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-5200 | Equivalence Induced by Epimorphism is Congruence Relation | Let $\struct {S, \oplus}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \oplus} \to \struct {T, *}$ be an epimorphism.
Let $\RR_\phi$ be the equivalence induced by $\phi$.
Then the induced equivalence $\RR_\phi$ is a congruence relation for $\oplus$. | Let $x, x', y, y' \in S$ such that:
:$x \mathrel {\RR_\phi} x' \land y \mathrel {\RR_\phi} y'$
By definition of induced equivalence:
{{begin-eqn}}
{{eqn | l = x \mathrel {\RR_\phi} x'
| o = \leadsto
| r = \map \phi x = \map \phi {x'}
| c =
}}
{{eqn | l = y \mathrel {\RR_\phi} y'
| o = \leadsto
... | Let $\struct {S, \oplus}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \oplus} \to \struct {T, *}$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]].
Let $\RR_\phi$ be the [[Definition:Equivalence Relation Induced by Mapping|e... | Let $x, x', y, y' \in S$ such that:
:$x \mathrel {\RR_\phi} x' \land y \mathrel {\RR_\phi} y'$
By definition of [[Definition:Equivalence Relation Induced by Mapping|induced equivalence]]:
{{begin-eqn}}
{{eqn | l = x \mathrel {\RR_\phi} x'
| o = \leadsto
| r = \map \phi x = \map \phi {x'}
| c =
}}
... | Equivalence Induced by Epimorphism is Congruence Relation | https://proofwiki.org/wiki/Equivalence_Induced_by_Epimorphism_is_Congruence_Relation | https://proofwiki.org/wiki/Equivalence_Induced_by_Epimorphism_is_Congruence_Relation | [
"Epimorphisms (Abstract Algebra)",
"Congruence Relations",
"Examples of Equivalence Relations"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Congruence Relation"
] | [
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Congruence Relation"
] |
proofwiki-5201 | Unique Isomorphism from Quotient Mapping to Epimorphism Domain | Let $\struct {S, \oplus}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \oplus} \to \struct {T, *}$ be an epimorphism.
Let $\RR_\phi$ be the equivalence induced by $\phi$.
Let $S / \RR_\phi$ be the quotient of $S$ by $\RR_\phi$.
Let $q_{\RR_\phi}: S \to S / \RR_\phi$ be the quotient mapping induc... | From the Quotient Theorem for Surjections, there is a unique bijection from $S / \RR_\phi$ onto $T$ satisfying $\psi \circ q_{\RR_\phi} = \phi$.
Also:
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
| l = \map \psi {\eqclass x {\RR_\phi} \oplus_{\RR_\phi} \eqclass y {\RR_\phi} }
| r = \map \psi {\eqclass {x \o... | Let $\struct {S, \oplus}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \oplus} \to \struct {T, *}$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]].
Let $\RR_\phi$ be the [[Definition:Equivalence Relation Induced by Mapping|e... | From the [[Quotient Theorem for Surjections]], there is a unique [[Definition:Bijection|bijection]] from $S / \RR_\phi$ onto $T$ satisfying $\psi \circ q_{\RR_\phi} = \phi$.
Also:
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
| l = \map \psi {\eqclass x {\RR_\phi} \oplus_{\RR_\phi} \eqclass y {\RR_\phi} }
... | Unique Isomorphism from Quotient Mapping to Epimorphism Domain | https://proofwiki.org/wiki/Unique_Isomorphism_from_Quotient_Mapping_to_Epimorphism_Domain | https://proofwiki.org/wiki/Unique_Isomorphism_from_Quotient_Mapping_to_Epimorphism_Domain | [
"Epimorphisms (Abstract Algebra)",
"Isomorphisms (Abstract Algebra)",
"Quotient Mappings"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Quotient Set",
"Definition:Quotient Mapping",
"Definition:Quotient Structure",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Composi... | [
"Quotient Theorem for Surjections",
"Definition:Bijection",
"Definition:Isomorphism (Abstract Algebra)"
] |
proofwiki-5202 | Identity is in Kernel of Group Homomorphism | Let $G$ and $H$ be groups.
Let $e_G$ and $e_H$ be the identity elements of $G$ and $H$ respectively.
Let $\phi: G \to H$ be a (group) homomorphism from $G$ to $H$.
Then:
:$e_G \in \map \ker \phi$
where $\map \ker \phi$ is the kernel of $\phi$. | From the definition of kernel:
:$\map \ker \phi = \set {x \in G: \map \phi x = e_H}$
From Group Homomorphism Preserves Identity we have that:
:$\map \phi {e_G} = e_H$
Hence the result.
{{Qed}} | Let $G$ and $H$ be [[Definition:Group|groups]].
Let $e_G$ and $e_H$ be the [[Definition:Identity Element|identity elements]] of $G$ and $H$ respectively.
Let $\phi: G \to H$ be a [[Definition:Group Homomorphism|(group) homomorphism]] from $G$ to $H$.
Then:
:$e_G \in \map \ker \phi$
where $\map \ker \phi$ is the [[D... | From the definition of [[Definition:Kernel of Group Homomorphism|kernel]]:
:$\map \ker \phi = \set {x \in G: \map \phi x = e_H}$
From [[Group Homomorphism Preserves Identity]] we have that:
:$\map \phi {e_G} = e_H$
Hence the result.
{{Qed}} | Identity is in Kernel of Group Homomorphism | https://proofwiki.org/wiki/Identity_is_in_Kernel_of_Group_Homomorphism | https://proofwiki.org/wiki/Identity_is_in_Kernel_of_Group_Homomorphism | [
"Kernels of Group Homomorphisms"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group Homomorphism",
"Definition:Kernel of Group Homomorphism"
] | [
"Definition:Kernel of Group Homomorphism",
"Group Homomorphism Preserves Identity"
] |
proofwiki-5203 | Union is Dominated by Disjoint Union | Let $I$ be an indexing set.
For all $i \in I$, let $S_i$ be a set.
Then:
:$\ds \bigcup_{i \mathop \in I} S_i \preccurlyeq \bigsqcup_{i \mathop \in I} S_i$
where $\preccurlyeq$ denotes domination, $\bigcup$ denotes union, and $\bigsqcup$ denotes disjoint union. | For all $\ds x \in \bigcup_{i \mathop \in I} S_i$, there exists a $\map i x \in I$ such that $x \in S_{\map i x}$.
Thus the mapping $\ds \iota : \bigcup_{i \mathop \in I} S_i \to \bigsqcup_{i \mathop \in I} S_i$ defined by:
:$\map \iota x = \tuple {x, \map i x}$
is an injection.
{{handwaving}}
{{qed}}
Category:Set Theo... | Let $I$ be an [[Definition:Indexing Set|indexing set]].
For all $i \in I$, let $S_i$ be a [[Definition:Set|set]].
Then:
:$\ds \bigcup_{i \mathop \in I} S_i \preccurlyeq \bigsqcup_{i \mathop \in I} S_i$
where $\preccurlyeq$ denotes [[Definition:Dominate (Set Theory)|domination]], $\bigcup$ denotes [[Definition:Set Uni... | For all $\ds x \in \bigcup_{i \mathop \in I} S_i$, there exists a $\map i x \in I$ such that $x \in S_{\map i x}$.
Thus the [[Definition:Mapping|mapping]] $\ds \iota : \bigcup_{i \mathop \in I} S_i \to \bigsqcup_{i \mathop \in I} S_i$ defined by:
:$\map \iota x = \tuple {x, \map i x}$
is an [[Definition:Injection|inje... | Union is Dominated by Disjoint Union | https://proofwiki.org/wiki/Union_is_Dominated_by_Disjoint_Union | https://proofwiki.org/wiki/Union_is_Dominated_by_Disjoint_Union | [
"Set Theory"
] | [
"Definition:Indexing Set",
"Definition:Set",
"Definition:Dominate (Set Theory)",
"Definition:Set Union",
"Definition:Disjoint Union (Set Theory)"
] | [
"Definition:Mapping",
"Definition:Injection",
"Category:Set Theory"
] |
proofwiki-5204 | Disjoint Union Preserves Domination | Let $I$ be an indexing set.
For all $i \in I$, let $A_i$ and $B_i$ be sets such that $A_i \preccurlyeq B_i$.
Here, $\preccurlyeq$ denotes domination.
Then:
:$\ds \bigsqcup_{i \mathop \in I} A_i \preccurlyeq \bigsqcup_{i \mathop \in I} B_i$
where $\bigsqcup$ denotes disjoint union. | By definition of domination, for all $i \in I$, there exists an injection $\iota_i: A_i \to B_i$.
Thus the mapping $\ds \iota : \bigsqcup_{i \mathop \in I} A_i \to \bigsqcup_{i \mathop \in I} B_i$ defined by:
:$\map \iota {x, i} = \tuple {\map {\iota_i} x, i}$
is an injection.
{{handwaving|That statement needs proof. P... | Let $I$ be an [[Definition:Indexing Set|indexing set]].
For all $i \in I$, let $A_i$ and $B_i$ be [[Definition:Set|sets]] such that $A_i \preccurlyeq B_i$.
Here, $\preccurlyeq$ denotes [[Definition:Dominate (Set Theory)|domination]].
Then:
:$\ds \bigsqcup_{i \mathop \in I} A_i \preccurlyeq \bigsqcup_{i \mathop \in ... | By definition of [[Definition:Dominate (Set Theory)|domination]], for all $i \in I$, there exists an [[Definition:Injection|injection]] $\iota_i: A_i \to B_i$.
Thus the mapping $\ds \iota : \bigsqcup_{i \mathop \in I} A_i \to \bigsqcup_{i \mathop \in I} B_i$ defined by:
:$\map \iota {x, i} = \tuple {\map {\iota_i} x, ... | Disjoint Union Preserves Domination | https://proofwiki.org/wiki/Disjoint_Union_Preserves_Domination | https://proofwiki.org/wiki/Disjoint_Union_Preserves_Domination | [
"Set Theory"
] | [
"Definition:Indexing Set",
"Definition:Set",
"Definition:Dominate (Set Theory)",
"Definition:Disjoint Union (Set Theory)"
] | [
"Definition:Dominate (Set Theory)",
"Definition:Injection",
"Definition:Injection",
"Category:Set Theory"
] |
proofwiki-5205 | Epimorphism from Real Numbers to Circle Group | Let $\struct {K, \times}$ be the circle group, that is:
:$K = \set {z \in \C: \cmod z = 1}$
under complex multiplication.
Let $f: \R \to K$ be the mapping from the real numbers to $K$ defined as:
:$\forall x \in \R: \map f x = \cos x + i \sin x$
Then $f: \struct {\R, +} \to \struct {K, \times}$ is a group epimorphism.
... | $f$ is a surjection from ...
{{link wanted|Needs a link to a result specifying that $f$ is surjective (may already exist).}}
{{qed|lemma}}
Then:
{{begin-eqn}}
{{eqn | l = \map f x \times \map f y
| r = \paren {\cos x + i \sin x} \paren {\cos y + i \sin y}
| c =
}}
{{eqn | r = \cos x \cos y + i \sin x \cos y... | Let $\struct {K, \times}$ be the [[Definition:Circle Group|circle group]], that is:
:$K = \set {z \in \C: \cmod z = 1}$
under [[Definition:Complex Multiplication|complex multiplication]].
Let $f: \R \to K$ be the [[Definition:Mapping|mapping]] from the [[Definition:Real Number|real numbers]] to $K$ defined as:
:$\fora... | $f$ is a [[Definition:Surjection|surjection]] from ...
{{link wanted|Needs a link to a result specifying that $f$ is surjective (may already exist).}}
{{qed|lemma}}
Then:
{{begin-eqn}}
{{eqn | l = \map f x \times \map f y
| r = \paren {\cos x + i \sin x} \paren {\cos y + i \sin y}
| c =
}}
{{eqn | r = \c... | Epimorphism from Real Numbers to Circle Group | https://proofwiki.org/wiki/Epimorphism_from_Real_Numbers_to_Circle_Group | https://proofwiki.org/wiki/Epimorphism_from_Real_Numbers_to_Circle_Group | [
"Circle Group",
"Group Epimorphisms"
] | [
"Definition:Circle Group",
"Definition:Multiplication/Complex Numbers",
"Definition:Mapping",
"Definition:Real Number",
"Definition:Group Epimorphism",
"Definition:Kernel of Group Homomorphism"
] | [
"Definition:Surjection",
"Cosine of Sum",
"Sine of Sum",
"Definition:Group Homomorphism",
"Definition:Surjection",
"Definition:Group Homomorphism",
"Definition:Group Epimorphism",
"Cosine of Integer Multiple of Pi",
"Sine of Integer Multiple of Pi",
"Sine and Cosine are Periodic on Reals",
"Defi... |
proofwiki-5206 | Increasing Sequence of Sets induces Partition on Limit | Let $\sequence {S_n}_{n \mathop \in \N} \uparrow S$ be an increasing sequence of sets with limit $S$.
Define $T_1 = S_1$, and, for $n \in \N$, $T_{n + 1} = S_{n + 1} \setminus S_n$, where $\setminus$ denotes set difference.
Then $\sequence {T_n}_{n \mathop \in \N}$ is a countable partition of $S$. | That $\sequence {T_n}_{n \mathop \in \N}$ partitions $S$, means precisely that:
:$(1): \quad$ The $T_n$ are pairwise disjoint
:$(2): \quad \ds \bigcup_{n \mathop \in \N} T_n = S$
It is more convenient to prove $(1)$ and $(2)$ separately: | Let $\sequence {S_n}_{n \mathop \in \N} \uparrow S$ be an [[Definition:Increasing Sequence of Sets|increasing sequence of sets]] with [[Definition:Limit of Increasing Sequence of Sets|limit]] $S$.
Define $T_1 = S_1$, and, for $n \in \N$, $T_{n + 1} = S_{n + 1} \setminus S_n$, where $\setminus$ denotes [[Definition:Set... | That $\sequence {T_n}_{n \mathop \in \N}$ [[Definition:Set Partition|partitions]] $S$, means precisely that:
:$(1): \quad$ The $T_n$ are [[Definition:Pairwise Disjoint|pairwise disjoint]]
:$(2): \quad \ds \bigcup_{n \mathop \in \N} T_n = S$
It is more convenient to prove $(1)$ and $(2)$ separately: | Increasing Sequence of Sets induces Partition on Limit | https://proofwiki.org/wiki/Increasing_Sequence_of_Sets_induces_Partition_on_Limit | https://proofwiki.org/wiki/Increasing_Sequence_of_Sets_induces_Partition_on_Limit | [
"Increasing Sequences of Sets",
"Set Theory",
"Increasing Sequences of Sets"
] | [
"Definition:Increasing Sequence of Sets",
"Definition:Limit of Increasing Sequence of Sets",
"Definition:Set Difference",
"Definition:Countable Set",
"Definition:Set Partition"
] | [
"Definition:Set Partition",
"Definition:Pairwise Disjoint",
"Definition:Pairwise Disjoint"
] |
proofwiki-5207 | Power Function on Complex Numbers is Epimorphism | Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $\struct {\C_{\ne 0}, \times}$ be the multiplicative group of complex numbers.
Let $f_n: \C_{\ne 0} \to \C_{\ne 0}$ be the mapping from the set of complex numbers less zero to itself defined as:
:$\forall z \in \C_{\ne 0}: \map {f_n} z = z^n$
Then $f_n: \struct {\... | From Non-Zero Complex Numbers under Multiplication form Group, $\struct {\C_{\ne 0}, \times}$ is a group.
Therefore $\struct {\C_{\ne 0}, \times}$ is closed by {{Group-axiom|0}}.
Let $w, z \in \C_{\ne 0}$.
{{begin-eqn}}
{{eqn | l = \map {f_n} {w \times z}
| r = \paren {w \times z}^n
| c =
}}
{{eqn | r = w^... | Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Let $\struct {\C_{\ne 0}, \times}$ be the [[Definition:Multiplicative Group of Complex Numbers|multiplicative group of complex numbers]].
Let $f_n: \C_{\ne 0} \to \C_{\ne 0}$ be the [[Definition:Mapping|mapping]] from the [[D... | From [[Non-Zero Complex Numbers under Multiplication form Group]], $\struct {\C_{\ne 0}, \times}$ is a [[Definition:Group|group]].
Therefore $\struct {\C_{\ne 0}, \times}$ is [[Definition:Closed Algebraic Structure|closed]] by {{Group-axiom|0}}.
Let $w, z \in \C_{\ne 0}$.
{{begin-eqn}}
{{eqn | l = \map {f_n} {w \ti... | Power Function on Complex Numbers is Epimorphism | https://proofwiki.org/wiki/Power_Function_on_Complex_Numbers_is_Epimorphism | https://proofwiki.org/wiki/Power_Function_on_Complex_Numbers_is_Epimorphism | [
"Complex Numbers",
"Group Epimorphisms",
"Power Function on Complex Numbers is Epimorphism"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Multiplicative Group of Complex Numbers",
"Definition:Mapping",
"Definition:Complex Number",
"Definition:Group Epimorphism",
"Definition:Kernel of Group Homomorphism",
"Definition:Root of Unity/Complex"
] | [
"Non-Zero Complex Numbers under Multiplication form Group",
"Definition:Group",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Power of Product of Commutative Elements in Group",
"Definition:Group Homomorphism",
"Definition:Complex Number/Polar Form",
"Definition:Surjection",
"Definitio... |
proofwiki-5208 | Real Part as Mapping is Endomorphism for Complex Addition | Let $\struct {\C, +}$ be the additive group of complex numbers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Let $f: \C \to \R$ be the mapping from the complex numbers to the real numbers defined as:
:$\forall z \in \C: \map f z = \map \Re z$
where $\map \Re z$ denotes the real part of $z$.
Then $f: \st... | From Real Part as Mapping is Surjection, $f$ is a surjection.
Let $z_1, z_2 \in \C$.
Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$.
Then:
{{begin-eqn}}
{{eqn | l = \map f {z_1 + z_2}
| r = \map \Re {z_1 + z_2}
| c = Definition of $f$
}}
{{eqn | r = \map \Re {x_1 + i y_1 + x_2 + i y_2}
| c = Definition of... | Let $\struct {\C, +}$ be the [[Definition:Additive Group of Complex Numbers|additive group of complex numbers]].
Let $\struct {\R, +}$ be the [[Definition:Additive Group of Real Numbers|additive group of real numbers]].
Let $f: \C \to \R$ be the [[Definition:Mapping|mapping]] from the [[Definition:Complex Number|comp... | From [[Real Part as Mapping is Surjection]], $f$ is a [[Definition:Surjection|surjection]].
Let $z_1, z_2 \in \C$.
Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$.
Then:
{{begin-eqn}}
{{eqn | l = \map f {z_1 + z_2}
| r = \map \Re {z_1 + z_2}
| c = Definition of $f$
}}
{{eqn | r = \map \Re {x_1 + i y_1 + x_2 ... | Real Part as Mapping is Endomorphism for Complex Addition | https://proofwiki.org/wiki/Real_Part_as_Mapping_is_Endomorphism_for_Complex_Addition | https://proofwiki.org/wiki/Real_Part_as_Mapping_is_Endomorphism_for_Complex_Addition | [
"Additive Group of Complex Numbers",
"Additive Group of Real Numbers",
"Real Parts",
"Complex Addition",
"Group Endomorphisms"
] | [
"Definition:Additive Group of Complex Numbers",
"Definition:Additive Group of Real Numbers",
"Definition:Mapping",
"Definition:Complex Number",
"Definition:Real Number",
"Definition:Complex Number/Real Part",
"Definition:Group Epimorphism",
"Definition:Kernel of Group Homomorphism",
"Definition:Set"... | [
"Real Part as Mapping is Surjection",
"Definition:Surjection",
"Definition:Group Homomorphism",
"Definition:Surjection",
"Definition:Group Homomorphism",
"Definition:Group Epimorphism",
"Complex Addition Identity is Zero"
] |
proofwiki-5209 | Imaginary Part as Mapping is Endomorphism for Complex Addition | Let $\struct {\C, +}$ be the additive group of complex numbers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Let $f: \C \to \R$ be the mapping from the complex numbers to the real numbers defined as:
:$\forall z \in \C: \map f z = \map \Im z$
where $\map \Im z$ denotes the imaginary part of $z$.
Then $f... | From Imaginary Part as Mapping is Surjection, $f$ is a surjection.
Let $z_1, z_2 \in \C$.
Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$.
Then:
{{begin-eqn}}
{{eqn | l = \map f {z_1 + z_2}
| r = \map \Im {z_1 + z_2}
| c = Definition of $f$
}}
{{eqn | r = \map \Im {x_1 + i y_1 + x_2 + i y_2}
| c = Definiti... | Let $\struct {\C, +}$ be the [[Definition:Additive Group of Complex Numbers|additive group of complex numbers]].
Let $\struct {\R, +}$ be the [[Definition:Additive Group of Real Numbers|additive group of real numbers]].
Let $f: \C \to \R$ be the [[Definition:Mapping|mapping]] from the [[Definition:Complex Number|comp... | From [[Imaginary Part as Mapping is Surjection]], $f$ is a [[Definition:Surjection|surjection]].
Let $z_1, z_2 \in \C$.
Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$.
Then:
{{begin-eqn}}
{{eqn | l = \map f {z_1 + z_2}
| r = \map \Im {z_1 + z_2}
| c = Definition of $f$
}}
{{eqn | r = \map \Im {x_1 + i y_1 +... | Imaginary Part as Mapping is Endomorphism for Complex Addition | https://proofwiki.org/wiki/Imaginary_Part_as_Mapping_is_Endomorphism_for_Complex_Addition | https://proofwiki.org/wiki/Imaginary_Part_as_Mapping_is_Endomorphism_for_Complex_Addition | [
"Complex Addition",
"Group Endomorphisms",
"Imaginary Parts"
] | [
"Definition:Additive Group of Complex Numbers",
"Definition:Additive Group of Real Numbers",
"Definition:Mapping",
"Definition:Complex Number",
"Definition:Real Number",
"Definition:Complex Number/Imaginary Part",
"Definition:Group Epimorphism",
"Definition:Kernel of Group Homomorphism",
"Definition... | [
"Imaginary Part as Mapping is Surjection",
"Definition:Surjection",
"Definition:Group Homomorphism",
"Definition:Surjection",
"Definition:Group Homomorphism",
"Definition:Group Epimorphism",
"Complex Addition Identity is Zero"
] |
proofwiki-5210 | Reduced Echelon Matrix is Unique | Every $m \times n$ matrix is row equivalent to exactly one $m \times n$ reduced echelon matrix.
That is, the reduced echelon form of a matrix is unique. | === Proof of Existence ===
Proved in Matrix is Row Equivalent to Reduced Echelon Matrix.
{{qed|lemma}} | Every [[Definition:Matrix|$m \times n$ matrix]] is [[Definition:Row Equivalence|row equivalent]] to [[Definition:Unique|exactly one]] $m \times n$ [[Definition:Reduced Echelon Matrix|reduced echelon matrix]].
That is, the [[Definition:Reduced Echelon Form|reduced echelon form]] of a [[Definition:Matrix|matrix]] is [[D... | === Proof of Existence ===
Proved in [[Matrix is Row Equivalent to Reduced Echelon Matrix]].
{{qed|lemma}} | Reduced Echelon Matrix is Unique | https://proofwiki.org/wiki/Reduced_Echelon_Matrix_is_Unique | https://proofwiki.org/wiki/Reduced_Echelon_Matrix_is_Unique | [
"Echelon Matrices"
] | [
"Definition:Matrix",
"Definition:Row Equivalence",
"Definition:Unique",
"Definition:Echelon Matrix/Reduced Echelon Form",
"Definition:Echelon Matrix/Reduced Echelon Form",
"Definition:Matrix",
"Definition:Unique"
] | [
"Matrix is Row Equivalent to Reduced Echelon Matrix"
] |
proofwiki-5211 | Measure of Set Difference with Subset | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $S, T \in \Sigma$ be such that $S \subseteq T$, and suppose that $\mu \paren S < +\infty$.
Then:
:$\mu \paren {T \setminus S} = \mu \paren T - \mu \paren S$
where $T \setminus S$ denotes set difference. | {{begin-eqn}}
{{eqn | l = T
| r = \paren {T \setminus S} \cup \paren {T \cap S}
| c = Set Difference Union Intersection
}}
{{eqn | r = \paren {T \setminus S} \cup S
| c = Intersection with Subset is Subset
}}
{{eqn | ll= \leadsto
| l = \mu \paren T
| r = \mu \paren {T \setminus S} + \mu \p... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $S, T \in \Sigma$ be such that $S \subseteq T$, and suppose that $\mu \paren S < +\infty$.
Then:
:$\mu \paren {T \setminus S} = \mu \paren T - \mu \paren S$
where $T \setminus S$ denotes [[Definition:Set Difference|set difference]]... | {{begin-eqn}}
{{eqn | l = T
| r = \paren {T \setminus S} \cup \paren {T \cap S}
| c = [[Set Difference Union Intersection]]
}}
{{eqn | r = \paren {T \setminus S} \cup S
| c = [[Intersection with Subset is Subset]]
}}
{{eqn | ll= \leadsto
| l = \mu \paren T
| r = \mu \paren {T \setminus S} ... | Measure of Set Difference with Subset | https://proofwiki.org/wiki/Measure_of_Set_Difference_with_Subset | https://proofwiki.org/wiki/Measure_of_Set_Difference_with_Subset | [
"Measure of Set Difference with Subset",
"Measures",
"Set Difference",
"Measure of Set Difference with Subset"
] | [
"Definition:Measure Space",
"Definition:Set Difference"
] | [
"Set Difference Union Intersection",
"Intersection with Subset is Subset",
"Measure is Finitely Additive Function",
"Set Difference Intersection with Second Set is Empty Set",
"Definition:Subtraction/Real Numbers"
] |
proofwiki-5212 | Subset of Codomain is Superset of Image of Preimage | Let $f: S \to T$ be a mapping.
Then:
:$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B \subseteq B$
where:
:$f \sqbrk B$ denotes the image of $B$ under $f$
:$f^{-1}$ denotes the inverse of $f$
:$f \circ f^{-1}$ denotes composition of $f$ and $f^{-1}$.
This can be expressed in the language and notation of direc... | From Image of Preimage under Mapping:
: $B \subseteq T \implies \left({f \circ f^{-1} }\right) \left[{B}\right] = B \cap f \left[{S}\right]$
The result follows from Intersection is Subset.
{{qed}} | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B \subseteq B$
where:
:$f \sqbrk B$ denotes the [[Definition:Image of Subset under Mapping|image of $B$ under $f$]]
:$f^{-1}$ denotes the [[Definition:Inverse of Mapping|inverse of $f$]]
:$f \circ f^... | From [[Image of Preimage under Mapping]]:
: $B \subseteq T \implies \left({f \circ f^{-1} }\right) \left[{B}\right] = B \cap f \left[{S}\right]$
The result follows from [[Intersection is Subset]].
{{qed}} | Subset of Codomain is Superset of Image of Preimage/Proof 1 | https://proofwiki.org/wiki/Subset_of_Codomain_is_Superset_of_Image_of_Preimage | https://proofwiki.org/wiki/Subset_of_Codomain_is_Superset_of_Image_of_Preimage/Proof_1 | [
"Composite Mappings",
"Preimages under Mappings",
"Subset of Codomain is Superset of Image of Preimage"
] | [
"Definition:Mapping",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Inverse of Mapping",
"Definition:Composition of Mappings",
"Definition:Direct Image Mapping",
"Definition:Inverse Image Mapping"
] | [
"Image of Preimage under Mapping",
"Intersection is Subset"
] |
proofwiki-5213 | Subset of Codomain is Superset of Image of Preimage | Let $f: S \to T$ be a mapping.
Then:
:$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B \subseteq B$
where:
:$f \sqbrk B$ denotes the image of $B$ under $f$
:$f^{-1}$ denotes the inverse of $f$
:$f \circ f^{-1}$ denotes composition of $f$ and $f^{-1}$.
This can be expressed in the language and notation of direc... | Let $y \in B$.
Then:
:$\exists x \in S: y = f \left({x}\right)$
Therefore by definition of preimage of subset:
:$\exists x \in f^{-1} \left[{B}\right]$
It follows by definition of image of subset that:
:$y \in f \left[{f^{-1} \left[{B}\right]}\right]$
Thus by definition of composition $f$ with $f^{-1}$:
:$y \in \left({... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B \subseteq B$
where:
:$f \sqbrk B$ denotes the [[Definition:Image of Subset under Mapping|image of $B$ under $f$]]
:$f^{-1}$ denotes the [[Definition:Inverse of Mapping|inverse of $f$]]
:$f \circ f^... | Let $y \in B$.
Then:
:$\exists x \in S: y = f \left({x}\right)$
Therefore by definition of [[Definition:Preimage of Subset under Mapping|preimage of subset]]:
:$\exists x \in f^{-1} \left[{B}\right]$
It follows by definition of [[Definition:Image of Subset under Mapping|image of subset]] that:
:$y \in f \left[{f^{-1... | Subset of Codomain is Superset of Image of Preimage/Proof 2 | https://proofwiki.org/wiki/Subset_of_Codomain_is_Superset_of_Image_of_Preimage | https://proofwiki.org/wiki/Subset_of_Codomain_is_Superset_of_Image_of_Preimage/Proof_2 | [
"Composite Mappings",
"Preimages under Mappings",
"Subset of Codomain is Superset of Image of Preimage"
] | [
"Definition:Mapping",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Inverse of Mapping",
"Definition:Composition of Mappings",
"Definition:Direct Image Mapping",
"Definition:Inverse Image Mapping"
] | [
"Definition:Preimage/Mapping/Subset",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Composition of Mappings",
"Definition:Subset"
] |
proofwiki-5214 | Subset of Codomain is Superset of Image of Preimage | Let $f: S \to T$ be a mapping.
Then:
:$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B \subseteq B$
where:
:$f \sqbrk B$ denotes the image of $B$ under $f$
:$f^{-1}$ denotes the inverse of $f$
:$f \circ f^{-1}$ denotes composition of $f$ and $f^{-1}$.
This can be expressed in the language and notation of direc... | Let $B \subseteq T$.
Then:
{{begin-eqn}}
{{eqn | l = y
| o = \in
| r = \paren {f \circ f^{-1} } \sqbrk B
}}
{{eqn | ll= \leadsto
| l = y
| o = \in
| r = f \sqbrk {f^{-1} \sqbrk B}
| c = {{Defof|Composition of Mappings}}
}}
{{eqn | ll= \leadsto
| q = \exists x \in f^{-1} \sqbrk ... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B \subseteq B$
where:
:$f \sqbrk B$ denotes the [[Definition:Image of Subset under Mapping|image of $B$ under $f$]]
:$f^{-1}$ denotes the [[Definition:Inverse of Mapping|inverse of $f$]]
:$f \circ f^... | Let $B \subseteq T$.
Then:
{{begin-eqn}}
{{eqn | l = y
| o = \in
| r = \paren {f \circ f^{-1} } \sqbrk B
}}
{{eqn | ll= \leadsto
| l = y
| o = \in
| r = f \sqbrk {f^{-1} \sqbrk B}
| c = {{Defof|Composition of Mappings}}
}}
{{eqn | ll= \leadsto
| q = \exists x \in f^{-1} \sqbr... | Subset of Codomain is Superset of Image of Preimage/Proof 3 | https://proofwiki.org/wiki/Subset_of_Codomain_is_Superset_of_Image_of_Preimage | https://proofwiki.org/wiki/Subset_of_Codomain_is_Superset_of_Image_of_Preimage/Proof_3 | [
"Composite Mappings",
"Preimages under Mappings",
"Subset of Codomain is Superset of Image of Preimage"
] | [
"Definition:Mapping",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Inverse of Mapping",
"Definition:Composition of Mappings",
"Definition:Direct Image Mapping",
"Definition:Inverse Image Mapping"
] | [
"Definition:Subset"
] |
proofwiki-5215 | Induced Outer Measure Restricted to Semiring is Pre-Measure | Let $\SS$ be a semiring over a set $X$.
Let $\mu: \SS \to \overline \R_{\ge 0}$ be a pre-measure on $\SS$, where $\overline \R_{\ge 0}$ denotes the set of positive extended real numbers.
Let $\mu^*: \powerset X \to \overline \R_{\ge 0}$ be the outer measure induced by $\mu$.
Then:
:$\ds \mu^*\restriction_\SS \, = \mu$
... | Let $S \in \SS$.
It follows immediately from the definition of the induced outer measure that $\map {\mu^*} S \le \map \mu S$.
Therefore, it suffices to show that if $\ds \sequence {A_n}_{n \mathop = 0}^\infty$ is a countable cover for $S$, then:
:$\ds \map \mu S \le \sum_{n \mathop = 0}^\infty \map \mu {A_n}$
If the a... | Let $\SS$ be a [[Definition:Semiring of Sets|semiring]] over a [[Definition:Set|set]] $X$.
Let $\mu: \SS \to \overline \R_{\ge 0}$ be a [[Definition:Pre-Measure|pre-measure]] on $\SS$, where $\overline \R_{\ge 0}$ denotes the set of [[Definition:Positive Real Number|positive]] [[Definition:Extended Real Number Line|ex... | Let $S \in \SS$.
It follows immediately from the [[Definition:Induced Outer Measure|definition of the induced outer measure]] that $\map {\mu^*} S \le \map \mu S$.
Therefore, it suffices to show that if $\ds \sequence {A_n}_{n \mathop = 0}^\infty$ is a [[Definition:Countable Cover|countable cover]] for $S$, then:
:$\... | Induced Outer Measure Restricted to Semiring is Pre-Measure | https://proofwiki.org/wiki/Induced_Outer_Measure_Restricted_to_Semiring_is_Pre-Measure | https://proofwiki.org/wiki/Induced_Outer_Measure_Restricted_to_Semiring_is_Pre-Measure | [
"Outer Measures",
"Pre-Measures"
] | [
"Definition:Semiring of Sets",
"Definition:Set",
"Definition:Pre-Measure",
"Definition:Positive/Real Number",
"Definition:Extended Real Number Line",
"Definition:Outer Measure",
"Definition:Induced Outer Measure",
"Definition:Restriction/Mapping"
] | [
"Definition:Induced Outer Measure",
"Definition:Cover of Set/Countable",
"Definition:Infimum of Set",
"Definition:Natural Numbers",
"Definition:Set Difference",
"Principle of Mathematical Induction",
"Definition:Natural Numbers",
"Definition:Set Union/Finite Union",
"Definition:Pairwise Disjoint",
... |
proofwiki-5216 | Ordering on Extended Real Numbers is Ordering | Denote with $\le$ the usual ordering on the extended real numbers $\overline \R$.
Then $\le$ is an ordering, and so $\overline \R$ is an ordered set. | === Transitive ===
Let $a, b, c \in \overline \R$.
Suppose that $a \le b$ and $b \le c$.
If $c = +\infty$, then by the definition of $\le$, $a \le c$.
If $b = +\infty$, then by Positive Infinity is Maximal, $b = c$, so $a \le c$.
If $a = +\infty$, then applying Positive Infinity is Maximal twice yields $a \le c$.
The c... | Denote with $\le$ the [[Definition:Ordering on Extended Real Numbers|usual ordering]] on the [[Definition:Extended Real Number Line|extended real numbers]] $\overline \R$.
Then $\le$ is an [[Definition:Ordering|ordering]], and so $\overline \R$ is an [[Definition:Ordered Set|ordered set]]. | === Transitive ===
Let $a, b, c \in \overline \R$.
Suppose that $a \le b$ and $b \le c$.
If $c = +\infty$, then by the definition of $\le$, $a \le c$.
If $b = +\infty$, then by [[Positive Infinity is Maximal]], $b = c$, so $a \le c$.
If $a = +\infty$, then applying [[Positive Infinity is Maximal]] twice yields $a ... | Ordering on Extended Real Numbers is Ordering | https://proofwiki.org/wiki/Ordering_on_Extended_Real_Numbers_is_Ordering | https://proofwiki.org/wiki/Ordering_on_Extended_Real_Numbers_is_Ordering | [
"Usual Ordering on Extended Real Numbers"
] | [
"Definition:Ordering on Extended Real Numbers",
"Definition:Extended Real Number Line",
"Definition:Ordering",
"Definition:Ordered Set"
] | [
"Positive Infinity is Maximal",
"Positive Infinity is Maximal",
"Definition:Transitive Relation",
"Positive Infinity is Maximal"
] |
proofwiki-5217 | Ordering on Extended Real Numbers is Total Ordering | Let $\le$ denote the ordering on the extended real numbers $\overline \R$.
Then $\le$ is a total ordering, and so $\overline \R$ is a toset. | {{ProofWanted}}
Category:Usual Ordering on Extended Real Numbers
0xw3mdk4joabiju4q94c91ov1lkxfiv | Let $\le$ denote the [[Definition:Ordering on Extended Real Numbers|ordering]] on the [[Definition:Extended Real Number Line|extended real numbers]] $\overline \R$.
Then $\le$ is a [[Definition:Total Ordering|total ordering]], and so $\overline \R$ is a [[Definition:Toset|toset]]. | {{ProofWanted}}
[[Category:Usual Ordering on Extended Real Numbers]]
0xw3mdk4joabiju4q94c91ov1lkxfiv | Ordering on Extended Real Numbers is Total Ordering | https://proofwiki.org/wiki/Ordering_on_Extended_Real_Numbers_is_Total_Ordering | https://proofwiki.org/wiki/Ordering_on_Extended_Real_Numbers_is_Total_Ordering | [
"Usual Ordering on Extended Real Numbers"
] | [
"Definition:Ordering on Extended Real Numbers",
"Definition:Extended Real Number Line",
"Definition:Total Ordering",
"Symbols:Abbreviations/T/Toset"
] | [
"Category:Usual Ordering on Extended Real Numbers"
] |
proofwiki-5218 | Extended Real Number Space is Compact | The extended real number space is compact. | {{ProofWanted|opens at each infinity, and a closed, bounded area remains}}
Category:Extended Real Number Space
r328iv05o9wxczee89g4akd3wy7jays | The [[Definition:Extended Real Number Space|extended real number space]] is [[Definition:Compact Topological Space|compact]]. | {{ProofWanted|opens at each infinity, and a closed, bounded area remains}}
[[Category:Extended Real Number Space]]
r328iv05o9wxczee89g4akd3wy7jays | Extended Real Number Space is Compact | https://proofwiki.org/wiki/Extended_Real_Number_Space_is_Compact | https://proofwiki.org/wiki/Extended_Real_Number_Space_is_Compact | [
"Extended Real Number Space"
] | [
"Definition:Topology on Extended Real Numbers",
"Definition:Compact Topological Space"
] | [
"Category:Extended Real Number Space"
] |
proofwiki-5219 | Euclidean Space is Subspace of Extended Real Number Space | Let $\struct {\overline \R, \tau}$ be the extended real number space.
Then $\tau {\restriction_\R}$, the subspace topology on $\R$, is the Euclidean topology.
That is, Euclidean $1$-space is a subspace of the extended real number space. | {{ProofWanted}}
Category:Extended Real Number Space
Category:Real Euclidean Spaces
0tb3hq0nyqarqlg8426ymnurzcg75kg | Let $\struct {\overline \R, \tau}$ be the [[Definition:Extended Real Number Space|extended real number space]].
Then $\tau {\restriction_\R}$, the [[Definition:Subspace Topology|subspace topology]] on $\R$, is the [[Definition:Euclidean Topology on Real Number Line|Euclidean topology]].
That is, [[Definition:Euclide... | {{ProofWanted}}
[[Category:Extended Real Number Space]]
[[Category:Real Euclidean Spaces]]
0tb3hq0nyqarqlg8426ymnurzcg75kg | Euclidean Space is Subspace of Extended Real Number Space | https://proofwiki.org/wiki/Euclidean_Space_is_Subspace_of_Extended_Real_Number_Space | https://proofwiki.org/wiki/Euclidean_Space_is_Subspace_of_Extended_Real_Number_Space | [
"Extended Real Number Space",
"Real Euclidean Spaces"
] | [
"Definition:Topology on Extended Real Numbers",
"Definition:Topological Subspace",
"Definition:Euclidean Space/Euclidean Topology/Real Number Line",
"Definition:Euclidean Space",
"Definition:Topological Subspace",
"Definition:Topology on Extended Real Numbers"
] | [
"Category:Extended Real Number Space",
"Category:Real Euclidean Spaces"
] |
proofwiki-5220 | Transcendental Slope | The slope of a line may be transcendental. | The slope form of any number $x$ may be produced by:
{{begin-eqn}}
{{eqn|l = {\mathrm m}
|r = \frac {x} {1}
|c = (Slope Form of $x$)
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn|l = {\mathrm m}
|r = {x}
|c =
}}
{{end-eqn}}
If $x$ is transcendental, then the slope of a line $\mathrm m$ is transcendental. | The slope of a line may be transcendental. | The slope form of any number $x$ may be produced by:
{{begin-eqn}}
{{eqn|l = {\mathrm m}
|r = \frac {x} {1}
|c = (Slope Form of $x$)
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn|l = {\mathrm m}
|r = {x}
|c =
}}
{{end-eqn}}
If $x$ is transcendental, then the slope of a line $\mathrm m$ is transcendental. | Transcendental Slope | https://proofwiki.org/wiki/Transcendental_Slope | https://proofwiki.org/wiki/Transcendental_Slope | [
"Transcendental Number Theory"
] | [] | [] |
proofwiki-5221 | Lagrange's Formula | Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be vectors in a vector space $\mathbf V$ of $3$ dimensions.
Then:
:$\mathbf a \times \paren {\mathbf b \times \mathbf c} = \paren {\mathbf a \cdot \mathbf c} \mathbf b - \paren {\mathbf a \cdot \mathbf b} \mathbf c$ | Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be embedded in a Cartesian $3$-space:
:$\mathbf a = \begin{bmatrix} a_x \\ a_y \\ a_z \end{bmatrix}$, $\mathbf b = \begin{bmatrix} b_x \\ b_y \\ b_z \end{bmatrix}$, $\mathbf c = \begin{bmatrix} c_x \\ c_y \\ c_z \end{bmatrix}$
{{begin-eqn}}
{{eqn | l = \mathbf b \times \math... | Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be [[Definition:Vector (Linear Algebra)|vectors]] in a [[Definition:Vector Space|vector space]] $\mathbf V$ of [[Definition:Dimension of Vector Space|$3$ dimensions]].
Then:
:$\mathbf a \times \paren {\mathbf b \times \mathbf c} = \paren {\mathbf a \cdot \mathbf c} \mathb... | Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be embedded in a [[Definition:Cartesian 3-Space|Cartesian $3$-space]]:
:$\mathbf a = \begin{bmatrix} a_x \\ a_y \\ a_z \end{bmatrix}$, $\mathbf b = \begin{bmatrix} b_x \\ b_y \\ b_z \end{bmatrix}$, $\mathbf c = \begin{bmatrix} c_x \\ c_y \\ c_z \end{bmatrix}$
{{begin-eqn}... | Lagrange's Formula | https://proofwiki.org/wiki/Lagrange's_Formula | https://proofwiki.org/wiki/Lagrange's_Formula | [
"Vector Algebra",
"Vector Triple Product"
] | [
"Definition:Vector/Linear Algebra",
"Definition:Vector Space",
"Definition:Dimension of Vector Space"
] | [
"Definition:Cartesian 3-Space"
] |
proofwiki-5222 | Isomorphism of External Direct Products/General Result | Let:
:$(1): \quad \ds \struct {S, \circ} = \prod_{k \mathop = 1}^n S_k = \struct {S_1, \circ_1} \times \struct {S_2, \circ_2} \times \cdots \times \struct {S_n, \circ_n}$
:$(2): \quad \ds \struct {T, \ast} = \prod_{k \mathop = 1}^n T_k = \struct {T_1, \ast_1} \times \struct {T_2, \ast_2} \times \cdots \times \struct {T... | By definition of isomorphism, each $\phi_k$ is a homomorphism which is a bijection.
From Cartesian Product of Bijections is Bijection: General Result, $\phi$ is a bijection.
From Homomorphism of External Direct Products: General Result, $\phi$ is a homomorphism.
Hence the result.
{{qed}} | Let:
:$(1): \quad \ds \struct {S, \circ} = \prod_{k \mathop = 1}^n S_k = \struct {S_1, \circ_1} \times \struct {S_2, \circ_2} \times \cdots \times \struct {S_n, \circ_n}$
:$(2): \quad \ds \struct {T, \ast} = \prod_{k \mathop = 1}^n T_k = \struct {T_1, \ast_1} \times \struct {T_2, \ast_2} \times \cdots \times \struct {... | By definition of [[Definition:Isomorphism (Abstract Algebra)|isomorphism]], each $\phi_k$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] which is a [[Definition:Bijection|bijection]].
From [[Cartesian Product of Bijections is Bijection/General Result|Cartesian Product of Bijections is Bijection: Gene... | Isomorphism of External Direct Products/General Result | https://proofwiki.org/wiki/Isomorphism_of_External_Direct_Products/General_Result | https://proofwiki.org/wiki/Isomorphism_of_External_Direct_Products/General_Result | [
"Isomorphism of External Direct Products"
] | [
"Definition:External Direct Product/General Definition",
"Definition:Algebraic Structure/One Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Bijection",
"Cartesian Product of Bijections is Bijection/General Result",
"Definition:Bijection",
"Homomorphism of External Direct Products/General Result",
"Definition:Homomorphism (Abstract Algebra)... |
proofwiki-5223 | External Direct Product of Groups is Group/Finite Product | The external direct product of a finite sequence of groups is itself a group. | Let $\struct {G_1, \circ_1}, \struct {G_2, \circ_2}, \ldots, \struct {G_n, \circ_n}$ be groups.
Let $\ds \struct {G, \circ} = \prod_{k \mathop = 1}^n G_k$ be the external direct product of $\struct {G_1, \circ_1}, \struct {G_2, \circ_2}, \ldots, \struct {G_n, \circ_n}$.
Taking the group axioms in turn: | The [[Definition:Group Direct Product|external direct product]] of a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Group|groups]] is itself a [[Definition:Group|group]]. | Let $\struct {G_1, \circ_1}, \struct {G_2, \circ_2}, \ldots, \struct {G_n, \circ_n}$ be [[Definition:Group|groups]].
Let $\ds \struct {G, \circ} = \prod_{k \mathop = 1}^n G_k$ be the [[Definition:External Direct Product|external direct product]] of $\struct {G_1, \circ_1}, \struct {G_2, \circ_2}, \ldots, \struct {G_n,... | External Direct Product of Groups is Group/Finite Product | https://proofwiki.org/wiki/External_Direct_Product_of_Groups_is_Group/Finite_Product | https://proofwiki.org/wiki/External_Direct_Product_of_Groups_is_Group/Finite_Product | [
"Group Theory",
"Group Direct Products"
] | [
"Definition:Group Direct Product",
"Definition:Finite Sequence",
"Definition:Group",
"Definition:Group"
] | [
"Definition:Group",
"Definition:External Direct Product",
"Axiom:Group Axioms",
"Axiom:Group Axioms",
"Definition:Group"
] |
proofwiki-5224 | External Direct Product Inverses/General Result | Let $\ds \struct {\SS, \circ} = \prod_{k \mathop = 1}^n S_k$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$.
Let $\tuple {x_1, x_2, \ldots, x_n} \in S$.
Let $y_k$ be an inverse of $x_k$ in $\struct {S_k, \circ_k}$ for each of $... | Let $e_1, e_2, \ldots, e_n$ be the identity elements of $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$ respectively.
Let $x := \tuple {x_1, x_2, \ldots, x_n}$.
Let $y := \tuple {x_1, x_2, \ldots, x_n}$.
From External Direct Product Identity, $e := \tuple {e_1, e_2, \ldots, e_n}$ is the... | Let $\ds \struct {\SS, \circ} = \prod_{k \mathop = 1}^n S_k$ be the [[Definition:External Direct Product/General Definition|external direct product]] of the [[Definition:Algebraic Structure with One Operation|algebraic structures]] $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$.
Let $... | Let $e_1, e_2, \ldots, e_n$ be the [[Definition:Identity Element|identity elements]] of $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$ respectively.
Let $x := \tuple {x_1, x_2, \ldots, x_n}$.
Let $y := \tuple {x_1, x_2, \ldots, x_n}$.
From [[External Direct Product Identity/General ... | External Direct Product Inverses/General Result | https://proofwiki.org/wiki/External_Direct_Product_Inverses/General_Result | https://proofwiki.org/wiki/External_Direct_Product_Inverses/General_Result | [
"External Direct Product Inverses"
] | [
"Definition:External Direct Product/General Definition",
"Definition:Algebraic Structure/One Operation",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"External Direct Product Identity/General Result",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"External Direct Product Identity/General Result",
"External Direct Product Identity/General Result"
] |
proofwiki-5225 | External Direct Product Identity/General Result | Let $\ds \struct {\SS, \circ} = \prod_{k \mathop = 1}^n S_k$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$.
Let $e_1, e_2, \ldots, e_n$ be the identity elements of $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \stru... | Let $s := \tuple {s_1, s_2, \ldots, s_n}$ be an arbitrary element of $\struct {S_1, \circ_1} \times \struct {S_2, \circ_2} \times \cdots \times \struct {S_n, \circ_n}$.
Let $e := \tuple {e_1, e_2, \ldots, e_n}$.
Then:
{{begin-eqn}}
{{eqn | l = s \circ e
| r = \tuple {s_1, s_2, \ldots, s_n} \circ \tuple {e_1, e_2,... | Let $\ds \struct {\SS, \circ} = \prod_{k \mathop = 1}^n S_k$ be the [[Definition:External Direct Product/General Definition|external direct product]] of the [[Definition:Algebraic Structure with One Operation|algebraic structures]] $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$.
Let ... | Let $s := \tuple {s_1, s_2, \ldots, s_n}$ be an arbitrary element of $\struct {S_1, \circ_1} \times \struct {S_2, \circ_2} \times \cdots \times \struct {S_n, \circ_n}$.
Let $e := \tuple {e_1, e_2, \ldots, e_n}$.
Then:
{{begin-eqn}}
{{eqn | l = s \circ e
| r = \tuple {s_1, s_2, \ldots, s_n} \circ \tuple {e_1, e_... | External Direct Product Identity/General Result | https://proofwiki.org/wiki/External_Direct_Product_Identity/General_Result | https://proofwiki.org/wiki/External_Direct_Product_Identity/General_Result | [
"External Direct Product Identity"
] | [
"Definition:External Direct Product/General Definition",
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [] |
proofwiki-5226 | External Direct Product Commutativity/General Result | Let $\ds \struct {S, \circ} = \prod_{k \mathop = 1}^n S_k$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$.
If $\circ_1, \ldots, \circ_n$ are all commutative, then so is $\circ$. | Suppose that, for all $k \in \N^*_n$, $\circ_k$ is commutative.
Let $\tuple {s_1, s_2, \ldots, s_n}$ and $\tuple {t_1, t_2, \ldots, t_n}$ be elements of $\struct {S_1, \circ_1} \times \struct {S_2, \circ_2} \times \cdots \times \struct {S_n, \circ_n}$.
{{begin-eqn}}
{{eqn | l = \tuple {s_1, s_2, \ldots, s_n} \circ \tup... | Let $\ds \struct {S, \circ} = \prod_{k \mathop = 1}^n S_k$ be the [[Definition:External Direct Product/General Definition|external direct product]] of the [[Definition:Algebraic Structure with One Operation|algebraic structures]] $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$.
If $\c... | Suppose that, for all $k \in \N^*_n$, $\circ_k$ is [[Definition:Commutative Operation|commutative]].
Let $\tuple {s_1, s_2, \ldots, s_n}$ and $\tuple {t_1, t_2, \ldots, t_n}$ be elements of $\struct {S_1, \circ_1} \times \struct {S_2, \circ_2} \times \cdots \times \struct {S_n, \circ_n}$.
{{begin-eqn}}
{{eqn | l = \t... | External Direct Product Commutativity/General Result | https://proofwiki.org/wiki/External_Direct_Product_Commutativity/General_Result | https://proofwiki.org/wiki/External_Direct_Product_Commutativity/General_Result | [
"External Direct Product Commutativity"
] | [
"Definition:External Direct Product/General Definition",
"Definition:Algebraic Structure/One Operation",
"Definition:Commutative/Operation"
] | [
"Definition:Commutative/Operation"
] |
proofwiki-5227 | External Direct Product Associativity/General Result | Let $\ds \struct {S, \circ} = \prod_{k \mathop = 1}^n S_k$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$.
If $\circ_1, \ldots, \circ_n$ are all associative, then so is $\circ$. | Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:If $\circ_1, \ldots, \circ_n$ are all associative, then so is the external direct product $\circ$ of $\circ_1, \ldots, \circ_n$. | Let $\ds \struct {S, \circ} = \prod_{k \mathop = 1}^n S_k$ be the [[Definition:External Direct Product/General Definition|external direct product]] of the [[Definition:Algebraic Structure with One Operation|algebraic structures]] $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$.
If $\c... | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:If $\circ_1, \ldots, \circ_n$ are all [[Definition:Associative Operation|associative]], then so is the [[Definition:External Direct Product/General Definition|external di... | External Direct Product Associativity/General Result | https://proofwiki.org/wiki/External_Direct_Product_Associativity/General_Result | https://proofwiki.org/wiki/External_Direct_Product_Associativity/General_Result | [
"External Direct Product Associativity"
] | [
"Definition:External Direct Product/General Definition",
"Definition:Algebraic Structure/One Operation",
"Definition:Associative Operation"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Associative Operation",
"Definition:External Direct Product/General Definition",
"Definition:Associative Operation",
"Definition:Associative Operation",
"Definition:External Direct Product/General Definition",
"Definition:Ass... |
proofwiki-5228 | Vector Cross Product is not Associative | The vector cross product is ''not'' associative.
That is, in general:
:$\mathbf a \times \paren {\mathbf b \times \mathbf c} \ne \paren {\mathbf a \times \mathbf b} \times \mathbf c$
for $\mathbf a, \mathbf b, \mathbf c \in \R^3$. | Proof by Counterexample:
Let:
{{begin-eqn}}
{{eqn | l = \mathbf a
| r = \begin {bmatrix} 1 \\ 0 \\ 0 \end {bmatrix}
}}
{{eqn | l = \mathbf b
| r = \begin {bmatrix} 1 \\ 1 \\ 0 \end {bmatrix}
}}
{{eqn | l = \mathbf c
| r = \begin {bmatrix} 1 \\ 1 \\ 1 \end {bmatrix}
}}
{{end-eqn}}
be vectors in the rea... | The [[Definition:Vector Cross Product|vector cross product]] is ''not'' [[Definition:Associative Operation|associative]].
That is, in general:
:$\mathbf a \times \paren {\mathbf b \times \mathbf c} \ne \paren {\mathbf a \times \mathbf b} \times \mathbf c$
for $\mathbf a, \mathbf b, \mathbf c \in \R^3$. | [[Proof by Counterexample]]:
Let:
{{begin-eqn}}
{{eqn | l = \mathbf a
| r = \begin {bmatrix} 1 \\ 0 \\ 0 \end {bmatrix}
}}
{{eqn | l = \mathbf b
| r = \begin {bmatrix} 1 \\ 1 \\ 0 \end {bmatrix}
}}
{{eqn | l = \mathbf c
| r = \begin {bmatrix} 1 \\ 1 \\ 1 \end {bmatrix}
}}
{{end-eqn}}
be [[Definition... | Vector Cross Product is not Associative | https://proofwiki.org/wiki/Vector_Cross_Product_is_not_Associative | https://proofwiki.org/wiki/Vector_Cross_Product_is_not_Associative | [
"Vector Cross Product"
] | [
"Definition:Vector Cross Product",
"Definition:Associative Operation"
] | [
"Proof by Counterexample",
"Definition:Vector/Real Euclidean Space/Space Vector",
"Definition:Euclidean Space/Real"
] |
proofwiki-5229 | Vector Cross Product is Anticommutative | The vector cross product is anticommutative:
:$\forall \mathbf a, \mathbf b \in \R^3: \mathbf a \times \mathbf b = -\paren {\mathbf b \times \mathbf a}$ | {{begin-eqn}}
{{eqn | l = \mathbf b \times \mathbf a
| r = \begin {bmatrix} b_i \\ b_j \\ b_k \end {bmatrix} \times \begin {bmatrix} a_i \\ a_j \\ a_k \end {bmatrix}
}}
{{eqn | r = \begin {bmatrix} b_j a_k - a_j b_k \\ b_k a_i - b_i a_k \\ b_i a_j - a_i b_j \end {bmatrix}
}}
{{eqn | l = \mathbf a \times \mathbf b... | The [[Definition:Vector Cross Product|vector cross product]] is [[Definition:Anticommutative|anticommutative]]:
:$\forall \mathbf a, \mathbf b \in \R^3: \mathbf a \times \mathbf b = -\paren {\mathbf b \times \mathbf a}$ | {{begin-eqn}}
{{eqn | l = \mathbf b \times \mathbf a
| r = \begin {bmatrix} b_i \\ b_j \\ b_k \end {bmatrix} \times \begin {bmatrix} a_i \\ a_j \\ a_k \end {bmatrix}
}}
{{eqn | r = \begin {bmatrix} b_j a_k - a_j b_k \\ b_k a_i - b_i a_k \\ b_i a_j - a_i b_j \end {bmatrix}
}}
{{eqn | l = \mathbf a \times \mathbf b... | Vector Cross Product is Anticommutative/Proof 1 | https://proofwiki.org/wiki/Vector_Cross_Product_is_Anticommutative | https://proofwiki.org/wiki/Vector_Cross_Product_is_Anticommutative/Proof_1 | [
"Vector Cross Product is Anticommutative",
"Vector Cross Product",
"Examples of Anticommutativity"
] | [
"Definition:Vector Cross Product",
"Definition:Anticommutative"
] | [] |
proofwiki-5230 | Vector Cross Product is Anticommutative | The vector cross product is anticommutative:
:$\forall \mathbf a, \mathbf b \in \R^3: \mathbf a \times \mathbf b = -\paren {\mathbf b \times \mathbf a}$ | {{begin-eqn}}
{{eqn | l = \paren {\mathbf a + \mathbf b} \times \paren {\mathbf a + \mathbf b}
| r = \mathbf a \times \mathbf a + \mathbf a \times \mathbf b + \mathbf b \times \mathbf a + \mathbf b \times \mathbf b
| c = Vector Cross Product Operator is Bilinear
}}
{{eqn | l = 0
| r = 0 + \mathbf a \t... | The [[Definition:Vector Cross Product|vector cross product]] is [[Definition:Anticommutative|anticommutative]]:
:$\forall \mathbf a, \mathbf b \in \R^3: \mathbf a \times \mathbf b = -\paren {\mathbf b \times \mathbf a}$ | {{begin-eqn}}
{{eqn | l = \paren {\mathbf a + \mathbf b} \times \paren {\mathbf a + \mathbf b}
| r = \mathbf a \times \mathbf a + \mathbf a \times \mathbf b + \mathbf b \times \mathbf a + \mathbf b \times \mathbf b
| c = [[Vector Cross Product Operator is Bilinear]]
}}
{{eqn | l = 0
| r = 0 + \mathbf ... | Vector Cross Product is Anticommutative/Proof 2 | https://proofwiki.org/wiki/Vector_Cross_Product_is_Anticommutative | https://proofwiki.org/wiki/Vector_Cross_Product_is_Anticommutative/Proof_2 | [
"Vector Cross Product is Anticommutative",
"Vector Cross Product",
"Examples of Anticommutativity"
] | [
"Definition:Vector Cross Product",
"Definition:Anticommutative"
] | [
"Vector Cross Product Operator is Bilinear",
"Cross Product of Vector with Itself is Zero"
] |
proofwiki-5231 | Vector Cross Product is Anticommutative | The vector cross product is anticommutative:
:$\forall \mathbf a, \mathbf b \in \R^3: \mathbf a \times \mathbf b = -\paren {\mathbf b \times \mathbf a}$ | {{begin-eqn}}
{{eqn | l = \mathbf a \times \mathbf b
| r = \begin {vmatrix} \mathbf i & \mathbf j & \mathbf k \\ a_i & a_j & a_k \\ b_i & b_j & b_k \end {vmatrix}
| c = {{Defof|Vector Cross Product}}
}}
{{eqn | r = -\begin {vmatrix} \mathbf i & \mathbf j & \mathbf k \\ b_i & b_j & b_k \\ a_i & a_j & a_k \en... | The [[Definition:Vector Cross Product|vector cross product]] is [[Definition:Anticommutative|anticommutative]]:
:$\forall \mathbf a, \mathbf b \in \R^3: \mathbf a \times \mathbf b = -\paren {\mathbf b \times \mathbf a}$ | {{begin-eqn}}
{{eqn | l = \mathbf a \times \mathbf b
| r = \begin {vmatrix} \mathbf i & \mathbf j & \mathbf k \\ a_i & a_j & a_k \\ b_i & b_j & b_k \end {vmatrix}
| c = {{Defof|Vector Cross Product}}
}}
{{eqn | r = -\begin {vmatrix} \mathbf i & \mathbf j & \mathbf k \\ b_i & b_j & b_k \\ a_i & a_j & a_k \en... | Vector Cross Product is Anticommutative/Proof 3 | https://proofwiki.org/wiki/Vector_Cross_Product_is_Anticommutative | https://proofwiki.org/wiki/Vector_Cross_Product_is_Anticommutative/Proof_3 | [
"Vector Cross Product is Anticommutative",
"Vector Cross Product",
"Examples of Anticommutativity"
] | [
"Definition:Vector Cross Product",
"Definition:Anticommutative"
] | [
"Determinant with Rows Transposed"
] |
proofwiki-5232 | Vector Cross Product is Anticommutative | The vector cross product is anticommutative:
:$\forall \mathbf a, \mathbf b \in \R^3: \mathbf a \times \mathbf b = -\paren {\mathbf b \times \mathbf a}$ | From the definition of the vector cross product:
{{:Definition:Vector Cross Product/Definition 2}}
Hence we have that:
:$\mathbf a \times \mathbf b$ is a vector whose direction is specified according to the right-hand rule
while:
:$\mathbf b \times \mathbf a$ is a vector whose direction, also specified according to the... | The [[Definition:Vector Cross Product|vector cross product]] is [[Definition:Anticommutative|anticommutative]]:
:$\forall \mathbf a, \mathbf b \in \R^3: \mathbf a \times \mathbf b = -\paren {\mathbf b \times \mathbf a}$ | From the definition of the [[Definition:Vector Cross Product/Definition 2|vector cross product]]:
{{:Definition:Vector Cross Product/Definition 2}}
Hence we have that:
:$\mathbf a \times \mathbf b$ is a [[Definition:Vector Quantity|vector]] whose [[Definition:Direction|direction]] is specified according to the [[Defin... | Vector Cross Product is Anticommutative/Proof 4 | https://proofwiki.org/wiki/Vector_Cross_Product_is_Anticommutative | https://proofwiki.org/wiki/Vector_Cross_Product_is_Anticommutative/Proof_4 | [
"Vector Cross Product is Anticommutative",
"Vector Cross Product",
"Examples of Anticommutativity"
] | [
"Definition:Vector Cross Product",
"Definition:Anticommutative"
] | [
"Definition:Vector Cross Product/Definition 2",
"Definition:Vector Quantity",
"Definition:Direction",
"Definition:Right-Hand Rule/Cross Product",
"Definition:Vector Quantity",
"Definition:Direction",
"Definition:Right-Hand Rule/Cross Product",
"Definition:Opposite Direction"
] |
proofwiki-5233 | Projection is Surjection/General Version | For all non-empty sets $S_1, S_2, \ldots, S_j, \ldots, S_n$, the $j$th projection $\pr_j$ on $\ds \prod_{i \mathop = 1}^n S_i$ is a surjection. | Consider the $j$th projection.
As long as none of $S_1, S_2, \ldots, S_n$ is the empty set, then:
:$\ds \forall x \in S_j: \exists \tuple {s_1, s_2, \ldots, s_{j - 1}, x, s_{j + 1}, \ldots, s_n} \in \prod_{k \mathop = 1}^n S_k: \map {\pr_j} {\tuple {s_1, s_2, \ldots, s_{j - 1}, x, s_{j + 1}, \ldots, s_n} } = x$
Hence t... | For all non-[[Definition:Empty Set|empty sets]] $S_1, S_2, \ldots, S_j, \ldots, S_n$, the [[Definition:Projection (Mapping Theory)|$j$th projection]] $\pr_j$ on $\ds \prod_{i \mathop = 1}^n S_i$ is a [[Definition:Surjection|surjection]]. | Consider the [[Definition:Projection (Mapping Theory)|$j$th projection]].
As long as none of $S_1, S_2, \ldots, S_n$ is the [[Definition:Empty Set|empty set]], then:
:$\ds \forall x \in S_j: \exists \tuple {s_1, s_2, \ldots, s_{j - 1}, x, s_{j + 1}, \ldots, s_n} \in \prod_{k \mathop = 1}^n S_k: \map {\pr_j} {\tuple {... | Projection is Surjection/General Version | https://proofwiki.org/wiki/Projection_is_Surjection/General_Version | https://proofwiki.org/wiki/Projection_is_Surjection/General_Version | [
"Surjections",
"Projections"
] | [
"Definition:Empty Set",
"Definition:Projection (Mapping Theory)",
"Definition:Surjection"
] | [
"Definition:Projection (Mapping Theory)",
"Definition:Empty Set",
"Category:Surjections",
"Category:Projections"
] |
proofwiki-5234 | Canonical Injection is Monomorphism/General Result | Let $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \dotsc, \struct {S_j, \circ_j}, \dotsc, \struct {S_n, \circ_n}$ be algebraic structures with identities $e_1, e_2, \dotsc, e_j, \dotsc, e_n$ respectively.
Then the canonical injection:
:$\ds \inj_j: \struct {S_j, \circ_j} \to \prod_{i \mathop = 1}^n \struct {S_i, \ci... | From Canonical Injection is Injection we have that the canonical injections are in fact injective.
It remains to prove the morphism property.
Let $x, y \in \struct {S_j, \circ_j}$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\inj_j} {x \circ_j y}
| r = \tuple {e_1, e_2, \dotsc, e_{j - 1}, x \circ_j y, e_{j + 1}, \dotsc... | Let $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \dotsc, \struct {S_j, \circ_j}, \dotsc, \struct {S_n, \circ_n}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]] with [[Definition:Identity Element|identities]] $e_1, e_2, \dotsc, e_j, \dotsc, e_n$ respectively.
Then the [[Definition:Can... | From [[Canonical Injection is Injection/General Result|Canonical Injection is Injection]] we have that the [[Definition:Canonical Injection (Abstract Algebra)/General Definition|canonical injections]] are in fact [[Definition:Injection|injective]].
It remains to prove the [[Definition:Morphism Property|morphism prope... | Canonical Injection is Monomorphism/General Result | https://proofwiki.org/wiki/Canonical_Injection_is_Monomorphism/General_Result | https://proofwiki.org/wiki/Canonical_Injection_is_Monomorphism/General_Result | [
"Canonical Injection is Monomorphism"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Canonical Injection (Abstract Algebra)/General Definition",
"Definition:Monomorphism (Abstract Algebra)"
] | [
"Canonical Injection is Injection/General Result",
"Definition:Canonical Injection (Abstract Algebra)/General Definition",
"Definition:Injection",
"Definition:Morphism Property",
"Definition:Morphism Property",
"Definition:Injection",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Monomorph... |
proofwiki-5235 | Orthogonality of Solutions to the Sturm-Liouville Equation with Distinct Eigenvalues | Let $\map f x$ and $\map g x$ be solutions of the Sturm-Liouville equation:
:$(1): \quad -\map {\dfrac \d {\d x} } {\map p x \dfrac {\d y} {\d x} } + \map q x y = \lambda \map w x y$
where $y$ is a function of the free variable $x$.
The functions $\map p x$, $\map q x$ and $\map w x$ are specified.
In the simplest case... | Multiply the equation for $\map g x$ by $\overline {\map f x}$ (the complex conjugate of $\map f x$) to get:
:$-\overline {\map f x} \dfrac {\map \d {\map p x \map {\dfrac {\d g} {\d x} } x} } {\d x} + \overline {\map f x} \map q x \map g x = \mu \overline {\map f x} \map w x \map g x$
Only $\map f x$, $\map g x$, $\la... | Let $\map f x$ and $\map g x$ be solutions of the [[Definition:Sturm-Liouville Equation|Sturm-Liouville equation]]:
:$(1): \quad -\map {\dfrac \d {\d x} } {\map p x \dfrac {\d y} {\d x} } + \map q x y = \lambda \map w x y$
where $y$ is a [[Definition:Real Function|function]] of the [[Definition:Free Variable|free var... | Multiply the equation for $\map g x$ by $\overline {\map f x}$ (the complex conjugate of $\map f x$) to get:
:$-\overline {\map f x} \dfrac {\map \d {\map p x \map {\dfrac {\d g} {\d x} } x} } {\d x} + \overline {\map f x} \map q x \map g x = \mu \overline {\map f x} \map w x \map g x$
Only $\map f x$, $\map g x$, $\... | Orthogonality of Solutions to the Sturm-Liouville Equation with Distinct Eigenvalues | https://proofwiki.org/wiki/Orthogonality_of_Solutions_to_the_Sturm-Liouville_Equation_with_Distinct_Eigenvalues | https://proofwiki.org/wiki/Orthogonality_of_Solutions_to_the_Sturm-Liouville_Equation_with_Distinct_Eigenvalues | [
"Sturm-Liouville Theory"
] | [
"Definition:Sturm-Liouville Equation",
"Definition:Real Function",
"Definition:Free Variable",
"Definition:Real Function",
"Definition:Continuous Real Function",
"Definition:Real Interval/Closed",
"Definition:Continuous Real Function",
"Definition:Derivative",
"Definition:Boundary Condition",
"Def... | [
"Category:Sturm-Liouville Theory"
] |
proofwiki-5236 | Row Equivalent Matrix for Homogeneous System has same Solutions | Let $\mathbf A$ be a matrix in the matrix space $\map {\MM_\R} {m, n}$ such that:
:$\mathbf A \mathbf x = \mathbf 0$
represents a homogeneous system of linear equations.
Let $\mathbf H$ be row equivalent to $\mathbf A$.
Then the solution set of $\mathbf H \mathbf x = \mathbf 0$ equals the solution set of $\mathbf A \ma... | Let:
{{begin-eqn}}
{{eqn | l = \alpha_{1 1} x_1 + \alpha_{1 2} x_2 + \ldots + \alpha_{1 n} x_n
| r = 0
| c =
}}
{{eqn | l = \alpha_{2 1} x_1 + \alpha_{2 2} x_2 + \ldots + \alpha_{2 n} x_n
| r = 0
| c =
}}
{{eqn | o = \vdots
}}
{{eqn | l = \alpha_{m 1} x_1 + \alpha_{m 2} x_2 + \ldots + \alpha_{... | Let $\mathbf A$ be a [[Definition:Matrix|matrix]] in the [[Definition:Matrix Space|matrix space]] $\map {\MM_\R} {m, n}$ such that:
:$\mathbf A \mathbf x = \mathbf 0$
represents a [[Definition:Homogeneous Linear Equations|homogeneous system of linear equations]].
Let $\mathbf H$ be [[Definition:Row Equivalence|row e... | Let:
{{begin-eqn}}
{{eqn | l = \alpha_{1 1} x_1 + \alpha_{1 2} x_2 + \ldots + \alpha_{1 n} x_n
| r = 0
| c =
}}
{{eqn | l = \alpha_{2 1} x_1 + \alpha_{2 2} x_2 + \ldots + \alpha_{2 n} x_n
| r = 0
| c =
}}
{{eqn | o = \vdots
}}
{{eqn | l = \alpha_{m 1} x_1 + \alpha_{m 2} x_2 + \ldots + \alpha_... | Row Equivalent Matrix for Homogeneous System has same Solutions | https://proofwiki.org/wiki/Row_Equivalent_Matrix_for_Homogeneous_System_has_same_Solutions | https://proofwiki.org/wiki/Row_Equivalent_Matrix_for_Homogeneous_System_has_same_Solutions | [
"Linear Algebra"
] | [
"Definition:Matrix",
"Definition:Matrix Space",
"Definition:Homogeneous Simultaneous Linear Equations",
"Definition:Row Equivalence",
"Definition:Simultaneous Equations/Solution Set",
"Definition:Simultaneous Equations/Solution Set",
"Definition:Row Equivalence"
] | [
"Definition:Elementary Operation/Row",
"Definition:Simultaneous Linear Equations/Matrix Representation",
"Definition:Logical Equivalence",
"Definition:Elementary Operation/Row",
"Definition:Simultaneous Linear Equations/Matrix Representation",
"Definition:Logical Equivalence",
"Definition:Elementary Ope... |
proofwiki-5237 | Null Space of Reduced Echelon Form | Let $\mathbf A$ be a matrix in the matrix space $\map {\MM_\R} {m, n}$ such that:
:$\mathbf A \mathbf x = \mathbf 0$
represents a homogeneous system of linear equations.
The null space of $\mathbf A$ is the same as that of the null space of the reduced row echelon form of $\mathbf A$:
:$\map {\mathrm N} {\mathbf A} = \... | By the definition of null space:
:$\mathbf x \in \map {\mathrm N} {\mathbf A} \iff \mathbf A \mathbf x = \mathbf 0$
From {{Corollary|Row Equivalent Matrix for Homogeneous System has same Solutions}}:
:$\mathbf A \mathbf x = \mathbf 0 \iff \map {\mathrm {rref} } {\mathbf A} \mathbf x = \mathbf 0$
Hence the result, by th... | Let $\mathbf A$ be a [[Definition:Matrix|matrix]] in the [[Definition:Matrix Space|matrix space]] $\map {\MM_\R} {m, n}$ such that:
:$\mathbf A \mathbf x = \mathbf 0$
represents a [[Definition:Homogeneous Linear Equations|homogeneous system of linear equations]].
The [[Definition:Null Space|null space]] of $\mathbf ... | By the definition of [[Definition:Null Space|null space]]:
:$\mathbf x \in \map {\mathrm N} {\mathbf A} \iff \mathbf A \mathbf x = \mathbf 0$
From {{Corollary|Row Equivalent Matrix for Homogeneous System has same Solutions}}:
:$\mathbf A \mathbf x = \mathbf 0 \iff \map {\mathrm {rref} } {\mathbf A} \mathbf x = \math... | Null Space of Reduced Echelon Form | https://proofwiki.org/wiki/Null_Space_of_Reduced_Echelon_Form | https://proofwiki.org/wiki/Null_Space_of_Reduced_Echelon_Form | [
"Linear Algebra",
"Null Spaces",
"Echelon Matrices"
] | [
"Definition:Matrix",
"Definition:Matrix Space",
"Definition:Homogeneous Simultaneous Linear Equations",
"Definition:Null Space",
"Definition:Echelon Matrix/Reduced Echelon Form"
] | [
"Definition:Null Space",
"Definition:Set Equality"
] |
proofwiki-5238 | Null Space Contains Only Zero Vector iff Columns are Independent | Let $\mathbb F$ be a field.
Let:
{{begin-eqn}}
{{eqn | l = \mathbf A_{m \times n}
| r = \begin{bmatrix} \mathbf a_1 & \mathbf a_2 & \cdots & \mathbf a_n \end{bmatrix}
}}
{{end-eqn}}
be a matrix over $\mathbb F$ where:
:$\forall i: 1 \le i \le n: \mathbf a_i = \begin{bmatrix} a_{1i} \\ a_{2i} \\ \vdots \\ a_{mi} \... | Let $\mathbf x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \in \mathbb F^m$.
We have that:
{{begin-eqn}}
{{eqn | l = \mathbf x
| o = \in
| r = \map {\mathrm N} {\mathbf A}
}}
{{eqn | ll= \leadstoandfrom
| l = \mathbf A \mathbf x_{n \times 1}
| r = \mathbf 0_{m \times 1}
}}
{{eqn... | Let $\mathbb F$ be a [[Definition:Field (Abstract Algebra)|field]].
Let:
{{begin-eqn}}
{{eqn | l = \mathbf A_{m \times n}
| r = \begin{bmatrix} \mathbf a_1 & \mathbf a_2 & \cdots & \mathbf a_n \end{bmatrix}
}}
{{end-eqn}}
be a [[Definition:Matrix|matrix]] over $\mathbb F$ where:
:$\forall i: 1 \le i \le n: \m... | Let $\mathbf x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \in \mathbb F^m$.
We have that:
{{begin-eqn}}
{{eqn | l = \mathbf x
| o = \in
| r = \map {\mathrm N} {\mathbf A}
}}
{{eqn | ll= \leadstoandfrom
| l = \mathbf A \mathbf x_{n \times 1}
| r = \mathbf 0_{m \times 1}
}}
{{e... | Null Space Contains Only Zero Vector iff Columns are Independent | https://proofwiki.org/wiki/Null_Space_Contains_Only_Zero_Vector_iff_Columns_are_Independent | https://proofwiki.org/wiki/Null_Space_Contains_Only_Zero_Vector_iff_Columns_are_Independent | [
"Linear Independence",
"Null Spaces"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Matrix",
"Definition:Vector/Real Euclidean Space",
"Definition:Linearly Independent/Set",
"Definition:Null Space"
] | [] |
proofwiki-5239 | Axiom of Dependent Choice Implies Axiom of Countable Choice | The axiom of dependent choice implies the axiom of countable choice. | Let $\sequence {S_n}_{n \mathop \in \N}$ be a sequence of non-empty sets.
Define:
:$\ds S = \bigsqcup_{n \mathop \in \N} S_n$
where $\bigsqcup$ denotes disjoint union.
Let $\RR$ be the binary endorelation on $S$ defined by:
:$\tuple {x, m} \mathrel \RR \tuple {y, n} \iff n = m + 1$
Note that $\RR$ satisfies:
:$\forall ... | The [[Axiom:Axiom of Dependent Choice|axiom of dependent choice]] implies the [[Axiom:Axiom of Countable Choice|axiom of countable choice]]. | Let $\sequence {S_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|sets]].
Define:
:$\ds S = \bigsqcup_{n \mathop \in \N} S_n$
where $\bigsqcup$ denotes [[Definition:Disjoint Union (Set Theory)|disjoint union]].
Let $\RR$ be the [[Definition:Endor... | Axiom of Dependent Choice Implies Axiom of Countable Choice | https://proofwiki.org/wiki/Axiom_of_Dependent_Choice_Implies_Axiom_of_Countable_Choice | https://proofwiki.org/wiki/Axiom_of_Dependent_Choice_Implies_Axiom_of_Countable_Choice | [
"Axiom of Dependent Choice",
"Axiom of Countable Choice"
] | [
"Axiom:Axiom of Dependent Choice",
"Axiom:Axiom of Countable Choice"
] | [
"Definition:Sequence",
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Disjoint Union (Set Theory)",
"Definition:Endorelation",
"Axiom:Axiom of Dependent Choice",
"Definition:Sequence",
"Principle of Mathematical Induction",
"Definition:Cartesian Product",
"Finite Cartesian Product of No... |
proofwiki-5240 | Axiom of Choice Implies Axiom of Dependent Choice | The axiom of choice implies the axiom of dependent choice. | Let $\RR$ be a binary endorelation on a non-empty set $S$ such that:
:$\forall a \in S: \exists b \in S: a \mathrel \RR b$
For an element $x \in S$, define:
:$\map R x = \set {y \in S: x \mathrel \RR y}$
By assumption, $\map R x$ is non-empty for all $x \in S$.
Now, consider the indexed family of sets:
:$\family {\map ... | The [[Axiom:Axiom of Choice|axiom of choice]] implies the [[Axiom:Axiom of Dependent Choice|axiom of dependent choice]]. | Let $\RR$ be a [[Definition:Endorelation|binary endorelation]] on a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]] $S$ such that:
:$\forall a \in S: \exists b \in S: a \mathrel \RR b$
For an [[Definition:Element|element]] $x \in S$, define:
:$\map R x = \set {y \in S: x \mathrel \RR y}$
By assumption, ... | Axiom of Choice Implies Axiom of Dependent Choice | https://proofwiki.org/wiki/Axiom_of_Choice_Implies_Axiom_of_Dependent_Choice | https://proofwiki.org/wiki/Axiom_of_Choice_Implies_Axiom_of_Dependent_Choice | [
"Axiom of Dependent Choice",
"Axiom of Choice"
] | [
"Axiom:Axiom of Choice",
"Axiom:Axiom of Dependent Choice"
] | [
"Definition:Endorelation",
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Element",
"Definition:Non-Empty Set",
"Definition:Indexing Set/Family of Sets",
"Axiom:Axiom of Choice",
"Definition:Mapping",
"Definition:Sequence",
"Definition:Composition of Mappings",
"Definition:Sequence"
] |
proofwiki-5241 | Infimum of Empty Set is Greatest Element | Let $\struct {S, \preceq}$ be an ordered set.
Suppose that $\map \inf \O$, the infimum of the empty set, exists in $S$.
Then $\forall s \in S: s \preceq \map \inf \O$.
That is, $\map \inf \O$ is the greatest element of $S$. | Observe that, vacuously, any $s \in S$ is a lower bound for $\O$.
But for any lower bound $s$ of $\O$, $s \preceq \map \inf \O$ by definition of infimum.
Hence:
:$\forall s \in S: s \preceq \map \inf \O$
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Suppose that $\map \inf \O$, the [[Definition:Infimum of Set|infimum]] of the [[Definition:Empty Set|empty set]], exists in $S$.
Then $\forall s \in S: s \preceq \map \inf \O$.
That is, $\map \inf \O$ is the [[Definition:Greatest Element|great... | Observe that, [[Definition:Vacuous Truth|vacuously]], any $s \in S$ is a [[Definition:Lower Bound of Set|lower bound]] for $\O$.
But for any [[Definition:Lower Bound of Set|lower bound]] $s$ of $\O$, $s \preceq \map \inf \O$ by definition of [[Definition:Infimum of Set|infimum]].
Hence:
:$\forall s \in S: s \preceq ... | Infimum of Empty Set is Greatest Element | https://proofwiki.org/wiki/Infimum_of_Empty_Set_is_Greatest_Element | https://proofwiki.org/wiki/Infimum_of_Empty_Set_is_Greatest_Element | [
"Empty Set",
"Infima"
] | [
"Definition:Ordered Set",
"Definition:Infimum of Set",
"Definition:Empty Set",
"Definition:Greatest Element"
] | [
"Definition:Vacuous Truth",
"Definition:Lower Bound of Set",
"Definition:Lower Bound of Set",
"Definition:Infimum of Set"
] |
proofwiki-5242 | Supremum of Empty Set is Smallest Element | Let $\struct {S, \preceq}$ be an ordered set.
Then:
:the supremum in $S$ of the empty set exists {{iff}} the smallest element of $S$ exists
in which case:
:$\map \sup \O$ is the smallest element of $S$ | Observe that, vacuously, any $s \in S$ is an upper bound for $\O$. | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Then:
:the [[Definition:Supremum of Set|supremum]] in $S$ of the [[Definition:Empty Set|empty set]] exists {{iff}} the [[Definition:Smallest Element|smallest element]] of $S$ exists
in which case:
:$\map \sup \O$ is the [[Definition:Smallest Ele... | Observe that, [[Definition:Vacuous Truth|vacuously]], any $s \in S$ is an [[Definition:Upper Bound of Set|upper bound]] for $\O$. | Supremum of Empty Set is Smallest Element | https://proofwiki.org/wiki/Supremum_of_Empty_Set_is_Smallest_Element | https://proofwiki.org/wiki/Supremum_of_Empty_Set_is_Smallest_Element | [
"Suprema",
"Empty Set"
] | [
"Definition:Ordered Set",
"Definition:Supremum of Set",
"Definition:Empty Set",
"Definition:Smallest Element",
"Definition:Smallest Element"
] | [
"Definition:Vacuous Truth",
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set"
] |
proofwiki-5243 | Countable Union of Countable Sets is Countable | Let the {{Axiom-link|Countable Choice}} be accepted.
Then it can be proved that a countable union of countable sets is countable. | Consider the countable sets $S_0, S_1, S_2, \ldots$ where $\ds S = \bigcup_{i \mathop \in \N} {S_i}$.
Assume that none of these sets have any elements in common.
Otherwise, we can consider the sets $S_0' = S_0, S_1' = S_1 \setminus S_0, S_2' = S_2 \setminus \paren {S_0 \cup S_1}, \ldots$
All of these are countable by S... | Let the {{Axiom-link|Countable Choice}} be accepted.
Then it can be proved that a [[Definition:Countable Union|countable union]] of [[Definition:Countable Set|countable sets]] is [[Definition:Countable|countable]]. | Consider the [[Definition:Countable|countable sets]] $S_0, S_1, S_2, \ldots$ where $\ds S = \bigcup_{i \mathop \in \N} {S_i}$.
Assume that none of these sets have any elements in common.
Otherwise, we can consider the sets $S_0' = S_0, S_1' = S_1 \setminus S_0, S_2' = S_2 \setminus \paren {S_0 \cup S_1}, \ldots$
All... | Countable Union of Countable Sets is Countable/Informal Proof | https://proofwiki.org/wiki/Countable_Union_of_Countable_Sets_is_Countable | https://proofwiki.org/wiki/Countable_Union_of_Countable_Sets_is_Countable/Informal_Proof | [
"Set Union",
"Countable Sets",
"Countable Union of Countable Sets is Countable"
] | [
"Definition:Set Union/Countable Union",
"Definition:Countable Set",
"Definition:Countable Set"
] | [
"Definition:Countable Set",
"Definition:Countable Set",
"Subset of Countable Set is Countable",
"Definition:Set Union",
"Definition:Infinite Set",
"Definition:Injection",
"Cartesian Product of Countable Sets is Countable",
"Definition:Injection",
"Definition:Injection",
"Composite of Injections is... |
proofwiki-5244 | Countable Union of Countable Sets is Countable | Let the {{Axiom-link|Countable Choice}} be accepted.
Then it can be proved that a countable union of countable sets is countable. | Let $\sequence {S_n}_{n \mathop \in \N}$ be a sequence of countable sets.
Define:
:$\ds S = \bigcup_{n \mathop \in \N} S_n$
For all $n \in \N$, let $\FF_n$ denote the set of all injections from $S_n$ to $\N$.
Since $S_n$ is countable, $\FF_n$ is non-empty.
Using the {{Axiom-link|Countable Choice}}, there exists a seque... | Let the {{Axiom-link|Countable Choice}} be accepted.
Then it can be proved that a [[Definition:Countable Union|countable union]] of [[Definition:Countable Set|countable sets]] is [[Definition:Countable|countable]]. | Let $\sequence {S_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Countable Set|countable sets]].
Define:
:$\ds S = \bigcup_{n \mathop \in \N} S_n$
For all $n \in \N$, let $\FF_n$ denote the set of all [[Definition:Injection|injections]] from $S_n$ to $\N$.
Since $S_n$ is [[Definition:C... | Countable Union of Countable Sets is Countable/Proof 1 | https://proofwiki.org/wiki/Countable_Union_of_Countable_Sets_is_Countable | https://proofwiki.org/wiki/Countable_Union_of_Countable_Sets_is_Countable/Proof_1 | [
"Set Union",
"Countable Sets",
"Countable Union of Countable Sets is Countable"
] | [
"Definition:Set Union/Countable Union",
"Definition:Countable Set",
"Definition:Countable Set"
] | [
"Definition:Sequence",
"Definition:Countable Set",
"Definition:Injection",
"Definition:Countable Set",
"Definition:Non-Empty Set",
"Definition:Sequence",
"Definition:Cartesian Product",
"Definition:Mapping",
"Smallest Element is Unique",
"Definition:Smallest Element",
"Definition:Natural Numbers... |
proofwiki-5245 | Countable Union of Countable Sets is Countable | Let the {{Axiom-link|Countable Choice}} be accepted.
Then it can be proved that a countable union of countable sets is countable. | Let $\sequence {S_n}_{n \mathop \in \N}$ be a sequence of countable sets.
Define:
:$\ds S = \bigcup_{n \mathop \in \N} S_n$.
For all $n \in \N$, let $\FF_n$ be the set of all surjections from $\N$ to $S_n$.
Since $S_n$ is countable, it follows by Surjection from Natural Numbers iff Countable that $\FF_n$ is non-empty.
... | Let the {{Axiom-link|Countable Choice}} be accepted.
Then it can be proved that a [[Definition:Countable Union|countable union]] of [[Definition:Countable Set|countable sets]] is [[Definition:Countable|countable]]. | Let $\sequence {S_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Countable Set|countable sets]].
Define:
:$\ds S = \bigcup_{n \mathop \in \N} S_n$.
For all $n \in \N$, let $\FF_n$ be the set of all [[Definition:Surjection|surjections]] from $\N$ to $S_n$.
Since $S_n$ is [[Definition:Co... | Countable Union of Countable Sets is Countable/Proof 2 | https://proofwiki.org/wiki/Countable_Union_of_Countable_Sets_is_Countable | https://proofwiki.org/wiki/Countable_Union_of_Countable_Sets_is_Countable/Proof_2 | [
"Set Union",
"Countable Sets",
"Countable Union of Countable Sets is Countable"
] | [
"Definition:Set Union/Countable Union",
"Definition:Countable Set",
"Definition:Countable Set"
] | [
"Definition:Sequence",
"Definition:Countable Set",
"Definition:Surjection",
"Definition:Countable Set",
"Surjection from Natural Numbers iff Countable",
"Definition:Non-Empty Set",
"Definition:Existential Quantifier",
"Definition:Sequence",
"Definition:Universal Quantifier",
"Definition:Cartesian ... |
proofwiki-5246 | Projection is Epimorphism/General Result | Let $\struct {\SS, \circ}$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_k, \circ_k}, \ldots, \struct {S_n, \circ_n}$.
Then:
:for each $j \in \closedint 1 n$, $\pr_j$ is an epimorphism from $\struct {\SS, \circ}$ to $\struct {S_j, \circ_j}... | From Projection is Surjection, $\pr_j$ is a surjection for all $j$.
We now need to show it is a homomorphism.
Let $\mathbf s, \mathbf t \in \struct {\SS, \circ}$ where $\mathbf s = \tuple {s_1, s_2, \ldots, s_j, \ldots, s_n}$ and $\mathbf t = \tuple {t_1, t_2, \ldots, t_j, \ldots, t_n}$.
Then:
{{begin-eqn}}
{{eqn | l =... | Let $\struct {\SS, \circ}$ be the [[Definition:External Direct Product|external direct product]] of the [[Definition:Algebraic Structure with One Operation|algebraic structures]] $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_k, \circ_k}, \ldots, \struct {S_n, \circ_n}$.
Then:
:for each $j \in \cl... | From [[Projection is Surjection]], $\pr_j$ is a [[Definition:Surjection|surjection]] for all $j$.
We now need to show it is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
Let $\mathbf s, \mathbf t \in \struct {\SS, \circ}$ where $\mathbf s = \tuple {s_1, s_2, \ldots, s_j, \ldots, s_n}$ and $\mathbf t ... | Projection is Epimorphism/General Result | https://proofwiki.org/wiki/Projection_is_Epimorphism/General_Result | https://proofwiki.org/wiki/Projection_is_Epimorphism/General_Result | [
"Projection is Epimorphism"
] | [
"Definition:External Direct Product",
"Definition:Algebraic Structure/One Operation",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Projection (Mapping Theory)"
] | [
"Projection is Surjection",
"Definition:Surjection",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Morphism Property"
] |
proofwiki-5247 | External Direct Product of Projection with Canonical Injection/General Result | Let $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \dotsc, \struct {S_j, \circ_j}, \dotsc, \struct {S_n, \circ_n}$ be algebraic structures with identities $e_1, e_2, \dotsc, e_j, \dotsc, e_n$ respectively.
Let $\ds \struct {S, \circ} = \prod_{i \mathop = 1}^n \struct {S_i, \circ_i}$ be the external direct product of ... | Let $\ds \tuple {s_1, s_2, \dotsc, s_{j - 1}, s_j, s_{j + 1}, \dotsc, s_n} \in \prod_{i \mathop = 1}^n \struct {S_i, \circ_i}$.
So:
:$s_j \in S_j$
From the definition of the canonical injection:
:$\map {\inj_j} {s_j} = \tuple {e_1, e_2, \dotsc, e_{j - 1}, s_j, e_{j + 1}, \dotsc, e_n}$
So from the definition of the $j$t... | Let $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \dotsc, \struct {S_j, \circ_j}, \dotsc, \struct {S_n, \circ_n}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]] with [[Definition:Identity Element|identities]] $e_1, e_2, \dotsc, e_j, \dotsc, e_n$ respectively.
Let $\ds \struct {S, \cir... | Let $\ds \tuple {s_1, s_2, \dotsc, s_{j - 1}, s_j, s_{j + 1}, \dotsc, s_n} \in \prod_{i \mathop = 1}^n \struct {S_i, \circ_i}$.
So:
:$s_j \in S_j$
From the definition of the [[Definition:Canonical Injection (Abstract Algebra)/General Definition|canonical injection]]:
:$\map {\inj_j} {s_j} = \tuple {e_1, e_2, \dotsc, ... | External Direct Product of Projection with Canonical Injection/General Result | https://proofwiki.org/wiki/External_Direct_Product_of_Projection_with_Canonical_Injection/General_Result | https://proofwiki.org/wiki/External_Direct_Product_of_Projection_with_Canonical_Injection/General_Result | [
"External Direct Products",
"Canonical Injections"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:External Direct Product/General Definition",
"Definition:Projection (Mapping Theory)",
"Definition:Canonical Injection (Abstract Algebra)/General Definition",
"Definition:Identity Mappi... | [
"Definition:Canonical Injection (Abstract Algebra)/General Definition",
"Definition:Projection (Mapping Theory)",
"Definition:Identity Mapping"
] |
proofwiki-5248 | Extended Real Addition is Commutative | Extended real addition $+_{\overline{\R}}$ is commutative.
That is, for all $x, y \in \overline{\R}$:
:$x +_{\overline{\R}} y = y +_{\overline{\R}} x$
whenever at least one of the sides is defined. | When $x,y \in \R$, then $x +_{\overline{\R}} y = y +_{\overline{\R}} x$ follows from Real Addition is Commutative.
The remaining cases where the expressions are defined, are already imposed in the definition of $+_{\overline{\R}}$.
{{qed}}
Category:Extended Real Numbers
re0bgbsugf5tynnbcvx6y0icrntr5fa | [[Definition:Extended Real Addition|Extended real addition]] $+_{\overline{\R}}$ is [[Definition:Commutative Operation|commutative]].
That is, for all $x, y \in \overline{\R}$:
:$x +_{\overline{\R}} y = y +_{\overline{\R}} x$
whenever at least one of the sides is defined. | When $x,y \in \R$, then $x +_{\overline{\R}} y = y +_{\overline{\R}} x$ follows from [[Real Addition is Commutative]].
The remaining cases where the expressions are defined, are already imposed in the [[Definition:Extended Real Addition|definition of $+_{\overline{\R}}$]].
{{qed}}
[[Category:Extended Real Numbers]]
r... | Extended Real Addition is Commutative | https://proofwiki.org/wiki/Extended_Real_Addition_is_Commutative | https://proofwiki.org/wiki/Extended_Real_Addition_is_Commutative | [
"Extended Real Numbers"
] | [
"Definition:Extended Real Addition",
"Definition:Commutative/Operation"
] | [
"Real Addition is Commutative",
"Definition:Extended Real Addition",
"Category:Extended Real Numbers"
] |
proofwiki-5249 | Extended Real Addition is Associative | Extended real addition $+_{\overline{\R}}$ is commutative.
That is, for all $x, y, z \in \overline \R$:
:$(1): \quad x +_{\overline \R} \paren {y +_{\overline \R} z} = \paren {x +_{\overline \R} y} +_{\overline \R} z$
whenever at least one of the sides is defined. | When $x, y, z \in \R$, then $(1)$ follows from Real Addition is Associative.
From the definition of $+_{\overline \R}$, it follows that either expression in $(1)$ is defined {{iff}} at most one of $+\infty$ and $-\infty$ occurs.
The case where neither occurs is already covered above; now assume that $+\infty$ occurs (t... | [[Definition:Extended Real Addition|Extended real addition]] $+_{\overline{\R}}$ is [[Definition:Commutative Operation|commutative]].
That is, for all $x, y, z \in \overline \R$:
:$(1): \quad x +_{\overline \R} \paren {y +_{\overline \R} z} = \paren {x +_{\overline \R} y} +_{\overline \R} z$
whenever at least one o... | When $x, y, z \in \R$, then $(1)$ follows from [[Real Addition is Associative]].
From the [[Definition:Extended Real Addition|definition of $+_{\overline \R}$]], it follows that either expression in $(1)$ is defined {{iff}} at most one of $+\infty$ and $-\infty$ occurs.
The case where neither occurs is already cover... | Extended Real Addition is Associative | https://proofwiki.org/wiki/Extended_Real_Addition_is_Associative | https://proofwiki.org/wiki/Extended_Real_Addition_is_Associative | [
"Extended Real Numbers"
] | [
"Definition:Extended Real Addition",
"Definition:Commutative/Operation"
] | [
"Real Addition is Associative",
"Definition:Extended Real Addition",
"Extended Real Addition is Commutative",
"Proof by Cases",
"Category:Extended Real Numbers"
] |
proofwiki-5250 | Extended Real Multiplication is Commutative | Extended real multiplication $\times_{\overline \R}$ is commutative.
That is, for all $x, y \in \overline \R$:
:$x \times_{\overline \R} y = y \times_{\overline \R} x$ | Let $x, y \in \R$.
Then from Real Multiplication is Commutative:
:$x \times_{\overline \R} y = y \times_{\overline \R} x$
The remaining cases are explicitly imposed in the definition of $\times_{\overline \R}$.
{{qed}}
Category:Extended Real Numbers
ers5dtbwrqxqxdxuunybf70906gsovl | [[Definition:Extended Real Multiplication|Extended real multiplication]] $\times_{\overline \R}$ is [[Definition:Commutative Operation|commutative]].
That is, for all $x, y \in \overline \R$:
:$x \times_{\overline \R} y = y \times_{\overline \R} x$ | Let $x, y \in \R$.
Then from [[Real Multiplication is Commutative]]:
:$x \times_{\overline \R} y = y \times_{\overline \R} x$
The remaining cases are explicitly imposed in the [[Definition:Extended Real Addition|definition of $\times_{\overline \R}$]].
{{qed}}
[[Category:Extended Real Numbers]]
ers5dtbwrqxqxdxuunybf... | Extended Real Multiplication is Commutative | https://proofwiki.org/wiki/Extended_Real_Multiplication_is_Commutative | https://proofwiki.org/wiki/Extended_Real_Multiplication_is_Commutative | [
"Extended Real Numbers"
] | [
"Definition:Extended Real Multiplication",
"Definition:Commutative/Operation"
] | [
"Real Multiplication is Commutative",
"Definition:Extended Real Addition",
"Category:Extended Real Numbers"
] |
proofwiki-5251 | Extended Real Multiplication is Associative | Extended real multiplication $\cdot_{\overline \R}$ is commutative.
That is, for all $x, y, z \in \overline \R$:
:$(1): \quad x \cdot_{\overline \R} \left({y \cdot_{\overline \R} z}\right) = \left({x \cdot_{\overline \R} y}\right) \cdot_{\overline \R} z$ | When $x, y, z \in \R$, then $(1)$ follows from Real Multiplication is Associative.
Next, the cases where at least one of $+\infty$ and $-\infty$ occurs need to be dealt with.
{{ProofWanted|when someone thinks of a nice way to deal with the case distinctions, go ahead}}
Category:Extended Real Numbers
42pagy9miv7xxla3wtf... | [[Definition:Extended Real Multiplication|Extended real multiplication]] $\cdot_{\overline \R}$ is [[Definition:Commutative Operation|commutative]].
That is, for all $x, y, z \in \overline \R$:
:$(1): \quad x \cdot_{\overline \R} \left({y \cdot_{\overline \R} z}\right) = \left({x \cdot_{\overline \R} y}\right) \cdot... | When $x, y, z \in \R$, then $(1)$ follows from [[Real Multiplication is Associative]].
Next, the cases where at least one of $+\infty$ and $-\infty$ occurs need to be dealt with.
{{ProofWanted|when someone thinks of a nice way to deal with the case distinctions, go ahead}}
[[Category:Extended Real Numbers]]
42pagy9... | Extended Real Multiplication is Associative | https://proofwiki.org/wiki/Extended_Real_Multiplication_is_Associative | https://proofwiki.org/wiki/Extended_Real_Multiplication_is_Associative | [
"Extended Real Numbers"
] | [
"Definition:Extended Real Multiplication",
"Definition:Commutative/Operation"
] | [
"Real Multiplication is Associative",
"Category:Extended Real Numbers"
] |
proofwiki-5252 | Extended Real Numbers under Multiplication form Monoid | Denote with $\overline \R$ the extended real numbers.
Denote with $\cdot_{\overline \R}$ the extended real multiplication.
The algebraic structure $\struct {\overline \R, \cdot_{\overline \R} }$ is a monoid. | Checking the axioms for a monoid in turn: | Denote with $\overline \R$ the [[Definition:Extended Real Number Line|extended real numbers]].
Denote with $\cdot_{\overline \R}$ the [[Definition:Extended Real Multiplication|extended real multiplication]].
The [[Definition:Algebraic Structure with One Operation|algebraic structure]] $\struct {\overline \R, \cdot_{... | Checking the axioms for a [[Definition:Monoid|monoid]] in turn: | Extended Real Numbers under Multiplication form Monoid | https://proofwiki.org/wiki/Extended_Real_Numbers_under_Multiplication_form_Monoid | https://proofwiki.org/wiki/Extended_Real_Numbers_under_Multiplication_form_Monoid | [
"Extended Real Numbers",
"Examples of Monoids"
] | [
"Definition:Extended Real Number Line",
"Definition:Extended Real Multiplication",
"Definition:Algebraic Structure/One Operation",
"Definition:Monoid"
] | [
"Definition:Monoid",
"Definition:Monoid"
] |
proofwiki-5253 | Extended Real Numbers under Multiplication form Commutative Monoid | Denote with $\overline \R$ the extended real numbers.
Denote with $\cdot_{\overline \R}$ the extended real multiplication.
The algebraic structure $\struct {\overline \R, \cdot_{\overline \R} }$ is a commutative monoid. | By Extended Real Numbers under Multiplication form Monoid, $\struct {\overline \R, \cdot_{\overline \R} }$ is a monoid.
By Extended Real Multiplication is Commutative, $\cdot_{\overline \R}$ is commutative.
Hence $\struct {\overline \R, \cdot_{\overline \R} }$ is a commutative monoid.
{{qed}}
Category:Extended Real Num... | Denote with $\overline \R$ the [[Definition:Extended Real Number Line|extended real numbers]].
Denote with $\cdot_{\overline \R}$ the [[Definition:Extended Real Multiplication|extended real multiplication]].
The [[Definition:Algebraic Structure with One Operation|algebraic structure]] $\struct {\overline \R, \cdot_{... | By [[Extended Real Numbers under Multiplication form Monoid]], $\struct {\overline \R, \cdot_{\overline \R} }$ is a [[Definition:Monoid|monoid]].
By [[Extended Real Multiplication is Commutative]], $\cdot_{\overline \R}$ is [[Definition:Commutative Operation|commutative]].
Hence $\struct {\overline \R, \cdot_{\overl... | Extended Real Numbers under Multiplication form Commutative Monoid | https://proofwiki.org/wiki/Extended_Real_Numbers_under_Multiplication_form_Commutative_Monoid | https://proofwiki.org/wiki/Extended_Real_Numbers_under_Multiplication_form_Commutative_Monoid | [
"Extended Real Numbers",
"Examples of Commutative Monoids"
] | [
"Definition:Extended Real Number Line",
"Definition:Extended Real Multiplication",
"Definition:Algebraic Structure/One Operation",
"Definition:Commutative Monoid"
] | [
"Extended Real Numbers under Multiplication form Monoid",
"Definition:Monoid",
"Extended Real Multiplication is Commutative",
"Definition:Commutative/Operation",
"Definition:Commutative Monoid",
"Category:Extended Real Numbers",
"Category:Examples of Commutative Monoids"
] |
proofwiki-5254 | Translation in Euclidean Space is Measurable Mapping | Let $\BB$ be the Borel $\sigma$-algebra on $\R^n$.
Let $\mathbf x \in \R^n$, and denote with $\tau_{\mathbf x}: \R^n \to \R^n$ translation by $\mathbf x$.
Then $\tau_{\mathbf x}$ is $\BB \, / \, \BB$-measurable. | By Characterization of Euclidean Borel Sigma-Algebra, $\BB = \map \sigma {\JJ_{ho}^n}$.
Here, $\JJ_{ho}^n$ is the set of half-open $n$-rectangles, and $\sigma$ denotes generated $\sigma$-algebra.
Now, for any half-open $n$-rectangle $\horectr {\mathbf a} {\mathbf b}$, it is trivial that:
:$\map {\tau_{\mathbf x}^{-1} }... | Let $\BB$ be the [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]] on $\R^n$.
Let $\mathbf x \in \R^n$, and denote with $\tau_{\mathbf x}: \R^n \to \R^n$ [[Definition:Translation in Euclidean Space|translation by $\mathbf x$]].
Then $\tau_{\mathbf x}$ is [[Definition:Measurable Mapping|$\BB \, / \, \BB$-meas... | By [[Characterization of Euclidean Borel Sigma-Algebra]], $\BB = \map \sigma {\JJ_{ho}^n}$.
Here, $\JJ_{ho}^n$ is the set of [[Definition:Half-Open Rectangle|half-open $n$-rectangles]], and $\sigma$ denotes [[Definition:Generated Sigma-Algebra|generated $\sigma$-algebra]].
Now, for any [[Definition:Half-Open Rectang... | Translation in Euclidean Space is Measurable Mapping | https://proofwiki.org/wiki/Translation_in_Euclidean_Space_is_Measurable_Mapping | https://proofwiki.org/wiki/Translation_in_Euclidean_Space_is_Measurable_Mapping | [
"Measure Theory",
"Translation Mappings"
] | [
"Definition:Borel Sigma-Algebra",
"Definition:Translation Mapping/Euclidean Space",
"Definition:Measurable Mapping"
] | [
"Characterization of Euclidean Borel Sigma-Algebra",
"Definition:Half-Open Rectangle",
"Definition:Generated Sigma-Algebra",
"Definition:Half-Open Rectangle",
"Definition:Preimage/Mapping/Subset",
"Definition:Half-Open Rectangle",
"Definition:Half-Open Rectangle",
"Mapping Measurable iff Measurable on... |
proofwiki-5255 | Conditions for Internal Group Direct Product | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $H$ and $K$ be subgroups of $G$.
Let the mapping $\phi: H \times K \to G$ defined as:
:$\forall h \in H, k \in K: \map \phi {h, k} = h \circ k$
be a group isomorphism from the cartesian product $\struct {H, \circ {\restriction_H} } \times \struct {K, \circ ... | === Necessary Condition ===
Let the mapping $\phi: H \times K \to G$ defined as:
:$\forall h \in H, k \in K: \map \phi {h, k} = h \circ k$
be a group isomorphism from the cartesian product $\struct {H, \circ {\restriction_H} } \times \struct {K, \circ {\restriction_K} }$ onto $\struct {G, \circ}$.
Let the symbol $\circ... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $H$ and $K$ be [[Definition:Subgroup|subgroups]] of $G$.
Let the [[Definition:Mapping|mapping]] $\phi: H \times K \to G$ defined as:
:$\forall h \in H, k \in K: \map \phi {h, k} = h \circ k$
be a [[Def... | === Necessary Condition ===
Let the [[Definition:Mapping|mapping]] $\phi: H \times K \to G$ defined as:
:$\forall h \in H, k \in K: \map \phi {h, k} = h \circ k$
be a [[Definition:Group Isomorphism|group isomorphism]] from the [[Definition:Cartesian Product|cartesian product]] $\struct {H, \circ {\restriction_H} } \ti... | Conditions for Internal Group Direct Product | https://proofwiki.org/wiki/Conditions_for_Internal_Group_Direct_Product | https://proofwiki.org/wiki/Conditions_for_Internal_Group_Direct_Product | [
"Internal Group Direct Products"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subgroup",
"Definition:Mapping",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Cartesian Product",
"Definition:Internal Group Direct Product"
] | [
"Definition:Mapping",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Cartesian Product",
"Definition:Operation Induced by Direct Product",
"Internal Group Direct Product Commutativity",
"Codomain of Internal Direct Isomorphism is Subset Product of Factors",
"Intersection of S... |
proofwiki-5256 | Internal Group Direct Product Commutativity/General Result | Let $\struct {G, \circ}$ be the internal group direct product of $H_1, H_2, \ldots, H_n$.
Let $h_i$ and $h_j$ be elements of $H_i$ and $H_j$ respectively, $i \ne j$.
Then $h_i \circ h_j = h_j \circ h_i$. | Let $g = h_i \circ h_j \circ h_i^{-1} \circ h_j^{-1}$.
From the Internal Direct Product Theorem: General Result, $H_i$ and $H_j$ are normal in $G$.
Hence $h_i \circ h_j \circ h_i^{-1} \in H_j$ and thus $g \in H_j$.
Similarly, $g \in H_i$ and thus $g \in H_i \cap H_j$.
But:
{{begin-eqn}}
{{eqn | l = H_i \cap H_j
|... | Let $\struct {G, \circ}$ be the [[Definition:Internal Group Direct Product/General Definition|internal group direct product]] of $H_1, H_2, \ldots, H_n$.
Let $h_i$ and $h_j$ be [[Definition:Element|elements]] of $H_i$ and $H_j$ respectively, $i \ne j$.
Then $h_i \circ h_j = h_j \circ h_i$. | Let $g = h_i \circ h_j \circ h_i^{-1} \circ h_j^{-1}$.
From the [[Internal Direct Product Theorem/General Result|Internal Direct Product Theorem: General Result]], $H_i$ and $H_j$ are [[Definition:Normal Subgroup|normal in $G$]].
Hence $h_i \circ h_j \circ h_i^{-1} \in H_j$ and thus $g \in H_j$.
Similarly, $g \in H_... | Internal Group Direct Product Commutativity/General Result | https://proofwiki.org/wiki/Internal_Group_Direct_Product_Commutativity/General_Result | https://proofwiki.org/wiki/Internal_Group_Direct_Product_Commutativity/General_Result | [
"Internal Group Direct Product Commutativity"
] | [
"Definition:Internal Group Direct Product/General Definition",
"Definition:Element"
] | [
"Internal Direct Product Theorem/General Result",
"Definition:Normal Subgroup"
] |
proofwiki-5257 | Pre-Image Sigma-Algebra on Codomain is Sigma-Algebra | Let $X, X'$ be sets, and let $f: X \to X'$ be a mapping.
Let $\Sigma$ be a $\sigma$-algebra on $X$.
Denote with $\Sigma'$ the pre-image $\sigma$-algebra on the domain of $f$.
Then $\Sigma'$ is a $\sigma$-algebra on $X'$. | Verify the axioms for a $\sigma$-algebra in turn: | Let $X, X'$ be [[Definition:Set|sets]], and let $f: X \to X'$ be a [[Definition:Mapping|mapping]].
Let $\Sigma$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $X$.
Denote with $\Sigma'$ the [[Definition:Pre-Image Sigma-Algebra on Codomain|pre-image $\sigma$-algebra]] on the [[Definition:Codomain of Mapping|dom... | Verify the axioms for a [[Definition:Sigma-Algebra|$\sigma$-algebra]] in turn: | Pre-Image Sigma-Algebra on Codomain is Sigma-Algebra | https://proofwiki.org/wiki/Pre-Image_Sigma-Algebra_on_Codomain_is_Sigma-Algebra | https://proofwiki.org/wiki/Pre-Image_Sigma-Algebra_on_Codomain_is_Sigma-Algebra | [
"Sigma-Algebras",
"Examples of Sigma-Algebras"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Sigma-Algebra",
"Definition:Pre-Image Sigma-Algebra/Codomain",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Sigma-Algebra"
] | [
"Definition:Sigma-Algebra",
"Definition:Sigma-Algebra",
"Definition:Sigma-Algebra",
"Definition:Sigma-Algebra"
] |
proofwiki-5258 | Mapping Measurable iff Measurable on Generator | Let $\struct {X, \Sigma}$ and $\struct {X', \Sigma'}$ be measurable spaces.
Suppose that $\Sigma'$ is generated by $\GG'$.
Then a mapping $f: X \to X'$ is $\Sigma \, / \, \Sigma'$-measurable {{iff}}:
:$\forall G' \in \GG': \map {f^{-1} } {G'} \in \Sigma$
That is, {{iff}} the preimage of every generator under $f$ is a m... | === Necessary Condition ===
Let $f$ be $\Sigma \, / \, \Sigma'$-measurable.
By definition of generated $\sigma$-algebra $\GG' \subseteq \Sigma'$.
Hence, in particular, $f$ satisfies:
:$\forall G' \in \GG': \map {f^{-1} } {G'} \in \Sigma$
{{qed|lemma}} | Let $\struct {X, \Sigma}$ and $\struct {X', \Sigma'}$ be [[Definition:Measurable Space|measurable spaces]].
Suppose that $\Sigma'$ is [[Definition:Sigma-Algebra Generated by Collection of Subsets|generated by]] $\GG'$.
Then a [[Definition:Mapping|mapping]] $f: X \to X'$ is [[Definition:Measurable Mapping|$\Sigma \, ... | === Necessary Condition ===
Let $f$ be [[Definition:Measurable Mapping|$\Sigma \, / \, \Sigma'$-measurable]].
By definition of [[Definition:Sigma-Algebra Generated by Collection of Subsets|generated $\sigma$-algebra]] $\GG' \subseteq \Sigma'$.
Hence, in particular, $f$ satisfies:
:$\forall G' \in \GG': \map {f^{-1}... | Mapping Measurable iff Measurable on Generator | https://proofwiki.org/wiki/Mapping_Measurable_iff_Measurable_on_Generator | https://proofwiki.org/wiki/Mapping_Measurable_iff_Measurable_on_Generator | [
"Measure Theory"
] | [
"Definition:Measurable Space",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Mapping",
"Definition:Measurable Mapping",
"Definition:Preimage/Mapping/Subset",
"Definition:Sigma-Algebra Generated by Collection of Subsets/Generator",
"Definition:Measurable Set"
] | [
"Definition:Measurable Mapping",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Measurable Mapping"
] |
proofwiki-5259 | Internal Direct Product Theorem/General Result | Let $G$ be a group whose identity is $e$.
Let $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of subgroups of $G$.
Then $G$ is the internal group direct product of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ {{iff}}:
{{begin-itemize}}
{{item|(1):|$G {{=}} H_1 H_2 \cdots H_n$}}
{{item|(2):|$\sequen... | By definition, $G$ is the internal group direct product of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ {{iff}} the mapping:
:$\ds C: \prod_{k \mathop = 1}^n H_k \to G: \map C {h_1, \ldots, h_n} = \prod_{k \mathop = 1}^n h_k$
is a group isomorphism from the cartesian product $\struct {H_1, \circ {\restriction_{H_1... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Subgroup|subgroups]] of $G$.
Then $G$ is the [[Definition:Internal Group Direct Product/General Definition|interna... | By definition, $G$ is the [[Definition:Internal Group Direct Product/General Definition|internal group direct product]] of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ {{iff}} the [[Definition:Mapping|mapping]]:
:$\ds C: \prod_{k \mathop = 1}^n H_k \to G: \map C {h_1, \ldots, h_n} = \prod_{k \mathop = 1}^n h_k$
... | Internal Direct Product Theorem/General Result/Proof 1 | https://proofwiki.org/wiki/Internal_Direct_Product_Theorem/General_Result | https://proofwiki.org/wiki/Internal_Direct_Product_Theorem/General_Result/Proof_1 | [
"Internal Direct Product Theorem"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Sequence",
"Definition:Subgroup",
"Definition:Internal Group Direct Product/General Definition",
"Definition:Sequence",
"Definition:Independent Subgroups",
"Definition:Normal Subgroup"
] | [
"Definition:Internal Group Direct Product/General Definition",
"Definition:Mapping",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Cartesian Product/Finite",
"Definition:Internal Group Direct Product/General Definition",
"Subgroup Product is Internal Group Direct Product iff S... |
proofwiki-5260 | Internal Direct Product Theorem/General Result | Let $G$ be a group whose identity is $e$.
Let $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of subgroups of $G$.
Then $G$ is the internal group direct product of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ {{iff}}:
{{begin-itemize}}
{{item|(1):|$G {{=}} H_1 H_2 \cdots H_n$}}
{{item|(2):|$\sequen... | It is to be shown that:
{{:Definition:Internal Group Direct Product/General Definition/Definition by Unique Expression}}
by definition of Internal Group Direct Product.
Condition $(3)$ already gives that $H_i$ is normal for all $i \in \set {1, 2, \ldots, n}$.
Condition $(1)$ gives us that each element $g$ of $G$ can be... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Subgroup|subgroups]] of $G$.
Then $G$ is the [[Definition:Internal Group Direct Product/General Definition|interna... | It is to be shown that:
{{:Definition:Internal Group Direct Product/General Definition/Definition by Unique Expression}}
by [[Definition:Internal Group Direct Product/General Definition/Definition by Unique Expression|definition of Internal Group Direct Product]].
Condition $(3)$ already gives that $H_i$ is [[Defin... | Internal Direct Product Theorem/General Result/Proof 2 | https://proofwiki.org/wiki/Internal_Direct_Product_Theorem/General_Result | https://proofwiki.org/wiki/Internal_Direct_Product_Theorem/General_Result/Proof_2 | [
"Internal Direct Product Theorem"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Sequence",
"Definition:Subgroup",
"Definition:Internal Group Direct Product/General Definition",
"Definition:Sequence",
"Definition:Independent Subgroups",
"Definition:Normal Subgroup"
] | [
"Definition:Internal Group Direct Product/General Definition/Definition by Unique Expression",
"Definition:Normal Subgroup",
"Definition:Element",
"Definition:Unique",
"Definition:Integer",
"Definition:Independent Subgroups",
"Definition:Contradiction"
] |
proofwiki-5261 | Structure Induced by Group Operation is Group | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $S$ be a set.
Let $\struct {G^S, \oplus}$ be the structure on $G^S$ induced by $\circ$.
Then $\struct {G^S, \oplus}$ is a group. | Taking the group axioms in turn: | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $S$ be a [[Definition:Set|set]].
Let $\struct {G^S, \oplus}$ be the [[Definition:Induced Structure|structure on $G^S$ induced]] by $\circ$.
Then $\struct {G^S, \oplus}$ is a [[Definition:Group|group]... | Taking the [[Axiom:Group Axioms|group axioms]] in turn: | Structure Induced by Group Operation is Group | https://proofwiki.org/wiki/Structure_Induced_by_Group_Operation_is_Group | https://proofwiki.org/wiki/Structure_Induced_by_Group_Operation_is_Group | [
"Group Theory",
"Pointwise Operations"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Set",
"Definition:Pointwise Operation/Induced Structure",
"Definition:Group"
] | [
"Axiom:Group Axioms",
"Axiom:Group Axioms"
] |
proofwiki-5262 | Continuous Mapping is Measurable | Let $\struct {X, \tau}$ and $\struct {X', \tau'}$ be topological spaces.
Denote with $\map \BB {X, \tau}$ and $\map \BB {X', \tau'}$ their respective Borel $\sigma$-algebras.
Let $f: X \to X'$ be a continuous mapping.
Then $f$ is $\map \BB {X, \tau} \, / \, \map \BB {X', \tau'}$-measurable. | As $f$ is a continuous mapping, by definition, it holds that:
:$\forall S' \in \tau': \map {f^{-1} } {S'} \in \tau$
Now, by definition, $\map \BB {X, \tau} = \map \sigma \tau$, and so $\tau \subseteq \map \BB {X, \tau}$.
Also, $\map \BB {X', \tau'} = \map \sigma {\tau'}$.
Hence Mapping Measurable iff Measurable on Gene... | Let $\struct {X, \tau}$ and $\struct {X', \tau'}$ be [[Definition:Topological Space|topological spaces]].
Denote with $\map \BB {X, \tau}$ and $\map \BB {X', \tau'}$ their respective [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebras]].
Let $f: X \to X'$ be a [[Definition:Continuous Mapping|continuous mapping]]... | As $f$ is a [[Definition:Continuous Mapping|continuous mapping]], by definition, it holds that:
:$\forall S' \in \tau': \map {f^{-1} } {S'} \in \tau$
Now, [[Definition:Borel Sigma-Algebra|by definition]], $\map \BB {X, \tau} = \map \sigma \tau$, and so $\tau \subseteq \map \BB {X, \tau}$.
Also, $\map \BB {X', \tau'}... | Continuous Mapping is Measurable | https://proofwiki.org/wiki/Continuous_Mapping_is_Measurable | https://proofwiki.org/wiki/Continuous_Mapping_is_Measurable | [
"Measurable Functions",
"Continuous Mappings"
] | [
"Definition:Topological Space",
"Definition:Borel Sigma-Algebra",
"Definition:Continuous Mapping",
"Definition:Measurable Mapping"
] | [
"Definition:Continuous Mapping",
"Definition:Borel Sigma-Algebra",
"Mapping Measurable iff Measurable on Generator",
"Definition:Measurable Mapping"
] |
proofwiki-5263 | Composition of Measurable Mappings is Measurable | Let $\struct {X_1, \Sigma_1}$, $\struct {X_2, \Sigma_2}$ and $\struct {X_3, \Sigma_3}$ be measurable spaces.
Let $f: X_1 \to X_2$ be a $\Sigma_1 \, / \, \Sigma_2$-measurable mapping.
Let $g: X_2 \to X_3$ be a $\Sigma_2 \, / \, \Sigma_3$-measurable mapping.
Then their composition $g \circ f: X_1 \to X_3$ is $\Sigma_1 \,... | Let $E_3 \in \Sigma_3$.
Then $\map {g^{-1} } {E_3} \in \Sigma_2$, and $\map {f^{-1} } {\map {g^{-1} } {E_3} } \in \Sigma_1$ as $f, g$ are measurable.
That is, $\map {\paren {g \circ f}^{-1} } {E_3} \in \Sigma_1$ for all $E_3 \in \Sigma_3$.
Hence, $g \circ f$ is $\Sigma_1 \, / \, \Sigma_3$-measurable.
{{qed}} | Let $\struct {X_1, \Sigma_1}$, $\struct {X_2, \Sigma_2}$ and $\struct {X_3, \Sigma_3}$ be [[Definition:Measurable Space|measurable spaces]].
Let $f: X_1 \to X_2$ be a [[Definition:Measurable Mapping|$\Sigma_1 \, / \, \Sigma_2$-measurable mapping]].
Let $g: X_2 \to X_3$ be a [[Definition:Measurable Mapping|$\Sigma_2 \... | Let $E_3 \in \Sigma_3$.
Then $\map {g^{-1} } {E_3} \in \Sigma_2$, and $\map {f^{-1} } {\map {g^{-1} } {E_3} } \in \Sigma_1$ as $f, g$ are [[Definition:Measurable Mapping|measurable]].
That is, $\map {\paren {g \circ f}^{-1} } {E_3} \in \Sigma_1$ for all $E_3 \in \Sigma_3$.
Hence, $g \circ f$ is [[Definition:Measura... | Composition of Measurable Mappings is Measurable | https://proofwiki.org/wiki/Composition_of_Measurable_Mappings_is_Measurable | https://proofwiki.org/wiki/Composition_of_Measurable_Mappings_is_Measurable | [
"Measure Theory",
"Composite Mappings"
] | [
"Definition:Measurable Space",
"Definition:Measurable Mapping",
"Definition:Measurable Mapping",
"Definition:Composition of Mappings",
"Definition:Measurable Mapping"
] | [
"Definition:Measurable Mapping",
"Definition:Measurable Mapping"
] |
proofwiki-5264 | Restriction of Equivalence Relation is Equivalence | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be an equivalence relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is an equivalence relation on $T$. | Let $\RR$ be an equivalence relation on $S$.
Then by definition:
:$\RR$ is a reflexive relation on $S$
:$\RR$ is a symmetric relation on $S$
:$\RR$ is a transitive relation on $S$.
Then:
:from Restriction of Reflexive Relation is Reflexive, $\RR {\restriction_T}$ is a reflexive relation on $T$
:from Restriction of Symm... | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be an [[Definition:Equivalence Relation|equivalence relation]] on $S$.
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the [[Definition:Restriction of Relation|restriction]] of $\RR$ ... | Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on $S$.
Then by definition:
:$\RR$ is a [[Definition:Reflexive Relation|reflexive relation]] on $S$
:$\RR$ is a [[Definition:Symmetric Relation|symmetric relation]] on $S$
:$\RR$ is a [[Definition:Transitive Relation|transitive relation]] on $S$.... | Restriction of Equivalence Relation is Equivalence | https://proofwiki.org/wiki/Restriction_of_Equivalence_Relation_is_Equivalence | https://proofwiki.org/wiki/Restriction_of_Equivalence_Relation_is_Equivalence | [
"Equivalence Relations",
"Restrictions"
] | [
"Definition:Set",
"Definition:Equivalence Relation",
"Definition:Subset",
"Definition:Restriction/Relation",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Reflexive Relation",
"Definition:Symmetric Relation",
"Definition:Transitive Relation",
"Restriction of Reflexive Relation is Reflexive",
"Definition:Reflexive Relation",
"Restriction of Symmetric Relation is Symmetric",
"Definition:Symmetric Relation",
... |
proofwiki-5265 | Restriction of Reflexive Relation is Reflexive | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a reflexive relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is a reflexive relation on $T$. | Suppose $\RR$ is reflexive on $S$.
So:
:$\forall x \in S: \tuple {x, x} \in \RR$
We are given $T$ is a subset of $S$, so:
:$\forall x \in T: \tuple {x, x} \in \RR$
By definition of restriction:
:$\forall x \in T: \tuple {x, x} \in \RR {\restriction_T}$
Hence $\RR {\restriction_T}$ is by definition reflexive on $T$.
{{... | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be a [[Definition:Reflexive Relation|reflexive relation]] on $S$.
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the [[Definition:Restriction of Relation|restriction]] of $\RR$ to $T... | Suppose $\RR$ is [[Definition:Reflexive Relation|reflexive]] on $S$.
So:
:$\forall x \in S: \tuple {x, x} \in \RR$
We are [[Definition:Given|given]] $T$ is a [[Definition:Subset|subset]] of $S$, so:
:$\forall x \in T: \tuple {x, x} \in \RR$
By definition of [[Definition:Restriction of Relation|restriction]]:
:$\for... | Restriction of Reflexive Relation is Reflexive | https://proofwiki.org/wiki/Restriction_of_Reflexive_Relation_is_Reflexive | https://proofwiki.org/wiki/Restriction_of_Reflexive_Relation_is_Reflexive | [
"Reflexive Relations",
"Restrictions"
] | [
"Definition:Set",
"Definition:Reflexive Relation",
"Definition:Subset",
"Definition:Restriction/Relation",
"Definition:Reflexive Relation"
] | [
"Definition:Reflexive Relation",
"Definition:Given",
"Definition:Subset",
"Definition:Restriction/Relation",
"Definition:Reflexive Relation"
] |
proofwiki-5266 | Restriction of Symmetric Relation is Symmetric | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a symmetric relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is a symmetric relation on $T$. | Suppose $\RR$ is symmetric on $S$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {x, y}
| o = \in
| r = \RR {\restriction_T}
| c =
}}
{{eqn | ll= \leadsto
| l = \tuple {x, y}
| o = \in
| r = \paren {T \times T} \cap \RR
| c = {{Defof|Restriction of Relation}}
}}
{{eqn | ll= \leadsto
... | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be a [[Definition:Symmetric Relation|symmetric relation]] on $S$.
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the [[Definition:Restriction of Relation|restriction]] of $\RR$ to $T... | Suppose $\RR$ is [[Definition:Symmetric Relation|symmetric]] on $S$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {x, y}
| o = \in
| r = \RR {\restriction_T}
| c =
}}
{{eqn | ll= \leadsto
| l = \tuple {x, y}
| o = \in
| r = \paren {T \times T} \cap \RR
| c = {{Defof|Restriction of... | Restriction of Symmetric Relation is Symmetric | https://proofwiki.org/wiki/Restriction_of_Symmetric_Relation_is_Symmetric | https://proofwiki.org/wiki/Restriction_of_Symmetric_Relation_is_Symmetric | [
"Symmetric Relations",
"Restrictions"
] | [
"Definition:Set",
"Definition:Symmetric Relation",
"Definition:Subset",
"Definition:Restriction/Relation",
"Definition:Symmetric Relation"
] | [
"Definition:Symmetric Relation",
"Definition:Symmetric Relation",
"Definition:Symmetric Relation"
] |
proofwiki-5267 | Restriction of Transitive Relation is Transitive | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a transitive relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is a transitive relation on $T$. | Suppose $\RR$ is transitive on $S$.
Then by definition:
:$\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR \implies \tuple {x, z} \in \RR$
So:
{{begin-eqn}}
{{eqn | l = \set {\tuple {x, y}, \tuple {y, z} }
| o = \subseteq
| r = \RR {\restriction_T}
| c =
}}
{{eqn | ll= \leadsto
| l = \set {\t... | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be a [[Definition:Transitive Relation|transitive relation]] on $S$.
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the [[Definition:Restriction of Relation|restriction]] of $\RR$ to ... | Suppose $\RR$ is [[Definition:Transitive Relation|transitive]] on $S$.
Then by definition:
:$\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR \implies \tuple {x, z} \in \RR$
So:
{{begin-eqn}}
{{eqn | l = \set {\tuple {x, y}, \tuple {y, z} }
| o = \subseteq
| r = \RR {\restriction_T}
| c =
}}
{{e... | Restriction of Transitive Relation is Transitive | https://proofwiki.org/wiki/Restriction_of_Transitive_Relation_is_Transitive | https://proofwiki.org/wiki/Restriction_of_Transitive_Relation_is_Transitive | [
"Transitive Relations",
"Restrictions"
] | [
"Definition:Set",
"Definition:Transitive Relation",
"Definition:Subset",
"Definition:Restriction/Relation",
"Definition:Transitive Relation"
] | [
"Definition:Transitive Relation",
"Definition:Transitive Relation",
"Definition:Transitive Relation"
] |
proofwiki-5268 | Restriction of Asymmetric Relation is Asymmetric | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a asymmetric relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is a asymmetric relation on $T$. | Suppose $\RR$ is asymmetric on $S$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {x, y}
| o = \in
| r = \RR {\restriction_T}
| c =
}}
{{eqn | ll= \leadsto
| l = \tuple {x, y}
| o = \in
| r = \paren {T \times T} \cap \RR
| c = {{Defof|Restriction of Relation}}
}}
{{eqn | ll= \leadsto... | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be a [[Definition:Asymmetric Relation|asymmetric relation]] on $S$.
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the [[Definition:Restriction of Relation|restriction]] of $\RR$ to ... | Suppose $\RR$ is [[Definition:Asymmetric Relation|asymmetric]] on $S$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {x, y}
| o = \in
| r = \RR {\restriction_T}
| c =
}}
{{eqn | ll= \leadsto
| l = \tuple {x, y}
| o = \in
| r = \paren {T \times T} \cap \RR
| c = {{Defof|Restriction ... | Restriction of Asymmetric Relation is Asymmetric | https://proofwiki.org/wiki/Restriction_of_Asymmetric_Relation_is_Asymmetric | https://proofwiki.org/wiki/Restriction_of_Asymmetric_Relation_is_Asymmetric | [
"Asymmetric Relations",
"Restrictions"
] | [
"Definition:Set",
"Definition:Asymmetric Relation",
"Definition:Subset",
"Definition:Restriction/Relation",
"Definition:Asymmetric Relation"
] | [
"Definition:Asymmetric Relation",
"Intersection is Subset",
"Definition:Asymmetric Relation",
"Proof by Contraposition",
"Intersection is Subset",
"Definition:Asymmetric Relation"
] |
proofwiki-5269 | Restriction of Antisymmetric Relation is Antisymmetric | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be an antisymmetric relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is an antisymmetric relation on $T$. | Suppose $\RR$ is antisymmetric on $S$.
Then:
{{begin-eqn}}
{{eqn | l = \set {\tuple {x, y}, \tuple {y, x} }
| o = \subseteq
| r = \RR {\restriction_T}
| c =
}}
{{eqn | ll= \leadsto
| l = \set {\tuple {x, y}, \tuple {y, x} }
| o = \subseteq
| r = \paren {T \times T} \cap \RR
| ... | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be an [[Definition:Antisymmetric Relation|antisymmetric relation]] on $S$.
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the [[Definition:Restriction of Relation|restriction]] of $\... | Suppose $\RR$ is [[Definition:Antisymmetric Relation|antisymmetric]] on $S$.
Then:
{{begin-eqn}}
{{eqn | l = \set {\tuple {x, y}, \tuple {y, x} }
| o = \subseteq
| r = \RR {\restriction_T}
| c =
}}
{{eqn | ll= \leadsto
| l = \set {\tuple {x, y}, \tuple {y, x} }
| o = \subseteq
| r ... | Restriction of Antisymmetric Relation is Antisymmetric | https://proofwiki.org/wiki/Restriction_of_Antisymmetric_Relation_is_Antisymmetric | https://proofwiki.org/wiki/Restriction_of_Antisymmetric_Relation_is_Antisymmetric | [
"Antisymmetric Relations",
"Restrictions"
] | [
"Definition:Set",
"Definition:Antisymmetric Relation",
"Definition:Subset",
"Definition:Restriction/Relation",
"Definition:Antisymmetric Relation"
] | [
"Definition:Antisymmetric Relation",
"Intersection is Subset",
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation"
] |
proofwiki-5270 | Restriction of Antireflexive Relation is Antireflexive | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be an antireflexive relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is an antireflexive relation on $T$. | Suppose $\RR$ is antireflexive on $S$.
Then by definition of antireflexive:
:$\forall x \in S: \tuple {x, x} \notin \RR$
We are given $T$ is a subset of $S$, so:
:$\forall x \in T: \tuple {x, x} \notin \RR$
We are given $\RR {\restriction_T} \subseteq T \times T$, so:
:$\forall x \in T: \tuple {x, x} \notin \RR \restr... | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be an [[Definition:Antireflexive Relation|antireflexive relation]] on $S$.
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the [[Definition:Restriction of Relation|restriction]] of $\... | Suppose $\RR$ is [[Definition:Antireflexive Relation|antireflexive]] on $S$.
Then by definition of [[Definition:Antireflexive Relation|antireflexive]]:
:$\forall x \in S: \tuple {x, x} \notin \RR$
We are [[Definition:Given|given]] $T$ is a [[Definition:Subset|subset]] of $S$, so:
:$\forall x \in T: \tuple {x, x} \no... | Restriction of Antireflexive Relation is Antireflexive | https://proofwiki.org/wiki/Restriction_of_Antireflexive_Relation_is_Antireflexive | https://proofwiki.org/wiki/Restriction_of_Antireflexive_Relation_is_Antireflexive | [
"Antireflexive Relations",
"Restrictions"
] | [
"Definition:Set",
"Definition:Antireflexive Relation",
"Definition:Subset",
"Definition:Restriction/Relation",
"Definition:Antireflexive Relation"
] | [
"Definition:Antireflexive Relation",
"Definition:Antireflexive Relation",
"Definition:Given",
"Definition:Subset",
"Definition:Given",
"Definition:Antireflexive Relation"
] |
proofwiki-5271 | Restriction of Antitransitive Relation is Antitransitive | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be an antitransitive relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is an antitransitive relation on $T$. | Suppose $\RR$ is antitransitive on $S$.
Then by definition:
:$\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR \implies \tuple {x, z} \notin \RR$
So:
{{begin-eqn}}
{{eqn | l = \set {\tuple {x, y}, \tuple {y, z} }
| o = \subseteq
| r = \RR {\restriction_T}
| c =
}}
{{eqn | ll= \leadsto
| l = \... | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be an [[Definition:Antitransitive Relation|antitransitive relation]] on $S$.
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the [[Definition:Restriction of Relation|restriction]] of ... | Suppose $\RR$ is [[Definition:Antitransitive Relation|antitransitive]] on $S$.
Then by definition:
:$\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR \implies \tuple {x, z} \notin \RR$
So:
{{begin-eqn}}
{{eqn | l = \set {\tuple {x, y}, \tuple {y, z} }
| o = \subseteq
| r = \RR {\restriction_T}
| ... | Restriction of Antitransitive Relation is Antitransitive | https://proofwiki.org/wiki/Restriction_of_Antitransitive_Relation_is_Antitransitive | https://proofwiki.org/wiki/Restriction_of_Antitransitive_Relation_is_Antitransitive | [
"Antitransitive Relations",
"Restrictions"
] | [
"Definition:Set",
"Definition:Antitransitive Relation",
"Definition:Subset",
"Definition:Restriction/Relation",
"Definition:Antitransitive Relation"
] | [
"Definition:Antitransitive Relation",
"Definition:Antitransitive Relation",
"Definition:Antitransitive Relation"
] |
proofwiki-5272 | Relation is Total iff Union with Inverse is Trivial Relation | Let $\RR$ be a relation on $S$.
Then $\RR$ is a total relation {{iff}}:
:$\RR \cup \RR^{-1} = S \times S$
where:
:$\RR^{-1}$ is the inverse of $\RR$.
:$S \times S$ is the trivial relation on $S$. | === Necessary Condition ===
Let $\RR$ be a total relation.
By definition of relation, both $\RR \subseteq S \times S$ and $\RR^{-1} \subseteq S \times S$.
So from Union is Smallest Superset (and indeed, trivially) $\RR \cup \RR^{-1} \subseteq S \times S$.
Let $\tuple {a, b} \in S \times S$.
As $\RR$ is total, either:
:... | Let $\RR$ be a [[Definition:Relation|relation]] on $S$.
Then $\RR$ is a [[Definition:Total Relation|total relation]] {{iff}}:
:$\RR \cup \RR^{-1} = S \times S$
where:
:$\RR^{-1}$ is the [[Definition:Inverse Relation|inverse]] of $\RR$.
:$S \times S$ is the [[Definition:Trivial Relation|trivial relation]] on $S$. | === Necessary Condition ===
Let $\RR$ be a [[Definition:Total Relation|total relation]].
By definition of [[Definition:Relation|relation]], both $\RR \subseteq S \times S$ and $\RR^{-1} \subseteq S \times S$.
So from [[Union is Smallest Superset]] (and indeed, trivially) $\RR \cup \RR^{-1} \subseteq S \times S$.
L... | Relation is Total iff Union with Inverse is Trivial Relation | https://proofwiki.org/wiki/Relation_is_Total_iff_Union_with_Inverse_is_Trivial_Relation | https://proofwiki.org/wiki/Relation_is_Total_iff_Union_with_Inverse_is_Trivial_Relation | [
"Total Relations",
"Inverse Relations",
"Trivial Relation"
] | [
"Definition:Relation",
"Definition:Total Relation",
"Definition:Inverse Relation",
"Definition:Trivial Relation"
] | [
"Definition:Total Relation",
"Definition:Relation",
"Union is Smallest Superset",
"Definition:Total Relation",
"Definition:Inverse Relation",
"Definition:Subset",
"Definition:Set Equality/Definition 2",
"Definition:Inverse Relation",
"Definition:Total Relation"
] |
proofwiki-5273 | Restriction of Connected Relation is Connected | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a connected relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is a connected relation on $T$. | Suppose $\RR$ is connected on $S$.
That is:
:$\forall a, b \in S: a \ne b \implies \tuple {a, b} \in \RR \lor \tuple {b, a} \in \RR$
So:
{{begin-eqn}}
{{eqn | l = a, b
| o = \in
| r = T
| c =
}}
{{eqn | ll= \leadsto
| l = \tuple {a, b}
| o = \in
| r = T \times T
| c =
}}
{{eq... | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be a [[Definition:Connected Relation|connected relation]] on $S$.
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the [[Definition:Restriction of Relation|restriction]] of $\RR$ to $T... | Suppose $\RR$ is [[Definition:Connected Relation|connected]] on $S$.
That is:
:$\forall a, b \in S: a \ne b \implies \tuple {a, b} \in \RR \lor \tuple {b, a} \in \RR$
So:
{{begin-eqn}}
{{eqn | l = a, b
| o = \in
| r = T
| c =
}}
{{eqn | ll= \leadsto
| l = \tuple {a, b}
| o = \in
... | Restriction of Connected Relation is Connected | https://proofwiki.org/wiki/Restriction_of_Connected_Relation_is_Connected | https://proofwiki.org/wiki/Restriction_of_Connected_Relation_is_Connected | [
"Connected Relations",
"Restrictions"
] | [
"Definition:Set",
"Definition:Connected Relation",
"Definition:Subset",
"Definition:Restriction/Relation",
"Definition:Connected Relation"
] | [
"Definition:Connected Relation",
"Definition:Connected Relation",
"Definition:Connected Relation"
] |
proofwiki-5274 | Restriction of Ordering is Ordering | Let $S$ be a set.
Let $\preceq$ be an ordering on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\preceq \restriction_T$ be the restriction of $\preceq$ to $T$.
Then $\preceq \restriction_T$ is an ordering on $T$. | Let $\preceq$ be an ordering on $S$.
Then, by definition:
:$\preceq$ is a reflexive relation on $S$
:$\preceq$ is an antisymmetric relation on $S$
:$\preceq$ is a transitive relation on $S$.
Then:
:from Restriction of Reflexive Relation is Reflexive, $\preceq \restriction_T$ is a reflexive relation on $T$
:from Restric... | Let $S$ be a [[Definition:Set|set]].
Let $\preceq$ be an [[Definition:Ordering|ordering]] on $S$.
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\preceq \restriction_T$ be the [[Definition:Restriction of Ordering|restriction]] of $\preceq$ to $T$.
Then $\preceq \restriction_T$ is an [[Definiti... | Let $\preceq$ be an [[Definition:Ordering|ordering]] on $S$.
Then, by definition:
:$\preceq$ is a [[Definition:Reflexive Relation|reflexive relation]] on $S$
:$\preceq$ is an [[Definition:Antisymmetric Relation|antisymmetric relation]] on $S$
:$\preceq$ is a [[Definition:Transitive Relation|transitive relation]] on $S... | Restriction of Ordering is Ordering | https://proofwiki.org/wiki/Restriction_of_Ordering_is_Ordering | https://proofwiki.org/wiki/Restriction_of_Ordering_is_Ordering | [
"Orderings",
"Restrictions"
] | [
"Definition:Set",
"Definition:Ordering",
"Definition:Subset",
"Definition:Restriction of Ordering",
"Definition:Ordering"
] | [
"Definition:Ordering",
"Definition:Reflexive Relation",
"Definition:Antisymmetric Relation",
"Definition:Transitive Relation",
"Restriction of Reflexive Relation is Reflexive",
"Definition:Reflexive Relation",
"Restriction of Antisymmetric Relation is Antisymmetric",
"Definition:Antisymmetric Relation... |
proofwiki-5275 | Properties of Relation Not Preserved by Restriction | If a relation is:
* serial,
* non-reflexive,
* non-symmetric,
* non-transitive or
* non-connected
it is impossible to state without further information whether or not any restriction of that relation has the same properties. | === Restriction of Serial Relation is Not Necessarily Serial ===
{{:Restriction of Serial Relation is Not Necessarily Serial}} | If a [[Definition:Relation|relation]] is:
* [[Definition:Serial Relation|serial]],
* [[Definition:Non-Reflexive Relation|non-reflexive]],
* [[Definition:Non-Symmetric Relation|non-symmetric]],
* [[Definition:Non-Transitive Relation|non-transitive]] or
* non-[[Definition:Connected Relation|connected]]
it is impossible ... | === [[Restriction of Serial Relation is Not Necessarily Serial]] ===
{{:Restriction of Serial Relation is Not Necessarily Serial}} | Properties of Relation Not Preserved by Restriction | https://proofwiki.org/wiki/Properties_of_Relation_Not_Preserved_by_Restriction | https://proofwiki.org/wiki/Properties_of_Relation_Not_Preserved_by_Restriction | [
"Restrictions"
] | [
"Definition:Relation",
"Definition:Serial Relation",
"Definition:Non-Reflexive Relation",
"Definition:Non-Symmetric Relation",
"Definition:Non-Transitive Relation",
"Definition:Connected Relation",
"Definition:Restriction/Relation"
] | [
"Restriction of Serial Relation is Not Necessarily Serial"
] |
proofwiki-5276 | Restriction of Serial Relation is Not Necessarily Serial | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a serial relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is not necessarily a serial relation on $T$. | Proof by Counterexample:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {a, b}, \tuple {b, b} }$.
$\RR$ is a serial relation, as can be seen by definition.
Now let $T = \set a$.
Then:
:$\RR {\restriction_T} = \O$
So:
:$\not \exists y \in T: \tuple {a, y} \in \RR {\restriction_T}$
That is, $\RR {\restriction_T}$ is not ... | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be a [[Definition:Serial Relation|serial relation]] on $S$.
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the [[Definition:Restriction of Relation|restriction]] of $\RR$ to $T$.
T... | [[Proof by Counterexample]]:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {a, b}, \tuple {b, b} }$.
$\RR$ is a [[Definition:Serial Relation|serial relation]], as can be seen by definition.
Now let $T = \set a$.
Then:
:$\RR {\restriction_T} = \O$
So:
:$\not \exists y \in T: \tuple {a, y} \in \RR {\restriction_T... | Restriction of Serial Relation is Not Necessarily Serial | https://proofwiki.org/wiki/Restriction_of_Serial_Relation_is_Not_Necessarily_Serial | https://proofwiki.org/wiki/Restriction_of_Serial_Relation_is_Not_Necessarily_Serial | [
"Serial Relations",
"Restrictions"
] | [
"Definition:Set",
"Definition:Serial Relation",
"Definition:Subset",
"Definition:Restriction/Relation",
"Definition:Serial Relation"
] | [
"Proof by Counterexample",
"Definition:Serial Relation",
"Definition:Serial Relation"
] |
proofwiki-5277 | Restriction of Non-Reflexive Relation is Not Necessarily Non-Reflexive | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a non-reflexive relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is not necessarily a non-reflexive relation on $T$. | Proof by Counterexample:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {b, b} }$.
$\RR$ is a non-reflexive relation, as can be seen by definition:
:$\tuple {a, a} \notin \RR$
:$\tuple {b, b} \in \RR$
Now let $T = \set a$.
Then $\RR {\restriction_T} = \O$.
So:
:$\forall x \in T: \tuple {x, x} \notin \RR {\restriction_T... | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be a [[Definition:Non-Reflexive Relation|non-reflexive relation]] on $S$.
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the [[Definition:Restriction of Relation|restriction]] of $\R... | [[Proof by Counterexample]]:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {b, b} }$.
$\RR$ is a [[Definition:Non-Reflexive Relation|non-reflexive relation]], as can be seen by definition:
:$\tuple {a, a} \notin \RR$
:$\tuple {b, b} \in \RR$
Now let $T = \set a$.
Then $\RR {\restriction_T} = \O$.
So:
:$\forall x... | Restriction of Non-Reflexive Relation is Not Necessarily Non-Reflexive | https://proofwiki.org/wiki/Restriction_of_Non-Reflexive_Relation_is_Not_Necessarily_Non-Reflexive | https://proofwiki.org/wiki/Restriction_of_Non-Reflexive_Relation_is_Not_Necessarily_Non-Reflexive | [
"Non-Reflexive Relations",
"Restrictions"
] | [
"Definition:Set",
"Definition:Non-Reflexive Relation",
"Definition:Subset",
"Definition:Restriction/Relation",
"Definition:Non-Reflexive Relation"
] | [
"Proof by Counterexample",
"Definition:Non-Reflexive Relation",
"Definition:Antireflexive Relation",
"Definition:Non-Reflexive Relation"
] |
proofwiki-5278 | Restriction of Non-Symmetric Relation is Not Necessarily Non-Symmetric | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a non-symmetric relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \ \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is not necessarily a non-symmetric relation on $T$. | Proof by Counterexample:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {a, b}, \tuple {b, b} }$.
$\RR$ is a non-symmetric relation, as can be seen by definition.
Now let $T = \set b$.
Then $\RR {\restriction_T} \ = \set {\tuple {b, b} }$.
So:
:$\forall x, y \in T: \tuple {x, y} \in \RR {\restriction_T} \implies \tuple... | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be a [[Definition:Non-Symmetric Relation|non-symmetric relation]] on $S$.
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\RR {\restriction_T} \ \subseteq T \times T$ be the [[Definition:Restriction of Relation|restriction]] of $... | [[Proof by Counterexample]]:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {a, b}, \tuple {b, b} }$.
$\RR$ is a [[Definition:Non-Symmetric Relation|non-symmetric relation]], as can be seen by definition.
Now let $T = \set b$.
Then $\RR {\restriction_T} \ = \set {\tuple {b, b} }$.
So:
:$\forall x, y \in T: \tuple... | Restriction of Non-Symmetric Relation is Not Necessarily Non-Symmetric | https://proofwiki.org/wiki/Restriction_of_Non-Symmetric_Relation_is_Not_Necessarily_Non-Symmetric | https://proofwiki.org/wiki/Restriction_of_Non-Symmetric_Relation_is_Not_Necessarily_Non-Symmetric | [
"Non-Symmetric Relations",
"Restrictions"
] | [
"Definition:Set",
"Definition:Non-Symmetric Relation",
"Definition:Subset",
"Definition:Restriction/Relation",
"Definition:Non-Symmetric Relation"
] | [
"Proof by Counterexample",
"Definition:Non-Symmetric Relation",
"Definition:Symmetric Relation",
"Definition:Non-Symmetric Relation"
] |
proofwiki-5279 | Restriction of Non-Transitive Relation is Not Necessarily Non-Transitive | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a non-transitive relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is not necessarily a non-transitive relation on $T$. | Proof by Counterexample:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {a, b}, \tuple {b, a}, \tuple {b, b} }$.
$\RR$ is a non-transitive relation, as can be seen by definition.
Now let $T = \set b$.
Then:
:$\RR {\restriction_T} = \set {\tuple {b, b} }$
So:
:$\forall x, y \in T: \tuple {x, y} \in \RR {\restriction_T} ... | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be a [[Definition:Non-Transitive Relation|non-transitive relation]] on $S$.
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the [[Definition:Restriction of Relation|restriction]] of $... | [[Proof by Counterexample]]:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {a, b}, \tuple {b, a}, \tuple {b, b} }$.
$\RR$ is a [[Definition:Non-Transitive Relation|non-transitive relation]], as can be seen by definition.
Now let $T = \set b$.
Then:
:$\RR {\restriction_T} = \set {\tuple {b, b} }$
So:
:$\forall x,... | Restriction of Non-Transitive Relation is Not Necessarily Non-Transitive | https://proofwiki.org/wiki/Restriction_of_Non-Transitive_Relation_is_Not_Necessarily_Non-Transitive | https://proofwiki.org/wiki/Restriction_of_Non-Transitive_Relation_is_Not_Necessarily_Non-Transitive | [
"Non-Transitive Relations",
"Restrictions"
] | [
"Definition:Set",
"Definition:Non-Transitive Relation",
"Definition:Subset",
"Definition:Restriction/Relation",
"Definition:Non-Transitive Relation"
] | [
"Proof by Counterexample",
"Definition:Non-Transitive Relation",
"Definition:Transitive Relation",
"Definition:Non-Transitive Relation"
] |
proofwiki-5280 | Existence and Uniqueness of Sigma-Algebra Generated by Collection of Mappings | Let $I$ be an indexing set.
Let $\family {\struct {X_i, \Sigma_i} }_{i \mathop \in I}$ be a family of measurable spaces.
Let $X$ be a set.
Let $\family {f_i: X \to X_i}_{i \mathop \in I}$ be a family of mappings.
Then $\map \sigma {f_i: i \in I}$, the $\sigma$-algebra generated by $\family {f_i}_{i \mathop \in I}$, exi... | By Characterization of Sigma-Algebra Generated by Collection of Mappings:
:$\ds \map \sigma {f_i: i \in I} = \map \sigma {\bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} }$
where the second is a $\sigma$-algebra generated by a collection of subsets.
The result follows from applying Existence and Uniqueness of Sig... | Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\family {\struct {X_i, \Sigma_i} }_{i \mathop \in I}$ be a [[Definition:Indexed Family|family]] of [[Definition:Measurable Space|measurable spaces]].
Let $X$ be a [[Definition:Set|set]].
Let $\family {f_i: X \to X_i}_{i \mathop \in I}$ be a [[Definition:In... | By [[Characterization of Sigma-Algebra Generated by Collection of Mappings]]:
:$\ds \map \sigma {f_i: i \in I} = \map \sigma {\bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} }$
where the second is a [[Definition:Sigma-Algebra Generated by Collection of Subsets|$\sigma$-algebra generated by a collection of subse... | Existence and Uniqueness of Sigma-Algebra Generated by Collection of Mappings | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Sigma-Algebra_Generated_by_Collection_of_Mappings | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Sigma-Algebra_Generated_by_Collection_of_Mappings | [
"Sigma-Algebras",
"Sigma-Algebras Generated by Collection of Mappings",
"Sigma-Algebras Generated by Collection of Mappings"
] | [
"Definition:Indexing Set",
"Definition:Indexing Set/Family",
"Definition:Measurable Space",
"Definition:Set",
"Definition:Indexing Set/Family",
"Definition:Mapping",
"Definition:Sigma-Algebra Generated by Collection of Mappings"
] | [
"Characterization of Sigma-Algebra Generated by Collection of Mappings",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Existence and Uniqueness of Sigma-Algebra Generated by Collection of Subsets"
] |
proofwiki-5281 | Characterization of Sigma-Algebra Generated by Collection of Mappings | Let $\struct {X_i, \Sigma_i}$ be measurable spaces, with $i \in I$ for some index set $I$.
Let $X$ be a set, and let, for $i \in I$, $f_i: X \to X_i$ be a mapping.
Then:
:$\map \sigma {f_i: i \in I} = \map \sigma {\ds \bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} }$
where:
:$\map \sigma {f_i: i \in I}$ is the $... | For each $i \in I$, one has by definition of generated $\sigma$-algebra:
:$\ds \map {f_i^{-1} } {\Sigma_i} \subseteq \bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} \subseteq \map \sigma {\bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} }$
which shows that each of the $f_i$ is measurable.
Next, suppose that ... | Let $\struct {X_i, \Sigma_i}$ be [[Definition:Measurable Space|measurable spaces]], with $i \in I$ for some [[Definition:Index Set|index set]] $I$.
Let $X$ be a [[Definition:Set|set]], and let, for $i \in I$, $f_i: X \to X_i$ be a [[Definition:Mapping|mapping]].
Then:
:$\map \sigma {f_i: i \in I} = \map \sigma {\ds... | For each $i \in I$, one has by definition of [[Definition:Sigma-Algebra Generated by Collection of Subsets|generated $\sigma$-algebra]]:
:$\ds \map {f_i^{-1} } {\Sigma_i} \subseteq \bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} \subseteq \map \sigma {\bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} }$
whi... | Characterization of Sigma-Algebra Generated by Collection of Mappings | https://proofwiki.org/wiki/Characterization_of_Sigma-Algebra_Generated_by_Collection_of_Mappings | https://proofwiki.org/wiki/Characterization_of_Sigma-Algebra_Generated_by_Collection_of_Mappings | [
"Sigma-Algebras",
"Sigma-Algebras Generated by Collection of Mappings",
"Sigma-Algebras Generated by Collection of Mappings"
] | [
"Definition:Measurable Space",
"Definition:Indexing Set",
"Definition:Set",
"Definition:Mapping",
"Definition:Sigma-Algebra Generated by Collection of Mappings",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Pre-Image Sigma-Algebra/Domain"
] | [
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Measurable Mapping",
"Definition:Sigma-Algebra",
"Definition:Measurable Mapping",
"Union is Smallest Superset/Family of Sets",
"Generated Sigma-Algebra Preserves Subset",
"Definition:Sigma-Algebra Generated by Collection of Mappi... |
proofwiki-5282 | Restriction of Non-Connected Relation is Not Necessarily Non-Connected | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a non-connected relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is not necessarily a non-connected relation on $T$. | Proof by Counterexample:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {a, a}, \tuple {b, b} }$.
$\RR$ is a non-connected relation, as can be seen by definition: neither $a \mathrel \RR b$ nor $b \mathrel \RR a$.
Now let $T = \set a$.
Then $\RR {\restriction_T} = \set {\tuple {a, a} }$.
Then $\RR {\restriction_T}$ is ... | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be a [[Definition:Connected Relation|non-connected relation]] on $S$.
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the [[Definition:Restriction of Relation|restriction]] of $\RR$ t... | [[Proof by Counterexample]]:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {a, a}, \tuple {b, b} }$.
$\RR$ is a [[Definition:Connected Relation|non-connected relation]], as can be seen by definition: neither $a \mathrel \RR b$ nor $b \mathrel \RR a$.
Now let $T = \set a$.
Then $\RR {\restriction_T} = \set {\tuple... | Restriction of Non-Connected Relation is Not Necessarily Non-Connected | https://proofwiki.org/wiki/Restriction_of_Non-Connected_Relation_is_Not_Necessarily_Non-Connected | https://proofwiki.org/wiki/Restriction_of_Non-Connected_Relation_is_Not_Necessarily_Non-Connected | [
"Connected Relations",
"Restrictions"
] | [
"Definition:Set",
"Definition:Connected Relation",
"Definition:Subset",
"Definition:Restriction/Relation",
"Definition:Connected Relation"
] | [
"Proof by Counterexample",
"Definition:Connected Relation",
"Definition:Connected Relation"
] |
proofwiki-5283 | Duals of Isomorphic Ordered Sets are Isomorphic | Let $\struct {S, \preccurlyeq_1}$ and $\struct {T, \preccurlyeq_2}$ be ordered sets.
Let $\struct {S, \succcurlyeq_1}$ and $\struct {T, \succcurlyeq_2}$ be the dual ordered sets of $\struct {S, \preccurlyeq_1}$ and $\struct {T, \preccurlyeq_2}$ respectively.
Let $f: \struct {S, \preccurlyeq_1} \to \struct {T, \preccurl... | {{begin-eqn}}
{{eqn | q =\forall x, y \in S
| l = x
| o = \succcurlyeq_1
| r = y
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = y
| o = \preccurlyeq_1
| r = x
| c = {{Defof|Dual Ordering}}
}}
{{eqn | ll= \leadstoandfrom
| l = \map f y
| o = \preccurlyeq_2
|... | Let $\struct {S, \preccurlyeq_1}$ and $\struct {T, \preccurlyeq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $\struct {S, \succcurlyeq_1}$ and $\struct {T, \succcurlyeq_2}$ be the [[Definition:Dual Ordered Set|dual ordered sets]] of $\struct {S, \preccurlyeq_1}$ and $\struct {T, \preccurlyeq_2}$ respectively.
... | {{begin-eqn}}
{{eqn | q =\forall x, y \in S
| l = x
| o = \succcurlyeq_1
| r = y
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = y
| o = \preccurlyeq_1
| r = x
| c = {{Defof|Dual Ordering}}
}}
{{eqn | ll= \leadstoandfrom
| l = \map f y
| o = \preccurlyeq_2
|... | Duals of Isomorphic Ordered Sets are Isomorphic | https://proofwiki.org/wiki/Duals_of_Isomorphic_Ordered_Sets_are_Isomorphic | https://proofwiki.org/wiki/Duals_of_Isomorphic_Ordered_Sets_are_Isomorphic | [
"Order Isomorphisms",
"Dual Orderings"
] | [
"Definition:Ordered Set",
"Definition:Dual Ordering/Dual Ordered Set",
"Definition:Order Isomorphism",
"Definition:Order Isomorphism"
] | [] |
proofwiki-5284 | Pushforward Measure is Measure | Let $\struct {X, \Sigma}$ and $\struct {X', \Sigma'}$ be measurable spaces.
Let $\mu$ be a measure on $\struct {X, \Sigma}$.
Let $f: X \to X'$ be a $\Sigma \, / \, \Sigma'$-measurable mapping.
Then the pushforward measure $f_* \mu: \Sigma' \to \overline \R$ is a measure. | To show that $f_* \mu$ is a measure, it will suffice to check the axioms $(1)$, $(2)$ and $(3')$ for a measure. | Let $\struct {X, \Sigma}$ and $\struct {X', \Sigma'}$ be [[Definition:Measurable Space|measurable spaces]].
Let $\mu$ be a [[Definition:Measure (Measure Theory)|measure]] on $\struct {X, \Sigma}$.
Let $f: X \to X'$ be a [[Definition:Measurable Mapping|$\Sigma \, / \, \Sigma'$-measurable mapping]].
Then the [[Defini... | To show that $f_* \mu$ is a [[Definition:Measure (Measure Theory)|measure]], it will suffice to check the axioms $(1)$, $(2)$ and $(3')$ for a [[Definition:Measure (Measure Theory)|measure]]. | Pushforward Measure is Measure | https://proofwiki.org/wiki/Pushforward_Measure_is_Measure | https://proofwiki.org/wiki/Pushforward_Measure_is_Measure | [
"Measures",
"Pushforward Measures"
] | [
"Definition:Measurable Space",
"Definition:Measure (Measure Theory)",
"Definition:Measurable Mapping",
"Definition:Pushforward Measure",
"Definition:Measure (Measure Theory)"
] | [
"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-5285 | Lebesgue Measure Invariant under Orthogonal Group | Let $M \in \map {\mathrm O} {n, \R}$ be an orthogonal matrix.
Let $\lambda^n$ be $n$-dimensional Lebesgue measure.
Then the pushforward measure $M_* \lambda^n$ equals $\lambda^n$. | By Orthogonal Group is Subgroup of General Linear Group, $M \in \GL {n, \R}$.
From Pushforward of Lebesgue Measure under General Linear Group, it follows that:
:$M_* \lambda^n = \size {\det M^{-1} } \lambda^n$
Since $M^{-1} \in \map {\mathrm O} {n, \R}$ by Orthogonal Group is Group, Determinant of Orthogonal Matrix app... | Let $M \in \map {\mathrm O} {n, \R}$ be an [[Definition:Orthogonal Matrix|orthogonal matrix]].
Let $\lambda^n$ be [[Definition:Lebesgue Measure|$n$-dimensional Lebesgue measure]].
Then the [[Definition:Pushforward Measure|pushforward measure]] $M_* \lambda^n$ equals $\lambda^n$. | By [[Orthogonal Group is Subgroup of General Linear Group]], $M \in \GL {n, \R}$.
From [[Pushforward of Lebesgue Measure under General Linear Group]], it follows that:
:$M_* \lambda^n = \size {\det M^{-1} } \lambda^n$
Since $M^{-1} \in \map {\mathrm O} {n, \R}$ by [[Orthogonal Group is Group]], [[Determinant of Orth... | Lebesgue Measure Invariant under Orthogonal Group | https://proofwiki.org/wiki/Lebesgue_Measure_Invariant_under_Orthogonal_Group | https://proofwiki.org/wiki/Lebesgue_Measure_Invariant_under_Orthogonal_Group | [
"Measure Theory",
"Lebesgue Measure",
"Lebesgue Measure",
"Orthogonal Groups"
] | [
"Definition:Orthogonal Matrix",
"Definition:Lebesgue Measure",
"Definition:Pushforward Measure"
] | [
"Orthogonal Group is Subgroup of General Linear Group",
"Pushforward of Lebesgue Measure under General Linear Group",
"Orthogonal Group is Group",
"Determinant of Orthogonal Matrix",
"Pushforward of Lebesgue Measure under General Linear Group",
"Determinant as Volume of Parallelotope"
] |
proofwiki-5286 | Pushforward of Lebesgue Measure under General Linear Group | Let $M \in \GL {n, \R}$ be a nonsingular matrix.
Let $\lambda^n$ be $n$-dimensional Lebesgue measure.
Then the pushforward measure $M_* \lambda^n$ satisfies:
:$M_* \lambda^n = \size {\det M^{-1} } \cdot \lambda^n$ | From Linear Transformation on Euclidean Space is Continuous, $M^{-1}$ is a continuous mapping.
Thus from Continuous Mapping is Measurable, it is measurable, and so $M_* \lambda^n$ is defined.
Now let $B \in \map \BB {\R^n}$ be a Borel measurable set, and let $\mathbf x \in \R^n$.
Then:
{{begin-eqn}}
{{eqn | l = \map {M... | Let $M \in \GL {n, \R}$ be a [[Definition:Nonsingular Matrix|nonsingular matrix]].
Let $\lambda^n$ be [[Definition:Lebesgue Measure|$n$-dimensional Lebesgue measure]].
Then the [[Definition:Pushforward Measure|pushforward measure]] $M_* \lambda^n$ satisfies:
:$M_* \lambda^n = \size {\det M^{-1} } \cdot \lambda^n$ | From [[Linear Transformation on Euclidean Space is Continuous]], $M^{-1}$ is a [[Definition:Continuous Mapping (Topology)|continuous mapping]].
Thus from [[Continuous Mapping is Measurable]], it is [[Definition:Measurable Mapping|measurable]], and so $M_* \lambda^n$ is defined.
Now let $B \in \map \BB {\R^n}$ be a [... | Pushforward of Lebesgue Measure under General Linear Group | https://proofwiki.org/wiki/Pushforward_of_Lebesgue_Measure_under_General_Linear_Group | https://proofwiki.org/wiki/Pushforward_of_Lebesgue_Measure_under_General_Linear_Group | [
"Lebesgue Measure"
] | [
"Definition:Nonsingular Matrix",
"Definition:Lebesgue Measure",
"Definition:Pushforward Measure"
] | [
"Linear Transformation on Euclidean Space is Continuous",
"Definition:Continuous Mapping (Topology)",
"Continuous Mapping is Measurable",
"Definition:Measurable Mapping",
"Definition:Borel Sigma-Algebra",
"Definition:Linear Transformation",
"Lebesgue Measure is Invariant under Translations",
"Definiti... |
proofwiki-5287 | Pre-Image Sigma-Algebra on Domain is Generated by Mapping | Let $X, X'$ be sets, and let $f: X \to X'$ be a mapping.
Let $\Sigma'$ be a $\sigma$-algebra on $X'$.
Let $f: X \to X'$ be a mapping.
Then:
:$\map \sigma f = \map {f^{-1} } {\Sigma'}$
where
:$\map \sigma f$ denotes the $\sigma$-algebra generated by $f$
:$\map {f^{-1} } {\Sigma'}$ denotes the pre-image $\sigma$-algebra ... | By Characterization of Sigma-Algebra Generated by Collection of Mappings:
:$\map \sigma f = \map \sigma {\map {f^{-1} } {\Sigma'} }$
where the latter $\sigma$ denotes a $\sigma$-algebra generated by a collection of subsets.
By Pre-Image Sigma-Algebra on Domain is Sigma-Algebra, $\map {f^{-1} } {\Sigma'}$ is a $\sigma$-... | Let $X, X'$ be [[Definition:Set|sets]], and let $f: X \to X'$ be a [[Definition:Mapping|mapping]].
Let $\Sigma'$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $X'$.
Let $f: X \to X'$ be a [[Definition:Mapping|mapping]].
Then:
:$\map \sigma f = \map {f^{-1} } {\Sigma'}$
where
:$\map \sigma f$ denotes the ... | By [[Characterization of Sigma-Algebra Generated by Collection of Mappings]]:
:$\map \sigma f = \map \sigma {\map {f^{-1} } {\Sigma'} }$
where the latter $\sigma$ denotes a [[Definition:Sigma-Algebra Generated by Collection of Subsets|$\sigma$-algebra generated by a collection of subsets]].
By [[Pre-Image Sigma-Alge... | Pre-Image Sigma-Algebra on Domain is Generated by Mapping | https://proofwiki.org/wiki/Pre-Image_Sigma-Algebra_on_Domain_is_Generated_by_Mapping | https://proofwiki.org/wiki/Pre-Image_Sigma-Algebra_on_Domain_is_Generated_by_Mapping | [
"Measure Theory"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Sigma-Algebra",
"Definition:Mapping",
"Definition:Sigma-Algebra Generated by Collection of Mappings",
"Definition:Pre-Image Sigma-Algebra/Domain"
] | [
"Characterization of Sigma-Algebra Generated by Collection of Mappings",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Pre-Image Sigma-Algebra on Domain is Sigma-Algebra",
"Definition:Sigma-Algebra"
] |
proofwiki-5288 | Mapping between Euclidean Spaces Measurable iff Components Measurable | Let $\R^n$ and $\R^m$ be Euclidean spaces.
Denote by $\BB^n$ and $\BB^m$ their respective Borel $\sigma$-algebras.
Denote with $\BB$ the Borel $\sigma$-algebra on $\R$.
Let $f: \R^n \to \R^m$ be a mapping, and write:
:$\map f {\mathbf x} = \begin{bmatrix} \map {f_1} {\mathbf x} \\ \vdots \\ \map {f_m} {\mathbf x} \end{... | === Necessary Condition ===
Suppose that $f$ is $\BB^n \, / \, \BB^m$-measurable.
It is to be shown that for $1 \le i \le m$, $f_i: \R^n \to \R$ is $\BB^n \, / \, \BB$-measurable.
By Mapping Measurable iff Measurable on Generator and Characterization of Euclidean Borel Sigma-Algebra, it suffices to show that:
:$\map {f... | Let $\R^n$ and $\R^m$ be [[Definition:Euclidean Space|Euclidean spaces]].
Denote by $\BB^n$ and $\BB^m$ their respective [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebras]].
Denote with $\BB$ the [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]] on $\R$.
Let $f: \R^n \to \R^m$ be a [[Definition:Mappin... | === Necessary Condition ===
Suppose that $f$ is [[Definition:Measurable Mapping|$\BB^n \, / \, \BB^m$-measurable]].
It is to be shown that for $1 \le i \le m$, $f_i: \R^n \to \R$ is [[Definition:Measurable Mapping|$\BB^n \, / \, \BB$-measurable]].
By [[Mapping Measurable iff Measurable on Generator]] and [[Characte... | Mapping between Euclidean Spaces Measurable iff Components Measurable | https://proofwiki.org/wiki/Mapping_between_Euclidean_Spaces_Measurable_iff_Components_Measurable | https://proofwiki.org/wiki/Mapping_between_Euclidean_Spaces_Measurable_iff_Components_Measurable | [
"Measure Theory"
] | [
"Definition:Euclidean Space",
"Definition:Borel Sigma-Algebra",
"Definition:Borel Sigma-Algebra",
"Definition:Mapping",
"Definition:Measurable Mapping",
"Definition:Measurable Mapping"
] | [
"Definition:Measurable Mapping",
"Definition:Measurable Mapping",
"Mapping Measurable iff Measurable on Generator",
"Characterization of Euclidean Borel Sigma-Algebra",
"Definition:Real Interval/Open",
"Preimage of Union under Mapping",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
... |
proofwiki-5289 | Pre-Image Sigma-Algebra of Generated Sigma-Algebra | Let $f: X \to Y$ be a mapping.
Let $\GG \subseteq \powerset Y$ be a collection of subsets of $Y$.
Then the following equality of $\sigma$-algebras on $X$ holds:
:$f^{-1} \sqbrk {\map \sigma \GG} = \map \sigma {f^{-1} \sqbrk \GG}$
where
:$\sigma$ denotes a generated $\sigma$-algebra
:$f^{-1} \sqbrk {\map \sigma \GG}$ de... | Since $\GG \subseteq \map \sigma \GG$, it follows immediately that:
:$f^{-1} \sqbrk \GG \subseteq f^{-1} \sqbrk {\map \sigma \GG}$
By Pre-Image Sigma-Algebra on Domain is Sigma-Algebra, the latter is a $\sigma$-algebra, and so by Generated Sigma-Algebra Preserves Subset, it follows that:
:$\map \sigma {f^{-1} \sqbrk \G... | Let $f: X \to Y$ be a [[Definition:Mapping|mapping]].
Let $\GG \subseteq \powerset Y$ be a collection of [[Definition:Subset|subsets]] of $Y$.
Then the following equality of [[Definition:Sigma-Algebra|$\sigma$-algebras]] on $X$ holds:
:$f^{-1} \sqbrk {\map \sigma \GG} = \map \sigma {f^{-1} \sqbrk \GG}$
where
:$\s... | Since $\GG \subseteq \map \sigma \GG$, it follows immediately that:
:$f^{-1} \sqbrk \GG \subseteq f^{-1} \sqbrk {\map \sigma \GG}$
By [[Pre-Image Sigma-Algebra on Domain is Sigma-Algebra]], the latter is a [[Definition:Sigma-Algebra|$\sigma$-algebra]], and so by [[Generated Sigma-Algebra Preserves Subset]], it follow... | Pre-Image Sigma-Algebra of Generated Sigma-Algebra | https://proofwiki.org/wiki/Pre-Image_Sigma-Algebra_of_Generated_Sigma-Algebra | https://proofwiki.org/wiki/Pre-Image_Sigma-Algebra_of_Generated_Sigma-Algebra | [
"Sigma-Algebras"
] | [
"Definition:Mapping",
"Definition:Subset",
"Definition:Sigma-Algebra",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Pre-Image Sigma-Algebra/Domain",
"Definition:Preimage/Mapping/Subset"
] | [
"Pre-Image Sigma-Algebra on Domain is Sigma-Algebra",
"Definition:Sigma-Algebra",
"Generated Sigma-Algebra Preserves Subset",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Mapping Measurable iff Measurable on Generator",
"Definition:Measurable Mapping",
"Definition:Measurable Mapping",... |
proofwiki-5290 | Stieltjes Function of Measure is Stieltjes Function | Let $\mu$ be a measure on $\R$ with the Borel $\sigma$-algebra $\map \BB \R$.
Suppose that for every $n \in \N$:
:$\map \mu {\closedint {-n} n} < +\infty$
Then $F_\mu: \R \to \overline \R$, the Stieltjes function of $\mu$, is a Stieltjes function. | By definition, $F_\mu$ is a Stieltjes function {{iff}} it is increasing and left-continuous. | Let $\mu$ be a [[Definition:Measure (Measure Theory)|measure]] on $\R$ with the [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]] $\map \BB \R$.
Suppose that for every $n \in \N$:
:$\map \mu {\closedint {-n} n} < +\infty$
Then $F_\mu: \R \to \overline \R$, the [[Definition:Stieltjes Function of Measure on R... | By definition, $F_\mu$ is a [[Definition:Stieltjes Function|Stieltjes function]] {{iff}} it is [[Definition:Increasing Real Function|increasing]] and [[Definition:Left-Continuous at Point|left-continuous]]. | Stieltjes Function of Measure is Stieltjes Function | https://proofwiki.org/wiki/Stieltjes_Function_of_Measure_is_Stieltjes_Function | https://proofwiki.org/wiki/Stieltjes_Function_of_Measure_is_Stieltjes_Function | [
"Stieltjes Functions",
"Measure Theory"
] | [
"Definition:Measure (Measure Theory)",
"Definition:Borel Sigma-Algebra",
"Definition:Stieltjes Function of Measure on Real Numbers",
"Definition:Stieltjes Function"
] | [
"Definition:Stieltjes Function",
"Definition:Increasing/Real Function",
"Definition:Continuous Real Function/Left-Continuous",
"Definition:Increasing/Real Function"
] |
proofwiki-5291 | Characterization of Differentiability | {{refactor|one-dimensional case deserves its own page, perhaps|level = medium}}
Let $\mathbb X$ be an open rectangle of $\R^n$.
Let $f: \mathbb X \to \R, \mathbf x \mapsto \map f {\mathbf x}$ be a real-valued function.
Let $\mathbf x = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix} \in \R^n$.
Let $\map {\... | Suppose $f: \R^1 \to \R$.
Define:
:$\map f x = y$
:$\Delta y = \map f {x + \Delta x} - \Delta x$
From the definition of the derivative of a real function, we can say that $f$ is differentiable {{iff}}:
:$\dfrac {\Delta y} {\Delta x} \to \dfrac {\d y} {\d x}$
as $\Delta x \to 0$.
Clearly, this is equivalent to saying th... | {{refactor|one-dimensional case deserves its own page, perhaps|level = medium}}
Let $\mathbb X$ be an [[Definition:Open Rectangle|open rectangle]] of $\R^n$.
Let $f: \mathbb X \to \R, \mathbf x \mapsto \map f {\mathbf x}$ be a [[Definition:Real-Valued Function|real-valued function]].
Let $\mathbf x = \begin {bmatrix... | Suppose $f: \R^1 \to \R$.
Define:
:$\map f x = y$
:$\Delta y = \map f {x + \Delta x} - \Delta x$
From the definition of the [[Definition:Derivative of Real Function|derivative of a real function]], we can say that $f$ is differentiable {{iff}}:
:$\dfrac {\Delta y} {\Delta x} \to \dfrac {\d y} {\d x}$
as $\Delta x \... | Characterization of Differentiability | https://proofwiki.org/wiki/Characterization_of_Differentiability | https://proofwiki.org/wiki/Characterization_of_Differentiability | [
"Differential Calculus"
] | [
"Definition:Open Rectangle",
"Definition:Real-Valued Function",
"Definition:Partial Derivative",
"Definition:Differentiable Mapping/Real-Valued Function"
] | [
"Definition:Derivative/Real Function",
"Definition:Logical Equivalence",
"Definition:Real Number",
"Definition:Real Function",
"Definition:Differentiable Mapping/Real-Valued Function",
"Definition:Differentiable Mapping/Real-Valued Function"
] |
proofwiki-5292 | Chain Rule for Real-Valued Functions | Let $f: \R^n \to \R, \mathbf x \mapsto z$ be a differentiable real-valued function.
Let $\mathbf x = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix} \in \R^n$.
Further, let every element $x_i: 1 \le i \le n$ represent an implicitly defined differentiable real function of $t$.
Then $z$ is itself differentiab... | $f$ is {{hypothesis}} differentiable.
From Characterization of Differentiability:
{{begin-eqn}}
{{eqn | l = \Delta z
| r = \sum_{i \mathop = 1}^n \frac {\partial z} {\partial x_i} \Delta x_i + \sum_{i \mathop = 1}^n \epsilon_i \Delta x_i
| c = $\forall i: 1 \le i \le n: \epsilon_i \to 0$ as $\Delta x_i \to ... | Let $f: \R^n \to \R, \mathbf x \mapsto z$ be a [[Definition:Differentiable Real-Valued Function|differentiable real-valued function]].
Let $\mathbf x = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix} \in \R^n$.
Further, let every [[Definition:Element|element]] $x_i: 1 \le i \le n$ represent an [[Definitio... | $f$ is {{hypothesis}} [[Definition:Differentiable Real-Valued Function|differentiable]].
From [[Characterization of Differentiability]]:
{{begin-eqn}}
{{eqn | l = \Delta z
| r = \sum_{i \mathop = 1}^n \frac {\partial z} {\partial x_i} \Delta x_i + \sum_{i \mathop = 1}^n \epsilon_i \Delta x_i
| c = $\foral... | Chain Rule for Real-Valued Functions | https://proofwiki.org/wiki/Chain_Rule_for_Real-Valued_Functions | https://proofwiki.org/wiki/Chain_Rule_for_Real-Valued_Functions | [
"Derivative of Composite Function"
] | [
"Definition:Differentiable Mapping/Real-Valued Function",
"Definition:Element",
"Definition:Implicit Function",
"Definition:Differentiable Mapping/Real Function",
"Definition:Differentiable Mapping/Real Function",
"Definition:Partial Derivative"
] | [
"Definition:Differentiable Mapping/Real-Valued Function",
"Characterization of Differentiability",
"Definition:Differentiable Mapping/Real Function"
] |
proofwiki-5293 | Totally Ordered Set is Well-Ordered iff Subsets Contain Infima | Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Then $\struct {S, \preccurlyeq}$ is a well-ordered set {{iff}} every non-empty subset of $T \subseteq S$ has an infimum such that $\map \inf T \in T$. | === Sufficient Condition ===
Let $T \subseteq S$ such that $m := \map \inf T \in T$.
By definition, $m$ is a lower bound of $T$ and so:
:$\forall x \in T: m \preccurlyeq x$
That is:
:$\neg \exists x \in T: x \prec m$
Thus by definition $x$ is a minimal element of $T$.
Thus by definition $S$ is well-founded and so is a ... | Let $\struct {S, \preccurlyeq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Then $\struct {S, \preccurlyeq}$ is a [[Definition:Well-Ordered Set|well-ordered set]] {{iff}} every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $T \subseteq S$ has an [[Definition:Infimum of Set|in... | === Sufficient Condition ===
Let $T \subseteq S$ such that $m := \map \inf T \in T$.
By definition, $m$ is a [[Definition:Lower Bound of Set|lower bound]] of $T$ and so:
:$\forall x \in T: m \preccurlyeq x$
That is:
:$\neg \exists x \in T: x \prec m$
Thus by definition $x$ is a [[Definition:Minimal Element|minimal ... | Totally Ordered Set is Well-Ordered iff Subsets Contain Infima | https://proofwiki.org/wiki/Totally_Ordered_Set_is_Well-Ordered_iff_Subsets_Contain_Infima | https://proofwiki.org/wiki/Totally_Ordered_Set_is_Well-Ordered_iff_Subsets_Contain_Infima | [
"Total Orderings",
"Well-Orderings"
] | [
"Definition:Totally Ordered Set",
"Definition:Well-Ordered Set",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Infimum of Set"
] | [
"Definition:Lower Bound of Set",
"Definition:Minimal/Element",
"Definition:Well-Founded Relation",
"Definition:Well-Ordered Set",
"Definition:Well-Ordered Set",
"Definition:Well-Ordered Set"
] |
proofwiki-5294 | Pre-Measure of Finite Stieltjes Function is Pre-Measure | Let $\JJ_{ho}$ denote the collection of half-open intervals in $\R$.
Let $f: \R \to \R$ be a finite Stieltjes function.
Then the pre-measure of $f$, $\mu_f: \JJ_{ho} \to \overline \R_{\ge 0}$ is a pre-measure.
Here, $\overline \R_{\ge 0}$ denotes the set of positive extended real numbers. | It is immediate from the definition of $\mu_f$ that:
:$\map {\mu_f} \O = 0$
Now suppose that for some half-open interval $\hointr a b$ one has:
:$\ds \hointr a b = \bigcup_{n \mathop \in \N} \hointr {b_n} {b_{n + 1} }$
where $b_0 = a$ and $\ds \lim_{n \mathop \to \infty} b_n = b$.
Then we compute:
{{begin-eqn}}
{{eqn |... | Let $\JJ_{ho}$ denote the collection of [[Definition:Half-Open Real Interval|half-open intervals]] in $\R$.
Let $f: \R \to \R$ be a [[Definition:Finite Stieltjes Function|finite Stieltjes function]].
Then the [[Definition:Pre-Measure of Finite Stieltjes Function|pre-measure of $f$]], $\mu_f: \JJ_{ho} \to \overline \... | It is immediate from the [[Definition:Pre-Measure of Finite Stieltjes Function|definition of $\mu_f$]] that:
:$\map {\mu_f} \O = 0$
Now suppose that for some [[Definition:Half-Open Real Interval|half-open interval]] $\hointr a b$ one has:
:$\ds \hointr a b = \bigcup_{n \mathop \in \N} \hointr {b_n} {b_{n + 1} }$
wh... | Pre-Measure of Finite Stieltjes Function is Pre-Measure | https://proofwiki.org/wiki/Pre-Measure_of_Finite_Stieltjes_Function_is_Pre-Measure | https://proofwiki.org/wiki/Pre-Measure_of_Finite_Stieltjes_Function_is_Pre-Measure | [
"Stieltjes Functions",
"Measure Theory",
"Pre-Measures",
"Pre-Measures",
"Stieltjes Functions"
] | [
"Definition:Real Interval/Half-Open",
"Definition:Stieltjes Function",
"Definition:Pre-Measure of Finite Stieltjes Function",
"Definition:Pre-Measure",
"Definition:Positive/Real Number",
"Definition:Extended Real Number Line"
] | [
"Definition:Pre-Measure of Finite Stieltjes Function",
"Definition:Real Interval/Half-Open",
"Telescoping Series/Example 2",
"Definition:Pre-Measure",
"Definition:Pre-Measure",
"Category:Pre-Measures",
"Category:Stieltjes Functions"
] |
proofwiki-5295 | Pre-Measure of Finite Stieltjes Function Extends to Unique Measure | Let $\JJ_{ho}$ denote the collection of half-open intervals in $\R$.
Let $f: \R \to \R$ be a finite Stieltjes function.
Then the pre-measure of $f$, $\mu_f$, extends uniquely to a measure $\mu$ on $\map \BB \R$, the Borel $\sigma$-algebra on $\R$.
This unique measure $\mu$ is the measure of $f$. | We intend to use Carathéodory's Theorem (Measure Theory).
To this end, observe that by Characterization of Euclidean Borel Sigma-Algebra, we have:
:$\map \BB \R = \map \sigma {\JJ_{ho} }$
From Pre-Measure of Finite Stieltjes Function is Pre-Measure, $\mu_f$ is a pre-measure.
Note that for all $n \in \N$, we have:
:$\ma... | Let $\JJ_{ho}$ denote the collection of [[Definition:Half-Open Real Interval|half-open intervals]] in $\R$.
Let $f: \R \to \R$ be a [[Definition:Finite Stieltjes Function|finite Stieltjes function]].
Then the [[Definition:Pre-Measure of Finite Stieltjes Function|pre-measure of $f$]], $\mu_f$, [[Definition:Extension ... | We intend to use [[Carathéodory's Theorem (Measure Theory)]].
To this end, observe that by [[Characterization of Euclidean Borel Sigma-Algebra]], we have:
:$\map \BB \R = \map \sigma {\JJ_{ho} }$
From [[Pre-Measure of Finite Stieltjes Function is Pre-Measure]], $\mu_f$ is a [[Definition:Pre-Measure|pre-measure]].
... | Pre-Measure of Finite Stieltjes Function Extends to Unique Measure | https://proofwiki.org/wiki/Pre-Measure_of_Finite_Stieltjes_Function_Extends_to_Unique_Measure | https://proofwiki.org/wiki/Pre-Measure_of_Finite_Stieltjes_Function_Extends_to_Unique_Measure | [
"Stieltjes Functions",
"Measure Theory",
"Pre-Measures",
"Pre-Measures",
"Stieltjes Functions"
] | [
"Definition:Real Interval/Half-Open",
"Definition:Stieltjes Function",
"Definition:Pre-Measure of Finite Stieltjes Function",
"Definition:Extension (Measure Theory)",
"Definition:Measure (Measure Theory)",
"Definition:Borel Sigma-Algebra",
"Definition:Measure (Measure Theory)",
"Definition:Measure of ... | [
"Carathéodory's Theorem (Measure Theory)",
"Characterization of Euclidean Borel Sigma-Algebra",
"Pre-Measure of Finite Stieltjes Function is Pre-Measure",
"Definition:Pre-Measure",
"Definition:Increasing Sequence of Sets",
"Carathéodory's Theorem (Measure Theory)"
] |
proofwiki-5296 | Stieltjes Function of Measure of Finite Stieltjes Function | Let $f: \R \to \R$ be a finite Stieltjes function.
Let $\mu_f$ be the measure of $f$.
Let $f_{\mu_f}$ be the Stieltjes function of $\mu_f$.
Then $f_{\mu_f} = f$. | {{ProofWanted}}
Category:Measure Theory
2av83obmotj1dfnhu3llbfontcqut8d | Let $f: \R \to \R$ be a [[Definition:Finite Stieltjes Function|finite Stieltjes function]].
Let $\mu_f$ be the [[Definition:Measure of Finite Stieltjes Function|measure of $f$]].
Let $f_{\mu_f}$ be the [[Definition:Stieltjes Function of Measure|Stieltjes function of $\mu_f$]].
Then $f_{\mu_f} = f$. | {{ProofWanted}}
[[Category:Measure Theory]]
2av83obmotj1dfnhu3llbfontcqut8d | Stieltjes Function of Measure of Finite Stieltjes Function | https://proofwiki.org/wiki/Stieltjes_Function_of_Measure_of_Finite_Stieltjes_Function | https://proofwiki.org/wiki/Stieltjes_Function_of_Measure_of_Finite_Stieltjes_Function | [
"Measure Theory"
] | [
"Definition:Stieltjes Function",
"Definition:Measure of Finite Stieltjes Function",
"Definition:Stieltjes Function of Measure on Real Numbers"
] | [
"Category:Measure Theory"
] |
proofwiki-5297 | Measure of Stieltjes Function of Measure | Let $\mu$ be a measure on $\map \BB \R$, the Borel $\sigma$-algebra on $\R$.
Suppose that for all $n \in \N$, $\mu$ satisfies:
:$\map \mu {\hointr {-n} n} < +\infty$
Let $f_\mu$ be the Stieltjes function of $\mu$.
Let $\mu_{f_\mu}$ be the measure of $f_\mu$.
Then $\mu_{f_\mu} = \mu$. | From Pre-Measure of Finite Stieltjes Function Extends to Unique Measure, it suffices to verify that:
:$\map {\mu_{f_\mu} } {\hointr a b} = \map \mu {\hointr a b}$
for all half-open intervals $\hointr a b$.
Now we have:
{{begin-eqn}}
{{eqn|l = \map {\mu_{f_\mu} } {\hointr a b}
|r = \map {f_\mu} b - \map {f_\mu} a
... | Let $\mu$ be a [[Definition:Measure (Measure Theory)|measure]] on $\map \BB \R$, the [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]] on $\R$.
Suppose that for all $n \in \N$, $\mu$ satisfies:
:$\map \mu {\hointr {-n} n} < +\infty$
Let $f_\mu$ be the [[Definition:Stieltjes Function of Measure|Stieltjes fun... | From [[Pre-Measure of Finite Stieltjes Function Extends to Unique Measure]], it suffices to verify that:
:$\map {\mu_{f_\mu} } {\hointr a b} = \map \mu {\hointr a b}$
for all [[Definition:Half-Open Real Interval|half-open intervals]] $\hointr a b$.
Now we have:
{{begin-eqn}}
{{eqn|l = \map {\mu_{f_\mu} } {\hointr ... | Measure of Stieltjes Function of Measure | https://proofwiki.org/wiki/Measure_of_Stieltjes_Function_of_Measure | https://proofwiki.org/wiki/Measure_of_Stieltjes_Function_of_Measure | [
"Measure Theory"
] | [
"Definition:Measure (Measure Theory)",
"Definition:Borel Sigma-Algebra",
"Definition:Stieltjes Function of Measure on Real Numbers",
"Definition:Measure of Finite Stieltjes Function"
] | [
"Pre-Measure of Finite Stieltjes Function Extends to Unique Measure",
"Definition:Real Interval/Half-Open",
"Definition:Measure of Finite Stieltjes Function",
"Definition:Stieltjes Function of Measure on Real Numbers",
"Definition:Stieltjes Function of Measure on Real Numbers",
"Measure of Set Difference ... |
proofwiki-5298 | Cantor Set has Zero Lebesgue Measure | Let $\CC$ be the Cantor set.
Let $\lambda$ denote the Lebesgue measure on the Borel $\sigma$-algebra $\map \BB \R$ on $\R$.
Then $\CC$ is $\map \BB \R$-measurable, and $\map \lambda \CC = 0$.
That is, $\CC$ is a $\lambda$-null set. | Consider the definition of $\CC$ as a limit of a decreasing sequence.
In the notation as introduced there, we see that each $S_n$ is a collection of disjoint closed intervals.
From Closed Set Measurable in Borel Sigma-Algebra, these are measurable sets.
Furthermore, each $S_n$ is finite.
Hence by Sigma-Algebra Closed u... | Let $\CC$ be the [[Definition:Cantor Set|Cantor set]].
Let $\lambda$ denote the [[Definition:Lebesgue Measure|Lebesgue measure]] on the [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]] $\map \BB \R$ on $\R$.
Then $\CC$ is [[Definition:Measurable Set|$\map \BB \R$-measurable]], and $\map \lambda \CC = 0$.
T... | Consider the definition of $\CC$ [[Definition:Cantor Set/Limit of Decreasing Sequence|as a limit of a decreasing sequence]].
In the notation as introduced there, we see that each $S_n$ is a collection of [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Real Interval|closed intervals]].
From [[Closed Set Meas... | Cantor Set has Zero Lebesgue Measure | https://proofwiki.org/wiki/Cantor_Set_has_Zero_Lebesgue_Measure | https://proofwiki.org/wiki/Cantor_Set_has_Zero_Lebesgue_Measure | [
"Cantor Set",
"Lebesgue Measure"
] | [
"Definition:Cantor Set",
"Definition:Lebesgue Measure",
"Definition:Borel Sigma-Algebra",
"Definition:Measurable Set",
"Definition:Null Set"
] | [
"Definition:Cantor Set/Limit of Decreasing Sequence",
"Definition:Disjoint Sets",
"Definition:Real Interval/Closed",
"Closed Set Measurable in Borel Sigma-Algebra",
"Definition:Measurable Set",
"Definition:Finite Set",
"Sigma-Algebra Closed under Union",
"Definition:Measurable Set",
"Sigma-Algebra C... |
proofwiki-5299 | Factorization Lemma/Real-Valued Function | Then a mapping $g: X \to \R$ is $\map \sigma f \, / \, \map \BB \R$-measurable {{iff}}:
:There exists a $\Sigma \, / \, \map \BB \R$-measurable mapping $\tilde g: Y \to \R$ such that $g = \tilde g \circ f$
where:
:$\map \sigma f$ denotes the $\sigma$-algebra generated by $f$
:$\map \BB \R$ denotes the Borel $\sigma$-al... | === Necessary Condition ===
Let $g$ be a $\map \sigma f \, / \, \map \BB \R$-measurable function.
We need to construct a measurable $\tilde g$ such that $g = \tilde g \circ f$.
Let us proceed in the following fashion:
:Establish the result for $g$ a characteristic function;
:Establish the result for $g$ a simple functi... | Then a [[Definition:Mapping|mapping]] $g: X \to \R$ is [[Definition:Measurable Mapping|$\map \sigma f \, / \, \map \BB \R$-measurable]] {{iff}}:
:There exists a [[Definition:Measurable Mapping|$\Sigma \, / \, \map \BB \R$-measurable mapping]] $\tilde g: Y \to \R$ such that $g = \tilde g \circ f$
where:
:$\map \sigma... | === Necessary Condition ===
Let $g$ be a [[Definition:Measurable Mapping|$\map \sigma f \, / \, \map \BB \R$-measurable function]].
We need to construct a [[Definition:Measurable Mapping|measurable]] $\tilde g$ such that $g = \tilde g \circ f$.
Let us proceed in the following fashion:
:Establish the result for $g$... | Factorization Lemma/Real-Valued Function | https://proofwiki.org/wiki/Factorization_Lemma/Real-Valued_Function | https://proofwiki.org/wiki/Factorization_Lemma/Real-Valued_Function | [
"Measure Theory"
] | [
"Definition:Mapping",
"Definition:Measurable Mapping",
"Definition:Measurable Mapping",
"Definition:Sigma-Algebra Generated by Collection of Mappings",
"Definition:Borel Sigma-Algebra"
] | [
"Definition:Measurable Mapping",
"Definition:Measurable Mapping",
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Simple Function",
"Definition:Characteristic Function (Set Theory)/Set",
"Characteristic Function Measurable iff Set Measurable",
"Definition:Measurable Set",
"Character... |
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