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-19700 | Index Laws/Product of Indices/Field | :$(a):\quad \forall a \in \F^* : \forall n, m \in \Z : \paren{a^m}^n = a^\paren{mn}$
:$(b):\quad \forall a \in \F : \forall n, m \in \Z_{\ge 0} : \paren{a^m}^n = a^\paren{mn}$ | === Statement $(a)$ ===
By {{Defof|Field (Abstract Algebra)|Field}}:
:$\struct{F^*, \circ}$ is an Abelian group
By {{Defof|Power of Field Element}}:
:For all $a \in F^*$ and $n \in \Z$, $a^n$ is defined as the $n$th power of $a$ with respect to the Abelian group $\struct {F^*, \circ}$
From Product of Powers of Group El... | :$(a):\quad \forall a \in \F^* : \forall n, m \in \Z : \paren{a^m}^n = a^\paren{mn}$
:$(b):\quad \forall a \in \F : \forall n, m \in \Z_{\ge 0} : \paren{a^m}^n = a^\paren{mn}$ | === Statement $(a)$ ===
By {{Defof|Field (Abstract Algebra)|Field}}:
:$\struct{F^*, \circ}$ is an [[Definition:Abelian Group|Abelian group]]
By {{Defof|Power of Field Element}}:
:For all $a \in F^*$ and $n \in \Z$, $a^n$ is defined as the [[Definition:Power of Group Element|$n$th power of $a$]] with respect to the [[D... | Index Laws/Product of Indices/Field | https://proofwiki.org/wiki/Index_Laws/Product_of_Indices/Field | https://proofwiki.org/wiki/Index_Laws/Product_of_Indices/Field | [
"Index Laws"
] | [] | [
"Definition:Abelian Group",
"Definition:Power of Element/Group",
"Definition:Abelian Group",
"Powers of Group Elements/Product of Indices"
] |
proofwiki-19701 | Index Laws/Common Index/Field | {{begin-eqn}}
{{eqn | n = $a$
| q = \forall a, b \in \F^*
| qq= \forall n \in \Z
| l = a^n \circ b^n
| r = \paren {a b}^n
}}
{{eqn | n = $b$
| q = \forall a, b \in \F
| qq= \forall n \in \Z_{\ge 0}
| l = a^n \circ b^n
| r = \paren {a b}^n
}}
{{end-eqn}} | === Statement $(a)$ ===
By {{Defof|Field (Abstract Algebra)|Field}}:
:$\struct{F^*, \circ}$ is an Abelian group
By {{Defof|Power of Field Element}}:
:For all $a \in F^*$ and $n \in \Z$, $a^n$ is defined as the $n$th power of $a$ with respect to the Abelian group $\struct {F^*, \circ}$
From Sum of Powers of Group Elemen... | {{begin-eqn}}
{{eqn | n = $a$
| q = \forall a, b \in \F^*
| qq= \forall n \in \Z
| l = a^n \circ b^n
| r = \paren {a b}^n
}}
{{eqn | n = $b$
| q = \forall a, b \in \F
| qq= \forall n \in \Z_{\ge 0}
| l = a^n \circ b^n
| r = \paren {a b}^n
}}
{{end-eqn}} | === Statement $(a)$ ===
By {{Defof|Field (Abstract Algebra)|Field}}:
:$\struct{F^*, \circ}$ is an [[Definition:Abelian Group|Abelian group]]
By {{Defof|Power of Field Element}}:
:For all $a \in F^*$ and $n \in \Z$, $a^n$ is defined as the [[Definition:Power of Group Element|$n$th power of $a$]] with respect to the [[... | Index Laws/Common Index/Field | https://proofwiki.org/wiki/Index_Laws/Common_Index/Field | https://proofwiki.org/wiki/Index_Laws/Common_Index/Field | [
"Index Laws"
] | [] | [
"Definition:Abelian Group",
"Definition:Power of Element/Group",
"Definition:Abelian Group",
"Powers of Group Elements/Sum of Indices"
] |
proofwiki-19702 | G-Tower is G-Ordered | Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Then $M$ is $g$-ordered. | Recall the definition of a $g$-ordered class:
$M$ is a $g$-ordered class {{iff}} $M$ is well-ordered by the subset relation such that:
{{begin-axiom}}
{{axiom | n = 1
| t = the smallest element of $M$ is $\O$
}}
{{axiom | n = 2
| t = every immediate successor $y$ is $\map g x$, where $x$ is the immediat... | Let $M$ be a [[Definition:Class (Class Theory)|class]].
Let $g: M \to M$ be a [[Definition:Progressing Mapping|progressing mapping]] on $M$.
Let $M$ be a [[Definition:G-Tower|$g$-tower]].
Then $M$ is [[Definition:G-Ordered Class|$g$-ordered]]. | Recall the definition of a [[Definition:G-Ordered Class|$g$-ordered class]]:
$M$ is a [[Definition:G-Ordered Class|$g$-ordered class]] {{iff}} $M$ is [[Definition:Well-Ordered Class|well-ordered]] by the [[Definition:Subset Relation|subset relation]] such that:
{{begin-axiom}}
{{axiom | n = 1
| t = the [[Defi... | G-Tower is G-Ordered | https://proofwiki.org/wiki/G-Tower_is_G-Ordered | https://proofwiki.org/wiki/G-Tower_is_G-Ordered | [
"G-Towers",
"G-Ordered Classes"
] | [
"Definition:Class (Class Theory)",
"Definition:Progressing Mapping",
"Definition:G-Tower",
"Definition:G-Ordered Class"
] | [
"Definition:G-Ordered Class",
"Definition:G-Ordered Class",
"Definition:Well-Ordered Class",
"Definition:Subset Relation",
"Definition:Smallest Element/Class Theory",
"Definition:Immediate Successor Element/Class Theory",
"Definition:Immediate Predecessor Element/Class Theory",
"Definition:Limit Eleme... |
proofwiki-19703 | Integer Power of Root of Unity is Root of Unity | Let $n \in \Z_{> 0}$ be a strictly positive integer.
Let $F$ be a field.
Let $\alpha$ be an $n$-th root of unity.
Let $k \in \Z$.
Then:
:$\alpha^k$ is an $n$-th root of unity. | We have:
{{begin-eqn}}
{{eqn | l = \alpha^n
| r = 1
| c = By Assumption
}}
{{eqn | l = \paren{\alpha^n}^k
| r = 1^k
| c =
}}
{{eqn | l = \paren{\alpha^k}^n
| r = 1
| c = Product of Indices Law for Field
}}
{{end-eqn}}
{{qed}}
Category:Roots of Unity
kogr9c6m0ik1uo6e99ev1bkenfhu0qc | Let $n \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $\alpha$ be an [[Definition:Root of Unity|$n$-th root of unity]].
Let $k \in \Z$.
Then:
:$\alpha^k$ is an [[Definition:Root of Unity|$n$-th root of unity]]. | We have:
{{begin-eqn}}
{{eqn | l = \alpha^n
| r = 1
| c = By Assumption
}}
{{eqn | l = \paren{\alpha^n}^k
| r = 1^k
| c =
}}
{{eqn | l = \paren{\alpha^k}^n
| r = 1
| c = [[Product of Indices Law for Field]]
}}
{{end-eqn}}
{{qed}}
[[Category:Roots of Unity]]
kogr9c6m0ik1uo6e99ev1bke... | Integer Power of Root of Unity is Root of Unity | https://proofwiki.org/wiki/Integer_Power_of_Root_of_Unity_is_Root_of_Unity | https://proofwiki.org/wiki/Integer_Power_of_Root_of_Unity_is_Root_of_Unity | [
"Roots of Unity"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Field (Abstract Algebra)",
"Definition:Root of Unity",
"Definition:Root of Unity"
] | [
"Index Laws/Product of Indices/Field",
"Category:Roots of Unity"
] |
proofwiki-19704 | Set is G-Set iff Element of G-Ordered Set | Let $g$ be a progressing mapping.
Let $x$ be a set.
Then:
:$x$ is a $g$-set
{{iff}}:
:$x$ is an element of a $g$-ordered set. | Let $M$ be the class of all $g$-sets.
Then $M$ is a $g$-tower. | Let $g$ be a [[Definition:Progressing Mapping|progressing mapping]].
Let $x$ be a [[Definition:Set|set]].
Then:
:$x$ is a [[Definition:G-Set|$g$-set]]
{{iff}}:
:$x$ is an [[Definition:Element|element]] of a [[Definition:G-Ordered Class|$g$-ordered set]]. | Let $M$ be the [[Definition:Class (Class Theory)|class]] of all [[Definition:G-Set|$g$-sets]].
Then $M$ is a [[Definition:G-Tower|$g$-tower]]. | Set is G-Set iff Element of G-Ordered Set | https://proofwiki.org/wiki/Set_is_G-Set_iff_Element_of_G-Ordered_Set | https://proofwiki.org/wiki/Set_is_G-Set_iff_Element_of_G-Ordered_Set | [
"G-Sets",
"G-Ordered Classes"
] | [
"Definition:Progressing Mapping",
"Definition:Set",
"Definition:G-Set",
"Definition:Element",
"Definition:G-Ordered Class"
] | [
"Definition:Class (Class Theory)",
"Definition:G-Set",
"Definition:G-Tower",
"Definition:G-Set"
] |
proofwiki-19705 | Congruent Powers of Root of Unity are Equal | Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $F$ be a field.
Let $\alpha$ be an $n$th root of unity.
Let $k, l \in \Z$ such that $k \equiv l \pmod n$.
Then:
:$\alpha^k = \alpha^l$ | By {{Defof|Congruence Modulo Integer}}:
:$\exists c \in \Z : k = l + c n$
We have:
{{begin-eqn}}
{{eqn | l = \alpha^k
| r = \alpha^{\paren {l + c n} }
}}
{{eqn | r = \alpha^l \alpha^{\paren {c n} }
| c = Sum of Indices Law for Field
}}
{{eqn | r = \alpha^l \cdot 1
| c = Integer Power of Root of Unity ... | Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $\alpha$ be an [[Definition:Root of Unity|$n$th root of unity]].
Let $k, l \in \Z$ such that $k \equiv l \pmod n$.
Then:
:$\alpha^k = \alpha^l$ | By {{Defof|Congruence Modulo Integer}}:
:$\exists c \in \Z : k = l + c n$
We have:
{{begin-eqn}}
{{eqn | l = \alpha^k
| r = \alpha^{\paren {l + c n} }
}}
{{eqn | r = \alpha^l \alpha^{\paren {c n} }
| c = [[Sum of Indices Law for Field]]
}}
{{eqn | r = \alpha^l \cdot 1
| c = [[Integer Power of Root of... | Congruent Powers of Root of Unity are Equal | https://proofwiki.org/wiki/Congruent_Powers_of_Root_of_Unity_are_Equal | https://proofwiki.org/wiki/Congruent_Powers_of_Root_of_Unity_are_Equal | [
"Roots of Unity"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Field (Abstract Algebra)",
"Definition:Root of Unity"
] | [
"Index Laws/Sum of Indices/Field",
"Integer Power of Root of Unity is Root of Unity",
"Category:Roots of Unity"
] |
proofwiki-19706 | Power of Root of Unity Equals Power of Remainder | Let $n \in \Z_{> 0}$ be a strictly positive integer.
Let $F$ be a field.
Let $\alpha$ be an $n$-th root of unity.
Let $k \in \Z$.
Then:
:$\alpha^k = \alpha^r$
where $0 \le r < n$ is the remainder of $k$ on division by $n$. | From Division Theorem:
:$\exists r, c \in \Z : 0 \le r < n : k = r + c n$
By {{Defof|Congruence Modulo Integer}}:
:$k \equiv r \pmod n$
From Congruent Powers of Root of Unity are Equal
:$\alpha^k = \alpha^r$
{{qed}}
Category:Roots of Unity
o141p9qnp7h8wekn6jxfsr1dr8xmzlg | Let $n \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $\alpha$ be an [[Definition:Root of Unity|$n$-th root of unity]].
Let $k \in \Z$.
Then:
:$\alpha^k = \alpha^r$
where $0 \le r < n$ is the [[Definition:Remain... | From [[Division Theorem]]:
:$\exists r, c \in \Z : 0 \le r < n : k = r + c n$
By {{Defof|Congruence Modulo Integer}}:
:$k \equiv r \pmod n$
From [[Congruent Powers of Root of Unity are Equal]]
:$\alpha^k = \alpha^r$
{{qed}}
[[Category:Roots of Unity]]
o141p9qnp7h8wekn6jxfsr1dr8xmzlg | Power of Root of Unity Equals Power of Remainder | https://proofwiki.org/wiki/Power_of_Root_of_Unity_Equals_Power_of_Remainder | https://proofwiki.org/wiki/Power_of_Root_of_Unity_Equals_Power_of_Remainder | [
"Roots of Unity"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Field (Abstract Algebra)",
"Definition:Root of Unity",
"Definition:Remainder"
] | [
"Division Theorem",
"Congruent Powers of Root of Unity are Equal",
"Category:Roots of Unity"
] |
proofwiki-19707 | Conditional Entropy Given Trivial Sigma-Algebra is Entropy | Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\AA \subseteq \Sigma$ be a finite sub-$\sigma$-algebra.
Let $\NN := \set {\O, \Omega}$ be the trivial $\sigma$-algebra.
Then:
:$\ds \map H {\AA \mid \NN} = \map H \AA$
where:
:$\map H {\cdot \mid \cdot}$ denotes the conditional entropy
:$\map H {\, \cdot ... | {{begin-eqn}}
{{eqn | l = \map H {\AA \mid \NN}
| r = \map H {\map \xi \AA \mid \map \xi \NN}
| c = {{Defof|Conditional Entropy of Finite Sub-Sigma-Algebra|Conditional Entropy of Finite Sub-$\sigma$-Algebra}}
}}
{{eqn | r = \sum_{\substack {B \mathop \in {\map \xi \NN } \\ \map \Pr B \mathop > 0} } \sum_{... | Let $\struct {\Omega, \Sigma, \Pr}$ be a [[Definition:Probability Space|probability space]].
Let $\AA \subseteq \Sigma$ be a [[Definition:Finite Sub-Sigma-Algebra|finite sub-$\sigma$-algebra]].
Let $\NN := \set {\O, \Omega}$ be the [[Sigma-Algebra/Examples/Trivial Sigma-Algebra|trivial $\sigma$-algebra]].
Then:
:$\... | {{begin-eqn}}
{{eqn | l = \map H {\AA \mid \NN}
| r = \map H {\map \xi \AA \mid \map \xi \NN}
| c = {{Defof|Conditional Entropy of Finite Sub-Sigma-Algebra|Conditional Entropy of Finite Sub-$\sigma$-Algebra}}
}}
{{eqn | r = \sum_{\substack {B \mathop \in {\map \xi \NN } \\ \map \Pr B \mathop > 0} } \sum_{... | Conditional Entropy Given Trivial Sigma-Algebra is Entropy | https://proofwiki.org/wiki/Conditional_Entropy_Given_Trivial_Sigma-Algebra_is_Entropy | https://proofwiki.org/wiki/Conditional_Entropy_Given_Trivial_Sigma-Algebra_is_Entropy | [
"Probability Theory",
"Ergodic Theory"
] | [
"Definition:Probability Space",
"Definition:Finite Sub-Sigma-Algebra",
"Sigma-Algebra/Examples/Trivial Sigma-Algebra",
"Definition:Conditional Entropy of Finite Sub-Sigma-Algebra",
"Definition:Entropy of Finite Sub-Sigma-Algebra"
] | [
"Definition:Finite Partition Generated by Finite Sub-Sigma-Algebra",
"Definition:Probability Measure"
] |
proofwiki-19708 | Preimage of Union Mapping is Union of Preimages | Let $A$, $B$ and $Y$ be sets.
Let $f: A \to Y$ and $g: B \to Y$ be mappings that agree on $A \cap B$.
Let $f \cup g$ be the union of the mappings $f$ and $g$:
:$\forall x \in A \cup B: \map {f \cup g} x = \begin {cases} \map f x & : x \in A \\ \map g x & : x \in B \end {cases}$
Let $S \subseteq Y$ be a subset of $Y$.
T... | From Union of Mappings which Agree is Mapping:
:$f \cup g$ is well-defined
Let $x \in \paren {f \cup g}^{-1} \sqbrk S$
By {{Defof|Inverse Image}}:
:$\paren {f \cup g}^{-1} \sqbrk S \subseteq A \cup B$
Let $x \in A$.
We have:
:$\map f x = \map {\paren {f \cup g} } x \in S$
By {{Defof|Inverse Image}}:
:$x \in f^{-1} \sqb... | Let $A$, $B$ and $Y$ be [[Definition:Set|sets]].
Let $f: A \to Y$ and $g: B \to Y$ be [[Definition:Mapping|mappings]] that [[Definition:Agreement of Mappings|agree]] on $A \cap B$.
Let $f \cup g$ be the [[Definition:Set Union|union]] of the [[Definition:Mapping|mappings]] $f$ and $g$:
:$\forall x \in A \cup B: \map... | From [[Union of Mappings which Agree is Mapping]]:
:$f \cup g$ is [[Definition:Well-Defined|well-defined]]
Let $x \in \paren {f \cup g}^{-1} \sqbrk S$
By {{Defof|Inverse Image}}:
:$\paren {f \cup g}^{-1} \sqbrk S \subseteq A \cup B$
Let $x \in A$.
We have:
:$\map f x = \map {\paren {f \cup g} } x \in S$
By {{Def... | Preimage of Union Mapping is Union of Preimages | https://proofwiki.org/wiki/Preimage_of_Union_Mapping_is_Union_of_Preimages | https://proofwiki.org/wiki/Preimage_of_Union_Mapping_is_Union_of_Preimages | [
"Preimages under Mappings",
"Union Mappings"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Agreement/Mappings",
"Definition:Set Union",
"Definition:Mapping",
"Definition:Subset"
] | [
"Union of Mappings which Agree is Mapping",
"Definition:Well-Defined",
"Category:Preimages under Mappings",
"Category:Union Mappings"
] |
proofwiki-19709 | Conditional Entropy of Join as Sum | Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\AA, \CC, \DD \subseteq \Sigma$ be finite sub-$\sigma$-algebras.
Then:
:$\ds \map H {\AA \vee \CC \mid \DD} = \map H {\AA \mid \DD} + \map H {\CC \mid \AA \vee \DD} $
where:
:$\map H {\cdot \mid \cdot}$ denotes the conditional entropy
:$\vee$ denotes the ... | Consider the generated finite partitions:
:$\xi := \map \xi \AA$
:$\eta := \map \xi \CC$
:$\gamma := \map \xi \DD$
By definition of conditional entropy of finite sub-sigma-algebra, we shall show:
:$\map H {\xi \vee \eta \mid \gamma} = \map H {\xi \mid \gamma} + \map H {\eta \mid \xi \vee \gamma}$
Then:
{{begin-eqn}}
{{... | Let $\struct {\Omega, \Sigma, \Pr}$ be a [[Definition:Probability Space|probability space]].
Let $\AA, \CC, \DD \subseteq \Sigma$ be [[Definition:Finite Sub-Sigma-Algebra|finite sub-$\sigma$-algebras]].
Then:
:$\ds \map H {\AA \vee \CC \mid \DD} = \map H {\AA \mid \DD} + \map H {\CC \mid \AA \vee \DD} $
where:
:$\ma... | Consider the [[Definition:Finite Partition Generated by Finite Sub-Sigma-Algebra|generated finite partitions]]:
:$\xi := \map \xi \AA$
:$\eta := \map \xi \CC$
:$\gamma := \map \xi \DD$
By definition of [[Definition:Conditional Entropy of Finite Sub-Sigma-Algebra|conditional entropy of finite sub-sigma-algebra]], we sh... | Conditional Entropy of Join as Sum | https://proofwiki.org/wiki/Conditional_Entropy_of_Join_as_Sum | https://proofwiki.org/wiki/Conditional_Entropy_of_Join_as_Sum | [
"Conditional Entropy of Join as Sum",
"Probability Theory",
"Ergodic Theory"
] | [
"Definition:Probability Space",
"Definition:Finite Sub-Sigma-Algebra",
"Definition:Conditional Entropy of Finite Sub-Sigma-Algebra",
"Definition:Join of Finite Sub-Sigma-Algebras"
] | [
"Definition:Finite Partition Generated by Finite Sub-Sigma-Algebra",
"Definition:Conditional Entropy of Finite Sub-Sigma-Algebra",
"Real Logarithm is Completely Additive"
] |
proofwiki-19710 | Square of Norm of Vector Cross Product | Let $\mathbf a$ and $\mathbf b$ be vectors in the real Euclidean space $\R^3$.
Let $\times$ denote the vector cross product.
Then:
:$\norm {\mathbf a \times \mathbf b}^2 = \norm {\mathbf a}^2 \norm {\mathbf b}^2 - \paren {\mathbf a \cdot \mathbf b}^2$ | Let $\mathbf a = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$, and $\mathbf b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {\mathbf a \times \mathbf b }^2
| r = \paren {\mathbf a \times \mathbf b } \cdot \paren {\mathbf a \times \mathbf b}
| c = {{Defof|Eucli... | Let $\mathbf a$ and $\mathbf b$ be [[Definition:Space Vector|vectors]] in the [[Definition:Real Euclidean Space|real Euclidean space]] $\R^3$.
Let $\times$ denote the [[Definition:Vector Cross Product|vector cross product]].
Then:
:$\norm {\mathbf a \times \mathbf b}^2 = \norm {\mathbf a}^2 \norm {\mathbf b}^2 - \p... | Let $\mathbf a = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$, and $\mathbf b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {\mathbf a \times \mathbf b }^2
| r = \paren {\mathbf a \times \mathbf b } \cdot \paren {\mathbf a \times \mathbf b}
| c = {{Defof|Euc... | Square of Norm of Vector Cross Product | https://proofwiki.org/wiki/Square_of_Norm_of_Vector_Cross_Product | https://proofwiki.org/wiki/Square_of_Norm_of_Vector_Cross_Product | [
"Vector Cross Product"
] | [
"Definition:Vector/Real Euclidean Space/Space Vector",
"Definition:Euclidean Space/Real",
"Definition:Vector Cross Product"
] | [] |
proofwiki-19711 | Class of All Ordinals is Minimally Superinductive over Successor Mapping | The class of all ordinals $\On$ is the unique class which is minimally superinductive under the successor mapping. | We need to show that:
:$\On$ is a superinductive class under the successor mapping
and:
:no proper subclass of $\On$ is superinductive class under the successor mapping.
We recall immediately that Successor Mapping is Progressing.
This validates the definition of superinductive class under the successor mapping.
By def... | The [[Definition:Class of All Ordinals|class of all ordinals]] $\On$ is the [[Definition:Unique|unique]] [[Definition:Class (Class Theory)|class]] which is [[Definition:Minimally Superinductive Class|minimally superinductive]] under the [[Definition:Successor Mapping|successor mapping]]. | We need to show that:
:$\On$ is a [[Definition:Superinductive Class|superinductive class]] under the [[Definition:Successor Mapping|successor mapping]]
and:
:no [[Definition:Proper Subclass|proper subclass]] of $\On$ is [[Definition:Superinductive Class|superinductive class]] under the [[Definition:Successor Mapping|su... | Class of All Ordinals is Minimally Superinductive over Successor Mapping | https://proofwiki.org/wiki/Class_of_All_Ordinals_is_Minimally_Superinductive_over_Successor_Mapping | https://proofwiki.org/wiki/Class_of_All_Ordinals_is_Minimally_Superinductive_over_Successor_Mapping | [
"Class of All Ordinals",
"Minimally Superinductive Classes",
"Successor Mapping"
] | [
"Definition:Class of All Ordinals",
"Definition:Unique",
"Definition:Class (Class Theory)",
"Definition:Minimally Superinductive Class",
"Definition:Successor Mapping"
] | [
"Definition:Superinductive Class",
"Definition:Successor Mapping",
"Definition:Proper Subclass",
"Definition:Superinductive Class",
"Definition:Successor Mapping",
"Successor Mapping is Progressing",
"Definition:Superinductive Class",
"Definition:Successor Mapping",
"Definition:Ordinal",
"Definiti... |
proofwiki-19712 | Properties of Class of All Ordinals/Zero is Ordinal | The natural number $0$ is an element of $\On$. | We have the result that Class of All Ordinals is Minimally Superinductive over Successor Mapping.
Hence $\On$ is {{afortiori}} a superinductive class {{WRT}} the successor mapping.
Hence, by definition of superinductive class:
:$\O \in \On$
We identify the natural number $0$ via the von Neumann construction of the natu... | The [[Definition:Zero (Number)|natural number $0$]] is an [[Definition:Element|element]] of $\On$. | We have the result that [[Class of All Ordinals is Minimally Superinductive over Successor Mapping]].
Hence $\On$ is {{afortiori}} a [[Definition:Superinductive Class|superinductive class]] {{WRT}} the [[Definition:Successor Mapping|successor mapping]].
Hence, by definition of [[Definition:Superinductive Class|superi... | Properties of Class of All Ordinals/Zero is Ordinal | https://proofwiki.org/wiki/Properties_of_Class_of_All_Ordinals/Zero_is_Ordinal | https://proofwiki.org/wiki/Properties_of_Class_of_All_Ordinals/Zero_is_Ordinal | [
"Properties of Class of All Ordinals"
] | [
"Definition:Zero (Number)",
"Definition:Element"
] | [
"Class of All Ordinals is Minimally Superinductive over Successor Mapping",
"Definition:Superinductive Class",
"Definition:Successor Mapping",
"Definition:Superinductive Class",
"Definition:Zero (Number)",
"Definition:Natural Numbers/Von Neumann Construction"
] |
proofwiki-19713 | Properties of Class of All Ordinals/Union of Chain of Ordinals is Ordinal | Let $C$ be a chain of elements of $\On$.
Then its union $\bigcup C$ is also an element of $\On$. | We have the result that Class of All Ordinals is Minimally Superinductive over Successor Mapping.
Hence $\On$ is {{afortiori}} a superinductive class {{WRT}} the successor mapping.
Hence, by definition of superinductive class:
:$\On$ is closed under chain unions.
That is:
:$\forall C \in \On: \bigcup C \in \On$
where:
... | Let $C$ be a [[Definition:Chain of Sets|chain]] of [[Definition:Element|elements]] of $\On$.
Then its [[Definition:Union of Set of Sets|union]] $\bigcup C$ is also an [[Definition:Element|element]] of $\On$. | We have the result that [[Class of All Ordinals is Minimally Superinductive over Successor Mapping]].
Hence $\On$ is {{afortiori}} a [[Definition:Superinductive Class|superinductive class]] {{WRT}} the [[Definition:Successor Mapping|successor mapping]].
Hence, by definition of [[Definition:Superinductive Class|superi... | Properties of Class of All Ordinals/Union of Chain of Ordinals is Ordinal | https://proofwiki.org/wiki/Properties_of_Class_of_All_Ordinals/Union_of_Chain_of_Ordinals_is_Ordinal | https://proofwiki.org/wiki/Properties_of_Class_of_All_Ordinals/Union_of_Chain_of_Ordinals_is_Ordinal | [
"Properties of Class of All Ordinals"
] | [
"Definition:Chain (Order Theory)/Subset Relation",
"Definition:Element",
"Definition:Set Union/Set of Sets",
"Definition:Element"
] | [
"Class of All Ordinals is Minimally Superinductive over Successor Mapping",
"Definition:Superinductive Class",
"Definition:Successor Mapping",
"Definition:Superinductive Class",
"Definition:Closure under Chain Unions",
"Definition:Chain (Order Theory)/Subset Relation",
"Definition:Element/Class",
"Def... |
proofwiki-19714 | Properties of Class of All Ordinals/Superinduction Principle | Let $A$ be a class which satisfies the following $3$ conditions:
{{begin-axiom}}
{{axiom | n = 1
| lc= $A$ contains the zero ordinal $0$:
| q =
| m = 0 \in A
}}
{{axiom | n = 2
| lc= $A$ is closed under successor mapping:
| q = \forall \alpha
| m = \paren {\alpha \in A \... | We note that the zero ordinal, denoted $0$, is identified as the empty set:
:$0 \:= \O$
Hence by definition $A$ is indeed a superinductive class under the successor mapping.
By the definition of ordinal:
:$\alpha$ is an '''ordinal''' {{iff}} $\alpha$ is an element of every superinductive class.
Hence $\On$ is a subclas... | Let $A$ be a [[Definition:Class (Class Theory)|class]] which satisfies the following $3$ conditions:
{{begin-axiom}}
{{axiom | n = 1
| lc= $A$ contains the [[Definition:Zero Ordinal|zero ordinal]] $0$:
| q =
| m = 0 \in A
}}
{{axiom | n = 2
| lc= $A$ is [[Definition:Closed Class under ... | We note that the [[Definition:Zero Ordinal|zero ordinal]], denoted $0$, is identified as the [[Definition:Empty Set|empty set]]:
:$0 \:= \O$
Hence by definition $A$ is indeed a [[Definition:Superinductive Class|superinductive class]] under the [[Definition:Successor Mapping|successor mapping]].
By the definition of ... | Properties of Class of All Ordinals/Superinduction Principle | https://proofwiki.org/wiki/Properties_of_Class_of_All_Ordinals/Superinduction_Principle | https://proofwiki.org/wiki/Properties_of_Class_of_All_Ordinals/Superinduction_Principle | [
"Properties of Class of All Ordinals"
] | [
"Definition:Class (Class Theory)",
"Definition:Zero (Ordinal)",
"Definition:Closed under Mapping/Class Theory",
"Definition:Successor Mapping",
"Definition:Closure under Chain Unions",
"Definition:Chain (Order Theory)/Subset Relation",
"Definition:Element/Class",
"Definition:Superinductive Class",
"... | [
"Definition:Zero (Ordinal)",
"Definition:Empty Set",
"Definition:Superinductive Class",
"Definition:Successor Mapping",
"Definition:Ordinal",
"Definition:Ordinal",
"Definition:Element/Class",
"Definition:Superinductive Class",
"Definition:Subclass",
"Definition:Superinductive Class"
] |
proofwiki-19715 | Class of All Ordinals is G-Tower | Let $\On$ denote the class of all ordinals.
Let $g$ be the successor mapping:
:$\forall x \in \On: \map g x = x \cup \set x$
Then $\On$ is a $g$-tower. | From Successor Mapping is Progressing, $g$ is a progressing mapping.
From Class of All Ordinals is Minimally Superinductive over Successor Mapping, $\On$ is superinductive over $g$.
Hence the result by definition of $g$-tower.
{{qed}}
Category:G-Towers
Category:Class of All Ordinals
42w18hcsmtascq5stlvt4kp9n3i9fq5 | Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
Let $g$ be the [[Definition:Successor Mapping|successor mapping]]:
:$\forall x \in \On: \map g x = x \cup \set x$
Then $\On$ is a [[Definition:G-Tower|$g$-tower]]. | From [[Successor Mapping is Progressing]], $g$ is a [[Definition:Progressing Mapping|progressing mapping]].
From [[Class of All Ordinals is Minimally Superinductive over Successor Mapping]], $\On$ is [[Definition:Superinductive Class|superinductive]] over $g$.
Hence the result by definition of [[Definition:G-Tower|$g... | Class of All Ordinals is G-Tower | https://proofwiki.org/wiki/Class_of_All_Ordinals_is_G-Tower | https://proofwiki.org/wiki/Class_of_All_Ordinals_is_G-Tower | [
"G-Towers",
"Class of All Ordinals"
] | [
"Definition:Class of All Ordinals",
"Definition:Successor Mapping",
"Definition:G-Tower"
] | [
"Successor Mapping is Progressing",
"Definition:Progressing Mapping",
"Class of All Ordinals is Minimally Superinductive over Successor Mapping",
"Definition:Superinductive Class",
"Definition:G-Tower",
"Category:G-Towers",
"Category:Class of All Ordinals"
] |
proofwiki-19716 | Natural Number is Ordinal | Let $n \in \N$ be a natural number.
Then $n$ is an ordinal. | From the von Neumann construction of the natural numbers, $\N$ is identified with the minimally inductive set $\omega$.
From Superinductive Class under Successor Mapping contains All Ordinals, it follows by the Principle of Mathematical Induction that every natural number is an ordinal.
{{qed}} | Let $n \in \N$ be a [[Definition:Natural Number|natural number]].
Then $n$ is an [[Definition:Ordinal|ordinal]]. | From the [[Definition:Von Neumann Construction of Natural Numbers|von Neumann construction of the natural numbers]], $\N$ is identified with the [[Definition:Minimally Inductive Set|minimally inductive set]] $\omega$.
From [[Superinductive Class under Successor Mapping contains All Ordinals]], it follows by the [[Prin... | Natural Number is Ordinal/Proof 2 | https://proofwiki.org/wiki/Natural_Number_is_Ordinal | https://proofwiki.org/wiki/Natural_Number_is_Ordinal/Proof_2 | [
"Natural Number is Ordinal",
"Ordinals",
"Natural Numbers"
] | [
"Definition:Natural Numbers",
"Definition:Ordinal"
] | [
"Definition:Natural Numbers/Von Neumann Construction",
"Definition:Minimally Inductive Set",
"Properties of Class of All Ordinals/Superinduction Principle",
"Principle of Mathematical Induction",
"Definition:Natural Numbers",
"Definition:Ordinal"
] |
proofwiki-19717 | Zero is Smallest Ordinal | The natural number $0$ is the smallest ordinal. | Let $\On$ denote the class of all ordinals.
By Zero is Ordinal, $0$ is an element of $\On$.
We identify the natural number $0$ via the von Neumann construction of the natural numbers as:
:$0 := \O$
By Empty Class is Subclass of All Classes:
:$\forall \alpha \in \On: \O \subseteq \alpha$
Hence the result by definition o... | The [[Definition:Zero (Number)|natural number $0$]] is the [[Definition:Smallest Element (Class Theory)|smallest]] [[Definition:Ordinal|ordinal]]. | Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
By [[Zero is Ordinal]], $0$ is an [[Definition:Element|element]] of $\On$.
We identify the [[Definition:Zero (Number)|natural number $0$]] via the [[Definition:Von Neumann Construction of Natural Numbers|von Neumann construction of the n... | Zero is Smallest Ordinal | https://proofwiki.org/wiki/Zero_is_Smallest_Ordinal | https://proofwiki.org/wiki/Zero_is_Smallest_Ordinal | [
"Ordinals",
"Zero"
] | [
"Definition:Zero (Number)",
"Definition:Smallest Element/Class Theory",
"Definition:Ordinal"
] | [
"Definition:Class of All Ordinals",
"Properties of Class of All Ordinals/Zero is Ordinal",
"Definition:Element",
"Definition:Zero (Number)",
"Definition:Natural Numbers/Von Neumann Construction",
"Empty Class is Subclass of All Classes",
"Definition:Smallest Element/Class Theory",
"Category:Ordinals",... |
proofwiki-19718 | Empty Set is Ordinary | The empty set is an ordinary set:
:$\O \notin \O$ | By definition:
:$\forall x: x \notin \O$
and so in particular:
:$\O \notin \O$
Hence the result.
{{qed}}
Category:Empty Set
Category:Ordinary Sets
rgi3smowkehxpv2nnbxd3ahvbwl0v2m | The [[Definition:Empty Set|empty set]] is an [[Definition:Ordinary Set|ordinary set]]:
:$\O \notin \O$ | By definition:
:$\forall x: x \notin \O$
and so in particular:
:$\O \notin \O$
Hence the result.
{{qed}}
[[Category:Empty Set]]
[[Category:Ordinary Sets]]
rgi3smowkehxpv2nnbxd3ahvbwl0v2m | Empty Set is Ordinary | https://proofwiki.org/wiki/Empty_Set_is_Ordinary | https://proofwiki.org/wiki/Empty_Set_is_Ordinary | [
"Empty Set",
"Ordinary Sets",
"Empty Set",
"Ordinary Sets"
] | [
"Definition:Empty Set",
"Definition:Ordinary Set"
] | [
"Category:Empty Set",
"Category:Ordinary Sets"
] |
proofwiki-19719 | Successor Set of Ordinary Transitive Set is Ordinary | Let $x$ be a transitive set which is also ordinary.
Let $x^+$ denote the successor set of $x$:
:$x^+ = x \cup \set x$
Then $x^+$ is also an ordinary set. | By definition of ordinary set:
:$x \notin x$
{{AimForCont}} $x^+ \in x^+$.
Then we have:
{{begin-eqn}}
{{eqn | l = x^+
| r = x \cup \set x
| c = {{Defof|Successor Set}}
}}
{{eqn | ll= \leadsto
| l = x^+
| r = x
| c = {{Defof|Set Union}}
}}
{{eqn | lo= \lor
| l = x^+
| o = \in
... | Let $x$ be a [[Definition:Transitive Set|transitive set]] which is also [[Definition:Ordinary Set|ordinary]].
Let $x^+$ denote the [[Definition:Successor Set|successor set]] of $x$:
:$x^+ = x \cup \set x$
Then $x^+$ is also an [[Definition:Ordinary Set|ordinary set]]. | By definition of [[Definition:Ordinary Set|ordinary set]]:
:$x \notin x$
{{AimForCont}} $x^+ \in x^+$.
Then we have:
{{begin-eqn}}
{{eqn | l = x^+
| r = x \cup \set x
| c = {{Defof|Successor Set}}
}}
{{eqn | ll= \leadsto
| l = x^+
| r = x
| c = {{Defof|Set Union}}
}}
{{eqn | lo= \lor
... | Successor Set of Ordinary Transitive Set is Ordinary | https://proofwiki.org/wiki/Successor_Set_of_Ordinary_Transitive_Set_is_Ordinary | https://proofwiki.org/wiki/Successor_Set_of_Ordinary_Transitive_Set_is_Ordinary | [
"Ordinary Sets",
"Transitive Classes",
"Successor Mapping"
] | [
"Definition:Transitive Class",
"Definition:Ordinary Set",
"Definition:Successor Mapping/Successor Set",
"Definition:Ordinary Set"
] | [
"Definition:Ordinary Set",
"Definition:Transitive Class",
"Definition:Set Union/Set of Sets",
"Proof by Cases",
"Definition:Contradiction",
"Proof by Counterexample"
] |
proofwiki-19720 | Ordinal is Proper Subset of Successor | Let $\alpha$ be an ordinal.
Then:
:$\alpha \subsetneqq \alpha^+$
where $\alpha^+$ denotes the successor set of $\alpha$.
That is:
:$\alpha$ is a proper subset of $\alpha^+$. | {{AimForCont}} $\alpha = \alpha^+$.
By definition:
:$\alpha^+ = \alpha \cup \set \alpha$
and so:
:$\alpha \subseteq \alpha^+$
and:
:$\alpha \in \alpha^+$
which leads to:
:$\alpha \in \alpha$
But from Ordinal is not Element of Itself:
:$\alpha \notin \alpha$
Hence by Proof by Contradiction:
:$\alpha \ne \alpha^+$
But we... | Let $\alpha$ be an [[Definition:Ordinal|ordinal]].
Then:
:$\alpha \subsetneqq \alpha^+$
where $\alpha^+$ denotes the [[Definition:Successor Set|successor set]] of $\alpha$.
That is:
:$\alpha$ is a [[Definition:Proper Subset|proper subset]] of $\alpha^+$. | {{AimForCont}} $\alpha = \alpha^+$.
By definition:
:$\alpha^+ = \alpha \cup \set \alpha$
and so:
:$\alpha \subseteq \alpha^+$
and:
:$\alpha \in \alpha^+$
which leads to:
:$\alpha \in \alpha$
But from [[Ordinal is not Element of Itself]]:
:$\alpha \notin \alpha$
Hence by [[Proof by Contradiction]]:
:$\alpha \ne \al... | Ordinal is Proper Subset of Successor | https://proofwiki.org/wiki/Ordinal_is_Proper_Subset_of_Successor | https://proofwiki.org/wiki/Ordinal_is_Proper_Subset_of_Successor | [
"Ordinals",
"Successor Mapping"
] | [
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set",
"Definition:Proper Subset"
] | [
"Ordinal is not Element of Itself",
"Proof by Contradiction"
] |
proofwiki-19721 | Successor Mapping on Ordinals is Strictly Progressing | Let $\On$ denote the class of all ordinals.
Let $s$ denote the successor mapping on $\On$:
:$\forall \alpha \in \On: \map s \alpha := \alpha \cup \set \alpha$
where $\alpha$ is an ordinal.
Then $s$ is a strictly progressing mapping. | From Ordinal is Proper Subset of Successor:
:$\alpha \subsetneqq \alpha^+$
where $\alpha^+$ is identified as $\map s \alpha$.
The result follows by definition of strictly progressing mapping.
{{qed}} | Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
Let $s$ denote the [[Definition:Successor Mapping|successor mapping]] on $\On$:
:$\forall \alpha \in \On: \map s \alpha := \alpha \cup \set \alpha$
where $\alpha$ is an [[Definition:Ordinal|ordinal]].
Then $s$ is a [[Definition:Strictly... | From [[Ordinal is Proper Subset of Successor]]:
:$\alpha \subsetneqq \alpha^+$
where $\alpha^+$ is identified as $\map s \alpha$.
The result follows by definition of [[Definition:Strictly Progressing Mapping|strictly progressing mapping]].
{{qed}} | Successor Mapping on Ordinals is Strictly Progressing | https://proofwiki.org/wiki/Successor_Mapping_on_Ordinals_is_Strictly_Progressing | https://proofwiki.org/wiki/Successor_Mapping_on_Ordinals_is_Strictly_Progressing | [
"Successor Mapping",
"Strictly Progressing Mappings",
"Ordinals"
] | [
"Definition:Class of All Ordinals",
"Definition:Successor Mapping",
"Definition:Ordinal",
"Definition:Strictly Progressing Mapping"
] | [
"Ordinal is Proper Subset of Successor",
"Definition:Strictly Progressing Mapping"
] |
proofwiki-19722 | Class of All Ordinals is Proper Class | Let $\On$ denote the class of all ordinals.
Then $\On$ is a proper class.
That is, $\On$ is not a set. | We have that Successor Mapping on Ordinals is Strictly Progressing.
The result follows from Superinductive Class under Strictly Progressing Mapping is Proper Class.
{{qed}} | Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
Then $\On$ is a [[Definition:Proper Class|proper class]].
That is, $\On$ is not a [[Definition:Set|set]]. | We have that [[Successor Mapping on Ordinals is Strictly Progressing]].
The result follows from [[Superinductive Class under Strictly Progressing Mapping is Proper Class]].
{{qed}} | Class of All Ordinals is Proper Class/Proof 1 | https://proofwiki.org/wiki/Class_of_All_Ordinals_is_Proper_Class | https://proofwiki.org/wiki/Class_of_All_Ordinals_is_Proper_Class/Proof_1 | [
"Class of All Ordinals is Proper Class",
"Class of All Ordinals"
] | [
"Definition:Class of All Ordinals",
"Definition:Class (Class Theory)/Proper Class",
"Definition:Set"
] | [
"Successor Mapping on Ordinals is Strictly Progressing",
"Superinductive Class under Strictly Progressing Mapping is Proper Class"
] |
proofwiki-19723 | Nilpotent Element is Contained in Prime Ideals | Let $R$ be a commutative ring with unity.
Let $\mathfrak p \subset R$ be a prime ideal of $R$.
Let $x \in R$ be a nilpotent element of $R$.
Then $x \in \mathfrak p$. | Let $x$ be nilpotent in $R$ as asserted.
Then by definition:
:$\exists n \in \Z_{>0}: x^n = 0$
But $0 \in \mathfrak p$ so:
:$x^n \in \mathfrak p$
{{begin-eqn}}
{{eqn | l = x
| r = \map \Rad {\mathfrak p}
| o = \in
| c = {{Defof|Radical of Ideal of Ring|index=1}} }}
{{eqn | r = \mathfrak p
| c = ... | Let $R$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let $\mathfrak p \subset R$ be a [[Definition:Prime Ideal of Commutative and Unitary Ring|prime ideal]] of $R$.
Let $x \in R$ be a [[Definition:Nilpotent Ring Element|nilpotent element]] of $R$.
Then $x \in \mathfrak p$. | Let $x$ be [[Definition:Nilpotent Ring Element|nilpotent]] in $R$ as asserted.
Then by definition:
:$\exists n \in \Z_{>0}: x^n = 0$
But $0 \in \mathfrak p$ so:
:$x^n \in \mathfrak p$
{{begin-eqn}}
{{eqn | l = x
| r = \map \Rad {\mathfrak p}
| o = \in
| c = {{Defof|Radical of Ideal of Ring|index=1... | Nilpotent Element is Contained in Prime Ideals | https://proofwiki.org/wiki/Nilpotent_Element_is_Contained_in_Prime_Ideals | https://proofwiki.org/wiki/Nilpotent_Element_is_Contained_in_Prime_Ideals | [
"Prime Ideals of Rings"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Prime Ideal of Ring/Commutative and Unitary Ring",
"Definition:Nilpotent Ring Element"
] | [
"Definition:Nilpotent Ring Element",
"Radical of Prime Ideal",
"Category:Prime Ideals of Rings"
] |
proofwiki-19724 | Well-Ordering of Class of All Ordinals under Subset Relation | Let $\On$ denote the class of all ordinals.
$\On$ is well-ordered by the subset relation such that the following $3$ conditions hold:
{{begin-axiom}}
{{axiom | n = 1
| lc= the smallest ordinal is $0$
}}
{{axiom | n = 2
| lc= for $\alpha \in \On$, the immediate successor of $\alpha$ is its successor set ... | We have that Class of All Ordinals is $g$-Tower.
By Zero is Smallest Ordinal, $0$ is the smallest element of $\On$.
We identify the natural number $0$ via the von Neumann construction of the natural numbers as:
:$0 := \O$
The result then follows directly from $g$-Tower is Well-Ordered under Subset Relation.
{{qed}} | Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
$\On$ is [[Definition:Well-Ordered Class|well-ordered]] by the [[Definition:Subset Relation|subset relation]] such that the following $3$ conditions hold:
{{begin-axiom}}
{{axiom | n = 1
| lc= the [[Definition:Smallest Element (C... | We have that [[Class of All Ordinals is G-Tower|Class of All Ordinals is $g$-Tower]].
By [[Zero is Smallest Ordinal]], $0$ is the [[Definition:Smallest Element (Class Theory)|smallest element]] of $\On$.
We identify the [[Definition:Zero (Number)|natural number $0$]] via the [[Definition:Von Neumann Construction of N... | Well-Ordering of Class of All Ordinals under Subset Relation | https://proofwiki.org/wiki/Well-Ordering_of_Class_of_All_Ordinals_under_Subset_Relation | https://proofwiki.org/wiki/Well-Ordering_of_Class_of_All_Ordinals_under_Subset_Relation | [
"Class of All Ordinals",
"Well-Orderings"
] | [
"Definition:Class of All Ordinals",
"Definition:Well-Ordered Class",
"Definition:Subset Relation",
"Definition:Smallest Element/Class Theory",
"Definition:Ordinal",
"Definition:Immediate Successor Element",
"Definition:Successor Mapping/Successor Set",
"Definition:Limit Ordinal",
"Definition:Set Uni... | [
"Class of All Ordinals is G-Tower",
"Zero is Smallest Ordinal",
"Definition:Smallest Element/Class Theory",
"Definition:Zero (Number)",
"Definition:Natural Numbers/Von Neumann Construction",
"G-Tower is Well-Ordered under Subset Relation"
] |
proofwiki-19725 | Strict Ordering of Ordinals is Equivalent to Membership Relation | Let $\On$ denote the class of all ordinals.
Let $<$ denote the (strict) usual ordering of $\On$.
Then:
:$\forall \alpha, \beta \in \On: \alpha < \beta \iff \alpha \in \beta$ | === Necessary Condition ===
Let $\alpha \in \beta$.
Then from Ordinal is Transitive:
:$\alpha \subseteq \beta$
But if $\alpha = \beta$ we would have $\alpha \in \alpha$.
This is contrary to Ordinal is not Element of Itself.
Hence we have:
:$\alpha \subseteq \beta$
and:
:$\alpha \ne \beta$
That is:
:$\alpha \subsetneqq ... | Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
Let $<$ denote the [[Definition:Usual Ordering of Ordinals|(strict) usual ordering of $\On$]].
Then:
:$\forall \alpha, \beta \in \On: \alpha < \beta \iff \alpha \in \beta$ | === Necessary Condition ===
Let $\alpha \in \beta$.
Then from [[Ordinal is Transitive]]:
:$\alpha \subseteq \beta$
But if $\alpha = \beta$ we would have $\alpha \in \alpha$.
This is contrary to [[Ordinal is not Element of Itself]].
Hence we have:
:$\alpha \subseteq \beta$
and:
:$\alpha \ne \beta$
That is:
:$\alph... | Strict Ordering of Ordinals is Equivalent to Membership Relation | https://proofwiki.org/wiki/Strict_Ordering_of_Ordinals_is_Equivalent_to_Membership_Relation | https://proofwiki.org/wiki/Strict_Ordering_of_Ordinals_is_Equivalent_to_Membership_Relation | [
"Ordinals"
] | [
"Definition:Class of All Ordinals",
"Definition:Usual Ordering of Ordinals"
] | [
"Ordinal is Transitive",
"Ordinal is not Element of Itself",
"Definition:Usual Ordering of Ordinals"
] |
proofwiki-19726 | Conditional Entropy Decreases if More Given | :$\CC \subseteq \DD \implies \map H {\AA \mid \CC} \ge \map H {\AA \mid \DD}$ | Consider the generated finite partitions:
:$\xi := \map \xi \AA$
:$\eta := \map \xi \CC$
:$\gamma := \map \xi \DD$
By Generating Partition Preserves Order, $\CC \subseteq \DD$ implies:
:$\eta \le \gamma$
where $\le$ denotes the refinement.
By {{Defof|Conditional Entropy of Finite Sub-Sigma-Algebra}}, we shall show:
:$\... | :$\CC \subseteq \DD \implies \map H {\AA \mid \CC} \ge \map H {\AA \mid \DD}$ | Consider the [[Definition:Finite Partition Generated by Finite Sub-Sigma-Algebra|generated finite partitions]]:
:$\xi := \map \xi \AA$
:$\eta := \map \xi \CC$
:$\gamma := \map \xi \DD$
By [[Generating Partition Preserves Order]], $\CC \subseteq \DD$ implies:
:$\eta \le \gamma$
where $\le$ denotes the [[Definition:Refi... | Conditional Entropy Decreases if More Given | https://proofwiki.org/wiki/Conditional_Entropy_Decreases_if_More_Given | https://proofwiki.org/wiki/Conditional_Entropy_Decreases_if_More_Given | [
"Probability Theory",
"Ergodic Theory"
] | [] | [
"Definition:Finite Partition Generated by Finite Sub-Sigma-Algebra",
"Generating Partition Preserves Order",
"Definition:Refinement of Partition (Probability Theory)",
"Definition:Real Function",
"Definition:Concave Real Function",
"Real Function with Strictly Negative Second Derivative is Strictly Concav... |
proofwiki-19727 | Ordinal Membership is Transitive | Let $\On$ denote the class of all ordinals.
Then:
:$\forall \alpha, \beta, \gamma \in \On: \paren {\alpha \in \beta} \land \paren {\beta \in \gamma} \implies \alpha \in \gamma$ | By Strict Ordering of Ordinals is Equivalent to Membership Relation the statement to be proved is equivalent to:
:$\forall \alpha, \beta, \gamma \in \On: \paren {\alpha \subsetneqq \beta} \land \paren {\beta \subsetneqq \gamma} \implies \alpha \subsetneqq \gamma$
which follows (indirectly) from Subset Relation is Trans... | Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
Then:
:$\forall \alpha, \beta, \gamma \in \On: \paren {\alpha \in \beta} \land \paren {\beta \in \gamma} \implies \alpha \in \gamma$ | By [[Strict Ordering of Ordinals is Equivalent to Membership Relation]] the statement to be proved is [[Definition:Logical Equivalence|equivalent]] to:
:$\forall \alpha, \beta, \gamma \in \On: \paren {\alpha \subsetneqq \beta} \land \paren {\beta \subsetneqq \gamma} \implies \alpha \subsetneqq \gamma$
which follows (i... | Ordinal Membership is Transitive | https://proofwiki.org/wiki/Ordinal_Membership_is_Transitive | https://proofwiki.org/wiki/Ordinal_Membership_is_Transitive | [
"Ordinals",
"Transitive Relations"
] | [
"Definition:Class of All Ordinals"
] | [
"Strict Ordering of Ordinals is Equivalent to Membership Relation",
"Definition:Logical Equivalence",
"Subset Relation is Transitive"
] |
proofwiki-19728 | Equality of Successors implies Equality of Ordinals | Let $\On$ denote the class of all ordinals.
Then:
:$\forall \alpha, \beta \in \On: \alpha^+ = \beta^+ \implies \alpha = \beta$ | From Class of All Ordinals is Well-Ordered by Subset Relation:
:$\alpha^+$ is the immediate successor of $\alpha$
:$\beta^+$ is the immediate successor of $\beta$
and no two distinct elements of $\On$ can have the same immediate successor.
The result follows.
{{qed}} | Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
Then:
:$\forall \alpha, \beta \in \On: \alpha^+ = \beta^+ \implies \alpha = \beta$ | From [[Class of All Ordinals is Well-Ordered by Subset Relation]]:
:$\alpha^+$ is the [[Definition:Immediate Successor Element|immediate successor]] of $\alpha$
:$\beta^+$ is the [[Definition:Immediate Successor Element|immediate successor]] of $\beta$
and no two [[Definition:Distinct Elements|distinct elements]] of $... | Equality of Successors implies Equality of Ordinals | https://proofwiki.org/wiki/Equality_of_Successors_implies_Equality_of_Ordinals | https://proofwiki.org/wiki/Equality_of_Successors_implies_Equality_of_Ordinals | [
"Ordinals"
] | [
"Definition:Class of All Ordinals"
] | [
"Class of All Ordinals is Well-Ordered by Subset Relation",
"Definition:Immediate Successor Element",
"Definition:Immediate Successor Element",
"Definition:Distinct/Plural",
"Definition:Immediate Successor Element"
] |
proofwiki-19729 | Exists Ordinal Greater than Set of Ordinals | Let $S$ be a set of ordinals.
Then there exists an ordinal greater than every element of $S$:
:If $S$ contains a greatest ordinal $\alpha$, then $\alpha^+$ is greater than every element of $S$
:If $S$ does not contain a greatest ordinal, then $\bigcup S$ is greater than every element of $S$. | Recall that Class of All Ordinals is Well-Ordered by Subset Relation.
Suppose $S$ contains a greatest ordinal $\alpha$.
Because $\alpha^+$ is greater than $\alpha$ by definition, it follows {{apriori}} that $\alpha^+$ is greater than every element of $S$.
{{qed|lemma}}
Suppose $S$ does not contain a greatest ordinal.
C... | Let $S$ be a [[Definition:Set|set]] of [[Definition:Ordinal|ordinals]].
Then there exists an [[Definition:Ordinal|ordinal]] greater than every [[Definition:Element|element]] of $S$:
:If $S$ contains a [[Definition:Greatest Element|greatest ordinal]] $\alpha$, then $\alpha^+$ is greater than every [[Definition:Element... | Recall that [[Class of All Ordinals is Well-Ordered by Subset Relation]].
Suppose $S$ contains a [[Definition:Greatest Element|greatest ordinal]] $\alpha$.
Because $\alpha^+$ is greater than $\alpha$ by definition, it follows {{apriori}} that $\alpha^+$ is greater than every [[Definition:Element|element]] of $S$.
{{q... | Exists Ordinal Greater than Set of Ordinals | https://proofwiki.org/wiki/Exists_Ordinal_Greater_than_Set_of_Ordinals | https://proofwiki.org/wiki/Exists_Ordinal_Greater_than_Set_of_Ordinals | [
"Ordinals"
] | [
"Definition:Set",
"Definition:Ordinal",
"Definition:Ordinal",
"Definition:Element",
"Definition:Greatest Element",
"Definition:Element",
"Definition:Greatest Element",
"Definition:Element"
] | [
"Class of All Ordinals is Well-Ordered by Subset Relation",
"Definition:Greatest Element",
"Definition:Element",
"Definition:Greatest Element",
"Definition:Set Union/Set of Sets",
"Definition:Greatest Element",
"Union of Set of Ordinals is Ordinal",
"Definition:Ordinal",
"Definition:Usual Ordering o... |
proofwiki-19730 | Successor of Ordinal Smaller than Limit Ordinal is also Smaller | Let $\On$ denote the class of all ordinals.
Let $\lambda \in \On$ be a limit ordinal.
Then:
:$\forall \alpha \in \On: \alpha < \lambda \implies \alpha^+ < \lambda$ | Let $\lambda$ be a limit ordinal such that $\alpha < \lambda$.
From Successor of Element of Ordinal is Subset
Then as $\alpha^+$ is the successor set of $\alpha$ it follows that:
:$\alpha^+ \le \lambda$
{{explain|Find the result that says $\alpha < \beta \implies \alpha^+ \le \beta$}}
But as $\lambda$ is not a successo... | Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
Let $\lambda \in \On$ be a [[Definition:Limit Ordinal|limit ordinal]].
Then:
:$\forall \alpha \in \On: \alpha < \lambda \implies \alpha^+ < \lambda$ | Let $\lambda$ be a [[Definition:Limit Ordinal|limit ordinal]] such that $\alpha < \lambda$.
From [[Successor of Element of Ordinal is Subset]]
Then as $\alpha^+$ is the [[Definition:Successor Set|successor set]] of $\alpha$ it follows that:
:$\alpha^+ \le \lambda$
{{explain|Find the result that says $\alpha < \beta ... | Successor of Ordinal Smaller than Limit Ordinal is also Smaller/Proof 1 | https://proofwiki.org/wiki/Successor_of_Ordinal_Smaller_than_Limit_Ordinal_is_also_Smaller | https://proofwiki.org/wiki/Successor_of_Ordinal_Smaller_than_Limit_Ordinal_is_also_Smaller/Proof_1 | [
"Successor of Ordinal Smaller than Limit Ordinal is also Smaller",
"Ordinals",
"Limit Ordinals",
"Successor Mapping"
] | [
"Definition:Class of All Ordinals",
"Definition:Limit Ordinal"
] | [
"Definition:Limit Ordinal",
"Successor of Element of Ordinal is Subset",
"Definition:Successor Mapping/Successor Set",
"Definition:Successor Ordinal"
] |
proofwiki-19731 | Unit Ideal is Principal Ideal Generated by Unity | Let $A$ be a commutative ring with unity.
Then:
:$A = \ideal 1$
where:
:$A$ is called the unit ideal of $A$
:$\ideal 1$ denotes the principal ideal generated by the unity of $A$ | $\ideal 1 \subseteq A$ is clear by definition of principal ideal.
To see $A \subseteq \ideal 1$, let $a \in A$ be an arbitrary element.
Then:
{{begin-eqn}}
{{eqn | l = a
| r = a 1
| c = {{Defof|Identity Element}}
}}
{{eqn | r = \ideal 1
| o = \in
| c = {{Defof|Principal Ideal of Ring}}
}}
{{end-... | Let $A$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Then:
:$A = \ideal 1$
where:
:$A$ is called the [[Definition:Unit Ideal|unit ideal]] of $A$
:$\ideal 1$ denotes the [[Definition:Principal Ideal of Ring|principal ideal]] generated by the [[Definition:Unity of Ring|unity]] of $A$ | $\ideal 1 \subseteq A$ is clear by definition of [[Definition:Principal Ideal of Ring|principal ideal]].
To see $A \subseteq \ideal 1$, let $a \in A$ be an arbitrary [[Definition:Element|element]].
Then:
{{begin-eqn}}
{{eqn | l = a
| r = a 1
| c = {{Defof|Identity Element}}
}}
{{eqn | r = \ideal 1
... | Unit Ideal is Principal Ideal Generated by Unity | https://proofwiki.org/wiki/Unit_Ideal_is_Principal_Ideal_Generated_by_Unity | https://proofwiki.org/wiki/Unit_Ideal_is_Principal_Ideal_Generated_by_Unity | [
"Commutative Algebra",
"Ideal Theory"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Unit Ideal",
"Definition:Principal Ideal of Ring",
"Definition:Unity (Abstract Algebra)/Ring"
] | [
"Definition:Principal Ideal of Ring",
"Definition:Element",
"Category:Commutative Algebra",
"Category:Ideal Theory"
] |
proofwiki-19732 | Union of Ordinal is Subset of Itself | Let $\alpha$ be an ordinal.
Then:
:$\bigcup \alpha \subseteq \alpha$
where $\bigcup \alpha$ denotes the union of $\alpha$. | Let $x \in \bigcup \alpha$.
Then:
:$\exists \beta \in \alpha: x \in \beta$
By Element of Ordinal is Ordinal, $\beta$ is an ordinal.
Thus:
:$x \in \beta$ and $\beta \in \alpha$
Hence by Ordinal Membership is Transitive:
:$x \in \alpha$
{{qed}} | Let $\alpha$ be an [[Definition:Ordinal|ordinal]].
Then:
:$\bigcup \alpha \subseteq \alpha$
where $\bigcup \alpha$ denotes the [[Definition:Union of Set of Sets|union]] of $\alpha$. | Let $x \in \bigcup \alpha$.
Then:
:$\exists \beta \in \alpha: x \in \beta$
By [[Element of Ordinal is Ordinal]], $\beta$ is an [[Definition:Ordinal|ordinal]].
Thus:
:$x \in \beta$ and $\beta \in \alpha$
Hence by [[Ordinal Membership is Transitive]]:
:$x \in \alpha$
{{qed}} | Union of Ordinal is Subset of Itself | https://proofwiki.org/wiki/Union_of_Ordinal_is_Subset_of_Itself | https://proofwiki.org/wiki/Union_of_Ordinal_is_Subset_of_Itself | [
"Ordinals",
"Set Union",
"Subsets"
] | [
"Definition:Ordinal",
"Definition:Set Union/Set of Sets"
] | [
"Element of Ordinal is Ordinal",
"Definition:Ordinal",
"Ordinal Membership is Transitive"
] |
proofwiki-19733 | Ideal is Unit Ideal iff Includes Unity | Let $A$ be a commutative ring with unity.
Let $\mathfrak a$ be an ideal of $A$.
Then:
:$\mathfrak a = A \iff 1 \in \mathfrak a$
where $1$ denotes the unity of $A$. | $\implies$ is trivial.
To see $\impliedby$, suppose $1 \in \mathfrak a$.
Let $a \in A$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = a
| r = a 1
| c = {{Defof|Identity Element}}
}}
{{eqn | r = \mathfrak a
| o = \in
| c = since $1 \in \mathfrak a$, by {{Defof|Ideal of Ring}}
}}
{{end-eqn}}
{{qed}... | Let $A$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Let $\mathfrak a$ be an [[Definition:Ideal of Ring|ideal]] of $A$.
Then:
:$\mathfrak a = A \iff 1 \in \mathfrak a$
where $1$ denotes the [[Definition:Unity of Ring|unity]] of $A$. | $\implies$ is trivial.
To see $\impliedby$, suppose $1 \in \mathfrak a$.
Let $a \in A$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = a
| r = a 1
| c = {{Defof|Identity Element}}
}}
{{eqn | r = \mathfrak a
| o = \in
| c = since $1 \in \mathfrak a$, by {{Defof|Ideal of Ring}}
}}
{{end-eqn}}
{{q... | Ideal is Unit Ideal iff Includes Unity | https://proofwiki.org/wiki/Ideal_is_Unit_Ideal_iff_Includes_Unity | https://proofwiki.org/wiki/Ideal_is_Unit_Ideal_iff_Includes_Unity | [
"Commutative Algebra",
"Ideal Theory"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Ideal of Ring",
"Definition:Unity (Abstract Algebra)/Ring"
] | [
"Category:Commutative Algebra",
"Category:Ideal Theory"
] |
proofwiki-19734 | Unit Ideal iff Radical is Unit Ideal | Let $A$ be a commutative ring with unity.
Let $\mathfrak a$ be an ideal of $A$.
Then:
:$ \mathfrak a = A \iff \map \Rad {\mathfrak a} = A$
where:
:$A$ is called the unit ideal of $A$
:$\map \Rad {\mathfrak a}$ denotes the radical of $\mathfrak a$ | {{begin-eqn}}
{{eqn | l = \mathfrak a
| r = A
}}
{{eqn | ll= \leadstoandfrom
| l = 1
| r = A
| o = \in
| c = Ideal is Unit Ideal iff Includes Unity
}}
{{eqn | ll= \leadstoandfrom
| q = \exists n \in \Z_{>0}
| l = 1^n
| r = A
| o = \in
| c = since $ 1^n = 1$ b... | Let $A$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Let $\mathfrak a$ be an [[Definition:Ideal of Ring|ideal]] of $A$.
Then:
:$ \mathfrak a = A \iff \map \Rad {\mathfrak a} = A$
where:
:$A$ is called the [[Definition:Unit Ideal|unit ideal]] of $A$
:$\map \Rad {\mathfrak a}$ denotes t... | {{begin-eqn}}
{{eqn | l = \mathfrak a
| r = A
}}
{{eqn | ll= \leadstoandfrom
| l = 1
| r = A
| o = \in
| c = [[Ideal is Unit Ideal iff Includes Unity]]
}}
{{eqn | ll= \leadstoandfrom
| q = \exists n \in \Z_{>0}
| l = 1^n
| r = A
| o = \in
| c = since $ 1^n = ... | Unit Ideal iff Radical is Unit Ideal | https://proofwiki.org/wiki/Unit_Ideal_iff_Radical_is_Unit_Ideal | https://proofwiki.org/wiki/Unit_Ideal_iff_Radical_is_Unit_Ideal | [
"Radical of Ideals"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Ideal of Ring",
"Definition:Unit Ideal",
"Definition:Radical of Ideal of Ring"
] | [
"Ideal is Unit Ideal iff Includes Unity",
"Ideal is Unit Ideal iff Includes Unity",
"Category:Radical of Ideals"
] |
proofwiki-19735 | Ordinal equals Successor of its Union | Let $\alpha$ be an ordinal.
Then:
:$\bigcup \alpha^+ = \alpha$
where:
:$\alpha^+$ denotes the successor set of $\alpha$
:$\bigcup \alpha^+$ denotes the union of $\alpha^+$. | Let $x \in \bigcup \alpha^+$.
Then there exists an ordinal $\beta$ such that:
:$x \in \beta$
and:
:$\beta \in \alpha^+$
By definition of the usual ordering of ordinals:
:$\beta < \alpha^+$
Thus:
:$\beta \le \alpha$
Hence because $x < \beta$:
:$x < \alpha$
and thus:
:$x \in \alpha$
That is:
:$\bigcup \alpha^+ \subseteq ... | Let $\alpha$ be an [[Definition:Ordinal|ordinal]].
Then:
:$\bigcup \alpha^+ = \alpha$
where:
:$\alpha^+$ denotes the [[Definition:Successor Set|successor set]] of $\alpha$
:$\bigcup \alpha^+$ denotes the [[Definition:Union of Set of Sets|union]] of $\alpha^+$. | Let $x \in \bigcup \alpha^+$.
Then there exists an [[Definition:Ordinal|ordinal]] $\beta$ such that:
:$x \in \beta$
and:
:$\beta \in \alpha^+$
By definition of the [[Definition:Usual Ordering of Ordinals|usual ordering of ordinals]]:
:$\beta < \alpha^+$
Thus:
:$\beta \le \alpha$
Hence because $x < \beta$:
:$x < \al... | Ordinal equals Successor of its Union | https://proofwiki.org/wiki/Ordinal_equals_Successor_of_its_Union | https://proofwiki.org/wiki/Ordinal_equals_Successor_of_its_Union | [
"Ordinals",
"Set Union",
"Successor Mapping"
] | [
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set",
"Definition:Set Union/Set of Sets"
] | [
"Definition:Ordinal",
"Definition:Usual Ordering of Ordinals",
"Definition:Set Equality"
] |
proofwiki-19736 | Transitive Class of Ordinals is Subset of Ordinal not in it | Let $A$ be a transitive class of ordinals.
Let $\alpha$ be an ordinal which is not an element of $A$.
Then:
:$A \subseteq \alpha$ | Let $A$ and $\alpha$ be {{hypothesis}}.
Let $\beta \in A$ be arbitrary.
Because $\beta \in A$ and $\alpha \notin A$ we have that:
:$\beta \ne \alpha$
Because $\beta \in A$ and $A$ is transitive:
:all elements of $\beta$ are in $A$
But because $\alpha \notin A$:
:$\alpha \notin \beta$
Thus we have:
:$\alpha \notin \beta... | Let $A$ be a [[Definition:Transitive Class|transitive class]] of [[Definition:Ordinal|ordinals]].
Let $\alpha$ be an [[Definition:Ordinal|ordinal]] which is not an [[Definition:Element of Class|element]] of $A$.
Then:
:$A \subseteq \alpha$ | Let $A$ and $\alpha$ be {{hypothesis}}.
Let $\beta \in A$ be arbitrary.
Because $\beta \in A$ and $\alpha \notin A$ we have that:
:$\beta \ne \alpha$
Because $\beta \in A$ and $A$ is [[Definition:Transitive Class|transitive]]:
:all [[Definition:Element|elements]] of $\beta$ are in $A$
But because $\alpha \notin A$:... | Transitive Class of Ordinals is Subset of Ordinal not in it | https://proofwiki.org/wiki/Transitive_Class_of_Ordinals_is_Subset_of_Ordinal_not_in_it | https://proofwiki.org/wiki/Transitive_Class_of_Ordinals_is_Subset_of_Ordinal_not_in_it | [
"Ordinals",
"Transitive Classes"
] | [
"Definition:Transitive Class",
"Definition:Ordinal",
"Definition:Ordinal",
"Definition:Element/Class"
] | [
"Definition:Transitive Class",
"Definition:Element",
"Ordinal Membership is Trichotomy"
] |
proofwiki-19737 | Transitive Set of Ordinals is Ordinal | Let $x$ be a transitive set of ordinals.
Then $x$ is itself an ordinal. | Let $x$ be a transitive set of ordinals according to the statement of the theorem.
We have from Class of All Ordinals is Well-Ordered by Subset Relation that $\On$ is well-ordered by $\subseteq$.
By Exists Ordinal Greater than Set of Ordinals there exists $\alpha$ such that $\alpha \notin x$.
Hence let $\alpha$ be the ... | Let $x$ be a [[Definition:Transitive Set|transitive set]] of [[Definition:Ordinal|ordinals]].
Then $x$ is itself an [[Definition:Ordinal|ordinal]]. | Let $x$ be a [[Definition:Transitive Set|transitive set]] of [[Definition:Ordinal|ordinals]] according to the statement of the theorem.
We have from [[Class of All Ordinals is Well-Ordered by Subset Relation]] that $\On$ is [[Definition:Well-Ordering|well-ordered]] by $\subseteq$.
By [[Exists Ordinal Greater than Set... | Transitive Set of Ordinals is Ordinal | https://proofwiki.org/wiki/Transitive_Set_of_Ordinals_is_Ordinal | https://proofwiki.org/wiki/Transitive_Set_of_Ordinals_is_Ordinal | [
"Ordinals",
"Transitive Classes"
] | [
"Definition:Transitive Class",
"Definition:Ordinal",
"Definition:Ordinal"
] | [
"Definition:Transitive Class",
"Definition:Ordinal",
"Class of All Ordinals is Well-Ordered by Subset Relation",
"Definition:Well-Ordering",
"Exists Ordinal Greater than Set of Ordinals",
"Definition:Smallest Element",
"Definition:Ordinal",
"Definition:Ordinal",
"Ordinal equals its Initial Segment",... |
proofwiki-19738 | Class of All Ordinals is Only Proper Class of Ordinals | Let $A$ be a transitive proper class of ordinals.
Then $A$ is the class of all ordinals $\On$. | Let $A$ be a transitive class of ordinals.
Let there exist $\alpha \in \On$ such that $\alpha \notin A$.
Then by Transitive Class of Ordinals is Subset of Ordinal not in it:
:$A \subseteq \alpha$
But that makes $A$ a set.
So if $A$ is a proper class, it must contain all ordinals.
That is:
:$A = \On$
{{qed}} | Let $A$ be a [[Definition:Transitive Class|transitive]] [[Definition:Proper Class|proper class]] of [[Definition:Ordinal|ordinals]].
Then $A$ is the [[Definition:Class of All Ordinals|class of all ordinals]] $\On$. | Let $A$ be a [[Definition:Transitive Class|transitive class]] of [[Definition:Ordinal|ordinals]].
Let there exist $\alpha \in \On$ such that $\alpha \notin A$.
Then by [[Transitive Class of Ordinals is Subset of Ordinal not in it]]:
:$A \subseteq \alpha$
But that makes $A$ a [[Definition:Set|set]].
So if $A$ is a [... | Class of All Ordinals is Only Proper Class of Ordinals | https://proofwiki.org/wiki/Class_of_All_Ordinals_is_Only_Proper_Class_of_Ordinals | https://proofwiki.org/wiki/Class_of_All_Ordinals_is_Only_Proper_Class_of_Ordinals | [
"Class of All Ordinals"
] | [
"Definition:Transitive Class",
"Definition:Class (Class Theory)/Proper Class",
"Definition:Ordinal",
"Definition:Class of All Ordinals"
] | [
"Definition:Transitive Class",
"Definition:Ordinal",
"Transitive Class of Ordinals is Subset of Ordinal not in it",
"Definition:Set",
"Definition:Class (Class Theory)/Proper Class",
"Definition:Ordinal"
] |
proofwiki-19739 | Set is Ordinal iff Every Transitive Proper Subset is Element of it | Let $x$ be a set.
Then:
:$x$ is an ordinal
{{iff}}
:every transitive proper subset of $x$ is an element of $x$. | === Necessary Condition ===
Let $\alpha$ be an arbitrary ordinal.
Let $y$ be a proper subset of $\alpha$ such that $y$ is transitive.
From Transitive Set of Ordinals is Ordinal it follows that $y$ is an ordinal.
By definition of usual ordering of ordinals:
:$y < \alpha$
Hence from Strict Ordering of Ordinals is Equival... | Let $x$ be a [[Definition:Set|set]].
Then:
:$x$ is an [[Definition:Ordinal|ordinal]]
{{iff}}
:every [[Definition:Transitive Set|transitive]] [[Definition:Proper Subset|proper subset]] of $x$ is an [[Definition:Element|element]] of $x$. | === Necessary Condition ===
Let $\alpha$ be an arbitrary [[Definition:Ordinal|ordinal]].
Let $y$ be a [[Definition:Proper Subset|proper subset]] of $\alpha$ such that $y$ is [[Definition:Transitive Set|transitive]].
From [[Transitive Set of Ordinals is Ordinal]] it follows that $y$ is an [[Definition:Ordinal|ordinal... | Set is Ordinal iff Every Transitive Proper Subset is Element of it | https://proofwiki.org/wiki/Set_is_Ordinal_iff_Every_Transitive_Proper_Subset_is_Element_of_it | https://proofwiki.org/wiki/Set_is_Ordinal_iff_Every_Transitive_Proper_Subset_is_Element_of_it | [
"Ordinals",
"Transitive Classes"
] | [
"Definition:Set",
"Definition:Ordinal",
"Definition:Transitive Class",
"Definition:Proper Subset",
"Definition:Element"
] | [
"Definition:Ordinal",
"Definition:Proper Subset",
"Definition:Transitive Class",
"Transitive Set of Ordinals is Ordinal",
"Definition:Ordinal",
"Definition:Usual Ordering of Ordinals",
"Definition:Ordinal",
"Strict Ordering of Ordinals is Equivalent to Membership Relation",
"Definition:Transitive Cl... |
proofwiki-19740 | Class such that Every Transitive Subset is Element of it Contains All Ordinals | Let $K$ be a class.
Let $K$ be such that every transitive subset of $K$ is an element of $K$.
Then every ordinal is an element of $K$. | Let us assume the hypothesis.
Let $\On$ denote the class of all ordinals.
From Well-Ordering of Class of All Ordinals under Subset Relation, $\On$ is well-ordered under $\subseteq$.
Hence we can use the First Principle of Transfinite Induction.
Let $\alpha$ be an ordinal such that every ordinal less than $\alpha$ is an... | Let $K$ be a [[Definition:Class (Class Theory)|class]].
Let $K$ be such that every [[Definition:Transitive Set|transitive]] [[Definition:Subset|subset]] of $K$ is an [[Definition:Element of Class|element]] of $K$.
Then every [[Definition:Ordinal|ordinal]] is an [[Definition:Element of Class|element]] of $K$. | Let us assume the hypothesis.
Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
From [[Well-Ordering of Class of All Ordinals under Subset Relation]], $\On$ is [[Definition:Well-Ordered Class|well-ordered]] under $\subseteq$.
Hence we can use the [[First Principle of Transfinite Induct... | Class such that Every Transitive Subset is Element of it Contains All Ordinals | https://proofwiki.org/wiki/Class_such_that_Every_Transitive_Subset_is_Element_of_it_Contains_All_Ordinals | https://proofwiki.org/wiki/Class_such_that_Every_Transitive_Subset_is_Element_of_it_Contains_All_Ordinals | [
"Ordinals",
"Transitive Classes"
] | [
"Definition:Class (Class Theory)",
"Definition:Transitive Class",
"Definition:Subset",
"Definition:Element/Class",
"Definition:Ordinal",
"Definition:Element/Class"
] | [
"Definition:Class of All Ordinals",
"Well-Ordering of Class of All Ordinals under Subset Relation",
"Definition:Well-Ordered Class",
"First Principle of Transfinite Induction",
"Definition:Ordinal",
"Definition:Ordinal",
"Definition:Element/Class",
"Ordinal is Transitive",
"Definition:Ordinal",
"D... |
proofwiki-19741 | Element of Every Transitive-Closed Class is Ordinal | Let $x$ be a set such that $x$ is an element of every transitive-closed class.
Then $x$ is an ordinal. | {{ProofWanted|Not even sure if it actually is true yet. The exercise as set in S&F asks the question whether it is or not. If it turns out not true, we rename this page.}} | Let $x$ be a [[Definition:Set|set]] such that $x$ is an [[Definition:Element of Class|element]] of every [[Definition:Transitive-Closed Class|transitive-closed class]].
Then $x$ is an [[Definition:Ordinal|ordinal]]. | {{ProofWanted|Not even sure if it actually is true yet. The exercise as set in S&F asks the question whether it is or not. If it turns out not true, we rename this page.}} | Element of Every Transitive-Closed Class is Ordinal | https://proofwiki.org/wiki/Element_of_Every_Transitive-Closed_Class_is_Ordinal | https://proofwiki.org/wiki/Element_of_Every_Transitive-Closed_Class_is_Ordinal | [
"Transitive-Closed Classes",
"Ordinals"
] | [
"Definition:Set",
"Definition:Element/Class",
"Definition:Transitive-Closed Class",
"Definition:Ordinal"
] | [] |
proofwiki-19742 | Ideal Quotient is Ideal | Let $R$ be a commutative ring with unity.
Let $\mathfrak a, \mathfrak b$ be ideals of $R$.
Then the ideal quotient $\ideal {\mathfrak a : \mathfrak b}$ is indeed an ideal. | We shall check $(1)$-$(3)$ of Test for Ideal. | Let $R$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let $\mathfrak a, \mathfrak b$ be [[Definition:Ideal of Ring|ideals]] of $R$.
Then the [[Definition:Ideal Quotient of Commutative Ring|ideal quotient]] $\ideal {\mathfrak a : \mathfrak b}$ is indeed an [[Definition:Ideal of Ring|id... | We shall check $(1)$-$(3)$ of [[Test for Ideal]]. | Ideal Quotient is Ideal | https://proofwiki.org/wiki/Ideal_Quotient_is_Ideal | https://proofwiki.org/wiki/Ideal_Quotient_is_Ideal | [
"Commutative Algebra",
"Ideal Theory"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Ideal of Ring",
"Definition:Ideal Quotient of Commutative Ring",
"Definition:Ideal of Ring"
] | [
"Test for Ideal"
] |
proofwiki-19743 | Ordinal Addition/Examples/Ordinal Addition by Two | Let $2$ denote the successor of the ordinal $1$.
Then:
:$x + 2 = x^{++}$ | {{begin-eqn}}
{{eqn | l = x + 2
| r = x + 1^+
| c =
}}
{{eqn | r = \paren {x + 1}^+
| c = {{Defof|Ordinal Addition}}
}}
{{eqn | r = x^{++}
| c = Ordinal Addition by One
}}
{{end-eqn}}
{{qed}} | Let $2$ denote the [[Definition:Successor Set|successor]] of the [[Definition:One (Ordinal)|ordinal $1$]].
Then:
:$x + 2 = x^{++}$ | {{begin-eqn}}
{{eqn | l = x + 2
| r = x + 1^+
| c =
}}
{{eqn | r = \paren {x + 1}^+
| c = {{Defof|Ordinal Addition}}
}}
{{eqn | r = x^{++}
| c = [[Ordinal Addition by One]]
}}
{{end-eqn}}
{{qed}} | Ordinal Addition/Examples/Ordinal Addition by Two | https://proofwiki.org/wiki/Ordinal_Addition/Examples/Ordinal_Addition_by_Two | https://proofwiki.org/wiki/Ordinal_Addition/Examples/Ordinal_Addition_by_Two | [
"Ordinal Addition",
"2"
] | [
"Definition:Successor Mapping/Successor Set",
"Definition:One (Ordinal)"
] | [
"Ordinal Addition/Examples/Ordinal Addition by One"
] |
proofwiki-19744 | Ordinal Addition/Examples/Ordinal Addition by Natural Number | Let $n$ be a natural number.
Then:
:$x + \paren {n + 1} = \paren {x + n}^+$ | {{begin-eqn}}
{{eqn | l = x + \paren {n + 1}
| r = x + n^+
| c = Ordinal Addition by One
}}
{{eqn | r = \paren {x + n}^+
| c = {{Defof|Ordinal Addition}}
}}
{{end-eqn}}
{{qed}} | Let $n$ be a [[Definition:Natural Number|natural number]].
Then:
:$x + \paren {n + 1} = \paren {x + n}^+$ | {{begin-eqn}}
{{eqn | l = x + \paren {n + 1}
| r = x + n^+
| c = [[Ordinal Addition by One]]
}}
{{eqn | r = \paren {x + n}^+
| c = {{Defof|Ordinal Addition}}
}}
{{end-eqn}}
{{qed}} | Ordinal Addition/Examples/Ordinal Addition by Natural Number | https://proofwiki.org/wiki/Ordinal_Addition/Examples/Ordinal_Addition_by_Natural_Number | https://proofwiki.org/wiki/Ordinal_Addition/Examples/Ordinal_Addition_by_Natural_Number | [
"Ordinal Addition",
"Natural Numbers"
] | [
"Definition:Natural Numbers"
] | [
"Ordinal Addition/Examples/Ordinal Addition by One"
] |
proofwiki-19745 | Set of Natural Numbers is Ordinal | The set of natural numbers $\N$ is an ordinal. | From Natural Number is Ordinal, every element of $\N$ is an ordinal.
From Union of Set of Ordinals is Ordinal, $\bigcup \N$ is therefore itself an ordinal.
From Set of Natural Numbers Equals its Union:
:$\bigcup \N = \N$
Hence the result.
{{qed}} | The [[Definition:Natural Numbers|set of natural numbers]] $\N$ is an [[Definition:Ordinal|ordinal]]. | From [[Natural Number is Ordinal]], every [[Definition:Element|element]] of $\N$ is an [[Definition:Ordinal|ordinal]].
From [[Union of Set of Ordinals is Ordinal]], $\bigcup \N$ is therefore itself an [[Definition:Ordinal|ordinal]].
From [[Set of Natural Numbers Equals its Union]]:
:$\bigcup \N = \N$
Hence the resul... | Set of Natural Numbers is Ordinal | https://proofwiki.org/wiki/Set_of_Natural_Numbers_is_Ordinal | https://proofwiki.org/wiki/Set_of_Natural_Numbers_is_Ordinal | [
"Natural Numbers",
"Ordinals"
] | [
"Definition:Natural Numbers",
"Definition:Ordinal"
] | [
"Natural Number is Ordinal",
"Definition:Element",
"Definition:Ordinal",
"Union of Set of Ordinals is Ordinal",
"Definition:Ordinal",
"Set of Natural Numbers Equals its Union"
] |
proofwiki-19746 | Set of Natural Numbers is Smallest Ordinal Greater than All Natural Numbers | The set of natural numbers $\N$ is the smallest ordinal which is greater than all natural numbers. | From Set of Natural Numbers is Ordinal, $\N$ is an ordinal.
{{finish}} | The [[Definition:Natural Numbers|set of natural numbers]] $\N$ is the [[Definition:Smallest Element|smallest]] [[Definition:Ordinal|ordinal]] which is greater than all [[Definition:Natural Number|natural numbers]]. | From [[Set of Natural Numbers is Ordinal]], $\N$ is an [[Definition:Ordinal|ordinal]].
{{finish}} | Set of Natural Numbers is Smallest Ordinal Greater than All Natural Numbers | https://proofwiki.org/wiki/Set_of_Natural_Numbers_is_Smallest_Ordinal_Greater_than_All_Natural_Numbers | https://proofwiki.org/wiki/Set_of_Natural_Numbers_is_Smallest_Ordinal_Greater_than_All_Natural_Numbers | [
"Natural Numbers",
"Ordinals"
] | [
"Definition:Natural Numbers",
"Definition:Smallest Element",
"Definition:Ordinal",
"Definition:Natural Numbers"
] | [
"Set of Natural Numbers is Ordinal",
"Definition:Ordinal"
] |
proofwiki-19747 | Set of Natural Numbers is Limit Ordinal | The set of natural numbers $\N$ is a limit ordinal. | From Set of Natural Numbers is Ordinal, $\N$ is an ordinal.
{{finish}} | The [[Definition:Natural Numbers|set of natural numbers]] $\N$ is a [[Definition:Limit Ordinal|limit ordinal]]. | From [[Set of Natural Numbers is Ordinal]], $\N$ is an [[Definition:Ordinal|ordinal]].
{{finish}} | Set of Natural Numbers is Limit Ordinal | https://proofwiki.org/wiki/Set_of_Natural_Numbers_is_Limit_Ordinal | https://proofwiki.org/wiki/Set_of_Natural_Numbers_is_Limit_Ordinal | [
"Natural Numbers",
"Limit Ordinals"
] | [
"Definition:Natural Numbers",
"Definition:Limit Ordinal"
] | [
"Set of Natural Numbers is Ordinal",
"Definition:Ordinal"
] |
proofwiki-19748 | Radical of Primary Ideal is Smallest Larger Prime Ideal | Let $R$ be a commutative ring with unity.
Let $\mathfrak q$ be a primary ideal of $R$.
Let $\map \Rad {\mathfrak q}$ be the radical of $\mathfrak q$.
Then $\map \Rad {\mathfrak q}$ is the smallest prime ideal containing $\mathfrak q$. | First, we show that $\map \Rad {\mathfrak q}$ is a prime ideal.
Let $x y \in \map \Rad {\mathfrak q}$.
Then by definition of radical of ideal:
:$\exists n \in \N_{>0} : \paren {xy}^n \in \mathfrak q$
By Commutativity $(\text M 2)$:
:$x^n y^n \in \mathfrak q$
By definition of primary ideal:
:$x^n \in \mathfrak q \lor \e... | Let $R$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let $\mathfrak q$ be a [[Definition:Primary Ideal|primary ideal]] of $R$.
Let $\map \Rad {\mathfrak q}$ be the [[Definition:Radical of Ideal of Ring|radical]] of $\mathfrak q$.
Then $\map \Rad {\mathfrak q}$ is the smallest [[Defi... | First, we show that $\map \Rad {\mathfrak q}$ is a [[Definition:Prime Ideal|prime ideal]].
Let $x y \in \map \Rad {\mathfrak q}$.
Then by definition of [[Definition:Radical of Ideal of Ring|radical of ideal]]:
:$\exists n \in \N_{>0} : \paren {xy}^n \in \mathfrak q$
By [[Axiom:Commutative and Unitary Ring Axioms|Com... | Radical of Primary Ideal is Smallest Larger Prime Ideal | https://proofwiki.org/wiki/Radical_of_Primary_Ideal_is_Smallest_Larger_Prime_Ideal | https://proofwiki.org/wiki/Radical_of_Primary_Ideal_is_Smallest_Larger_Prime_Ideal | [
"Radical of Ideals",
"Primary Ideals",
"Prime Ideals of Rings"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Primary Ideal",
"Definition:Radical of Ideal of Ring",
"Definition:Prime Ideal of Ring"
] | [
"Definition:Prime Ideal",
"Definition:Radical of Ideal of Ring",
"Axiom:Commutative and Unitary Ring Axioms",
"Definition:Primary Ideal",
"Definition:Radical of Ideal of Ring",
"Definition:Prime Ideal",
"Prime Ideal Including Ideal Includes Radical",
"Definition:Prime Ideal",
"Definition:Prime Ideal... |
proofwiki-19749 | Image of Class under Mapping is Image of Restriction of Mapping to Class | Let $V$ be a basic universe
Let $f: V \to V$ be a mapping.
Let $A$ be a class.
Let $f \sqbrk A$ denote the image of $A$ under $f$.
Then $f \sqbrk A$ is the image of the restriction of $f$ to $A$:
:$f \sqbrk A = \Img {f {\restriction} A}$ | By definition, $f {\restriction} A$ is class of all ordered pairs $\tuple {a, \map f a}$, where $a \in A$.
The result follows from Restriction of Mapping is Subset of Cartesian Product.
{{qed}} | Let $V$ be a [[Definition:Basic Universe|basic universe]]
Let $f: V \to V$ be a [[Definition:Class Mapping|mapping]].
Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $f \sqbrk A$ denote the [[Definition:Image of Class under Mapping|image of $A$ under $f$]].
Then $f \sqbrk A$ is the [[Definition:Image o... | By definition, $f {\restriction} A$ is [[Definition:Class (Class Theory)|class]] of all [[Definition:Ordered Pair|ordered pairs]] $\tuple {a, \map f a}$, where $a \in A$.
The result follows from [[Restriction of Mapping is Subset of Cartesian Product]].
{{qed}} | Image of Class under Mapping is Image of Restriction of Mapping to Class | https://proofwiki.org/wiki/Image_of_Class_under_Mapping_is_Image_of_Restriction_of_Mapping_to_Class | https://proofwiki.org/wiki/Image_of_Class_under_Mapping_is_Image_of_Restriction_of_Mapping_to_Class | [
"Images",
"Restrictions"
] | [
"Definition:Basic Universe",
"Definition:Mapping/Class Theory",
"Definition:Class (Class Theory)",
"Definition:Image of Class under Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Restriction/Relation/Class Theory"
] | [
"Definition:Class (Class Theory)",
"Definition:Ordered Pair",
"Restriction of Mapping is Subset of Cartesian Product"
] |
proofwiki-19750 | Image of Set under Mapping is Set iff Restriction is Set | Let $V$ be a basic universe
Let $f: V \to V$ be a mapping.
Let $x$ be a set.
Let $f \sqbrk x$ denote the image of $x$ under $f$.
Let $f {\restriction} x$ denote the restriction of $f$ to $x$.
Then $f \sqbrk x$ is a set {{iff}} $f {\restriction} x$ is a set. | Let $f {\restriction} x$ be a set.
Then its image is also a set.
But then:
:$\Img {f {\restriction} x} = f \sqbrk x$
Conversely let $f \sqbrk x$ be a set.
From Cartesian Product of Sets is Set:
:$x \times f \sqbrk x$ is a set.
From Restriction of Mapping is Subclass of Cartesian Product:
:$f {\restriction} A \subseteq ... | Let $V$ be a [[Definition:Basic Universe|basic universe]]
Let $f: V \to V$ be a [[Definition:Class Mapping|mapping]].
Let $x$ be a [[Definition:Set|set]].
Let $f \sqbrk x$ denote the [[Definition:Image of Class under Mapping|image of $x$ under $f$]].
Let $f {\restriction} x$ denote the [[Definition:Restriction of R... | Let $f {\restriction} x$ be a [[Definition:Set|set]].
Then its [[Definition:Image of Relation (Class Theory)|image]] is also a [[Definition:Set|set]].
But then:
:$\Img {f {\restriction} x} = f \sqbrk x$
Conversely let $f \sqbrk x$ be a [[Definition:Set|set]].
From [[Cartesian Product of Sets is Set]]:
:$x \times f... | Image of Set under Mapping is Set iff Restriction is Set | https://proofwiki.org/wiki/Image_of_Set_under_Mapping_is_Set_iff_Restriction_is_Set | https://proofwiki.org/wiki/Image_of_Set_under_Mapping_is_Set_iff_Restriction_is_Set | [
"Images",
"Restrictions"
] | [
"Definition:Basic Universe",
"Definition:Mapping/Class Theory",
"Definition:Set",
"Definition:Image of Class under Mapping",
"Definition:Restriction/Relation/Class Theory",
"Definition:Set",
"Definition:Set"
] | [
"Definition:Set",
"Definition:Image (Set Theory)/Relation/Relation/Class Theory",
"Definition:Set",
"Definition:Set",
"Cartesian Product of Sets is Set",
"Definition:Set",
"Restriction of Mapping is Subclass of Cartesian Product",
"Subclass of Set is Set"
] |
proofwiki-19751 | Restriction of Mapping is Subclass of Cartesian Product | Let $V$ be a basic universe
Let $f: V \to V$ be a mapping.
Let $A$ be a class.
Let $f \sqbrk A$ denote the image of $A$ under $f$.
Let $f {\restriction} A$ denote the restriction of $f$ to $A$.
Then $f {\restriction} A$ is a subclass of the cartesian product of $A$ with its image:
:$f {\restriction} A \subseteq A \time... | {{Proofread|Missing explicit mention of Cartesian Product in the proof}}
Follows directly from:
:the definition of restriction
:the definition of mapping.
{{qed}} | Let $V$ be a [[Definition:Basic Universe|basic universe]]
Let $f: V \to V$ be a [[Definition:Class Mapping|mapping]].
Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $f \sqbrk A$ denote the [[Definition:Image of Class under Mapping|image of $A$ under $f$]].
Let $f {\restriction} A$ denote the [[Definiti... | {{Proofread|Missing explicit mention of Cartesian Product in the proof}}
Follows directly from:
:the definition of [[Definition:Restriction of Relation (Class Theory)|restriction]]
:the definition of [[Definition:Class Mapping|mapping]].
{{qed}} | Restriction of Mapping is Subclass of Cartesian Product | https://proofwiki.org/wiki/Restriction_of_Mapping_is_Subclass_of_Cartesian_Product | https://proofwiki.org/wiki/Restriction_of_Mapping_is_Subclass_of_Cartesian_Product | [
"Images",
"Cartesian Product"
] | [
"Definition:Basic Universe",
"Definition:Mapping/Class Theory",
"Definition:Class (Class Theory)",
"Definition:Image of Class under Mapping",
"Definition:Restriction/Relation/Class Theory",
"Definition:Subclass",
"Definition:Cartesian Product/Class Theory",
"Definition:Image of Class under Mapping"
] | [
"Definition:Restriction/Relation/Class Theory",
"Definition:Mapping/Class Theory"
] |
proofwiki-19752 | Radical of Prime Ideal | Let $R$ be a commutative ring with unity.
Let $\mathfrak p$ be a prime ideal of $R$.
Let $\map \Rad {\mathfrak p}$ be the radical of $\mathfrak p$.
Then:
:$\map \Rad {\mathfrak p} = \mathfrak p$ | === $\supseteq$ ===
Let $x \in \mathfrak p$.
Since $x = x^1$, by {{Defof|Radical of Ideal of Ring|index=1}}:
:$x \in \map \Rad {\mathfrak p}$
{{qed|lemma}} | Let $R$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Let $\mathfrak p$ be a [[Definition:Prime Ideal of Ring|prime ideal]] of $R$.
Let $\map \Rad {\mathfrak p}$ be the [[Definition:Radical of Ideal of Ring|radical]] of $\mathfrak p$.
Then:
:$\map \Rad {\mathfrak p} = \mathfrak p$ | === $\supseteq$ ===
Let $x \in \mathfrak p$.
Since $x = x^1$, by {{Defof|Radical of Ideal of Ring|index=1}}:
:$x \in \map \Rad {\mathfrak p}$
{{qed|lemma}} | Radical of Prime Ideal | https://proofwiki.org/wiki/Radical_of_Prime_Ideal | https://proofwiki.org/wiki/Radical_of_Prime_Ideal | [
"Radical of Ideals",
"Ideal Theory",
"Commutative Algebra",
"Ring Theory"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Prime Ideal of Ring",
"Definition:Radical of Ideal of Ring"
] | [] |
proofwiki-19753 | Prime Ideal Including Ideal Includes Radical | Let $R$ be a commutative ring with unity.
Let $\mathfrak p$ be a prime ideal.
Let $\mathfrak a$ be an ideal of $R$ such that:
:$\mathfrak a \subseteq \mathfrak p$
Let $\map \Rad {\mathfrak a}$ be the radical of $\mathfrak a$.
Then:
:$\map \Rad {\mathfrak a} \subseteq \mathfrak p$ | Let $x \in \relcomp R {\mathfrak p}$.
By {{Defof|Prime Ideal of Ring/Commutative and Unitary Ring|Prime Ideal|index=3}}:
{{begin-eqn}}
{{eqn | ll = \forall n \in \N_{>0} :
| l = x^n
| r = \mathfrak p
| o = \not\in
}}
{{eqn | ll = \leadsto
| l = x^n
| r = \mathfrak a
| o = \not\in
... | Let $R$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let $\mathfrak p$ be a [[Definition:Prime Ideal of Ring|prime ideal]].
Let $\mathfrak a$ be an [[Definition:Ideal of Ring|ideal]] of $R$ such that:
:$\mathfrak a \subseteq \mathfrak p$
Let $\map \Rad {\mathfrak a}$ be the [[Definit... | Let $x \in \relcomp R {\mathfrak p}$.
By {{Defof|Prime Ideal of Ring/Commutative and Unitary Ring|Prime Ideal|index=3}}:
{{begin-eqn}}
{{eqn | ll = \forall n \in \N_{>0} :
| l = x^n
| r = \mathfrak p
| o = \not\in
}}
{{eqn | ll = \leadsto
| l = x^n
| r = \mathfrak a
| o = \not\in
... | Prime Ideal Including Ideal Includes Radical | https://proofwiki.org/wiki/Prime_Ideal_Including_Ideal_Includes_Radical | https://proofwiki.org/wiki/Prime_Ideal_Including_Ideal_Includes_Radical | [
"Radical of Ideals",
"Ideal Theory",
"Commutative Algebra",
"Ring Theory"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Prime Ideal of Ring",
"Definition:Ideal of Ring",
"Definition:Radical of Ideal of Ring"
] | [
"Relative Complement inverts Subsets",
"Category:Radical of Ideals",
"Category:Ideal Theory",
"Category:Commutative Algebra",
"Category:Ring Theory"
] |
proofwiki-19754 | Composite of Mapping with Inverse of Another is Identity implies Mappings are Equal | Let $A$ and $B$ be classes.
Let $f$ and $g$ be mappings on $A \times B$.
Let $f$ and $g$ be such that:
:$f \circ g^{-1} = I_B$
where:
:$g^{-1}$ denotes the inverse of $g$
:$I_B$ denotes the identity mapping on $B$
:$\circ$ denotes composition of mappings.
Then:
:$f = g$
{{questionable|Each of the definitions inverse, i... | Let $\tuple {a, b} \in f$.
Then because $\tuple {b, b} \in I_B$ we must have:
:$\tuple {b, a} \in g^{-1}$
and so by definition of inverse of mapping:
:$\tuple {a, b} \in g$
Hence:
:$f \subseteq g$
Let $\tuple {a, b} \in g$.
Then by definition of inverse of mapping:
:$\tuple {b, a} \in g^{-1}$
Then because $\tuple {b, b... | Let $A$ and $B$ be [[Definition:Class (Class Theory)|classes]].
Let $f$ and $g$ be [[Definition:Class Mapping|mappings]] on $A \times B$.
Let $f$ and $g$ be such that:
:$f \circ g^{-1} = I_B$
where:
:$g^{-1}$ denotes the [[Definition:Inverse of Mapping|inverse]] of $g$
:$I_B$ denotes the [[Definition:Identity Mapping... | Let $\tuple {a, b} \in f$.
Then because $\tuple {b, b} \in I_B$ we must have:
:$\tuple {b, a} \in g^{-1}$
and so by definition of [[Definition:Inverse of Mapping|inverse of mapping]]:
:$\tuple {a, b} \in g$
Hence:
:$f \subseteq g$
Let $\tuple {a, b} \in g$.
Then by definition of [[Definition:Inverse of Mapping|in... | Composite of Mapping with Inverse of Another is Identity implies Mappings are Equal | https://proofwiki.org/wiki/Composite_of_Mapping_with_Inverse_of_Another_is_Identity_implies_Mappings_are_Equal | https://proofwiki.org/wiki/Composite_of_Mapping_with_Inverse_of_Another_is_Identity_implies_Mappings_are_Equal | [
"Composite Mappings",
"Inverse Mappings",
"Identity Mappings"
] | [
"Definition:Class (Class Theory)",
"Definition:Mapping/Class Theory",
"Definition:Inverse of Mapping",
"Definition:Identity Mapping",
"Definition:Composition of Mappings",
"Definition:Inverse of Mapping",
"Definition:Identity Mapping",
"Definition:Composition of Mappings"
] | [
"Definition:Inverse of Mapping",
"Definition:Inverse of Mapping",
"Definition:Set Equality"
] |
proofwiki-19755 | Ideal Contains Extension of Contraction | Let $A$ and $B$ be commutative rings with unity.
Let $f : A \to B$ be a ring homomorphism.
Let $\mathfrak b \subseteq B$ be an ideal.
Then $\mathfrak b$ contains the extension of its contraction by $f$:
:$\mathfrak b^{ce} \subseteq \mathfrak b$ | {{begin-eqn}}
{{eqn | l = f \sqbrk {\mathfrak b ^c}
| r = f \sqbrk {f^{-1} \sqbrk {\mathfrak b} }
| c = {{Defof|Contraction of Ideal}}
}}
{{eqn | r = \mathfrak b
| o = \subseteq
| c = Image of Preimage under Mapping
}}
{{end-eqn}}
Since $\mathfrak b^{ce}$ is generated by $f \sqbrk {\mathfrak b ^... | Let $A$ and $B$ be [[Definition:Commutative Ring with Unity|commutative rings with unity]].
Let $f : A \to B$ be a [[Definition:unital Ring Homomorphism|ring homomorphism]].
Let $\mathfrak b \subseteq B$ be an [[Definition:Ideal of Ring|ideal]].
Then $\mathfrak b$ [[Definition:Set Containment|contains]] the [[Defin... | {{begin-eqn}}
{{eqn | l = f \sqbrk {\mathfrak b ^c}
| r = f \sqbrk {f^{-1} \sqbrk {\mathfrak b} }
| c = {{Defof|Contraction of Ideal}}
}}
{{eqn | r = \mathfrak b
| o = \subseteq
| c = [[Image of Preimage under Mapping]]
}}
{{end-eqn}}
Since $\mathfrak b^{ce}$ is [[Definition:Generated Ideal of ... | Ideal Contains Extension of Contraction | https://proofwiki.org/wiki/Ideal_Contains_Extension_of_Contraction | https://proofwiki.org/wiki/Ideal_Contains_Extension_of_Contraction | [
"Ideal Theory",
"Commutative Algebra",
"Ring Theory"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:unital Ring Homomorphism",
"Definition:Ideal of Ring",
"Definition:Subset",
"Definition:Extension of Ideal",
"Definition:Contraction of Ideal"
] | [
"Image of Preimage under Mapping",
"Definition:Generated Ideal of Ring/Commutative and Unitary",
"Category:Ideal Theory",
"Category:Commutative Algebra",
"Category:Ring Theory"
] |
proofwiki-19756 | Strict Lower Closure of Element is Proper Lower Section | Let $A$ be a class under an ordering $\preccurlyeq$.
Let $a \in A$.
Let $a^\prec$ denote the strict lower closure of $a$ in $A$.
Then $a^\prec$ is a proper lower section of $A$. | By definition of strict lower closure:
:$a^\prec := \set {b \in S: b \prec a}$
By definition of $\prec$:
:$a \not \prec a$
and so:
:$a \notin a^\prec$
and so:
:$a \in A \setminus a^\prec$
Let $x \in a^\prec$.
Then by definition of $a^\prec$:
:$x \prec a$
and so:
:$\exists a \in A \setminus a^\prec: x \prec a$
and so $x... | Let $A$ be a [[Definition:Class (Class Theory)|class]] under an [[Definition:Ordering (Class Theory)|ordering]] $\preccurlyeq$.
Let $a \in A$.
Let $a^\prec$ denote the [[Definition:Strict Lower Closure of Element (Class Theory)|strict lower closure of $a$ in $A$]].
Then $a^\prec$ is a [[Definition:Proper Lower Sect... | By definition of [[Definition:Strict Lower Closure of Element (Class Theory)|strict lower closure]]:
:$a^\prec := \set {b \in S: b \prec a}$
By definition of $\prec$:
:$a \not \prec a$
and so:
:$a \notin a^\prec$
and so:
:$a \in A \setminus a^\prec$
Let $x \in a^\prec$.
Then by definition of $a^\prec$:
:$x \prec a... | Strict Lower Closure of Element is Proper Lower Section | https://proofwiki.org/wiki/Strict_Lower_Closure_of_Element_is_Proper_Lower_Section | https://proofwiki.org/wiki/Strict_Lower_Closure_of_Element_is_Proper_Lower_Section | [
"Lower Closures",
"Lower Sections"
] | [
"Definition:Class (Class Theory)",
"Definition:Ordering/Class Theory",
"Definition:Strict Lower Closure/Element/Class Theory",
"Definition:Proper Lower Section (Class Theory)"
] | [
"Definition:Strict Lower Closure/Element/Class Theory",
"Definition:Proper Lower Section (Class Theory)",
"Definition:Element/Class",
"Definition:Proper Lower Section (Class Theory)",
"Definition:Proper Lower Section (Class Theory)"
] |
proofwiki-19757 | Proper Lower Section under Well-Ordering is Initial Segment | Let $A$ be a class under a well-ordering $\preccurlyeq$.
Let $B$ be a proper lower section of $A$.
Let $b$ be the smallest element of $A \setminus B$.
Then $B$ is the initial segment of $b$ in $A$. | By definition of proper lower section of $A$:
:$\forall x \in B: \forall a \in A \setminus B: x \preccurlyeq a$
while:
:$B \ne \O$
:$B \ne A$
Because $B \ne A$ we have that $A \setminus B \ne \O$.
Because $\preccurlyeq$ is a well-ordering, $A \setminus B$ does indeed have a smallest element, which we can indeed call $b... | Let $A$ be a [[Definition:Class (Class Theory)|class]] under a [[Definition:Well-Ordering (Class Theory)|well-ordering]] $\preccurlyeq$.
Let $B$ be a [[Definition:Proper Lower Section (Class Theory)|proper lower section]] of $A$.
Let $b$ be the [[Definition:Smallest Element (Class Theory)|smallest element]] of $A \se... | By definition of [[Definition:Proper Lower Section (Class Theory)|proper lower section]] of $A$:
:$\forall x \in B: \forall a \in A \setminus B: x \preccurlyeq a$
while:
:$B \ne \O$
:$B \ne A$
Because $B \ne A$ we have that $A \setminus B \ne \O$.
Because $\preccurlyeq$ is a [[Definition:Well-Ordering (Class Theory)|... | Proper Lower Section under Well-Ordering is Initial Segment | https://proofwiki.org/wiki/Proper_Lower_Section_under_Well-Ordering_is_Initial_Segment | https://proofwiki.org/wiki/Proper_Lower_Section_under_Well-Ordering_is_Initial_Segment | [
"Initial Segments",
"Lower Sections",
"Well-Orderings"
] | [
"Definition:Class (Class Theory)",
"Definition:Well-Ordering/Class Theory",
"Definition:Proper Lower Section (Class Theory)",
"Definition:Smallest Element/Class Theory",
"Definition:Initial Segment/Class Theory"
] | [
"Definition:Proper Lower Section (Class Theory)",
"Definition:Well-Ordering/Class Theory",
"Definition:Smallest Element/Class Theory",
"Definition:Initial Segment/Class Theory"
] |
proofwiki-19758 | Well-Ordering on Set is Proper Well-Ordering | Let $\struct {S, \preccurlyeq}$ be a well-ordered set.
Then $\preccurlyeq$ is a proper well-ordering. | By definition, a proper well-ordering is a well-ordering on a class such that:
:every proper lower section of $S$ is a set.
We have {{afortiori}} that a proper lower section of $S$ is a subclass of $S$.
But here we have that $S$ is a set.
The result follows from Subclass of Set is Set.
{{qed}} | Let $\struct {S, \preccurlyeq}$ be a [[Definition:Well-Ordered Set|well-ordered set]].
Then $\preccurlyeq$ is a [[Definition:Proper Well-Ordering|proper well-ordering]]. | By definition, a [[Definition:Proper Well-Ordering|proper well-ordering]] is a [[Definition:Well-Ordering|well-ordering]] on a [[Definition:Class (Class Theory)|class]] such that:
:every [[Definition:Proper Lower Section (Class Theory)|proper lower section]] of $S$ is a [[Definition:Set|set]].
We have {{afortiori}} ... | Well-Ordering on Set is Proper Well-Ordering | https://proofwiki.org/wiki/Well-Ordering_on_Set_is_Proper_Well-Ordering | https://proofwiki.org/wiki/Well-Ordering_on_Set_is_Proper_Well-Ordering | [
"Proper Well-Orderings"
] | [
"Definition:Well-Ordered Set",
"Definition:Proper Well-Ordering"
] | [
"Definition:Proper Well-Ordering",
"Definition:Well-Ordering",
"Definition:Class (Class Theory)",
"Definition:Proper Lower Section (Class Theory)",
"Definition:Set",
"Definition:Proper Lower Section (Class Theory)",
"Definition:Subclass",
"Definition:Set",
"Subclass of Set is Set"
] |
proofwiki-19759 | G-Tower is Properly Well-Ordered under Subset Relation | Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Then $M$ is properly well-ordered under the subset relation. | From $g$-Tower is Well-Ordered under Subset Relation, $\subseteq$ is a well-ordering on $M$.
Let $L$ be a proper lower section of $M$.
From Proper Lower Section under Well-Ordering is Initial Segment, $L$ is an initial segment $x^\subset$ of $M$ for some $x \in M$.
By the definition of the structure of a $g$-tower, eac... | Let $M$ be a [[Definition:Class (Class Theory)|class]].
Let $g: M \to M$ be a [[Definition:Progressing Mapping|progressing mapping]] on $M$.
Let $M$ be a [[Definition:G-Tower|$g$-tower]].
Then $M$ is [[Definition:Proper Well-Ordering|properly well-ordered]] under the [[Definition:Subset Relation|subset relation]]. | From [[G-Tower is Well-Ordered under Subset Relation|$g$-Tower is Well-Ordered under Subset Relation]], $\subseteq$ is a [[Definition:Well-Ordering (Class Theory)|well-ordering]] on $M$.
Let $L$ be a [[Definition:Proper Lower Section (Class Theory)|proper lower section]] of $M$.
From [[Proper Lower Section under Well... | G-Tower is Properly Well-Ordered under Subset Relation | https://proofwiki.org/wiki/G-Tower_is_Properly_Well-Ordered_under_Subset_Relation | https://proofwiki.org/wiki/G-Tower_is_Properly_Well-Ordered_under_Subset_Relation | [
"Proper Well-Orderings",
"G-Towers"
] | [
"Definition:Class (Class Theory)",
"Definition:Progressing Mapping",
"Definition:G-Tower",
"Definition:Proper Well-Ordering",
"Definition:Subset Relation"
] | [
"G-Tower is Well-Ordered under Subset Relation",
"Definition:Well-Ordering/Class Theory",
"Definition:Proper Lower Section (Class Theory)",
"Proper Lower Section under Well-Ordering is Initial Segment",
"Definition:Initial Segment/Class Theory",
"Definition:G-Tower",
"Definition:Element",
"Definition:... |
proofwiki-19760 | Extension of Contraction of Extension of Ideal is Extension | Let $A$ and $B$ be commutative rings with unity.
Let $f : A \to B$ be a ring homomorphism.
Let $\mathfrak a$ be an ideal of $A$.
Let ${\mathfrak a}^e$ be the extension of $\mathfrak a$ by $f$.
Let ${\mathfrak a}^{ec}$ be the contraction of ${\mathfrak a}^e$ by $f$.
Let ${\mathfrak a}^{ece}$ be the extension of ${\mathf... | Since $\mathfrak a \subseteq {\mathfrak a}^{ec}$ by Ideal is Contained in Contraction of Extension:
{{begin-eqn}}
{{eqn | l = {\mathfrak a}^e
| r = \paren { {\mathfrak a}^{ec} }^e
| o = \subseteq
| c = Extension of Ideals Preserves Inclusion Order }}
{{eqn | r = {\mathfrak a}^{ece}
| c = Composi... | Let $A$ and $B$ be [[Definition:Commutative Ring with Unity|commutative rings with unity]].
Let $f : A \to B$ be a [[Definition:unital Ring Homomorphism|ring homomorphism]].
Let $\mathfrak a$ be an [[Definition:Ideal of Ring|ideal]] of $A$.
Let ${\mathfrak a}^e$ be the [[Definition:Extension of Ideal|extension]] of ... | Since $\mathfrak a \subseteq {\mathfrak a}^{ec}$ by [[Ideal is Contained in Contraction of Extension]]:
{{begin-eqn}}
{{eqn | l = {\mathfrak a}^e
| r = \paren { {\mathfrak a}^{ec} }^e
| o = \subseteq
| c = [[Extension of Ideals Preserves Inclusion Order]] }}
{{eqn | r = {\mathfrak a}^{ece}
| c =... | Extension of Contraction of Extension of Ideal is Extension | https://proofwiki.org/wiki/Extension_of_Contraction_of_Extension_of_Ideal_is_Extension | https://proofwiki.org/wiki/Extension_of_Contraction_of_Extension_of_Ideal_is_Extension | [
"Ideal Theory",
"Commutative Algebra",
"Ring Theory"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:unital Ring Homomorphism",
"Definition:Ideal of Ring",
"Definition:Extension of Ideal",
"Definition:Contraction of Ideal",
"Definition:Extension of Ideal"
] | [
"Ideal is Contained in Contraction of Extension",
"Extension of Ideals Preserves Inclusion Order",
"Composition of Mappings is Associative",
"Composition of Mappings is Associative",
"Ideal Contains Extension of Contraction",
"Category:Ideal Theory",
"Category:Commutative Algebra",
"Category:Ring Theo... |
proofwiki-19761 | Contraction of Extension of Contraction of Ideal is Contraction | Let $A$ and $B$ be commutative rings with unity.
Let $f : A \to B$ be a ring homomorphism.
Let $\mathfrak b$ be an ideal of $B$.
Let ${\mathfrak b}^c$ be the contraction of $\mathfrak b$ by $f$.
Let ${\mathfrak b}^{ce}$ be the extension of ${\mathfrak b}^c$ by $f$.
Let ${\mathfrak b}^{cec}$ be the contraction of ${\mat... | Since $\mathfrak b \supseteq {\mathfrak b}^{ce}$ by Ideal Contains Extension of Contraction:
{{begin-eqn}}
{{eqn | l = {\mathfrak b}^c
| r = \paren { {\mathfrak b}^{ce} }^c
| o = \supseteq
| c = Contraction of Ideals Preserves Inclusion Order
}}
{{eqn | r = {\mathfrak b}^{cec}
| c = Composition ... | Let $A$ and $B$ be [[Definition:Commutative Ring with Unity|commutative rings with unity]].
Let $f : A \to B$ be a [[Definition:unital Ring Homomorphism|ring homomorphism]].
Let $\mathfrak b$ be an [[Definition:Ideal of Ring|ideal]] of $B$.
Let ${\mathfrak b}^c$ be the [[Definition:Contraction of Ideal|contraction]]... | Since $\mathfrak b \supseteq {\mathfrak b}^{ce}$ by [[Ideal Contains Extension of Contraction]]:
{{begin-eqn}}
{{eqn | l = {\mathfrak b}^c
| r = \paren { {\mathfrak b}^{ce} }^c
| o = \supseteq
| c = [[Contraction of Ideals Preserves Inclusion Order]]
}}
{{eqn | r = {\mathfrak b}^{cec}
| c = [[C... | Contraction of Extension of Contraction of Ideal is Contraction | https://proofwiki.org/wiki/Contraction_of_Extension_of_Contraction_of_Ideal_is_Contraction | https://proofwiki.org/wiki/Contraction_of_Extension_of_Contraction_of_Ideal_is_Contraction | [
"Ideal Theory",
"Commutative Algebra",
"Ring Theory"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:unital Ring Homomorphism",
"Definition:Ideal of Ring",
"Definition:Contraction of Ideal",
"Definition:Extension of Ideal",
"Definition:Contraction of Ideal"
] | [
"Ideal Contains Extension of Contraction",
"Contraction of Ideals Preserves Inclusion Order",
"Composition of Mappings is Associative",
"Composition of Mappings is Associative",
"Ideal is Contained in Contraction of Extension",
"Category:Ideal Theory",
"Category:Commutative Algebra",
"Category:Ring Th... |
proofwiki-19762 | Measure-Preserving Transformation Preserves Conditional Entropy | Let $\struct {X, \BB, \mu}$ be a probability space.
Let $T: X \to X$ be a $\mu$-preserving transformation.
Let $\AA, \DD \subseteq \Sigma$ be finite sub-$\sigma$-algebras.
Then:
:$\map H {T^{-1} \AA \mid T^{-1} \DD} = \map H {\AA \mid \DD}$
where:
:$\map H {\cdot \mid \cdot}$ denotes the conditional entropy
:$T^{-1} \A... | By {{Defof|Finite Partition Generated by Finite Sub-Sigma-Algebra}}, we have:
:$\map \xi {T^{-1} \AA} = T^{-1} {\map \xi \AA}$
for each finite sub-$\sigma$-algebras $\AA \subseteq \Sigma$.
Thus it suffices to show that the entropy of finite partition satisfies:
:$\map H {T^{-1} \xi \mid T^{-1} \eta} = \map H {\xi \mid ... | Let $\struct {X, \BB, \mu}$ be a [[Definition:Probability Space|probability space]].
Let $T: X \to X$ be a $\mu$-[[Definition:Measure-Preserving Transformation|preserving transformation]].
Let $\AA, \DD \subseteq \Sigma$ be [[Definition:Finite Sub-Sigma-Algebra|finite sub-$\sigma$-algebras]].
Then:
:$\map H {T^{-1}... | By {{Defof|Finite Partition Generated by Finite Sub-Sigma-Algebra}}, we have:
:$\map \xi {T^{-1} \AA} = T^{-1} {\map \xi \AA}$
for each [[Definition:Finite Sub-Sigma-Algebra|finite sub-$\sigma$-algebras]] $\AA \subseteq \Sigma$.
Thus it suffices to show that the [[Definition:Conditional Entropy of Finite Partitions|en... | Measure-Preserving Transformation Preserves Conditional Entropy | https://proofwiki.org/wiki/Measure-Preserving_Transformation_Preserves_Conditional_Entropy | https://proofwiki.org/wiki/Measure-Preserving_Transformation_Preserves_Conditional_Entropy | [
"Probability Theory",
"Ergodic Theory"
] | [
"Definition:Probability Space",
"Definition:Measure-Preserving Transformation",
"Definition:Finite Sub-Sigma-Algebra",
"Definition:Conditional Entropy of Finite Sub-Sigma-Algebra",
"Definition:Pre-Image Sigma-Algebra/Domain",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Pre-Image Sigma-Algebra/... | [
"Definition:Finite Sub-Sigma-Algebra",
"Definition:Conditional Entropy of Finite Partitions",
"Definition:Finite Partition (Probability Theory)",
"Definition:Measure-Preserving Transformation",
"Category:Probability Theory",
"Category:Ergodic Theory"
] |
proofwiki-19763 | Union of Nest of Mappings is Mapping | Let $N$ be a nest of mappings.
Then:
:$\bigcup N$ is a mapping
where $\bigcup N$ denotes the union of $N$. | By definition of union, the elements of $\bigcup N$ are elements of elements of $N$.
That is, the elements of $\bigcup N$ are ordered pairs of sets.
Let $x \in \bigcup N$.
Then:
:$\exists f \subseteq \bigcup N: \exists y: \tuple {x, y} \in f$
{{AimForCont}}:
:$\exists y_1, y_2: \tuple {x, y_1} \in \bigcup N \land \tupl... | Let $N$ be a [[Definition:Nest of Mappings|nest of mappings]].
Then:
:$\bigcup N$ is a [[Definition:Class Mapping|mapping]]
where $\bigcup N$ denotes the [[Definition:Union of Class|union]] of $N$. | By definition of [[Definition:Union of Class|union]], the [[Definition:Element of Class|elements]] of $\bigcup N$ are [[Definition:Element of Class|elements]] of [[Definition:Element of Class|elements]] of $N$.
That is, the [[Definition:Element of Class|elements]] of $\bigcup N$ are [[Definition:Ordered Pair|ordered p... | Union of Nest of Mappings is Mapping/Proof | https://proofwiki.org/wiki/Union_of_Nest_of_Mappings_is_Mapping | https://proofwiki.org/wiki/Union_of_Nest_of_Mappings_is_Mapping/Proof | [
"Union of Nest of Mappings is Mapping",
"Nests of Mappings"
] | [
"Definition:Nest of Mappings",
"Definition:Mapping/Class Theory",
"Definition:Class Union/General Definition"
] | [
"Definition:Class Union/General Definition",
"Definition:Element/Class",
"Definition:Element/Class",
"Definition:Element/Class",
"Definition:Element/Class",
"Definition:Ordered Pair",
"Definition:Set",
"Definition:Nest of Mappings",
"Definition:Mapping/Class Theory",
"Definition:Contradiction",
... |
proofwiki-19764 | Domain of Union of Nest of Mappings is Union of Class of Domains | Let $N$ be a nest of mappings.
Let $\bigcup N$ denote the union of $N$.
Then:
:$\Dom {\bigcup N} = \ds \bigcup_{f \mathop \in N} \Dom f$
where $\Dom f$ denotes the domain of $f$. | From Union of Nest of Mappings is Mapping we have that $\bigcup N$ is a mapping.
Let $x \in \Dom {\bigcup N}$.
Then by definition of mapping:
:$\exists \tuple {x, y} \in \bigcup N$
Then by definition of union of class:
:$\exists f \subseteq \bigcup N: \tuple {x, y} \in f$
Hence:
:$\exists f \subseteq \bigcup N: x \in \... | Let $N$ be a [[Definition:Nest of Mappings|nest of mappings]].
Let $\bigcup N$ denote the [[Definition:Union of Class|union]] of $N$.
Then:
:$\Dom {\bigcup N} = \ds \bigcup_{f \mathop \in N} \Dom f$
where $\Dom f$ denotes the [[Definition:Domain of Relation (Class Theory)|domain]] of $f$. | From [[Union of Nest of Mappings is Mapping]] we have that $\bigcup N$ is a [[Definition:Class Mapping|mapping]].
Let $x \in \Dom {\bigcup N}$.
Then by definition of [[Definition:Class Mapping|mapping]]:
:$\exists \tuple {x, y} \in \bigcup N$
Then by definition of [[Definition:Union of Class|union of class]]:
:$\ex... | Domain of Union of Nest of Mappings is Union of Class of Domains/Proof | https://proofwiki.org/wiki/Domain_of_Union_of_Nest_of_Mappings_is_Union_of_Class_of_Domains | https://proofwiki.org/wiki/Domain_of_Union_of_Nest_of_Mappings_is_Union_of_Class_of_Domains/Proof | [
"Domain of Union of Nest of Mappings is Union of Class of Domains",
"Nests of Mappings"
] | [
"Definition:Nest of Mappings",
"Definition:Class Union/General Definition",
"Definition:Domain (Set Theory)/Relation/Class Theory"
] | [
"Union of Nest of Mappings is Mapping",
"Definition:Mapping/Class Theory",
"Definition:Mapping/Class Theory",
"Definition:Class Union/General Definition",
"Definition:Mapping/Class Theory",
"Definition:Class Union/General Definition",
"Definition:Set Equality"
] |
proofwiki-19765 | Image of Union of Nest of Mappings is Union of Class of Images | Let $N$ be a nest of mappings.
Let $\bigcup N$ denote the union of $N$.
Then:
:$\Img {\bigcup N} = \ds \bigcup_{f \mathop \in N} \Img f$
where $\Img f$ denotes the image of $f$. | From Union of Nest of Mappings is Mapping we have that $\bigcup N$ is a mapping.
Let $y \in \Img {\bigcup N}$.
Then by definition of mapping:
:$\exists \tuple {x, y} \in \bigcup N$
Then by definition of union of class:
:$\exists f \subseteq \bigcup N: \tuple {x, y} \in f$
Hence:
:$\exists f \subseteq \bigcup N: y \in \... | Let $N$ be a [[Definition:Nest of Mappings|nest of mappings]].
Let $\bigcup N$ denote the [[Definition:Union of Class|union]] of $N$.
Then:
:$\Img {\bigcup N} = \ds \bigcup_{f \mathop \in N} \Img f$
where $\Img f$ denotes the [[Definition:Image of Relation (Class Theory)|image]] of $f$. | From [[Union of Nest of Mappings is Mapping]] we have that $\bigcup N$ is a [[Definition:Class Mapping|mapping]].
Let $y \in \Img {\bigcup N}$.
Then by definition of [[Definition:Class Mapping|mapping]]:
:$\exists \tuple {x, y} \in \bigcup N$
Then by definition of [[Definition:Union of Class|union of class]]:
:$\exi... | Image of Union of Nest of Mappings is Union of Class of Images/Proof | https://proofwiki.org/wiki/Image_of_Union_of_Nest_of_Mappings_is_Union_of_Class_of_Images | https://proofwiki.org/wiki/Image_of_Union_of_Nest_of_Mappings_is_Union_of_Class_of_Images/Proof | [
"Image of Union of Nest of Mappings is Union of Class of Images",
"Nests of Mappings"
] | [
"Definition:Nest of Mappings",
"Definition:Class Union/General Definition",
"Definition:Image (Set Theory)/Relation/Relation/Class Theory"
] | [
"Union of Nest of Mappings is Mapping",
"Definition:Mapping/Class Theory",
"Definition:Mapping/Class Theory",
"Definition:Class Union/General Definition",
"Definition:Mapping/Class Theory",
"Definition:Class Union/General Definition",
"Definition:Set Equality"
] |
proofwiki-19766 | Union of Nest of Injections is Injection | Let $N$ be a nest of mappings which are all injections.
Then:
:$\bigcup N$ is an injection
where $\bigcup N$ denotes the union of $N$. | From Union of Nest of Mappings is Mapping we have that $\bigcup N$ is a mapping.
{{AimForCont}}:
:$\exists x_1, x_2: \tuple {x_1, y} \in \bigcup N \land \tuple {x_2, y} \in \bigcup N$
such that:
:$x_1 \ne x_2$
Then:
:$\exists f \subseteq \bigcup N: \tuple {x_1, y} \in f$
and:
:$\exists g \subseteq \bigcup N: \tuple {x_... | Let $N$ be a [[Definition:Nest of Mappings|nest of mappings]] which are all [[Definition:Injection (Class Theory)|injections]].
Then:
:$\bigcup N$ is an [[Definition:Injection (Class Theory)|injection]]
where $\bigcup N$ denotes the [[Definition:Union of Class|union]] of $N$. | From [[Union of Nest of Mappings is Mapping]] we have that $\bigcup N$ is a [[Definition:Class Mapping|mapping]].
{{AimForCont}}:
:$\exists x_1, x_2: \tuple {x_1, y} \in \bigcup N \land \tuple {x_2, y} \in \bigcup N$
such that:
:$x_1 \ne x_2$
Then:
:$\exists f \subseteq \bigcup N: \tuple {x_1, y} \in f$
and:
:$\exist... | Union of Nest of Injections is Injection/Proof | https://proofwiki.org/wiki/Union_of_Nest_of_Injections_is_Injection | https://proofwiki.org/wiki/Union_of_Nest_of_Injections_is_Injection/Proof | [
"Union of Nest of Injections is Injection",
"Nests of Mappings"
] | [
"Definition:Nest of Mappings",
"Definition:Injection/Class Theory",
"Definition:Injection/Class Theory",
"Definition:Class Union/General Definition"
] | [
"Union of Nest of Mappings is Mapping",
"Definition:Mapping/Class Theory",
"Definition:Nest of Mappings",
"Definition:Injection/Class Theory",
"Definition:Contradiction",
"Definition:Injection/Class Theory"
] |
proofwiki-19767 | Even Functions in 2-Lebesgue Space form Closed Subspace of 2-Lebesgue Space | Let $\map {\LL^2} \R$ be the Lebesgue $2$-space.
For all $f \in \map {\LL^2} \R$ denote $\map {\check f} x = \map f {-x}$.
Let $Y = \set {f \in \map {\LL^2} \R : f = \check f}$ be the set of all even functions in $\map {\LL^2} \R$.
Then $Y$ is a closed subspace of $\map {\LL^2} \R$. | Let $R : \map {\LL^2} \R \to \map {\LL^2} \R$ be the reflection mapping such that:
:$\map R f = \check f$
Then:
{{begin-eqn}}
{{eqn | ll = \forall f, g \in \map {\LL^2} \R : \forall x \in \R :
| l = \map R {\map f x + \map g x}
| r = \map R {\map {\paren {f + g} } x}
| c = {{Defof|Pointwise Addition ... | Let $\map {\LL^2} \R$ be the [[Definition:Lebesgue Space|Lebesgue $2$-space]].
For all $f \in \map {\LL^2} \R$ denote $\map {\check f} x = \map f {-x}$.
Let $Y = \set {f \in \map {\LL^2} \R : f = \check f}$ be the [[Definition:Set|set]] of all [[Definition:Even Function|even functions]] in $\map {\LL^2} \R$.
Then $... | Let $R : \map {\LL^2} \R \to \map {\LL^2} \R$ be the [[Definition:Plane Reflection|reflection mapping]] such that:
:$\map R f = \check f$
Then:
{{begin-eqn}}
{{eqn | ll = \forall f, g \in \map {\LL^2} \R : \forall x \in \R :
| l = \map R {\map f x + \map g x}
| r = \map R {\map {\paren {f + g} } x}
... | Even Functions in 2-Lebesgue Space form Closed Subspace of 2-Lebesgue Space | https://proofwiki.org/wiki/Even_Functions_in_2-Lebesgue_Space_form_Closed_Subspace_of_2-Lebesgue_Space | https://proofwiki.org/wiki/Even_Functions_in_2-Lebesgue_Space_form_Closed_Subspace_of_2-Lebesgue_Space | [
"Even Functions",
"Lebesgue Spaces",
"Examples of Vector Subspaces"
] | [
"Definition:Lebesgue Space",
"Definition:Set",
"Definition:Even Function",
"Definition:Closed Set/Normed Vector Space",
"Definition:Vector Subspace"
] | [
"Definition:Reflection (Geometry)/Plane",
"Definition:Set of All Linear Transformations",
"Definition:Reflection (Geometry)/Plane",
"Definition:Axis/Y-Axis",
"Definition:Area",
"Definition:Line/Curve",
"Continuity of Linear Transformation/Normed Vector Space",
"Definition:Continuous Mapping (Normed Ve... |
proofwiki-19768 | Well-Ordering on Class is not necessarily Proper | Let $A$ be a class.
Let $\preccurlyeq$ be a well-ordering on $A$.
Then it is not necessarily the case that $\preccurlyeq$ is a proper well-ordering. | Proof by Counterexample:
{{ProofWanted}} | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $\preccurlyeq$ be a [[Definition:Well-Ordering (Class Theory)|well-ordering]] on $A$.
Then it is not necessarily the case that $\preccurlyeq$ is a [[Definition:Proper Well-Ordering|proper well-ordering]]. | [[Proof by Counterexample]]:
{{ProofWanted}} | Well-Ordering on Class is not necessarily Proper | https://proofwiki.org/wiki/Well-Ordering_on_Class_is_not_necessarily_Proper | https://proofwiki.org/wiki/Well-Ordering_on_Class_is_not_necessarily_Proper | [
"Well-Orderings",
"Proper Well-Orderings"
] | [
"Definition:Class (Class Theory)",
"Definition:Well-Ordering/Class Theory",
"Definition:Proper Well-Ordering"
] | [
"Proof by Counterexample"
] |
proofwiki-19769 | Order Automorphism on Well-Ordered Class is Forward Moving | Let $\struct {A, \preccurlyeq}$ be a well-ordered class.
Let $\phi$ be an order isomorphism on $\struct {A, \preccurlyeq}$.
Then:
:$\forall a \in A: a \preccurlyeq \map \phi a$ | Let us define an element $a$ of $A$ such that:
:$\map \phi a \prec a$
as '''moving backwards'''.
{{AimForCont}} there exists an element $a$ of $A$ that '''moves backwards''':
:$\map \phi a \prec a$
for some $a \in A$.
Then applying $\phi$ to both sides:
:$\map \phi {\map \phi a} \prec \map \phi a$
That is:
:$\map \phi ... | Let $\struct {A, \preccurlyeq}$ be a [[Definition:Well-Ordered Class|well-ordered class]].
Let $\phi$ be an [[Definition:Order Isomorphism on Well-Orderings (Class Theory)|order isomorphism]] on $\struct {A, \preccurlyeq}$.
Then:
:$\forall a \in A: a \preccurlyeq \map \phi a$ | Let us define an [[Definition:Element of Class|element]] $a$ of $A$ such that:
:$\map \phi a \prec a$
as '''moving backwards'''.
{{AimForCont}} there exists an [[Definition:Element of Class|element]] $a$ of $A$ that '''moves backwards''':
:$\map \phi a \prec a$
for some $a \in A$.
Then applying $\phi$ to both sides:... | Order Automorphism on Well-Ordered Class is Forward Moving | https://proofwiki.org/wiki/Order_Automorphism_on_Well-Ordered_Class_is_Forward_Moving | https://proofwiki.org/wiki/Order_Automorphism_on_Well-Ordered_Class_is_Forward_Moving | [
"Order Isomorphisms",
"Well-Orderings"
] | [
"Definition:Well-Ordered Class",
"Definition:Order Isomorphism/Well-Orderings/Class Theory"
] | [
"Definition:Element/Class",
"Definition:Element/Class",
"Definition:Precede",
"Definition:Smallest Element",
"Definition:Element/Class",
"Definition:Contradiction",
"Definition:Well-Ordered Class",
"Definition:Non-Empty Set/Class Theory",
"Definition:Subclass",
"Definition:Smallest Element/Class T... |
proofwiki-19770 | Order Automorphism on Well-Ordered Class is Identity Mapping | Let $\struct {A, \preccurlyeq}$ be a well-ordered class.
Let $\phi$ be an order isomorphism on $\struct {A, \preccurlyeq}$.
Then $\phi$ is the identity mapping:
:$\forall a \in A: \map \phi a = a$ | Let $\phi$ be an order isomorphism.
Then from Inverse of Order Isomorphism is Order Isomorphism, the inverse mapping $\phi^{-1}$ is also an order isomorphism.
From Order Automorphism on Well-Ordered Class is Forward Moving:
:$\forall a \in A: a \preccurlyeq \map \phi a$
and:
:$\forall a \in A: a \preccurlyeq \map {\phi... | Let $\struct {A, \preccurlyeq}$ be a [[Definition:Well-Ordered Class|well-ordered class]].
Let $\phi$ be an [[Definition:Order Isomorphism on Well-Orderings (Class Theory)|order isomorphism]] on $\struct {A, \preccurlyeq}$.
Then $\phi$ is the [[Definition:Identity Mapping|identity mapping]]:
:$\forall a \in A: \map \... | Let $\phi$ be an [[Definition:Order Isomorphism on Well-Orderings (Class Theory)|order isomorphism]].
Then from [[Inverse of Order Isomorphism is Order Isomorphism]], the [[Definition:Inverse Mapping|inverse mapping]] $\phi^{-1}$ is also an [[Definition:Order Isomorphism on Well-Orderings (Class Theory)|order isomorph... | Order Automorphism on Well-Ordered Class is Identity Mapping | https://proofwiki.org/wiki/Order_Automorphism_on_Well-Ordered_Class_is_Identity_Mapping | https://proofwiki.org/wiki/Order_Automorphism_on_Well-Ordered_Class_is_Identity_Mapping | [
"Order Isomorphisms",
"Well-Orderings"
] | [
"Definition:Well-Ordered Class",
"Definition:Order Isomorphism/Well-Orderings/Class Theory",
"Definition:Identity Mapping"
] | [
"Definition:Order Isomorphism/Well-Orderings/Class Theory",
"Inverse of Order Isomorphism is Order Isomorphism",
"Definition:Inverse Mapping",
"Definition:Order Isomorphism/Well-Orderings/Class Theory",
"Order Automorphism on Well-Ordered Class is Forward Moving"
] |
proofwiki-19771 | Well-Ordered Classes are Isomorphic at Most Uniquely | Let $\struct {A, \preccurlyeq_A}$ and $\struct {B, \preccurlyeq_B}$ be well-ordered classes.
Then there exists at most one order isomorphism from $\struct {A, \preccurlyeq_A}$ to $\struct {B, \preccurlyeq_B}$ | {{questionable|Set-theoretic definitions and proofs are used in the below. They need to be expanded to class-theoretical versions.}}
Let $\phi$ and $\psi$ be order isomorphisms from $A$ to $B$.
Then from Inverse of Order Isomorphism is Order Isomorphism:
:the inverse mapping $\psi^{-1}$ is an order isomorphism from $B$... | Let $\struct {A, \preccurlyeq_A}$ and $\struct {B, \preccurlyeq_B}$ be [[Definition:Well-Ordered Class|well-ordered classes]].
Then there exists at most one [[Definition:Order Isomorphism on Well-Orderings (Class Theory)|order isomorphism]] from $\struct {A, \preccurlyeq_A}$ to $\struct {B, \preccurlyeq_B}$ | {{questionable|Set-theoretic definitions and proofs are used in the below. They need to be expanded to class-theoretical versions.}}
Let $\phi$ and $\psi$ be [[Definition:Order Isomorphism on Well-Orderings (Class Theory)|order isomorphisms]] from $A$ to $B$.
Then from [[Inverse of Order Isomorphism is Order Isomorph... | Well-Ordered Classes are Isomorphic at Most Uniquely | https://proofwiki.org/wiki/Well-Ordered_Classes_are_Isomorphic_at_Most_Uniquely | https://proofwiki.org/wiki/Well-Ordered_Classes_are_Isomorphic_at_Most_Uniquely | [
"Order Isomorphisms",
"Well-Orderings"
] | [
"Definition:Well-Ordered Class",
"Definition:Order Isomorphism/Well-Orderings/Class Theory"
] | [
"Definition:Order Isomorphism/Well-Orderings/Class Theory",
"Inverse of Order Isomorphism is Order Isomorphism",
"Definition:Inverse Mapping",
"Definition:Order Isomorphism/Well-Orderings/Class Theory",
"Composite of Order Isomorphisms is Order Isomorphism",
"Definition:Order Isomorphism/Well-Orderings/Cl... |
proofwiki-19772 | Well-Ordered Class is not Isomorphic to Initial Segment | Let $\struct {A, \preccurlyeq}$ be a well-ordered class.
There exists no order isomorphism from $\struct {A, \preccurlyeq}$ to an initial segment of $A$. | Let $a \in A$.
Let $A_a$ be the initial segment of $A$ determined by $a$.
{{AimForCont}} $\phi: A \to A_a$ is an order isomorphism from $A$ to $A_a$.
From Order Automorphism on Well-Ordered Class is Forward Moving:
:$a \preccurlyeq \map \phi a$
But by definition of initial segment:
:$\map \phi a \notin A_a$
Hence $\phi... | Let $\struct {A, \preccurlyeq}$ be a [[Definition:Well-Ordered Class|well-ordered class]].
There exists no [[Definition:Order Isomorphism on Well-Orderings (Class Theory)|order isomorphism]] from $\struct {A, \preccurlyeq}$ to an [[Definition:Initial Segment (Class Theory)|initial segment]] of $A$. | Let $a \in A$.
Let $A_a$ be the [[Definition:Initial Segment (Class Theory)|initial segment]] of $A$ determined by $a$.
{{AimForCont}} $\phi: A \to A_a$ is an [[Definition:Order Isomorphism on Well-Orderings (Class Theory)|order isomorphism]] from $A$ to $A_a$.
From [[Order Automorphism on Well-Ordered Class is Forw... | Well-Ordered Class is not Isomorphic to Initial Segment | https://proofwiki.org/wiki/Well-Ordered_Class_is_not_Isomorphic_to_Initial_Segment | https://proofwiki.org/wiki/Well-Ordered_Class_is_not_Isomorphic_to_Initial_Segment | [
"Well-Ordered Class is not Isomorphic to Initial Segment",
"Order Isomorphisms",
"Well-Orderings",
"Initial Segments"
] | [
"Definition:Well-Ordered Class",
"Definition:Order Isomorphism/Well-Orderings/Class Theory",
"Definition:Initial Segment/Class Theory"
] | [
"Definition:Initial Segment/Class Theory",
"Definition:Order Isomorphism/Well-Orderings/Class Theory",
"Order Automorphism on Well-Ordered Class is Forward Moving",
"Definition:Initial Segment/Class Theory",
"Definition:Order Isomorphism/Well-Orderings/Class Theory",
"Proof by Contradiction",
"Definitio... |
proofwiki-19773 | Distinct Ordinals are not Order Isomorphic | Let $\alpha$ and $\beta$ be ordinals such that $\alpha \ne \beta$.
Then $\alpha$ and $\beta$ are not order isomorphic. | By definition, an ordinal is well-ordered by the subset relation.
From Class of All Ordinals is Well-Ordered by Subset Relation, the class of all ordinals is a nest.
Hence:
:$\paren {\alpha \subsetneqq \beta} \lor \paren {\beta \subsetneqq \alpha}$
{{finish|there should be a result somewhere that explicitly states that... | Let $\alpha$ and $\beta$ be [[Definition:Ordinal|ordinals]] such that $\alpha \ne \beta$.
Then $\alpha$ and $\beta$ are not [[Definition:Order Isomorphic Well-Orderings|order isomorphic]]. | By definition, an [[Definition:Ordinal|ordinal]] is [[Definition:Well-Ordered Set|well-ordered]] by the [[Definition:Subset Relation|subset relation]].
From [[Class of All Ordinals is Well-Ordered by Subset Relation]], the [[Definition:Class of All Ordinals|class of all ordinals]] is a [[Definition:Nest (Class Theory)... | Distinct Ordinals are not Order Isomorphic | https://proofwiki.org/wiki/Distinct_Ordinals_are_not_Order_Isomorphic | https://proofwiki.org/wiki/Distinct_Ordinals_are_not_Order_Isomorphic | [
"Ordinals",
"Well-Orderings",
"Initial Segments"
] | [
"Definition:Ordinal",
"Definition:Order Isomorphism/Well-Orderings"
] | [
"Definition:Ordinal",
"Definition:Well-Ordered Set",
"Definition:Subset Relation",
"Class of All Ordinals is Well-Ordered by Subset Relation",
"Definition:Class of All Ordinals",
"Definition:Nest/Class Theory",
"Well-Ordered Class is not Isomorphic to Initial Segment"
] |
proofwiki-19774 | Distinct Lower Sections of Well-Ordered Class are not Order Isomorphic | Let $\struct {A, \preccurlyeq}$ be a well-ordered class.
Let $L_1$ and $L_2$ be distinct lower sections of $\struct {A, \preccurlyeq}$.
Then $L_1$ and $L_2$ are not order isomorphic {{WRT}} $\preccurlyeq$. | A lower section of $A$ is a subclass of $A$.
Hence by definition of well-ordered class. $L_1$ and $L_2$ are themselves well-ordered classes.
We have $L_1 \ne L_2$.
The result follows from
{{finish|Result needed here to the effect that distinct well-ordered classes are not isomorphic. We have this for sets but not clas... | Let $\struct {A, \preccurlyeq}$ be a [[Definition:Well-Ordered Class|well-ordered class]].
Let $L_1$ and $L_2$ be [[Definition:Distinct Elements|distinct]] [[Definition:Lower Section (Class Theory)|lower sections]] of $\struct {A, \preccurlyeq}$.
Then $L_1$ and $L_2$ are not [[Definition:Order Isomorphism on Well-Or... | A [[Definition:Lower Section (Class Theory)|lower section]] of $A$ is a [[Definition:Subclass|subclass]] of $A$.
Hence by definition of [[Definition:Well-Ordered Class|well-ordered class]]. $L_1$ and $L_2$ are themselves [[Definition:Well-Ordered Class|well-ordered classes]].
We have $L_1 \ne L_2$.
The result follow... | Distinct Lower Sections of Well-Ordered Class are not Order Isomorphic | https://proofwiki.org/wiki/Distinct_Lower_Sections_of_Well-Ordered_Class_are_not_Order_Isomorphic | https://proofwiki.org/wiki/Distinct_Lower_Sections_of_Well-Ordered_Class_are_not_Order_Isomorphic | [
"Well-Orderings",
"Lower Sections",
"Order Isomorphisms"
] | [
"Definition:Well-Ordered Class",
"Definition:Distinct/Plural",
"Definition:Lower Section/Class Theory",
"Definition:Order Isomorphism/Well-Orderings/Class Theory"
] | [
"Definition:Lower Section/Class Theory",
"Definition:Subclass",
"Definition:Well-Ordered Class",
"Definition:Well-Ordered Class"
] |
proofwiki-19775 | Isomorphisms between Lower Sections of Well-Ordered Classes are Nested | Let $\struct {A, \preccurlyeq_A}$ and $\struct {B, \preccurlyeq_B}$ be well-ordered classes.
Let $\phi_1$ and $\phi_2$ be order isomorphisms from a lower section of $\struct {A, \preccurlyeq_A}$ to a lower section of $\struct {B, \preccurlyeq_B}$.
Then either:
:$\phi_1 \subseteq \phi_2$
or:
:$\phi_2 \subseteq \phi_1$
w... | Let us label the domains of $\phi_1$ and $\phi_2$:
:$D_1 = \Dom {\phi_1}$
:$D_2 = \Dom {\phi_2}$
Because $D_1$ and $D_2$ are both lower sections of $\struct {A, \preccurlyeq_A}$, they are comparable under the subset relation.
{{WLOG}}, suppose $D_1 \subseteq D_2$.
Then the restriction $\phi_2 {\restriction_{D_1} }$ is ... | Let $\struct {A, \preccurlyeq_A}$ and $\struct {B, \preccurlyeq_B}$ be [[Definition:Well-Ordered Class|well-ordered classes]].
Let $\phi_1$ and $\phi_2$ be [[Definition:Order Isomorphism on Well-Orderings (Class Theory)|order isomorphisms]] from a [[Definition:Lower Section (Class Theory)|lower section]] of $\struct {... | Let us label the [[Definition:Domain of Relation (Class Theory)|domains]] of $\phi_1$ and $\phi_2$:
:$D_1 = \Dom {\phi_1}$
:$D_2 = \Dom {\phi_2}$
Because $D_1$ and $D_2$ are both [[Definition:Lower Section (Class Theory)|lower sections]] of $\struct {A, \preccurlyeq_A}$, they are [[Definition:Comparable Elements|compa... | Isomorphisms between Lower Sections of Well-Ordered Classes are Nested | https://proofwiki.org/wiki/Isomorphisms_between_Lower_Sections_of_Well-Ordered_Classes_are_Nested | https://proofwiki.org/wiki/Isomorphisms_between_Lower_Sections_of_Well-Ordered_Classes_are_Nested | [
"Well-Orderings",
"Lower Sections",
"Order Isomorphisms",
"Nests"
] | [
"Definition:Well-Ordered Class",
"Definition:Order Isomorphism/Well-Orderings/Class Theory",
"Definition:Lower Section/Class Theory",
"Definition:Lower Section/Class Theory",
"Definition:Subset Relation on Mappings"
] | [
"Definition:Domain (Set Theory)/Relation/Class Theory",
"Definition:Lower Section/Class Theory",
"Definition:Comparable Elements",
"Definition:Subset Relation",
"Definition:Restriction/Relation/Class Theory",
"Definition:Order Isomorphism/Well-Orderings/Class Theory",
"Definition:Lower Section/Class The... |
proofwiki-19776 | Sum of Larger Ideals is Larger | Let $R$ be a commutative ring with unity.
Let $\mathfrak a_1, \mathfrak a_2, \mathfrak b_1, \mathfrak b_2$ be ideals of $R$ such that:
:$\mathfrak a_1 \subseteq \mathfrak a_2$
and:
:$\mathfrak b_1 \subseteq \mathfrak b_2$
Then their sums satisfy:
:$\mathfrak a_1 + \mathfrak b_1 \subseteq \mathfrak a_2 + \mathfrak b_2$ | Let $x \in \mathfrak a_1 + \mathfrak b_1$.
By {{Defof|Sum of Ideals of Ring|Sum of Ideals}}, there are $a \in \mathfrak a_1$ and $a \in \mathfrak b_1$ such that:
:$x = a + b$
Then:
:$a \in \mathfrak a_1 \subseteq \mathfrak a_2$
and:
:$b \in \mathfrak b_1 \subseteq \mathfrak b_2$
Thus, by {{Defof|Sum of Ideals of Ring|S... | Let $R$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Let $\mathfrak a_1, \mathfrak a_2, \mathfrak b_1, \mathfrak b_2$ be [[Definition:Ideal of Ring|ideals]] of $R$ such that:
:$\mathfrak a_1 \subseteq \mathfrak a_2$
and:
:$\mathfrak b_1 \subseteq \mathfrak b_2$
Then their [[Definition... | Let $x \in \mathfrak a_1 + \mathfrak b_1$.
By {{Defof|Sum of Ideals of Ring|Sum of Ideals}}, there are $a \in \mathfrak a_1$ and $a \in \mathfrak b_1$ such that:
:$x = a + b$
Then:
:$a \in \mathfrak a_1 \subseteq \mathfrak a_2$
and:
:$b \in \mathfrak b_1 \subseteq \mathfrak b_2$
Thus, by {{Defof|Sum of Ideals of Rin... | Sum of Larger Ideals is Larger | https://proofwiki.org/wiki/Sum_of_Larger_Ideals_is_Larger | https://proofwiki.org/wiki/Sum_of_Larger_Ideals_is_Larger | [
"Radical of Ideals"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Ideal of Ring",
"Definition:Sum of Ideals of Ring"
] | [
"Category:Radical of Ideals"
] |
proofwiki-19777 | Prime Ideal is Primary Ideal | Let $R$ be a commutative ring with unity.
Let $\mathfrak p$ be a prime ideal of $R$.
Then $\mathfrak p$ is a primary ideal of $R$. | Let $xy \in \mathfrak p$.
Let $x \not \in \mathfrak p$
By definition of prime ideal:
:$y^1 = y \in \mathfrak p$
Thus, by definition, $\mathfrak p$ is a primary ideal.
{{qed}}
Category:Prime Ideals of Rings
Category:Primary Ideals
2e1qxhlf2bpm647btf4cvstgy1xgiku | Let $R$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let $\mathfrak p$ be a [[Definition:Prime Ideal of Ring|prime ideal]] of $R$.
Then $\mathfrak p$ is a [[Definition:Primary Ideal|primary ideal]] of $R$. | Let $xy \in \mathfrak p$.
Let $x \not \in \mathfrak p$
By definition of [[Definition:Prime Ideal of Commutative and Unitary Ring|prime ideal]]:
:$y^1 = y \in \mathfrak p$
Thus, by definition, $\mathfrak p$ is a [[Definition:Primary Ideal|primary ideal]].
{{qed}}
[[Category:Prime Ideals of Rings]]
[[Category:Primary... | Prime Ideal is Primary Ideal | https://proofwiki.org/wiki/Prime_Ideal_is_Primary_Ideal | https://proofwiki.org/wiki/Prime_Ideal_is_Primary_Ideal | [
"Prime Ideals of Rings",
"Primary Ideals"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Prime Ideal of Ring",
"Definition:Primary Ideal"
] | [
"Definition:Prime Ideal of Ring/Commutative and Unitary Ring",
"Definition:Primary Ideal",
"Category:Prime Ideals of Rings",
"Category:Primary Ideals"
] |
proofwiki-19778 | Contraction of Primary Ideal is Primary Ideal | Let $A$ and $B$ be commutative rings with unity.
Let $f : A \to B$ be a ring homomorphism.
Let $\mathfrak b$ be a primary ideal of $B$.
Let $\mathfrak b^c$ be the contraction of $\mathfrak b$ by $f$.
Then $\mathfrak b^c$ is a primary ideal of $A$. | Let $x,y \in A$ such that:
:$xy \in \mathfrak b^c$.
That is:
:$\map f {xy} = \map f x \map f y \in \mathfrak b$
Suppose that $x \ne \mathfrak b^c$.
That is:
:$\map f x \not \in \mathfrak b$
Since $\mathfrak b$ is primary:
:$\exists n \in \N_{>0} : \map f {y^n} = \paren {\map f y}^n \in \mathfrak b$
That is:
:$y^n \in \... | Let $A$ and $B$ be [[Definition:Commutative Ring with Unity|commutative rings with unity]].
Let $f : A \to B$ be a [[Definition:unital Ring Homomorphism|ring homomorphism]].
Let $\mathfrak b$ be a [[Definition:Primary Ideal|primary ideal]] of $B$.
Let $\mathfrak b^c$ be the [[Definition:Contraction of Ideal|contract... | Let $x,y \in A$ such that:
:$xy \in \mathfrak b^c$.
That is:
:$\map f {xy} = \map f x \map f y \in \mathfrak b$
Suppose that $x \ne \mathfrak b^c$.
That is:
:$\map f x \not \in \mathfrak b$
Since $\mathfrak b$ is [[Definition:Primary Ideal|primary]]:
:$\exists n \in \N_{>0} : \map f {y^n} = \paren {\map f y}^n \in... | Contraction of Primary Ideal is Primary Ideal | https://proofwiki.org/wiki/Contraction_of_Primary_Ideal_is_Primary_Ideal | https://proofwiki.org/wiki/Contraction_of_Primary_Ideal_is_Primary_Ideal | [
"Primary Ideals"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:unital Ring Homomorphism",
"Definition:Primary Ideal",
"Definition:Contraction of Ideal",
"Definition:Primary Ideal"
] | [
"Definition:Primary Ideal",
"Category:Primary Ideals"
] |
proofwiki-19779 | Fundamental Property of Norm on Bounded Linear Functional | Let $\Bbb F \in \set {\R, \C}$.
Let $\struct {X, \norm \cdot_X}$ be a normed vector space over $\Bbb F$.
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual of $\struct {X, \norm \cdot_X}$.
Then for all $x \in X$ and $f \in X^\ast$:
:$\cmod {\map f x} \le \norm f_{X^\ast} \norm x_X$ | Let $x \in X$ and $f \in X^\ast$.
If $x = 0$, the claim is trivial, since by the linearity of $f$:
:$\cmod {\map f x} = \cmod {\map f 0} = 0$
Let $x \ne 0$.
Recall by {{Defof|Norm/Bounded Linear Functional/Normed Vector Space|Norm of Bounded Linear Functional|index = 3}}:
:$\norm f_{X^\ast} = \sup \set {\dfrac {\cmod {... | Let $\Bbb F \in \set {\R, \C}$.
Let $\struct {X, \norm \cdot_X}$ be a [[Definition:Normed Vector Space|normed vector space]] over $\Bbb F$.
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the [[Definition:Normed Dual Space|normed dual]] of $\struct {X, \norm \cdot_X}$.
Then for all $x \in X$ and $f \in X^\ast$:
... | Let $x \in X$ and $f \in X^\ast$.
If $x = 0$, the claim is trivial, since by the [[Definition:Linear Functional|linearity]] of $f$:
:$\cmod {\map f x} = \cmod {\map f 0} = 0$
Let $x \ne 0$.
Recall by {{Defof|Norm/Bounded Linear Functional/Normed Vector Space|Norm of Bounded Linear Functional|index = 3}}:
:$\norm f_... | Fundamental Property of Norm on Bounded Linear Functional | https://proofwiki.org/wiki/Fundamental_Property_of_Norm_on_Bounded_Linear_Functional | https://proofwiki.org/wiki/Fundamental_Property_of_Norm_on_Bounded_Linear_Functional | [
"Second Normed Duals",
"Evaluation Linear Transformations (Normed Vector Spaces)",
"Bounded Linear Functionals",
"Bounded Linear Functionals"
] | [
"Definition:Normed Vector Space",
"Definition:Normed Dual Space"
] | [
"Definition:Linear Functional",
"Category:Bounded Linear Functionals"
] |
proofwiki-19780 | Fundamental Theorem of Well-Ordering | Let $\struct {A, \preccurlyeq_A}$ and $\struct {B, \preccurlyeq_B}$ be well-ordered classes.
Then either:
:$\struct {A, \preccurlyeq_A}$ is order isomorphic to a lower section of $\struct {B, \preccurlyeq_B}$, or perhaps all of $\struct {B, \preccurlyeq_B}$
or:
:$\struct {B, \preccurlyeq_B}$ is order isomorphic to a lo... | Let $N$ be the class of all order isomorphisms from lower sections of $\struct {A, \preccurlyeq_A}$ which are sets to lower sections of $\struct {B, \preccurlyeq_B}$ which are sets.
Each such order isomorphism is itself a set.
By Isomorphisms between Lower Sections of Well-Ordered Classes are Nested, $N$ is a nest.
$N$... | Let $\struct {A, \preccurlyeq_A}$ and $\struct {B, \preccurlyeq_B}$ be [[Definition:Well-Ordered Class|well-ordered classes]].
Then either:
:$\struct {A, \preccurlyeq_A}$ is [[Definition:Order Isomorphism on Well-Orderings (Class Theory)|order isomorphic]] to a [[Definition:Lower Section (Class Theory)|lower section]]... | Let $N$ be the [[Definition:Class (Class Theory)|class]] of all [[Definition:Order Isomorphism on Well-Orderings (Class Theory)|order isomorphisms]] from [[Definition:Lower Section (Class Theory)|lower sections]] of $\struct {A, \preccurlyeq_A}$ which are [[Definition:Set|sets]] to [[Definition:Lower Section (Class The... | Fundamental Theorem of Well-Ordering | https://proofwiki.org/wiki/Fundamental_Theorem_of_Well-Ordering | https://proofwiki.org/wiki/Fundamental_Theorem_of_Well-Ordering | [
"Fundamental Theorem of Well-Ordering",
"Well-Orderings",
"Fundamental Theorems",
"Order Isomorphisms"
] | [
"Definition:Well-Ordered Class",
"Definition:Order Isomorphism/Well-Orderings/Class Theory",
"Definition:Lower Section/Class Theory",
"Definition:Order Isomorphism/Well-Orderings/Class Theory",
"Definition:Lower Section/Class Theory"
] | [
"Definition:Class (Class Theory)",
"Definition:Order Isomorphism/Well-Orderings/Class Theory",
"Definition:Lower Section/Class Theory",
"Definition:Set",
"Definition:Lower Section/Class Theory",
"Definition:Set",
"Definition:Order Isomorphism/Well-Orderings/Class Theory",
"Definition:Set",
"Isomorph... |
proofwiki-19781 | Axiom of Swelledness is implied by Axiom of Replacement | Let the {{axiom-link|Replacement|Class Theory}} (in the context of class theory) be accepted.
Then the {{axiom-link|Swelledness}} holds. | Recall the Axiom of Replacement:
{{:Axiom:Axiom of Replacement/Class Theory/Formulation 1}}
Recall the {{axiom-link|Swelledness}}:
{{:Axiom:Axiom of Swelledness}}
That is:
:Every subclass of a set is a set.
Let $x$ be a set.
Let $A$ be a class such that $A \subseteq x$.
Suppose $A$ is the empty class.
Then by the {{axi... | Let the {{axiom-link|Replacement|Class Theory}} (in the context of [[Definition:Class Theory|class theory]]) be accepted.
Then the {{axiom-link|Swelledness}} holds. | Recall the [[Axiom:Axiom of Replacement/Class Theory/Formulation 1|Axiom of Replacement]]:
{{:Axiom:Axiom of Replacement/Class Theory/Formulation 1}}
Recall the {{axiom-link|Swelledness}}:
{{:Axiom:Axiom of Swelledness}}
That is:
:Every [[Definition:Subclass|subclass]] of a [[Definition:Set|set]] is a [[Definition:Set... | Axiom of Swelledness is implied by Axiom of Replacement | https://proofwiki.org/wiki/Axiom_of_Swelledness_is_implied_by_Axiom_of_Replacement | https://proofwiki.org/wiki/Axiom_of_Swelledness_is_implied_by_Axiom_of_Replacement | [
"Axiom of Swelledness",
"Axiom of Replacement"
] | [
"Definition:Class Theory"
] | [
"Axiom:Axiom of Replacement/Class Theory/Formulation 1",
"Definition:Subclass",
"Definition:Set",
"Definition:Set",
"Definition:Set",
"Definition:Class (Class Theory)",
"Definition:Empty Class (Class Theory)",
"Definition:Set",
"Definition:Non-Empty Set/Class Theory",
"Definition:Class (Class Theo... |
proofwiki-19782 | Continuous Linear Transformations form Subspace of Linear Transformations | The space of continuous linear transformations is a subspace of the space of linear transformations. | Let $\struct {X, \norm \cdot }$ and $\struct {Y, \norm \cdot }$ be normed vector spaces. | The [[Definition:Continuous Linear Transformation Space|space of continuous linear transformations]] is a [[Definition:Vector Subspace|subspace]] of the [[Definition:Set of All Linear Transformations/Vector Space|space of linear transformations]]. | Let $\struct {X, \norm \cdot }$ and $\struct {Y, \norm \cdot }$ be [[Definition:Normed Vector Space|normed vector spaces]]. | Continuous Linear Transformations form Subspace of Linear Transformations | https://proofwiki.org/wiki/Continuous_Linear_Transformations_form_Subspace_of_Linear_Transformations | https://proofwiki.org/wiki/Continuous_Linear_Transformations_form_Subspace_of_Linear_Transformations | [
"Linear Transformations",
"Continuous Linear Transformations",
"Continuous Mappings",
"Examples of Vector Subspaces"
] | [
"Definition:Continuous Linear Transformation Space",
"Definition:Vector Subspace",
"Definition:Set of All Linear Transformations/Vector Space"
] | [
"Definition:Normed Vector Space"
] |
proofwiki-19783 | Axiom of Replacement implies Image of Bijection on Set is Set | Let the Axiom of Replacement (in the context of class theory) be accepted.
Let $A$ be a class which can be put into one-to-one correspondence with a set.
Then $A$ is a set. | Let $x$ be a set such that $A$ can be put into one-to-one correspondence with $x$.
Then by definition there exists a bijection from $x$ to $A$.
Let $f: x \to A$ be such a bijection.
Then by the Axiom of Replacement, $f \sqbrk x$ is a set.
But:
:$f \sqbrk x = A$
Hence the result.
{{qed}} | Let the [[Axiom:Axiom of Replacement (Class Theory)|Axiom of Replacement]] (in the context of [[Definition:Class Theory|class theory]]) be accepted.
Let $A$ be a [[Definition:Class (Class Theory)|class]] which can be put into [[Definition:One-to-One Correspondence|one-to-one correspondence]] with a [[Definition:Set|se... | Let $x$ be a [[Definition:Set|set]] such that $A$ can be put into [[Definition:One-to-One Correspondence|one-to-one correspondence]] with $x$.
Then by definition there exists a [[Definition:Class Bijection|bijection]] from $x$ to $A$.
Let $f: x \to A$ be such a [[Definition:Class Bijection|bijection]].
Then by the [... | Axiom of Replacement implies Image of Bijection on Set is Set | https://proofwiki.org/wiki/Axiom_of_Replacement_implies_Image_of_Bijection_on_Set_is_Set | https://proofwiki.org/wiki/Axiom_of_Replacement_implies_Image_of_Bijection_on_Set_is_Set | [
"Axiom of Replacement"
] | [
"Axiom:Axiom of Replacement/Class Theory",
"Definition:Class Theory",
"Definition:Class (Class Theory)",
"Definition:Bijection",
"Definition:Set",
"Definition:Set"
] | [
"Definition:Set",
"Definition:Bijection",
"Definition:Bijection/Class Theory",
"Definition:Bijection/Class Theory",
"Axiom:Axiom of Replacement/Class Theory",
"Definition:Set"
] |
proofwiki-19784 | Characterization of Convergence in Locally Convex Space | Let $\Bbb F \in \set {\R, \C}$.
Let $\struct {X, \mathcal P}$ be a locally convex space over $\Bbb F$ with standard topology $\tau$.
Let $x \in X$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$.
Then:
:$x_n \to x$ in $\struct {X, \tau}$
{{iff}}:
:$\map p {x_n - x} \to 0$ as $n \to \infty$ as a real s... | === Necessary Condition ===
Suppose that:
:$x_n \to x$ in $\struct {X, \tau}$
Let $p \in \mathcal P$.
Let $\epsilon > 0$.
We aim to show that there exists $N \in \N$ such that:
:$\map p {x_n - x} < \epsilon$ for $n \ge N$
showing that:
:$\map p {x_n - x} \to 0$ as $n \to \infty$.
From the definition of a converge... | Let $\Bbb F \in \set {\R, \C}$.
Let $\struct {X, \mathcal P}$ be a [[Definition:Locally Convex Space|locally convex space]] over $\Bbb F$ with [[Definition:Locally Convex Space/Standard Topology|standard topology]] $\tau$.
Let $x \in X$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]]... | === Necessary Condition ===
Suppose that:
:$x_n \to x$ in $\struct {X, \tau}$
Let $p \in \mathcal P$.
Let $\epsilon > 0$.
We aim to show that there exists $N \in \N$ such that:
:$\map p {x_n - x} < \epsilon$ for $n \ge N$
showing that:
:$\map p {x_n - x} \to 0$ as $n \to \infty$.
From the definition of a... | Characterization of Convergence in Locally Convex Space | https://proofwiki.org/wiki/Characterization_of_Convergence_in_Locally_Convex_Space | https://proofwiki.org/wiki/Characterization_of_Convergence_in_Locally_Convex_Space | [
"Locally Convex Spaces",
"Convergent Sequences"
] | [
"Definition:Locally Convex Space",
"Definition:Locally Convex Space/Standard Topology",
"Definition:Sequence",
"Definition:Convergent Sequence/Real Numbers"
] | [
"Definition:Convergent Sequence/Topology",
"Definition:Locally Convex Space/Standard Topology",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology"
] |
proofwiki-19785 | Characterization of Continuous Linear Transformations between Locally Convex Spaces | Let $\Bbb F \in \set {\R, \C}$.
Let $\struct {X, \mathcal P}$ and $\struct {Y, \mathcal Q}$ be locally convex spaces over $\Bbb F$.
Let $T : X \to Y$ be a linear transformation.
{{TFAE}}
:$(1) \quad$ $T$ is everywhere continuous
:$(2) \quad$ $T$ is continuous at ${\mathbf 0}_X$
:$(3) \quad$ for each $q \in \mathcal Q$ ... | === $(1)$ iff $(2)$ ===
This follows from Linear Transformation between Topological Vector Spaces Continuous iff Continuous at Origin.
{{qed|lemma}} | Let $\Bbb F \in \set {\R, \C}$.
Let $\struct {X, \mathcal P}$ and $\struct {Y, \mathcal Q}$ be [[Definition:Locally Convex Space|locally convex spaces]] over $\Bbb F$.
Let $T : X \to Y$ be a [[Definition:Linear Transformation|linear transformation]].
{{TFAE}}
:$(1) \quad$ $T$ is [[Definition:Everywhere Continuous ... | === $(1)$ iff $(2)$ ===
This follows from [[Linear Transformation between Topological Vector Spaces Continuous iff Continuous at Origin]].
{{qed|lemma}} | Characterization of Continuous Linear Transformations between Locally Convex Spaces | https://proofwiki.org/wiki/Characterization_of_Continuous_Linear_Transformations_between_Locally_Convex_Spaces | https://proofwiki.org/wiki/Characterization_of_Continuous_Linear_Transformations_between_Locally_Convex_Spaces | [
"Linear Transformations",
"Locally Convex Spaces"
] | [
"Definition:Locally Convex Space",
"Definition:Linear Transformation",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Continuous Mapping (Topology)/Point",
"Definition:Real Number",
"Definition:Finite Set",
"Definition:Continuous Mapping (Topology)",
"Definition:Locally Convex Spac... | [
"Linear Transformation between Topological Vector Spaces Continuous iff Continuous at Origin"
] |
proofwiki-19786 | Translation of Open Set in Topological Vector Space is Open | Let $\struct {X, \tau}$ be a topological vector space.
Let $U \subseteq X$ be an open set.
Let $x \in X$.
Then:
:$U + x$ is open. | Let $T_{-x}$ be the translation by $x$ mapping.
From Translation Mapping on Topological Vector Space is Homeomorphism, we have:
:$T_{-x}$ is a homeomorphism.
So, since $U$ is open, we have:
:$T_{-x} \sqbrk U$ is open.
That is:
:$U - \paren {-x} = U + x$ is open.
{{qed}}
{{explain|Review notation -- there is a cavea... | Let $\struct {X, \tau}$ be a [[Definition:Topological Vector Space|topological vector space]].
Let $U \subseteq X$ be an [[Definition:Open Set (Topology)|open set]].
Let $x \in X$.
Then:
:$U + x$ is [[Definition:Open Set (Topology)|open]]. | Let $T_{-x}$ be the [[Definition:Translation in Vector Space|translation by $x$ mapping]].
From [[Translation Mapping on Topological Vector Space is Homeomorphism]], we have:
:$T_{-x}$ is a [[Definition:Homeomorphism|homeomorphism]].
So, since $U$ is [[Definition:Open Set (Topology)|open]], we have:
:$T_{-x} \sq... | Translation of Open Set in Topological Vector Space is Open | https://proofwiki.org/wiki/Translation_of_Open_Set_in_Topological_Vector_Space_is_Open | https://proofwiki.org/wiki/Translation_of_Open_Set_in_Topological_Vector_Space_is_Open | [
"Topological Vector Spaces",
"Translation of Subsets of Vector Spaces"
] | [
"Definition:Topological Vector Space",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology"
] | [
"Definition:Translation Mapping/Vector Space",
"Translation Mapping on Topological Vector Space is Homeomorphism",
"Definition:Homeomorphism",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Translation Mapping/Vector Space"
] |
proofwiki-19787 | Metric Induced by Norm is Invariant Metric | Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $d$ be the metric induced by $\norm {\, \cdot \,}$.
Then $d$ is invariant. | Let $x, y, z \in X$.
Then, we have:
{{begin-eqn}}
{{eqn | l = \map d {x + z, y + z}
| r = \norm {\paren {x + z} - \paren {y + z} }
| c = {{Defof|Metric Induced by Norm}}
}}
{{eqn | r = \norm {x - y + z - z}
}}
{{eqn | r = \norm {x - y}
}}
{{eqn | r = \map d {x, y}
| c = {{defof|Metric Induced by Norm}}
}}
{{end... | Let $\struct {X, \norm {\, \cdot \,} }$ be a [[Definition:Normed Vector Space|normed vector space]].
Let $d$ be the [[Definition:Metric Induced by Norm|metric induced by $\norm {\, \cdot \,}$]].
Then $d$ is [[Definition:Invariant Metric on Vector Space|invariant]]. | Let $x, y, z \in X$.
Then, we have:
{{begin-eqn}}
{{eqn | l = \map d {x + z, y + z}
| r = \norm {\paren {x + z} - \paren {y + z} }
| c = {{Defof|Metric Induced by Norm}}
}}
{{eqn | r = \norm {x - y + z - z}
}}
{{eqn | r = \norm {x - y}
}}
{{eqn | r = \map d {x, y}
| c = {{defof|Metric Induced by Norm}}
}}
{{e... | Metric Induced by Norm is Invariant Metric | https://proofwiki.org/wiki/Metric_Induced_by_Norm_is_Invariant_Metric | https://proofwiki.org/wiki/Metric_Induced_by_Norm_is_Invariant_Metric | [
"Metric Spaces",
"Normed Vector Spaces"
] | [
"Definition:Normed Vector Space",
"Definition:Metric Induced by Norm",
"Definition:Invariant Metric on Vector Space"
] | [
"Category:Metric Spaces",
"Category:Normed Vector Spaces"
] |
proofwiki-19788 | Equivalence of Definitions of Primary Ideal | Let $R$ be a commutative ring with unity.
{{TFAE|def = Primary Ideal}} | === Definition 1 implies Definition 2 ===
Let $x + \mathfrak q$ be a zero-divisor of $R / \mathfrak q$.
That is, there is a $y \not \in \mathfrak q$ such that:
:$\paren {x + \mathfrak q} \paren {y + \mathfrak q} = 0 + \mathfrak q$
Thus:
:$xy + \mathfrak q = 0 + \mathfrak q$
which means:
:$xy \in \mathfrak q$
Then, {{hy... | Let $R$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
{{TFAE|def = Primary Ideal}} | === Definition 1 implies Definition 2 ===
Let $x + \mathfrak q$ be a [[Definition:Zero Divisor of Ring|zero-divisor]] of $R / \mathfrak q$.
That is, there is a $y \not \in \mathfrak q$ such that:
:$\paren {x + \mathfrak q} \paren {y + \mathfrak q} = 0 + \mathfrak q$
Thus:
:$xy + \mathfrak q = 0 + \mathfrak q$
which m... | Equivalence of Definitions of Primary Ideal | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Primary_Ideal | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Primary_Ideal | [
"Primary Ideals"
] | [
"Definition:Commutative and Unitary Ring"
] | [
"Definition:Zero Divisor/Ring",
"Definition:Nilpotent Ring Element",
"Definition:Nilpotent Ring Element"
] |
proofwiki-19789 | Minimally Superinductive Class is Well-Ordered under Subset Relation | Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be minimally superinductive under $g$.
Then $M$ is well-ordered under the subset relation. | We have {{apriori}} that $M$ is a $g$-tower.
The result follows from $g$-Tower is Well-Ordered under Subset Relation.
{{qed}}
Category:Minimally Superinductive Classes
Category:Well-Orderings
Category:Subset Relation
8ynmqqm4swiiis0usxx5rbpz7h0s5tx | Let $M$ be a [[Definition:Class (Class Theory)|class]].
Let $g: M \to M$ be a [[Definition:Progressing Mapping|progressing mapping]] on $M$.
Let $M$ be [[Definition:Minimally Superinductive Class|minimally superinductive]] under $g$.
Then $M$ is [[Definition:Well-Ordered Class|well-ordered]] under the [[Definition:... | We have {{apriori}} that $M$ is a [[Definition:G-Tower|$g$-tower]].
The result follows from [[G-Tower is Well-Ordered under Subset Relation|$g$-Tower is Well-Ordered under Subset Relation]].
{{qed}}
[[Category:Minimally Superinductive Classes]]
[[Category:Well-Orderings]]
[[Category:Subset Relation]]
8ynmqqm4swiiis0u... | Minimally Superinductive Class is Well-Ordered under Subset Relation | https://proofwiki.org/wiki/Minimally_Superinductive_Class_is_Well-Ordered_under_Subset_Relation | https://proofwiki.org/wiki/Minimally_Superinductive_Class_is_Well-Ordered_under_Subset_Relation | [
"Minimally Superinductive Classes",
"Well-Orderings",
"Subset Relation"
] | [
"Definition:Class (Class Theory)",
"Definition:Progressing Mapping",
"Definition:Minimally Superinductive Class",
"Definition:Well-Ordered Class",
"Definition:Subset Relation"
] | [
"Definition:G-Tower",
"G-Tower is Well-Ordered under Subset Relation",
"Category:Minimally Superinductive Classes",
"Category:Well-Orderings",
"Category:Subset Relation"
] |
proofwiki-19790 | Balanced Subset of Real Numbers is Bounded or Entire Space | Consider $\R$ as a vector space over $\R$.
Let $E$ be a balanced subset of $\R$.
Then $E$ is bounded, or $E = \R$. | Suppose that $E$ is not bounded.
Then, for each $M > 0$ there exists some $x_M \in E$ such that $\size {x_M} > M$.
We show that:
:$\closedint {-M} M \subseteq E$ for each $M > 0$.
Let:
:$t \in \closedint {-M} M$
Then, we have:
:$\ds \size {\frac t {x_M} } < 1$
So, since $E$ is balanced, we have:
:$\ds x_M \cdot \par... | Consider $\R$ as a [[Definition:Vector Space|vector space]] over $\R$.
Let $E$ be a [[Definition:Balanced Set|balanced]] [[Definition:Subset|subset]] of $\R$.
Then $E$ is [[Definition:Bounded Subset of Real Numbers|bounded]], or $E = \R$. | Suppose that $E$ is not [[Definition:Bounded Subset of Real Numbers|bounded]].
Then, for each $M > 0$ there exists some $x_M \in E$ such that $\size {x_M} > M$.
We show that:
:$\closedint {-M} M \subseteq E$ for each $M > 0$.
Let:
:$t \in \closedint {-M} M$
Then, we have:
:$\ds \size {\frac t {x_M} } < 1$
So,... | Balanced Subset of Real Numbers is Bounded or Entire Space | https://proofwiki.org/wiki/Balanced_Subset_of_Real_Numbers_is_Bounded_or_Entire_Space | https://proofwiki.org/wiki/Balanced_Subset_of_Real_Numbers_is_Bounded_or_Entire_Space | [] | [
"Definition:Vector Space",
"Definition:Balanced Set",
"Definition:Subset",
"Definition:Bounded Set/Real Numbers"
] | [
"Definition:Bounded Set/Real Numbers",
"Definition:Balanced Set",
"Union of Subsets is Subset",
"CAtegory:Balanced Sets"
] |
proofwiki-19791 | Balanced Subset of Complex Plane is Bounded or Entire Space | Consider $\C$ as a vector space over $\C$.
Let $E$ be a balanced subset of $\C$.
Then $E$ is bounded, or $E = \C$. | Suppose that $E$ is not bounded.
Then, for each $R > 0$ there exists some $z_R \in E$ such that $\size {z_R} > R$.
We show that:
:$\map {B_R} 0 \subseteq E$ for each $R > 0$
where $\map {B_R} 0$ is the open ball centered at $0$ with radius $R$.
Let:
:$w \in \map {B_R} 0$
Then:
:$\ds \cmod {\frac w {z_R} } < 1$
So, si... | Consider $\C$ as a [[Definition:Vector Space|vector space]] over $\C$.
Let $E$ be a [[Definition:Balanced Set|balanced]] [[Definition:Subset|subset]] of $\C$.
Then $E$ is [[Definition:Bounded Subset of Complex Plane|bounded]], or $E = \C$. | Suppose that $E$ is not [[Definition:Bounded Subset of Real Numbers|bounded]].
Then, for each $R > 0$ there exists some $z_R \in E$ such that $\size {z_R} > R$.
We show that:
:$\map {B_R} 0 \subseteq E$ for each $R > 0$
where $\map {B_R} 0$ is the [[Definition:Open Ball|open ball]] [[Definition:Center of Open Ball... | Balanced Subset of Complex Plane is Bounded or Entire Space | https://proofwiki.org/wiki/Balanced_Subset_of_Complex_Plane_is_Bounded_or_Entire_Space | https://proofwiki.org/wiki/Balanced_Subset_of_Complex_Plane_is_Bounded_or_Entire_Space | [
"Balanced Sets"
] | [
"Definition:Vector Space",
"Definition:Balanced Set",
"Definition:Subset",
"Definition:Bounded Metric Space/Complex"
] | [
"Definition:Bounded Set/Real Numbers",
"Definition:Open Ball",
"Definition:Open Ball/Center",
"Definition:Open Ball/Radius",
"Definition:Balanced Set",
"Union of Subsets is Subset",
"Category:Balanced Sets"
] |
proofwiki-19792 | Image of Balanced Set under Linear Transformation is Balanced | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ and $Y$ be vector spaces over $\Bbb F$.
Let $E \subseteq X$ be balanced.
Let $T : X \to Y$ be a linear transformation.
Then $\map T E \subseteq Y$ is balanced. | We aim to show that for all $s \in \R$ with $\cmod s \le 1$, we have:
:$s \map T E \subseteq \map T E$
Let $y \in s \map T E$.
Then there exists $x \in E$ such that $y = s T x$.
From the linearity of $T$, we have:
:$y = \map T {s x}$
Since $x \in E$, we have $s x \in s E$ from the definition of dilation of $E$ by $s$... | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ and $Y$ be [[Definition:Vector Space|vector spaces]] over $\Bbb F$.
Let $E \subseteq X$ be [[Definition:Balanced Set|balanced]].
Let $T : X \to Y$ be a [[Definition:Linear Transformation|linear transformation]].
Then $\map T E \subseteq Y$ is [[Definition:Balanced Set|balanc... | We aim to show that for all $s \in \R$ with $\cmod s \le 1$, we have:
:$s \map T E \subseteq \map T E$
Let $y \in s \map T E$.
Then there exists $x \in E$ such that $y = s T x$.
From the [[Definition:Linear Transformation|linearity]] of $T$, we have:
:$y = \map T {s x}$
Since $x \in E$, we have $s x \in s E$ fr... | Image of Balanced Set under Linear Transformation is Balanced | https://proofwiki.org/wiki/Image_of_Balanced_Set_under_Linear_Transformation_is_Balanced | https://proofwiki.org/wiki/Image_of_Balanced_Set_under_Linear_Transformation_is_Balanced | [
"Balanced Sets",
"Linear Transformations"
] | [
"Definition:Vector Space",
"Definition:Balanced Set",
"Definition:Linear Transformation",
"Definition:Balanced Set"
] | [
"Definition:Linear Transformation",
"Definition:Linear Combination of Subsets of Vector Space/Dilation",
"Definition:Balanced Set",
"Definition:Balanced Set",
"Category:Balanced Sets",
"Category:Linear Transformations"
] |
proofwiki-19793 | Extending Operation is a Slowly Progressing Mapping | Let $S$ denote the class of all ordinal sequences.
Let $E: S \to S$ be an extending operation on $S$.
Then $E$ is a slowly progressing mapping. | Let $\theta \in S$ be an $\alpha$-sequence.
By definition of extending operation:
:$\map E \theta = \theta \cup \tuple {\alpha, x}$
where $x$ is arbitrary.
Thus:
:$\theta \subseteq \map E \theta$
and it is seen that $E$ is by definition a progressing mapping.
It is also seen that:
:$\card \theta = \card \alpha$
while:
... | Let $S$ denote the [[Definition:Class (Class Theory)|class]] of all [[Definition:Ordinal Sequence|ordinal sequences]].
Let $E: S \to S$ be an [[Definition:Extending Operation|extending operation]] on $S$.
Then $E$ is a [[Definition:Slowly Progressing Mapping|slowly progressing mapping]]. | Let $\theta \in S$ be an [[Definition:Ordinal Sequence|$\alpha$-sequence]].
By definition of [[Definition:Extending Operation|extending operation]]:
:$\map E \theta = \theta \cup \tuple {\alpha, x}$
where $x$ is arbitrary.
Thus:
:$\theta \subseteq \map E \theta$
and it is seen that $E$ is by definition a [[Definitio... | Extending Operation is a Slowly Progressing Mapping | https://proofwiki.org/wiki/Extending_Operation_is_a_Slowly_Progressing_Mapping | https://proofwiki.org/wiki/Extending_Operation_is_a_Slowly_Progressing_Mapping | [
"Extending Operations",
"Slowly Progressing Mappings"
] | [
"Definition:Class (Class Theory)",
"Definition:Ordinal Sequence",
"Definition:Extending Operation",
"Definition:Slowly Progressing Mapping"
] | [
"Definition:Ordinal Sequence",
"Definition:Extending Operation",
"Definition:Progressing Mapping",
"Definition:Slowly Progressing Mapping"
] |
proofwiki-19794 | Extending Operation is a Strictly Progressing Mapping | Let $S$ denote the class of all ordinal sequences.
Let $E: S \to S$ be an extending operation on $S$.
Then $E$ is a strictly progressing mapping. | Let $\theta \in S$ be an $\alpha$-sequence.
By definition of extending operation:
:$\map E \theta = \theta \cup \tuple {\alpha, x}$
where $x$ is arbitrary.
From Extending Operation is a Slowly Progressing Mapping we have {{afortiori}} that $E$ is a progressing mapping.
By definition of $\alpha$-sequence:
:$\alpha \noti... | Let $S$ denote the [[Definition:Class (Class Theory)|class]] of all [[Definition:Ordinal Sequence|ordinal sequences]].
Let $E: S \to S$ be an [[Definition:Extending Operation|extending operation]] on $S$.
Then $E$ is a [[Definition:Strictly Progressing Mapping|strictly progressing mapping]]. | Let $\theta \in S$ be an [[Definition:Ordinal Sequence|$\alpha$-sequence]].
By definition of [[Definition:Extending Operation|extending operation]]:
:$\map E \theta = \theta \cup \tuple {\alpha, x}$
where $x$ is arbitrary.
From [[Extending Operation is a Slowly Progressing Mapping]] we have {{afortiori}} that $E$ is ... | Extending Operation is a Strictly Progressing Mapping | https://proofwiki.org/wiki/Extending_Operation_is_a_Strictly_Progressing_Mapping | https://proofwiki.org/wiki/Extending_Operation_is_a_Strictly_Progressing_Mapping | [
"Extending Operations",
"Strictly Progressing Mappings"
] | [
"Definition:Class (Class Theory)",
"Definition:Ordinal Sequence",
"Definition:Extending Operation",
"Definition:Strictly Progressing Mapping"
] | [
"Definition:Ordinal Sequence",
"Definition:Extending Operation",
"Extending Operation is a Slowly Progressing Mapping",
"Definition:Progressing Mapping",
"Definition:Ordinal Sequence",
"Definition:Strictly Progressing Mapping"
] |
proofwiki-19795 | Dilation of Union of Subsets of Vector Space | Let $K$ be a field.
Let $X$ be a vector space over $K$.
Let $\family {E_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $X$.
Let $\lambda \in K$.
Then:
:$\ds \lambda \bigcup_{\alpha \mathop \in I} E_\alpha = \bigcup_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$
where $\lambda E_\alpha$ denotes ... | We have:
:$\ds v \in \lambda \bigcup_{\alpha \mathop \in I} E_\alpha$
{{iff}}:
:$v = \lambda x$ for some $\ds x \in \bigcup_{\alpha \mathop \in I} E_\alpha$.
This is equivalent to:
:there exists some $\alpha \in I$ such that $v = \lambda x$ for some $x \in E_\alpha$.
This is equivalent to:
:there exists some $\alpha... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $X$ be a [[Definition:Vector Space|vector space]] over $K$.
Let $\family {E_\alpha}_{\alpha \mathop \in I}$ be an [[Definition:Indexed Family of Subsets|indexed family]] of [[Definition:Subset|subsets]] of $X$.
Let $\lambda \in K$.
Then:
:$\ds \lam... | We have:
:$\ds v \in \lambda \bigcup_{\alpha \mathop \in I} E_\alpha$
{{iff}}:
:$v = \lambda x$ for some $\ds x \in \bigcup_{\alpha \mathop \in I} E_\alpha$.
This is equivalent to:
:there exists some $\alpha \in I$ such that $v = \lambda x$ for some $x \in E_\alpha$.
This is equivalent to:
:there exists some ... | Dilation of Union of Subsets of Vector Space | https://proofwiki.org/wiki/Dilation_of_Union_of_Subsets_of_Vector_Space | https://proofwiki.org/wiki/Dilation_of_Union_of_Subsets_of_Vector_Space | [
"Dilations of Subsets of Vector Spaces",
"Set Union"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Vector Space",
"Definition:Indexing Set/Family of Subsets",
"Definition:Subset",
"Definition:Linear Combination of Subsets of Vector Space/Dilation"
] | [
"Definition:Set Equality",
"Category:Dilations of Subsets of Vector Spaces",
"Category:Set Union"
] |
proofwiki-19796 | Dilation of Intersection of Subsets of Vector Space | Let $K$ be a field.
Let $X$ be a vector space over $K$.
Let $\family {E_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $X$.
Let $\lambda \in K$.
Then:
:$\ds \lambda \bigcap_{\alpha \mathop \in I} E_\alpha = \bigcap_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$
where $\lambda E_\alpha$ denotes t... | First, if $\lambda = 0_K$ then we have:
:$\lambda E_\alpha = \set { {\mathbf 0}_X}$
and:
:$\ds \lambda \bigcap_{\alpha \mathop \in I} E_\alpha = \set { {\mathbf 0}_X}$
so that:
:$\ds \bigcap_{\alpha \mathop \in I} \paren {\lambda E_\alpha} = \set { {\mathbf 0}_X} = \lambda \bigcap_{\alpha \mathop \in I} E_\alpha$
Now t... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $X$ be a [[Definition:Vector Space|vector space]] over $K$.
Let $\family {E_\alpha}_{\alpha \mathop \in I}$ be an [[Definition:Indexed Family of Subsets|indexed family]] of [[Definition:Subset|subsets]] of $X$.
Let $\lambda \in K$.
Then:
:$\ds \lambd... | First, if $\lambda = 0_K$ then we have:
:$\lambda E_\alpha = \set { {\mathbf 0}_X}$
and:
:$\ds \lambda \bigcap_{\alpha \mathop \in I} E_\alpha = \set { {\mathbf 0}_X}$
so that:
:$\ds \bigcap_{\alpha \mathop \in I} \paren {\lambda E_\alpha} = \set { {\mathbf 0}_X} = \lambda \bigcap_{\alpha \mathop \in I} E_\alpha$
Now... | Dilation of Intersection of Subsets of Vector Space | https://proofwiki.org/wiki/Dilation_of_Intersection_of_Subsets_of_Vector_Space | https://proofwiki.org/wiki/Dilation_of_Intersection_of_Subsets_of_Vector_Space | [
"Dilations of Subsets of Vector Spaces",
"Set Intersection"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Vector Space",
"Definition:Indexing Set/Family of Subsets",
"Definition:Subset",
"Definition:Linear Combination of Subsets of Vector Space/Dilation"
] | [
"Category:Dilations of Subsets of Vector Spaces",
"Category:Set Intersection"
] |
proofwiki-19797 | Vertical Section of Continuous Function is Continuous | Let $X$, $Y$ and $T$ be topological spaces.
Equip the Cartesian product $X \times Y$ with the product topology.
Let $f : X \times Y \to T$ be a continuous mapping.
Let $x \in X$.
Then the $x$-vertical section $f_x : Y \to T$ is continuous. | From the definition of the $x$-vertical section, we have:
:$\map {f_x} y = \map f {x, y}$
for each $y \in Y$.
Define the map $p_x : Y \to X \times Y$ by:
:$\map {p_x} y = \tuple {x, y}$
for each $y \in Y$.
We have that:
:$f_x = f \circ p_x$
From Composite of Continuous Mappings is Continuous, since $f$ is continu... | Let $X$, $Y$ and $T$ be [[Definition:Topological Space|topological spaces]].
Equip the [[Definition:Cartesian Product|Cartesian product]] $X \times Y$ with the [[Definition:Product Topology|product topology]].
Let $f : X \times Y \to T$ be a [[Definition:Continuous Mapping (Topology)|continuous mapping]].
Let $x \i... | From the definition of the [[Definition:Vertical Section of Function|$x$-vertical section]], we have:
:$\map {f_x} y = \map f {x, y}$
for each $y \in Y$.
Define the map $p_x : Y \to X \times Y$ by:
:$\map {p_x} y = \tuple {x, y}$
for each $y \in Y$.
We have that:
:$f_x = f \circ p_x$
From [[Composite of C... | Vertical Section of Continuous Function is Continuous | https://proofwiki.org/wiki/Vertical_Section_of_Continuous_Function_is_Continuous | https://proofwiki.org/wiki/Vertical_Section_of_Continuous_Function_is_Continuous | [
"Vertical Section of Functions",
"Continuous Mappings",
"Product Topology"
] | [
"Definition:Topological Space",
"Definition:Cartesian Product",
"Definition:Product Topology",
"Definition:Continuous Mapping (Topology)",
"Definition:Vertical Section of Function",
"Definition:Continuous Mapping (Topology)"
] | [
"Definition:Vertical Section of Function",
"Composite of Continuous Mappings is Continuous",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)",
"Box Topology on Finite Product Space is Product Topology",
"Definition:Product Topology",
"Definition:Box Topology",
"Co... |
proofwiki-19798 | Dilation Mapping on Topological Vector Space is Continuous | Let $\struct {K, +_K, \circ_K, \tau_K}$ be a topological field.
Let $\struct {X, +_X, \circ_X, \tau_X}$ be a vector space over $K$.
Let $\lambda \in K$.
Let $c_\lambda$ be the dilation by $\lambda$ mapping.
Then $c_\lambda$ is continuous. | Let $f: K \times X \to X$ be the map defined by:
:$\tuple {\lambda, x} \mapsto \lambda \circ_X x$
for any $x \in X$
From the definition of a topological vector space, the mapping $f$ is continuous.
We are given that $c_\lambda$ is the dilation by $\lambda$ mapping.
That is, $c_\lambda : X \to X$ is the map:
:$\forall... | Let $\struct {K, +_K, \circ_K, \tau_K}$ be a [[Definition:Topological Field|topological field]].
Let $\struct {X, +_X, \circ_X, \tau_X}$ be a [[Definition:Vector Space|vector space]] over $K$.
Let $\lambda \in K$.
Let $c_\lambda$ be the [[Definition:Dilation Mapping|dilation by $\lambda$ mapping]].
Then $c_\lamb... | Let $f: K \times X \to X$ be the [[Definition:Mapping|map]] defined by:
:$\tuple {\lambda, x} \mapsto \lambda \circ_X x$
for any $x \in X$
From the definition of a [[Definition:Topological Vector Space|topological vector space]], the [[Definition:Mapping|mapping]] $f$ is [[Definition:Continuous Mapping (Topology)|cont... | Dilation Mapping on Topological Vector Space is Continuous | https://proofwiki.org/wiki/Dilation_Mapping_on_Topological_Vector_Space_is_Continuous | https://proofwiki.org/wiki/Dilation_Mapping_on_Topological_Vector_Space_is_Continuous | [
"Dilation Mappings",
"Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Vector Space",
"Definition:Central Dilatation Mapping",
"Definition:Continuous Mapping (Topology)"
] | [
"Definition:Mapping",
"Definition:Topological Vector Space",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)",
"Definition:Given",
"Definition:Central Dilatation Mapping",
"Definition:Mapping",
"Definition:Vertical Section of Function",
"Vertical Section of Continuous Function is Con... |
proofwiki-19799 | Dilation Mapping on Topological Vector Space is Homeomorphism | Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $\lambda \in K \setminus \set {0_K}$.
Let $c_\lambda$ be the dilation by $\lambda$ mapping.
Then $c_\lambda$ is a homeomorphism. | From Dilation Mapping on Topological Vector Space is Continuous, both $c_{\lambda}$ and $c_{1/\lambda}$ are continuous.
It is therefore sufficient to establish that $c_{1/\lambda}$ is the inverse mapping of $c_\lambda$.
For all $x \in X$, we have:
{{begin-eqn}}
{{eqn | l = \map {\paren {c_\lambda \circ c_{1/\lambda} } ... | Let $K$ be a [[Definition:Topological Field|topological field]].
Let $X$ be a [[Definition:Topological Vector Space|topological vector space]] over $K$.
Let $\lambda \in K \setminus \set {0_K}$.
Let $c_\lambda$ be the [[Definition:Dilation Mapping|dilation by $\lambda$ mapping]].
Then $c_\lambda$ is a [[Definiti... | From [[Dilation Mapping on Topological Vector Space is Continuous]], both $c_{\lambda}$ and $c_{1/\lambda}$ are [[Definition:Continuous Mapping (Topology)|continuous]].
It is therefore sufficient to establish that $c_{1/\lambda}$ is the [[Definition:Inverse Mapping|inverse mapping]] of $c_\lambda$.
For all $x \in X$,... | Dilation Mapping on Topological Vector Space is Homeomorphism | https://proofwiki.org/wiki/Dilation_Mapping_on_Topological_Vector_Space_is_Homeomorphism | https://proofwiki.org/wiki/Dilation_Mapping_on_Topological_Vector_Space_is_Homeomorphism | [
"Dilation Mappings",
"Topological Vector Spaces"
] | [
"Definition:Topological Field",
"Definition:Topological Vector Space",
"Definition:Central Dilatation Mapping",
"Definition:Homeomorphism"
] | [
"Dilation Mapping on Topological Vector Space is Continuous",
"Definition:Continuous Mapping (Topology)",
"Definition:Inverse Mapping",
"Definition:Identity Mapping",
"Definition:Inverse Mapping"
] |
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