki.org/wiki/Order_Embedding_into_Image_is_Isomorphism | https://proofwiki.org/wiki/Order_Embedding_into_Image_is_Isomorphism | [
"Order Embeddings",
"Order Isomorphisms"
] | [
"Definition:Ordered Set",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Mapping",
"Definition:Order Embedding",
"Definition:Order Isomorphism"
] | [
"Definition:Order Embedding",
"Definition:Injection",
"Restriction of Mapping to Image is Surjection",
"Definition:Mapping",
"Definition:Set",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Mapping",
"Definition:Surjection",
"Definition:Surjective Restriction",
"Definition:Order Embe... |
proofwiki-486 | Strictly Monotone Mapping with Totally Ordered Domain is Injective | Let $\struct {S, \preceq_1}$ be a totally ordered set.
Let $\struct {T, \preceq_2}$ be an ordered set.
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be a strictly monotone mapping.
Then $\phi$ is injective. | {{begin-eqn}}
{{eqn | l = x, y
| o = \in
| r = S
| c =
}}
{{eqn | lo= \land
| l = x
| o = \ne
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \prec_1
| r = y
}}
{{eqn | lo= \lor
| l = y
| o = \prec_1
| r = x
| c = Trichotomy Law
}... | Let $\struct {S, \preceq_1}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $\struct {T, \preceq_2}$ be an [[Definition:Ordered Set|ordered set]].
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be a [[Definition:Strictly Monotone Mapping|strictly monotone mapping]].
Then $\phi$ is [... | {{begin-eqn}}
{{eqn | l = x, y
| o = \in
| r = S
| c =
}}
{{eqn | lo= \land
| l = x
| o = \ne
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \prec_1
| r = y
}}
{{eqn | lo= \lor
| l = y
| o = \prec_1
| r = x
| c = [[Trichotomy Law... | Strictly Monotone Mapping with Totally Ordered Domain is Injective | https://proofwiki.org/wiki/Strictly_Monotone_Mapping_with_Totally_Ordered_Domain_is_Injective | https://proofwiki.org/wiki/Strictly_Monotone_Mapping_with_Totally_Ordered_Domain_is_Injective | [
"Strictly Monotone Mappings",
"Total Orderings",
"Injections"
] | [
"Definition:Totally Ordered Set",
"Definition:Ordered Set",
"Definition:Strictly Monotone/Mapping",
"Definition:Injection"
] | [
"Trichotomy Law (Ordering)",
"Definition:Strictly Monotone/Mapping"
] |
proofwiki-487 | Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing | Let $\struct {S, \preceq_1}$ be a totally ordered set.
Let $\struct {T, \preceq_2}$ be an ordered set.
Let $\phi: S \to T$ be a mapping.
Then $\phi$ is an order embedding {{iff}} $\phi$ is strictly increasing.
That is:
:$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
{{iff}}:
:$\forall x, y \... | === Forward Implication ===
Let $\phi$ be an order embedding.
Let $x, y \in S$ with $x \prec_1 y$, where $\prec_1$ is the strict predecessor relation.
Then:
{{begin-eqn}}
{{eqn | o =
| r = x \prec_1 y \land x \ne y
| c = {{Defof|Strictly Precedes|index = 2}}
}}
{{eqn | o = \leadsto
| r = \map \phi x \... | Let $\struct {S, \preceq_1}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $\struct {T, \preceq_2}$ be an [[Definition:Ordered Set|ordered set]].
Let $\phi: S \to T$ be a [[Definition:Mapping|mapping]].
Then $\phi$ is an [[Definition:Order Embedding|order embedding]] {{iff}} $\phi$ is [[Definitio... | === Forward Implication ===
Let $\phi$ be an [[Definition:Order Embedding|order embedding]].
Let $x, y \in S$ with $x \prec_1 y$, where $\prec_1$ is the [[Definition:Strictly Precedes/Definition 2|strict predecessor]] relation.
Then:
{{begin-eqn}}
{{eqn | o =
| r = x \prec_1 y \land x \ne y
| c = {{Defof... | Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing | https://proofwiki.org/wiki/Mapping_from_Totally_Ordered_Set_is_Order_Embedding_iff_Strictly_Increasing | https://proofwiki.org/wiki/Mapping_from_Totally_Ordered_Set_is_Order_Embedding_iff_Strictly_Increasing | [
"Order Embeddings",
"Total Orderings",
"Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing"
] | [
"Definition:Totally Ordered Set",
"Definition:Ordered Set",
"Definition:Mapping",
"Definition:Order Embedding",
"Definition:Strictly Increasing/Mapping"
] | [
"Definition:Order Embedding",
"Definition:Strictly Precede/Definition 2",
"Definition:Injection",
"Definition:Strictly Increasing/Mapping"
] |
proofwiki-488 | Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing | Let $\struct {S, \preceq_1}$ be a totally ordered set.
Let $\struct {T, \preceq_2}$ be an ordered set.
Let $\phi: S \to T$ be a mapping.
Then $\phi$ is an order embedding {{iff}} $\phi$ is strictly increasing.
That is:
:$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
{{iff}}:
:$\forall x, y \... | Let $x \preceq_1 y$.
Then $x = y$ or $x \prec_1 y$.
Let $x = y$.
Then
:$\map \phi x = \map \phi y$
so:
:$\map \phi x \preceq_2 \map \phi y$
Let $x \prec_1 y$.
Then by the definition of strictly increasing mapping:
:$\map \phi x \prec_2 \map \phi y$
so by the definition of $\prec_2$:
:$\map \phi x \preceq_2 \map \phi y$... | Let $\struct {S, \preceq_1}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $\struct {T, \preceq_2}$ be an [[Definition:Ordered Set|ordered set]].
Let $\phi: S \to T$ be a [[Definition:Mapping|mapping]].
Then $\phi$ is an [[Definition:Order Embedding|order embedding]] {{iff}} $\phi$ is [[Definitio... | Let $x \preceq_1 y$.
Then $x = y$ or $x \prec_1 y$.
Let $x = y$.
Then
:$\map \phi x = \map \phi y$
so:
:$\map \phi x \preceq_2 \map \phi y$
Let $x \prec_1 y$.
Then by the definition of [[Definition:Strictly Increasing Mapping|strictly increasing mapping]]:
:$\map \phi x \prec_2 \map \phi y$
so by the definition ... | Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication/Proof 1 | https://proofwiki.org/wiki/Mapping_from_Totally_Ordered_Set_is_Order_Embedding_iff_Strictly_Increasing | https://proofwiki.org/wiki/Mapping_from_Totally_Ordered_Set_is_Order_Embedding_iff_Strictly_Increasing/Reverse_Implication/Proof_1 | [
"Order Embeddings",
"Total Orderings",
"Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing"
] | [
"Definition:Totally Ordered Set",
"Definition:Ordered Set",
"Definition:Mapping",
"Definition:Order Embedding",
"Definition:Strictly Increasing/Mapping"
] | [
"Definition:Strictly Increasing/Mapping",
"Definition:Total Ordering",
"Definition:Strictly Increasing/Mapping",
"Rule of Transposition"
] |
proofwiki-489 | Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing | Let $\struct {S, \preceq_1}$ be a totally ordered set.
Let $\struct {T, \preceq_2}$ be an ordered set.
Let $\phi: S \to T$ be a mapping.
Then $\phi$ is an order embedding {{iff}} $\phi$ is strictly increasing.
That is:
:$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
{{iff}}:
:$\forall x, y \... | Let $\phi$ be strictly increasing.
Let $\map \phi x \preceq_2 \map \phi y$.
As $\struct {S, \prec_1}$ is a strictly totally ordered set:
:Either $y \prec_1 x$, $y = x$, or $x \prec_1 y$.
{{AimForCont}} that $y \prec_1 x$.
By the definition of a strictly increasing mapping:
:$\map \phi y \prec_2 \map \phi x$
which contr... | Let $\struct {S, \preceq_1}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $\struct {T, \preceq_2}$ be an [[Definition:Ordered Set|ordered set]].
Let $\phi: S \to T$ be a [[Definition:Mapping|mapping]].
Then $\phi$ is an [[Definition:Order Embedding|order embedding]] {{iff}} $\phi$ is [[Definitio... | Let $\phi$ be [[Definition:Strictly Increasing Mapping|strictly increasing]].
Let $\map \phi x \preceq_2 \map \phi y$.
As $\struct {S, \prec_1}$ is a [[Definition:Strictly Totally Ordered Set|strictly totally ordered set]]:
:Either $y \prec_1 x$, $y = x$, or $x \prec_1 y$.
{{AimForCont}} that $y \prec_1 x$.
By the... | Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication/Proof 2 | https://proofwiki.org/wiki/Mapping_from_Totally_Ordered_Set_is_Order_Embedding_iff_Strictly_Increasing | https://proofwiki.org/wiki/Mapping_from_Totally_Ordered_Set_is_Order_Embedding_iff_Strictly_Increasing/Reverse_Implication/Proof_2 | [
"Order Embeddings",
"Total Orderings",
"Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing"
] | [
"Definition:Totally Ordered Set",
"Definition:Ordered Set",
"Definition:Mapping",
"Definition:Order Embedding",
"Definition:Strictly Increasing/Mapping"
] | [
"Definition:Strictly Increasing/Mapping",
"Definition:Strictly Totally Ordered Set",
"Definition:Strictly Increasing/Mapping",
"Definition:Contradiction",
"Definition:Order Embedding"
] |
proofwiki-490 | Finite Totally Ordered Set is Well-Ordered | Every finite totally ordered set is well-ordered. | Let $\struct {S, \preceq}$ be a finite totally ordered set.
{{AimForCont}} $\preceq$ is not a well-founded relation.
From Infinite Sequence Property of Well-Founded Relation, $\struct {S, \preceq}$ is well-founded {{iff}} there is no infinite sequence $\sequence {a_n}$ of elements of $S$ such that $\forall n \in \N: a_... | Every [[Definition:Finite Set|finite]] [[Definition:Totally Ordered Set|totally ordered set]] is [[Definition:Well-Ordered Set|well-ordered]]. | Let $\struct {S, \preceq}$ be a [[Definition:Finite Set|finite]] [[Definition:Totally Ordered Set|totally ordered set]].
{{AimForCont}} $\preceq$ is not a [[Definition:Well-Founded Relation|well-founded relation]].
From [[Infinite Sequence Property of Well-Founded Relation]], $\struct {S, \preceq}$ is [[Definition:We... | Finite Totally Ordered Set is Well-Ordered | https://proofwiki.org/wiki/Finite_Totally_Ordered_Set_is_Well-Ordered | https://proofwiki.org/wiki/Finite_Totally_Ordered_Set_is_Well-Ordered | [
"Total Orderings",
"Well-Orderings"
] | [
"Definition:Finite Set",
"Definition:Totally Ordered Set",
"Definition:Well-Ordered Set"
] | [
"Definition:Finite Set",
"Definition:Totally Ordered Set",
"Definition:Well-Founded Relation",
"Infinite Sequence Property of Well-Founded Relation",
"Definition:Well-Founded Relation",
"Definition:Sequence/Infinite Sequence",
"Definition:Element",
"Definition:Sequence/Infinite Sequence",
"Definitio... |
proofwiki-491 | Subset of Well-Ordered Set is Well-Ordered | Let $\struct {S, \preceq}$ be a well-ordered set.
Let $T \subseteq S$ be a subset of $S$.
Let $\preceq'$ be the restriction of $\preceq$ to $T$.
Then the relational structure $\struct {T, \preceq'}$ is a well-ordered set. | First suppose that $T = \O$.
From Empty Set is Subset of All Sets, $T$ is a subset of $S$.
By Empty Set is Well-Ordered, $\struct {\O, \preceq'}$ is a well-ordered set.
Otherwise, let $T$ be non-empty.
Let $X \subseteq T$ such that $X \ne \O$ be arbitrary.
Such a subset exists, as from Set is Subset of Itself, $T$ its... | Let $\struct {S, \preceq}$ be a [[Definition:Well-Ordered Set|well-ordered set]].
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\preceq'$ be the [[Definition:Restriction of Ordering|restriction]] of $\preceq$ to $T$.
Then the [[Definition:Relational Structure|relational structure]] $\struct {T,... | First suppose that $T = \O$.
From [[Empty Set is Subset of All Sets]], $T$ is a [[Definition:Subset|subset]] of $S$.
By [[Empty Set is Well-Ordered]], $\struct {\O, \preceq'}$ is a [[Definition:Well-Ordered Set|well-ordered set]].
Otherwise, let $T$ be [[Definition:Non-Empty Set|non-empty]].
Let $X \subseteq T$ s... | Subset of Well-Ordered Set is Well-Ordered/Proof 1 | https://proofwiki.org/wiki/Subset_of_Well-Ordered_Set_is_Well-Ordered | https://proofwiki.org/wiki/Subset_of_Well-Ordered_Set_is_Well-Ordered/Proof_1 | [
"Well-Orderings",
"Subset of Well-Ordered Set is Well-Ordered"
] | [
"Definition:Well-Ordered Set",
"Definition:Subset",
"Definition:Restriction of Ordering",
"Definition:Relational Structure",
"Definition:Well-Ordered Set"
] | [
"Empty Set is Subset of All Sets",
"Definition:Subset",
"Empty Set is Well-Ordered",
"Definition:Well-Ordered Set",
"Definition:Non-Empty Set",
"Definition:Subset",
"Set is Subset of Itself",
"Definition:Subset",
"Subset Relation is Transitive",
"Definition:Well-Ordered Set",
"Definition:Smalles... |
proofwiki-492 | Subset of Well-Ordered Set is Well-Ordered | Let $\struct {S, \preceq}$ be a well-ordered set.
Let $T \subseteq S$ be a subset of $S$.
Let $\preceq'$ be the restriction of $\preceq$ to $T$.
Then the relational structure $\struct {T, \preceq'}$ is a well-ordered set. | By definition of well-ordered set, $\struct {S, \preceq}$ is:
:a totally ordered set
and:
:a well-founded set.
By Subset of Toset is Toset, $\struct {T, \preceq'}$ is a totally ordered set.
By Subset of Well-Founded Relation is Well-Founded, $\preceq'$ is a well-founded relation.
Hence the result.
{{qed}} | Let $\struct {S, \preceq}$ be a [[Definition:Well-Ordered Set|well-ordered set]].
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\preceq'$ be the [[Definition:Restriction of Ordering|restriction]] of $\preceq$ to $T$.
Then the [[Definition:Relational Structure|relational structure]] $\struct {T,... | By definition of [[Definition:Well-Ordered Set|well-ordered set]], $\struct {S, \preceq}$ is:
:a [[Definition:Totally Ordered Set|totally ordered set]]
and:
:a [[Definition:Well-Founded Set|well-founded set]].
By [[Subset of Toset is Toset]], $\struct {T, \preceq'}$ is a [[Definition:Totally Ordered Set|totally ordere... | Subset of Well-Ordered Set is Well-Ordered/Proof 2 | https://proofwiki.org/wiki/Subset_of_Well-Ordered_Set_is_Well-Ordered | https://proofwiki.org/wiki/Subset_of_Well-Ordered_Set_is_Well-Ordered/Proof_2 | [
"Well-Orderings",
"Subset of Well-Ordered Set is Well-Ordered"
] | [
"Definition:Well-Ordered Set",
"Definition:Subset",
"Definition:Restriction of Ordering",
"Definition:Relational Structure",
"Definition:Well-Ordered Set"
] | [
"Definition:Well-Ordered Set",
"Definition:Totally Ordered Set",
"Definition:Well-Founded Set",
"Subset of Toset is Toset",
"Definition:Totally Ordered Set",
"Subset of Well-Founded Relation is Well-Founded",
"Definition:Well-Founded Relation"
] |
proofwiki-493 | Subset of Well-Ordered Set is Well-Ordered | Let $\struct {S, \preceq}$ be a well-ordered set.
Let $T \subseteq S$ be a subset of $S$.
Let $\preceq'$ be the restriction of $\preceq$ to $T$.
Then the relational structure $\struct {T, \preceq'}$ is a well-ordered set. | Let $V$ be a basic universe.
By definition of basic universe, $S$ and $T$ are all elements of $V$.
By the {{axiom-link|Transitivity}}, $S$ and $T$ are both classes.
Thus $T$ is a subclass of $S$.
We have {{hypothesis}} that $\preceq$ is a well-ordering on $S$.
So from Subclass of Well-Ordered Class is Well-Ordered, $\p... | Let $\struct {S, \preceq}$ be a [[Definition:Well-Ordered Set|well-ordered set]].
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\preceq'$ be the [[Definition:Restriction of Ordering|restriction]] of $\preceq$ to $T$.
Then the [[Definition:Relational Structure|relational structure]] $\struct {T,... | Let $V$ be a [[Definition:Basic Universe|basic universe]].
By definition of [[Definition:Basic Universe|basic universe]], $S$ and $T$ are all [[Definition:Element|elements]] of $V$.
By the {{axiom-link|Transitivity}}, $S$ and $T$ are both [[Definition:Class (Class Theory)|classes]].
Thus $T$ is a [[Definition:Subcla... | Subset of Well-Ordered Set is Well-Ordered/Proof 3 | https://proofwiki.org/wiki/Subset_of_Well-Ordered_Set_is_Well-Ordered | https://proofwiki.org/wiki/Subset_of_Well-Ordered_Set_is_Well-Ordered/Proof_3 | [
"Well-Orderings",
"Subset of Well-Ordered Set is Well-Ordered"
] | [
"Definition:Well-Ordered Set",
"Definition:Subset",
"Definition:Restriction of Ordering",
"Definition:Relational Structure",
"Definition:Well-Ordered Set"
] | [
"Definition:Basic Universe",
"Definition:Basic Universe",
"Definition:Element",
"Definition:Class (Class Theory)",
"Definition:Subclass",
"Definition:Well-Ordering/Class Theory",
"Subclass of Well-Ordered Class is Well-Ordered",
"Definition:Well-Ordering/Class Theory"
] |
proofwiki-494 | Element Commutes with Product of Commuting Elements | Let $\struct {S, \circ}$ be a semigroup.
Let $x, y, z \in S$.
Let $x$ commute with both $y$ and $z$.
Then $x$ commutes with $y \circ z$. | {{begin-eqn}}
{{eqn | l = x \circ \paren {y \circ z}
| r = \paren {x \circ y} \circ z
| c = {{Semigroup-axiom|1}}
}}
{{eqn | r = \paren {y \circ x} \circ z
| c = $x$ commutes with $y$
}}
{{eqn | r = y \circ \paren {x \circ z}
| c = {{Semigroup-axiom|1}}
}}
{{eqn | r = y \circ \paren {z \circ x}
... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $x, y, z \in S$.
Let $x$ [[Definition:Commuting Elements|commute]] with both $y$ and $z$.
Then $x$ [[Definition:Commuting Elements|commutes]] with $y \circ z$. | {{begin-eqn}}
{{eqn | l = x \circ \paren {y \circ z}
| r = \paren {x \circ y} \circ z
| c = {{Semigroup-axiom|1}}
}}
{{eqn | r = \paren {y \circ x} \circ z
| c = $x$ [[Definition:Commuting Elements|commutes]] with $y$
}}
{{eqn | r = y \circ \paren {x \circ z}
| c = {{Semigroup-axiom|1}}
}}
{{eqn... | Element Commutes with Product of Commuting Elements | https://proofwiki.org/wiki/Element_Commutes_with_Product_of_Commuting_Elements | https://proofwiki.org/wiki/Element_Commutes_with_Product_of_Commuting_Elements | [
"Element Commutes with Product of Commuting Elements",
"Commutativity",
"Semigroups"
] | [
"Definition:Semigroup",
"Definition:Commutative/Elements",
"Definition:Commutative/Elements"
] | [
"Definition:Commutative/Elements",
"Definition:Commutative/Elements"
] |
proofwiki-495 | Associative Idempotent Anticommutative | Let $\circ$ be a binary operation on a non-empty set $S$.
Let $\circ$ be associative.
Then $\circ$ is anticommutative {{iff}}:
:$(1): \quad \circ$ is idempotent
and:
:$(2): \quad \forall a, b \in S: a \circ b \circ a = a$ | Let $\circ$ be an associative operation on $S$. | Let $\circ$ be a [[Definition:Binary Operation|binary operation]] on a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]] $S$.
Let $\circ$ be [[Definition:Associative Operation|associative]].
Then $\circ$ is [[Definition:Anticommutative|anticommutative]] {{iff}}:
:$(1): \quad \circ$ is [[Definition:Idempo... | Let $\circ$ be an [[Definition:Associative Operation|associative operation]] on $S$. | Associative Idempotent Anticommutative | https://proofwiki.org/wiki/Associative_Idempotent_Anticommutative | https://proofwiki.org/wiki/Associative_Idempotent_Anticommutative | [
"Abstract Algebra",
"Anticommutativity",
"Associativity",
"Idempotence"
] | [
"Definition:Operation/Binary Operation",
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Associative Operation",
"Definition:Anticommutative",
"Definition:Idempotence/Operation"
] | [
"Definition:Associative Operation",
"Definition:Associative Operation",
"Definition:Associative Operation",
"Definition:Associative Operation",
"Definition:Associative Operation",
"Definition:Associative Operation",
"Definition:Associative Operation",
"Definition:Associative Operation",
"Definition:... |
proofwiki-496 | Associative and Anticommutative | Let $\circ$ be a binary operation on a set $S$.
Let $\circ$ be both associative and anticommutative.
Then:
:$\forall x, y, z \in S: x \circ y \circ z = x \circ z$ | Let $\circ$ be both associative and anticommutative.
Then from Associative Idempotent Anticommutative:
:$\forall x, z \in S: x \circ z \circ x = x$
and $\circ$ is idempotent.
Consider $x \circ y \circ z \circ x \circ z$.
We have:
{{begin-eqn}}
{{eqn | l = x \circ y \circ z \circ x \circ z
| r = x \circ \paren {y ... | Let $\circ$ be a [[Definition:Binary Operation|binary operation]] on a set $S$.
Let $\circ$ be both [[Definition:Associative Operation|associative]] and [[Definition:Anticommutative|anticommutative]].
Then:
:$\forall x, y, z \in S: x \circ y \circ z = x \circ z$ | Let $\circ$ be both [[Definition:Associative Operation|associative]] and [[Definition:Anticommutative|anticommutative]].
Then from [[Associative Idempotent Anticommutative]]:
:$\forall x, z \in S: x \circ z \circ x = x$
and $\circ$ is [[Definition:Idempotent Operation|idempotent]].
Consider $x \circ y \circ z \circ ... | Associative and Anticommutative | https://proofwiki.org/wiki/Associative_and_Anticommutative | https://proofwiki.org/wiki/Associative_and_Anticommutative | [
"Anticommutativity",
"Associativity"
] | [
"Definition:Operation/Binary Operation",
"Definition:Associative Operation",
"Definition:Anticommutative"
] | [
"Definition:Associative Operation",
"Definition:Anticommutative",
"Associative Idempotent Anticommutative",
"Definition:Idempotence/Operation"
] |
proofwiki-497 | Constant Operation is Commutative | Let $S$ be a set.
Let $x \sqbrk c y = c$ be a constant operation on $S$.
Then $\sqbrk c$ is a commutative operation:
:$\forall x, y \in S: x \sqbrk c y = y \sqbrk c x$ | {{begin-eqn}}
{{eqn | l = x \sqbrk c y
| r = c
| c = {{Defof|Constant Operation}}
}}
{{eqn | l = y \sqbrk c x
| r = c
| c = {{Defof|Constant Operation}}
}}
{{eqn | ll= \leadsto
| l = x \sqbrk c y
| r = y \sqbrk c x
| c =
}}
{{end-eqn}}
Hence the result by definition of commuta... | Let $S$ be a [[Definition:Set|set]].
Let $x \sqbrk c y = c$ be a [[Definition:Constant Operation|constant operation]] on $S$.
Then $\sqbrk c$ is a [[Definition:Commutative Operation|commutative operation]]:
:$\forall x, y \in S: x \sqbrk c y = y \sqbrk c x$ | {{begin-eqn}}
{{eqn | l = x \sqbrk c y
| r = c
| c = {{Defof|Constant Operation}}
}}
{{eqn | l = y \sqbrk c x
| r = c
| c = {{Defof|Constant Operation}}
}}
{{eqn | ll= \leadsto
| l = x \sqbrk c y
| r = y \sqbrk c x
| c =
}}
{{end-eqn}}
Hence the result by definition of [[Def... | Constant Operation is Commutative | https://proofwiki.org/wiki/Constant_Operation_is_Commutative | https://proofwiki.org/wiki/Constant_Operation_is_Commutative | [
"Constant Operation",
"Examples of Commutative Operations"
] | [
"Definition:Set",
"Definition:Constant Operation",
"Definition:Commutative/Operation"
] | [
"Definition:Commutative/Operation",
"Category:Constant Operation",
"Category:Examples of Commutative Operations"
] |
proofwiki-498 | Constant Operation is Associative | Let $S$ be a set.
Let $x \sqbrk c y = c$ be a constant operation on $S$.
Then $\sqbrk c$ is an associative operation:
:$\forall x, y, z \in S: \paren {x \sqbrk c y} \sqbrk c z = x \sqbrk c \paren {y \sqbrk c z}$ | {{begin-eqn}}
{{eqn | l = \paren {x \sqbrk c y} \sqbrk c z
| r = c \sqbrk c z
| c = {{Defof|Constant Operation}}
}}
{{eqn | r = c
| c = {{Defof|Constant Operation}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = x \sqbrk c \paren {y \sqbrk c z}
| r = x \sqbrk c c
| c = {{Defof|Constant Operation... | Let $S$ be a [[Definition:Set|set]].
Let $x \sqbrk c y = c$ be a [[Definition:Constant Operation|constant operation]] on $S$.
Then $\sqbrk c$ is an [[Definition:Associative Operation|associative operation]]:
:$\forall x, y, z \in S: \paren {x \sqbrk c y} \sqbrk c z = x \sqbrk c \paren {y \sqbrk c z}$ | {{begin-eqn}}
{{eqn | l = \paren {x \sqbrk c y} \sqbrk c z
| r = c \sqbrk c z
| c = {{Defof|Constant Operation}}
}}
{{eqn | r = c
| c = {{Defof|Constant Operation}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = x \sqbrk c \paren {y \sqbrk c z}
| r = x \sqbrk c c
| c = {{Defof|Constant Operati... | Constant Operation is Associative | https://proofwiki.org/wiki/Constant_Operation_is_Associative | https://proofwiki.org/wiki/Constant_Operation_is_Associative | [
"Constant Operation",
"Examples of Associative Operations"
] | [
"Definition:Set",
"Definition:Constant Operation",
"Definition:Associative Operation"
] | [
"Category:Constant Operation",
"Category:Examples of Associative Operations"
] |
proofwiki-499 | Left Operation is Idempotent | The left operation is idempotent:
:$\forall x: x \leftarrow x = x$ | Immediate from the definition of the left operation.
{{Qed}} | The [[Definition:Left Operation|left operation]] is [[Definition:Idempotent Operation|idempotent]]:
:$\forall x: x \leftarrow x = x$ | Immediate from the definition of the [[Definition:Left Operation|left operation]].
{{Qed}} | Left Operation is Idempotent | https://proofwiki.org/wiki/Left_Operation_is_Idempotent | https://proofwiki.org/wiki/Left_Operation_is_Idempotent | [
"Left Operation",
"Examples of Idempotence"
] | [
"Definition:Left Operation",
"Definition:Idempotence/Operation"
] | [
"Definition:Left Operation"
] |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.