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-5100 | Set of Invertible Mappings forms Symmetric Group | Let $S$ be a set.
Let $\GG$ be the set of all invertible mappings from $S$ to $S$.
Then $\struct {\GG, \circ}$ is the symmetric group on $S$. | Let $\struct {S^S, \circ}$ be the algebraic structure formed from the set of all mappings from $S$ to itself.
From Set of all Self-Maps under Composition forms Monoid, $\struct {S^S, \circ}$ is a monoid.
By Inverse of Permutation is Permutation, if $f$ is a permutation of $S$, then so is its inverse $f^{-1}$.
By Biject... | Let $S$ be a [[Definition:Set|set]].
Let $\GG$ be the [[Definition:Set|set]] of all [[Definition:Invertible Mapping|invertible mappings]] from $S$ to $S$.
Then $\struct {\GG, \circ}$ is the [[Definition:Symmetric Group|symmetric group]] on $S$. | Let $\struct {S^S, \circ}$ be the [[Definition:Algebraic Structure|algebraic structure]] formed from the [[Definition:Set of All Mappings|set of all mappings from $S$ to itself]].
From [[Set of all Self-Maps under Composition forms Monoid]], $\struct {S^S, \circ}$ is a [[Definition:Monoid|monoid]].
By [[Inverse of Pe... | Set of Invertible Mappings forms Symmetric Group | https://proofwiki.org/wiki/Set_of_Invertible_Mappings_forms_Symmetric_Group | https://proofwiki.org/wiki/Set_of_Invertible_Mappings_forms_Symmetric_Group | [
"Symmetric Groups",
"Inverse Mappings"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Inverse Mapping",
"Definition:Symmetric Group"
] | [
"Definition:Algebraic Structure",
"Definition:Set of All Mappings",
"Set of all Self-Maps under Composition forms Monoid",
"Definition:Monoid",
"Inverse of Permutation is Permutation",
"Definition:Permutation",
"Definition:Inverse of Mapping",
"Inverse of Bijection is Bijection",
"Definition:Inverti... |
proofwiki-5101 | Group of Units of Field | Let $k^\times$ denote the group of units of a field $\struct {k, +, \times}$.
Then:
:$k^\times = k \setminus \set 0$
where:
:$\setminus$ denotes set difference
:$\set 0$ denotes the set containing only the zero of $k$.
That is, $0$ is the only element of $k$ which does not have a multiplicative inverse in $k$. | $0$ is not invertible in $k$ since $0 a = 0$ for all $a \in k$.
Thus $0 \notin k^\times$.
Consider now $a \in k$ such that $a \ne 0$.
From {{Field-axiom|M4}} it follows that there exists $a^{-1}$ such that:
:$a \times a^{-1} = 1 = a^{-1} \times a$
where $1$ is the unity of $k$.
So:
:$a \in k^\times$
Hence we have demon... | Let $k^\times$ denote the [[Definition:Group of Units|group of units]] of a [[Definition:Field (Abstract Algebra)|field]] $\struct {k, +, \times}$.
Then:
:$k^\times = k \setminus \set 0$
where:
:$\setminus$ denotes [[Definition:Set Difference|set difference]]
:$\set 0$ denotes the [[Definition:Singleton|set]] containi... | $0$ is not invertible in $k$ since $0 a = 0$ for all $a \in k$.
Thus $0 \notin k^\times$.
Consider now $a \in k$ such that $a \ne 0$.
From {{Field-axiom|M4}} it follows that there exists $a^{-1}$ such that:
:$a \times a^{-1} = 1 = a^{-1} \times a$
where $1$ is the [[Definition:Unity of Field|unity]] of $k$.
So:
:$... | Group of Units of Field | https://proofwiki.org/wiki/Group_of_Units_of_Field | https://proofwiki.org/wiki/Group_of_Units_of_Field | [
"Field Theory"
] | [
"Definition:Group of Units",
"Definition:Field (Abstract Algebra)",
"Definition:Set Difference",
"Definition:Singleton",
"Definition:Field Zero",
"Definition:Element",
"Definition:Multiplicative Inverse"
] | [
"Definition:Multiplicative Identity",
"Category:Field Theory"
] |
proofwiki-5102 | Automorphism Group is Subgroup of Symmetric Group | Let $\struct {S, *}$ be an algebraic structure.
Let $\Aut S$ be the automorphism group of $\struct {S, *}$.
Then $\Aut S$ is a subgroup of the symmetric group $\struct {\Gamma \paren S, \circ}$ on $S$. | An automorphism is an isomorphism $\phi: S \to S$ from an algebraic structure $S$ to itself.
The Identity Mapping is Automorphism, so $\Aut S$ is not empty.
The composite of isomorphisms is itself an isomorphism, as demonstrated on Isomorphism is Equivalence Relation.
So:
:$\phi_1, \phi_2 \in \Aut S \implies \phi_1 \ci... | Let $\struct {S, *}$ be an [[Definition:Algebraic Structure|algebraic structure]].
Let $\Aut S$ be the [[Definition:Automorphism Group of Group|automorphism group]] of $\struct {S, *}$.
Then $\Aut S$ is a [[Definition:Subgroup|subgroup]] of the [[Definition:Symmetric Group|symmetric group ]] $\struct {\Gamma \paren ... | An [[Definition:Automorphism (Abstract Algebra)|automorphism]] is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] $\phi: S \to S$ from an [[Definition:Algebraic Structure|algebraic structure]] $S$ to itself.
The [[Identity Mapping is Automorphism]], so $\Aut S$ is not [[Definition:Empty Set|empty]].
The ... | Automorphism Group is Subgroup of Symmetric Group | https://proofwiki.org/wiki/Automorphism_Group_is_Subgroup_of_Symmetric_Group | https://proofwiki.org/wiki/Automorphism_Group_is_Subgroup_of_Symmetric_Group | [
"Examples of Groups",
"Automorphism Groups",
"Group Automorphisms"
] | [
"Definition:Algebraic Structure",
"Definition:Automorphism Group/Group",
"Definition:Subgroup",
"Definition:Symmetric Group"
] | [
"Definition:Automorphism (Abstract Algebra)",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Algebraic Structure",
"Identity Mapping is Automorphism",
"Definition:Empty Set",
"Definition:Composition of Mappings",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Isomorphism (Abstract ... |
proofwiki-5103 | Differentiation of Vector-Valued Function Componentwise | Let:
:$\mathbf r: t \mapsto \tuple {\map {r_1} t, \map {r_2} t, \ldots, \map {r_n} t}$
be a differentiable vector-valued function.
The derivative of a vector-valued function can be calculated by differentiating each of its component functions:
:$\dfrac {\d \map {\mathbf r} t} {\d t} = \tuple {\dfrac \d {\d t} \map {r_... | {{begin-eqn}}
{{eqn | l = \frac {\d \map {\mathbf r} t} {\d t}
| r = \lim_{\Delta t \mathop \to 0} \frac {\map {\mathbf r} {t + \Delta t} - \map {\mathbf r} t} {\Delta t}
| c = {{Defof|Derivative of Vector-Valued Function}}
}}
{{eqn | r = <nowiki>\begin {bmatrix}
\ds \lim_{\Delta t \mathop \to 0} \dfrac {\m... | Let:
:$\mathbf r: t \mapsto \tuple {\map {r_1} t, \map {r_2} t, \ldots, \map {r_n} t}$
be a [[Definition:Differentiable Vector-Valued Function|differentiable]] [[Definition:Vector-Valued Function|vector-valued function]].
The [[Definition:Derivative of Vector-Valued Function|derivative]] of a [[Definition:Vector-Va... | {{begin-eqn}}
{{eqn | l = \frac {\d \map {\mathbf r} t} {\d t}
| r = \lim_{\Delta t \mathop \to 0} \frac {\map {\mathbf r} {t + \Delta t} - \map {\mathbf r} t} {\Delta t}
| c = {{Defof|Derivative of Vector-Valued Function}}
}}
{{eqn | r = <nowiki>\begin {bmatrix}
\ds \lim_{\Delta t \mathop \to 0} \dfrac {\m... | Differentiation of Vector-Valued Function Componentwise | https://proofwiki.org/wiki/Differentiation_of_Vector-Valued_Function_Componentwise | https://proofwiki.org/wiki/Differentiation_of_Vector-Valued_Function_Componentwise | [
"Differential Calculus",
"Vector Calculus"
] | [
"Definition:Differentiable Mapping/Vector-Valued Function",
"Definition:Vector-Valued Function",
"Definition:Derivative/Vector-Valued Function",
"Definition:Vector-Valued Function",
"Definition:Vector-Valued Function/Component Function"
] | [] |
proofwiki-5104 | Derivative of Dot Product of Vector-Valued Functions | Let $\mathbf a: \R \to \R^n$ and $\mathbf b: \R \to \R^n$ be differentiable vector-valued functions.
The derivative of their dot product is given by:
:$\map {\dfrac \d {\d x} } {\mathbf a \cdot \mathbf b} = \dfrac {\d \mathbf a} {\d x} \cdot \mathbf b + \mathbf a \cdot \dfrac {\d \mathbf b} {\d x}$ | Let:
:$\mathbf a: x \mapsto \tuple {\map {a_1} x, \map {a_2} x, \ldots, \map {a_n} x}$
:$\mathbf b: x \mapsto \tuple {\map {b_1} x, \map {b_2} x, \ldots, \map {b_n} x}$
Then:
{{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {\mathbf a \cdot \mathbf b}
| r = \map {\frac \d {\d x} } {\sum_{i \mathop = 1}^n a_i b_i... | Let $\mathbf a: \R \to \R^n$ and $\mathbf b: \R \to \R^n$ be [[Definition:Differentiable Vector-Valued Function|differentiable]] [[Definition:Vector-Valued Function|vector-valued functions]].
The [[Definition:Derivative of Vector-Valued Function|derivative]] of their [[Definition:Dot Product|dot product]] is given by... | Let:
:$\mathbf a: x \mapsto \tuple {\map {a_1} x, \map {a_2} x, \ldots, \map {a_n} x}$
:$\mathbf b: x \mapsto \tuple {\map {b_1} x, \map {b_2} x, \ldots, \map {b_n} x}$
Then:
{{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {\mathbf a \cdot \mathbf b}
| r = \map {\frac \d {\d x} } {\sum_{i \mathop = 1}^n a_i b... | Derivative of Dot Product of Vector-Valued Functions/Proof 1 | https://proofwiki.org/wiki/Derivative_of_Dot_Product_of_Vector-Valued_Functions | https://proofwiki.org/wiki/Derivative_of_Dot_Product_of_Vector-Valued_Functions/Proof_1 | [
"Derivative of Dot Product of Vector-Valued Functions",
"Dot Product",
"Product Rule for Derivatives",
"Vector Calculus"
] | [
"Definition:Differentiable Mapping/Vector-Valued Function",
"Definition:Vector-Valued Function",
"Definition:Derivative/Vector-Valued Function",
"Definition:Dot Product"
] | [
"Sum Rule for Derivatives/General Result",
"Product Rule for Derivatives",
"Summation is Linear"
] |
proofwiki-5105 | Derivative of Dot Product of Vector-Valued Functions | Let $\mathbf a: \R \to \R^n$ and $\mathbf b: \R \to \R^n$ be differentiable vector-valued functions.
The derivative of their dot product is given by:
:$\map {\dfrac \d {\d x} } {\mathbf a \cdot \mathbf b} = \dfrac {\d \mathbf a} {\d x} \cdot \mathbf b + \mathbf a \cdot \dfrac {\d \mathbf b} {\d x}$ | Let $v = \mathbf a \cdot \mathbf b$.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {\d v} {\d x}
| r = \lim_{h \mathop \to 0} \dfrac {\map v {x + h} - \map v x} h
| c = {{Defof|Derivative of Real Function}}
}}
{{eqn | r = \lim_{h \mathop \to 0} \dfrac {\map {\mathbf a} {x + h} \cdot \map {\mathbf b} {x + h} - \map... | Let $\mathbf a: \R \to \R^n$ and $\mathbf b: \R \to \R^n$ be [[Definition:Differentiable Vector-Valued Function|differentiable]] [[Definition:Vector-Valued Function|vector-valued functions]].
The [[Definition:Derivative of Vector-Valued Function|derivative]] of their [[Definition:Dot Product|dot product]] is given by... | Let $v = \mathbf a \cdot \mathbf b$.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {\d v} {\d x}
| r = \lim_{h \mathop \to 0} \dfrac {\map v {x + h} - \map v x} h
| c = {{Defof|Derivative of Real Function}}
}}
{{eqn | r = \lim_{h \mathop \to 0} \dfrac {\map {\mathbf a} {x + h} \cdot \map {\mathbf b} {x + h} - \ma... | Derivative of Dot Product of Vector-Valued Functions/Proof 2 | https://proofwiki.org/wiki/Derivative_of_Dot_Product_of_Vector-Valued_Functions | https://proofwiki.org/wiki/Derivative_of_Dot_Product_of_Vector-Valued_Functions/Proof_2 | [
"Derivative of Dot Product of Vector-Valued Functions",
"Dot Product",
"Product Rule for Derivatives",
"Vector Calculus"
] | [
"Definition:Differentiable Mapping/Vector-Valued Function",
"Definition:Vector-Valued Function",
"Definition:Derivative/Vector-Valued Function",
"Definition:Dot Product"
] | [] |
proofwiki-5106 | Borel Sigma-Algebra of Subset is Trace Sigma-Algebra | Let $\struct {X, \tau}$ be a topological space, and let $A \subseteq X$ be a subset of $X$.
Let $\tau_A$ be the subspace topology on $A$.
Then the following equality of $\sigma$-algebras on $A$ holds:
:$\map \BB {A, \tau_A} = \map \BB {X, \tau}_A$
where $\BB$ signifies Borel $\sigma$-algebra, and $\map \BB {X, \tau}_A$... | By definition of Borel $\sigma$-algebra, it holds that:
:$\map \BB {X, \tau} = \map \sigma \tau$
and also, by definition of subspace topology:
:$\tau_A = A \cap \tau = \set {A \cap U: U \in \tau}$
Thus, it follows that:
:$\map \BB {A, \tau_A} = \map \sigma {A \cap \tau}$
Thereby, the desired equality:
:$\map \BB {A, \t... | Let $\struct {X, \tau}$ be a [[Definition:Topological Space|topological space]], and let $A \subseteq X$ be a [[Definition:Subset|subset]] of $X$.
Let $\tau_A$ be the [[Definition:Subspace Topology|subspace topology]] on $A$.
Then the following equality of [[Definition:Sigma-Algebra|$\sigma$-algebras]] on $A$ holds:... | By definition of [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]], it holds that:
:$\map \BB {X, \tau} = \map \sigma \tau$
and also, by definition of [[Definition:Subspace Topology|subspace topology]]:
:$\tau_A = A \cap \tau = \set {A \cap U: U \in \tau}$
Thus, it follows that:
:$\map \BB {A, \tau_A} = \m... | Borel Sigma-Algebra of Subset is Trace Sigma-Algebra | https://proofwiki.org/wiki/Borel_Sigma-Algebra_of_Subset_is_Trace_Sigma-Algebra | https://proofwiki.org/wiki/Borel_Sigma-Algebra_of_Subset_is_Trace_Sigma-Algebra | [
"Trace Sigma-Algebras",
"Sigma-Algebras",
"Borel Sigma-Algebras",
"Borel Sigma-Algebras",
"Trace Sigma-Algebras"
] | [
"Definition:Topological Space",
"Definition:Subset",
"Definition:Topological Subspace",
"Definition:Sigma-Algebra",
"Definition:Borel Sigma-Algebra",
"Definition:Trace Sigma-Algebra"
] | [
"Definition:Borel Sigma-Algebra",
"Definition:Topological Subspace",
"Trace Sigma-Algebra of Generated Sigma-Algebra"
] |
proofwiki-5107 | Derivative of Vector Cross Product of Vector-Valued Functions | Let $\mathbf a: \R \to \R^3$ and $\mathbf b: \R \to \R^3$ be differentiable vector-valued functions in Cartesian $3$-space.
The derivative of their vector cross product is given by:
:$\map {\dfrac \d {\d x} } {\mathbf a \times \mathbf b} = \dfrac {\d \mathbf a} {\d x} \times \mathbf b + \mathbf a \times \dfrac {\d \mat... | Let:
:$\mathbf a: x \mapsto \begin {bmatrix} a_1 \\ a_2 \\ a_3 \end {bmatrix}$
:$\mathbf b: x \mapsto \begin {bmatrix} b_1 \\ b_2 \\ b_3 \end {bmatrix}$
Then:
{{begin-eqn}}
{{eqn | l = \mathbf a \times \mathbf b
| r = \begin {bmatrix} a_1 \\ a_2 \\ a_3 \end {bmatrix} \times \begin {bmatrix} b_1 \\ b_2 \\ b_3 \end... | Let $\mathbf a: \R \to \R^3$ and $\mathbf b: \R \to \R^3$ be [[Definition:Differentiable Vector-Valued Function|differentiable]] [[Definition:Vector-Valued Function|vector-valued functions]] in [[Definition:Cartesian 3-Space|Cartesian $3$-space]].
The [[Definition:Derivative of Vector-Valued Function|derivative]] of ... | Let:
:$\mathbf a: x \mapsto \begin {bmatrix} a_1 \\ a_2 \\ a_3 \end {bmatrix}$
:$\mathbf b: x \mapsto \begin {bmatrix} b_1 \\ b_2 \\ b_3 \end {bmatrix}$
Then:
{{begin-eqn}}
{{eqn | l = \mathbf a \times \mathbf b
| r = \begin {bmatrix} a_1 \\ a_2 \\ a_3 \end {bmatrix} \times \begin {bmatrix} b_1 \\ b_2 \\ b_3... | Derivative of Vector Cross Product of Vector-Valued Functions/Proof 1 | https://proofwiki.org/wiki/Derivative_of_Vector_Cross_Product_of_Vector-Valued_Functions | https://proofwiki.org/wiki/Derivative_of_Vector_Cross_Product_of_Vector-Valued_Functions/Proof_1 | [
"Product Rule for Derivatives",
"Vector Calculus",
"Vector Cross Product",
"Derivative of Vector Cross Product of Vector-Valued Functions"
] | [
"Definition:Differentiable Mapping/Vector-Valued Function",
"Definition:Vector-Valued Function",
"Definition:Cartesian 3-Space",
"Definition:Derivative/Vector-Valued Function",
"Definition:Vector Cross Product"
] | [
"Differentiation of Vector-Valued Function Componentwise",
"Product Rule for Derivatives",
"Linear Combination of Derivatives"
] |
proofwiki-5108 | Existence and Uniqueness of Monotone Class Generated by Collection of Subsets | Let $X$ be a set.
Let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Then $\map {\mathfrak m} \GG$, the monotone class generated by $\GG$, exists and is unique. | === Existence ===
By Power Set is Monotone Class, there is at least one monotone class containing $\GG$.
Now let $\Bbb M$ be the collection of monotone classes containing $\GG$:
:$\Bbb M := \set {\mathfrak m': \GG \subseteq \mathfrak m', \mathfrak m' \text{ is a monotone class} }$
By Intersection of Monotone Classes is... | Let $X$ be a [[Definition:Set|set]].
Let $\GG \subseteq \powerset X$ be a collection of [[Definition:Subset|subsets]] of $X$.
Then $\map {\mathfrak m} \GG$, the [[Definition:Monotone Class Generated by Collection of Subsets|monotone class generated by $\GG$]], exists and is unique. | === Existence ===
By [[Power Set is Monotone Class]], there is at least one [[Definition:Monotone Class|monotone class]] containing $\GG$.
Now let $\Bbb M$ be the collection of [[Definition:Monotone Class|monotone classes]] containing $\GG$:
:$\Bbb M := \set {\mathfrak m': \GG \subseteq \mathfrak m', \mathfrak m' \t... | Existence and Uniqueness of Monotone Class Generated by Collection of Subsets | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Monotone_Class_Generated_by_Collection_of_Subsets | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Monotone_Class_Generated_by_Collection_of_Subsets | [
"Monotone Classes"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Monotone Class Generated by Collection of Subsets"
] | [
"Power Set is Monotone Class",
"Definition:Monotone Class",
"Definition:Monotone Class",
"Intersection of Monotone Classes is Monotone Class",
"Definition:Monotone Class",
"Set Intersection Preserves Subsets/Families of Sets",
"Definition:Monotone Class",
"Intersection is Subset/General Result",
"De... |
proofwiki-5109 | Derivative of Product of Real Function and Vector-Valued Function | Let:
:$\mathbf z:\R \to \R^n$
be a differentiable vector-valued function, where:
:$\mathbf z = \begin {bmatrix} z_1 \\ z_2 \\ \vdots \\ z_n \end {bmatrix}$
such that:
:$z_1, z_2, \cdots, z_n$
are differentiable real functions.
Let:
:$y: \R \to \R$
be a differentiable real function.
Then:
:$\map {D_x} {y \, \mathbf z} =... | {{begin-eqn}}
{{eqn | l = \map {D_x} {y \, \mathbf z}
| r = \map {D_x} {\begin {bmatrix} y \ z_1 \\ y \ z_2 \\ \vdots \\ y \ z_n \end {bmatrix} }
}}
{{eqn | r = \begin {bmatrix} \map {D_x} {y \ z_1} \\ \map {D_x} {y \ z_2} \\ \vdots \\ \map {D_x} {y \ z_n} \end{bmatrix}
| c = Differentiation of Vector-Value... | Let:
:$\mathbf z:\R \to \R^n$
be a [[Definition:Differentiable Vector-Valued Function|differentiable]] [[Definition:Vector-Valued Function|vector-valued function]], where:
:$\mathbf z = \begin {bmatrix} z_1 \\ z_2 \\ \vdots \\ z_n \end {bmatrix}$
such that:
:$z_1, z_2, \cdots, z_n$
are [[Definition:Differentiable Real ... | {{begin-eqn}}
{{eqn | l = \map {D_x} {y \, \mathbf z}
| r = \map {D_x} {\begin {bmatrix} y \ z_1 \\ y \ z_2 \\ \vdots \\ y \ z_n \end {bmatrix} }
}}
{{eqn | r = \begin {bmatrix} \map {D_x} {y \ z_1} \\ \map {D_x} {y \ z_2} \\ \vdots \\ \map {D_x} {y \ z_n} \end{bmatrix}
| c = [[Differentiation of Vector-Val... | Derivative of Product of Real Function and Vector-Valued Function | https://proofwiki.org/wiki/Derivative_of_Product_of_Real_Function_and_Vector-Valued_Function | https://proofwiki.org/wiki/Derivative_of_Product_of_Real_Function_and_Vector-Valued_Function | [
"Product Rule for Derivatives",
"Vector Calculus"
] | [
"Definition:Differentiable Mapping/Vector-Valued Function",
"Definition:Vector-Valued Function",
"Definition:Differentiable Mapping/Real Function",
"Definition:Real Function",
"Definition:Differentiable Mapping/Real Function",
"Definition:Real Function"
] | [
"Differentiation of Vector-Valued Function Componentwise",
"Product Rule for Derivatives",
"Differentiation of Vector-Valued Function Componentwise"
] |
proofwiki-5110 | Generated Sigma-Algebra by Generated Monotone Class | Let $X$ be a set, and let $\GG \subseteq \powerset X$ be a nonempty collection of subsets of $X$.
Suppose that $\GG$ satisfies the following condition:
:$(1):\quad A \in \GG \implies \relcomp X A \in \GG$
that is, $\GG$ is closed under complement in $X$.
Then:
:$\map {\mathfrak m} \GG = \map \sigma \GG$
where $\mathfra... | By Sigma-Algebra is Monotone Class, and the definition of generated monotone class, it follows that:
:$\map {\mathfrak m} \GG \subseteq \map \sigma \GG$
Next, define $\Sigma$ by:
:$\Sigma := \set {M \in \map {\mathfrak m} \GG: X \setminus M \in \map {\mathfrak m} \GG}$
By $(1)$, it follows that $\GG \subseteq \Sigma$.
... | Let $X$ be a [[Definition:Set|set]], and let $\GG \subseteq \powerset X$ be a [[Definition:Empty Set|nonempty]] collection of [[Definition:Subset|subsets]] of $X$.
Suppose that $\GG$ satisfies the following condition:
:$(1):\quad A \in \GG \implies \relcomp X A \in \GG$
that is, $\GG$ is closed under [[Definition:Re... | By [[Sigma-Algebra is Monotone Class]], and the definition of [[Definition:Generated Monotone Class|generated monotone class]], it follows that:
:$\map {\mathfrak m} \GG \subseteq \map \sigma \GG$
Next, define $\Sigma$ by:
:$\Sigma := \set {M \in \map {\mathfrak m} \GG: X \setminus M \in \map {\mathfrak m} \GG}$
... | Generated Sigma-Algebra by Generated Monotone Class | https://proofwiki.org/wiki/Generated_Sigma-Algebra_by_Generated_Monotone_Class | https://proofwiki.org/wiki/Generated_Sigma-Algebra_by_Generated_Monotone_Class | [
"Monotone Classes",
"Sigma-Algebras Generated by Collection of Subsets",
"Generated Sigma-Algebra by Generated Monotone Class"
] | [
"Definition:Set",
"Definition:Empty Set",
"Definition:Subset",
"Definition:Relative Complement",
"Definition:Monotone Class Generated by Collection of Subsets",
"Definition:Sigma-Algebra Generated by Collection of Subsets"
] | [
"Sigma-Algebra is Monotone Class",
"Definition:Monotone Class Generated by Collection of Subsets",
"Definition:Sigma-Algebra",
"Definition:Monotone Class",
"Set Difference Intersection with Second Set is Empty Set",
"Set Difference and Intersection form Partition",
"Set Difference with Set Difference",
... |
proofwiki-5111 | Totally Bounded Metric Space is Second-Countable | Let $M = \struct {A, d}$ be a metric space which is totally bounded.
Then $M$ is second-countable. | Let $M = \struct {A, d}$ be totally bounded.
Let $\epsilon = 1, \dfrac 1 2, \dfrac 1 3, \ldots$
As $M$ is totally bounded, for each $\epsilon$ there exists a finite $\epsilon$-net $\CC$ for $M$.
From Net forms Basis for Metric Space, $\CC$ is a countable basis for $M$.
That is, $M$ is second-countable.
{{qed}} | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]] which is [[Definition:Totally Bounded Metric Space|totally bounded]].
Then $M$ is [[Definition:Second-Countable Space|second-countable]]. | Let $M = \struct {A, d}$ be [[Definition:Totally Bounded Metric Space|totally bounded]].
Let $\epsilon = 1, \dfrac 1 2, \dfrac 1 3, \ldots$
As $M$ is [[Definition:Totally Bounded Metric Space|totally bounded]], for each $\epsilon$ there exists a [[Definition:Finite Epsilon-Net|finite $\epsilon$-net]] $\CC$ for $M$.
... | Totally Bounded Metric Space is Second-Countable/Proof 1 | https://proofwiki.org/wiki/Totally_Bounded_Metric_Space_is_Second-Countable | https://proofwiki.org/wiki/Totally_Bounded_Metric_Space_is_Second-Countable/Proof_1 | [
"Metric Spaces",
"Second-Countable Spaces",
"Totally Bounded Metric Spaces",
"Totally Bounded Metric Space is Second-Countable"
] | [
"Definition:Metric Space",
"Definition:Totally Bounded Metric Space",
"Definition:Second-Countable Space"
] | [
"Definition:Totally Bounded Metric Space",
"Definition:Totally Bounded Metric Space",
"Definition:Epsilon-Net/Finite Net",
"Net forms Basis for Metric Space",
"Definition:Countable Basis",
"Definition:Second-Countable Space"
] |
proofwiki-5112 | Totally Bounded Metric Space is Second-Countable | Let $M = \struct {A, d}$ be a metric space which is totally bounded.
Then $M$ is second-countable. | Follows directly from:
: Totally Bounded Metric Space is Separable
: Separable Metric Space is Second-Countable
{{qed}} | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]] which is [[Definition:Totally Bounded Metric Space|totally bounded]].
Then $M$ is [[Definition:Second-Countable Space|second-countable]]. | Follows directly from:
: [[Totally Bounded Metric Space is Separable]]
: [[Separable Metric Space is Second-Countable]]
{{qed}} | Totally Bounded Metric Space is Second-Countable/Proof 2 | https://proofwiki.org/wiki/Totally_Bounded_Metric_Space_is_Second-Countable | https://proofwiki.org/wiki/Totally_Bounded_Metric_Space_is_Second-Countable/Proof_2 | [
"Metric Spaces",
"Second-Countable Spaces",
"Totally Bounded Metric Spaces",
"Totally Bounded Metric Space is Second-Countable"
] | [
"Definition:Metric Space",
"Definition:Totally Bounded Metric Space",
"Definition:Second-Countable Space"
] | [
"Totally Bounded Metric Space is Separable",
"Separable Metric Space is Second-Countable"
] |
proofwiki-5113 | Heine-Borel Theorem/Metric Space | A metric space is compact {{iff}} it is both complete and totally bounded. | === Sufficient Condition ===
This follows directly from:
* Compact Metric Space is Complete
* Compact Metric Space is Totally Bounded
{{qed|lemma}} | A [[Definition:Metric Space|metric space]] is [[Definition:Compact Metric Space|compact]] {{iff}} it is both [[Definition:Complete Metric Space|complete]] and [[Definition:Totally Bounded Metric Space|totally bounded]]. | === Sufficient Condition ===
This follows directly from:
* [[Compact Metric Space is Complete]]
* [[Compact Metric Space is Totally Bounded]]
{{qed|lemma}} | Heine-Borel Theorem/Metric Space | https://proofwiki.org/wiki/Heine-Borel_Theorem/Metric_Space | https://proofwiki.org/wiki/Heine-Borel_Theorem/Metric_Space | [
"Heine-Borel Theorem",
"Totally Bounded Metric Spaces",
"Complete Metric Spaces",
"Compact Metric Spaces"
] | [
"Definition:Metric Space",
"Definition:Compact Space/Metric Space",
"Definition:Complete Metric Space",
"Definition:Totally Bounded Metric Space"
] | [
"Compact Metric Space is Complete",
"Compact Metric Space is Totally Bounded"
] |
proofwiki-5114 | Borel Sigma-Algebra on Euclidean Space by Monotone Class | Let $\sqbrk {\R^n, \tau}$ be the $n$-dimensional Euclidean space.
Then:
:$\map \BB {\R^n, \tau} = \map {\mathfrak m} \tau$
where $\BB$ denotes Borel $\sigma$-algebra, and $\mathfrak m$ denotes generated monotone class. | Let $U \in \tau$ be an open set, and define $C$ by:
:$C := X \setminus U$
hence $C$ is a closed set.
Further, define, for all $n \in \N$:
:$C_n := \ds \bigcup_{c \mathop \in C} \map B {c; \frac 1 n}$
where $B$ denotes open ball.
The $C_n$ are open sets, being the union of open balls.
It is clear that $C \subseteq C_n$ ... | Let $\sqbrk {\R^n, \tau}$ be the $n$-dimensional [[Definition:Euclidean Space|Euclidean space]].
Then:
:$\map \BB {\R^n, \tau} = \map {\mathfrak m} \tau$
where $\BB$ denotes [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]], and $\mathfrak m$ denotes [[Definition:Generated Monotone Class|generated monotone ... | Let $U \in \tau$ be an [[Definition:Open Set (Topology)|open set]], and define $C$ by:
:$C := X \setminus U$
hence $C$ is a [[Definition:Closed Set (Topology)|closed set]].
Further, define, for all $n \in \N$:
:$C_n := \ds \bigcup_{c \mathop \in C} \map B {c; \frac 1 n}$
where $B$ denotes [[Definition:Open Ball|o... | Borel Sigma-Algebra on Euclidean Space by Monotone Class | https://proofwiki.org/wiki/Borel_Sigma-Algebra_on_Euclidean_Space_by_Monotone_Class | https://proofwiki.org/wiki/Borel_Sigma-Algebra_on_Euclidean_Space_by_Monotone_Class | [
"Borel Sigma-Algebras",
"Monotone Classes"
] | [
"Definition:Euclidean Space",
"Definition:Borel Sigma-Algebra",
"Definition:Monotone Class Generated by Collection of Subsets"
] | [
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Open Ball",
"Definition:Open Set/Topology",
"Definition:Set Union",
"Definition:Open Ball",
"Definition:Open Set/Metric Space",
"Definition:Set Intersection",
"Generated Monotone Class Preserves Subset",
"Generated Sigma... |
proofwiki-5115 | Measure is Strongly Additive | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then $\mu$ is strongly additive, that is:
:$\forall E, F \in \Sigma: \map \mu {E \cap F} + \map \mu {E \cup F} = \map \mu E + \map \mu F$ | Combine Measure is Finitely Additive Function with Additive Function is Strongly Additive.
{{qed}} | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Then $\mu$ is [[Definition:Strongly Additive Function|strongly additive]], that is:
:$\forall E, F \in \Sigma: \map \mu {E \cap F} + \map \mu {E \cup F} = \map \mu E + \map \mu F$ | Combine [[Measure is Finitely Additive Function]] with [[Additive Function is Strongly Additive]].
{{qed}} | Measure is Strongly Additive | https://proofwiki.org/wiki/Measure_is_Strongly_Additive | https://proofwiki.org/wiki/Measure_is_Strongly_Additive | [
"Measures"
] | [
"Definition:Measure Space",
"Definition:Strongly Additive Function"
] | [
"Measure is Finitely Additive Function",
"Additive Function is Strongly Additive"
] |
proofwiki-5116 | Measure is Subadditive | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then $\mu$ is subadditive, that is:
:$\forall E, F \in \Sigma: \map \mu {E \cup F} \le \map \mu E + \map \mu F$ | A measure is an additive function, and, by definition, nowhere negative.
So Additive Nowhere Negative Function is Subadditive applies.
Hence the result directly:
:$\map \mu {E \cup F} \le \map \mu E + \map \mu F$
{{qed}} | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Then $\mu$ is [[Definition:Subadditive Function (Measure Theory)|subadditive]], that is:
:$\forall E, F \in \Sigma: \map \mu {E \cup F} \le \map \mu E + \map \mu F$ | A [[Definition:Measure (Measure Theory)|measure]] is an [[Definition:Additive Function (Measure Theory)|additive function]], and, by definition, nowhere negative.
So [[Additive Nowhere Negative Function is Subadditive]] applies.
Hence the result directly:
:$\map \mu {E \cup F} \le \map \mu E + \map \mu F$
{{qed}} | Measure is Subadditive | https://proofwiki.org/wiki/Measure_is_Subadditive | https://proofwiki.org/wiki/Measure_is_Subadditive | [
"Measures",
"Measure is Subadditive"
] | [
"Definition:Measure Space",
"Definition:Subadditive Function (Measure Theory)"
] | [
"Definition:Measure (Measure Theory)",
"Definition:Additive Function (Measure Theory)",
"Additive Nowhere Negative Function is Subadditive"
] |
proofwiki-5117 | Representation of Degree One is Irreducible | Let $\struct {G, \cdot}$ be a finite group.
Let $\rho: G \to \GL V$ be a linear representation of $G$ on $V$ of degree $1$.
Then $\rho$ is an irreducible linear representation. | By the definition of degree of a linear representation, it is known that $\dim V = 1$.
Let $W$ be a proper vector subspace of $V$.
It follows from Dimension of Proper Subspace is Less Than its Superspace that:
:$\dim W < 1$
and hence $\dim W = 0$.
Now from Trivial Vector Space iff Zero Dimension, it follows that:
:$W ... | Let $\struct {G, \cdot}$ be a [[Definition:Finite Group|finite group]].
Let $\rho: G \to \GL V$ be a [[Definition:Linear Representation|linear representation]] of $G$ on $V$ of [[Definition:Dimension (Representation Theory)|degree]] $1$.
Then $\rho$ is an [[Definition:Irreducible Linear Representation|irreducible ... | By the definition of [[Definition:Degree of Linear Representation|degree]] of a [[Definition:Linear Representation|linear representation]], it is known that $\dim V = 1$.
Let $W$ be a [[Definition:Proper Vector Subspace|proper vector subspace]] of $V$.
It follows from [[Dimension of Proper Subspace is Less Than its S... | Representation of Degree One is Irreducible | https://proofwiki.org/wiki/Representation_of_Degree_One_is_Irreducible | https://proofwiki.org/wiki/Representation_of_Degree_One_is_Irreducible | [
"Representation Theory"
] | [
"Definition:Finite Group",
"Definition:Linear Representation",
"Definition:Dimension (Representation Theory)",
"Definition:Irreducible (Representation Theory)/Linear Representation"
] | [
"Definition:Dimension (Representation Theory)",
"Definition:Linear Representation",
"Definition:Vector Subspace/Proper Subspace",
"Dimension of Proper Subspace is Less Than its Superspace",
"Trivial Vector Space iff Zero Dimension",
"Definition:Zero Subspace",
"Definition:Vector Subspace/Proper Subspace... |
proofwiki-5118 | Irreducible Representations of Abelian Group | Let $\struct {G, \cdot}$ be a finite abelian group.
Let $V$ be a non-null vector space over an algebraically closed field $K$.
Let $\rho: G \to \GL V$ be a linear representation.
Then $\rho$ is irreducible {{iff}} $\map {\dim_K} V = 1$, where, $\dim_K$ denotes dimension. | === Sufficient Condition ===
Suppose that $\map {\dim_K} V = 1$.
That $\rho$ is irreducible is shown on Representation of Degree One is Irreducible.
{{qed|lemma}} | Let $\struct {G, \cdot}$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]].
Let $V$ be a non-[[Definition:Null Module|null]] [[Definition:Vector Space|vector space]] over an [[Definition:Algebraically Closed Field|algebraically closed field]] $K$.
Let $\rho: G \to \GL V$ be a [[Defini... | === Sufficient Condition ===
Suppose that $\map {\dim_K} V = 1$.
That $\rho$ is [[Definition:Irreducible Linear Representation|irreducible]] is shown on [[Representation of Degree One is Irreducible]].
{{qed|lemma}} | Irreducible Representations of Abelian Group | https://proofwiki.org/wiki/Irreducible_Representations_of_Abelian_Group | https://proofwiki.org/wiki/Irreducible_Representations_of_Abelian_Group | [
"Representation Theory"
] | [
"Definition:Finite Group",
"Definition:Abelian Group",
"Definition:Null Module",
"Definition:Vector Space",
"Definition:Algebraically Closed Field",
"Definition:Linear Representation",
"Definition:Irreducible (Representation Theory)/Linear Representation",
"Definition:Dimension (Linear Algebra)"
] | [
"Definition:Irreducible (Representation Theory)/Linear Representation",
"Representation of Degree One is Irreducible",
"Definition:Irreducible (Representation Theory)/Linear Representation",
"Definition:Irreducible (Representation Theory)/Linear Representation"
] |
proofwiki-5119 | Schur's Lemma (Representation Theory) | Let $\struct {G, \cdot}$ be a finite group.
Let $V$ and $V'$ be two irreducible $G$-modules.
Let $f: V \to V'$ be a homomorphism of $G$-modules.
Then either:
:$\map f v = 0$ for all $v \in V$
or:
:$f$ is an isomorphism. | From Kernel is G-Module, $\map \ker f$ is a $G$-submodule of $V$.
From Image is G-Module, $\Img f$ is a $G$-submodule of $V'$.
By the definition of irreducible:
:$\map \ker f = \set 0$
or:
:$\map \ker f = V$
{{explain|Link to a result which shows this. While it does indeed follow from the definition, it would be useful... | Let $\struct {G, \cdot}$ be a [[Definition:Finite Group|finite group]].
Let $V$ and $V'$ be two [[Definition:Irreducible Linear Representation|irreducible]] [[Definition:G-Module|$G$-modules]].
Let $f: V \to V'$ be a [[Definition:G-Module Homomorphism|homomorphism of $G$-modules]].
Then either:
:$\map f v = 0$ for a... | From [[Kernel is G-Module]], $\map \ker f$ is a [[Definition:G-Submodule|$G$-submodule]] of $V$.
From [[Image is G-Module]], $\Img f$ is a [[Definition:G-Submodule|$G$-submodule]] of $V'$.
By the definition of [[Definition:Reducible Linear Representation|irreducible]]:
:$\map \ker f = \set 0$
or:
:$\map \ker f = V$
... | Schur's Lemma (Representation Theory) | https://proofwiki.org/wiki/Schur's_Lemma_(Representation_Theory) | https://proofwiki.org/wiki/Schur's_Lemma_(Representation_Theory) | [
"Representation Theory"
] | [
"Definition:Finite Group",
"Definition:Irreducible (Representation Theory)/Linear Representation",
"Definition:Module over Group",
"Definition:G-Module Homomorphism",
"Definition:Isomorphism (Abstract Algebra)/R-Algebraic Structure Isomorphism/Module Isomorphism"
] | [
"Kernel is G-Module",
"Definition:G-Submodule",
"Image is G-Module",
"Definition:G-Submodule",
"Definition:Reducible Linear Representation",
"Linear Transformation is Injective iff Kernel Contains Only Zero",
"Definition:Injection",
"Definition:Surjection",
"Definition:Injection",
"Definition:Bije... |
proofwiki-5120 | G-Module is Irreducible iff no Non-Trivial Proper Submodules | Let $\struct {G, \circ}$ be a finite group.
Let $\struct {V, \phi}$ be a $G$-module.
Then $V$ is an irreducible $G$-module {{iff}} $V$ has no non-trivial proper $G$-submodules. | === Necessary Condition ===
{{AimForCont}} that $V$ is an irreducible $G$-module, but it has a non-trivial proper $G$-submodule.
By the definition irreducible, its associated representation is irreducible.
Let this representation be denoted:
:$\tilde \phi = \rho: G \to \GL V$
In Correspondence between Linear Group Acti... | Let $\struct {G, \circ}$ be a [[Definition:Finite Group|finite group]].
Let $\struct {V, \phi}$ be a [[Definition:G-Module|$G$-module]].
Then $V$ is an [[Definition:Irreducible G-Module|irreducible $G$-module]] {{iff}} $V$ has no [[Definition:Trivial G-Module|non-trivial]] [[Definition:Proper G-Submodule|proper $G$-... | === Necessary Condition ===
{{AimForCont}} that $V$ is an [[Definition:Irreducible G-Module|irreducible]] [[Definition:G-Module|$G$-module]], but it has a [[Definition:Trivial G-Module|non-trivial]] [[Definition:Proper G-Submodule|proper $G$-submodule]].
By the definition [[Definition:Irreducible Linear Representatio... | G-Module is Irreducible iff no Non-Trivial Proper Submodules | https://proofwiki.org/wiki/G-Module_is_Irreducible_iff_no_Non-Trivial_Proper_Submodules | https://proofwiki.org/wiki/G-Module_is_Irreducible_iff_no_Non-Trivial_Proper_Submodules | [
"Representation Theory"
] | [
"Definition:Finite Group",
"Definition:Module over Group",
"Definition:Irreducible (Representation Theory)",
"Definition:Trivial G-Module",
"Definition:Proper G-Submodule"
] | [
"Definition:Irreducible (Representation Theory)",
"Definition:Module over Group",
"Definition:Trivial G-Module",
"Definition:Proper G-Submodule",
"Definition:Irreducible (Representation Theory)/Linear Representation",
"Definition:Linear Representation",
"Definition:Irreducible (Representation Theory)/Li... |
proofwiki-5121 | Kernel is G-Module | Let $\struct {G, \cdot}$ be a group.
Let $f: \struct {V, \phi} \to \struct {V', \mu}$ be a homomorphism of $G$-modules.
Then its kernel $\map \ker f$ is a $G$-submodule of $V$. | From G-Submodule Test it suffices to prove that $\phi \sqbrk {\struct {G, \map \ker f} } \subseteq \map \ker f$.
That is, it is to be shown that, if $g \in G$ and $v \in \map \ker f$, then $\map \phi {g, v} \in \map \ker f$.
Assume that $g \in G$ and $v \in \map \ker f$.
{{begin-eqn}}
{{eqn | l = \map f {\map \phi {g,... | Let $\struct {G, \cdot}$ be a [[Definition:Group|group]].
Let $f: \struct {V, \phi} \to \struct {V', \mu}$ be a [[Definition:G-Module Homomorphism |homomorphism of $G$-modules]].
Then its [[Definition:Kernel of Linear Transformation|kernel]] $\map \ker f$ is a [[Definition:G-Submodule|$G$-submodule]] of $V$. | From [[G-Submodule Test]] it suffices to prove that $\phi \sqbrk {\struct {G, \map \ker f} } \subseteq \map \ker f$.
That is, it is to be shown that, if $g \in G$ and $v \in \map \ker f$, then $\map \phi {g, v} \in \map \ker f$.
Assume that $g \in G$ and $v \in \map \ker f$.
{{begin-eqn}}
{{eqn | l = \map f {\map ... | Kernel is G-Module | https://proofwiki.org/wiki/Kernel_is_G-Module | https://proofwiki.org/wiki/Kernel_is_G-Module | [
"Representation Theory"
] | [
"Definition:Group",
"Definition:G-Module Homomorphism ",
"Definition:Kernel of Linear Transformation",
"Definition:G-Submodule"
] | [
"G-Submodule Test",
"Definition:G-Module Homomorphism",
"Definition:Linear Group Action",
"Definition:G-Submodule",
"Category:Representation Theory"
] |
proofwiki-5122 | Image is G-Module | Let $\struct {G, \cdot}$ be a group.
Let $f: \struct {V, \phi} \to \struct {V', \mu}$ be a $G$-Module Homomorphism.
Then $\Img f$ is a $G$-Submodule of $V'$. | From $G$-Submodule Test it suffices to prove that $\map \mu {G, \Img f} \subseteq \Img f$.
In other words: for any $g \in G$ and $w \in \Img f$, it is to be shown that $\map \mu {g, w} \in \Img f$.
Assume that $g \in G$ and $w \in \Img f$.
Then:
:$\exists v \in V: \map f v = w$
By definition of homomorphism:
:$\map \mu... | Let $\struct {G, \cdot}$ be a [[Definition:Group|group]].
Let $f: \struct {V, \phi} \to \struct {V', \mu}$ be a [[Definition:G-Module Homomorphism|$G$-Module Homomorphism]].
Then $\Img f$ is a [[Definition:G-Submodule|$G$-Submodule]] of $V'$. | From [[G-Submodule Test|$G$-Submodule Test]] it suffices to prove that $\map \mu {G, \Img f} \subseteq \Img f$.
In other words: for any $g \in G$ and $w \in \Img f$, it is to be shown that $\map \mu {g, w} \in \Img f$.
Assume that $g \in G$ and $w \in \Img f$.
Then:
:$\exists v \in V: \map f v = w$
By definition ... | Image is G-Module | https://proofwiki.org/wiki/Image_is_G-Module | https://proofwiki.org/wiki/Image_is_G-Module | [
"Representation Theory"
] | [
"Definition:Group",
"Definition:G-Module Homomorphism",
"Definition:G-Submodule"
] | [
"G-Submodule Test",
"Definition:G-Module Homomorphism",
"G-Submodule Test",
"Definition:G-Submodule",
"Category:Representation Theory"
] |
proofwiki-5123 | Set Intersection Preserves Subsets/Families of Sets | Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be indexed families of subsets of a set $S$.
Let:
:$\forall \beta \in I: A_\beta \subseteq B_\beta$
Then:
:$\ds \bigcap_{\alpha \mathop \in I} A_\alpha \subseteq \bigcap_{\alpha \mathop \in I} B_\... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcap_{\alpha \mathop \in I} A_\alpha
| c =
}}
{{eqn | ll= \leadsto
| q = \forall \alpha \in I
| l = x
| o = \in
| r = A_\alpha
| c = {{Defof|Intersection of Family}}
}}
{{eqn | ll= \leadsto
| q = \forall \alpha \in I
... | Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be [[Definition:Indexed Family of Subsets|indexed families of subsets]] of a [[Definition:Set|set]] $S$.
Let:
:$\forall \beta \in I: A_\beta \subseteq B_\beta$
Then... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcap_{\alpha \mathop \in I} A_\alpha
| c =
}}
{{eqn | ll= \leadsto
| q = \forall \alpha \in I
| l = x
| o = \in
| r = A_\alpha
| c = {{Defof|Intersection of Family}}
}}
{{eqn | ll= \leadsto
| q = \forall \alpha \in I
... | Set Intersection Preserves Subsets/Families of Sets | https://proofwiki.org/wiki/Set_Intersection_Preserves_Subsets/Families_of_Sets | https://proofwiki.org/wiki/Set_Intersection_Preserves_Subsets/Families_of_Sets | [
"Set Intersection Preserves Subsets",
"Indexed Families"
] | [
"Definition:Indexing Set",
"Definition:Indexing Set/Family of Subsets",
"Definition:Set"
] | [
"Definition:Subset"
] |
proofwiki-5124 | Set Intersection Preserves Subsets | Let $A, B, S, T$ be sets.
Then:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$ | Let $A \subseteq B$ and $S \subseteq T$.
Then:
{{begin-eqn}}
{{eqn| l = x \in A
| o = \implies
| r = x \in B
| c = {{Defof|Subset}}
}}
{{eqn| l = x \in S
| o = \implies
| r = x \in T
| c = {{Defof|Subset}}
}}
{{end-eqn}}
Now we invoke the Praeclarum Theorema of propositional logic:
:$... | Let $A, B, S, T$ be [[Definition:Set|sets]].
Then:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$ | Let $A \subseteq B$ and $S \subseteq T$.
Then:
{{begin-eqn}}
{{eqn| l = x \in A
| o = \implies
| r = x \in B
| c = {{Defof|Subset}}
}}
{{eqn| l = x \in S
| o = \implies
| r = x \in T
| c = {{Defof|Subset}}
}}
{{end-eqn}}
Now we invoke the [[Praeclarum Theorema]] of [[Definition:Pr... | Set Intersection Preserves Subsets | https://proofwiki.org/wiki/Set_Intersection_Preserves_Subsets | https://proofwiki.org/wiki/Set_Intersection_Preserves_Subsets | [
"Set Intersection",
"Subsets",
"Set Intersection Preserves Subsets"
] | [
"Definition:Set"
] | [
"Praeclarum Theorema",
"Definition:Propositional Logic",
"Definition:Set Intersection",
"Definition:Subset",
"Category:Set Intersection",
"Category:Subsets",
"Category:Set Intersection Preserves Subsets"
] |
proofwiki-5125 | Set Intersection Preserves Subsets | Let $A, B, S, T$ be sets.
Then:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$ | Let $A \subseteq B$, and let $S$ be any set.
From Set Intersection Preserves Subsets:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$
for arbitrary sets $S$ and $T$.
Substituting $S$ for $T$:
:$A \subseteq B, \ S \subseteq S \implies A \cap S \subseteq B \cap S$
From Set is Subset of Itself, $S \... | Let $A, B, S, T$ be [[Definition:Set|sets]].
Then:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$ | Let $A \subseteq B$, and let $S$ be any set.
From [[Set Intersection Preserves Subsets]]:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$
for arbitrary sets $S$ and $T$.
Substituting $S$ for $T$:
:$A \subseteq B, \ S \subseteq S \implies A \cap S \subseteq B \cap S$
From [[Set is Subset of It... | Set Intersection Preserves Subsets/Corollary/Proof 1 | https://proofwiki.org/wiki/Set_Intersection_Preserves_Subsets | https://proofwiki.org/wiki/Set_Intersection_Preserves_Subsets/Corollary/Proof_1 | [
"Set Intersection",
"Subsets",
"Set Intersection Preserves Subsets"
] | [
"Definition:Set"
] | [
"Set Intersection Preserves Subsets",
"Set is Subset of Itself"
] |
proofwiki-5126 | Set Intersection Preserves Subsets | Let $A, B, S, T$ be sets.
Then:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$ | Recall the Factor Principles, themselves a corollary of the Praeclarum Theorema:
:$\paren {p \implies q} \vdash \paren {p \land r} \implies \paren {q \land r}$
This is applied as:
{{begin-eqn}}
{{eqn | o =
| r = A \subseteq B
| c =
}}
{{eqn | o = \leadsto
| r = \paren {x \in A \implies x \in B}
... | Let $A, B, S, T$ be [[Definition:Set|sets]].
Then:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$ | Recall the [[Factor Principles]], themselves a [[Definition:Corollary|corollary]] of the [[Praeclarum Theorema]]:
:$\paren {p \implies q} \vdash \paren {p \land r} \implies \paren {q \land r}$
This is applied as:
{{begin-eqn}}
{{eqn | o =
| r = A \subseteq B
| c =
}}
{{eqn | o = \leadsto
| r = \p... | Set Intersection Preserves Subsets/Corollary/Proof 2 | https://proofwiki.org/wiki/Set_Intersection_Preserves_Subsets | https://proofwiki.org/wiki/Set_Intersection_Preserves_Subsets/Corollary/Proof_2 | [
"Set Intersection",
"Subsets",
"Set Intersection Preserves Subsets"
] | [
"Definition:Set"
] | [
"Factor Principles",
"Definition:Corollary",
"Praeclarum Theorema",
"Factor Principles"
] |
proofwiki-5127 | Null Space Contains Zero Vector | Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n: \mathbf A \mathbf x = \mathbf 0}$
be the null space of $\mathbf A$, where:
$\quad \mathbf A_{m \times n} = \begin {bmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m 1} & a_... | {{begin-eqn}}
{{eqn | l = \mathbf A \mathbf 0
| r = <nowiki>\begin {bmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m 1} & a_{m 2} & \cdots & a_{m n} \\
\end {bmatrix} \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}</nowiki>... | Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n: \mathbf A \mathbf x = \mathbf 0}$
be the [[Definition:Null Space|null space]] of $\mathbf A$, where:
$\quad \mathbf A_{m \times n} = \begin {bmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\
\vdots & \vdots & \ddot... | {{begin-eqn}}
{{eqn | l = \mathbf A \mathbf 0
| r = <nowiki>\begin {bmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m 1} & a_{m 2} & \cdots & a_{m n} \\
\end {bmatrix} \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}</nowiki>... | Null Space Contains Zero Vector/Proof 1 | https://proofwiki.org/wiki/Null_Space_Contains_Zero_Vector | https://proofwiki.org/wiki/Null_Space_Contains_Zero_Vector/Proof_1 | [
"Linear Algebra",
"Null Spaces",
"Null Space Contains Zero Vector"
] | [
"Definition:Null Space",
"Definition:Matrix",
"Definition:Matrix Space",
"Definition:Null Space",
"Definition:Zero Vector"
] | [
"Definition:Matrix/Order",
"Definition:Null Space"
] |
proofwiki-5128 | Null Space Contains Zero Vector | Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n: \mathbf A \mathbf x = \mathbf 0}$
be the null space of $\mathbf A$, where:
$\quad \mathbf A_{m \times n} = \begin {bmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m 1} & a_... | From Matrix Product as Linear Transformation, $\mathbf {Ax} = \mathbf 0$ defines a linear transformation from $\R^m$ to $\R^n$.
The result then follows from Linear Transformation Maps Zero Vector to Zero Vector. | Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n: \mathbf A \mathbf x = \mathbf 0}$
be the [[Definition:Null Space|null space]] of $\mathbf A$, where:
$\quad \mathbf A_{m \times n} = \begin {bmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\
\vdots & \vdots & \ddot... | From [[Matrix Product as Linear Transformation]], $\mathbf {Ax} = \mathbf 0$ defines a [[Definition:Linear Transformation on Vector Space|linear transformation]] from $\R^m$ to $\R^n$.
The result then follows from [[Linear Transformation Maps Zero Vector to Zero Vector]]. | Null Space Contains Zero Vector/Proof 2 | https://proofwiki.org/wiki/Null_Space_Contains_Zero_Vector | https://proofwiki.org/wiki/Null_Space_Contains_Zero_Vector/Proof_2 | [
"Linear Algebra",
"Null Spaces",
"Null Space Contains Zero Vector"
] | [
"Definition:Null Space",
"Definition:Matrix",
"Definition:Matrix Space",
"Definition:Null Space",
"Definition:Zero Vector"
] | [
"Matrix Product as Linear Transformation",
"Definition:Linear Transformation/Vector Space",
"Linear Transformation Maps Zero Vector to Zero Vector"
] |
proofwiki-5129 | Homogeneous System has Zero Vector as Solution | Every homogeneous system of linear equations has the zero vector as a solution. | By the definition of null space, $\mathbf 0$ is a solution {{iff}} the null space contains the zero vector.
The result follows from Null Space Contains Zero Vector.
{{qed}}
Category:Linear Algebra
Category:Null Spaces
mp2ftq9clu23hsbzit9h6vwj15zfend | Every [[Definition:Homogeneous Linear Equations|homogeneous system of linear equations]] has the [[Definition:Zero Vector|zero vector]] as a [[Definition:Solution to System of Simultaneous Equations|solution]]. | By the definition of [[Definition:Null Space|null space]], $\mathbf 0$ is a [[Definition:Solution to System of Simultaneous Equations|solution]] {{iff}} the [[Definition:Null Space|null space]] contains the [[Definition:Zero Vector|zero vector]].
The result follows from [[Null Space Contains Zero Vector]].
{{qed}}
[[... | Homogeneous System has Zero Vector as Solution | https://proofwiki.org/wiki/Homogeneous_System_has_Zero_Vector_as_Solution | https://proofwiki.org/wiki/Homogeneous_System_has_Zero_Vector_as_Solution | [
"Linear Algebra",
"Null Spaces"
] | [
"Definition:Homogeneous Simultaneous Linear Equations",
"Definition:Zero Vector",
"Definition:Simultaneous Equations/Solution"
] | [
"Definition:Null Space",
"Definition:Simultaneous Equations/Solution",
"Definition:Null Space",
"Definition:Zero Vector",
"Null Space Contains Zero Vector",
"Category:Linear Algebra",
"Category:Null Spaces"
] |
proofwiki-5130 | Commutative Linear Transformation is G-Module Homomorphism | Let $\rho: G \to \GL V$ be a representation.
Let $f: V \to V$ be a linear transformation.
Let:
:$\forall g \in G: \map \rho g \circ f = f \circ \map \rho g$
Then $f: V \to V$ is a $G$-module homomorphism. | Let:
:$\forall g \in G: \map \rho g \circ f = f \circ \map \rho g$
Let $v$ be a vector $v \in V$.
Then:
:$\map {\map \rho g} {\map f v} = \map f {\map {\map \rho g} v}$
Using the properties from Correspondence between Linear Group Actions and Linear Representations:
:there exists a $G$-module $\struct {V, \phi}$ assoc... | Let $\rho: G \to \GL V$ be a [[Definition:Linear Representation|representation]].
Let $f: V \to V$ be a [[Definition:Linear Transformation|linear transformation]].
Let:
:$\forall g \in G: \map \rho g \circ f = f \circ \map \rho g$
Then $f: V \to V$ is a [[Definition:G-Module Homomorphism|$G$-module homomorphism]]. | Let:
:$\forall g \in G: \map \rho g \circ f = f \circ \map \rho g$
Let $v$ be a [[Definition:Vector Space|vector]] $v \in V$.
Then:
:$\map {\map \rho g} {\map f v} = \map f {\map {\map \rho g} v}$
Using the properties from [[Correspondence between Linear Group Actions and Linear Representations]]:
:there exists a [[... | Commutative Linear Transformation is G-Module Homomorphism | https://proofwiki.org/wiki/Commutative_Linear_Transformation_is_G-Module_Homomorphism | https://proofwiki.org/wiki/Commutative_Linear_Transformation_is_G-Module_Homomorphism | [
"Representation Theory"
] | [
"Definition:Linear Representation",
"Definition:Linear Transformation",
"Definition:G-Module Homomorphism"
] | [
"Definition:Vector Space",
"Correspondence between Linear Group Actions and Linear Representations",
"Definition:Module over Group",
"Definition:G-Module Homomorphism",
"Definition:G-Module Homomorphism",
"Category:Representation Theory"
] |
proofwiki-5131 | Characterization of Measures | Let $\struct {X, \Sigma}$ be a measurable space.
Denote $\overline \R_{\ge 0}$ for the set of positive extended real numbers.
A mapping $\mu: \Sigma \to \overline \R_{\ge 0}$ is a measure {{iff}}:
{{begin-itemize}}
{{item|(1):|$\map \mu \O {{=}} 0$}}
{{item|(2):|$\mu$ is finitely additive}}
{{item|(3):|For every increa... | === Necessary Condition ===
To show is that a measure $\mu$ has the properties $(1)$, $(2)$, $(3)$, $(3')$ and $(3' ')$.
Property $(1)$ is also part of the definition of measure, and hence is immediate.
Property $(2)$ is precisely the statement of Measure is Finitely Additive Function.
Property $(3)$ is precisely the s... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Denote $\overline \R_{\ge 0}$ for the set of [[Definition:Positive Real Number|positive]] [[Definition:Extended Real Number Line|extended real numbers]].
A [[Definition:Mapping|mapping]] $\mu: \Sigma \to \overline \R_{\ge 0}$ is a [[Def... | === Necessary Condition ===
To show is that a [[Definition:Measure (Measure Theory)|measure]] $\mu$ has the properties $(1)$, $(2)$, $(3)$, $(3')$ and $(3' ')$.
Property $(1)$ is also part of the definition of [[Definition:Measure (Measure Theory)|measure]], and hence is immediate.
Property $(2)$ is precisely the s... | Characterization of Measures | https://proofwiki.org/wiki/Characterization_of_Measures | https://proofwiki.org/wiki/Characterization_of_Measures | [
"Measure Theory"
] | [
"Definition:Measurable Space",
"Definition:Positive/Real Number",
"Definition:Extended Real Number Line",
"Definition:Mapping",
"Definition:Measure (Measure Theory)",
"Definition:Additive Function (Measure Theory)",
"Definition:Increasing Sequence of Sets",
"Definition:Limit of Increasing Sequence of ... | [
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)",
"Measure is Finitely Additive Function",
"Measure of Limit of Increasing Sequence of Measurable Sets",
"Definition:Decreasing Sequence of Sets",
"Measure is Monotone",
"Measure of Set Difference with Subset",
"Measure of Set... |
proofwiki-5132 | Measure is Countably Subadditive | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then $\mu$ is a countably subadditive function. | Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of sets in $\Sigma$.
It is required to show that:
:$\ds \map \mu {\bigcup_{n \mathop \in \N} E_n} \le \sum_{n \mathop \in \N} \map \mu {E_n}$
Now define the sequence $\sequence {F_n}_{n \mathop \in \N}$ in $\Sigma$ by:
:$F_n := \ds \bigcup_{k \mathop = 0}^n E_n$
By... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Then $\mu$ is a [[Definition:Countably Subadditive Function|countably subadditive function]]. | Let $\sequence {E_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Set|sets]] in $\Sigma$.
It is required to show that:
:$\ds \map \mu {\bigcup_{n \mathop \in \N} E_n} \le \sum_{n \mathop \in \N} \map \mu {E_n}$
Now define the [[Definition:Sequence|sequence]] $\sequence {F_n}_{n \mathop ... | Measure is Countably Subadditive | https://proofwiki.org/wiki/Measure_is_Countably_Subadditive | https://proofwiki.org/wiki/Measure_is_Countably_Subadditive | [
"Measures"
] | [
"Definition:Measure Space",
"Definition:Countably Subadditive Function"
] | [
"Definition:Sequence",
"Definition:Set",
"Definition:Sequence",
"Set is Subset of Union",
"Definition:Increasing Sequence of Sets",
"Definition:Limit of Increasing Sequence of Sets",
"Measure of Limit of Increasing Sequence of Measurable Sets",
"Measure is Subadditive/Corollary"
] |
proofwiki-5133 | Dirac Measure is Measure | Let $\struct {X, \Sigma}$ be a measurable space.
Let $x \in X$, and let $\delta_x$ be the Dirac measure at $x$.
Then $\delta_x$ is a measure. | Let us verify in turn that $\delta_x$ satisfies the axioms for a measure. | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $x \in X$, and let $\delta_x$ be the [[Definition:Dirac Measure|Dirac measure at $x$]].
Then $\delta_x$ is a [[Definition:Measure (Measure Theory)|measure]]. | Let us verify in turn that $\delta_x$ satisfies the axioms for a [[Definition:Measure (Measure Theory)|measure]]. | Dirac Measure is Measure | https://proofwiki.org/wiki/Dirac_Measure_is_Measure | https://proofwiki.org/wiki/Dirac_Measure_is_Measure | [
"Dirac Measure",
"Measures",
"Dirac Measure"
] | [
"Definition:Measurable Space",
"Definition:Dirac Measure",
"Definition:Measure (Measure Theory)"
] | [
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)"
] |
proofwiki-5134 | Dirac Measure is Probability Measure | Let $\struct {X, \AA}$ be a measurable space.
Let $x \in X$, and let $\delta_x$ be the Dirac measure at $x$.
Then $\delta_x$ is a probability measure. | By Dirac Measure is Measure, $\delta_x$ is a measure.
Also, $\map {\delta_x} X = 1$ because $x \in X$.
Hence $\delta_x$ is a probability measure.
{{qed}}
Category:Probability Measures
gtm3ifhzjlvmmqvlih66r361v1se0cu | Let $\struct {X, \AA}$ be a [[Definition:Measurable Space|measurable space]].
Let $x \in X$, and let $\delta_x$ be the [[Definition:Dirac Measure|Dirac measure at $x$]].
Then $\delta_x$ is a [[Definition:Probability Measure|probability measure]]. | By [[Dirac Measure is Measure]], $\delta_x$ is a [[Definition:Measure (Measure Theory)|measure]].
Also, $\map {\delta_x} X = 1$ because $x \in X$.
Hence $\delta_x$ is a [[Definition:Probability Measure|probability measure]].
{{qed}}
[[Category:Probability Measures]]
gtm3ifhzjlvmmqvlih66r361v1se0cu | Dirac Measure is Probability Measure | https://proofwiki.org/wiki/Dirac_Measure_is_Probability_Measure | https://proofwiki.org/wiki/Dirac_Measure_is_Probability_Measure | [
"Measure Theory",
"Probability Measures",
"Probability Measures"
] | [
"Definition:Measurable Space",
"Definition:Dirac Measure",
"Definition:Probability Measure"
] | [
"Dirac Measure is Measure",
"Definition:Measure (Measure Theory)",
"Definition:Probability Measure",
"Category:Probability Measures"
] |
proofwiki-5135 | Co-Countable Measure is Measure | Let $X$ be an uncountable set.
Let $\Sigma$ be the $\sigma$-algebra of countable sets on $X$.
Then the co-countable measure $\mu$ on $X$ is a measure. | Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\mu$. | Let $X$ be an [[Definition:Uncountable Set|uncountable set]].
Let $\Sigma$ be the [[Sigma-Algebra of Countable Sets|$\sigma$-algebra of countable sets]] on $X$.
Then the [[Definition:Co-Countable Measure|co-countable measure]] $\mu$ on $X$ is a [[Definition:Measure (Measure Theory)|measure]]. | Let us verify the [[Definition:Measure (Measure Theory)|measure axioms]] $(1)$, $(2)$ and $(3')$ for $\mu$. | Co-Countable Measure is Measure | https://proofwiki.org/wiki/Co-Countable_Measure_is_Measure | https://proofwiki.org/wiki/Co-Countable_Measure_is_Measure | [
"Measure Theory",
"Measures",
"Measures"
] | [
"Definition:Uncountable/Set",
"Sigma-Algebra of Countable Sets",
"Definition:Co-Countable Measure",
"Definition:Measure (Measure Theory)"
] | [
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)"
] |
proofwiki-5136 | Co-Countable Measure is Probability Measure | Let $X$ be an uncountable set.
Let $\AA$ be the $\sigma$-algebra of countable sets on $X$.
Then the co-countable measure $\mu$ on $X$ is a probability measure. | By Co-Countable Measure is Measure, $\mu$ is a measure.
By Relative Complement with Self is Empty Set, have $\relcomp X X = \O$.
As $\O$ is countable, it follows that $X$ is co-countable.
Hence $\map \mu X = 1$, and so $\mu$ is a probability measure.
{{qed}}
Category:Probability Measures
l6l99ophv7dghlzykdqlc2lxwpgphka | Let $X$ be an [[Definition:Uncountable Set|uncountable set]].
Let $\AA$ be the [[Sigma-Algebra of Countable Sets|$\sigma$-algebra of countable sets]] on $X$.
Then the [[Definition:Co-Countable Measure|co-countable measure]] $\mu$ on $X$ is a [[Definition:Probability Measure|probability measure]]. | By [[Co-Countable Measure is Measure]], $\mu$ is a [[Definition:Measure (Measure Theory)|measure]].
By [[Relative Complement with Self is Empty Set]], have $\relcomp X X = \O$.
As $\O$ is [[Definition:Countable Set|countable]], it follows that $X$ is [[Definition:Co-Countable Set|co-countable]].
Hence $\map \mu X =... | Co-Countable Measure is Probability Measure | https://proofwiki.org/wiki/Co-Countable_Measure_is_Probability_Measure | https://proofwiki.org/wiki/Co-Countable_Measure_is_Probability_Measure | [
"Measure Theory",
"Probability Measures",
"Probability Measures"
] | [
"Definition:Uncountable/Set",
"Sigma-Algebra of Countable Sets",
"Definition:Co-Countable Measure",
"Definition:Probability Measure"
] | [
"Co-Countable Measure is Measure",
"Definition:Measure (Measure Theory)",
"Relative Complement with Self is Empty Set",
"Definition:Countable Set",
"Definition:Co-Countable Set",
"Definition:Probability Measure",
"Category:Probability Measures"
] |
proofwiki-5137 | Counting Measure is Measure | Let $\struct {X, \Sigma}$ be a measurable space.
Then the counting measure $\size {\,\cdot\,}$ on $\struct {X, \Sigma}$ is a measure. | Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\size {\,\cdot\,}$. | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Then the [[Definition:Counting Measure|counting measure]] $\size {\,\cdot\,}$ on $\struct {X, \Sigma}$ is a [[Definition:Measure (Measure Theory)|measure]]. | Let us verify the [[Definition:Measure (Measure Theory)|measure axioms]] $(1)$, $(2)$ and $(3')$ for $\size {\,\cdot\,}$. | Counting Measure is Measure | https://proofwiki.org/wiki/Counting_Measure_is_Measure | https://proofwiki.org/wiki/Counting_Measure_is_Measure | [
"Counting Measure",
"Measures",
"Counting Measure"
] | [
"Definition:Measurable Space",
"Definition:Counting Measure",
"Definition:Measure (Measure Theory)"
] | [
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)"
] |
proofwiki-5138 | Trivial Vector Space iff Zero Dimension | Let $V$ be a vector space.
Then $V = \set \bszero$ {{iff}} $\map \dim V = 0$, where $\dim$ signifies dimension of vector space. | === Necessary Condition ===
Suppose $V = \set \bszero$.
We have that $V$ has no $\mathbf v \in V, \mathbf v \ne \bszero$.
Thus there exists no $\set {\mathbf v} \subseteq V$ such that $\mathbf v \ne \bszero$.
Such a $\set {\mathbf v}$ would be a linearly independent set.
Hence $\O$ is the only possible basis for $V$.
H... | Let $V$ be a [[Definition:Vector Space|vector space]].
Then $V = \set \bszero$ {{iff}} $\map \dim V = 0$, where $\dim$ signifies [[Definition:Dimension of Vector Space|dimension of vector space]]. | === Necessary Condition ===
Suppose $V = \set \bszero$.
We have that $V$ has no $\mathbf v \in V, \mathbf v \ne \bszero$.
Thus there exists no $\set {\mathbf v} \subseteq V$ such that $\mathbf v \ne \bszero$.
Such a $\set {\mathbf v}$ would be a [[Definition:Linearly Independent Set|linearly independent set]].
Hen... | Trivial Vector Space iff Zero Dimension | https://proofwiki.org/wiki/Trivial_Vector_Space_iff_Zero_Dimension | https://proofwiki.org/wiki/Trivial_Vector_Space_iff_Zero_Dimension | [
"Vector Spaces",
"Dimension of Vector Space"
] | [
"Definition:Vector Space",
"Definition:Dimension of Vector Space"
] | [
"Definition:Linearly Independent/Set",
"Definition:Basis (Linear Algebra)",
"Definition:Basis (Linear Algebra)",
"Definition:Linearly Independent/Set"
] |
proofwiki-5139 | G-Submodule Test | Let $\struct {V, \phi}$ be a $G$-module over a field $k$.
Let $W$ be a vector subspace of $V$.
Let $\phi_W: G \times W \to V$ denote the restriction of $\phi$ to $G \times W$.
Then:
:$\struct {W, \phi_W}$ is a $G$-submodule of $V$
{{iff}}:
:$\map \phi {G, W} \subseteq W$ | === Necessary Condition ===
Let $W$ be a $G$-submodule of $V$.
Hence by definition $\phi_W: G \times W \to W$ is a linear action on $W$.
Also by definition:
:$\map {\phi_W} {G, W} = \map \phi {G, W} \subseteq W$
{{qed|lemma}} | Let $\struct {V, \phi}$ be a [[Definition:G-Module|$G$-module]] over a [[Definition:Field (Abstract Algebra)|field]] $k$.
Let $W$ be a [[Definition:Vector Subspace|vector subspace]] of $V$.
Let $\phi_W: G \times W \to V$ denote the [[Definition:Restriction of Mapping|restriction]] of $\phi$ to $G \times W$.
Then:
:... | === Necessary Condition ===
Let $W$ be a [[Definition:G-Submodule|$G$-submodule]] of $V$.
Hence by definition $\phi_W: G \times W \to W$ is a [[Definition:Linear Group Action|linear action]] on $W$.
Also by definition:
:$\map {\phi_W} {G, W} = \map \phi {G, W} \subseteq W$
{{qed|lemma}} | G-Submodule Test | https://proofwiki.org/wiki/G-Submodule_Test | https://proofwiki.org/wiki/G-Submodule_Test | [
"Representation Theory"
] | [
"Definition:Module over Group",
"Definition:Field (Abstract Algebra)",
"Definition:Vector Subspace",
"Definition:Restriction/Mapping",
"Definition:G-Submodule"
] | [
"Definition:G-Submodule",
"Definition:Linear Group Action",
"Definition:Linear Group Action",
"Definition:Linear Group Action",
"Definition:Linear Group Action",
"Definition:G-Submodule"
] |
proofwiki-5140 | Composite of Continuous Mappings is Continuous/Point | Let $T_1, T_2, T_3$ be topological spaces.
Let the mapping $f : T_1 \to T_2$ be continuous at $x$.
Let the mapping $g : T_2 \to T_3$ be continuous at $\map f x$.
Then the composite mapping $g \circ f : T_1 \to T_3$ is continuous at $x$. | Let $N$ be any neighborhood of $\map {\paren {g \circ f} } x$.
By the definition of continuity at a point:
:there exists a neighborhood $L$ of $\map f x$ such that $g \sqbrk L \subseteq N$
and
:there exists a neighborhood $M$ of $x$ such that $f \sqbrk M \subseteq L$.
Thus $\paren {g \circ f} \sqbrk M \subseteq g \sqbr... | Let $T_1, T_2, T_3$ be [[Definition:Topological Space|topological spaces]].
Let the [[Definition:Mapping|mapping]] $f : T_1 \to T_2$ be [[Definition:Continuous Mapping at Point (Topology)|continuous at $x$]].
Let the [[Definition:Mapping|mapping]] $g : T_2 \to T_3$ be [[Definition:Continuous Mapping at Point (Topolog... | Let $N$ be any [[Definition:Neighborhood (Topology)|neighborhood]] of $\map {\paren {g \circ f} } x$.
By the [[Definition:Continuous Mapping at Point (Topology)|definition of continuity at a point]]:
:there exists a [[Definition:Neighborhood (Topology)|neighborhood]] $L$ of $\map f x$ such that $g \sqbrk L \subseteq N... | Composite of Continuous Mappings is Continuous/Point | https://proofwiki.org/wiki/Composite_of_Continuous_Mappings_is_Continuous/Point | https://proofwiki.org/wiki/Composite_of_Continuous_Mappings_is_Continuous/Point | [
"Composite of Continuous Mappings is Continuous",
"Composite Mappings",
"Continuous Mappings (Topology)"
] | [
"Definition:Topological Space",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)/Point",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)/Point",
"Definition:Composition of Mappings",
"Definition:Continuous Mapping (Topology)/Point"
] | [
"Definition:Neighborhood (Topology)",
"Definition:Continuous Mapping (Topology)/Point",
"Definition:Neighborhood (Topology)",
"Definition:Neighborhood (Topology)"
] |
proofwiki-5141 | Null Space Closed under Vector Addition | Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n : \mathbf A \mathbf x = \mathbf 0}$
be the null space of $\mathbf A$, where:
:<nowiki>$\mathbf A_{m \times n} = \begin {bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} &... | Let $\mathbf v, \mathbf w \in \map {\mathrm N} {\mathbf A}$.
By the definition of null space:
{{begin-eqn}}
{{eqn | l = \mathbf A \mathbf v
| r = \mathbf 0
}}
{{eqn | l = \mathbf A \mathbf w
| r = \mathbf 0
}}
{{end-eqn}}
Next, observe that:
{{begin-eqn}}
{{eqn | l = \mathbf A \paren {\mathbf v + \mathbf w}... | Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n : \mathbf A \mathbf x = \mathbf 0}$
be the [[Definition:Null Space|null space]] of $\mathbf A$, where:
:<nowiki>$\mathbf A_{m \times n} = \begin {bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & ... | Let $\mathbf v, \mathbf w \in \map {\mathrm N} {\mathbf A}$.
By the definition of [[Definition:Null Space|null space]]:
{{begin-eqn}}
{{eqn | l = \mathbf A \mathbf v
| r = \mathbf 0
}}
{{eqn | l = \mathbf A \mathbf w
| r = \mathbf 0
}}
{{end-eqn}}
Next, observe that:
{{begin-eqn}}
{{eqn | l = \mathbf A ... | Null Space Closed under Vector Addition | https://proofwiki.org/wiki/Null_Space_Closed_under_Vector_Addition | https://proofwiki.org/wiki/Null_Space_Closed_under_Vector_Addition | [
"Linear Algebra",
"Null Spaces"
] | [
"Definition:Null Space",
"Definition:Matrix",
"Definition:Column Matrix",
"Definition:Vector/Real Euclidean Space",
"Definition:Euclidean Space/Real",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Vector Sum"
] | [
"Definition:Null Space",
"Matrix Multiplication Distributes over Matrix Addition",
"Definition:Matrix/Order",
"Definition:By Hypothesis",
"Definition:Null Space"
] |
proofwiki-5142 | Null Space Closed under Scalar Multiplication | Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n : \mathbf {A x} = \mathbf 0}$
be the null space of $\mathbf A$, where:
:<nowiki>$\mathbf A_{m \times n} = \begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} & \cdots... | Let $\mathbf v \in \map {\mathrm N} {\mathbf A}$, $\lambda \in \R$.
By the definition of null space:
{{begin-eqn}}
{{eqn | l = \mathbf {A v}
| r = \mathbf {0}
}}
{{end-eqn}}
Observe that:
{{begin-eqn}}
{{eqn | l = \mathbf A \paren {\lambda \mathbf v}
| r = \lambda \paren {\mathbf {A v} }
| c = Matrix ... | Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n : \mathbf {A x} = \mathbf 0}$
be the [[Definition:Null Space|null space]] of $\mathbf A$, where:
:<nowiki>$\mathbf A_{m \times n} = \begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots ... | Let $\mathbf v \in \map {\mathrm N} {\mathbf A}$, $\lambda \in \R$.
By the definition of [[Definition:Null Space|null space]]:
{{begin-eqn}}
{{eqn | l = \mathbf {A v}
| r = \mathbf {0}
}}
{{end-eqn}}
Observe that:
{{begin-eqn}}
{{eqn | l = \mathbf A \paren {\lambda \mathbf v}
| r = \lambda \paren {\math... | Null Space Closed under Scalar Multiplication | https://proofwiki.org/wiki/Null_Space_Closed_under_Scalar_Multiplication | https://proofwiki.org/wiki/Null_Space_Closed_under_Scalar_Multiplication | [
"Linear Algebra",
"Null Spaces"
] | [
"Definition:Null Space",
"Definition:Matrix",
"Definition:Matrix/Column",
"Definition:Element",
"Definition:Real Vector Space",
"Definition:Closure (Abstract Algebra)/Scalar Product",
"Definition:Scalar Multiplication/Vector Space"
] | [
"Definition:Null Space",
"Matrix Multiplication is Homogeneous of Degree 1",
"Definition:Null Space"
] |
proofwiki-5143 | Null Space is Subspace | Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n: \mathbf {A x} = \mathbf 0}$
be the null space of $\mathbf A$, where:
:<nowiki>$\mathbf A_{m \times n} = \begin {bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} & \cdots... | $\map {\mathrm N} {\mathbf A} \subseteq \R^n$, by construction.
We have:
:$\mathbf 0 \in \map {\mathrm N} {\mathbf A}$, from Null Space Contains Zero Vector
:$\forall \mathbf v, \mathbf w \in \map {\mathrm N} {\mathbf A}: \mathbf v + \mathbf w \in \map {\mathrm N} {\mathbf A}$, from Null Space Closed under Vector Addit... | Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n: \mathbf {A x} = \mathbf 0}$
be the [[Definition:Null Space|null space]] of $\mathbf A$, where:
:<nowiki>$\mathbf A_{m \times n} = \begin {bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdot... | $\map {\mathrm N} {\mathbf A} \subseteq \R^n$, by construction.
We have:
:$\mathbf 0 \in \map {\mathrm N} {\mathbf A}$, from [[Null Space Contains Zero Vector]]
:$\forall \mathbf v, \mathbf w \in \map {\mathrm N} {\mathbf A}: \mathbf v + \mathbf w \in \map {\mathrm N} {\mathbf A}$, from [[Null Space Closed under Vec... | Null Space is Subspace | https://proofwiki.org/wiki/Null_Space_is_Subspace | https://proofwiki.org/wiki/Null_Space_is_Subspace | [
"Linear Algebra",
"Null Spaces"
] | [
"Definition:Null Space",
"Definition:Matrix",
"Definition:Vector Subspace"
] | [
"Null Space Contains Zero Vector",
"Null Space Closed under Vector Addition",
"Null Space Closed under Scalar Multiplication",
"Vector Subspace of Real Vector Space"
] |
proofwiki-5144 | Norm is Continuous | Let $\struct {V, \norm {\,\cdot\,} }$ be a normed vector space.
Then the mapping $x \mapsto \norm x$ is continuous.
Here, the metric used is the metric $d$ induced by $\norm {\,\cdot\,}$. | Since $\norm x = \map d {x, \mathbf 0}$, the result follows directly from Distance Function of Metric Space is Continuous.
{{qed}} | Let $\struct {V, \norm {\,\cdot\,} }$ be a [[Definition:Normed Vector Space|normed]] [[Definition:Vector Space|vector space]].
Then the [[Definition:Mapping|mapping]] $x \mapsto \norm x$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous]].
Here, the [[Definition:Metric|metric]] used is the metric $d$ [[D... | Since $\norm x = \map d {x, \mathbf 0}$, the result follows directly from [[Distance Function of Metric Space is Continuous]].
{{qed}} | Norm is Continuous | https://proofwiki.org/wiki/Norm_is_Continuous | https://proofwiki.org/wiki/Norm_is_Continuous | [
"Norm Theory",
"Continuous Mappings"
] | [
"Definition:Normed Vector Space",
"Definition:Vector Space",
"Definition:Mapping",
"Definition:Continuous Mapping (Metric Space)",
"Definition:Metric Space/Metric",
"Definition:Metric Induced by Norm"
] | [
"Distance Function of Metric Space is Continuous"
] |
proofwiki-5145 | P-adic Valuation of Rational Number is Well Defined | The $p$-adic valuation:
:$\nu_p: \Q \to \Z \cup \set {+\infty}$
is well defined. | Let $\dfrac a b = \dfrac c d \in \Q$.
Thus:
:$a d = b c \in \Z$
By definition of rational number:
:$b, d \ne 0$
By definition of restricted $p$-adic valuation:
:$\map {\nu_p^\Z} b, \map {\nu_p^\Z} d < +\infty$ | The [[Definition:P-adic Valuation on Rational Numbers|$p$-adic valuation]]:
:$\nu_p: \Q \to \Z \cup \set {+\infty}$
is [[Definition:Well-Defined Operation|well defined]]. | Let $\dfrac a b = \dfrac c d \in \Q$.
Thus:
:$a d = b c \in \Z$
By definition of [[Definition:Rational Number|rational number]]:
:$b, d \ne 0$
By definition of [[Definition:Restricted P-adic Valuation|restricted $p$-adic valuation]]:
:$\map {\nu_p^\Z} b, \map {\nu_p^\Z} d < +\infty$ | P-adic Valuation of Rational Number is Well Defined | https://proofwiki.org/wiki/P-adic_Valuation_of_Rational_Number_is_Well_Defined | https://proofwiki.org/wiki/P-adic_Valuation_of_Rational_Number_is_Well_Defined | [
"P-adic Valuations"
] | [
"Definition:P-adic Valuation/Rational Numbers",
"Definition:Well-Defined/Operation"
] | [
"Definition:Rational Number",
"Definition:P-adic Valuation/Integers",
"Definition:P-adic Valuation/Integers",
"Definition:P-adic Valuation/Integers"
] |
proofwiki-5146 | P-adic Valuation is Valuation | The $p$-adic valuation $\nu_p: \Q \to \Z \cup \set {+\infty}$ is a valuation on $\Q$. | To prove that $\nu_p$ is a valuation it is necessary to demonstrate:
{{begin-axiom}}
{{axiom | n = \text V 1
| q = \forall q, r \in \Q
| ml= \map {\nu_p} {q r}
| mo= =
| mr= \map {\nu_p} q + \map {\nu_p} r
}}
{{axiom | n = \text V 2
| q = \forall q \in \Q
| ml= \map {\nu_... | The [[Definition:P-adic Valuation on Rational Numbers|$p$-adic valuation]] $\nu_p: \Q \to \Z \cup \set {+\infty}$ is a [[Definition:Valuation|valuation]] on $\Q$. | To prove that $\nu_p$ is a [[Definition:Valuation|valuation]] it is necessary to demonstrate:
{{begin-axiom}}
{{axiom | n = \text V 1
| q = \forall q, r \in \Q
| ml= \map {\nu_p} {q r}
| mo= =
| mr= \map {\nu_p} q + \map {\nu_p} r
}}
{{axiom | n = \text V 2
| q = \forall q \in \... | P-adic Valuation is Valuation | https://proofwiki.org/wiki/P-adic_Valuation_is_Valuation | https://proofwiki.org/wiki/P-adic_Valuation_is_Valuation | [
"P-adic Valuations"
] | [
"Definition:P-adic Valuation/Rational Numbers",
"Definition:Valuation"
] | [
"Definition:Valuation",
"Definition:Valuation"
] |
proofwiki-5147 | P-adic Norm not Complete on Rational Numbers | Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
Then:
:the valued field $\struct {\Q, \norm {\,\cdot\,}_p}$ is not complete.
That is, there exists a Cauchy sequence in $\struct {\Q, \norm{\,\cdot\,}_p}$ which does not converge to a limit in $\Q$. | === Case: $p \gt 3$ ===
{{:P-adic Norm not Complete on Rational Numbers/Proof 1/Case 1}}{{qed|lemma}}
=== Case: $p = 2$ or $3$ ===
{{:P-adic Norm not Complete on Rational Numbers/Proof 1/Case 2}}{{qed}} | Let $\norm {\,\cdot\,}_p$ be the [[Definition:P-adic Norm|$p$-adic norm]] on the [[Definition:Rational Numbers|rationals $\Q$]] for some [[Definition:Prime Number|prime]] $p$.
Then:
:the [[Definition:Valued Field|valued field]] $\struct {\Q, \norm {\,\cdot\,}_p}$ is not [[Definition:Complete Normed Division Ring|comp... | === [[P-adic Norm not Complete on Rational Numbers/Proof 1/Case 1|Case: $p \gt 3$]] ===
{{:P-adic Norm not Complete on Rational Numbers/Proof 1/Case 1}}{{qed|lemma}}
=== [[P-adic Norm not Complete on Rational Numbers/Proof 1/Case 2|Case: $p = 2$ or $3$]] ===
{{:P-adic Norm not Complete on Rational Numbers/Proof 1/Cas... | P-adic Norm not Complete on Rational Numbers/Proof 1 | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers/Proof_1 | [
"Metric Spaces",
"Normed Spaces",
"P-adic Number Theory",
"P-adic Norm not Complete on Rational Numbers"
] | [
"Definition:P-adic Norm",
"Definition:Rational Number",
"Definition:Prime Number",
"Definition:Valued Field",
"Definition:Complete Normed Division Ring",
"Definition:Cauchy Sequence/Normed Division Ring",
"Definition:Convergent Sequence/Normed Division Ring",
"Definition:Limit of Sequence/Normed Divis... | [
"P-adic Norm not Complete on Rational Numbers/Proof 1/Case 1",
"P-adic Norm not Complete on Rational Numbers/Proof 1/Case 2"
] |
proofwiki-5148 | P-adic Norm not Complete on Rational Numbers | Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
Then:
:the valued field $\struct {\Q, \norm {\,\cdot\,}_p}$ is not complete.
That is, there exists a Cauchy sequence in $\struct {\Q, \norm{\,\cdot\,}_p}$ which does not converge to a limit in $\Q$. | Hensel's Lemma is used to prove the existence of a Cauchy sequence that does not converge.
==== Lemma 1 ====
{{:P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 1}}{{qed|lemma}}
Let $x_1 \in \Z_{>0}: p \nmid x_1, x_1 \ge \dfrac {p + 1} 2$
Let $k$ be a positive integer such that $k \ge 2, p \nmid k$.
Let $a = ... | Let $\norm {\,\cdot\,}_p$ be the [[Definition:P-adic Norm|$p$-adic norm]] on the [[Definition:Rational Numbers|rationals $\Q$]] for some [[Definition:Prime Number|prime]] $p$.
Then:
:the [[Definition:Valued Field|valued field]] $\struct {\Q, \norm {\,\cdot\,}_p}$ is not [[Definition:Complete Normed Division Ring|comp... | [[Hensel's Lemma/First Form|Hensel's Lemma]] is used to prove the existence of a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] that does not [[Definition:Convergent Sequence in Normed Division Ring|converge]].
==== [[P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 1|Lemma 1]] ====
... | P-adic Norm not Complete on Rational Numbers/Proof 2 | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers/Proof_2 | [
"Metric Spaces",
"Normed Spaces",
"P-adic Number Theory",
"P-adic Norm not Complete on Rational Numbers"
] | [
"Definition:P-adic Norm",
"Definition:Rational Number",
"Definition:Prime Number",
"Definition:Valued Field",
"Definition:Complete Normed Division Ring",
"Definition:Cauchy Sequence/Normed Division Ring",
"Definition:Convergent Sequence/Normed Division Ring",
"Definition:Limit of Sequence/Normed Divis... | [
"Hensel's Lemma/First Form",
"Definition:Cauchy Sequence/Normed Division Ring",
"Definition:Convergent Sequence/Normed Division Ring",
"P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 1",
"Definition:Positive/Integer",
"P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 2",
"Definitio... |
proofwiki-5149 | P-adic Norm not Complete on Rational Numbers | Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
Then:
:the valued field $\struct {\Q, \norm {\,\cdot\,}_p}$ is not complete.
That is, there exists a Cauchy sequence in $\struct {\Q, \norm{\,\cdot\,}_p}$ which does not converge to a limit in $\Q$. | By Rational Numbers are Countably Infinite, the set of rational numbers is countably infinite.
By $p$-adic Numbers are Uncountable, the set of $p$-adic numbers $\Q_p$ is uncountably infinite.
Let $\CC$ be the commutative ring of Cauchy sequences over $\struct {\Q, \norm {\,\cdot\,}_p}$.
Let $\NN$ be the set of null seq... | Let $\norm {\,\cdot\,}_p$ be the [[Definition:P-adic Norm|$p$-adic norm]] on the [[Definition:Rational Numbers|rationals $\Q$]] for some [[Definition:Prime Number|prime]] $p$.
Then:
:the [[Definition:Valued Field|valued field]] $\struct {\Q, \norm {\,\cdot\,}_p}$ is not [[Definition:Complete Normed Division Ring|comp... | By [[Rational Numbers are Countably Infinite]], the [[Definition:Set|set]] of [[Definition:Rational Number|rational numbers]] is [[Definition:Countably Infinite Set|countably infinite]].
By [[P-adic Numbers are Uncountable|$p$-adic Numbers are Uncountable]], the [[Definition:Set|set]] of [[Definition:P-adic Numbers|$p... | P-adic Norm not Complete on Rational Numbers/Proof 3 | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers | https://proofwiki.org/wiki/P-adic_Norm_not_Complete_on_Rational_Numbers/Proof_3 | [
"Metric Spaces",
"Normed Spaces",
"P-adic Number Theory",
"P-adic Norm not Complete on Rational Numbers"
] | [
"Definition:P-adic Norm",
"Definition:Rational Number",
"Definition:Prime Number",
"Definition:Valued Field",
"Definition:Complete Normed Division Ring",
"Definition:Cauchy Sequence/Normed Division Ring",
"Definition:Convergent Sequence/Normed Division Ring",
"Definition:Limit of Sequence/Normed Divis... | [
"Rational Numbers are Countably Infinite",
"Definition:Set",
"Definition:Rational Number",
"Definition:Countably Infinite/Set",
"P-adic Numbers are Uncountable",
"Definition:Set",
"Definition:P-adic Number",
"Definition:Uncountable/Set",
"Definition:Ring of Cauchy Sequences",
"Definition:Set",
"... |
proofwiki-5150 | Restricted P-adic Valuation is Valuation | Let $\nu_p^\Z: \Z \to \Z \cup \set {+\infty}$ be the $p$-adic valuation restricted to the integers.
Then $\nu_p^\Z$ is a valuation. | To prove that $\nu_p^\Z$ is a valuation it is necessary to demonstrate:
{{begin-axiom}}
{{axiom | n = \text V 1
| q = \forall m, n \in \Z
| ml= \map {\nu_p^\Z} {m n}
| mo= =
| mr= \map {\nu_p^\Z} m + \map {\nu_p^\Z} n
}}
{{axiom | n = \text V 2
| q = \forall n \in \Z
| ml... | Let $\nu_p^\Z: \Z \to \Z \cup \set {+\infty}$ be the [[Definition:Restricted P-adic Valuation|$p$-adic valuation restricted to the integers]].
Then $\nu_p^\Z$ is a [[Definition:Valuation|valuation]]. | To prove that $\nu_p^\Z$ is a [[Definition:Valuation|valuation]] it is necessary to demonstrate:
{{begin-axiom}}
{{axiom | n = \text V 1
| q = \forall m, n \in \Z
| ml= \map {\nu_p^\Z} {m n}
| mo= =
| mr= \map {\nu_p^\Z} m + \map {\nu_p^\Z} n
}}
{{axiom | n = \text V 2
| q = \fo... | Restricted P-adic Valuation is Valuation | https://proofwiki.org/wiki/Restricted_P-adic_Valuation_is_Valuation | https://proofwiki.org/wiki/Restricted_P-adic_Valuation_is_Valuation | [
"P-adic Valuation on Integers",
"P-adic Valuations"
] | [
"Definition:P-adic Valuation/Integers",
"Definition:Valuation"
] | [
"Definition:Valuation"
] |
proofwiki-5151 | Null Measure is Measure | Let $\struct {X, \Sigma}$ be a measurable space.
Then the null measure $\mu$ on $\struct {X, \Sigma}$ is a measure. | Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\mu$. | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Then the [[Definition:Null Measure|null measure]] $\mu$ on $\struct {X, \Sigma}$ is a [[Definition:Measure (Measure Theory)|measure]]. | Let us verify the [[Definition:Measure (Measure Theory)|measure axioms]] $(1)$, $(2)$ and $(3')$ for $\mu$. | Null Measure is Measure | https://proofwiki.org/wiki/Null_Measure_is_Measure | https://proofwiki.org/wiki/Null_Measure_is_Measure | [
"Null Measure"
] | [
"Definition:Measurable Space",
"Definition:Null Measure",
"Definition:Measure (Measure Theory)"
] | [
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)"
] |
proofwiki-5152 | Infinite Measure is Measure | Let $\struct {X, \Sigma}$ be a measurable space.
Then the infinite measure $\mu$ on $\struct {X, \Sigma}$ is a measure. | Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\mu$. | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Then the [[Definition:Infinite Measure|infinite measure]] $\mu$ on $\struct {X, \Sigma}$ is a [[Definition:Measure (Measure Theory)|measure]]. | Let us verify the [[Definition:Measure (Measure Theory)|measure axioms]] $(1)$, $(2)$ and $(3')$ for $\mu$. | Infinite Measure is Measure | https://proofwiki.org/wiki/Infinite_Measure_is_Measure | https://proofwiki.org/wiki/Infinite_Measure_is_Measure | [
"Measures"
] | [
"Definition:Measurable Space",
"Definition:Infinite Measure",
"Definition:Measure (Measure Theory)"
] | [
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)"
] |
proofwiki-5153 | Linear Combination of Measures | Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu, \nu$ be measures on $\struct {X, \Sigma}$.
Then for all positive real numbers $a, b \in \R_{\ge 0}$, the pointwise sum:
:$a \mu + b \nu: \Sigma \to \overline \R, \ \map {\paren {a \mu + b \nu} } E := a \map \mu E + b \map \nu E$
is also a measure on $\struct {X... | Verifying the axioms $(1)$, $(2)$ and $(3')$ for a measure in turn: | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $\mu, \nu$ be [[Definition:Measure (Measure Theory)|measures]] on $\struct {X, \Sigma}$.
Then for all [[Definition:Positive Real Number|positive real numbers]] $a, b \in \R_{\ge 0}$, the [[Definition:Pointwise Addition|pointwise sum... | Verifying the axioms $(1)$, $(2)$ and $(3')$ for a [[Definition:Measure (Measure Theory)|measure]] in turn: | Linear Combination of Measures | https://proofwiki.org/wiki/Linear_Combination_of_Measures | https://proofwiki.org/wiki/Linear_Combination_of_Measures | [
"Measures"
] | [
"Definition:Measurable Space",
"Definition:Measure (Measure Theory)",
"Definition:Positive/Real Number",
"Definition:Pointwise Addition",
"Definition:Measure (Measure Theory)"
] | [
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)"
] |
proofwiki-5154 | Series of Measures is Measure | Let $\struct {X, \Sigma}$ be a measurable space.
Let $\sequence {\mu_n}_{n \mathop \in \N}$ be a sequence of measures on $\left({X, \Sigma}\right)$.
Let $\sequence {a_n}_{n \mathop \in \N} \subseteq \R_{\ge 0}$ be a sequence of positive real numbers.
Then the series of measures $\mu: \Sigma \to \overline \R$, defined b... | Let us verify the conditions for a measure in turn.
Let $E \in \Sigma$.
Then:
:$\forall n \in \N: \map {\mu_n} E \ge 0$
and so:
:$a_n \map {\mu_n} E \ge 0$
Therefore, by Series of Positive Real Numbers has Positive Limit:
:$\ds \map \mu E = \sum_{n \mathop \in \N} a_n \map {\mu_n} E \ge 0$
For every $n \in \N$, also $\... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $\sequence {\mu_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Measure (Measure Theory)|measures]] on $\left({X, \Sigma}\right)$.
Let $\sequence {a_n}_{n \mathop \in \N} \subseteq \R_{\ge 0}$ be a [[Defi... | Let us verify the conditions for a [[Definition:Measure (Measure Theory)|measure]] in turn.
Let $E \in \Sigma$.
Then:
:$\forall n \in \N: \map {\mu_n} E \ge 0$
and so:
:$a_n \map {\mu_n} E \ge 0$
Therefore, by [[Series of Positive Real Numbers has Positive Limit]]:
:$\ds \map \mu E = \sum_{n \mathop \in \N} a_n... | Series of Measures is Measure | https://proofwiki.org/wiki/Series_of_Measures_is_Measure | https://proofwiki.org/wiki/Series_of_Measures_is_Measure | [
"Series of Measures",
"Measures",
"Series of Measures"
] | [
"Definition:Measurable Space",
"Definition:Sequence",
"Definition:Measure (Measure Theory)",
"Definition:Sequence",
"Definition:Positive/Real Number",
"Definition:Series of Measures",
"Definition:Measure (Measure Theory)"
] | [
"Definition:Measure (Measure Theory)",
"Series of Positive Real Numbers has Positive Limit",
"Definition:Sequence",
"Definition:Pairwise Disjoint",
"Definition:Measure (Measure Theory)",
"Tonelli's Theorem/Corollary",
"Definition:Measure (Measure Theory)"
] |
proofwiki-5155 | Relation Induced by Quotient Set is Equivalence | Let $S$ be a set.
Let $\RR$ be an equivalence relation on $S$.
Let $S / \RR$ be the quotient set of $S$ determined by $\RR$.
Let $\RR'$ be the equivalence relation induced by $S / \RR$ on $S$.
Then $\RR' = \RR$. | Let $\RR$ be an equivalence relation on $S$.
Let $\tuple {x, y} \in \RR$.
By definition of equivalence class, $y \in \eqclass x \RR$ and $x \in \eqclass x \RR$.
By definition of quotient set, $x$ and $y$ both belong to the same element of $S / \RR$.
So, by definition of $\RR'$, it follows that $\tuple {x, y} \in \RR'$.... | Let $S$ be a [[Definition:Set|set]].
Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on $S$.
Let $S / \RR$ be the [[Definition:Quotient Set|quotient set]] of $S$ determined by $\RR$.
Let $\RR'$ be the [[Definition:Equivalence Relation Induced by Partition|equivalence relation induced by $S /... | Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on $S$.
Let $\tuple {x, y} \in \RR$.
By definition of [[Definition:Equivalence Class|equivalence class]], $y \in \eqclass x \RR$ and $x \in \eqclass x \RR$.
By definition of [[Definition:Quotient Set|quotient set]], $x$ and $y$ both belong to t... | Relation Induced by Quotient Set is Equivalence | https://proofwiki.org/wiki/Relation_Induced_by_Quotient_Set_is_Equivalence | https://proofwiki.org/wiki/Relation_Induced_by_Quotient_Set_is_Equivalence | [
"Examples of Equivalence Relations",
"Quotient Sets",
"Equivalence Relations",
"Examples of Equivalence Relations"
] | [
"Definition:Set",
"Definition:Equivalence Relation",
"Definition:Quotient Set",
"Definition:Equivalence Relation Induced by Partition"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Class",
"Definition:Quotient Set",
"Definition:Element",
"Definition:Subset",
"Definition:Element",
"Definition:Subset",
"Definition:Set Equality/Definition 2"
] |
proofwiki-5156 | Quotient Set Determined by Relation Induced by Partition is That Partition | Let $S$ be a set.
Let $\PP$ be a partition of $S$.
Let $\RR$ be the equivalence relation induced by $\PP$.
Then the quotient set $S / \RR$ of $S$ is $\PP$ itself. | Let $P \subseteq S$ such that $P \in \PP$.
Let $x \in P$.
Then:
{{begin-eqn}}
{{eqn | l = y
| o = \in
| r = \eqclass x \RR
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \tuple {x, y}
| o = \in
| r = \RR
| c = {{Defof|Equivalence Class}}
}}
{{eqn | ll= \leadstoandfrom
| l = ... | Let $S$ be a [[Definition:Set|set]].
Let $\PP$ be a [[Definition:Partition (Set Theory)|partition]] of $S$.
Let $\RR$ be the [[Definition:Equivalence Relation Induced by Partition|equivalence relation induced by $\PP$]].
Then the [[Definition:Quotient Set|quotient set]] $S / \RR$ of $S$ is $\PP$ itself. | Let $P \subseteq S$ such that $P \in \PP$.
Let $x \in P$.
Then:
{{begin-eqn}}
{{eqn | l = y
| o = \in
| r = \eqclass x \RR
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \tuple {x, y}
| o = \in
| r = \RR
| c = {{Defof|Equivalence Class}}
}}
{{eqn | ll= \leadstoandfrom
| l ... | Quotient Set Determined by Relation Induced by Partition is That Partition | https://proofwiki.org/wiki/Quotient_Set_Determined_by_Relation_Induced_by_Partition_is_That_Partition | https://proofwiki.org/wiki/Quotient_Set_Determined_by_Relation_Induced_by_Partition_is_That_Partition | [
"Quotient Sets",
"Equivalence Relations"
] | [
"Definition:Set",
"Definition:Set Partition",
"Definition:Equivalence Relation Induced by Partition",
"Definition:Quotient Set"
] | [
"Definition:Set Partition",
"Definition:Set Equality/Definition 2"
] |
proofwiki-5157 | Null Sets Closed under Subset | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $N \in \Sigma$ be a $\mu$-null set, and let $M \in \Sigma$ be a subset of $N$.
Then $M$ is also a $\mu$-null set. | As $\mu$ is a measure, $\map \mu M \ge 0$.
Also, by Measure is Monotone, $\map \mu M \le \map \mu N = 0$.
Hence $\map \mu M = 0$, and $M$ is a $\mu$-null set.
{{qed}} | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $N \in \Sigma$ be a [[Definition:Null Set|$\mu$-null set]], and let $M \in \Sigma$ be a [[Definition:Subset|subset]] of $N$.
Then $M$ is also a [[Definition:Null Set|$\mu$-null set]]. | As $\mu$ is a [[Definition:Measure (Measure Theory)|measure]], $\map \mu M \ge 0$.
Also, by [[Measure is Monotone]], $\map \mu M \le \map \mu N = 0$.
Hence $\map \mu M = 0$, and $M$ is a [[Definition:Null Set|$\mu$-null set]].
{{qed}} | Null Sets Closed under Subset | https://proofwiki.org/wiki/Null_Sets_Closed_under_Subset | https://proofwiki.org/wiki/Null_Sets_Closed_under_Subset | [
"Null Sets"
] | [
"Definition:Measure Space",
"Definition:Null Set",
"Definition:Subset",
"Definition:Null Set"
] | [
"Definition:Measure (Measure Theory)",
"Measure is Monotone",
"Definition:Null Set"
] |
proofwiki-5158 | Null Sets Closed under Subset | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $N \in \Sigma$ be a $\mu$-null set, and let $M \in \Sigma$ be a subset of $N$.
Then $M$ is also a $\mu$-null set. | For each $x, y \in \HH$, define $\EE_{x, y} : \map \BB X \to \C$ by:
:$\map {\EE_{x, y} } A = \innerprod {\map \EE A x} y$
for each $A \in \map \BB X$.
We then have:
:$\map {\EE_{x, y} } {E'} = 0$
for each $x, y \in \HH$.
From Null Sets Closed under Subset, we have $\map {\EE_{x, y} } E = 0$ for each $x, y \in \HH$.
Th... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $N \in \Sigma$ be a [[Definition:Null Set|$\mu$-null set]], and let $M \in \Sigma$ be a [[Definition:Subset|subset]] of $N$.
Then $M$ is also a [[Definition:Null Set|$\mu$-null set]]. | For each $x, y \in \HH$, define $\EE_{x, y} : \map \BB X \to \C$ by:
:$\map {\EE_{x, y} } A = \innerprod {\map \EE A x} y$
for each $A \in \map \BB X$.
We then have:
:$\map {\EE_{x, y} } {E'} = 0$
for each $x, y \in \HH$.
From [[Null Sets Closed under Subset]], we have $\map {\EE_{x, y} } E = 0$ for each $x, y \in \H... | Null Sets Closed under Subset/Resolution of the Identity/Proof 1 | https://proofwiki.org/wiki/Null_Sets_Closed_under_Subset | https://proofwiki.org/wiki/Null_Sets_Closed_under_Subset/Resolution_of_the_Identity/Proof_1 | [
"Null Sets"
] | [
"Definition:Measure Space",
"Definition:Null Set",
"Definition:Subset",
"Definition:Null Set"
] | [
"Null Sets Closed under Subset",
"Linear Subspace Dense iff Zero Orthocomplement"
] |
proofwiki-5159 | Null Sets Closed under Subset | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $N \in \Sigma$ be a $\mu$-null set, and let $M \in \Sigma$ be a subset of $N$.
Then $M$ is also a $\mu$-null set. | Let $\le_{\map B \HH}$ be the canonical preordering on $\map B \HH$.
From Measure is Monotone: Resolution of the Identity, we have:
:$\map \EE E \le_{\map B \HH} \map \EE {E'} = {\mathbf 0}_{\map B \HH}$
On the other hand from Bounds on Projection in Unital C*-Algebra, we have:
:${\mathbf 0}_{\map B \HH} \le_{\map B \H... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $N \in \Sigma$ be a [[Definition:Null Set|$\mu$-null set]], and let $M \in \Sigma$ be a [[Definition:Subset|subset]] of $N$.
Then $M$ is also a [[Definition:Null Set|$\mu$-null set]]. | Let $\le_{\map B \HH}$ be the [[Definition:Canonical Preordering of C*-Algebra|canonical preordering]] on $\map B \HH$.
From [[Measure is Monotone/Resolution of the Identity|Measure is Monotone: Resolution of the Identity]], we have:
:$\map \EE E \le_{\map B \HH} \map \EE {E'} = {\mathbf 0}_{\map B \HH}$
On the other... | Null Sets Closed under Subset/Resolution of the Identity/Proof 2 | https://proofwiki.org/wiki/Null_Sets_Closed_under_Subset | https://proofwiki.org/wiki/Null_Sets_Closed_under_Subset/Resolution_of_the_Identity/Proof_2 | [
"Null Sets"
] | [
"Definition:Measure Space",
"Definition:Null Set",
"Definition:Subset",
"Definition:Null Set"
] | [
"Definition:Canonical Preordering of C*-Algebra",
"Measure is Monotone/Resolution of the Identity",
"Bounds on Projection in Unital C*-Algebra",
"Canonical Preordering of C*-Algebra is Antisymmetric"
] |
proofwiki-5160 | Null Sets Closed under Countable Union | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\sequence {N_n}_{n \mathop \in \N}$ be a sequence of $\mu$-null sets.
Then $N := \ds \bigcup_{n \mathop \in \N} N_n$ is also a $\mu$-null set. | As $\mu$ is a measure:
:$\map \mu N \ge 0$
Also:
{{begin-eqn}}
{{eqn | l = \map \mu N
| o = \le
| r = \sum_{n \mathop \in \N} \map \mu {N_n}
| c = Measure is Countably Subadditive
}}
{{eqn | r = \sum_{n \mathop \in \N} 0
| c = as the $N_n$ are $\mu$-null sets
}}
{{eqn | r = 0
}}
{{end-eqn}}
Henc... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $\sequence {N_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Null Set|$\mu$-null sets]].
Then $N := \ds \bigcup_{n \mathop \in \N} N_n$ is also a [[Definition:Null Set|$\mu$-null set]]. | As $\mu$ is a [[Definition:Measure (Measure Theory)|measure]]:
:$\map \mu N \ge 0$
Also:
{{begin-eqn}}
{{eqn | l = \map \mu N
| o = \le
| r = \sum_{n \mathop \in \N} \map \mu {N_n}
| c = [[Measure is Countably Subadditive]]
}}
{{eqn | r = \sum_{n \mathop \in \N} 0
| c = as the $N_n$ are [[Defi... | Null Sets Closed under Countable Union | https://proofwiki.org/wiki/Null_Sets_Closed_under_Countable_Union | https://proofwiki.org/wiki/Null_Sets_Closed_under_Countable_Union | [
"Null Sets Closed under Countable Union",
"Null Sets",
"Measure Theory"
] | [
"Definition:Measure Space",
"Definition:Sequence",
"Definition:Null Set",
"Definition:Null Set"
] | [
"Definition:Measure (Measure Theory)",
"Measure is Countably Subadditive",
"Definition:Null Set",
"Definition:Null Set"
] |
proofwiki-5161 | Absolute Value induces Equivalence Compatible with Integer Multiplication | Let $\Z$ be the set of integers.
Let $\RR$ be the relation on $\Z$ defined as:
:$\forall x, y \in \Z: \struct {x, y} \in \RR \iff \size x = \size y$
where $\size x$ denotes the absolute value of $x$.
Then $\RR$ is a congruence relation for integer multiplication. | From Absolute Value Function on Integers induces Equivalence Relation, $\RR$ is an equivalence relation.
Let:
:$\size {x_1} = \size {x_2}$
:$\size {y_1} = \size {y_2}$
Then by Absolute Value Function is Completely Multiplicative:
{{begin-eqn}}
{{eqn | l = \size {x_1 y_1}
| r = \size {x_1} \size {y_1}
| c = ... | Let $\Z$ be the [[Definition:Integer|set of integers]].
Let $\RR$ be the [[Definition:Relation|relation]] on $\Z$ defined as:
:$\forall x, y \in \Z: \struct {x, y} \in \RR \iff \size x = \size y$
where $\size x$ denotes the [[Definition:Absolute Value|absolute value]] of $x$.
Then $\RR$ is a [[Definition:Congruence ... | From [[Absolute Value Function on Integers induces Equivalence Relation]], $\RR$ is an [[Definition:Equivalence Relation|equivalence relation]].
Let:
:$\size {x_1} = \size {x_2}$
:$\size {y_1} = \size {y_2}$
Then by [[Absolute Value Function is Completely Multiplicative]]:
{{begin-eqn}}
{{eqn | l = \size {x_1 y_1}
... | Absolute Value induces Equivalence Compatible with Integer Multiplication | https://proofwiki.org/wiki/Absolute_Value_induces_Equivalence_Compatible_with_Integer_Multiplication | https://proofwiki.org/wiki/Absolute_Value_induces_Equivalence_Compatible_with_Integer_Multiplication | [
"Examples of Congruence Relations",
"Integer Multiplication",
"Absolute Value Function"
] | [
"Definition:Integer",
"Definition:Relation",
"Definition:Absolute Value",
"Definition:Congruence Relation",
"Definition:Multiplication/Integers"
] | [
"Absolute Value Function on Integers induces Equivalence Relation",
"Definition:Equivalence Relation",
"Absolute Value Function is Completely Multiplicative",
"Definition:Congruence Relation",
"Definition:Multiplication/Integers"
] |
proofwiki-5162 | Completion Theorem (Measure Space) | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then there exists a completion $\struct {X, \Sigma^*, \bar \mu}$ of $\struct {X, \Sigma, \mu}$. | We give an explicit construction of $\struct {X, \Sigma^*, \bar \mu}$.
To this end, define $\NN$ to be the collection of subsets of $\mu$-null sets:
:$\NN := \set {N \subseteq X: \exists M \in \Sigma: \map \mu M = 0, N \subseteq M}$
Now, we define:
:$\Sigma^* := \set {E \cup N: E \in \Sigma, N \in \NN}$
and assert $\Si... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Then there exists a [[Definition:Completion (Measure Space)|completion]] $\struct {X, \Sigma^*, \bar \mu}$ of $\struct {X, \Sigma, \mu}$. | We give an explicit construction of $\struct {X, \Sigma^*, \bar \mu}$.
To this end, define $\NN$ to be the collection of [[Definition:Subset|subsets]] of [[Definition:Null Set|$\mu$-null sets]]:
:$\NN := \set {N \subseteq X: \exists M \in \Sigma: \map \mu M = 0, N \subseteq M}$
Now, we define:
:$\Sigma^* := \set {... | Completion Theorem (Measure Space) | https://proofwiki.org/wiki/Completion_Theorem_(Measure_Space) | https://proofwiki.org/wiki/Completion_Theorem_(Measure_Space) | [
"Completion Theorem (Measure Space)",
"Complete Measure Spaces",
"Completion Theorem"
] | [
"Definition:Measure Space",
"Definition:Completion (Measure Space)"
] | [
"Definition:Subset",
"Definition:Null Set",
"Definition:Sigma-Algebra",
"Measure of Empty Set is Zero",
"Union with Empty Set",
"De Morgan's Laws (Set Theory)/Set Difference/Difference with Union",
"Union of Relative Complements of Nested Subsets",
"Intersection Distributes over Union",
"Sigma-Algeb... |
proofwiki-5163 | Restricted Measure is Measure | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\Sigma'$ be a sub-$\sigma$-algebra of $\Sigma$.
Then the restricted measure $\mu \restriction_{\Sigma'}$ is a measure on the measurable space $\struct {X, \Sigma'}$. | Verify the axioms for a measure in turn for $\mu \restriction_{\Sigma'}$: | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $\Sigma'$ be a [[Definition:Sub-Sigma-Algebra|sub-$\sigma$-algebra]] of $\Sigma$.
Then the [[Definition:Restricted Measure|restricted measure]] $\mu \restriction_{\Sigma'}$ is a [[Definition:Measure (Measure Theory)|measure]] on the ... | Verify the axioms for a [[Definition:Measure (Measure Theory)|measure]] in turn for $\mu \restriction_{\Sigma'}$: | Restricted Measure is Measure | https://proofwiki.org/wiki/Restricted_Measure_is_Measure | https://proofwiki.org/wiki/Restricted_Measure_is_Measure | [
"Restricted Measures"
] | [
"Definition:Measure Space",
"Definition:Sub-Sigma-Algebra",
"Definition:Restricted Measure",
"Definition:Measure (Measure Theory)",
"Definition:Measurable Space"
] | [
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)"
] |
proofwiki-5164 | Restricting Measure Preserves Finiteness | Let $\left({X, \Sigma, \mu}\right)$ be a measure space.
Let $\mu$ be a finite measure.
Let $\Sigma'$ be a sub-$\sigma$-algebra of $\Sigma$.
Then the restricted measure $\mu \restriction_{\Sigma'}$ is also a finite measure. | By Restricted Measure is Measure, $\mu \restriction_{\Sigma'}$ is a measure.
Now by definition of $\mu \restriction_{\Sigma'}$, have:
:$\mu \restriction_{\Sigma'} \left({X}\right) = \mu \left({X}\right) < \infty$
as $\mu$ is a finite measure.
Hence $\mu \restriction_{\Sigma'}$ is also a finite measure.
{{qed}} | Let $\left({X, \Sigma, \mu}\right)$ be a [[Definition:Measure Space|measure space]].
Let $\mu$ be a [[Definition:Finite Measure|finite measure]].
Let $\Sigma'$ be a [[Definition:Sub-Sigma-Algebra|sub-$\sigma$-algebra]] of $\Sigma$.
Then the [[Definition:Restricted Measure|restricted measure]] $\mu \restriction_{\Si... | By [[Restricted Measure is Measure]], $\mu \restriction_{\Sigma'}$ is a [[Definition:Measure (Measure Theory)|measure]].
Now by [[Definition:Restricted Measure|definition of $\mu \restriction_{\Sigma'}$]], have:
:$\mu \restriction_{\Sigma'} \left({X}\right) = \mu \left({X}\right) < \infty$
as $\mu$ is a [[Definition... | Restricting Measure Preserves Finiteness | https://proofwiki.org/wiki/Restricting_Measure_Preserves_Finiteness | https://proofwiki.org/wiki/Restricting_Measure_Preserves_Finiteness | [
"Finite Measures",
"Measure Theory",
"Restricted Measures",
"Restricted Measures",
"Finite Measures"
] | [
"Definition:Measure Space",
"Definition:Finite Measure",
"Definition:Sub-Sigma-Algebra",
"Definition:Restricted Measure",
"Definition:Finite Measure"
] | [
"Restricted Measure is Measure",
"Definition:Measure (Measure Theory)",
"Definition:Restricted Measure",
"Definition:Finite Measure",
"Definition:Finite Measure"
] |
proofwiki-5165 | Measure Space has Exhausting Sequence of Finite Measure iff Cover by Sets of Finite Measure | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then there exists an exhausting sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ such that:
:$\forall n \in \N: \map \mu {E_n} < \infty$
{{iff}} there exists a sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ such that:
:$(1): \quad \ds \bigcup_{n \ma... | === Necessary Condition ===
Suppose that there exists an exhausting sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ such that:
::$\forall n \in \N: \map \mu {E_n} < \infty$
Let $\sequence {F_n}_{n \mathop \in \N}$ be an exhausting sequence in $\Sigma$ such that:
:$\forall n \in \N: \map \mu {F_n} < +\infty$
T... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Then there exists an [[Definition:Exhausting Sequence of Sets|exhausting sequence]] $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ such that:
:$\forall n \in \N: \map \mu {E_n} < \infty$
{{iff}} there exists a [[Definition:Sequence|se... | === Necessary Condition ===
Suppose that there exists an [[Definition:Exhausting Sequence of Sets|exhausting sequence]] $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ such that:
::$\forall n \in \N: \map \mu {E_n} < \infty$
Let $\sequence {F_n}_{n \mathop \in \N}$ be an [[Definition:Exhausting Sequence of Sets|exh... | Measure Space has Exhausting Sequence of Finite Measure iff Cover by Sets of Finite Measure | https://proofwiki.org/wiki/Measure_Space_has_Exhausting_Sequence_of_Finite_Measure_iff_Cover_by_Sets_of_Finite_Measure | https://proofwiki.org/wiki/Measure_Space_has_Exhausting_Sequence_of_Finite_Measure_iff_Cover_by_Sets_of_Finite_Measure | [
"Sigma-Finite Measures"
] | [
"Definition:Measure Space",
"Definition:Exhausting Sequence of Sets",
"Definition:Sequence"
] | [
"Definition:Exhausting Sequence of Sets",
"Definition:Exhausting Sequence of Sets",
"Definition:Exhausting Sequence of Sets",
"Definition:Sequence",
"Definition:Sequence",
"Definition:Exhausting Sequence of Sets",
"Definition:Exhausting Sequence of Sets"
] |
proofwiki-5166 | Central Subgroup is Normal | Let $G$ be a group.
Let $H$ be a central subgroup of $G$.
Then $H$ is a normal subgroup of $G$. | Let $H$ be a central subgroup of $G$.
By definition of central subgroup:
:$H \subseteq \map Z G$
where $\map Z G$ is the center of $G$.
Thus we have that $H$ is a group which is a subset of $\map Z G$.
Therefore by definition $H$ is a subgroup of $\map Z G$.
We also have from Center of Group is Abelian Subgroup that $\... | Let $G$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Central Subgroup|central subgroup]] of $G$.
Then $H$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$. | Let $H$ be a [[Definition:Central Subgroup|central subgroup]] of $G$.
By definition of [[Definition:Central Subgroup|central subgroup]]:
:$H \subseteq \map Z G$
where $\map Z G$ is the [[Definition:Center of Group|center]] of $G$.
Thus we have that $H$ is a [[Definition:Group|group]] which is a [[Definition:Subset|s... | Central Subgroup is Normal/Proof 1 | https://proofwiki.org/wiki/Central_Subgroup_is_Normal | https://proofwiki.org/wiki/Central_Subgroup_is_Normal/Proof_1 | [
"Normal Subgroups",
"Central Subgroups",
"Central Subgroup is Normal"
] | [
"Definition:Group",
"Definition:Central Subgroup",
"Definition:Normal Subgroup"
] | [
"Definition:Central Subgroup",
"Definition:Central Subgroup",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Group",
"Definition:Subset",
"Definition:Subgroup",
"Center of Group is Abelian Subgroup",
"Definition:Abelian Group",
"Subgroup of Abelian Group is Normal",
"Definition:Normal S... |
proofwiki-5167 | Central Subgroup is Normal | Let $G$ be a group.
Let $H$ be a central subgroup of $G$.
Then $H$ is a normal subgroup of $G$. | Let $H$ be a central subgroup of $G$.
By definition of central subgroup:
:$H \subseteq \map Z G$
where $\map Z G$ is the center of $G$.
Then:
{{begin-eqn}}
{{eqn | q = \forall x \in G: \forall h \in H
| l = x h x^{-1}
| r = x x^{-1} h
| c = as $h \in H \implies h \in \map Z G$
}}
{{eqn | r = h
|... | Let $G$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Central Subgroup|central subgroup]] of $G$.
Then $H$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$. | Let $H$ be a [[Definition:Central Subgroup|central subgroup]] of $G$.
By definition of [[Definition:Central Subgroup|central subgroup]]:
:$H \subseteq \map Z G$
where $\map Z G$ is the [[Definition:Center of Group|center]] of $G$.
Then:
{{begin-eqn}}
{{eqn | q = \forall x \in G: \forall h \in H
| l = x h x^{-... | Central Subgroup is Normal/Proof 2 | https://proofwiki.org/wiki/Central_Subgroup_is_Normal | https://proofwiki.org/wiki/Central_Subgroup_is_Normal/Proof_2 | [
"Normal Subgroups",
"Central Subgroups",
"Central Subgroup is Normal"
] | [
"Definition:Group",
"Definition:Central Subgroup",
"Definition:Normal Subgroup"
] | [
"Definition:Central Subgroup",
"Definition:Central Subgroup",
"Definition:Center (Abstract Algebra)/Group"
] |
proofwiki-5168 | Angle Between Vectors in Terms of Dot Product | The angle between two non-zero vector quantities can be calculated by:
:$\theta = \arccos \dfrac {\mathbf v \cdot \mathbf w} {\norm {\mathbf v} \norm {\mathbf w} }$
where:
:$\mathbf v \cdot \mathbf w$ represents the dot product of $\mathbf v$ and $\mathbf w$
:$\norm {\, \cdot \,}$ represents vector length
:$\arccos$ r... | {{begin-eqn}}
{{eqn | l = \norm {\mathbf v} \norm {\mathbf w} \cos \theta
| r = \mathbf v \cdot \mathbf w
| c = Cosine Formula for Dot Product
}}
{{eqn | l = \cos \theta
| r = \frac {\mathbf v \cdot \mathbf w} {\norm {\mathbf v} \norm {\mathbf w} }
| c = because $\mathbf v, \mathbf w \ne \mathbf... | The [[Definition:Angle between Vectors|angle between]] two non-[[Definition:Zero Vector|zero]] [[Definition:Vector Quantity|vector quantities]] can be calculated by:
:$\theta = \arccos \dfrac {\mathbf v \cdot \mathbf w} {\norm {\mathbf v} \norm {\mathbf w} }$
where:
:$\mathbf v \cdot \mathbf w$ represents the [[Defini... | {{begin-eqn}}
{{eqn | l = \norm {\mathbf v} \norm {\mathbf w} \cos \theta
| r = \mathbf v \cdot \mathbf w
| c = [[Cosine Formula for Dot Product]]
}}
{{eqn | l = \cos \theta
| r = \frac {\mathbf v \cdot \mathbf w} {\norm {\mathbf v} \norm {\mathbf w} }
| c = because $\mathbf v, \mathbf w \ne \ma... | Angle Between Vectors in Terms of Dot Product | https://proofwiki.org/wiki/Angle_Between_Vectors_in_Terms_of_Dot_Product | https://proofwiki.org/wiki/Angle_Between_Vectors_in_Terms_of_Dot_Product | [
"Angle Between Vectors in Terms of Dot Product",
"Angles between Vectors",
"Dot Product",
"Vector Algebra"
] | [
"Definition:Angle between Vectors",
"Definition:Zero Vector",
"Definition:Vector Quantity",
"Definition:Dot Product",
"Definition:Vector Length",
"Definition:Inverse Cosine/Real/Arccosine"
] | [
"Cosine Formula for Dot Product",
"Definition:Angle between Vectors",
"Composite of Bijection with Inverse is Identity Mapping"
] |
proofwiki-5169 | Angle Between Non-Zero Vectors Always Defined | Let $\mathbf v$ and $\mathbf w$ be non-zero vectors in real Euclidean space $\R^n$.
Then the angle between $\mathbf v$ and $\mathbf w$ is always defined. | === Case 1 ===
Suppose that $\mathbf v$ and $\mathbf w$ are not scalar multiples of each other.
:400px
From Construction of Triangle from Given Lengths, it is sufficient to show that sum of the lengths of any two sides is greater than the third side.
Consider the side with length $\norm {\mathbf v}$.
From the triangle ... | Let $\mathbf v$ and $\mathbf w$ be non-[[Definition:Zero Vector|zero]] [[Definition:Vector (Real Euclidean Space)|vectors]] in [[Definition:Real Euclidean Space|real Euclidean space]] $\R^n$.
Then the [[Definition:Angle Between Vectors|angle between]] $\mathbf v$ and $\mathbf w$ is always defined. | === Case 1 ===
Suppose that $\mathbf v$ and $\mathbf w$ are not [[Definition:Scalar Multiplication on Vector Space|scalar multiples]] of each other.
:[[File:AngleBetweenTwoVectors.png|400px]]
From [[Construction of Triangle from Given Lengths]], it is [[Definition:Sufficient Condition|sufficient]] to show that sum o... | Angle Between Non-Zero Vectors Always Defined | https://proofwiki.org/wiki/Angle_Between_Non-Zero_Vectors_Always_Defined | https://proofwiki.org/wiki/Angle_Between_Non-Zero_Vectors_Always_Defined | [
"Angles between Vectors"
] | [
"Definition:Zero Vector",
"Definition:Vector/Real Euclidean Space",
"Definition:Euclidean Space/Real",
"Definition:Angle between Vectors"
] | [
"Definition:Scalar Multiplication/Vector Space",
"File:AngleBetweenTwoVectors.png",
"Construction of Triangle from Given Lengths",
"Definition:Conditional/Sufficient Condition",
"Triangle Inequality",
"Definition:Inequality/Strict",
"Triangle Inequality",
"Definition:Scalar Multiplication/Vector Space... |
proofwiki-5170 | Dynkin System Contains Empty Set | Let $X$ be a set, and let $\DD$ be a Dynkin system on $X$.
Then the empty set $\O$ is an element of $\DD$. | As $\DD$ is a Dynkin system, $X \in \DD$.
By Set Difference with Self is Empty Set, $X \setminus X = \O$.
Hence, by property $(2)$ of a Dynkin system, $\O = X \setminus X \in \DD$.
{{qed}} | Let $X$ be a [[Definition:Set|set]], and let $\DD$ be a [[Definition:Dynkin System|Dynkin system]] on $X$.
Then the [[Definition:Empty Set|empty set]] $\O$ is an element of $\DD$. | As $\DD$ is a [[Definition:Dynkin System|Dynkin system]], $X \in \DD$.
By [[Set Difference with Self is Empty Set]], $X \setminus X = \O$.
Hence, by property $(2)$ of a [[Definition:Dynkin System|Dynkin system]], $\O = X \setminus X \in \DD$.
{{qed}} | Dynkin System Contains Empty Set | https://proofwiki.org/wiki/Dynkin_System_Contains_Empty_Set | https://proofwiki.org/wiki/Dynkin_System_Contains_Empty_Set | [
"Dynkin Systems",
"Empty Set"
] | [
"Definition:Set",
"Definition:Dynkin System",
"Definition:Empty Set"
] | [
"Definition:Dynkin System",
"Set Difference with Self is Empty Set",
"Definition:Dynkin System"
] |
proofwiki-5171 | Dynkin System Closed under Disjoint Union | Let $X$ be a set, and let $\DD$ be a Dynkin system on $X$.
Let $D, E \in \DD$ be disjoint.
Then the union $D \cup E$ is also an element of $\DD$. | Define $D_1 = D, D_2 = E$, and for $n \ge 2$, $D_n = \O$.
Then by Dynkin System Contains Empty Set:
:$\forall n \in \N: D_n \in \DD$
Also, by Intersection with Empty Set, it follows that $\sequence {D_n}_{n \mathop \in \N}$ is a pairwise disjoint sequence.
Hence, by property $(3)$ of a Dynkin system:
:$\ds D \cup E = \... | Let $X$ be a [[Definition:Set|set]], and let $\DD$ be a [[Definition:Dynkin System|Dynkin system]] on $X$.
Let $D, E \in \DD$ be [[Definition:Disjoint Sets|disjoint]].
Then the [[Definition:Set Union|union]] $D \cup E$ is also an element of $\DD$. | Define $D_1 = D, D_2 = E$, and for $n \ge 2$, $D_n = \O$.
Then by [[Dynkin System Contains Empty Set]]:
:$\forall n \in \N: D_n \in \DD$
Also, by [[Intersection with Empty Set]], it follows that $\sequence {D_n}_{n \mathop \in \N}$ is a [[Definition:Pairwise Disjoint|pairwise disjoint]] [[Definition:Sequence|sequenc... | Dynkin System Closed under Disjoint Union | https://proofwiki.org/wiki/Dynkin_System_Closed_under_Disjoint_Union | https://proofwiki.org/wiki/Dynkin_System_Closed_under_Disjoint_Union | [
"Dynkin Systems"
] | [
"Definition:Set",
"Definition:Dynkin System",
"Definition:Disjoint Sets",
"Definition:Set Union"
] | [
"Dynkin System Contains Empty Set",
"Intersection with Empty Set",
"Definition:Pairwise Disjoint",
"Definition:Sequence",
"Definition:Dynkin System"
] |
proofwiki-5172 | Sigma-Algebra is Dynkin System | Let $X$ be a set, and let $\Sigma$ be a $\sigma$-algebra on $X$.
Then $\Sigma$ is a Dynkin system on $X$. | The axioms $(1)$ and $(2)$ for both $\sigma$-algebras and Dynkin systems are identical.
Dynkin system axiom $(3)$ is seen to be a specification of $\sigma$-algebra axiom $(3)$ to pairwise disjoint sequences.
Hence $\Sigma$ is trivially a Dynkin system on $X$.
{{qed}} | Let $X$ be a [[Definition:Set|set]], and let $\Sigma$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $X$.
Then $\Sigma$ is a [[Definition:Dynkin System|Dynkin system]] on $X$. | The axioms $(1)$ and $(2)$ for both [[Definition:Sigma-Algebra|$\sigma$-algebras]] and [[Definition:Dynkin System|Dynkin systems]] are identical.
[[Definition:Dynkin System|Dynkin system]] axiom $(3)$ is seen to be a specification of [[Definition:Sigma-Algebra|$\sigma$-algebra]] axiom $(3)$ to [[Definition:Pairwise Di... | Sigma-Algebra is Dynkin System | https://proofwiki.org/wiki/Sigma-Algebra_is_Dynkin_System | https://proofwiki.org/wiki/Sigma-Algebra_is_Dynkin_System | [
"Sigma-Algebras",
"Dynkin Systems"
] | [
"Definition:Set",
"Definition:Sigma-Algebra",
"Definition:Dynkin System"
] | [
"Definition:Sigma-Algebra",
"Definition:Dynkin System",
"Definition:Dynkin System",
"Definition:Sigma-Algebra",
"Definition:Pairwise Disjoint",
"Definition:Sequence",
"Definition:Dynkin System"
] |
proofwiki-5173 | Existence and Uniqueness of Dynkin System Generated by Collection of Subsets | Let $X$ be a set.
Let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Then $\map \delta \GG$, the Dynkin system generated by $\GG$, exists and is unique. | === Existence ===
By Power Set is Dynkin System, there exists at least one Dynkin system containing $\GG$.
Next, let $\Bbb D$ be the collection of Dynkin systems containing $\GG$:
:$\Bbb D := \set {\DD': \GG \subseteq \DD', \text{$\DD'$ is a Dynkin system} }$
By Intersection of Dynkin Systems is Dynkin System, $\DD := ... | Let $X$ be a [[Definition:Set|set]].
Let $\GG \subseteq \powerset X$ be a collection of [[Definition:Subset|subsets]] of $X$.
Then $\map \delta \GG$, the [[Definition:Dynkin System Generated by Collection of Subsets|Dynkin system generated by $\GG$]], exists and is unique. | === Existence ===
By [[Power Set is Dynkin System]], there exists at least one [[Definition:Dynkin System|Dynkin system]] containing $\GG$.
Next, let $\Bbb D$ be the collection of [[Definition:Dynkin System|Dynkin systems]] containing $\GG$:
:$\Bbb D := \set {\DD': \GG \subseteq \DD', \text{$\DD'$ is a Dynkin system... | Existence and Uniqueness of Dynkin System Generated by Collection of Subsets | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Dynkin_System_Generated_by_Collection_of_Subsets | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Dynkin_System_Generated_by_Collection_of_Subsets | [
"Dynkin Systems"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Dynkin System Generated by Collection of Subsets"
] | [
"Power Set is Dynkin System",
"Definition:Dynkin System",
"Definition:Dynkin System",
"Intersection of Dynkin Systems is Dynkin System",
"Definition:Dynkin System",
"Set Intersection Preserves Subsets/Families of Sets",
"Definition:Dynkin System",
"Intersection is Subset/General Result",
"Definition... |
proofwiki-5174 | Generated Sigma-Algebra Contains Generated Dynkin System | Let $X$ be a set.
Let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Then $\map \delta \GG \subseteq \map \sigma \GG$.
Here $\delta$ denotes generated Dynkin system, and $\sigma$ denotes generated $\sigma$-algebra. | By Sigma-Algebra is Dynkin System, $\map \sigma \GG$ is a Dynkin system.
The definition of $\map \delta \GG$ now ensures that $\map \delta \GG \subseteq \map \sigma \GG$.
{{qed}} | Let $X$ be a [[Definition:Set|set]].
Let $\GG \subseteq \powerset X$ be a collection of [[Definition:Subset|subsets]] of $X$.
Then $\map \delta \GG \subseteq \map \sigma \GG$.
Here $\delta$ denotes [[Definition:Generated Dynkin System|generated Dynkin system]], and $\sigma$ denotes [[Definition:Sigma-Algebra Genera... | By [[Sigma-Algebra is Dynkin System]], $\map \sigma \GG$ is a [[Definition:Dynkin System|Dynkin system]].
The [[Definition:Generated Dynkin System|definition of $\map \delta \GG$]] now ensures that $\map \delta \GG \subseteq \map \sigma \GG$.
{{qed}} | Generated Sigma-Algebra Contains Generated Dynkin System | https://proofwiki.org/wiki/Generated_Sigma-Algebra_Contains_Generated_Dynkin_System | https://proofwiki.org/wiki/Generated_Sigma-Algebra_Contains_Generated_Dynkin_System | [
"Sigma-Algebras",
"Sigma-Algebras Generated by Collection of Subsets",
"Dynkin Systems",
"Sigma-Algebras Generated by Collection of Subsets"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Dynkin System Generated by Collection of Subsets",
"Definition:Sigma-Algebra Generated by Collection of Subsets"
] | [
"Sigma-Algebra is Dynkin System",
"Definition:Dynkin System",
"Definition:Dynkin System Generated by Collection of Subsets"
] |
proofwiki-5175 | Dynkin System Closed under Intersections is Sigma-Algebra | Let $X$ be a set, and let $\DD$ be a Dynkin system on $X$.
Suppose that $\DD$ satisfies the following condition:
:$(1):\quad \forall D, E \in \DD: D \cap E \in \DD$
That is, $\DD$ is closed under intersection.
Then $\DD$ is a $\sigma$-algebra. | The first two conditions for a Dynkin system are identical to those for a $\sigma$-algebra.
Hence it is only required to verify that $(1)$ implies that $\DD$ is closed under arbitrary countable unions.
So let $\sequence {D_n}_{n \mathop \in \N}$ be a sequence in $\DD$.
Now define the sequence $\sequence {E_n}_{n \matho... | Let $X$ be a [[Definition:Set|set]], and let $\DD$ be a [[Definition:Dynkin System|Dynkin system]] on $X$.
Suppose that $\DD$ satisfies the following condition:
:$(1):\quad \forall D, E \in \DD: D \cap E \in \DD$
That is, $\DD$ is closed under [[Definition:Set Intersection|intersection]].
Then $\DD$ is a [[Definit... | The first two conditions for a [[Definition:Dynkin System|Dynkin system]] are identical to those for a [[Definition:Sigma-Algebra|$\sigma$-algebra]].
Hence it is only required to verify that $(1)$ implies that $\DD$ is closed under arbitrary [[Definition:Countable Union|countable unions]].
So let $\sequence {D_n}_{n... | Dynkin System Closed under Intersections is Sigma-Algebra | https://proofwiki.org/wiki/Dynkin_System_Closed_under_Intersections_is_Sigma-Algebra | https://proofwiki.org/wiki/Dynkin_System_Closed_under_Intersections_is_Sigma-Algebra | [
"Dynkin Systems",
"Sigma-Algebras"
] | [
"Definition:Set",
"Definition:Dynkin System",
"Definition:Set Intersection",
"Definition:Sigma-Algebra"
] | [
"Definition:Dynkin System",
"Definition:Sigma-Algebra",
"Definition:Set Union/Countable Union",
"Definition:Sequence",
"Definition:Sequence",
"Dynkin System Closed under Union",
"Definition:Set Union/Countable Union",
"Definition:Sigma-Algebra"
] |
proofwiki-5176 | Dynkin System with Generator Closed under Intersection is Sigma-Algebra | Let $X$ be a set.
Let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Suppose that $\GG$ satisfies the following condition:
:$(1):\quad \forall G, H \in \GG: G \cap H \in \GG$
That is, $\GG$ is closed under intersection.
Then:
:$\map \delta \GG = \map \sigma \GG$
where $\delta$ denotes generated Dynkin s... | From Sigma-Algebra is Dynkin System and the definition of generated Dynkin system, it follows that:
:$\map \delta \GG \subseteq \map \sigma \GG$
Let $D \in \map \delta \GG$, and define:
:$\delta_D := \set {E \subseteq X: E \cap D \in \map \delta \GG}$
Let us verify that these $\delta_D$ form Dynkin systems.
First of al... | Let $X$ be a [[Definition:Set|set]].
Let $\GG \subseteq \powerset X$ be a collection of [[Definition:Subset|subsets]] of $X$.
Suppose that $\GG$ satisfies the following condition:
:$(1):\quad \forall G, H \in \GG: G \cap H \in \GG$
That is, $\GG$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definitio... | From [[Sigma-Algebra is Dynkin System]] and the definition of [[Definition:Generated Dynkin System|generated Dynkin system]], it follows that:
:$\map \delta \GG \subseteq \map \sigma \GG$
Let $D \in \map \delta \GG$, and define:
:$\delta_D := \set {E \subseteq X: E \cap D \in \map \delta \GG}$
Let us verify that t... | Dynkin System with Generator Closed under Intersection is Sigma-Algebra | https://proofwiki.org/wiki/Dynkin_System_with_Generator_Closed_under_Intersection_is_Sigma-Algebra | https://proofwiki.org/wiki/Dynkin_System_with_Generator_Closed_under_Intersection_is_Sigma-Algebra | [
"Dynkin Systems",
"Sigma-Algebras"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Set Intersection",
"Definition:Dynkin System Generated by Collection of Subsets",
"Definition:Sigma-Algebra Generated by Collection of Subsets"
] | [
"Sigma-Algebra is Dynkin System",
"Definition:Dynkin System Generated by Collection of Subsets",
"Definition:Dynkin System",
"Set Difference Intersection with Second Set is Empty Set",
"Union with Empty Set",
"Intersection Distributes over Union",
"De Morgan's Laws (Set Theory)/Set Difference/Difference... |
proofwiki-5177 | Carathéodory's Theorem (Measure Theory) | Let $X$ be a set.
Let $\SS \subseteq \powerset X$ be a semi-ring of subsets of $X$.
Let $\mu: \SS \to \overline \R$ be a pre-measure on $\SS$.
Let $\map \sigma \SS$ be the $\sigma$-algebra generated by $\SS$.
Then $\mu$ extends to a measure $\mu^*$ on $\map \sigma \SS$. | {{ProofWanted}}
{{Namedfor|Constantin Carathéodory|cat = Carathéodory}} | Let $X$ be a [[Definition:Set|set]].
Let $\SS \subseteq \powerset X$ be a [[Definition:Semiring of Sets|semi-ring]] of [[Definition:Subset|subsets]] of $X$.
Let $\mu: \SS \to \overline \R$ be a [[Definition:Pre-Measure|pre-measure]] on $\SS$.
Let $\map \sigma \SS$ be the [[Definition:Sigma-Algebra Generated by Colle... | {{ProofWanted}}
{{Namedfor|Constantin Carathéodory|cat = Carathéodory}} | Carathéodory's Theorem (Measure Theory) | https://proofwiki.org/wiki/Carathéodory's_Theorem_(Measure_Theory) | https://proofwiki.org/wiki/Carathéodory's_Theorem_(Measure_Theory) | [
"Measure Theory"
] | [
"Definition:Set",
"Definition:Semiring of Sets",
"Definition:Subset",
"Definition:Pre-Measure",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Extension (Measure Theory)",
"Definition:Measure (Measure Theory)"
] | [] |
proofwiki-5178 | Congruence Modulo Normal Subgroup is Congruence Relation | Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Then congruence modulo $N$ is a congruence relation for the group operation $\circ$. | Let $x \mathrel {\RR_N} y$ denote that $x$ and $y$ are in the same coset, that is:
:$x \mathrel {\RR_N} y \iff x \circ N = y \circ N$
as specified in the definition of congruence modulo $N$.
Let $x \mathrel {\RR_N} x'$ and $y \mathrel {\RR_N} y'$.
To demonstrate that $\RR_N$ is a congruence relation for $\circ$, we nee... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Then [[Definition:Congruence Modulo Subgroup|congruence modulo $N$]] is a [[Definition:Congruence Relation|congruence relation]] for the [[Definition:Group Operation|group operation]] $\circ$... | Let $x \mathrel {\RR_N} y$ denote that $x$ and $y$ are in the same [[Definition:Coset|coset]], that is:
:$x \mathrel {\RR_N} y \iff x \circ N = y \circ N$
as specified in the definition of [[Definition:Congruence Modulo Subgroup|congruence modulo $N$]].
Let $x \mathrel {\RR_N} x'$ and $y \mathrel {\RR_N} y'$.
To dem... | Congruence Modulo Normal Subgroup is Congruence Relation | https://proofwiki.org/wiki/Congruence_Modulo_Normal_Subgroup_is_Congruence_Relation | https://proofwiki.org/wiki/Congruence_Modulo_Normal_Subgroup_is_Congruence_Relation | [
"Normal Subgroups",
"Examples of Congruence Relations"
] | [
"Definition:Group",
"Definition:Normal Subgroup",
"Definition:Congruence Modulo Subgroup",
"Definition:Congruence Relation",
"Definition:Group Product/Group Law"
] | [
"Definition:Coset",
"Definition:Congruence Modulo Subgroup",
"Definition:Congruence Relation",
"Inverse of Group Product",
"Cosets are Equal iff Product with Inverse in Subgroup",
"Product of Subgroup with Itself",
"Definition:Congruence Relation"
] |
proofwiki-5179 | Intersection Measure is Measure | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $F \in \Sigma$ be a measurable set.
Then the intersection measure $\mu_F$ is a measure on the measurable space $\struct {X, \Sigma}$. | Verify the axioms for a measure in turn for $\mu_F$: | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $F \in \Sigma$ be a [[Definition:Measurable Set|measurable set]].
Then the [[Definition:Intersection Measure|intersection measure]] $\mu_F$ is a [[Definition:Measure (Measure Theory)|measure]] on the [[Definition:Measurable Space|mea... | Verify the axioms for a [[Definition:Measure (Measure Theory)|measure]] in turn for $\mu_F$: | Intersection Measure is Measure | https://proofwiki.org/wiki/Intersection_Measure_is_Measure | https://proofwiki.org/wiki/Intersection_Measure_is_Measure | [
"Intersection Measures",
"Measures"
] | [
"Definition:Measure Space",
"Definition:Measurable Set",
"Definition:Intersection Measure",
"Definition:Measure (Measure Theory)",
"Definition:Measurable Space"
] | [
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)"
] |
proofwiki-5180 | Lebesgue Measure is Invariant under Translations | Let $\lambda^n$ be the $n$-dimensional Lebesgue measure on $\R^n$ equipped with the Borel $\sigma$-algebra $\map \BB {\R^n}$.
Let $\mathbf x \in \R^n$.
Then $\lambda^n$ is translation invariant; that is, for all $B \in \map \BB {\R^n}$, have:
:$\map {\lambda^n} {\mathbf x + B} = \map {\lambda^n} B$
where $\mathbf x + B... | Denote with $\tau_{\mathbf x}: \R^n \to \R^n$ the translation by $\mathbf x$.
From Translation in Euclidean Space is Measurable Mapping, $\tau_{\mathbf x}$ is $\map \BB {\R^n} \, / \, \map \BB {\R^n}$-measurable.
Consider the pushforward measure $\lambda^n_{\mathbf x} := \paren {\tau_{\mathbf x} }_* \lambda^n$ on $\map... | Let $\lambda^n$ be the $n$-dimensional [[Definition:Lebesgue Measure|Lebesgue measure]] on $\R^n$ equipped with the [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]] $\map \BB {\R^n}$.
Let $\mathbf x \in \R^n$.
Then $\lambda^n$ is [[Definition:Translation Invariant Measure|translation invariant]]; that is, f... | Denote with $\tau_{\mathbf x}: \R^n \to \R^n$ the [[Definition:Translation in Euclidean Space|translation by $\mathbf x$]].
From [[Translation in Euclidean Space is Measurable Mapping]], $\tau_{\mathbf x}$ is [[Definition:Measurable Mapping|$\map \BB {\R^n} \, / \, \map \BB {\R^n}$-measurable]].
Consider the [[Defini... | Lebesgue Measure is Invariant under Translations | https://proofwiki.org/wiki/Lebesgue_Measure_is_Invariant_under_Translations | https://proofwiki.org/wiki/Lebesgue_Measure_is_Invariant_under_Translations | [
"Lebesgue Measure",
"Translation Mappings"
] | [
"Definition:Lebesgue Measure",
"Definition:Borel Sigma-Algebra",
"Definition:Translation Invariant Measure"
] | [
"Definition:Translation Mapping/Euclidean Space",
"Translation in Euclidean Space is Measurable Mapping",
"Definition:Measurable Mapping",
"Definition:Pushforward Measure",
"Characterization of Euclidean Borel Sigma-Algebra",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Hal... |
proofwiki-5181 | Translation Invariant Measure on Euclidean Space is Multiple of Lebesgue Measure | Let $\mu$ be a measure on $\R^n$ equipped with the Borel $\sigma$-algebra $\map \BB {\R^n}$.
Suppose that $\mu$ is translation invariant.
Also, suppose that $\kappa := \map \mu {\hointr 0 1^n} < +\infty$.
Then $\mu = \kappa \lambda^n$, where $\lambda^n$ is the $n$-dimensional Lebesgue measure. | From Characterization of Euclidean Borel Sigma-Algebra, we have:
:$\map \BB {\R^n} = \map \sigma {\JJ^n_{ho, \text {rat} } }$
where $\JJ^n_{ho, \text {rat} }$ denotes the collection of half-open $n$-rectangles with rational endpoints.
So let $J = \horectr {\mathbf a} {\mathbf b} \in \JJ^n_{ho, \text {rat} }$.
Let $M \i... | Let $\mu$ be a [[Definition:Measure (Measure Theory)|measure]] on $\R^n$ equipped with the [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]] $\map \BB {\R^n}$.
Suppose that $\mu$ is [[Definition:Translation Invariant Measure|translation invariant]].
Also, suppose that $\kappa := \map \mu {\hointr 0 1^n} < +\i... | From [[Characterization of Euclidean Borel Sigma-Algebra]], we have:
:$\map \BB {\R^n} = \map \sigma {\JJ^n_{ho, \text {rat} } }$
where $\JJ^n_{ho, \text {rat} }$ denotes the collection of [[Definition:Half-Open Rectangle|half-open $n$-rectangles]] with [[Definition:Rational Number|rational]] endpoints.
So let $J =... | Translation Invariant Measure on Euclidean Space is Multiple of Lebesgue Measure | https://proofwiki.org/wiki/Translation_Invariant_Measure_on_Euclidean_Space_is_Multiple_of_Lebesgue_Measure | https://proofwiki.org/wiki/Translation_Invariant_Measure_on_Euclidean_Space_is_Multiple_of_Lebesgue_Measure | [
"Measure Theory"
] | [
"Definition:Measure (Measure Theory)",
"Definition:Borel Sigma-Algebra",
"Definition:Translation Invariant Measure",
"Definition:Lebesgue Measure"
] | [
"Characterization of Euclidean Borel Sigma-Algebra",
"Definition:Half-Open Rectangle",
"Definition:Rational Number",
"Definition:Rational Number",
"Definition:Pairwise Disjoint",
"Definition:Half-Open Rectangle",
"Definition:Translation Invariant Measure",
"Lebesgue Measure is Invariant under Translat... |
proofwiki-5182 | Dynkin System Closed under Set Difference with Subset | Let $X$ be a set.
Let $\DD$ be a Dynkin system on $X$.
Let $D, E \in \DD$ and suppose that $E \subseteq D$.
Then the set difference $D \setminus E$ is also an element of $\DD$. | For brevity, write for example $E^\complement$ for $\relcomp X E = X \setminus E$.
We reason as follows:
{{begin-eqn}}
{{eqn | l = D \setminus E
| r = D \cap E^\complement
| c = Set Difference as Intersection with Relative Complement
}}
{{eqn | r = \paren {D^\complement \cup E}^\complement
| c = De Mo... | Let $X$ be a [[Definition:Set|set]].
Let $\DD$ be a [[Definition:Dynkin System|Dynkin system]] on $X$.
Let $D, E \in \DD$ and suppose that $E \subseteq D$.
Then the [[Definition:Set Difference|set difference]] $D \setminus E$ is also an element of $\DD$. | For brevity, write for example $E^\complement$ for $\relcomp X E = X \setminus E$.
We reason as follows:
{{begin-eqn}}
{{eqn | l = D \setminus E
| r = D \cap E^\complement
| c = [[Set Difference as Intersection with Relative Complement]]
}}
{{eqn | r = \paren {D^\complement \cup E}^\complement
| c =... | Dynkin System Closed under Set Difference with Subset | https://proofwiki.org/wiki/Dynkin_System_Closed_under_Set_Difference_with_Subset | https://proofwiki.org/wiki/Dynkin_System_Closed_under_Set_Difference_with_Subset | [
"Dynkin Systems"
] | [
"Definition:Set",
"Definition:Dynkin System",
"Definition:Set Difference"
] | [
"Set Difference as Intersection with Relative Complement",
"De Morgan's Laws (Set Theory)/Relative Complement/Complement of Union",
"Relative Complement of Relative Complement",
"Definition:Dynkin System",
"Dynkin System Closed under Disjoint Union"
] |
proofwiki-5183 | Open Rectangles Closed under Intersection | Let $\paren {\openint {\mathbf a} {\mathbf b} }$ and $\paren {\openint {\mathbf c} {\mathbf d} }$ be open $n$-rectangles.
Then $\paren {\openint {\mathbf a} {\mathbf b} } \cap \paren {\openint {\mathbf c} {\mathbf d} }$ is also an open $n$-rectangle. | From Cartesian Product of Intersections: General Case, we have:
:$\ds \paren {\openint {\mathbf a} {\mathbf b} } \cap \paren {\openint {\mathbf c} {\mathbf d} } = \prod_{i \mathop = 1}^n \openint {a_i} {b_i} \cap \openint {c_i} {d_i}$
Therefore, it suffices to show that the intersection of two open intervals is again a... | Let $\paren {\openint {\mathbf a} {\mathbf b} }$ and $\paren {\openint {\mathbf c} {\mathbf d} }$ be [[Definition:Open Rectangle|open $n$-rectangles]].
Then $\paren {\openint {\mathbf a} {\mathbf b} } \cap \paren {\openint {\mathbf c} {\mathbf d} }$ is also an [[Definition:Open Rectangle|open $n$-rectangle]]. | From [[Cartesian Product of Intersections/General Case|Cartesian Product of Intersections: General Case]], we have:
:$\ds \paren {\openint {\mathbf a} {\mathbf b} } \cap \paren {\openint {\mathbf c} {\mathbf d} } = \prod_{i \mathop = 1}^n \openint {a_i} {b_i} \cap \openint {c_i} {d_i}$
Therefore, it suffices to show... | Open Rectangles Closed under Intersection | https://proofwiki.org/wiki/Open_Rectangles_Closed_under_Intersection | https://proofwiki.org/wiki/Open_Rectangles_Closed_under_Intersection | [
"Analysis"
] | [
"Definition:Open Rectangle",
"Definition:Open Rectangle"
] | [
"Cartesian Product of Intersections/General Case",
"Definition:Set Intersection",
"Definition:Real Interval/Open",
"Definition:Real Interval/Open",
"Definition:Real Interval/Open",
"Definition:Open Rectangle",
"Category:Analysis"
] |
proofwiki-5184 | Euclidean Borel Sigma-Algebra Closed under Scalar Multiplication | Let $\map \BB {\R^n}$ be the Borel $\sigma$-algebra on $\R^n$.
Let $B \in \BB$, and let $t \in \R_{>0}$.
Then also $t \cdot B := \set {t \mathbf b: \mathbf b \in B} \in \BB$. | Define $f: \R^n \to \R^n$ by $\map f {\mathbf x} := \dfrac 1 t \mathbf x$.
Then for all $\mathbf x \in \R^n$, $\map {f^{-1} } {\mathbf x} = \set {t \mathbf x}$, where $f^{-1}$ denotes the preimage of $f$.
Thus $\ds t \cdot B = \bigcup_{\mathbf b \mathop \in B} \map {f^{-1} } {\mathbf b} = \map {f^{-1} } B$, where the l... | Let $\map \BB {\R^n}$ be the [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]] on $\R^n$.
Let $B \in \BB$, and let $t \in \R_{>0}$.
Then also $t \cdot B := \set {t \mathbf b: \mathbf b \in B} \in \BB$. | Define $f: \R^n \to \R^n$ by $\map f {\mathbf x} := \dfrac 1 t \mathbf x$.
Then for all $\mathbf x \in \R^n$, $\map {f^{-1} } {\mathbf x} = \set {t \mathbf x}$, where $f^{-1}$ denotes the [[Definition:Preimage of Mapping|preimage]] of $f$.
Thus $\ds t \cdot B = \bigcup_{\mathbf b \mathop \in B} \map {f^{-1} } {\mathb... | Euclidean Borel Sigma-Algebra Closed under Scalar Multiplication | https://proofwiki.org/wiki/Euclidean_Borel_Sigma-Algebra_Closed_under_Scalar_Multiplication | https://proofwiki.org/wiki/Euclidean_Borel_Sigma-Algebra_Closed_under_Scalar_Multiplication | [
"Measure Theory"
] | [
"Definition:Borel Sigma-Algebra"
] | [
"Definition:Preimage/Mapping/Mapping",
"Definition:Preimage/Mapping/Subset",
"Definition:Measurable Mapping",
"Characterization of Euclidean Borel Sigma-Algebra",
"Definition:Half-Open Rectangle",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Mapping Measurable iff Measurable on Genera... |
proofwiki-5185 | Lebesgue Measure of Scalar Multiple | Let $\lambda^n$ be the $n$-dimensional Lebesgue measure on $\R^n$ equipped with the Borel $\sigma$-algebra $\map \BB {\R^n}$.
Let $B \in \BB$.
Let $t \in \R_{>0}$.
Then:
:$\map {\lambda^n} {t \cdot B} = t^n \map {\lambda^n} B$
where $t \cdot B$ is the set $\set {t \mathbf b: \mathbf b \in B}$. | It follows from Rescaling is Linear Transformation that the mapping $\mathbf x \mapsto t \mathbf x$ is a linear transformation.
Denote $t \, \mathbf I_n$ for the matrix associated to this linear transformation by Linear Transformation as Matrix Product.
From Determinant of Rescaling Matrix:
:$\map \det {t \, \mathbf I_... | Let $\lambda^n$ be the $n$-dimensional [[Definition:Lebesgue Measure|Lebesgue measure]] on $\R^n$ equipped with the [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]] $\map \BB {\R^n}$.
Let $B \in \BB$.
Let $t \in \R_{>0}$.
Then:
:$\map {\lambda^n} {t \cdot B} = t^n \map {\lambda^n} B$
where $t \cdot B$ is t... | It follows from [[Rescaling is Linear Transformation]] that the [[Definition:Mapping|mapping]] $\mathbf x \mapsto t \mathbf x$ is a [[Definition:Linear Transformation|linear transformation]].
Denote $t \, \mathbf I_n$ for the [[Definition:Matrix|matrix]] associated to this [[Definition:Linear Transformation|linear tra... | Lebesgue Measure of Scalar Multiple | https://proofwiki.org/wiki/Lebesgue_Measure_of_Scalar_Multiple | https://proofwiki.org/wiki/Lebesgue_Measure_of_Scalar_Multiple | [
"Measure Theory",
"Lebesgue Measure",
"Lebesgue Measure"
] | [
"Definition:Lebesgue Measure",
"Definition:Borel Sigma-Algebra"
] | [
"Rescaling is Linear Transformation",
"Definition:Mapping",
"Definition:Linear Transformation",
"Definition:Matrix",
"Definition:Linear Transformation",
"Linear Transformation as Matrix Product",
"Determinant of Rescaling Matrix",
"Inverse of Rescaling Matrix",
"Definition:Inverse Matrix",
"Invers... |
proofwiki-5186 | Measure Invariant on Generator is Invariant | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\theta: X \to X$ be an $\Sigma / \Sigma$-measurable mapping.
Suppose that $\Sigma$ is generated by $\GG \subseteq \powerset X$.
Also, let $\GG$ satisfy the following:
:$(1): \quad \forall G, H \in \GG: G \cap H \in \GG$
:$(2): \quad$ There exists an exhausting seq... | Consider the pushforward measure $\theta_* \mu$ on $\struct {X, \Sigma}$.
By definition, this makes equation $(3)$ come down to:
:$\theta_* \mu \restriction_\GG = \mu \restriction_\GG$
where $\restriction$ denotes restriction.
The suppositions $(1)$, $(2)$ and $(3)$ together constitute precisely the prerequisites to Un... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $\theta: X \to X$ be an [[Definition:Measurable Mapping|$\Sigma / \Sigma$-measurable mapping]].
Suppose that $\Sigma$ is [[Definition:Sigma-Algebra Generated by Collection of Subsets|generated]] by $\GG \subseteq \powerset X$.
Also,... | Consider the [[Definition:Pushforward Measure|pushforward measure]] $\theta_* \mu$ on $\struct {X, \Sigma}$.
By definition, this makes equation $(3)$ come down to:
:$\theta_* \mu \restriction_\GG = \mu \restriction_\GG$
where $\restriction$ denotes [[Definition:Restriction of Mapping|restriction]].
The supposition... | Measure Invariant on Generator is Invariant | https://proofwiki.org/wiki/Measure_Invariant_on_Generator_is_Invariant | https://proofwiki.org/wiki/Measure_Invariant_on_Generator_is_Invariant | [
"Invariant Measures"
] | [
"Definition:Measure Space",
"Definition:Measurable Mapping",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Exhausting Sequence of Sets",
"Definition:Invariant Measure"
] | [
"Definition:Pushforward Measure",
"Definition:Restriction/Mapping",
"Uniqueness of Measures",
"Definition:Invariant Measure"
] |
proofwiki-5187 | Sigma-Algebras with Independent Generators are Independent | Let $\struct {\Omega, \EE, \Pr}$ be a probability space.
Let $\Sigma, \Sigma'$ be sub-$\sigma$-algebras of $\EE$.
Suppose that $\GG, \HH$ are $\cap$-stable generators for $\Sigma, \Sigma'$, respectively.
Suppose that, for all $G \in \GG, H \in \HH$:
:$(1): \quad \map \Pr {G \cap H} = \map \Pr G \map \Pr H$
Then $\Sigma... | Fix $H \in \HH$.
Define, for $E \in \Sigma$:
:$\map \mu E := \map \Pr {E \cap H}$
:$\map \nu E := \map \Pr E \map \Pr H$
Then by Intersection Measure is Measure and Restricted Measure is Measure, $\mu$ is a measure on $\Sigma$.
Namely, it is the intersection measure $\Pr_H$ restricted to $\Sigma$, that is $\Pr_H \restr... | Let $\struct {\Omega, \EE, \Pr}$ be a [[Definition:Probability Space|probability space]].
Let $\Sigma, \Sigma'$ be [[Definition:Sub-Sigma-Algebra|sub-$\sigma$-algebras]] of $\EE$.
Suppose that $\GG, \HH$ are [[Definition:Stable under Intersection|$\cap$-stable]] [[Definition:Sigma-Algebra Generated by Collection of ... | Fix $H \in \HH$.
Define, for $E \in \Sigma$:
:$\map \mu E := \map \Pr {E \cap H}$
:$\map \nu E := \map \Pr E \map \Pr H$
Then by [[Intersection Measure is Measure]] and [[Restricted Measure is Measure]], $\mu$ is a [[Definition:Measure (Measure Theory)|measure]] on $\Sigma$.
Namely, it is the [[Definition:Intersect... | Sigma-Algebras with Independent Generators are Independent | https://proofwiki.org/wiki/Sigma-Algebras_with_Independent_Generators_are_Independent | https://proofwiki.org/wiki/Sigma-Algebras_with_Independent_Generators_are_Independent | [
"Measure Theory"
] | [
"Definition:Probability Space",
"Definition:Sub-Sigma-Algebra",
"Definition:Stable under Intersection",
"Definition:Sigma-Algebra Generated by Collection of Subsets/Generator",
"Definition:Independent Sigma-Algebras"
] | [
"Intersection Measure is Measure",
"Restricted Measure is Measure",
"Definition:Measure (Measure Theory)",
"Definition:Intersection Measure",
"Definition:Restricted Measure",
"Linear Combination of Measures",
"Restricted Measure is Measure",
"Definition:Measure (Measure Theory)",
"Definition:Restric... |
proofwiki-5188 | Coset Product of Normal Subgroup is Consistent with Subset Product Definition | Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $a, b \in G$.
Let $a \circ N$ and $b \circ N$ be the left cosets of $a$ and $b$ by $N$.
Then the coset product:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is consistent with the definition of the coset product... | Consider the set:
:$\paren {a \circ N} \circ \paren {b \circ N} = \set {x \circ y: x \in a \circ N, y \in b \circ N}$
As $e \in N$, have:
:$\paren {a \circ b} \circ N = \paren {a \circ e} \circ \paren {b \circ N} \subseteq \paren {a \circ N} \circ \paren {b \circ N}$
by {{Group-axiom|1}} of $\circ$.
Hence $\paren {a \c... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Let $a, b \in G$.
Let $a \circ N$ and $b \circ N$ be the [[Definition:Left Coset|left cosets]] of $a$ and $b$ by $N$.
Then the [[Definition:Coset Product|coset product]]:
:$\paren {a \circ ... | Consider the [[Definition:Set|set]]:
:$\paren {a \circ N} \circ \paren {b \circ N} = \set {x \circ y: x \in a \circ N, y \in b \circ N}$
As $e \in N$, have:
:$\paren {a \circ b} \circ N = \paren {a \circ e} \circ \paren {b \circ N} \subseteq \paren {a \circ N} \circ \paren {b \circ N}$
by {{Group-axiom|1}} of $\cir... | Coset Product of Normal Subgroup is Consistent with Subset Product Definition | https://proofwiki.org/wiki/Coset_Product_of_Normal_Subgroup_is_Consistent_with_Subset_Product_Definition | https://proofwiki.org/wiki/Coset_Product_of_Normal_Subgroup_is_Consistent_with_Subset_Product_Definition | [
"Coset Product"
] | [
"Definition:Group",
"Definition:Normal Subgroup",
"Definition:Coset/Left Coset",
"Definition:Coset Product",
"Definition:Subset Product"
] | [
"Definition:Set",
"Definition:Subset Product",
"Definition:Subset Product",
"Definition:Coset Product",
"Category:Coset Product"
] |
proofwiki-5189 | Equivalence Relation is Congruence iff Compatible with Operation | Let $\struct {S, \circ}$ be an algebraic structure.
Let $\RR$ be an equivalence relation on $S$.
Then $\RR$ is a congruence relation for $\circ$ {{iff}}:
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in S
| l = x \mathrel \RR y
| o = \implies
| r = \paren {x \circ z} \mathrel \RR \paren {y \circ z}
}}
{{... | === Necessary Condition ===
Let $\RR$ be a congruence relation for $\circ$.
That is:
:$\forall x_1, x_2, y_1, y_2 \in S: x_1 \mathrel \RR x_2 \land y_1 \mathrel \RR y_2 \implies \paren {x_1 \circ y_1} \mathrel \RR \paren {x_2 \circ y_2}$
As $\RR$ is an equivalence relation it is by definition reflexive.
That is:
:$\for... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on $S$.
Then $\RR$ is a [[Definition:Congruence Relation|congruence relation]] for $\circ$ {{iff}}:
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in S
| ... | === Necessary Condition ===
Let $\RR$ be a [[Definition:Congruence Relation|congruence relation]] for $\circ$.
That is:
:$\forall x_1, x_2, y_1, y_2 \in S: x_1 \mathrel \RR x_2 \land y_1 \mathrel \RR y_2 \implies \paren {x_1 \circ y_1} \mathrel \RR \paren {x_2 \circ y_2}$
As $\RR$ is an [[Definition:Equivalence Rel... | Equivalence Relation is Congruence iff Compatible with Operation/Proof 1 | https://proofwiki.org/wiki/Equivalence_Relation_is_Congruence_iff_Compatible_with_Operation | https://proofwiki.org/wiki/Equivalence_Relation_is_Congruence_iff_Compatible_with_Operation/Proof_1 | [
"Equivalence Relation is Congruence iff Compatible with Operation",
"Equivalence Relations",
"Congruence Relations",
"Compatible Relations"
] | [
"Definition:Algebraic Structure",
"Definition:Equivalence Relation",
"Definition:Congruence Relation",
"Definition:Relation Compatible with Operation"
] | [
"Definition:Congruence Relation",
"Definition:Equivalence Relation",
"Definition:Reflexive Relation",
"Definition:Equivalence Relation",
"Definition:Transitive Relation"
] |
proofwiki-5190 | Equivalence Relation is Congruence iff Compatible with Operation | Let $\struct {S, \circ}$ be an algebraic structure.
Let $\RR$ be an equivalence relation on $S$.
Then $\RR$ is a congruence relation for $\circ$ {{iff}}:
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in S
| l = x \mathrel \RR y
| o = \implies
| r = \paren {x \circ z} \mathrel \RR \paren {y \circ z}
}}
{{... | We have that an equivalence relation is a (symmetric) preordering.
Thus the result Preordering of Products under Operation Compatible with Preordering can be applied directly.
{{qed}} | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on $S$.
Then $\RR$ is a [[Definition:Congruence Relation|congruence relation]] for $\circ$ {{iff}}:
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in S
| ... | We have that an [[Symmetric Preordering is Equivalence Relation|equivalence relation is a (symmetric) preordering]].
Thus the result [[Preordering of Products under Operation Compatible with Preordering]] can be applied directly.
{{qed}} | Equivalence Relation is Congruence iff Compatible with Operation/Proof 2 | https://proofwiki.org/wiki/Equivalence_Relation_is_Congruence_iff_Compatible_with_Operation | https://proofwiki.org/wiki/Equivalence_Relation_is_Congruence_iff_Compatible_with_Operation/Proof_2 | [
"Equivalence Relation is Congruence iff Compatible with Operation",
"Equivalence Relations",
"Congruence Relations",
"Compatible Relations"
] | [
"Definition:Algebraic Structure",
"Definition:Equivalence Relation",
"Definition:Congruence Relation",
"Definition:Relation Compatible with Operation"
] | [
"Symmetric Preordering is Equivalence Relation",
"Preordering of Products under Operation Compatible with Preordering"
] |
proofwiki-5191 | Outer Measure of Limit of Increasing Sequence of Sets | Let $\mu^*$ be an outer measure on a set $X$.
Let $\sequence {S_n}$ be an increasing sequence of $\mu^*$-measurable sets, and let $S_n \uparrow S$ (as $n \to \infty$).
Then for any subset $A \subseteq X$:
:$\ds \map {\mu^*} {A \cap S} = \lim_{n \mathop \to \infty} \map {\mu^*} {A \cap S_n}$ | By the monotonicity of $\mu^*$, it suffices to prove that:
:$\ds \map {\mu^*} {A \cap S} \le \lim_{n \mathop \to \infty} \map {\mu^*} {A \cap S_n}$
Assume that $\map {\mu^*} {A \cap S_n}$ is finite for all $n \in \N$, otherwise the statement is trivial by the monotonicity of $\mu^*$.
Let $S_0 = \O$.
Then $x \in S$ {{if... | Let $\mu^*$ be an [[Definition:Outer Measure|outer measure]] on a [[Definition:Set|set]] $X$.
Let $\sequence {S_n}$ be an [[Definition:Increasing Sequence of Sets|increasing sequence]] of [[Definition:Measurable Set#Measurable Sets of an Arbitrary Outer Measure|$\mu^*$-measurable sets]], and let $S_n \uparrow S$ (as $... | By the [[Definition:Monotone (Measure Theory)|monotonicity]] of $\mu^*$, it suffices to prove that:
:$\ds \map {\mu^*} {A \cap S} \le \lim_{n \mathop \to \infty} \map {\mu^*} {A \cap S_n}$
Assume that $\map {\mu^*} {A \cap S_n}$ is [[Definition:Finite Set|finite]] for all $n \in \N$, otherwise the statement is trivial... | Outer Measure of Limit of Increasing Sequence of Sets/Proof 1 | https://proofwiki.org/wiki/Outer_Measure_of_Limit_of_Increasing_Sequence_of_Sets | https://proofwiki.org/wiki/Outer_Measure_of_Limit_of_Increasing_Sequence_of_Sets/Proof_1 | [
"Measure Theory",
"Outer Measures",
"Outer Measures",
"Outer Measure of Limit of Increasing Sequence of Sets"
] | [
"Definition:Outer Measure",
"Definition:Set",
"Definition:Increasing Sequence of Sets",
"Definition:Measurable Set",
"Definition:Subset"
] | [
"Definition:Monotone (Measure Theory)",
"Definition:Finite Set",
"Definition:Monotone (Measure Theory)",
"Definition:Integer",
"Intersection Distributes over Union",
"Intersection with Subset is Subset",
"Telescoping Series/Example 2",
"Intersection with Empty Set"
] |
proofwiki-5192 | Half-Open Rectangles form Semiring of Sets | The half-open $n$-rectangles $\JJ_{ho}^n$ form a semiring of sets. | By definition, $\O$ is considered to be a half-open $n$-rectangle.
That $\JJ_{ho}^n$ is $\cap$-stable follows from Half-Open Rectangles Closed under Intersection.
Thus, it remains to show condition $(3')$ for a semiring of sets:
:$(3'):\quad$ If $A, B \in \JJ_{ho}^n$, then there exists a finite sequence of pairwise dis... | The [[Definition:Half-Open Rectangle|half-open $n$-rectangles]] $\JJ_{ho}^n$ form a [[Definition:Semiring of Sets|semiring of sets]]. | By definition, $\O$ is considered to be a [[Definition:Half-Open Rectangle|half-open $n$-rectangle]].
That $\JJ_{ho}^n$ is [[Definition:Stable under Intersection|$\cap$-stable]] follows from [[Half-Open Rectangles Closed under Intersection]].
Thus, it remains to show condition $(3')$ for a [[Definition:Semiring of S... | Half-Open Rectangles form Semiring of Sets | https://proofwiki.org/wiki/Half-Open_Rectangles_form_Semiring_of_Sets | https://proofwiki.org/wiki/Half-Open_Rectangles_form_Semiring_of_Sets | [
"Semirings of Sets"
] | [
"Definition:Half-Open Rectangle",
"Definition:Semiring of Sets"
] | [
"Definition:Half-Open Rectangle",
"Definition:Stable under Intersection",
"Half-Open Rectangles Closed under Intersection",
"Definition:Semiring of Sets",
"Definition:Finite Sequence",
"Definition:Pairwise Disjoint",
"Definition:Half-Open Rectangle",
"Principle of Mathematical Induction",
"Definitio... |
proofwiki-5193 | Lebesgue Pre-Measure is Pre-Measure | The Lebesgue pre-measure $\lambda^n$ on the half-open $n$-rectangles $\JJ_{ho}^n$ is a pre-measure. | We employ Characterization of Pre-Measures.
It is known that $\map {\lambda^n} \O = 0$ by definition of Lebesgue pre-measure.
The only possibility for two disjoint half-open $n$-rectangles to constitute a single, large half-open $n$-rectangle is when they are of the form:
:$\horectr {\mathbf a} {\mathbf b} \quad \horec... | The [[Definition:Lebesgue Pre-Measure|Lebesgue pre-measure]] $\lambda^n$ on the [[Definition:Half-Open Rectangle|half-open $n$-rectangles]] $\JJ_{ho}^n$ is a [[Definition:Pre-Measure|pre-measure]]. | We employ [[Characterization of Pre-Measures]].
It is known that $\map {\lambda^n} \O = 0$ by definition of [[Definition:Lebesgue Pre-Measure|Lebesgue pre-measure]].
The only possibility for two [[Definition:Disjoint Sets|disjoint]] [[Definition:Half-Open Rectangle|half-open $n$-rectangles]] to constitute a single, ... | Lebesgue Pre-Measure is Pre-Measure | https://proofwiki.org/wiki/Lebesgue_Pre-Measure_is_Pre-Measure | https://proofwiki.org/wiki/Lebesgue_Pre-Measure_is_Pre-Measure | [
"Measure Theory",
"Pre-Measures",
"Pre-Measures"
] | [
"Definition:Lebesgue Pre-Measure",
"Definition:Half-Open Rectangle",
"Definition:Pre-Measure"
] | [
"Characterization of Pre-Measures",
"Definition:Lebesgue Pre-Measure",
"Definition:Disjoint Sets",
"Definition:Half-Open Rectangle",
"Definition:Half-Open Rectangle",
"Definition:Additive Function (Measure Theory)",
"Definition:Decreasing Sequence of Sets",
"Definition:Limit of Decreasing Sequence of ... |
proofwiki-5194 | Existence and Uniqueness of Lebesgue Measure | Let $\lambda^n$ be the Lebesgue pre-measure on the half-open $n$-rectangles $\JJ_{ho}^n$.
Then Lebesgue measure, the extension of $\lambda^n$ to the Borel $\sigma$-algebra $\BB \left({\R^n}\right)$, exists and is unique. | From Lebesgue Pre-Measure is Pre-Measure, $\lambda^n$ is a pre-measure on $\JJ_{ho}^n$.
By Half-Open Rectangles form Semiring of Sets, $\JJ_{ho}^n$ is a semiring of sets.
Also, from Characterization of Euclidean Borel Sigma-Algebra, $\map \BB {\R^n} = \map \sigma {\JJ_{ho}^n}$.
Now observe that the half-open $n$-rectan... | Let $\lambda^n$ be the [[Definition:Lebesgue Pre-Measure|Lebesgue pre-measure]] on the [[Definition:Half-Open Rectangle|half-open $n$-rectangles]] $\JJ_{ho}^n$.
Then [[Definition:Lebesgue Measure|Lebesgue measure]], the [[Definition:Extension of Mapping|extension]] of $\lambda^n$ to the [[Definition:Borel Sigma-Algeb... | From [[Lebesgue Pre-Measure is Pre-Measure]], $\lambda^n$ is a [[Definition:Pre-Measure|pre-measure]] on $\JJ_{ho}^n$.
By [[Half-Open Rectangles form Semiring of Sets]], $\JJ_{ho}^n$ is a [[Definition:Semiring of Sets|semiring of sets]].
Also, from [[Characterization of Euclidean Borel Sigma-Algebra]], $\map \BB {\R^... | Existence and Uniqueness of Lebesgue Measure | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Lebesgue_Measure | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Lebesgue_Measure | [
"Lebesgue Measure"
] | [
"Definition:Lebesgue Pre-Measure",
"Definition:Half-Open Rectangle",
"Definition:Lebesgue Measure",
"Definition:Extension of Mapping",
"Definition:Borel Sigma-Algebra"
] | [
"Lebesgue Pre-Measure is Pre-Measure",
"Definition:Pre-Measure",
"Half-Open Rectangles form Semiring of Sets",
"Definition:Semiring of Sets",
"Characterization of Euclidean Borel Sigma-Algebra",
"Definition:Half-Open Rectangle",
"Definition:Increasing Sequence of Sets",
"Definition:Limit of Increasing... |
proofwiki-5195 | Lebesgue Measure is Diffuse | Let $\lambda^n$ be Lebesgue measure on $\R^n$.
Then $\lambda^n$ is a diffuse measure. | A singleton $\set {\mathbf x} \subseteq \R^n$ is seen to be closed by combining:
:Euclidean Space is Complete Metric Space
:Metric Space is Hausdorff
:{{Corollary|Compact Subspace of Hausdorff Space is Closed}}
Hence by Closed Set Measurable in Borel Sigma-Algebra:
:$\set {\mathbf x} \in \map \BB {\R^n}$
where $\map \B... | Let $\lambda^n$ be [[Definition:Lebesgue Measure|Lebesgue measure]] on $\R^n$.
Then $\lambda^n$ is a [[Definition:Diffuse Measure|diffuse measure]]. | A [[Definition:Singleton|singleton]] $\set {\mathbf x} \subseteq \R^n$ is seen to be [[Definition:Closed Set (Topology)|closed]] by combining:
:[[Euclidean Space is Complete Metric Space]]
:[[Metric Space is Hausdorff]]
:{{Corollary|Compact Subspace of Hausdorff Space is Closed}}
Hence by [[Closed Set Measurable in ... | Lebesgue Measure is Diffuse | https://proofwiki.org/wiki/Lebesgue_Measure_is_Diffuse | https://proofwiki.org/wiki/Lebesgue_Measure_is_Diffuse | [
"Diffuse Measures",
"Lebesgue Measure"
] | [
"Definition:Lebesgue Measure",
"Definition:Diffuse Measure"
] | [
"Definition:Singleton",
"Definition:Closed Set/Topology",
"Euclidean Space is Complete Metric Space",
"Metric Space is T2",
"Closed Set Measurable in Borel Sigma-Algebra",
"Definition:Borel Sigma-Algebra",
"Definition:Half-Open Rectangle",
"Definition:Lebesgue Measure",
"Characterization of Measures... |
proofwiki-5196 | Diffuse Measure of Countable Set | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Suppose that for all $x \in X$, the singleton $\set x$ is in $\Sigma$.
Suppose further that $\mu$ is a diffuse measure.
Let $E \in \Sigma$ be a countable measurable set.
Then $\map \mu E = 0$. | It holds trivially that:
:$\ds E = \bigcup_{e \mathop \in E} \set e$
and in particular, this union is countable.
Also, $\map \mu {\set e} = 0$ for all $e \in E$ as $\mu$ is diffuse.
Hence Null Sets Closed under Countable Union applies to yield:
:$\map \mu E = 0$
{{qed}} | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Suppose that for all $x \in X$, the [[Definition:Singleton|singleton]] $\set x$ is in $\Sigma$.
Suppose further that $\mu$ is a [[Definition:Diffuse Measure|diffuse measure]].
Let $E \in \Sigma$ be a [[Definition:Countable|countable]] [... | It holds trivially that:
:$\ds E = \bigcup_{e \mathop \in E} \set e$
and in particular, this union is [[Definition:Countable Union|countable]].
Also, $\map \mu {\set e} = 0$ for all $e \in E$ as $\mu$ is [[Definition:Diffuse Measure|diffuse]].
Hence [[Null Sets Closed under Countable Union]] applies to yield:
:$\... | Diffuse Measure of Countable Set | https://proofwiki.org/wiki/Diffuse_Measure_of_Countable_Set | https://proofwiki.org/wiki/Diffuse_Measure_of_Countable_Set | [
"Diffuse Measures"
] | [
"Definition:Measure Space",
"Definition:Singleton",
"Definition:Diffuse Measure",
"Definition:Countable Set",
"Definition:Measurable Set"
] | [
"Definition:Set Union/Countable Union",
"Definition:Diffuse Measure",
"Null Sets Closed under Countable Union"
] |
proofwiki-5197 | Half-Open Rectangles Closed under Intersection | Let $\horectr {\mathbf a} {\mathbf b}$ and $\horectr {\mathbf c} {\mathbf d}$ be half-open $n$-rectangles.
Then $\horectr {\mathbf a} {\mathbf b} \cap \horectr {\mathbf c} {\mathbf d}$ is also a half-open $n$-rectangle. | As $\O$ is trivially a half-open $n$-rectangle, let us assume that:
:$\horectr {\mathbf a} {\mathbf b} \cap \horectr {\mathbf c} {\mathbf d} \ne \O$
By Cartesian Product of Intersections: General Case, it follows that:
:$\ds \horectr {\mathbf a} {\mathbf b} \cap \horectr {\mathbf c} {\mathbf d} = \prod_{i \mathop = 1}^... | Let $\horectr {\mathbf a} {\mathbf b}$ and $\horectr {\mathbf c} {\mathbf d}$ be [[Definition:Half-Open Rectangle|half-open $n$-rectangles]].
Then $\horectr {\mathbf a} {\mathbf b} \cap \horectr {\mathbf c} {\mathbf d}$ is also a [[Definition:Half-Open Rectangle|half-open $n$-rectangle]]. | As $\O$ is trivially a [[Definition:Half-Open Rectangle|half-open $n$-rectangle]], let us assume that:
:$\horectr {\mathbf a} {\mathbf b} \cap \horectr {\mathbf c} {\mathbf d} \ne \O$
By [[Cartesian Product of Intersections/General Case|Cartesian Product of Intersections: General Case]], it follows that:
:$\ds \hor... | Half-Open Rectangles Closed under Intersection | https://proofwiki.org/wiki/Half-Open_Rectangles_Closed_under_Intersection | https://proofwiki.org/wiki/Half-Open_Rectangles_Closed_under_Intersection | [
"Analysis"
] | [
"Definition:Half-Open Rectangle",
"Definition:Half-Open Rectangle"
] | [
"Definition:Half-Open Rectangle",
"Cartesian Product of Intersections/General Case",
"Definition:Real Interval/Half-Open",
"Definition:Real Interval/Half-Open",
"Definition:Half-Open Rectangle"
] |
proofwiki-5198 | Decomposition of Probability Measures | Let $\struct {\Omega, \Sigma, P}$ be a probability space.
Suppose that for every $\omega \in \Omega$, it holds that:
:$\set \omega \in \Sigma$
that is, that $\Sigma$ contains all singletons.
Then there exist a unique diffuse measure $\mu$ and a unique discrete measure $\nu$ such that:
:$P = \mu + \nu$ | === Existence ===
For each $n \in \N$, define $\Omega_n$ by:
:$\Omega_n := \set {\omega \in \Omega: \map P {\set \omega} \ge \dfrac 1 n}$
{{AimForCont}} that $\Omega_n$ has more than $n$ elements, and define $\Omega'_n$ to be a finite subset of $\Omega_n$ with $n + 1$ elements.
From Measure is Monotone and Measure is F... | Let $\struct {\Omega, \Sigma, P}$ be a [[Definition:Probability Space|probability space]].
Suppose that for every $\omega \in \Omega$, it holds that:
:$\set \omega \in \Sigma$
that is, that $\Sigma$ contains all [[Definition:Singleton|singletons]].
Then there exist a unique [[Definition:Diffuse Measure|diffuse mea... | === Existence ===
For each $n \in \N$, define $\Omega_n$ by:
:$\Omega_n := \set {\omega \in \Omega: \map P {\set \omega} \ge \dfrac 1 n}$
{{AimForCont}} that $\Omega_n$ has more than $n$ elements, and define $\Omega'_n$ to be a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $\Omega_n$ with $n + 1$ ... | Decomposition of Probability Measures | https://proofwiki.org/wiki/Decomposition_of_Probability_Measures | https://proofwiki.org/wiki/Decomposition_of_Probability_Measures | [
"Probability Measures"
] | [
"Definition:Probability Space",
"Definition:Singleton",
"Definition:Diffuse Measure",
"Definition:Discrete Measure"
] | [
"Definition:Finite Set",
"Definition:Subset",
"Measure is Monotone",
"Measure is Finitely Additive Function",
"Definition:Finite Set",
"Sequence of Powers of Reciprocals is Null Sequence/Corollary",
"Countable Union of Countable Sets is Countable",
"Definition:Countable Set",
"Definition:Set Union/C... |
proofwiki-5199 | Homomorphism to Group Preserves Inverses | Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, *}$ be a group.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Let $x^{-1}$ be an inverse of $x$ for $\circ$.
Then $\map \phi {x^{-1} }$ is an inverse of $\map \phi x$ for $*$. | By hypothesis, $\struct {T, *}$ is a group.
By {{Group-axiom|2}}, $\struct {T, *}$ has an identity.
Thus Homomorphism with Identity Preserves Inverses can be applied.
{{Qed}} | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $\struct {T, *}$ be a [[Definition:Group|group]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
Let $\struct {S, \circ}$ have an [[Defini... | By hypothesis, $\struct {T, *}$ is a [[Definition:Group|group]].
By {{Group-axiom|2}}, $\struct {T, *}$ has an [[Definition:Identity Element|identity]].
Thus [[Homomorphism with Identity Preserves Inverses]] can be applied.
{{Qed}} | Homomorphism to Group Preserves Inverses | https://proofwiki.org/wiki/Homomorphism_to_Group_Preserves_Inverses | https://proofwiki.org/wiki/Homomorphism_to_Group_Preserves_Inverses | [
"Homomorphisms (Abstract Algebra)",
"Group Theory"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Group",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Homomorphism with Identity Preserves Inverses"
] |
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