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-5600 | Preimage of Intersection under Mapping/General Result | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $\powerset T$ be the power set of $T$.
Let $\mathbb T \subseteq \powerset T$.
Then:
:$\ds f^{-1} \sqbrk {\bigcap \mathbb T} = \bigcap_{X \mathop \in \mathbb T} f^{-1} \sqbrk X$ | $f$ is a mapping.
Therefore it is by definition a many-to-one relation.
It follows from Inverse of Many-to-One Relation is One-to-Many that its inverse $f^{-1}$ is a one-to-many relation.
Thus Image of Intersection under One-to-Many Relation: General Result applies:
:$\ds \RR \sqbrk {\bigcap \mathbb T} = \bigcap_{X \ma... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $\powerset T$ be the [[Definition:Power Set|power set]] of $T$.
Let $\mathbb T \subseteq \powerset T$.
Then:
:$\ds f^{-1} \sqbrk {\bigcap \mathbb T} = \bigcap_{X \mathop \in \mathbb T} f^{-1} \sqbrk X$ | $f$ is a [[Definition:Mapping|mapping]].
Therefore it is by definition a [[Definition:Many-to-One Relation|many-to-one relation]].
It follows from [[Inverse of Many-to-One Relation is One-to-Many]] that its [[Definition:Inverse of Mapping|inverse]] $f^{-1}$ is a [[Definition:One-to-Many Relation|one-to-many relation]... | Preimage of Intersection under Mapping/General Result | https://proofwiki.org/wiki/Preimage_of_Intersection_under_Mapping/General_Result | https://proofwiki.org/wiki/Preimage_of_Intersection_under_Mapping/General_Result | [
"Preimage of Intersection under Mapping"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Power Set"
] | [
"Definition:Mapping",
"Definition:Many-to-One Relation",
"Inverse of Many-to-One Relation is One-to-Many",
"Definition:Inverse of Mapping",
"Definition:One-to-Many Relation",
"Image of Intersection under One-to-Many Relation/General Result"
] |
proofwiki-5601 | Bijective Relation has Left and Right Inverse | Let $\RR \subseteq S \times T$ be a relation on a cartesian product $S \times T$.
Let:
:$I_S$ be the identity mapping on $S$
:$I_T$ be the identity mapping on $T$.
Let $\RR^{-1}$ be the inverse relation of $\RR$.
Let $\RR$ be a bijection.
Then:
:$\RR^{-1} \circ \RR = I_S$
and
:$\RR \circ \RR^{-1} = I_T$
where $\circ$ d... | Suppose $\RR$ is a bijection.
Then by definition:
:$(1): \quad \RR$ is a surjection and therefore right-total.
:$(2): \quad \RR$ is a mapping and therefore left-total.
:$(3): \quad \RR$ is a one-to-one relation and therefore also both a many-to-one relation and a one-to-many relation.
By Inverses of Right-Total and Lef... | Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] on a [[Definition:Cartesian Product|cartesian product]] $S \times T$.
Let:
:$I_S$ be the [[Definition:Identity Mapping|identity mapping]] on $S$
:$I_T$ be the [[Definition:Identity Mapping|identity mapping]] on $T$.
Let $\RR^{-1}$ be the [[Definitio... | Suppose $\RR$ is a [[Definition:Bijection|bijection]].
Then by definition:
:$(1): \quad \RR$ is a [[Definition:Surjection|surjection]] and therefore [[Definition:Right-Total Relation|right-total]].
:$(2): \quad \RR$ is a [[Definition:Mapping|mapping]] and therefore [[Definition:Left-Total Relation|left-total]].
:$(3... | Bijective Relation has Left and Right Inverse | https://proofwiki.org/wiki/Bijective_Relation_has_Left_and_Right_Inverse | https://proofwiki.org/wiki/Bijective_Relation_has_Left_and_Right_Inverse | [
"Inverse Relations",
"Bijections"
] | [
"Definition:Relation",
"Definition:Cartesian Product",
"Definition:Identity Mapping",
"Definition:Identity Mapping",
"Definition:Inverse Relation",
"Definition:Bijection",
"Definition:Composition of Relations"
] | [
"Definition:Bijection",
"Definition:Surjection",
"Definition:Right-Total Relation",
"Definition:Mapping",
"Definition:Left-Total Relation",
"Definition:One-to-One Relation",
"Definition:Many-to-One Relation",
"Definition:One-to-Many Relation",
"Inverses of Right-Total and Left-Total Relations",
"D... |
proofwiki-5602 | Left and Right Inverse Relations Implies Bijection | Let $\RR \subseteq S \times T$ be a relation on a cartesian product $S \times T$.
Let:
:$I_S$ be the identity mapping on $S$
:$I_T$ be the identity mapping on $T$.
Let $\RR^{-1}$ be the inverse relation of $\RR$.
Let $\RR$ be such that:
:$\RR^{-1} \circ \RR = I_S$ and
:$\RR \circ \RR^{-1} = I_T$
where $\circ$ denotes c... | Let $\RR \subseteq S \times T$ be such that:
:$\RR^{-1} \circ \RR = I_S$
and:
:$\RR \circ \RR^{-1} = I_T$.
From Condition for Composite Relation with Inverse to be Identity, we have that:
:$\RR$ is many-to-one
:$\RR$ is right-total
:$\RR^{-1}$ is many-to-one
:$\RR^{-1}$ is right-total.
From Inverse of Many-to-One Relat... | Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] on a [[Definition:Cartesian Product|cartesian product]] $S \times T$.
Let:
:$I_S$ be the [[Definition:Identity Mapping|identity mapping]] on $S$
:$I_T$ be the [[Definition:Identity Mapping|identity mapping]] on $T$.
Let $\RR^{-1}$ be the [[Definitio... | Let $\RR \subseteq S \times T$ be such that:
:$\RR^{-1} \circ \RR = I_S$
and:
:$\RR \circ \RR^{-1} = I_T$.
From [[Condition for Composite Relation with Inverse to be Identity]], we have that:
:$\RR$ is [[Definition:Many-to-One Relation|many-to-one]]
:$\RR$ is [[Definition:Right-Total Relation|right-total]]
:$\RR^{-1}... | Left and Right Inverse Relations Implies Bijection | https://proofwiki.org/wiki/Left_and_Right_Inverse_Relations_Implies_Bijection | https://proofwiki.org/wiki/Left_and_Right_Inverse_Relations_Implies_Bijection | [
"Inverse Relations",
"Bijections"
] | [
"Definition:Relation",
"Definition:Cartesian Product",
"Definition:Identity Mapping",
"Definition:Identity Mapping",
"Definition:Inverse Relation",
"Definition:Composition of Relations",
"Definition:Bijection"
] | [
"Condition for Composite Relation with Inverse to be Identity",
"Definition:Many-to-One Relation",
"Definition:Right-Total Relation",
"Definition:Many-to-One Relation",
"Definition:Right-Total Relation",
"Inverse of Many-to-One Relation is One-to-Many",
"Definition:One-to-One Relation",
"Inverses of R... |
proofwiki-5603 | Bijection has Left and Right Inverse | Let $f: S \to T$ be a bijection.
Let:
: $I_S$ be the identity mapping on $S$
: $I_T$ be the identity mapping on $T$.
Let $f^{-1}$ be the inverse of $f$.
Then:
: $f^{-1} \circ f = I_S$
and:
: $f \circ f^{-1} = I_T$
where $\circ$ denotes composition of mappings. | Let $f$ be a bijection.
Then it is both an injection and a surjection, thus both the described $g_1$ and $g_2$ must exist from Injection iff Left Inverse and Surjection iff Right Inverse.
The fact that $g_1 = g_2 = f^{-1}$ follows from Left and Right Inverses of Mapping are Inverse Mapping.
{{qed}} | Let $f: S \to T$ be a [[Definition:Bijection|bijection]].
Let:
: $I_S$ be the [[Definition:Identity Mapping|identity mapping]] on $S$
: $I_T$ be the [[Definition:Identity Mapping|identity mapping]] on $T$.
Let $f^{-1}$ be the [[Definition:Inverse Mapping|inverse]] of $f$.
Then:
: $f^{-1} \circ f = I_S$
and:
: $f \c... | Let $f$ be a [[Definition:Bijection|bijection]].
Then it is both an [[Definition:Injection|injection]] and a [[Definition:Surjection |surjection]], thus both the described $g_1$ and $g_2$ must exist from [[Injection iff Left Inverse]] and [[Surjection iff Right Inverse]].
The fact that $g_1 = g_2 = f^{-1}$ follows f... | Bijection has Left and Right Inverse/Proof 1 | https://proofwiki.org/wiki/Bijection_has_Left_and_Right_Inverse | https://proofwiki.org/wiki/Bijection_has_Left_and_Right_Inverse/Proof_1 | [
"Inverse Mappings",
"Bijections",
"Bijection has Left and Right Inverse"
] | [
"Definition:Bijection",
"Definition:Identity Mapping",
"Definition:Identity Mapping",
"Definition:Inverse Mapping",
"Definition:Composition of Mappings"
] | [
"Definition:Bijection",
"Definition:Injection",
"Definition:Surjection ",
"Injection iff Left Inverse",
"Surjection iff Right Inverse",
"Left and Right Inverses of Mapping are Inverse Mapping"
] |
proofwiki-5604 | Bijection has Left and Right Inverse | Let $f: S \to T$ be a bijection.
Let:
: $I_S$ be the identity mapping on $S$
: $I_T$ be the identity mapping on $T$.
Let $f^{-1}$ be the inverse of $f$.
Then:
: $f^{-1} \circ f = I_S$
and:
: $f \circ f^{-1} = I_T$
where $\circ$ denotes composition of mappings. | Suppose $f$ is a bijection.
From Bijection iff Inverse is Bijection and Composite of Bijection with Inverse is Identity Mapping, it is shown that the inverse mapping $f^{-1}$ such that:
: $f^{-1} \circ f = I_S$
: $f \circ f^{-1} = I_T$
is a bijection.
{{qed}} | Let $f: S \to T$ be a [[Definition:Bijection|bijection]].
Let:
: $I_S$ be the [[Definition:Identity Mapping|identity mapping]] on $S$
: $I_T$ be the [[Definition:Identity Mapping|identity mapping]] on $T$.
Let $f^{-1}$ be the [[Definition:Inverse Mapping|inverse]] of $f$.
Then:
: $f^{-1} \circ f = I_S$
and:
: $f \c... | Suppose $f$ is a [[Definition:Bijection|bijection]].
From [[Bijection iff Inverse is Bijection]] and [[Composite of Bijection with Inverse is Identity Mapping]], it is shown that the [[Definition:Inverse Mapping|inverse mapping]] $f^{-1}$ such that:
: $f^{-1} \circ f = I_S$
: $f \circ f^{-1} = I_T$
is a [[Definition:B... | Bijection has Left and Right Inverse/Proof 2 | https://proofwiki.org/wiki/Bijection_has_Left_and_Right_Inverse | https://proofwiki.org/wiki/Bijection_has_Left_and_Right_Inverse/Proof_2 | [
"Inverse Mappings",
"Bijections",
"Bijection has Left and Right Inverse"
] | [
"Definition:Bijection",
"Definition:Identity Mapping",
"Definition:Identity Mapping",
"Definition:Inverse Mapping",
"Definition:Composition of Mappings"
] | [
"Definition:Bijection",
"Inverse of Bijection is Bijection",
"Composite of Bijection with Inverse is Identity Mapping",
"Definition:Inverse Mapping",
"Definition:Bijection"
] |
proofwiki-5605 | Bijection has Left and Right Inverse | Let $f: S \to T$ be a bijection.
Let:
: $I_S$ be the identity mapping on $S$
: $I_T$ be the identity mapping on $T$.
Let $f^{-1}$ be the inverse of $f$.
Then:
: $f^{-1} \circ f = I_S$
and:
: $f \circ f^{-1} = I_T$
where $\circ$ denotes composition of mappings. | Let $f$ be a bijection.
By definition, $f$ is a mapping, and hence also by definition a relation.
Hence the result Bijective Relation has Left and Right Inverse applies directly and so:
:$f^{-1} \circ f = I_S$
and
:$f \circ f^{-1} = I_T$
{{qed}} | Let $f: S \to T$ be a [[Definition:Bijection|bijection]].
Let:
: $I_S$ be the [[Definition:Identity Mapping|identity mapping]] on $S$
: $I_T$ be the [[Definition:Identity Mapping|identity mapping]] on $T$.
Let $f^{-1}$ be the [[Definition:Inverse Mapping|inverse]] of $f$.
Then:
: $f^{-1} \circ f = I_S$
and:
: $f \c... | Let $f$ be a [[Definition:Bijection|bijection]].
By definition, $f$ is a [[Definition:Mapping|mapping]], and hence also by definition a [[Definition:Relation|relation]].
Hence the result [[Bijective Relation has Left and Right Inverse]] applies directly and so:
:$f^{-1} \circ f = I_S$
and
:$f \circ f^{-1} = I_T$
{{qe... | Bijection has Left and Right Inverse/Proof 3 | https://proofwiki.org/wiki/Bijection_has_Left_and_Right_Inverse | https://proofwiki.org/wiki/Bijection_has_Left_and_Right_Inverse/Proof_3 | [
"Inverse Mappings",
"Bijections",
"Bijection has Left and Right Inverse"
] | [
"Definition:Bijection",
"Definition:Identity Mapping",
"Definition:Identity Mapping",
"Definition:Inverse Mapping",
"Definition:Composition of Mappings"
] | [
"Definition:Bijection",
"Definition:Mapping",
"Definition:Relation",
"Bijective Relation has Left and Right Inverse"
] |
proofwiki-5606 | Symmetry in Space Implies Conservation of Momentum | The total derivative of the action $S_{12}$ from states $1$ to $2$ with regard to position is equal to the difference in momentum from states $1$ to $2$:
:$\dfrac {\d S_{1 2} } {\d x} = p_2 - p_1$ | From the definition of generalized momentum and the Euler-Lagrange Equations:
{{begin-eqn}}
{{eqn | l = 0
| r = \frac \d {\d t} \frac {\partial \LL} {\partial \dot x} - \frac {\partial \LL} {\partial x}
| c =
}}
{{eqn | r = \dot p_i - \frac {\partial \LL} {\partial x}
| c =
}}
{{eqn | ll= \leadsto
... | The [[Definition:Total Derivative|total derivative]] of the [[Definition:Action Applied by System|action]] $S_{12}$ from [[Definition:State of Thermodynamic System|states]] $1$ to $2$ with regard to [[Definition:Position|position]] is equal to the difference in [[Definition:Generalized Momentum|momentum]] from [[Defini... | From the definition of [[Definition:Generalized Momentum|generalized momentum]] and the [[Euler-Lagrange Equations]]:
{{begin-eqn}}
{{eqn | l = 0
| r = \frac \d {\d t} \frac {\partial \LL} {\partial \dot x} - \frac {\partial \LL} {\partial x}
| c =
}}
{{eqn | r = \dot p_i - \frac {\partial \LL} {\partial ... | Symmetry in Space Implies Conservation of Momentum | https://proofwiki.org/wiki/Symmetry_in_Space_Implies_Conservation_of_Momentum | https://proofwiki.org/wiki/Symmetry_in_Space_Implies_Conservation_of_Momentum | [
"Laws of Conservation"
] | [
"Definition:Total Derivative",
"Definition:Action Applied by System",
"Definition:State of Thermodynamic System",
"Definition:Position",
"Definition:Generalized Momentum",
"Definition:State of Thermodynamic System"
] | [
"Definition:Generalized Momentum",
"Euler-Lagrange Equations",
"Definition:Action Applied by System",
"Definite Integral of Partial Derivative",
"Fundamental Theorem of Calculus",
"Category:Laws of Conservation"
] |
proofwiki-5607 | Image of Intersection under Injection/General Result | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $\powerset S$ be the power set of $S$.
Then:
:$\ds \forall \mathbb S \subseteq \powerset S: f \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} f \sqbrk X$
{{iff}} $f$ is an injection.
This can be expressed in the language and notation of direct ... | An injection is a type of one-to-one relation, and therefore also a one-to-many relation.
Therefore Image of Intersection under One-to-Many Relation applies:
:$\ds \forall \mathbb S \subseteq \powerset S: \RR \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk {\mathbb S}$
{{iff}} $\RR$ is a one-t... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Then:
:$\ds \forall \mathbb S \subseteq \powerset S: f \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} f \sqbrk X$
{{iff}} $f$ is an [... | An [[Definition:Injection|injection]] is a type of [[Definition:One-to-One Relation|one-to-one relation]], and therefore also a [[Definition:One-to-Many Relation|one-to-many relation]].
Therefore [[Image of Intersection under One-to-Many Relation]] applies:
:$\ds \forall \mathbb S \subseteq \powerset S: \RR \sqbrk {... | Image of Intersection under Injection/General Result | https://proofwiki.org/wiki/Image_of_Intersection_under_Injection/General_Result | https://proofwiki.org/wiki/Image_of_Intersection_under_Injection/General_Result | [
"Image of Intersection under Injection"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Power Set",
"Definition:Injection",
"Definition:Direct Image Mapping"
] | [
"Definition:Injection",
"Definition:One-to-One Relation",
"Definition:One-to-Many Relation",
"Image of Intersection under One-to-Many Relation",
"Definition:One-to-Many Relation",
"Definition:Mapping",
"Definition:Many-to-One Relation",
"Definition:One-to-Many Relation",
"Definition:Injection",
"D... |
proofwiki-5608 | Mappings in Product of Sets are Surjections | Let $S$ and $T$ be sets.
Let $\struct {P, \phi_1, \phi_2}$ be a product of $S$ and $T$.
Then $\phi_1$ and $\phi_1$ are surjections. | From the definition:
:For all sets $X$ and all mappings $f_1: X \to S$ and $f_2: X \to T$ there exists a unique mapping $h: X \to P$ such that:
::$\phi_1 \circ h = f_1$
::$\phi_2 \circ h = f_2$
Let $X = S$ and let $f_1 = I_S$ where $I_S$ is the identity mapping on $S$.
Then we have:
:$\phi_1 \circ h = I_S$
We have from... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\struct {P, \phi_1, \phi_2}$ be a [[Definition:Set Product|product]] of $S$ and $T$.
Then $\phi_1$ and $\phi_1$ are [[Definition:Surjection|surjections]]. | From the definition:
:For all [[Definition:Set|sets]] $X$ and all [[Definition:Mapping|mappings]] $f_1: X \to S$ and $f_2: X \to T$ there exists a [[Definition:Unique|unique]] [[Definition:Mapping|mapping]] $h: X \to P$ such that:
::$\phi_1 \circ h = f_1$
::$\phi_2 \circ h = f_2$
Let $X = S$ and let $f_1 = I_S$ where... | Mappings in Product of Sets are Surjections | https://proofwiki.org/wiki/Mappings_in_Product_of_Sets_are_Surjections | https://proofwiki.org/wiki/Mappings_in_Product_of_Sets_are_Surjections | [
"Set Products",
"Surjections"
] | [
"Definition:Set",
"Definition:Set Product",
"Definition:Surjection"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Unique",
"Definition:Mapping",
"Definition:Identity Mapping",
"Identity Mapping is Surjection",
"Definition:Surjection",
"Surjection if Composite is Surjection",
"Definition:Surjection",
"Definition:Identity Mapping",
"Identity Mapping is Surje... |
proofwiki-5609 | Cartesian Product is Set Product | Let $S$ and $T$ be sets.
Let $S \times T$ be the Cartesian product of $S$ and $T$.
Let $\pr_1: S \times T \to S$ and $\pr_2: S \times T \to T$ be the first and second projections respectively on $S \times T$.
Then $\struct {S \times T, \pr_1, \pr_2}$ is a set product. | Consider any set $X$ and mappings $f_1: X \to S$ and $f_2: X \to T$.
Define $h: X \to S \times T$ by:
:$\forall x \in X: \map h x = \tuple {\map {f_1} x, \map {f_2} x}$
Then for all $x \in X$ we have:
:$\map {\paren {\pr_1 \mathop \circ h} } x = \pr_1 \tuple {\map {f_1} x, \map {f_2} x} = \map {f_1} x)$
and
:$\map {\pa... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $S \times T$ be the [[Definition:Cartesian Product|Cartesian product]] of $S$ and $T$.
Let $\pr_1: S \times T \to S$ and $\pr_2: S \times T \to T$ be the [[Definition:First Projection|first]] and [[Definition:Second Projection|second projections]] respectively on $S \ti... | Consider any set $X$ and [[Definition:Mapping|mappings]] $f_1: X \to S$ and $f_2: X \to T$.
Define $h: X \to S \times T$ by:
:$\forall x \in X: \map h x = \tuple {\map {f_1} x, \map {f_2} x}$
Then for all $x \in X$ we have:
:$\map {\paren {\pr_1 \mathop \circ h} } x = \pr_1 \tuple {\map {f_1} x, \map {f_2} x} = \map ... | Cartesian Product is Set Product | https://proofwiki.org/wiki/Cartesian_Product_is_Set_Product | https://proofwiki.org/wiki/Cartesian_Product_is_Set_Product | [
"Set Products",
"Cartesian Product"
] | [
"Definition:Set",
"Definition:Cartesian Product",
"Definition:Projection (Mapping Theory)/First Projection",
"Definition:Projection (Mapping Theory)/Second Projection",
"Definition:Set Product"
] | [
"Definition:Mapping",
"Definition:Mapping"
] |
proofwiki-5610 | Set Products on Same Set are Equivalent | Let $S$ and $T$ be sets.
Let $\struct {P, \phi_1, \phi_2}$ and $\struct {Q, \psi_1, \psi_2}$ be products of $S$ and $T$.
Then there exists a unique bijection $\chi: Q \to P$ such that:
:$\phi_1 \circ \chi = \psi_1$
:$\phi_2 \circ \chi = \psi_2$ | We have that $\struct {P, \phi_1, \phi_2}$ is a set product.
From the definition of set product $\chi: Q \to P$ is the unique mapping such that:
:$\phi_1 \circ \chi = \psi_1$
:$\phi_2 \circ \chi = \psi_2$
Similarly, we have that $\struct {Q, \psi_1, \psi_2}$ is a set product.
So from the definition of set product there... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\struct {P, \phi_1, \phi_2}$ and $\struct {Q, \psi_1, \psi_2}$ be [[Definition:Set Product|products]] of $S$ and $T$.
Then there exists a [[Definition:Unique|unique]] [[Definition:Bijection|bijection]] $\chi: Q \to P$ such that:
:$\phi_1 \circ \chi = \psi_1$
:$\phi_2 ... | We have that $\struct {P, \phi_1, \phi_2}$ is a [[Definition:Set Product|set product]].
From the definition of [[Definition:Set Product|set product]] $\chi: Q \to P$ is the [[Definition:Unique|unique]] [[Definition:Mapping|mapping]] such that:
:$\phi_1 \circ \chi = \psi_1$
:$\phi_2 \circ \chi = \psi_2$
Similarly, we... | Set Products on Same Set are Equivalent | https://proofwiki.org/wiki/Set_Products_on_Same_Set_are_Equivalent | https://proofwiki.org/wiki/Set_Products_on_Same_Set_are_Equivalent | [
"Set Products"
] | [
"Definition:Set",
"Definition:Set Product",
"Definition:Unique",
"Definition:Bijection"
] | [
"Definition:Set Product",
"Definition:Set Product",
"Definition:Unique",
"Definition:Mapping",
"Definition:Set Product",
"Definition:Set Product",
"Definition:Unique",
"Definition:Mapping",
"Definition:Unique",
"Definition:Mapping",
"Definition:Unique",
"Definition:Mapping",
"Definition:Iden... |
proofwiki-5611 | Positive Rational Number as Power of Number with Power of Itself | Every positive rational number can be written either as:
: $a^{a^a}$ for some irrational number $a$
or as:
: $n^{n^n}$ for some natural number $n$. | {{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {x^{x^x} }
| r = \map {\frac \d {\d x} } {e^{x^x \ln x} }
}}
{{eqn | r = e^{x^x \ln x} \map {\frac \d {\d x} } {x^x \ln x}
| c = Chain Rule for Derivatives
}}
{{eqn | r = x^{x^x} \paren {x^x \map {\frac \d {\d x} } {\ln x} + \map {\frac \d {\d x} } {x^x} \ln... | Every [[Definition:Positive Rational Number|positive rational number]] can be written either as:
: $a^{a^a}$ for some [[Definition:Irrational Number|irrational number]] $a$
or as:
: $n^{n^n}$ for some [[Definition:Natural Number|natural number]] $n$. | {{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {x^{x^x} }
| r = \map {\frac \d {\d x} } {e^{x^x \ln x} }
}}
{{eqn | r = e^{x^x \ln x} \map {\frac \d {\d x} } {x^x \ln x}
| c = [[Chain Rule for Derivatives]]
}}
{{eqn | r = x^{x^x} \paren {x^x \map {\frac \d {\d x} } {\ln x} + \map {\frac \d {\d x} } {x^x}... | Positive Rational Number as Power of Number with Power of Itself | https://proofwiki.org/wiki/Positive_Rational_Number_as_Power_of_Number_with_Power_of_Itself | https://proofwiki.org/wiki/Positive_Rational_Number_as_Power_of_Number_with_Power_of_Itself | [
"Number Theory"
] | [
"Definition:Positive/Rational Number",
"Definition:Irrational Number",
"Definition:Natural Numbers"
] | [
"Derivative of Composite Function",
"Product Rule for Derivatives",
"Derivative of Natural Logarithm Function",
"Derivative of x to the x",
"Definition:Positive/Real Number",
"Definition:Irrational Number",
"Definition:Rational Number",
"Definition:Natural Numbers",
"Rational Number as Power of Numb... |
proofwiki-5612 | Rational Number as Power of Number with Itself | Every rational number in the interval $\openint {\paren {\dfrac 1 e}^{\frac 1 e} }{+\infty}$ can be written either as:
: $a^a$ for some irrational number $a$
or as:
: $n^n$ for some natural number $n$. | $\dfrac \d {\d x} x^x = \dfrac \d {\d x} e^{x \ln x} = e^{x \ln x} \paren {\ln x + 1}$
So we have $\dfrac \d {\d x} x^x > 0$ for every $x > \dfrac 1 e$.
Thus $x^x: \openint {\dfrac 1 e} {+\infty} \to \openint {\paren {\dfrac 1 e}^{\frac 1 e} } {+\infty}$ is bijective.
For each $y \in \openint {\paren {\dfrac 1 e}^{\fra... | Every [[Definition:Rational Number|rational number]] in the [[Definition:Open Real Interval|interval]] $\openint {\paren {\dfrac 1 e}^{\frac 1 e} }{+\infty}$ can be written either as:
: $a^a$ for some [[Definition:Irrational Number|irrational number]] $a$
or as:
: $n^n$ for some [[Definition:Natural Number|natural numb... | $\dfrac \d {\d x} x^x = \dfrac \d {\d x} e^{x \ln x} = e^{x \ln x} \paren {\ln x + 1}$
So we have $\dfrac \d {\d x} x^x > 0$ for every $x > \dfrac 1 e$.
Thus $x^x: \openint {\dfrac 1 e} {+\infty} \to \openint {\paren {\dfrac 1 e}^{\frac 1 e} } {+\infty}$ is bijective.
For each $y \in \openint {\paren {\dfrac 1 e}^{... | Rational Number as Power of Number with Itself | https://proofwiki.org/wiki/Rational_Number_as_Power_of_Number_with_Itself | https://proofwiki.org/wiki/Rational_Number_as_Power_of_Number_with_Itself | [
"Number Theory"
] | [
"Definition:Rational Number",
"Definition:Real Interval/Open",
"Definition:Irrational Number",
"Definition:Natural Numbers"
] | [
"Definition:Irrational Number",
"Definition:Rational Number",
"Definition:Natural Numbers",
"Definition:Rational Number",
"Definition:Rational Number",
"Definition:Rational Number/Canonical Form",
"Canonical Form of Rational Number is Unique",
"Definition:Natural Numbers",
"Definition:Rational Numbe... |
proofwiki-5613 | Integral of Integrable Function is Homogeneous | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\mu$-integrable function.
Let $\lambda \in \R$.
Let $\lambda f$ be the pointwise $\lambda$-multiple of $f$.
Then $\lambda f$ is $\mu$-integrable, and:
:$\ds \int \lambda f \rd \mu = \lambda \int f \rd \mu$ | First suppose that $\lambda \ge 0$.
From Positive Part of Multiple of Function, we have:
:$\paren {\lambda f}^+ = \lambda f^+$
From Negative Part of Multiple of Function, we have:
:$\paren {\lambda f}^- = \lambda f^-$
From Function Measurable iff Positive and Negative Parts Measurable, we have:
:$f^-$ and $f^+$ are ... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f: X \to \overline \R$ be a [[Definition:Measure-Integrable Function|$\mu$-integrable function]].
Let $\lambda \in \R$.
Let $\lambda f$ be the [[Definition:Pointwise Scalar Multiplication|pointwise $\lambda$-multiple]] of $f$.
The... | First suppose that $\lambda \ge 0$.
From [[Positive Part of Multiple of Function]], we have:
:$\paren {\lambda f}^+ = \lambda f^+$
From [[Negative Part of Multiple of Function]], we have:
:$\paren {\lambda f}^- = \lambda f^-$
From [[Function Measurable iff Positive and Negative Parts Measurable]], we have:
:$f... | Integral of Integrable Function is Homogeneous | https://proofwiki.org/wiki/Integral_of_Integrable_Function_is_Homogeneous | https://proofwiki.org/wiki/Integral_of_Integrable_Function_is_Homogeneous | [
"Integrals of Integrable Functions"
] | [
"Definition:Measure Space",
"Definition:Integrable Function/Measure Space",
"Definition:Pointwise Scalar Multiplication of Mappings",
"Definition:Integrable Function/Measure Space"
] | [
"Positive Part of Multiple of Function",
"Negative Part of Multiple of Function",
"Function Measurable iff Positive and Negative Parts Measurable",
"Definition:Measurable Function",
"Pointwise Scalar Multiple of Measurable Function is Measurable",
"Definition:Measurable Function",
"Integral of Positive ... |
proofwiki-5614 | Pointwise Sum of Integrable Functions is Integrable Function | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \overline \R$ be $\mu$-integrable functions.
Suppose that their pointwise sum $f + g$ is well-defined.
Then $f + g$ is also a $\mu$-integrable function.
That is, the space of $\mu$-integrable functions $\LL^1_{\overline \R}$ is closed under pointwise a... | We are given $f, g: X \to \overline \R$ are $\mu$-integrable functions:
{{begin-eqn}}
{{eqn | l = \int f^+ \rd \mu
| o = <
| r = +\infty
}}
{{eqn | l = \int f^- \rd \mu
| o = <
| r = +\infty
}}
{{end-eqn}}
where $f^+$ and $f^-$ are the positive and negative parts of $f$ respectively.
Also:
{{beg... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f, g: X \to \overline \R$ be [[Definition:Measure-Integrable Function|$\mu$-integrable functions]].
Suppose that their [[Definition:Pointwise Addition of Extended Real-Valued Functions|pointwise sum]] $f + g$ is well-defined.
Then ... | We are [[Definition:Given|given]] $f, g: X \to \overline \R$ are [[Definition:Measure-Integrable Function|$\mu$-integrable functions]]:
{{begin-eqn}}
{{eqn | l = \int f^+ \rd \mu
| o = <
| r = +\infty
}}
{{eqn | l = \int f^- \rd \mu
| o = <
| r = +\infty
}}
{{end-eqn}}
where $f^+$ and $f^-$ ar... | Pointwise Sum of Integrable Functions is Integrable Function | https://proofwiki.org/wiki/Pointwise_Sum_of_Integrable_Functions_is_Integrable_Function | https://proofwiki.org/wiki/Pointwise_Sum_of_Integrable_Functions_is_Integrable_Function | [
"Measure-Integrable Functions",
"Pointwise Operations"
] | [
"Definition:Measure Space",
"Definition:Integrable Function/Measure Space",
"Definition:Pointwise Addition of Extended Real-Valued Functions",
"Definition:Integrable Function/Measure Space",
"Definition:Space of Integrable Functions",
"Definition:Closure (Abstract Algebra)",
"Definition:Pointwise Additi... | [
"Definition:Given",
"Definition:Integrable Function/Measure Space",
"Definition:Positive Part",
"Definition:Negative Part",
"Definition:Given",
"Definition:Pointwise Addition of Extended Real-Valued Functions",
"Bound for Positive Part of Pointwise Sum of Functions",
"Bound for Negative Part of Pointw... |
proofwiki-5615 | Integral of Integrable Function is Additive | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \R$ be $\mu$-integrable functions.
Then $f + g$ is $\mu$-integrable, with:
:$\ds \int \paren {f + g} \rd \mu = \int f \rd \mu + \int g \rd \mu$ | === Lemma ===
{{:Integral of Integrable Function is Additive/Lemma}}{{qed|lemma}}
From Pointwise Sum of Measurable Functions is Measurable:
:$f + g$ is $\Sigma$-measurable.
From Function Measurable iff Positive and Negative Parts Measurable we have that:
:$\paren {f + g}^+$ and $\paren {f + g}^-$ are $\Sigma$-measurab... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f, g: X \to \R$ be [[Definition:Measure-Integrable Function|$\mu$-integrable functions]].
Then $f + g$ is [[Definition:Measure-Integrable Function|$\mu$-integrable]], with:
:$\ds \int \paren {f + g} \rd \mu = \int f \rd \mu + \int ... | === [[Integral of Integrable Function is Additive/Lemma|Lemma]] ===
{{:Integral of Integrable Function is Additive/Lemma}}{{qed|lemma}}
From [[Pointwise Sum of Measurable Functions is Measurable]]:
:$f + g$ is [[Definition:Measurable Function|$\Sigma$-measurable]].
From [[Function Measurable iff Positive and Negati... | Integral of Integrable Function is Additive | https://proofwiki.org/wiki/Integral_of_Integrable_Function_is_Additive | https://proofwiki.org/wiki/Integral_of_Integrable_Function_is_Additive | [
"Integral of Integrable Function is Additive",
"Integrals of Integrable Functions",
"Measure-Integrable Functions"
] | [
"Definition:Measure Space",
"Definition:Integrable Function/Measure Space",
"Definition:Integrable Function/Measure Space"
] | [
"Integral of Integrable Function is Additive/Lemma",
"Pointwise Sum of Measurable Functions is Measurable",
"Definition:Measurable Function",
"Function Measurable iff Positive and Negative Parts Measurable",
"Definition:Measurable Function",
"Definition:Integrable Function/Measure Space",
"Definition:Me... |
proofwiki-5616 | Integral of Integrable Function is Monotone | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \overline \R$ be $\mu$-integrable functions.
Suppose that $f \le g$ pointwise.
Then:
:$\ds \int f \rd \mu \le \int g \rd \mu$ | Since:
:$f \le g$
we have that $g - f$ is well-defined with:
:$g - f \ge 0$
From {{Corollary|Integral of Integrable Function is Additive|2}}, we have:
:$g - f$ is $\mu$-integrable
with:
:$\ds \int \paren {g - f} \rd \mu = \int g \rd \mu - \int f \rd \mu$
Since:
:$\ds \int \paren {g - f} \rd \mu \ge 0$
we have:
:$\d... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f, g: X \to \overline \R$ be [[Definition:Measure-Integrable Function|$\mu$-integrable functions]].
Suppose that $f \le g$ [[Definition:Pointwise Inequality|pointwise]].
Then:
:$\ds \int f \rd \mu \le \int g \rd \mu$ | Since:
:$f \le g$
we have that $g - f$ is well-defined with:
:$g - f \ge 0$
From {{Corollary|Integral of Integrable Function is Additive|2}}, we have:
:$g - f$ is [[Definition:Measure-Integrable Function|$\mu$-integrable]]
with:
:$\ds \int \paren {g - f} \rd \mu = \int g \rd \mu - \int f \rd \mu$
Since:
:$... | Integral of Integrable Function is Monotone | https://proofwiki.org/wiki/Integral_of_Integrable_Function_is_Monotone | https://proofwiki.org/wiki/Integral_of_Integrable_Function_is_Monotone | [
"Integrals of Integrable Functions"
] | [
"Definition:Measure Space",
"Definition:Integrable Function/Measure Space",
"Definition:Pointwise Inequality"
] | [
"Definition:Integrable Function/Measure Space"
] |
proofwiki-5617 | Triangle Inequality for Integrals | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\mu$-integrable function.
Then:
:$\ds \size {\int_X f \rd \mu} \le \int_X \size f \rd \mu$ | We have:
{{begin-eqn}}
{{eqn | l = \size {\int f \rd \mu}
| r = \size {\int f^+ \rd \mu - \int f^- \rd \mu}
| c = {{Defof|Integral of Measure-Integrable Function}}
}}
{{eqn | o = \le
| r = \size {\int f^+ \rd \mu} + \size {-\int f^- \rd \mu}
| c = Triangle Inequality for Real Numbers, since $f$... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f: X \to \overline \R$ be a [[Definition:Measure-Integrable Function|$\mu$-integrable function]].
Then:
:$\ds \size {\int_X f \rd \mu} \le \int_X \size f \rd \mu$ | We have:
{{begin-eqn}}
{{eqn | l = \size {\int f \rd \mu}
| r = \size {\int f^+ \rd \mu - \int f^- \rd \mu}
| c = {{Defof|Integral of Measure-Integrable Function}}
}}
{{eqn | o = \le
| r = \size {\int f^+ \rd \mu} + \size {-\int f^- \rd \mu}
| c = [[Triangle Inequality for Real Numbers]], sinc... | Triangle Inequality for Integrals | https://proofwiki.org/wiki/Triangle_Inequality_for_Integrals | https://proofwiki.org/wiki/Triangle_Inequality_for_Integrals | [
"Triangle Inequality for Integrals",
"Triangle Inequality",
"Measure-Integrable Functions"
] | [
"Definition:Measure Space",
"Definition:Integrable Function/Measure Space"
] | [
"Triangle Inequality/Real Numbers",
"Definition:Integrable Function/Measure Space",
"Integral of Positive Measurable Function is Additive",
"Sum of Positive and Negative Parts"
] |
proofwiki-5618 | Measure with Density is Measure | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R_{\ge 0}$ be a positive $\Sigma$-measurable function.
Then the $f \mu$, the measure with density $f$ with respect to $\mu$ is a measure. | Note that for each $A \in \Sigma$, we have:
:$\ds \map {\paren {f \mu} } A = \int_A f \rd \mu$
We verify each of the three conditions for a measure. | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f: X \to \overline \R_{\ge 0}$ be a [[Definition:Positive Measurable Function|positive $\Sigma$-measurable function]].
Then the $f \mu$, the [[Definition:Measure with Density|measure with density $f$ with respect to $\mu$]] is a [[D... | Note that for each $A \in \Sigma$, we have:
:$\ds \map {\paren {f \mu} } A = \int_A f \rd \mu$
We verify each of the three conditions for a [[Definition:Measure (Measure Theory)|measure]]. | Measure with Density is Measure | https://proofwiki.org/wiki/Measure_with_Density_is_Measure | https://proofwiki.org/wiki/Measure_with_Density_is_Measure | [
"Measure with Density",
"Measures",
"Measure with Density"
] | [
"Definition:Measure Space",
"Definition:Measurable Function/Positive",
"Definition:Measure with Density",
"Definition:Measure (Measure Theory)"
] | [
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)"
] |
proofwiki-5619 | Measurable Function Zero A.E. iff Absolute Value has Zero Integral | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\Sigma$-measurable function.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$f {{=}} 0$ almost everywhere}}
{{item|(2):|$\ds \int \size f \rd \mu {{=}} 0$}}
{{end-itemize}} | Let $\EE^+$ be the space of positive simple functions. | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f: X \to \overline \R$ be a [[Definition:Measurable Function|$\Sigma$-measurable function]].
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$f {{=}} 0$ [[Definition:Almost Everywhere|almost everywhere]]}}
{{item|(2):|$\ds \int \size f \rd \... | Let $\EE^+$ be the [[Definition:Space of Positive Simple Functions|space of positive simple functions]]. | Measurable Function Zero A.E. iff Absolute Value has Zero Integral | https://proofwiki.org/wiki/Measurable_Function_Zero_A.E._iff_Absolute_Value_has_Zero_Integral | https://proofwiki.org/wiki/Measurable_Function_Zero_A.E._iff_Absolute_Value_has_Zero_Integral | [
"Measurable Functions",
"Measurable Function Zero A.E. iff Absolute Value has Zero Integral"
] | [
"Definition:Measure Space",
"Definition:Measurable Function",
"Definition:Almost Everywhere"
] | [
"Definition:Space of Simple Functions"
] |
proofwiki-5620 | Integral of Integrable Function over Null Set | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\mu$-integrable function.
Let $N$ be a $\mu$-null set.
Then:
:$\ds \int_N f \rd \mu = 0$
where $\ds \int_N$ signifies an integral over $N$. | We have, by definition:
:$\ds \int_N f \rd \mu = \int f \cdot \chi_N \rd \mu$
Note that if $x \in X \setminus N$, we have:
:$\map {\chi_N} x = 0$
So, if:
:$\map f x \map {\chi_N} x \ne 0$
we must have $x \in N$.
Since $N$ is a null set, this gives:
:$f \cdot \chi_N = 0$ $\mu$-almost everywhere.
From Measurable Fu... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f: X \to \overline \R$ be a [[Definition:Measure-Integrable Function|$\mu$-integrable function]].
Let $N$ be a [[Definition:Null Set|$\mu$-null set]].
Then:
:$\ds \int_N f \rd \mu = 0$
where $\ds \int_N$ signifies an [[Definition... | We have, by [[Definition:Integral of Measure-Integrable Function over Measurable Set|definition]]:
:$\ds \int_N f \rd \mu = \int f \cdot \chi_N \rd \mu$
Note that if $x \in X \setminus N$, we have:
:$\map {\chi_N} x = 0$
So, if:
:$\map f x \map {\chi_N} x \ne 0$
we must have $x \in N$.
Since $N$ is a [[Defi... | Integral of Integrable Function over Null Set | https://proofwiki.org/wiki/Integral_of_Integrable_Function_over_Null_Set | https://proofwiki.org/wiki/Integral_of_Integrable_Function_over_Null_Set | [
"Integrals of Measure-Integrable Functions"
] | [
"Definition:Measure Space",
"Definition:Integrable Function/Measure Space",
"Definition:Null Set",
"Definition:Integral of Measure-Integrable Function over Measurable Set"
] | [
"Definition:Integral of Measure-Integrable Function over Measurable Set",
"Definition:Null Set",
"Definition:Almost Everywhere",
"Measurable Function Zero A.E. iff Absolute Value has Zero Integral"
] |
proofwiki-5621 | A.E. Equal Positive Measurable Functions have Equal Integrals | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \overline \R_{\ge 0}$ be positive $\Sigma$-measurable functions.
Suppose that $f = g$ almost everywhere.
Then:
:$\ds \int f \rd \mu = \int g \rd \mu$ | Let:
:$A = \set {x \in X : \map f x \ne \map g x}$
From Measurable Functions Determine Measurable Sets, we have that:
:$A$ is $\Sigma$-measurable.
Define $h : X \to \overline \R$ by:
:$\map h x = \begin{cases}+\infty & x \in A \\ 0 & x \not \in A\end{cases}$
We can show that $h$ is $\Sigma$-measurable.
If $t < 0$, w... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f, g: X \to \overline \R_{\ge 0}$ be [[Definition:Positive Measurable Function|positive $\Sigma$-measurable functions]].
Suppose that $f = g$ [[Definition:Almost Everywhere|almost everywhere]].
Then:
:$\ds \int f \rd \mu = \int g ... | Let:
:$A = \set {x \in X : \map f x \ne \map g x}$
From [[Measurable Functions Determine Measurable Sets]], we have that:
:$A$ is [[Definition:Measurable Set|$\Sigma$-measurable]].
Define $h : X \to \overline \R$ by:
:$\map h x = \begin{cases}+\infty & x \in A \\ 0 & x \not \in A\end{cases}$
We can show that $... | A.E. Equal Positive Measurable Functions have Equal Integrals/Proof 2 | https://proofwiki.org/wiki/A.E._Equal_Positive_Measurable_Functions_have_Equal_Integrals | https://proofwiki.org/wiki/A.E._Equal_Positive_Measurable_Functions_have_Equal_Integrals/Proof_2 | [
"A.E. Equal Positive Measurable Functions have Equal Integrals",
"Integral of Positive Measurable Function",
"Measurable Functions"
] | [
"Definition:Measure Space",
"Definition:Measurable Function/Positive",
"Definition:Almost Everywhere"
] | [
"Measurable Functions Determine Measurable Sets",
"Definition:Measurable Set",
"Definition:Measurable Function",
"Sigma-Algebra Contains Empty Set",
"Definition:Measurable Set",
"Definition:Measurable Set",
"Definition:Measurable Function",
"Integral of Positive Measurable Function is Monotone",
"In... |
proofwiki-5622 | Integrable Function is A.E. Real-Valued | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\mu$-integrable function.
Then $\map f x \in \R$ for almost all $x \in X$. | From {{Corollary|Set of Points for which Measurable Function is Real-Valued is Measurable}}, we have that:
:$\set {x \in X : \size {\map f x} = +\infty}$ is $\Sigma$-measurable.
We now aim to show that this set is a null set.
Now note that we have:
:$\set {x \in X : \size {\map f x} = +\infty} \subseteq \set {x \in X... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f: X \to \overline \R$ be a [[Definition:Measure-Integrable Function|$\mu$-integrable function]].
Then $\map f x \in \R$ for [[Definition:Almost Everywhere|almost all]] $x \in X$. | From {{Corollary|Set of Points for which Measurable Function is Real-Valued is Measurable}}, we have that:
:$\set {x \in X : \size {\map f x} = +\infty}$ is [[Definition:Measurable Set|$\Sigma$-measurable]].
We now aim to show that this set is a [[Definition:Null Set|null set]].
Now note that we have:
:$\set {x \... | Integrable Function is A.E. Real-Valued | https://proofwiki.org/wiki/Integrable_Function_is_A.E._Real-Valued | https://proofwiki.org/wiki/Integrable_Function_is_A.E._Real-Valued | [
"Measure Theory"
] | [
"Definition:Measure Space",
"Definition:Integrable Function/Measure Space",
"Definition:Almost Everywhere"
] | [
"Definition:Measurable Set",
"Definition:Null Set",
"Measure is Monotone",
"Markov's Inequality",
"Lower and Upper Bounds for Sequences",
"Definition:Almost Everywhere"
] |
proofwiki-5623 | Integrable Functions with Equal Integrals on Sub-Sigma-Algebra are A.E. Equal | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\GG$ be a sub-$\sigma$-algebra of $\Sigma$.
Let $f, g: X \to \overline \R$ be $\GG$-integrable functions.
Suppose that, for all $G \in \GG$:
:$\ds \int_G f \rd \mu = \int_G g \rd \mu$
Then $f = g$ $\mu$-almost everywhere. | Define:
:$A = \set {x \in X : \map f x \in \R}$
and:
:$B = \set {x \in X : \map g x \in \R}$
From Set of Points for which Measurable Function is Real-Valued is Measurable, we have that:
:$A$ and $B$ are $\GG$-measurable.
From Sigma-Algebra Closed under Countable Intersection, we have that:
:$A \cap B$ is $\GG$-mea... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $\GG$ be a [[Definition:Sub-Sigma-Algebra|sub-$\sigma$-algebra]] of $\Sigma$.
Let $f, g: X \to \overline \R$ be [[Definition:Measure-Integrable Function|$\GG$-integrable functions]].
Suppose that, for all $G \in \GG$:
:$\ds \int_G ... | Define:
:$A = \set {x \in X : \map f x \in \R}$
and:
:$B = \set {x \in X : \map g x \in \R}$
From [[Set of Points for which Measurable Function is Real-Valued is Measurable]], we have that:
:$A$ and $B$ are [[Definition:Measurable Set|$\GG$-measurable]].
From [[Sigma-Algebra Closed under Countable Intersectio... | Integrable Functions with Equal Integrals on Sub-Sigma-Algebra are A.E. Equal | https://proofwiki.org/wiki/Integrable_Functions_with_Equal_Integrals_on_Sub-Sigma-Algebra_are_A.E._Equal | https://proofwiki.org/wiki/Integrable_Functions_with_Equal_Integrals_on_Sub-Sigma-Algebra_are_A.E._Equal | [
"Measure Theory"
] | [
"Definition:Measure Space",
"Definition:Sub-Sigma-Algebra",
"Definition:Integrable Function/Measure Space",
"Definition:Almost Everywhere"
] | [
"Set of Points for which Measurable Function is Real-Valued is Measurable",
"Definition:Measurable Set",
"Sigma-Algebra Closed under Countable Intersection",
"Definition:Measurable Set",
"Definition:Characteristic Function",
"Characteristic Function Measurable iff Set Measurable",
"Definition:Measurable... |
proofwiki-5624 | Measurable Functions with Equal Integrals on Sub-Sigma-Algebra are A.E. Equal | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\GG$ be a sub-$\sigma$-algebra of $\Sigma$.
Suppose that $\mu \restriction_\GG$, the restriction of $\mu$ to $\GG$, is $\sigma$-finite.
Let $f, g: X \to \overline \R$ be $\GG$-measurable functions.
Suppose that, for all $G \in \GG$:
:$\ds \int_G f \rd \mu = \int_G... | First, assume that $f$ and $g$ are $\mu$-integrable.
Observe:
{{begin-eqn}}
{{eqn | n = 1
| l = \set {f \ne g}
| r = \bigcup_{n \mathop = 1}^\infty \set {\size {f - g} \ge \dfrac 1 n }
}}
{{end-eqn}}
For each $n \in \N_{>0}$:
{{begin-eqn}}
{{eqn | l = \map \mu {\set {f - g \ge \dfrac 1 n} }
| r = \int... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $\GG$ be a [[Definition:Sub-Sigma-Algebra|sub-$\sigma$-algebra]] of $\Sigma$.
Suppose that $\mu \restriction_\GG$, the [[Definition:Restricted Measure|restriction]] of $\mu$ to $\GG$, is [[Definition:Sigma-Finite Measure|$\sigma$-fini... | First, assume that $f$ and $g$ are $\mu$-[[Definition:Measure-Integrable Function|integrable]].
Observe:
{{begin-eqn}}
{{eqn | n = 1
| l = \set {f \ne g}
| r = \bigcup_{n \mathop = 1}^\infty \set {\size {f - g} \ge \dfrac 1 n }
}}
{{end-eqn}}
For each $n \in \N_{>0}$:
{{begin-eqn}}
{{eqn | l = \map \mu {\... | Measurable Functions with Equal Integrals on Sub-Sigma-Algebra are A.E. Equal | https://proofwiki.org/wiki/Measurable_Functions_with_Equal_Integrals_on_Sub-Sigma-Algebra_are_A.E._Equal | https://proofwiki.org/wiki/Measurable_Functions_with_Equal_Integrals_on_Sub-Sigma-Algebra_are_A.E._Equal | [
"Measure Theory"
] | [
"Definition:Measure Space",
"Definition:Sub-Sigma-Algebra",
"Definition:Restricted Measure",
"Definition:Sigma-Finite Measure",
"Definition:Measurable Function",
"Definition:Almost-Everywhere Equality Relation/Measurable Functions/Extended Real-Valued Functions"
] | [
"Definition:Integrable Function/Measure Space",
"Definition:Almost-Everywhere Equality Relation/Measurable Functions/Extended Real-Valued Functions",
"Definition:Sigma-Finite Measure",
"Definition:Exhausting Sequence of Sets",
"Definition:Exhausting Sequence of Sets",
"Definition:Integrable Function/Measu... |
proofwiki-5625 | Intersection is Largest Subset/Family of Sets | Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Then for all sets $X$:
:$\ds \paren {\forall i \in I: X \subseteq S_i} \iff X \subseteq \bigcap_{i \mathop \in I} S_i$
where $\ds \bigcap_{i \mathop \in I} S_i$ is the intersection of $\family {S_i}$. | Let $X \subseteq S_i$ for all $i \in I$.
Then from Set is Subset of Intersection of Supersets: General Result:
:$\ds X \subseteq \bigcap_{i \mathop \in I} S_i$
{{qed|lemma}}
Now suppose that $\ds X \subseteq \bigcap_{i \mathop \in I} S_i$.
From Intersection is Subset: Family of Sets we have:
:$\ds \forall i \in I: \big... | Let $\family {S_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of sets indexed by $I$]].
Then for all [[Definition:Set|sets]] $X$:
:$\ds \paren {\forall i \in I: X \subseteq S_i} \iff X \subseteq \bigcap_{i \mathop \in I} S_i$
where $\ds \bigcap_{i \mathop \in I} S_i$ is the [[Definition:Inters... | Let $X \subseteq S_i$ for all $i \in I$.
Then from [[Set is Subset of Intersection of Supersets/General Result|Set is Subset of Intersection of Supersets: General Result]]:
:$\ds X \subseteq \bigcap_{i \mathop \in I} S_i$
{{qed|lemma}}
Now suppose that $\ds X \subseteq \bigcap_{i \mathop \in I} S_i$.
From [[Inters... | Intersection is Largest Subset/Family of Sets | https://proofwiki.org/wiki/Intersection_is_Largest_Subset/Family_of_Sets | https://proofwiki.org/wiki/Intersection_is_Largest_Subset/Family_of_Sets | [
"Set Intersection",
"Subsets",
"Indexed Families"
] | [
"Definition:Indexing Set/Family of Sets",
"Definition:Set",
"Definition:Set Intersection/Family of Sets"
] | [
"Set is Subset of Intersection of Supersets/General Result",
"Intersection is Subset/Family of Sets",
"Subset Relation is Transitive"
] |
proofwiki-5626 | Countable Set is Null Set under Lebesgue Measure | Let $S \subseteq \R$ be a countable set.
Then $\map \lambda S = 0$, where $\lambda$ is Lebesgue measure.
That is, $S$ is a $\lambda$-null set. | By Surjection from Natural Numbers iff Countable, there exists a surjection $f: \N \to S$.
It follows that:
:$S = \ds \bigcup_{n \mathop \in \N} \set{\map f n}$
As Lebesgue Measure is Diffuse, it holds that:
:$\forall n \in \N: \map \lambda {\set{\map f n}} = 0$
Thus, by Null Sets Closed under Countable Union, it follo... | Let $S \subseteq \R$ be a [[Definition:Countable Set|countable set]].
Then $\map \lambda S = 0$, where $\lambda$ is [[Definition:Lebesgue Measure|Lebesgue measure]].
That is, $S$ is a [[Definition:Null Set|$\lambda$-null set]]. | By [[Surjection from Natural Numbers iff Countable]], there exists a [[Definition:Surjection|surjection]] $f: \N \to S$.
It follows that:
:$S = \ds \bigcup_{n \mathop \in \N} \set{\map f n}$
As [[Lebesgue Measure is Diffuse]], it holds that:
:$\forall n \in \N: \map \lambda {\set{\map f n}} = 0$
Thus, by [[Null S... | Countable Set is Null Set under Lebesgue Measure | https://proofwiki.org/wiki/Countable_Set_is_Null_Set_under_Lebesgue_Measure | https://proofwiki.org/wiki/Countable_Set_is_Null_Set_under_Lebesgue_Measure | [
"Lebesgue Measure"
] | [
"Definition:Countable Set",
"Definition:Lebesgue Measure",
"Definition:Null Set"
] | [
"Surjection from Natural Numbers iff Countable",
"Definition:Surjection",
"Lebesgue Measure is Diffuse",
"Null Sets Closed under Countable Union"
] |
proofwiki-5627 | Intersection is Subset/Family of Sets | Let $\family {S_\alpha}_{\alpha \mathop \in I}$ be a family of sets indexed by $I$.
Then:
:$\ds \forall \beta \in I: \bigcap_{\alpha \mathop \in I} S_\alpha \subseteq S_\beta$
where $\ds \bigcap_{\alpha \mathop \in I} S_\alpha$ is the intersection of $\family {S_\alpha}_{\alpha \mathop \in I}$. | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcap_{\alpha \mathop \in I} S_\alpha
| c =
}}
{{eqn | ll= \leadsto
| q = \forall \beta \in I
| l = x
| o = \in
| r = S_\beta
| c = {{Defof|Intersection of Family}}
}}
{{eqn | ll= \leadsto
| q = \forall \beta \in I
... | Let $\family {S_\alpha}_{\alpha \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of sets indexed by $I$]].
Then:
:$\ds \forall \beta \in I: \bigcap_{\alpha \mathop \in I} S_\alpha \subseteq S_\beta$
where $\ds \bigcap_{\alpha \mathop \in I} S_\alpha$ is the [[Definition:Intersection of Family|intersect... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcap_{\alpha \mathop \in I} S_\alpha
| c =
}}
{{eqn | ll= \leadsto
| q = \forall \beta \in I
| l = x
| o = \in
| r = S_\beta
| c = {{Defof|Intersection of Family}}
}}
{{eqn | ll= \leadsto
| q = \forall \beta \in I
... | Intersection is Subset/Family of Sets | https://proofwiki.org/wiki/Intersection_is_Subset/Family_of_Sets | https://proofwiki.org/wiki/Intersection_is_Subset/Family_of_Sets | [
"Set Intersection",
"Subsets",
"Indexed Families"
] | [
"Definition:Indexing Set/Family of Sets",
"Definition:Set Intersection/Family of Sets"
] | [] |
proofwiki-5628 | Union is Smallest Superset/Family of Sets | Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Then for all sets $X$:
:$\ds \paren {\forall i \in I: S_i \subseteq X} \iff \bigcup_{i \mathop \in I} S_i \subseteq X$
where $\ds \bigcup_{i \mathop \in I} S_i$ is the union of $\family {S_i}$. | === Necessary Condition ===
From Union of Family of Subsets is Subset we have that:
:$\ds \paren {\forall i \in I: S_i \subseteq X} \implies \bigcup_{i \mathop \in I} S_i \subseteq X$
{{qed|lemma}} | Let $\family {S_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of sets indexed by $I$]].
Then for all [[Definition:Set|sets]] $X$:
:$\ds \paren {\forall i \in I: S_i \subseteq X} \iff \bigcup_{i \mathop \in I} S_i \subseteq X$
where $\ds \bigcup_{i \mathop \in I} S_i$ is the [[Definition:Union ... | === Necessary Condition ===
From [[Union of Family of Subsets is Subset]] we have that:
:$\ds \paren {\forall i \in I: S_i \subseteq X} \implies \bigcup_{i \mathop \in I} S_i \subseteq X$
{{qed|lemma}} | Union is Smallest Superset/Family of Sets | https://proofwiki.org/wiki/Union_is_Smallest_Superset/Family_of_Sets | https://proofwiki.org/wiki/Union_is_Smallest_Superset/Family_of_Sets | [
"Set Union",
"Subsets",
"Indexed Families"
] | [
"Definition:Indexing Set/Family of Sets",
"Definition:Set",
"Definition:Set Union/Family of Sets"
] | [
"Union of Subsets is Subset/Family of Sets"
] |
proofwiki-5629 | Fatou's Lemma for Integrals/Integrable Functions | Let $\sequence {f_n}_{n \mathop \in \N} \in \LL^1$, $f_n: X \to \R$ be a sequence of $\mu$-integrable functions.
Let $\ds \liminf_{n \mathop \to \infty} f_n: X \to \overline \R$ be the pointwise limit inferior of the $f_n$.
Suppose that there exists an $\mu$-integrable $f: X \to \R$ such that for all $n \in \N$, $f \le... | {{ProofWanted}}
{{Namedfor|Pierre Joseph Louis Fatou|cat = Fatou}} | Let $\sequence {f_n}_{n \mathop \in \N} \in \LL^1$, $f_n: X \to \R$ be a [[Definition:Sequence|sequence]] of [[Definition:Measure-Integrable Function|$\mu$-integrable functions]].
Let $\ds \liminf_{n \mathop \to \infty} f_n: X \to \overline \R$ be the [[Definition:Pointwise Limit Inferior|pointwise limit inferior]] of... | {{ProofWanted}}
{{Namedfor|Pierre Joseph Louis Fatou|cat = Fatou}} | Fatou's Lemma for Integrals/Integrable Functions | https://proofwiki.org/wiki/Fatou's_Lemma_for_Integrals/Integrable_Functions | https://proofwiki.org/wiki/Fatou's_Lemma_for_Integrals/Integrable_Functions | [
"Fatou's Lemma for Integrals",
"Measure Theory"
] | [
"Definition:Sequence",
"Definition:Integrable Function/Measure Space",
"Definition:Pointwise Limit Inferior",
"Definition:Integrable Function/Measure Space",
"Definition:Pointwise Inequality",
"Definition:Integral Sign",
"Definition:Integral of Measure-Integrable Function",
"Definition:Limit Inferior"... | [] |
proofwiki-5630 | Fatou's Lemma for Integrals/Positive Measurable Functions | Let $\sequence {f_n}_{n \mathop \in \N} \in \MM_{\overline \R}^+$, $f_n: X \to \overline \R$ be a sequence of positive measurable functions.
{{explain|What is $\MM_{\overline \R}^+$?}}
Let $\ds \liminf_{n \mathop \to \infty} f_n: X \to \overline \R$ be the pointwise limit inferior of the $f_n$.
Then:
:$\ds \int \liminf... | For each $n \in \N$, define $g_n : X \to \overline \R$ by:
:$\ds g_n = \inf_{k \mathop \ge n} f_k$
That is:
:$\map {g_n} x = \inf \set {\map {f_k} x : k \ge n}$
for each $x \in X$.
For each $n \in \N$, we have that:
:$\set {\map {f_k} x : k \ge n + 1} \subseteq \set {\map {f_k} x : k \ge n}$
From Infimum of Subset:... | Let $\sequence {f_n}_{n \mathop \in \N} \in \MM_{\overline \R}^+$, $f_n: X \to \overline \R$ be a [[Definition:Sequence|sequence]] of [[Definition:Positive Measurable Function|positive measurable functions]].
{{explain|What is $\MM_{\overline \R}^+$?}}
Let $\ds \liminf_{n \mathop \to \infty} f_n: X \to \overline \R$ b... | For each $n \in \N$, define $g_n : X \to \overline \R$ by:
:$\ds g_n = \inf_{k \mathop \ge n} f_k$
That is:
:$\map {g_n} x = \inf \set {\map {f_k} x : k \ge n}$
for each $x \in X$.
For each $n \in \N$, we have that:
:$\set {\map {f_k} x : k \ge n + 1} \subseteq \set {\map {f_k} x : k \ge n}$
From [[Infimum o... | Fatou's Lemma for Integrals/Positive Measurable Functions | https://proofwiki.org/wiki/Fatou's_Lemma_for_Integrals/Positive_Measurable_Functions | https://proofwiki.org/wiki/Fatou's_Lemma_for_Integrals/Positive_Measurable_Functions | [
"Measure Theory",
"Fatou's Lemma for Integrals"
] | [
"Definition:Sequence",
"Definition:Measurable Function/Positive",
"Definition:Pointwise Limit Inferior",
"Definition:Integral Sign",
"Definition:Integral of Positive Measurable Function",
"Definition:Limit Inferior",
"Definition:Extended Real Number Line"
] | [
"Infimum of Subset",
"Pointwise Infimum of Measurable Functions is Measurable",
"Definition:Measurable Function",
"Definition:Limit Inferior",
"Definition:Increasing Sequence of Real-Valued Functions",
"Definition:Sequence",
"Definition:Measurable Function",
"Monotone Convergence Theorem (Measure Theo... |
proofwiki-5631 | Reverse Fatou's Lemma/Integrable Functions | Let $\sequence {f_n}_{n \mathop \in \N} \in \LL^1$, $f_n: X \to \R$ be a sequence of $\mu$-integrable functions.
{{mistake|$f_n \in \LL^1$ for each $n$}}
Let $\ds \limsup_{n \mathop \to \infty} f_n: X \to \overline \R$ be the pointwise limit superior of the $f_n$.
Suppose that there exists an $\mu$-integrable $f: X \to... | {{ProofWanted}}
{{Namedfor|Pierre Joseph Louis Fatou|cat = Fatou}} | Let $\sequence {f_n}_{n \mathop \in \N} \in \LL^1$, $f_n: X \to \R$ be a [[Definition:Sequence|sequence]] of [[Definition:Measure-Integrable Function|$\mu$-integrable functions]].
{{mistake|$f_n \in \LL^1$ for each $n$}}
Let $\ds \limsup_{n \mathop \to \infty} f_n: X \to \overline \R$ be the [[Definition:Pointwise Lim... | {{ProofWanted}}
{{Namedfor|Pierre Joseph Louis Fatou|cat = Fatou}} | Reverse Fatou's Lemma/Integrable Functions | https://proofwiki.org/wiki/Reverse_Fatou's_Lemma/Integrable_Functions | https://proofwiki.org/wiki/Reverse_Fatou's_Lemma/Integrable_Functions | [
"Reverse Fatou's Lemma"
] | [
"Definition:Sequence",
"Definition:Integrable Function/Measure Space",
"Definition:Pointwise Limit Superior",
"Definition:Integrable Function/Measure Space",
"Definition:Pointwise Inequality",
"Definition:Integral Sign",
"Definition:Integral of Measure-Integrable Function",
"Definition:Limit Inferior"... | [] |
proofwiki-5632 | Set is Subset of Union/Family of Sets | Let $\family {S_\alpha}_{\alpha \mathop \in I}$ be a family of sets indexed by $I$.
Then:
:$\ds \forall \beta \in I: S_\beta \subseteq \bigcup_{\alpha \mathop \in I} S_\alpha$
where $\ds \bigcup_{\alpha \mathop \in I} S_\alpha$ is the union of $\family {S_\alpha}$. | Let $x \in S_\beta$ for some $\beta \in I$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S_\beta
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \set {x: \exists \alpha \in I: x \in S_\alpha}
| c = {{Defof|Indexed Family of Sets}}
}}
{{eqn | ll= \leadsto
| l =... | Let $\family {S_\alpha}_{\alpha \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of sets indexed by $I$]].
Then:
:$\ds \forall \beta \in I: S_\beta \subseteq \bigcup_{\alpha \mathop \in I} S_\alpha$
where $\ds \bigcup_{\alpha \mathop \in I} S_\alpha$ is the [[Definition:Union of Family|union of $\famil... | Let $x \in S_\beta$ for some $\beta \in I$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S_\beta
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \set {x: \exists \alpha \in I: x \in S_\alpha}
| c = {{Defof|Indexed Family of Sets}}
}}
{{eqn | ll= \leadsto
| l ... | Set is Subset of Union/Family of Sets/Proof 1 | https://proofwiki.org/wiki/Set_is_Subset_of_Union/Family_of_Sets | https://proofwiki.org/wiki/Set_is_Subset_of_Union/Family_of_Sets/Proof_1 | [
"Set is Subset of Union",
"Set Union",
"Subsets",
"Indexed Families"
] | [
"Definition:Indexing Set/Family of Sets",
"Definition:Set Union/Family of Sets"
] | [] |
proofwiki-5633 | Set is Subset of Union/Family of Sets | Let $\family {S_\alpha}_{\alpha \mathop \in I}$ be a family of sets indexed by $I$.
Then:
:$\ds \forall \beta \in I: S_\beta \subseteq \bigcup_{\alpha \mathop \in I} S_\alpha$
where $\ds \bigcup_{\alpha \mathop \in I} S_\alpha$ is the union of $\family {S_\alpha}$. | Let $\beta \in I$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \beta
| o = \in
| r = I
| c =
}}
{{eqn | ll= \leadsto
| l = \set \beta
| o = \subseteq
| r = I
| c = Singleton of Element is Subset
}}
{{eqn | ll= \leadsto
| l = \bigcup \set {S_\beta}
| o = \subseteq
... | Let $\family {S_\alpha}_{\alpha \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of sets indexed by $I$]].
Then:
:$\ds \forall \beta \in I: S_\beta \subseteq \bigcup_{\alpha \mathop \in I} S_\alpha$
where $\ds \bigcup_{\alpha \mathop \in I} S_\alpha$ is the [[Definition:Union of Family|union of $\famil... | Let $\beta \in I$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \beta
| o = \in
| r = I
| c =
}}
{{eqn | ll= \leadsto
| l = \set \beta
| o = \subseteq
| r = I
| c = [[Singleton of Element is Subset]]
}}
{{eqn | ll= \leadsto
| l = \bigcup \set {S_\beta}
| o = \subs... | Set is Subset of Union/Family of Sets/Proof 2 | https://proofwiki.org/wiki/Set_is_Subset_of_Union/Family_of_Sets | https://proofwiki.org/wiki/Set_is_Subset_of_Union/Family_of_Sets/Proof_2 | [
"Set is Subset of Union",
"Set Union",
"Subsets",
"Indexed Families"
] | [
"Definition:Indexing Set/Family of Sets",
"Definition:Set Union/Family of Sets"
] | [
"Singleton of Element is Subset",
"Union of Subset of Family is Subset of Union of Family"
] |
proofwiki-5634 | Reverse Fatou's Lemma/Positive Measurable Functions | Let $\sequence {f_n}_{n \mathop \in \N} \in \MM_{\overline \R}^+$, $f_n: X \to \overline \R$ be a sequence of positive measurable functions.
Suppose that there exists a positive measurable function $f: X \to \overline \R$ such that:
:$\ds \int f \rd \mu < +\infty$
:$\forall n \in \N: f_n \le f$
where $\le$ signifies a ... | {{ProofWanted}}
{{Namedfor|Pierre Joseph Louis Fatou|cat = Fatou}} | Let $\sequence {f_n}_{n \mathop \in \N} \in \MM_{\overline \R}^+$, $f_n: X \to \overline \R$ be a [[Definition:Sequence|sequence]] of [[Definition:Positive Measurable Function|positive measurable functions]].
Suppose that there exists a [[Definition:Positive Measurable Function|positive measurable function]] $f: X \to... | {{ProofWanted}}
{{Namedfor|Pierre Joseph Louis Fatou|cat = Fatou}} | Reverse Fatou's Lemma/Positive Measurable Functions | https://proofwiki.org/wiki/Reverse_Fatou's_Lemma/Positive_Measurable_Functions | https://proofwiki.org/wiki/Reverse_Fatou's_Lemma/Positive_Measurable_Functions | [
"Reverse Fatou's Lemma"
] | [
"Definition:Sequence",
"Definition:Measurable Function/Positive",
"Definition:Measurable Function/Positive",
"Definition:Pointwise Inequality of Extended Real-Valued Functions",
"Definition:Pointwise Limit Superior",
"Definition:Integral Sign",
"Definition:Integral of Positive Measurable Function",
"D... | [] |
proofwiki-5635 | Monotone Convergence Theorem (Measure Theory) | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $u : X \to \overline \R_{\ge 0}$ be a positive $\Sigma$-measurable function.
Let $\sequence {u_n}_{n \mathop \in \N}$ be an sequence of positive $\Sigma$-measurable functions $u_n : X \to \overline \R_{\ge 0}$ such that:
:$\map {u_i} x \le \map {u_j} x$ for all $... | First suppose that:
:$\map {u_i} x \le \map {u_j} x$ for all $i \le j$
and:
:$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$
for all $x \in X$.
From Integral of Positive Measurable Function is Monotone, we have that:
:$\ds \int u_i \rd \mu \le \int u_j \rd \mu$ for all $i \le j$.
From Monotone Convergence T... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $u : X \to \overline \R_{\ge 0}$ be a [[Definition:Positive Measurable Function|positive $\Sigma$-measurable function]].
Let $\sequence {u_n}_{n \mathop \in \N}$ be an [[Definition:Sequence|sequence]] of [[Definition:Positive Measura... | First suppose that:
:$\map {u_i} x \le \map {u_j} x$ for all $i \le j$
and:
:$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$
for all $x \in X$.
From [[Integral of Positive Measurable Function is Monotone]], we have that:
:$\ds \int u_i \rd \mu \le \int u_j \rd \mu$ for all $i \le j$.
From [[Monotone ... | Monotone Convergence Theorem (Measure Theory) | https://proofwiki.org/wiki/Monotone_Convergence_Theorem_(Measure_Theory) | https://proofwiki.org/wiki/Monotone_Convergence_Theorem_(Measure_Theory) | [
"Named Theorems",
"Measure Theory",
"Monotone Convergence Theorem",
"Monotone Convergence Theorem (Measure Theory)"
] | [
"Definition:Measure Space",
"Definition:Measurable Function/Positive",
"Definition:Sequence",
"Definition:Measurable Function/Positive",
"Definition:Almost All"
] | [
"Integral of Positive Measurable Function is Monotone",
"Monotone Convergence Theorem (Real Analysis)/Increasing Sequence",
"Integral of Positive Measurable Function is Monotone",
"Definition:Sequence",
"Definition:Increasing/Sequence",
"Monotone Convergence Theorem (Real Analysis)/Increasing Sequence",
... |
proofwiki-5636 | Intersection is Associative/Family of Sets | Let $\family {S_i}_{i \mathop \in I}$ and $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$ be indexed families of sets.
Let $\ds I = \bigcap_{\lambda \mathop \in \Lambda} I_\lambda$.
Then:
:$\ds \bigcap_{i \mathop \in I} S_i = \bigcap_{\lambda \mathop \in \Lambda} \paren {\bigcap_{i \mathop \in I_\lambda} S_i}$ | For every $\lambda \in \Lambda$, let $\ds T_\lambda = \bigcap_{i \mathop \in I_\lambda} S_i$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcap_{i \mathop \in I} S_i
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \forall i \in I
| l = x
| o = \in
| r = S_i
| c = {{De... | Let $\family {S_i}_{i \mathop \in I}$ and $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$ be [[Definition:Indexed Family of Sets|indexed families of sets]].
Let $\ds I = \bigcap_{\lambda \mathop \in \Lambda} I_\lambda$.
Then:
:$\ds \bigcap_{i \mathop \in I} S_i = \bigcap_{\lambda \mathop \in \Lambda} \paren {\bi... | For every $\lambda \in \Lambda$, let $\ds T_\lambda = \bigcap_{i \mathop \in I_\lambda} S_i$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcap_{i \mathop \in I} S_i
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \forall i \in I
| l = x
| o = \in
| r = S_i
| c = {... | Intersection is Associative/Family of Sets/Proof 1 | https://proofwiki.org/wiki/Intersection_is_Associative/Family_of_Sets | https://proofwiki.org/wiki/Intersection_is_Associative/Family_of_Sets/Proof_1 | [
"Intersection is Associative"
] | [
"Definition:Indexing Set/Family of Sets"
] | [] |
proofwiki-5637 | Intersection is Associative/Family of Sets | Let $\family {S_i}_{i \mathop \in I}$ and $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$ be indexed families of sets.
Let $\ds I = \bigcap_{\lambda \mathop \in \Lambda} I_\lambda$.
Then:
:$\ds \bigcap_{i \mathop \in I} S_i = \bigcap_{\lambda \mathop \in \Lambda} \paren {\bigcap_{i \mathop \in I_\lambda} S_i}$ | {{begin-eqn}}
{{eqn | l = \bigcap_{i \mathop \in I} S_i
| r = \map \complement {\map \complement {\bigcap_{i \mathop \in I} S_i} }
| c = Complement of Complement
}}
{{eqn | r = \map \complement {\bigcup_{i \mathop \in I} \map \complement {S_i} }
| c = De Morgan's Laws (Set Theory)/Set Complement/Famil... | Let $\family {S_i}_{i \mathop \in I}$ and $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$ be [[Definition:Indexed Family of Sets|indexed families of sets]].
Let $\ds I = \bigcap_{\lambda \mathop \in \Lambda} I_\lambda$.
Then:
:$\ds \bigcap_{i \mathop \in I} S_i = \bigcap_{\lambda \mathop \in \Lambda} \paren {\bi... | {{begin-eqn}}
{{eqn | l = \bigcap_{i \mathop \in I} S_i
| r = \map \complement {\map \complement {\bigcap_{i \mathop \in I} S_i} }
| c = [[Complement of Complement]]
}}
{{eqn | r = \map \complement {\bigcup_{i \mathop \in I} \map \complement {S_i} }
| c = [[De Morgan's Laws (Set Theory)/Set Complement... | Intersection is Associative/Family of Sets/Proof 2 | https://proofwiki.org/wiki/Intersection_is_Associative/Family_of_Sets | https://proofwiki.org/wiki/Intersection_is_Associative/Family_of_Sets/Proof_2 | [
"Intersection is Associative"
] | [
"Definition:Indexing Set/Family of Sets"
] | [
"Complement of Complement",
"De Morgan's Laws (Set Theory)/Set Complement/Family of Sets/Complement of Intersection",
"Union is Associative/Family of Sets",
"De Morgan's Laws (Set Theory)/Set Complement/Family of Sets/Complement of Intersection",
"De Morgan's Laws (Set Theory)/Set Complement/Family of Sets/... |
proofwiki-5638 | Union is Associative/Family of Sets | Let $\family {S_i}_{i \mathop \in I}$ and $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$ be indexed families of sets.
Let $\ds I = \bigcup_{\lambda \mathop \in \Lambda} I_\lambda$ denote the union of $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$.
Then:
:$\ds \bigcup_{i \mathop \in I} S_i = \bigcup_{\lambda ... | For every $\lambda \in \Lambda$, let $\ds T_\lambda = \bigcup_{i \mathop \in I_\lambda} S_i$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcup_{i \mathop \in I} S_i
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \exists i \in I
| l = x
| o = \in
| r = S_i
| c = {{De... | Let $\family {S_i}_{i \mathop \in I}$ and $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$ be [[Definition:Indexed Family of Sets|indexed families of sets]].
Let $\ds I = \bigcup_{\lambda \mathop \in \Lambda} I_\lambda$ denote the [[Definition:Union of Family|union]] of $\family {I_\lambda}_{\lambda \mathop \in \L... | For every $\lambda \in \Lambda$, let $\ds T_\lambda = \bigcup_{i \mathop \in I_\lambda} S_i$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcup_{i \mathop \in I} S_i
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \exists i \in I
| l = x
| o = \in
| r = S_i
| c = {... | Union is Associative/Family of Sets | https://proofwiki.org/wiki/Union_is_Associative/Family_of_Sets | https://proofwiki.org/wiki/Union_is_Associative/Family_of_Sets | [
"Union is Associative"
] | [
"Definition:Indexing Set/Family of Sets",
"Definition:Set Union/Family of Sets"
] | [] |
proofwiki-5639 | Lebesgue's Dominated Convergence Theorem | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f : X \to \overline \R$ be a $\Sigma$-measurable function.
Let $g : X \to \overline \R_{\ge 0}$ be a $\mu$-integrable function.
Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence of $\Sigma$-measurable functions $f_n : X \to \overline \R$ such that:
:$\ds \... | === Lemma ===
{{:Lebesgue's Dominated Convergence Theorem/Lemma}}{{qed|lemma}}
Since:
:$\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$
and:
:$\size {\map {f_n} x} \le \map g x$
hold for $\mu$-almost all $x \in X$, there exists a $\mu$-null set $N_1 \subseteq X$ such that whenever:
:either $\ds \lim_{n \math... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f : X \to \overline \R$ be a [[Definition:Measurable Function|$\Sigma$-measurable function]].
Let $g : X \to \overline \R_{\ge 0}$ be a [[Definition:Measure-Integrable Function|$\mu$-integrable function]].
Let $\sequence {f_n}_{n ... | === [[Lebesgue's Dominated Convergence Theorem/Lemma|Lemma]] ===
{{:Lebesgue's Dominated Convergence Theorem/Lemma}}{{qed|lemma}}
Since:
:$\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$
and:
:$\size {\map {f_n} x} \le \map g x$
hold for [[Definition:Almost All|$\mu$-almost all]] $x \in X$, there exists... | Lebesgue's Dominated Convergence Theorem | https://proofwiki.org/wiki/Lebesgue's_Dominated_Convergence_Theorem | https://proofwiki.org/wiki/Lebesgue's_Dominated_Convergence_Theorem | [
"Lebesgue's Dominated Convergence Theorem",
"Integrals of Integrable Functions"
] | [
"Definition:Measure Space",
"Definition:Measurable Function",
"Definition:Integrable Function/Measure Space",
"Definition:Sequence",
"Definition:Measurable Function",
"Definition:Almost All",
"Definition:Integrable Function/Measure Space",
"Definition:Integrable Function/Measure Space"
] | [
"Lebesgue's Dominated Convergence Theorem/Lemma",
"Definition:Almost All",
"Definition:Null Set",
"Integrable Function is A.E. Real-Valued",
"Definition:Null Set",
"Sigma-Algebra Closed under Countable Intersection",
"Definition:Measurable Set",
"Intersection is Subset",
"Null Sets Closed under Subs... |
proofwiki-5640 | Continuity under Integral Sign | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $U$ be a non-empty open set of a metric space.
Let $f: U \times X \to \R$ be a mapping satisfying:
:$(1): \quad$ For all $\lambda \in U$, the mapping $x \mapsto \map f {\lambda, x}$ is $\mu$-integrable
:$(2): \quad$ For $\mu$-almost all $x \in X$, the mapping $\lam... | Let $\lambda_0 \in U$ be arbitrary.
Let $\sequence {\lambda_n}_{n \mathop \ge 1}$ be a sequence in $U$ which converges to $\lambda_0$.
Define the sequence of functions $f_n: X \to \R$, for $n = 0$ and $n \ge 1$ by $\map {f_n} x = \map f {\lambda_n, x}$.
By hypothesis $(1)$, for each $n \ge 1$, the function $f_n$ is $\m... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $U$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Metric Space)|open set]] of a [[Definition:Metric Space|metric space]].
Let $f: U \times X \to \R$ be a [[Definition:Mapping|mapping]] satisfying:
:$(1): \quad$ F... | Let $\lambda_0 \in U$ be arbitrary.
Let $\sequence {\lambda_n}_{n \mathop \ge 1}$ be a [[Definition:Sequence|sequence]] in $U$ which [[Definition:Convergent Sequence (Metric Space)|converges]] to $\lambda_0$.
Define the [[Definition:Sequence|sequence]] of [[Definition:Real-Valued Function|functions]] $f_n: X \to \R$,... | Continuity under Integral Sign | https://proofwiki.org/wiki/Continuity_under_Integral_Sign | https://proofwiki.org/wiki/Continuity_under_Integral_Sign | [
"Measure Theory"
] | [
"Definition:Measure Space",
"Definition:Non-Empty Set",
"Definition:Open Set/Metric Space",
"Definition:Metric Space",
"Definition:Mapping",
"Definition:Mapping",
"Definition:Integrable Function/Measure Space",
"Definition:Almost Everywhere",
"Definition:Mapping",
"Definition:Continuous Mapping",
... | [
"Definition:Sequence",
"Definition:Convergent Sequence/Metric Space",
"Definition:Sequence",
"Definition:Real-Valued Function",
"Definition:Real-Valued Function",
"Definition:Integrable Function/Measure Space",
"Sequential Continuity is Equivalent to Continuity in Metric Space",
"Definition:Almost Eve... |
proofwiki-5641 | Riemann-Lebesgue Theorem | Let $f: \closedint a b \to \R$ be a bounded mapping.
Let $\mu$ be a one-dimensional Lebesgue measure.
{{explain|Link to definition of one-dimensional in this specific context}}
Then $f$ is Darboux integrable {{iff}} the set of all discontinuities of $f$ is a $\mu$-null set. | === Necessary Condition ===
Suppose that $f$ is Darboux integrable.
We need to prove that the set of all discontinuities of $f$ has measure $0$.
Let, for some positive real number $s$:
:$A_s = \set {x \in \closedint a b: \map {\omega_f} x > s}$
where
:$\map {\omega_f} x = \ds \inf \set {\map {\omega_f} {I \cap {\closed... | Let $f: \closedint a b \to \R$ be a [[Definition:Bounded Mapping|bounded mapping]].
Let $\mu$ be a one-dimensional [[Definition:Lebesgue Measure|Lebesgue measure]].
{{explain|Link to definition of one-dimensional in this specific context}}
Then $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] {{... | === Necessary Condition ===
Suppose that $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]].
We need to prove that the [[Definition:Set|set]] of all [[Definition:Discontinuous|discontinuities]] of $f$ has [[Definition:Null Measure|measure $0$]].
Let, for some [[Definition:Positive Real Number|posi... | Riemann-Lebesgue Theorem | https://proofwiki.org/wiki/Riemann-Lebesgue_Theorem | https://proofwiki.org/wiki/Riemann-Lebesgue_Theorem | [
"Integral Calculus",
"Measure Theory"
] | [
"Definition:Bounded Mapping",
"Definition:Lebesgue Measure",
"Definition:Darboux Integrable Function",
"Definition:Set",
"Definition:Discontinuous",
"Definition:Null Set"
] | [
"Definition:Darboux Integrable Function",
"Definition:Set",
"Definition:Discontinuous",
"Definition:Null Measure",
"Definition:Positive/Real Number",
"Definition:Set",
"Definition:Neighborhood (Real Analysis)/Open Subset",
"Definition:Oscillation/Real Space",
"Definition:Oscillation/Real Space/Oscil... |
proofwiki-5642 | Gamma Function is Continuous on Positive Reals | Let $\Gamma: \R_{>0} \to \R$ be the Gamma function, restricted to the strictly positive real numbers.
Then $\Gamma$ is continuous. | Let $0 < \alpha < a \le x \le y \le b < \beta$.
Let $0 < \delta < \Delta$.
Then:
:$(1): \quad \ds \size {\int_\delta^\Delta t^{x - 1} e^{-t} \rd t - \int_\delta^\Delta t^{y - 1} e^{-t} } \le \int_\delta^\Delta \paren {t^{x - 1} - t^{y - 1} }e^{-t} \rd t$
{{explain|why?}}
From the Mean Value Theorem:
:$(2): \quad \dfrac... | Let $\Gamma: \R_{>0} \to \R$ be the [[Definition:Gamma Function|Gamma function]], [[Definition:Restriction of Mapping|restricted]] to the [[Definition:Strictly Positive Real Number|strictly positive real numbers]].
Then $\Gamma$ is [[Definition:Continuous Mapping|continuous]]. | Let $0 < \alpha < a \le x \le y \le b < \beta$.
Let $0 < \delta < \Delta$.
Then:
:$(1): \quad \ds \size {\int_\delta^\Delta t^{x - 1} e^{-t} \rd t - \int_\delta^\Delta t^{y - 1} e^{-t} } \le \int_\delta^\Delta \paren {t^{x - 1} - t^{y - 1} }e^{-t} \rd t$
{{explain|why?}}
From the [[Mean Value Theorem]]:
:$(2): \qua... | Gamma Function is Continuous on Positive Reals | https://proofwiki.org/wiki/Gamma_Function_is_Continuous_on_Positive_Reals | https://proofwiki.org/wiki/Gamma_Function_is_Continuous_on_Positive_Reals | [
"Gamma Function"
] | [
"Definition:Gamma Function",
"Definition:Restriction/Mapping",
"Definition:Strictly Positive/Real Number",
"Definition:Continuous Mapping"
] | [
"Mean Value Theorem",
"Squeeze Theorem"
] |
proofwiki-5643 | Log of Gamma Function is Convex on Positive Reals | Let $\Gamma: \R_{>0} \to \R$ be the Gamma function, restricted to the strictly positive real numbers.
Let $\ln$ denote the natural logarithm function.
Then the composite mapping $\ln \circ \operatorname \Gamma$ is a convex function. | By definition, the Gamma function $\Gamma: \R_{> 0} \to \R$ is defined as:
:$\ds \map \Gamma z = \int_0^{\infty} t^{z - 1} e^{-t} \rd t$
:$\forall z > 0: \map \Gamma z > 0$, as an integral of a strictly positive function in $t$.
{{explain|A separate page is needed for the above statement}}
The function is smooth accord... | Let $\Gamma: \R_{>0} \to \R$ be the [[Definition:Gamma Function|Gamma function]], [[Definition:Restriction of Mapping|restricted]] to the [[Definition:Strictly Positive Real Number|strictly positive real numbers]].
Let $\ln$ denote the [[Definition:Natural Logarithm|natural logarithm function]].
Then the [[Definitio... | By definition, the [[Definition:Integral Form of Gamma Function|Gamma function]] $\Gamma: \R_{> 0} \to \R$ is defined as:
:$\ds \map \Gamma z = \int_0^{\infty} t^{z - 1} e^{-t} \rd t$
:$\forall z > 0: \map \Gamma z > 0$, as an integral of a strictly positive function in $t$.
{{explain|A separate page is needed for the... | Log of Gamma Function is Convex on Positive Reals/Proof 1 | https://proofwiki.org/wiki/Log_of_Gamma_Function_is_Convex_on_Positive_Reals | https://proofwiki.org/wiki/Log_of_Gamma_Function_is_Convex_on_Positive_Reals/Proof_1 | [
"Gamma Function",
"Log of Gamma Function is Convex on Positive Reals"
] | [
"Definition:Gamma Function",
"Definition:Restriction/Mapping",
"Definition:Strictly Positive/Real Number",
"Definition:Natural Logarithm",
"Definition:Composition of Mappings",
"Definition:Convex Real Function"
] | [
"Definition:Gamma Function/Integral Form",
"Gamma Function is Smooth on Positive Reals",
"Definition:Fraction/Numerator",
"Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces",
"Definition:Convex Real Function"
] |
proofwiki-5644 | Log of Gamma Function is Convex on Positive Reals | Let $\Gamma: \R_{>0} \to \R$ be the Gamma function, restricted to the strictly positive real numbers.
Let $\ln$ denote the natural logarithm function.
Then the composite mapping $\ln \circ \operatorname \Gamma$ is a convex function. | The strategy is to show that:
:$\map \ln {\map \Gamma {\dfrac x 2 + \dfrac y 2} } \le \dfrac 1 2 \map \ln {\map \Gamma x} + \dfrac 1 2 \map \ln {\map \Gamma y}$
Let $0 < \delta < \Delta$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {\int_\delta^\Delta t^{\paren {x + y - 2} / 2} e^{-t} \rd t}^2
| r = \paren {\int_\delt... | Let $\Gamma: \R_{>0} \to \R$ be the [[Definition:Gamma Function|Gamma function]], [[Definition:Restriction of Mapping|restricted]] to the [[Definition:Strictly Positive Real Number|strictly positive real numbers]].
Let $\ln$ denote the [[Definition:Natural Logarithm|natural logarithm function]].
Then the [[Definitio... | The strategy is to show that:
:$\map \ln {\map \Gamma {\dfrac x 2 + \dfrac y 2} } \le \dfrac 1 2 \map \ln {\map \Gamma x} + \dfrac 1 2 \map \ln {\map \Gamma y}$
Let $0 < \delta < \Delta$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {\int_\delta^\Delta t^{\paren {x + y - 2} / 2} e^{-t} \rd t}^2
| r = \paren {\int_\d... | Log of Gamma Function is Convex on Positive Reals/Proof 2 | https://proofwiki.org/wiki/Log_of_Gamma_Function_is_Convex_on_Positive_Reals | https://proofwiki.org/wiki/Log_of_Gamma_Function_is_Convex_on_Positive_Reals/Proof_2 | [
"Gamma Function",
"Log of Gamma Function is Convex on Positive Reals"
] | [
"Definition:Gamma Function",
"Definition:Restriction/Mapping",
"Definition:Strictly Positive/Real Number",
"Definition:Natural Logarithm",
"Definition:Composition of Mappings",
"Definition:Convex Real Function"
] | [
"Cauchy-Bunyakovsky-Schwarz Inequality/Definite Integrals",
"Definition:Convex Real Function"
] |
proofwiki-5645 | Log of Gamma Function is Convex on Positive Reals | Let $\Gamma: \R_{>0} \to \R$ be the Gamma function, restricted to the strictly positive real numbers.
Let $\ln$ denote the natural logarithm function.
Then the composite mapping $\ln \circ \operatorname \Gamma$ is a convex function. | The strategy is to use the Euler Form of the Gamma function and directly calculate the second derivative of $\ln \map \Gamma z$.
{{begin-eqn}}
{{eqn | l = \map \Gamma z
| r = \lim_{m \mathop \to \infty} \frac {m^z m!} {z \paren {z + 1} \paren {z + 2} \dotsm \paren {z + m} }
| c =
}}
{{eqn | ll= \leadsto
... | Let $\Gamma: \R_{>0} \to \R$ be the [[Definition:Gamma Function|Gamma function]], [[Definition:Restriction of Mapping|restricted]] to the [[Definition:Strictly Positive Real Number|strictly positive real numbers]].
Let $\ln$ denote the [[Definition:Natural Logarithm|natural logarithm function]].
Then the [[Definitio... | The strategy is to use the [[Definition:Euler Form of Gamma Function|Euler Form of the Gamma function]] and directly calculate the [[Definition:Second Derivative|second derivative]] of $\ln \map \Gamma z$.
{{begin-eqn}}
{{eqn | l = \map \Gamma z
| r = \lim_{m \mathop \to \infty} \frac {m^z m!} {z \paren {z + 1}... | Log of Gamma Function is Convex on Positive Reals/Proof 3 | https://proofwiki.org/wiki/Log_of_Gamma_Function_is_Convex_on_Positive_Reals | https://proofwiki.org/wiki/Log_of_Gamma_Function_is_Convex_on_Positive_Reals/Proof_3 | [
"Gamma Function",
"Log of Gamma Function is Convex on Positive Reals"
] | [
"Definition:Gamma Function",
"Definition:Restriction/Mapping",
"Definition:Strictly Positive/Real Number",
"Definition:Natural Logarithm",
"Definition:Composition of Mappings",
"Definition:Convex Real Function"
] | [
"Definition:Gamma Function/Euler Form",
"Definition:Derivative/Higher Derivatives/Second Derivative",
"Gamma Function is Smooth on Positive Reals",
"Real Function with Strictly Positive Second Derivative is Strictly Convex"
] |
proofwiki-5646 | Pratt's Lemma | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let:
:$\sequence {g_n}_{n \mathop \in \N}$
:$\sequence {G_n}_{n \mathop \in \N}$
:$\sequence {f_n}_{n \mathop \in \N}$
be sequences of $\mu$-integrable functions.
Let the pointwise limits:
:$\ds f := \lim_{n \mathop \to \infty} f_n$
:$\ds g := \lim_{n \mathop \to \inft... | {{ProofWanted}}
{{Namedfor|John Winsor Pratt|cat = Pratt}} | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let:
:$\sequence {g_n}_{n \mathop \in \N}$
:$\sequence {G_n}_{n \mathop \in \N}$
:$\sequence {f_n}_{n \mathop \in \N}$
be [[Definition:Sequence|sequences]] of [[Definition:Measure-Integrable Function|$\mu$-integrable functions]].
Let t... | {{ProofWanted}}
{{Namedfor|John Winsor Pratt|cat = Pratt}} | Pratt's Lemma | https://proofwiki.org/wiki/Pratt's_Lemma | https://proofwiki.org/wiki/Pratt's_Lemma | [
"Measure Theory"
] | [
"Definition:Measure Space",
"Definition:Sequence",
"Definition:Integrable Function/Measure Space",
"Definition:Pointwise Limit",
"Definition:Integrable Function/Measure Space",
"Definition:Finite Extended Real Number"
] | [] |
proofwiki-5647 | Hölder's Inequality for Integrals | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $p, q \in \R_{>0}$ such that $\dfrac 1 p + \dfrac 1 q = 1$.
{{improve|the assumption should read $p,q\in\R_{>0}\cup\set{+\infty}$.<br/>Suggestion: make a page for defining $p,q$ as satisfying this relation, including the pair $\tuple{1,\infty}$}}
Let $f \in \map {\... | Let $x \in X$.
Let:
:$a_x := \dfrac {\size {\map f x} } {\norm f_p}$
and:
:$b_x := \dfrac {\size {\map g x} } {\norm g_q}$
Applying Young's Inequality for Products to $a_x$ and $b_x$:
:$\dfrac {\size {\map f x \map g x} } {\norm f_p \cdot \norm g_q} \le \dfrac {\size {\map f x}^p} {p \norm f_p^p} + \dfrac {\size {\map ... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $p, q \in \R_{>0}$ such that $\dfrac 1 p + \dfrac 1 q = 1$.
{{improve|the assumption should read $p,q\in\R_{>0}\cup\set{+\infty}$.<br/>Suggestion: make a page for defining $p,q$ as satisfying this relation, including the pair $\tuple{1... | Let $x \in X$.
Let:
:$a_x := \dfrac {\size {\map f x} } {\norm f_p}$
and:
:$b_x := \dfrac {\size {\map g x} } {\norm g_q}$
Applying [[Young's Inequality for Products]] to $a_x$ and $b_x$:
:$\dfrac {\size {\map f x \map g x} } {\norm f_p \cdot \norm g_q} \le \dfrac {\size {\map f x}^p} {p \norm f_p^p} + \dfrac {\siz... | Hölder's Inequality for Integrals | https://proofwiki.org/wiki/Hölder's_Inequality_for_Integrals | https://proofwiki.org/wiki/Hölder's_Inequality_for_Integrals | [
"Hölder's Inequality for Integrals",
"Hölder's Inequality",
"Lebesgue Spaces",
"Inequalities"
] | [
"Definition:Measure Space",
"Definition:Lebesgue Space",
"Definition:Pointwise Multiplication",
"Definition:Integrable Function/Measure Space",
"Definition:Absolute Value",
"Definition:Pointwise Multiplication",
"Definition:P-Seminorm"
] | [
"Young's Inequality for Products",
"Integral of Positive Measurable Function is Monotone",
"Measurable Function Zero A.E. iff Absolute Value has Zero Integral",
"Definition:Almost Everywhere",
"Young's Inequality for Products"
] |
proofwiki-5648 | Hölder's Inequality for Integrals/Equality | :$\ds \int \size {f g} \rd \mu = \norm f_p \cdot \norm g_q$
holds {{iff}}, for $\mu$-almost all $x \in X$:
:$\dfrac {\size {\map f x}^p} {\norm f_p^p} = \dfrac {\size {\map g x}^q} {\norm g_q^q}$
{{questionable|What if e.g. $f{{=}}0$ and $g{{=}}1$? The equality holds, trivially, but how to read the necessary condition?... | {{ProofWanted}}
{{Namedfor|Otto Ludwig Hölder|cat = Hölder}} | :$\ds \int \size {f g} \rd \mu = \norm f_p \cdot \norm g_q$
holds {{iff}}, for $\mu$-[[Definition:Almost Everywhere|almost all]] $x \in X$:
:$\dfrac {\size {\map f x}^p} {\norm f_p^p} = \dfrac {\size {\map g x}^q} {\norm g_q^q}$
{{questionable|What if e.g. $f{{=}}0$ and $g{{=}}1$? The equality holds, trivially, but ho... | {{ProofWanted}}
{{Namedfor|Otto Ludwig Hölder|cat = Hölder}} | Hölder's Inequality for Integrals/Equality | https://proofwiki.org/wiki/Hölder's_Inequality_for_Integrals/Equality | https://proofwiki.org/wiki/Hölder's_Inequality_for_Integrals/Equality | [
"Hölder's Inequality for Integrals"
] | [
"Definition:Almost Everywhere"
] | [] |
proofwiki-5649 | Condition for Membership of Equivalence Class | Let $\RR$ be an equivalence relation on a set $S$.
Let $\eqclass x \RR$ denote the $\RR$-equivalence class of $x$.
Then:
:$\forall y \in S: y \in \eqclass x \RR \iff \tuple {x, y} \in \RR$ | From the definition of an equivalence class:
:$\eqclass x \RR = \set {y \in S: \tuple {x, y} \in \RR}$
Let $y \in S$ such that $y \in \eqclass x \RR$.
Then by definition $\tuple {x, y} \in \RR$.
Similarly, let $\tuple {x, y} \in \RR$.
Again by definition, $y \in \eqclass x \RR$.
{{qed}} | Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on a [[Definition:Set|set]] $S$.
Let $\eqclass x \RR$ denote the [[Definition:Equivalence Class|$\RR$-equivalence class of $x$]].
Then:
:$\forall y \in S: y \in \eqclass x \RR \iff \tuple {x, y} \in \RR$ | From the definition of an [[Definition:Equivalence Class|equivalence class]]:
:$\eqclass x \RR = \set {y \in S: \tuple {x, y} \in \RR}$
Let $y \in S$ such that $y \in \eqclass x \RR$.
Then by definition $\tuple {x, y} \in \RR$.
Similarly, let $\tuple {x, y} \in \RR$.
Again by definition, $y \in \eqclass x \RR$.
... | Condition for Membership of Equivalence Class | https://proofwiki.org/wiki/Condition_for_Membership_of_Equivalence_Class | https://proofwiki.org/wiki/Condition_for_Membership_of_Equivalence_Class | [
"Equivalence Classes"
] | [
"Definition:Equivalence Relation",
"Definition:Set",
"Definition:Equivalence Class"
] | [
"Definition:Equivalence Class"
] |
proofwiki-5650 | Linearly Ordered Space is Normal | Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Then $\struct {S, \tau}$ is normal. | From Linearly Ordered Space is Completely Normal, $\struct {S, \tau}$ is a completely normal space.
The result follows from Completely Normal Space is Normal.
{{qed}}
Category:Linearly Ordered Spaces
Category:Examples of Normal Spaces
j3o46fu7v2jy4y4a5ro1pxdq3n8ub2c | Let $T = \struct {S, \preceq, \tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
Then $\struct {S, \tau}$ is [[Definition:Normal Space|normal]]. | From [[Linearly Ordered Space is Completely Normal]], $\struct {S, \tau}$ is a [[Definition:Completely Normal Space|completely normal space]].
The result follows from [[Completely Normal Space is Normal]].
{{qed}}
[[Category:Linearly Ordered Spaces]]
[[Category:Examples of Normal Spaces]]
j3o46fu7v2jy4y4a5ro1pxdq3n8u... | Linearly Ordered Space is Normal | https://proofwiki.org/wiki/Linearly_Ordered_Space_is_Normal | https://proofwiki.org/wiki/Linearly_Ordered_Space_is_Normal | [
"Linearly Ordered Spaces",
"Examples of Normal Spaces"
] | [
"Definition:Linearly Ordered Space",
"Definition:Normal Space"
] | [
"Linearly Ordered Space is Completely Normal",
"Definition:Completely Normal Space",
"Completely Normal Space is Normal",
"Category:Linearly Ordered Spaces",
"Category:Examples of Normal Spaces"
] |
proofwiki-5651 | Mappings in Product of Sets are Surjections/Family of Sets | Let $\family {S_i}_{i \mathop \in I}$ be an indexed family of sets.
Let $\struct {P, \family {\phi_i} _{i \mathop \in I} }$ be a product of $S$ and $T$.
Then for all $i \in I$, $\phi_i$ is a surjection. | From the definition:
:For all sets $X$ and all indexed families $\family {f_i}_{i \mathop \in I}$ of mappings $f_i: X \to S_i$ there exists a unique mapping $h: X \to P$ such that:
::$\forall i \in I: \phi_i \circ h = f_i$
Let:
:$i \in I$
:$X = S_i$
:$f_i = I_{S_i}$
where $I_{S_i}$ is the identity mapping on $S_i$.
The... | Let $\family {S_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family of Sets|indexed family of sets]].
Let $\struct {P, \family {\phi_i} _{i \mathop \in I} }$ be a [[Definition:Set Product/Family of Sets|product]] of $S$ and $T$.
Then for all $i \in I$, $\phi_i$ is a [[Definition:Surjection|surjection]]. | From the definition:
:For all [[Definition:Set|sets]] $X$ and all [[Definition:Indexed Family|indexed families]] $\family {f_i}_{i \mathop \in I}$ of [[Definition:Mapping|mappings]] $f_i: X \to S_i$ there exists a [[Definition:Unique|unique]] [[Definition:Mapping|mapping]] $h: X \to P$ such that:
::$\forall i \in I: \p... | Mappings in Product of Sets are Surjections/Family of Sets | https://proofwiki.org/wiki/Mappings_in_Product_of_Sets_are_Surjections/Family_of_Sets | https://proofwiki.org/wiki/Mappings_in_Product_of_Sets_are_Surjections/Family_of_Sets | [
"Set Products",
"Surjections"
] | [
"Definition:Indexing Set/Family of Sets",
"Definition:Set Product/Family of Sets",
"Definition:Surjection"
] | [
"Definition:Set",
"Definition:Indexing Set/Family",
"Definition:Mapping",
"Definition:Unique",
"Definition:Mapping",
"Definition:Identity Mapping",
"Identity Mapping is Surjection",
"Definition:Surjection",
"Surjection if Composite is Surjection",
"Definition:Surjection"
] |
proofwiki-5652 | Cartesian Product is Set Product/Family of Sets | Let $\family {S_i}_{i \mathop \in I}$ be a family of sets.
For all $j \in I$, let $\pr_i: \ds \prod_{j \mathop \in I} \family {S_j} \to S_i$ be the $i$th projection from $\ds \prod_{j \mathop \in I} \family {S_j}$ to $S_i$.
Then $\struct {\ds \prod_{j \mathop \in I} \family {S_j}, \family {\pr_i}_{i \mathop \in I} }$ i... | Let $i \in I$.
Consider any set $X$ and any indexed family of mappings $\family {f_i: X \to S_i}_{i \mathop \in I}$.
Define $h: X \to \ds \prod_{j \mathop \in I} \family {S_j}$ by:
:$\forall x \in X: \map h x = \family {\map {f_i} x}_{i \mathop \in I}$
Then for all $x \in X$ and $i \in I$ we have:
:$\map {\paren {\pr_i... | Let $\family {S_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of sets]].
For all $j \in I$, let $\pr_i: \ds \prod_{j \mathop \in I} \family {S_j} \to S_i$ be the [[Definition:Projection on Family of Sets|$i$th projection]] from $\ds \prod_{j \mathop \in I} \family {S_j}$ to $S_i$.
Then $\stru... | Let $i \in I$.
Consider any set $X$ and any [[Definition:Indexed Family|indexed family]] of [[Definition:Mapping|mappings]] $\family {f_i: X \to S_i}_{i \mathop \in I}$.
Define $h: X \to \ds \prod_{j \mathop \in I} \family {S_j}$ by:
:$\forall x \in X: \map h x = \family {\map {f_i} x}_{i \mathop \in I}$
Then for al... | Cartesian Product is Set Product/Family of Sets | https://proofwiki.org/wiki/Cartesian_Product_is_Set_Product/Family_of_Sets | https://proofwiki.org/wiki/Cartesian_Product_is_Set_Product/Family_of_Sets | [
"Set Products",
"Cartesian Product"
] | [
"Definition:Indexing Set/Family of Sets",
"Definition:Projection (Mapping Theory)/Family of Sets",
"Definition:Set Product"
] | [
"Definition:Indexing Set/Family",
"Definition:Mapping",
"Definition:Mapping"
] |
proofwiki-5653 | Cauchy-Bunyakovsky-Schwarz Inequality/Lebesgue 2-Space | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \R$ be $\mu$-square integrable functions, that is $f, g \in \map {\LL^2} \mu$, Lebesgue $2$-space.
Then:
:$\ds \int \size {f g} \rd \mu \le \norm f_2^2 \cdot \norm g_2^2$
where $\norm {\, \cdot \,}_2$ is the $2$-norm. | Follows directly from Hölder's Inequality for Integrals with $p = q = 2$.
{{qed}} | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f, g: X \to \R$ be [[Definition:Square Integrable Function|$\mu$-square integrable functions]], that is $f, g \in \map {\LL^2} \mu$, [[Definition:Lebesgue Space|Lebesgue $2$-space]].
Then:
:$\ds \int \size {f g} \rd \mu \le \norm f... | Follows directly from [[Hölder's Inequality for Integrals]] with $p = q = 2$.
{{qed}} | Cauchy-Bunyakovsky-Schwarz Inequality/Lebesgue 2-Space | https://proofwiki.org/wiki/Cauchy-Bunyakovsky-Schwarz_Inequality/Lebesgue_2-Space | https://proofwiki.org/wiki/Cauchy-Bunyakovsky-Schwarz_Inequality/Lebesgue_2-Space | [
"Cauchy-Bunyakovsky-Schwarz Inequality",
"Lebesgue Spaces"
] | [
"Definition:Measure Space",
"Definition:Square Integrable Function",
"Definition:Lebesgue Space",
"Definition:P-Norm"
] | [
"Hölder's Inequality for Integrals"
] |
proofwiki-5654 | Preimage of Intersection under Relation/General Result | Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $\powerset T$ be the power set of $T$.
Let $\mathbb T \subseteq \powerset T$.
Then:
:$\ds \RR^{-1} \sqbrk {\bigcap \mathbb T} \subseteq \bigcap_{X \mathop \in \mathbb T} \RR^{-1} \sqbrk X$ | This follows from Image of Intersection under Relation: General Result, and the fact that $\RR^{-1}$ is itself a relation, and therefore obeys the same rules.
{{qed}}
Category:Relation Theory
Category:Set Intersection
4un5yucfff5xfy12bhj1clb6tbhxpt5 | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Let $\powerset T$ be the [[Definition:Power Set|power set]] of $T$.
Let $\mathbb T \subseteq \powerset T$.
Then:
:$\ds \RR^{-1} \sqbrk {\bigcap \mathbb T} \subseteq \bigcap_{X \mathop \in \mathbb T} \R... | This follows from [[Image of Intersection under Relation/General Result|Image of Intersection under Relation: General Result]], and the fact that $\RR^{-1}$ is itself a [[Definition:Relation|relation]], and therefore obeys the same rules.
{{qed}}
[[Category:Relation Theory]]
[[Category:Set Intersection]]
4un5yucfff5xf... | Preimage of Intersection under Relation/General Result | https://proofwiki.org/wiki/Preimage_of_Intersection_under_Relation/General_Result | https://proofwiki.org/wiki/Preimage_of_Intersection_under_Relation/General_Result | [
"Relation Theory",
"Set Intersection"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Power Set"
] | [
"Image of Intersection under Relation/General Result",
"Definition:Relation",
"Category:Relation Theory",
"Category:Set Intersection"
] |
proofwiki-5655 | Image of Union under Relation/Family of Sets | Let $S$ and $T$ be sets.
Let $\family {S_i}_{i \mathop \in I}$ be a family of subsets of $S$.
Let $\RR \subseteq S \times T$ be a relation.
Then:
:$\ds \RR \sqbrk {\bigcup_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} \RR \sqbrk {S_i}$
where $\ds \bigcup_{i \mathop \in I} S_i$ denotes the union of $\family {S_i}_{... | {{begin-eqn}}
{{eqn | l = t
| o = \in
| r = \RR \sqbrk {\bigcup_{i \mathop \in I} S_i}
}}
{{eqn | ll= \leadstoandfrom
| q = \exists s \in \bigcup_{i \mathop \in I} S_i
| l = t
| o = \in
| r = \map \RR s
| c = Image of Subset under Relation equals Union of Images of Elements
}}
... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\family {S_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Subsets|family of subsets]] of $S$.
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Then:
:$\ds \RR \sqbrk {\bigcup_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} \RR \sqbr... | {{begin-eqn}}
{{eqn | l = t
| o = \in
| r = \RR \sqbrk {\bigcup_{i \mathop \in I} S_i}
}}
{{eqn | ll= \leadstoandfrom
| q = \exists s \in \bigcup_{i \mathop \in I} S_i
| l = t
| o = \in
| r = \map \RR s
| c = [[Image of Subset under Relation equals Union of Images of Elements]]... | Image of Union under Relation/Family of Sets | https://proofwiki.org/wiki/Image_of_Union_under_Relation/Family_of_Sets | https://proofwiki.org/wiki/Image_of_Union_under_Relation/Family_of_Sets | [
"Image of Union under Relation",
"Indexed Families"
] | [
"Definition:Set",
"Definition:Indexing Set/Family of Subsets",
"Definition:Relation",
"Definition:Set Union/Family of Sets"
] | [
"Image of Subset under Relation equals Union of Images of Elements"
] |
proofwiki-5656 | Image of Intersection under Relation/Family of Sets | Let $S$ and $T$ be sets.
Let $\family {S_i}_{i \mathop \in I}$ be a family of subsets of $S$.
Let $\RR \subseteq S \times T$ be a relation.
Then:
:$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} \subseteq \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$
where $\ds \bigcap_{i \mathop \in I} S_i$ denotes the intersection of $... | {{begin-eqn}}
{{eqn | q = \forall i \in I
| l = \bigcap_{j \mathop \in I} S_j
| o = \subseteq
| r = S_i
| c = Intersection is Subset: Family of Sets
}}
{{eqn | ll= \leadsto
| q = \forall i \in I
| l = \RR \sqbrk {\bigcap_{j \mathop \in I} S_j}
| o = \subseteq
| r = \RR \s... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\family {S_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of subsets]] of $S$.
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Then:
:$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} \subseteq \bigcap_{i \mathop \in I} \RR ... | {{begin-eqn}}
{{eqn | q = \forall i \in I
| l = \bigcap_{j \mathop \in I} S_j
| o = \subseteq
| r = S_i
| c = [[Intersection is Subset/Family of Sets|Intersection is Subset: Family of Sets]]
}}
{{eqn | ll= \leadsto
| q = \forall i \in I
| l = \RR \sqbrk {\bigcap_{j \mathop \in I} S_j... | Image of Intersection under Relation/Family of Sets | https://proofwiki.org/wiki/Image_of_Intersection_under_Relation/Family_of_Sets | https://proofwiki.org/wiki/Image_of_Intersection_under_Relation/Family_of_Sets | [
"Image of Intersection under Relation",
"Indexed Families"
] | [
"Definition:Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Relation",
"Definition:Set Intersection/Family of Sets"
] | [
"Intersection is Subset/Family of Sets",
"Image of Subset under Relation is Subset of Image",
"Intersection is Largest Subset/Family of Sets"
] |
proofwiki-5657 | Preimage of Union under Relation/Family of Sets | Let $S$ and $T$ be sets.
Let $\family {T_i}_{i \mathop \in I}$ be a family of subsets of $T$.
Let $\RR \subseteq S \times T$ be a relation.
Then:
:$\ds \RR^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \RR^{-1} \sqbrk {T_i}$
where:
:$\ds \bigcup_{i \mathop \in I} T_i$ denotes the union of $\fa... | We have that $\RR^{-1}$ is itself a relation.
The result follows from Image of Union under Relation: Family of Sets.
{{qed}}
Category:Preimage of Union under Relation
Category:Indexed Families
g0l0paaenfqviok39mbmorxlo0x50lg | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\family {T_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of subsets]] of $T$.
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Then:
:$\ds \RR^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \RR^{-1... | We have that $\RR^{-1}$ is itself a [[Definition:Relation|relation]].
The result follows from [[Image of Union under Relation/Family of Sets|Image of Union under Relation: Family of Sets]].
{{qed}}
[[Category:Preimage of Union under Relation]]
[[Category:Indexed Families]]
g0l0paaenfqviok39mbmorxlo0x50lg | Preimage of Union under Relation/Family of Sets | https://proofwiki.org/wiki/Preimage_of_Union_under_Relation/Family_of_Sets | https://proofwiki.org/wiki/Preimage_of_Union_under_Relation/Family_of_Sets | [
"Preimage of Union under Relation",
"Indexed Families"
] | [
"Definition:Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Relation",
"Definition:Set Union/Family of Sets",
"Definition:Preimage/Relation/Subset"
] | [
"Definition:Relation",
"Image of Union under Relation/Family of Sets",
"Category:Preimage of Union under Relation",
"Category:Indexed Families"
] |
proofwiki-5658 | Preimage of Intersection under Relation/Family of Sets | Let $S$ and $T$ be sets.
Let $\family {T_i}_{i \mathop \in I}$ be a family of subsets of $T$.
Let $\RR \subseteq S \times T$ be a relation.
Then:
:$\ds \RR^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i} \subseteq \bigcap_{i \mathop \in I} \RR^{-1} \sqbrk {T_i}$
where $\ds \bigcap_{i \mathop \in I} T_i$ denotes the interse... | This follows from Image of Intersection under Relation: Family of Sets, and the fact that $\RR^{-1}$ is itself a relation, and therefore obeys the same rules.
{{qed}}
Category:Relation Theory
Category:Set Intersection
ay5y7m74ej87s8gd6b8vmfa30p39edj | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\family {T_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of subsets]] of $T$.
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Then:
:$\ds \RR^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i} \subseteq \bigcap_{i \mathop \in I}... | This follows from [[Image of Intersection under Relation/Family of Sets|Image of Intersection under Relation: Family of Sets]], and the fact that $\RR^{-1}$ is itself a [[Definition:Relation|relation]], and therefore obeys the same rules.
{{qed}}
[[Category:Relation Theory]]
[[Category:Set Intersection]]
ay5y7m74ej87s... | Preimage of Intersection under Relation/Family of Sets | https://proofwiki.org/wiki/Preimage_of_Intersection_under_Relation/Family_of_Sets | https://proofwiki.org/wiki/Preimage_of_Intersection_under_Relation/Family_of_Sets | [
"Relation Theory",
"Set Intersection"
] | [
"Definition:Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Relation",
"Definition:Set Intersection/Family of Sets"
] | [
"Image of Intersection under Relation/Family of Sets",
"Definition:Relation",
"Category:Relation Theory",
"Category:Set Intersection"
] |
proofwiki-5659 | Image of Intersection under Injection/Family of Sets | Let $S$ and $T$ be sets.
Let $\family {S_i}_{i \mathop \in I}$ be a family of subsets of $S$.
Let $f: S \to T$ be a mapping.
Then:
:$\ds f \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} f \sqbrk {S_i}$
{{iff}} $f$ is an injection.
This can be expressed in the language and notation of direct image ma... | An injection is a type of one-to-one relation, and therefore also a one-to-many relation.
Therefore Image of Intersection under One-to-Many Relation: Family of Sets applies:
:$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$
{{iff}} $\RR$ is a one-to-many relation.
We have th... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\family {S_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of subsets]] of $S$.
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$\ds f \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} f \sqbrk {S_i}$
{{iff}} $f$ is... | An [[Definition:Injection|injection]] is a type of [[Definition:One-to-One Relation|one-to-one relation]], and therefore also a [[Definition:One-to-Many Relation|one-to-many relation]].
Therefore [[Image of Intersection under One-to-Many Relation/Family of Sets|Image of Intersection under One-to-Many Relation: Family... | Image of Intersection under Injection/Family of Sets | https://proofwiki.org/wiki/Image_of_Intersection_under_Injection/Family_of_Sets | https://proofwiki.org/wiki/Image_of_Intersection_under_Injection/Family_of_Sets | [
"Image of Intersection under Injection"
] | [
"Definition:Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Mapping",
"Definition:Injection",
"Definition:Direct Image Mapping"
] | [
"Definition:Injection",
"Definition:One-to-One Relation",
"Definition:One-to-Many Relation",
"Image of Intersection under One-to-Many Relation/Family of Sets",
"Definition:One-to-Many Relation",
"Definition:Mapping",
"Definition:Many-to-One Relation",
"Definition:One-to-Many Relation",
"Definition:I... |
proofwiki-5660 | Image of Intersection under Mapping/Family of Sets | Let $S$ and $T$ be sets.
Let $\family {S_i}_{i \mathop \in I}$ be a family of subsets of $S$.
Let $f: S \to T$ be a mapping.
Then:
:$\ds f \sqbrk {\bigcap_{i \mathop \in I} S_i} \subseteq \bigcap_{i \mathop \in I} f \sqbrk {S_i}$
where $\ds \bigcap_{i \mathop \in I} S_i$ denotes the intersection of $\family {S_i}_{i \m... | As $f$, being a mapping, is also a relation, we can apply Image of Intersection under Relation: Family of Sets:
:$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} \subseteq \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$
{{qed}} | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\family {S_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of subsets]] of $S$.
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$\ds f \sqbrk {\bigcap_{i \mathop \in I} S_i} \subseteq \bigcap_{i \mathop \in I} f \sqbrk {S_i}$
where ... | As $f$, being a [[Definition:Mapping|mapping]], is also a [[Definition:Relation|relation]], we can apply [[Image of Intersection under Relation/Family of Sets|Image of Intersection under Relation: Family of Sets]]:
:$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} \subseteq \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$
{... | Image of Intersection under Mapping/Family of Sets | https://proofwiki.org/wiki/Image_of_Intersection_under_Mapping/Family_of_Sets | https://proofwiki.org/wiki/Image_of_Intersection_under_Mapping/Family_of_Sets | [
"Image of Intersection under Mapping"
] | [
"Definition:Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Mapping",
"Definition:Set Intersection/Family of Sets"
] | [
"Definition:Mapping",
"Definition:Relation",
"Image of Intersection under Relation/Family of Sets"
] |
proofwiki-5661 | Image of Union under Mapping/Family of Sets | Let $S$ and $T$ be sets.
Let $\family {S_i}_{i \mathop \in I}$ be a family of subsets of $S$.
Let $f: S \to T$ be a mapping.
Then:
:$\ds f \sqbrk {\bigcup_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} f \sqbrk {S_i}$
where $\ds \bigcup_{i \mathop \in I} S_i$ denotes the union of $\family {S_i}_{i \mathop \in I}$.
... | As $f$, being a mapping, is also a relation, we can apply Image of Union under Relation: Family of Sets:
:$\ds \RR \sqbrk {\bigcup_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} \RR \sqbrk {S_i}$
{{qed}} | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\family {S_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of subsets]] of $S$.
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$\ds f \sqbrk {\bigcup_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} f \sqbrk {S_i}$
where $\ds \bi... | As $f$, being a [[Definition:Mapping|mapping]], is also a [[Definition:Relation|relation]], we can apply [[Image of Union under Relation/Family of Sets|Image of Union under Relation: Family of Sets]]:
:$\ds \RR \sqbrk {\bigcup_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} \RR \sqbrk {S_i}$
{{qed}} | Image of Union under Mapping/Family of Sets | https://proofwiki.org/wiki/Image_of_Union_under_Mapping/Family_of_Sets | https://proofwiki.org/wiki/Image_of_Union_under_Mapping/Family_of_Sets | [
"Image of Union under Mapping"
] | [
"Definition:Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Mapping",
"Definition:Set Union/Family of Sets",
"Definition:Direct Image Mapping"
] | [
"Definition:Mapping",
"Definition:Relation",
"Image of Union under Relation/Family of Sets"
] |
proofwiki-5662 | Minkowski's Inequality/Lebesgue Spaces | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $p \in \closedint 1 \infty$.
Let $f, g: X \to \R$ be $p$-integrable, that is, elements of Lebesgue $p$-space $\map {\LL^p} \mu$.
Then their pointwise sum $f + g: X \to \R$ is also $p$-integrable, and:
:$\norm {f + g}_p \le \norm f_p + \norm g_p$
where $\norm {\, \... | We split into three cases. | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $p \in \closedint 1 \infty$.
Let $f, g: X \to \R$ be [[Definition:P-Integrable Function|$p$-integrable]], that is, elements of [[Definition:Lebesgue Space|Lebesgue $p$-space]] $\map {\LL^p} \mu$.
Then their [[Definition:Pointwise A... | We split into three cases. | Minkowski's Inequality/Lebesgue Spaces | https://proofwiki.org/wiki/Minkowski's_Inequality/Lebesgue_Spaces | https://proofwiki.org/wiki/Minkowski's_Inequality/Lebesgue_Spaces | [
"Minkowski's Inequality",
"Lebesgue Spaces"
] | [
"Definition:Measure Space",
"Definition:Integrable Function/p-Integrable",
"Definition:Lebesgue Space",
"Definition:Pointwise Addition",
"Definition:Integrable Function/p-Integrable",
"Definition:P-Seminorm"
] | [] |
proofwiki-5663 | Lebesgue Space is Vector Space | Let $\struct {X, \Sigma, \mu}$ be a measure space and let $p \in \closedint 1 \infty$.
Let $\map {\LL^p} {X, \Sigma, \mu}$ be Lebesgue $p$-space on $\struct {X, \Sigma, \mu}$.
Then $\map {\LL^p} {X, \Sigma, \mu}$ is a vector subspace of $\map \MM {X, \Sigma, \R}$, the space of real-valued $\Sigma$-measurable functions ... | From Space of Real-Valued Measurable Functions is Vector Space:
:$\map \MM {X, \Sigma, \R}$ forms a vector space with pointwise addition and pointwise scalar multiplication.
From construction, we have:
:$\map {\LL^p} {X, \Sigma, \mu} \subseteq \map \MM {X, \Sigma, \R}$
Since $0 \in \map {\LL^p} {X, \Sigma, \mu}$, we ... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]] and let $p \in \closedint 1 \infty$.
Let $\map {\LL^p} {X, \Sigma, \mu}$ be [[Definition:Lebesgue Space|Lebesgue $p$-space on $\struct {X, \Sigma, \mu}$]].
Then $\map {\LL^p} {X, \Sigma, \mu}$ is a [[Definition:Vector Subspace|vector subs... | From [[Space of Real-Valued Measurable Functions is Vector Space]]:
:$\map \MM {X, \Sigma, \R}$ forms a [[Definition:Vector Space|vector space]] with [[Definition:Pointwise Addition of Real-Valued Functions|pointwise addition]] and [[Definition:Pointwise Scalar Multiplication|pointwise scalar multiplication]].
From ... | Lebesgue Space is Vector Space | https://proofwiki.org/wiki/Lebesgue_Space_is_Vector_Space | https://proofwiki.org/wiki/Lebesgue_Space_is_Vector_Space | [
"Lebesgue Spaces"
] | [
"Definition:Measure Space",
"Definition:Lebesgue Space",
"Definition:Vector Subspace",
"Definition:Space of Measurable Functions/Real-Valued",
"Definition:Vector Space"
] | [
"Space of Real-Valued Measurable Functions is Vector Space",
"Definition:Vector Space",
"Definition:Pointwise Addition of Real-Valued Functions",
"Definition:Pointwise Scalar Multiplication of Mappings",
"One-Step Vector Subspace Test"
] |
proofwiki-5664 | Riesz-Fischer Theorem | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $p \in \R$, $p \ge 1$.
The Lebesgue $p$-space $\map {\LL^p} \mu$, endowed with the $p$-norm $\norm {\cdot}_p$, is a Banach space. | From Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach, to prove $\map {\LL^p} \mu$ is complete, it suffices to prove that every absolutely summable sequence in $\map {\LL^p} \mu$ is summable.
{{explain|Provide supporting information as to why this proves the result.}}
Let $\sequence... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $p \in \R$, $p \ge 1$.
The [[Definition:Lebesgue Space|Lebesgue $p$-space]] $\map {\LL^p} \mu$, endowed with the [[Definition:P-Norm|$p$-norm]] $\norm {\cdot}_p$, is a [[Definition:Banach Space|Banach space]]. | From [[Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach]], to prove $\map {\LL^p} \mu$ is complete, it suffices to prove that every [[Definition:Absolutely Convergent Series|absolutely summable sequence]] in $\map {\LL^p} \mu$ is [[Definition:Convergent Series|summable]].
{{explain... | Riesz-Fischer Theorem | https://proofwiki.org/wiki/Riesz-Fischer_Theorem | https://proofwiki.org/wiki/Riesz-Fischer_Theorem | [
"Riesz-Fischer Theorem",
"Lebesgue Spaces",
"Banach Spaces"
] | [
"Definition:Measure Space",
"Definition:Lebesgue Space",
"Definition:P-Norm",
"Definition:Banach Space"
] | [
"Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach",
"Definition:Absolutely Convergent Series",
"Definition:Convergent Series",
"Definition:Absolutely Convergent Series",
"Monotone Convergence Theorem (Measure Theory)",
"Minkowski's Inequality/Lebesgue Spaces",
"Defin... |
proofwiki-5665 | Pointwise Convergent Bounded Sequence in Lebesgue Space Converges in Norm | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $p \in \R_{\ge 1}$.
Let $\sequence {f_n}_{n \mathop \in \N}, f_n: X \to \R$ be a sequence in Lebesgue $p$-space $\map {\LL^p} \mu$.
Suppose that the pointwise limit $f := \ds \lim_{n \mathop \to \infty} f_n$ exists $\mu$-almost everywhere.
Suppose that for some $g ... | Since:
:$\ds \size f = \lim _{n \mathop \to \infty} \size {f_n} \le g$
$\mu$-almost everywhere, we have:
:$\ds \int \size f^p \rd \mu \le \int g^p \rd \mu < + \infty$
Thus:
:$f \in \map {\LL^p} \mu$
Furthermore, since:
:$\size {f_n - f} \le \size {f_n} + \size f \le 2 \size g$
we have:
:$\size {f_n - f}^p \le 2^p \size... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $p \in \R_{\ge 1}$.
Let $\sequence {f_n}_{n \mathop \in \N}, f_n: X \to \R$ be a [[Definition:Sequence|sequence]] in [[Definition:Lebesgue Space|Lebesgue $p$-space]] $\map {\LL^p} \mu$.
Suppose that the [[Definition:Pointwise Limit|... | Since:
:$\ds \size f = \lim _{n \mathop \to \infty} \size {f_n} \le g$
[[Definition:Almost Everywhere|$\mu$-almost everywhere]], we have:
:$\ds \int \size f^p \rd \mu \le \int g^p \rd \mu < + \infty$
Thus:
:$f \in \map {\LL^p} \mu$
Furthermore, since:
:$\size {f_n - f} \le \size {f_n} + \size f \le 2 \size g$
we hav... | Pointwise Convergent Bounded Sequence in Lebesgue Space Converges in Norm | https://proofwiki.org/wiki/Pointwise_Convergent_Bounded_Sequence_in_Lebesgue_Space_Converges_in_Norm | https://proofwiki.org/wiki/Pointwise_Convergent_Bounded_Sequence_in_Lebesgue_Space_Converges_in_Norm | [
"Lebesgue Spaces"
] | [
"Definition:Measure Space",
"Definition:Sequence",
"Definition:Lebesgue Space",
"Definition:Pointwise Limit",
"Definition:Almost Everywhere",
"Definition:Pointwise Inequality",
"Definition:P-Seminorm"
] | [
"Definition:Almost Everywhere",
"Definition:Integrable Function/Measure Space",
"Lebesgue's Dominated Convergence Theorem",
"Definition:Almost Everywhere"
] |
proofwiki-5666 | Riesz's Convergence Theorem | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $p \in \R$, $p \ge 1$.
Let $\sequence {f_n}_{n \mathop \in \N}, f_n: X \to \R$ be a sequence in Lebesgue $p$-space $\map {\LL^p} \mu$.
Suppose that the pointwise limit $f := \ds \lim_{n \mathop \to \infty} f_n$ exists $\mu$-almost everywhere, and that $f \in \map {... | === From $(1)$ to $(2)$ ===
This follows from the reverse triangle inequality:
:$\size {\norm f_p - \norm {f_n}_p} \le \norm {f - f_n}_p$
{{qed|lemma}} | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $p \in \R$, $p \ge 1$.
Let $\sequence {f_n}_{n \mathop \in \N}, f_n: X \to \R$ be a [[Definition:Sequence|sequence]] in [[Definition:Lebesgue Space|Lebesgue $p$-space]] $\map {\LL^p} \mu$.
Suppose that the [[Definition:Pointwise Lim... | === From $(1)$ to $(2)$ ===
This follows from the [[Reverse Triangle Inequality/Seminormed Vector Space|reverse triangle inequality]]:
:$\size {\norm f_p - \norm {f_n}_p} \le \norm {f - f_n}_p$
{{qed|lemma}} | Riesz's Convergence Theorem | https://proofwiki.org/wiki/Riesz's_Convergence_Theorem | https://proofwiki.org/wiki/Riesz's_Convergence_Theorem | [
"Lebesgue Spaces"
] | [
"Definition:Measure Space",
"Definition:Sequence",
"Definition:Lebesgue Space",
"Definition:Pointwise Limit",
"Definition:Almost Everywhere",
"Definition:P-Seminorm"
] | [
"Reverse Triangle Inequality/Seminormed Vector Space"
] |
proofwiki-5667 | Preimage of Union under Mapping/Family of Sets | Let $S$ and $T$ be sets.
Let $\family {T_i}_{i \mathop \in I}$ be a family of subsets of $T$.
Let $f: S \to T$ be a mapping.
Then:
:$\ds f^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} f^{-1} \sqbrk {T_i}$
where:
:$\ds \bigcup_{i \mathop \in I} T_i$ denotes the union of $\family {T_i}_{i \math... | As $f$, being a mapping, is also a relation, we can apply Preimage of Union under Relation: Family of Sets:
:$\ds \RR^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \RR^{-1} \sqbrk {T_i}$
where $\RR^{-1} \sqbrk {T_i}$ denotes the preimage of $T_i$ under $\RR^{-1}$.
{{qed}} | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\family {T_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of subsets]] of $T$.
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$\ds f^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} f^{-1} \sqbrk {T_i}$
whe... | As $f$, being a [[Definition:Mapping|mapping]], is also a [[Definition:Relation|relation]], we can apply [[Preimage of Union under Relation/Family of Sets|Preimage of Union under Relation: Family of Sets]]:
:$\ds \RR^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \RR^{-1} \sqbrk {T_i}$
where $... | Preimage of Union under Mapping/Family of Sets/Proof 1 | https://proofwiki.org/wiki/Preimage_of_Union_under_Mapping/Family_of_Sets | https://proofwiki.org/wiki/Preimage_of_Union_under_Mapping/Family_of_Sets/Proof_1 | [
"Preimage of Union under Mapping",
"Indexed Families"
] | [
"Definition:Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Mapping",
"Definition:Set Union/Family of Sets",
"Definition:Preimage/Mapping/Subset"
] | [
"Definition:Mapping",
"Definition:Relation",
"Preimage of Union under Relation/Family of Sets",
"Definition:Preimage/Relation/Subset"
] |
proofwiki-5668 | Preimage of Union under Mapping/Family of Sets | Let $S$ and $T$ be sets.
Let $\family {T_i}_{i \mathop \in I}$ be a family of subsets of $T$.
Let $f: S \to T$ be a mapping.
Then:
:$\ds f^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} f^{-1} \sqbrk {T_i}$
where:
:$\ds \bigcup_{i \mathop \in I} T_i$ denotes the union of $\family {T_i}_{i \math... | We have that $f$ is a mapping, and so also a relation.
Thus its inverse $f^{-1}$ is also a relation.
Hence we can apply Image of Union under Relation: Family of Sets:
:$\ds \RR \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \RR \sqbrk {T_i}$
where $\RR \sqbrk {T_i}$ denotes the image of $T_i$ under ... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\family {T_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of subsets]] of $T$.
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$\ds f^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} f^{-1} \sqbrk {T_i}$
whe... | We have that $f$ is a [[Definition:Mapping|mapping]], and so also a [[Definition:Relation|relation]].
Thus its [[Definition:Inverse of Mapping|inverse]] $f^{-1}$ is also a [[Definition:Relation|relation]].
Hence we can apply [[Image of Union under Relation/Family of Sets|Image of Union under Relation: Family of Sets]... | Preimage of Union under Mapping/Family of Sets/Proof 2 | https://proofwiki.org/wiki/Preimage_of_Union_under_Mapping/Family_of_Sets | https://proofwiki.org/wiki/Preimage_of_Union_under_Mapping/Family_of_Sets/Proof_2 | [
"Preimage of Union under Mapping",
"Indexed Families"
] | [
"Definition:Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Mapping",
"Definition:Set Union/Family of Sets",
"Definition:Preimage/Mapping/Subset"
] | [
"Definition:Mapping",
"Definition:Relation",
"Definition:Inverse of Mapping",
"Definition:Relation",
"Image of Union under Relation/Family of Sets",
"Definition:Image (Set Theory)/Relation/Subset"
] |
proofwiki-5669 | Image of Intersection under One-to-Many Relation/Family of Sets | Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Then $\RR$ is a one-to-many relation {{iff}}:
:$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$
where $\family {S_i}_{i \mathop \in I}$ is ''any'' family of subsets of $S$. | === Sufficient Condition ===
Suppose:
:$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$
where $\family {S_i}_{i \mathop \in I}$ is ''any'' family of subsets of $S$.
Then by definition of $\family {S_i}_{i \mathop \in I}$:
:$\forall i, j \in I: \RR \sqbrk {S_i \cap S_j} = \R... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Then $\RR$ is a [[Definition:One-to-Many Relation|one-to-many relation]] {{iff}}:
:$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$
where $\family {S_i}_{i \... | === Sufficient Condition ===
Suppose:
:$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$
where $\family {S_i}_{i \mathop \in I}$ is ''any'' [[Definition:Indexed Family of Sets|family of subsets]] of $S$.
Then by definition of $\family {S_i}_{i \mathop \in I}$:
:$\forall i,... | Image of Intersection under One-to-Many Relation/Family of Sets | https://proofwiki.org/wiki/Image_of_Intersection_under_One-to-Many_Relation/Family_of_Sets | https://proofwiki.org/wiki/Image_of_Intersection_under_One-to-Many_Relation/Family_of_Sets | [
"Image of Intersection under One-to-Many Relation",
"Indexed Families"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:One-to-Many Relation",
"Definition:Indexing Set/Family of Sets"
] | [
"Definition:Indexing Set/Family of Sets",
"Image of Intersection under One-to-Many Relation",
"Definition:One-to-Many Relation",
"Definition:One-to-Many Relation",
"Definition:One-to-Many Relation",
"Definition:One-to-Many Relation",
"Definition:One-to-Many Relation",
"Definition:Indexing Set/Family o... |
proofwiki-5670 | Preimage of Intersection under Mapping/Family of Sets | Let $S$ and $T$ be sets.
Let $\family {T_i}_{i \mathop \in I}$ be a family of subsets of $T$.
Let $f: S \to T$ be a mapping.
Then:
:$\ds f^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i} = \bigcap_{i \mathop \in I} f^{-1} \sqbrk {T_i}$
where:
:$\ds \bigcap_{i \mathop \in I} T_i$ denotes the intersection of $\family {T_i}_{... | As $f$ is a mapping, it is by definition also a many-to-one relation.
It follows from Inverse of Many-to-One Relation is One-to-Many that its inverse $f^{-1}$ is a one-to-many relation.
Thus Image of Intersection under One-to-Many Relation: Family of Sets can be applied for $\RR = f^{-1}$:
:$\ds \RR \sqbrk {\bigcap_{i ... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\family {T_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of subsets]] of $T$.
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$\ds f^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i} = \bigcap_{i \mathop \in I} f^{-1} \sqbrk {T_i}$
wher... | As $f$ is a [[Definition:Mapping|mapping]], it is by definition also a [[Definition:Many-to-One Relation|many-to-one relation]].
It follows from [[Inverse of Many-to-One Relation is One-to-Many]] that its [[Definition:Inverse of Mapping|inverse]] $f^{-1}$ is a [[Definition:One-to-Many Relation|one-to-many relation]].
... | Preimage of Intersection under Mapping/Family of Sets/Proof 1 | https://proofwiki.org/wiki/Preimage_of_Intersection_under_Mapping/Family_of_Sets | https://proofwiki.org/wiki/Preimage_of_Intersection_under_Mapping/Family_of_Sets/Proof_1 | [
"Indexed Families",
"Preimage of Intersection under Mapping"
] | [
"Definition:Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Mapping",
"Definition:Set Intersection/Family of Sets",
"Definition:Preimage/Mapping/Subset"
] | [
"Definition:Mapping",
"Definition:Many-to-One Relation",
"Inverse of Many-to-One Relation is One-to-Many",
"Definition:Inverse of Mapping",
"Definition:One-to-Many Relation",
"Image of Intersection under One-to-Many Relation/Family of Sets",
"Definition:Image (Set Theory)/Relation/Subset"
] |
proofwiki-5671 | Preimage of Intersection under Mapping/Family of Sets | Let $S$ and $T$ be sets.
Let $\family {T_i}_{i \mathop \in I}$ be a family of subsets of $T$.
Let $f: S \to T$ be a mapping.
Then:
:$\ds f^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i} = \bigcap_{i \mathop \in I} f^{-1} \sqbrk {T_i}$
where:
:$\ds \bigcap_{i \mathop \in I} T_i$ denotes the intersection of $\family {T_i}_{... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = f^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \map f x
| o = \in
| r = \bigcap_{i \mathop \in I} T_i
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \forall i \in I
| l = \map f x
... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\family {T_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of subsets]] of $T$.
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$\ds f^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i} = \bigcap_{i \mathop \in I} f^{-1} \sqbrk {T_i}$
wher... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = f^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \map f x
| o = \in
| r = \bigcap_{i \mathop \in I} T_i
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \forall i \in I
| l = \map f x
... | Preimage of Intersection under Mapping/Family of Sets/Proof 2 | https://proofwiki.org/wiki/Preimage_of_Intersection_under_Mapping/Family_of_Sets | https://proofwiki.org/wiki/Preimage_of_Intersection_under_Mapping/Family_of_Sets/Proof_2 | [
"Indexed Families",
"Preimage of Intersection under Mapping"
] | [
"Definition:Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Mapping",
"Definition:Set Intersection/Family of Sets",
"Definition:Preimage/Mapping/Subset"
] | [] |
proofwiki-5672 | Space of Simple P-Integrable Functions is Everywhere Dense in Lebesgue Space | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $p \in \R$, $p \ge 1$.
Let $\map {\LL^p} \mu$ be Lebesgue $p$-space for $\mu$.
Let $\map \EE \Sigma \cap \map {\LL^p} \mu$ be the space of $\Sigma$-simple, $p$-integrable functions.
Then $\map \EE \Sigma \cap \map {\LL^p} \mu$ is everywhere dense in $\map {\LL^p} \... | For $n \in \N$, we define $D_n : \R_{\ge 0} \to \R_{\ge 0}$ by:
$\quad \map {\Delta _n} y := \begin{cases}
\dfrac k {2^n} & : y \in \hointr {\dfrac k {2^n} } {\dfrac {k + 1} {2^n} }, k = 0, 1, \ldots, 2^{2 n} - 1 \\
0 & : y \ge 2^n
\end{cases}$
Clearly, for all $n \in \N$:
:$(1): \quad \forall y \in \R_{\ge 0} : 0 \le ... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $p \in \R$, $p \ge 1$.
Let $\map {\LL^p} \mu$ be [[Definition:Lebesgue Space|Lebesgue $p$-space for $\mu$]].
Let $\map \EE \Sigma \cap \map {\LL^p} \mu$ be the space of [[Definition:Simple Function|$\Sigma$-simple]], [[Definition:P-I... | For $n \in \N$, we define $D_n : \R_{\ge 0} \to \R_{\ge 0}$ by:
$\quad \map {\Delta _n} y := \begin{cases}
\dfrac k {2^n} & : y \in \hointr {\dfrac k {2^n} } {\dfrac {k + 1} {2^n} }, k = 0, 1, \ldots, 2^{2 n} - 1 \\
0 & : y \ge 2^n
\end{cases}$
Clearly, for all $n \in \N$:
:$(1): \quad \forall y \in \R_{\ge 0} : 0 \l... | Space of Simple P-Integrable Functions is Everywhere Dense in Lebesgue Space | https://proofwiki.org/wiki/Space_of_Simple_P-Integrable_Functions_is_Everywhere_Dense_in_Lebesgue_Space | https://proofwiki.org/wiki/Space_of_Simple_P-Integrable_Functions_is_Everywhere_Dense_in_Lebesgue_Space | [
"Lebesgue Spaces"
] | [
"Definition:Measure Space",
"Definition:Lebesgue Space",
"Definition:Simple Function",
"Definition:Integrable Function/p-Integrable",
"Definition:Everywhere Dense",
"Definition:Seminorm Topology",
"Definition:Singleton",
"Definition:Seminorm Topology",
"Definition:Locally Convex Topological Vector S... | [
"Definition:Signum Function",
"Definition:Absolute Value",
"Measurable Function is Simple Function iff Finite Image Set",
"Lebesgue's Dominated Convergence Theorem"
] |
proofwiki-5673 | Jensen's Inequality (Measure Theory)/Convex Functions | Let $V: \hointr 0 \infty \to \hointr 0 \infty$ be a convex function.
Then for all positive measurable functions $g: X \to \R$, $g \in \map {\MM^+} \Sigma$:
:$\map V {\dfrac {\ds \int g \cdot f \rd \mu} {\ds \int f \rd \mu} } \le \dfrac {\ds \int \paren {V \circ g} \cdot f \rd \mu} {\ds\int f \rd \mu}$
where $\circ$ den... | {{MissingLinks}}
Let $\d \map \nu x := \dfrac {\map f x} {\ds \int \map f s \rd \map \mu s} \rd \map \mu x$ be a probability measure.
{{explain|This proof invokes a probability measure. Needs to be for a measure space. Does the proof work for both?}}
Let $\ds x_0 := \int \map g s \rd \map \nu s$.
Then by convexity ther... | Let $V: \hointr 0 \infty \to \hointr 0 \infty$ be a [[Definition:Convex Real Function|convex function]].
Then for all [[Definition:Positive Measurable Function|positive measurable functions]] $g: X \to \R$, $g \in \map {\MM^+} \Sigma$:
:$\map V {\dfrac {\ds \int g \cdot f \rd \mu} {\ds \int f \rd \mu} } \le \dfrac {... | {{MissingLinks}}
Let $\d \map \nu x := \dfrac {\map f x} {\ds \int \map f s \rd \map \mu s} \rd \map \mu x$ be a probability measure.
{{explain|This proof invokes a probability measure. Needs to be for a measure space. Does the proof work for both?}}
Let $\ds x_0 := \int \map g s \rd \map \nu s$.
Then by convexity ... | Jensen's Inequality (Measure Theory)/Convex Functions | https://proofwiki.org/wiki/Jensen's_Inequality_(Measure_Theory)/Convex_Functions | https://proofwiki.org/wiki/Jensen's_Inequality_(Measure_Theory)/Convex_Functions | [
"Jensen's Inequality",
"Measure Theory"
] | [
"Definition:Convex Real Function",
"Definition:Measurable Function/Positive",
"Definition:Composition of Mappings",
"Definition:Pointwise Multiplication"
] | [
"Definition:Primitive (Calculus)/Integration",
"Category:Jensen's Inequality",
"Category:Measure Theory"
] |
proofwiki-5674 | Jensen's Inequality (Measure Theory)/Concave Functions | Let $\Lambda: \hointr 0 \infty \to \hointr 0 \infty$ be a concave function.
Then for all positive measurable functions $g: X \to \R$, $g \in \map {\MM^+} \Sigma$:
:$\dfrac {\int \paren {\Lambda \circ g} \cdot f \rd \mu} {\int f \rd \mu} \le \map \Lambda {\dfrac {\int g \cdot f \rd \mu} {\int f \rd \mu} }$
where $\circ$... | {{proof wanted}}
Category:Jensen's Inequality
Category:Measure Theory
axyoywtzza4yc9d9vlg2sxwq327so6n | Let $\Lambda: \hointr 0 \infty \to \hointr 0 \infty$ be a [[Definition:Concave Real Function|concave function]].
Then for all [[Definition:Positive Measurable Function|positive measurable functions]] $g: X \to \R$, $g \in \map {\MM^+} \Sigma$:
:$\dfrac {\int \paren {\Lambda \circ g} \cdot f \rd \mu} {\int f \rd \mu}... | {{proof wanted}}
[[Category:Jensen's Inequality]]
[[Category:Measure Theory]]
axyoywtzza4yc9d9vlg2sxwq327so6n | Jensen's Inequality (Measure Theory)/Concave Functions | https://proofwiki.org/wiki/Jensen's_Inequality_(Measure_Theory)/Concave_Functions | https://proofwiki.org/wiki/Jensen's_Inequality_(Measure_Theory)/Concave_Functions | [
"Jensen's Inequality",
"Measure Theory"
] | [
"Definition:Concave Real Function",
"Definition:Measurable Function/Positive",
"Definition:Composition of Mappings",
"Definition:Pointwise Multiplication"
] | [
"Category:Jensen's Inequality",
"Category:Measure Theory"
] |
proofwiki-5675 | Correspondence Theorem (Set Theory) | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be an equivalence relation on $S$.
Let $\mathscr A$ be the set of partitions of $S$ associated with equivalence relations $\RR'$ on $S$ such that:
:$\tuple {x, y} \in \RR \iff \tuple {x, y} \in \RR'$
Then there exists a bijection $\phi$ from $\mathscr A$ onto the set of ... | Denote the equivalence class of an element $x$ of $S$ by $\eqclass x {\RR}$ with respect to the relation $\RR$.
Consider the relation on $S$:
:$\phi = \set {\tuple {\eqclass x {\RR'}, \eqclass x \RR}: x \in S}$
We prove that $\phi$ is a bijection.
Let $\eqclass x \RR = \eqclass y \RR$.
Then:
:$\tuple {x, y} \in \RR$
... | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be an [[Definition:Equivalence Relation|equivalence relation]] on $S$.
Let $\mathscr A$ be the [[Definition:Set|set]] of [[Definition:Set Partition|partitions]] of $S$ associated with [[Definition:Equivalence Relation|equivalence relations]] $\RR'$ o... | Denote the [[Definition:Equivalence Class|equivalence class]] of an [[Definition:Element|element]] $x$ of $S$ by $\eqclass x {\RR}$ with respect to the [[Definition:Equivalence Relation|relation]] $\RR$.
Consider the [[Definition:Endorelation|relation]] on $S$:
:$\phi = \set {\tuple {\eqclass x {\RR'}, \eqclass x \RR... | Correspondence Theorem (Set Theory) | https://proofwiki.org/wiki/Correspondence_Theorem_(Set_Theory) | https://proofwiki.org/wiki/Correspondence_Theorem_(Set_Theory) | [
"Equivalence Relations",
"Set Theory",
"Quotient Sets",
"Named Theorems"
] | [
"Definition:Set",
"Definition:Equivalence Relation",
"Definition:Set",
"Definition:Set Partition",
"Definition:Equivalence Relation",
"Definition:Bijection",
"Definition:Set",
"Definition:Set Partition",
"Definition:Quotient Set"
] | [
"Definition:Equivalence Class",
"Definition:Element",
"Definition:Equivalence Relation",
"Definition:Endorelation",
"Definition:Bijection",
"Definition:One-to-Many Relation",
"Definition:Many-to-One Relation",
"Definition:One-to-Many Relation",
"Definition:Many-to-One Relation",
"Definition:One-to... |
proofwiki-5676 | Power Set is Lattice | Let $S$ be a set.
Let $\struct {\powerset S, \subseteq}$ be the relational structure defined on $\powerset S$ by the subset relation $\subseteq$.
Then $\struct {\powerset S, \subseteq}$ is a lattice. | From Subset Relation on Power Set is Partial Ordering, we have that $\subseteq$ is a partial ordering.
Let $X, Y \in \powerset S$.
Then from Union is Smallest Superset:
:$X \subseteq T, Y \subseteq T \iff X \cup Y \subseteq T$
and from Intersection is Largest Subset:
:$X \subseteq T, Y \subseteq T \iff T \subseteq X \c... | Let $S$ be a [[Definition:Set|set]].
Let $\struct {\powerset S, \subseteq}$ be the [[Definition:Relational Structure|relational structure]] defined on $\powerset S$ by the [[Definition:Subset|subset]] [[Definition:Relation|relation]] $\subseteq$.
Then $\struct {\powerset S, \subseteq}$ is a [[Definition:Lattice (Orde... | From [[Subset Relation on Power Set is Partial Ordering]], we have that $\subseteq$ is a [[Definition:Partial Ordering|partial ordering]].
Let $X, Y \in \powerset S$.
Then from [[Union is Smallest Superset]]:
:$X \subseteq T, Y \subseteq T \iff X \cup Y \subseteq T$
and from [[Intersection is Largest Subset]]:
:$X \... | Power Set is Lattice | https://proofwiki.org/wiki/Power_Set_is_Lattice | https://proofwiki.org/wiki/Power_Set_is_Lattice | [
"Lattice Theory",
"Power Set"
] | [
"Definition:Set",
"Definition:Relational Structure",
"Definition:Subset",
"Definition:Relation",
"Definition:Lattice (Order Theory)"
] | [
"Subset Relation on Power Set is Partial Ordering",
"Definition:Partial Ordering",
"Union is Smallest Superset",
"Intersection is Largest Subset",
"Definition:Infimum of Set",
"Definition:Supremum of Set",
"Definition:Lattice (Ordered Set)"
] |
proofwiki-5677 | Divisor Relation induces Lattice | Let $\struct {\Z_{> 0}, \divides}$ be the ordered set comprising:
:The set of positive integers $\Z_{> 0}$
:The divisor relation $\divides$ defined as:
::$a \divides b := \exists k \in \Z_{> 0}: b = ka$
Then $\struct {\Z_{> 0}, \divides}$ is a lattice. | It follows from Divisor Relation on Positive Integers is Partial Ordering that $\struct {\Z_{> 0}, \divides}$ is indeed an ordered set.
Let $a, b \in \Z_{>0}$.
Let $d = \gcd \set {a, b}$ be the greatest common divisor of $a$ and $b$.
By definition, $d$ is the infimum of $\set {a, b}$.
Similarly, let $m = \lcm \set {a, ... | Let $\struct {\Z_{> 0}, \divides}$ be the [[Definition:Ordered Set|ordered set]] comprising:
:The [[Definition:Positive Integer|set of positive integers $\Z_{> 0}$]]
:The [[Definition:Divisor of Integer|divisor relation]] $\divides$ defined as:
::$a \divides b := \exists k \in \Z_{> 0}: b = ka$
Then $\struct {\Z_{> 0... | It follows from [[Divisor Relation on Positive Integers is Partial Ordering]] that $\struct {\Z_{> 0}, \divides}$ is indeed an [[Definition:Ordered Set|ordered set]].
Let $a, b \in \Z_{>0}$.
Let $d = \gcd \set {a, b}$ be the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $a$ and $b$.
... | Divisor Relation induces Lattice | https://proofwiki.org/wiki/Divisor_Relation_induces_Lattice | https://proofwiki.org/wiki/Divisor_Relation_induces_Lattice | [
"Lattice Theory",
"Number Theory"
] | [
"Definition:Ordered Set",
"Definition:Positive/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Lattice (Ordered Set)"
] | [
"Divisor Relation on Positive Integers is Partial Ordering",
"Definition:Ordered Set",
"Definition:Greatest Common Divisor/Integers",
"Definition:Infimum of Set",
"Definition:Lowest Common Multiple/Integers",
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Supremum of Set",
"D... |
proofwiki-5678 | Generator for Product Sigma-Algebra | Let $\struct {X, \Sigma_1}$ and $\struct {Y, \Sigma_2}$ be measurable spaces.
Let $\GG_1$ and $\GG_2$ be generators for $\Sigma_1$ and $\Sigma_2$, respectively.
Then $\GG_1 \times \GG_2$ is a generator for the product $\sigma$-algebra $\Sigma_1 \otimes \Sigma_2$. | {{MissingLinks|throughout}}
{{handwaving|Much use of "clearly" in the below.}}
We begin by stating that this result is in fact incorrect.
For a simple counter-example, consider $X = Y = \set {1, 2}$ both equipped with the power set sigma algebra, that is, $\Sigma_1 = \Sigma_2 = \set {\O, \set 1, \set 2, \set {1, 2} }$.... | Let $\struct {X, \Sigma_1}$ and $\struct {Y, \Sigma_2}$ be [[Definition:Measurable Space|measurable spaces]].
Let $\GG_1$ and $\GG_2$ be [[Definition:Generator for Sigma-Algebra|generators]] for $\Sigma_1$ and $\Sigma_2$, respectively.
Then $\GG_1 \times \GG_2$ is a [[Definition:Generator for Sigma-Algebra|generato... | {{MissingLinks|throughout}}
{{handwaving|Much use of "clearly" in the below.}}
We begin by stating that this result is in fact incorrect.
For a simple [[Definition:Counterexample|counter-example]], consider $X = Y = \set {1, 2}$ both equipped with the power set sigma algebra, that is, $\Sigma_1 = \Sigma_2 = \set {\O... | Generator for Product Sigma-Algebra | https://proofwiki.org/wiki/Generator_for_Product_Sigma-Algebra | https://proofwiki.org/wiki/Generator_for_Product_Sigma-Algebra | [
"Sigma-Algebras",
"Product Sigma-Algebras",
"Product Sigma-Algebras"
] | [
"Definition:Measurable Space",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Product Sigma-Algebra"
] | [
"Definition:Counterexample"
] |
proofwiki-5679 | Existence of Product Measures | Let $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$ be $\sigma$-finite measure spaces.
Then there exists a measure $\rho$ on the product space $\paren {X \times Y, \Sigma_1 \otimes \Sigma_2}$ such that:
:$\forall E_1 \in \Sigma_X, E_2 \in \Sigma_Y: \map \rho {E_1 \times E_2} = \map \mu {E_1} \map \nu {E_2... | For each $E \in \Sigma_1 \otimes \Sigma_2$ define the function $f_E : X \to \overline \R$ by:
:$\map {f_E} x = \map \nu {E_x}$
for each $x \in X$.
Define the function $g_E : Y \to \overline \R$ by:
:$\map {g_E} y = \map \mu {E^y}$
for each $y \in Y$.
From Measure of Vertical Section of Measurable Set gives Measurab... | Let $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$ be [[Definition:Sigma-Finite Measure Space|$\sigma$-finite measure spaces]].
Then there exists a [[Definition:Measure (Measure Theory)|measure]] $\rho$ on the [[Definition:Product of Measurable Spaces|product space]] $\paren {X \times Y, \Sigma_1 \otim... | For each $E \in \Sigma_1 \otimes \Sigma_2$ define the [[Definition:Extended Real-Valued Function|function]] $f_E : X \to \overline \R$ by:
:$\map {f_E} x = \map \nu {E_x}$
for each $x \in X$.
Define the [[Definition:Extended Real-Valued Function|function]] $g_E : Y \to \overline \R$ by:
:$\map {g_E} y = \map \mu... | Existence of Product Measures | https://proofwiki.org/wiki/Existence_of_Product_Measures | https://proofwiki.org/wiki/Existence_of_Product_Measures | [
"Measure Theory",
"Product Measure"
] | [
"Definition:Sigma-Finite Measure Space",
"Definition:Measure (Measure Theory)",
"Definition:Product of Measurable Spaces",
"Definition:Measure (Measure Theory)",
"Definition:Sigma-Finite Measure",
"Definition:Measure (Measure Theory)"
] | [
"Definition:Extended Real-Valued Function",
"Definition:Extended Real-Valued Function",
"Measure of Vertical Section of Measurable Set gives Measurable Function",
"Definition:Measurable Function",
"Measure of Horizontal Section of Measurable Set gives Measurable Function",
"Definition:Measurable Function"... |
proofwiki-5680 | Tonelli's Theorem | Let $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$ be $\sigma$-finite measure spaces.
Let $\struct {X \times Y, \Sigma_X \otimes \Sigma_Y, \mu \times \nu}$ be the product measure space of $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$.
Let $f: X \times Y \to \overline \R_{\ge 0}$ be a posi... | We rewrite the demand as:
:$\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_Y \paren {\int_X f^y \rd \mu} \rd \nu = \int_X \paren {\int_Y f_x \rd \nu} \rd \mu$
where $f_x$ is the $x$-vertical section of $f$, and $f^y$ is the $y$-horizontal section of $f$.
From Horizontal Section of Measurable Function is Meas... | Let $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$ be [[Definition:Sigma-Finite Measure Space|$\sigma$-finite measure spaces]].
Let $\struct {X \times Y, \Sigma_X \otimes \Sigma_Y, \mu \times \nu}$ be the [[Definition:Product Measure Space|product measure space]] of $\struct {X, \Sigma_X, \mu}$ and $\st... | We rewrite the demand as:
:$\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_Y \paren {\int_X f^y \rd \mu} \rd \nu = \int_X \paren {\int_Y f_x \rd \nu} \rd \mu$
where $f_x$ is the [[Definition:Vertical Section of Function|$x$-vertical section]] of $f$, and $f^y$ is the [[Definition:Horizontal Section of Func... | Tonelli's Theorem | https://proofwiki.org/wiki/Tonelli's_Theorem | https://proofwiki.org/wiki/Tonelli's_Theorem | [
"Tonelli's Theorem",
"Integral of Positive Measurable Function",
"Product Measure"
] | [
"Definition:Sigma-Finite Measure Space",
"Definition:Product Measure Space",
"Definition:Measurable Function/Positive"
] | [
"Definition:Vertical Section of Function",
"Definition:Horizontal Section of Function",
"Horizontal Section of Measurable Function is Measurable",
"Definition:Measurable Function",
"Vertical Section of Measurable Function is Measurable",
"Definition:Measurable Function",
"Integral of Horizontal Section ... |
proofwiki-5681 | Condition for Power Set to be Totally Ordered | Let $\powerset S$ be the power set of a set $S$.
Let $\struct {\powerset S, \subseteq}$ be the set $\powerset S$ ordered by $\subseteq$.
Then $\struct {\powerset S, \subseteq}$ is totally ordered {{iff}} $S$ is either the empty set or a singleton. | From Subset Relation on Power Set is Partial Ordering we have that $\struct {\powerset S, \subseteq}$ is an ordered set.
We now need to show that $\struct {\powerset S, \subseteq}$ is a totally ordered set exactly when $S = \O$ or $S$ has exactly one element.
When $S = \O$ then $\powerset S = \set \O$ and it follows tr... | Let $\powerset S$ be the [[Definition:Power Set|power set]] of a [[Definition:Set|set]] $S$.
Let $\struct {\powerset S, \subseteq}$ be the [[Definition:Ordered Set|set $\powerset S$ ordered by $\subseteq$]].
Then $\struct {\powerset S, \subseteq}$ is [[Definition:Totally Ordered Set|totally ordered]] {{iff}} $S$ is ... | From [[Subset Relation on Power Set is Partial Ordering]] we have that $\struct {\powerset S, \subseteq}$ is an [[Definition:Ordered Set|ordered set]].
We now need to show that $\struct {\powerset S, \subseteq}$ is a [[Definition:Totally Ordered Set|totally ordered set]] exactly when $S = \O$ or $S$ has exactly one [[... | Condition for Power Set to be Totally Ordered | https://proofwiki.org/wiki/Condition_for_Power_Set_to_be_Totally_Ordered | https://proofwiki.org/wiki/Condition_for_Power_Set_to_be_Totally_Ordered | [
"Power Set",
"Subset Relation",
"Total Orderings"
] | [
"Definition:Power Set",
"Definition:Set",
"Definition:Ordered Set",
"Definition:Totally Ordered Set",
"Definition:Empty Set",
"Definition:Singleton"
] | [
"Subset Relation on Power Set is Partial Ordering",
"Definition:Ordered Set",
"Definition:Totally Ordered Set",
"Definition:Element",
"Definition:Totally Ordered Set",
"Empty Set is Subset of All Sets",
"Definition:Totally Ordered Set",
"Definition:Empty Set",
"Definition:Singleton",
"Definition:N... |
proofwiki-5682 | Fubini's Theorem | Let $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$ be $\sigma$-finite measure spaces.
Let $\struct {X \times Y, \Sigma_X \otimes \Sigma_Y, \mu \times \nu}$ be the product measure space of $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$.
Let $f: X \times Y \to \overline \R$ be a $\mu \times ... | We are given $f$ is $\mu \times \nu$-integrable.
That is:
:$\ds \int_{X \times Y} f^+ \map \rd {\mu \times \nu} < +\infty$
and
:$\ds \int_{X \times Y} f^- \map \rd {\mu \times \nu} < +\infty$
We have:
{{explain|Link to below conclusion should be included}}
:$\ds \int_{X \times Y} f \map \rd {\mu \times \nu} = \int_{X ... | Let $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$ be [[Definition:Sigma-Finite Measure Space|$\sigma$-finite measure spaces]].
Let $\struct {X \times Y, \Sigma_X \otimes \Sigma_Y, \mu \times \nu}$ be the [[Definition:Product Measure Space|product measure space]] of $\struct {X, \Sigma_X, \mu}$ and $\st... | We are [[Definition:Given|given]] $f$ is [[Definition:Measure-Integrable Function|$\mu \times \nu$-integrable]].
That is:
:$\ds \int_{X \times Y} f^+ \map \rd {\mu \times \nu} < +\infty$
and
:$\ds \int_{X \times Y} f^- \map \rd {\mu \times \nu} < +\infty$
We have:
{{explain|Link to below conclusion should be includ... | Fubini's Theorem/Proof | https://proofwiki.org/wiki/Fubini's_Theorem | https://proofwiki.org/wiki/Fubini's_Theorem/Proof | [
"Fubini's Theorem",
"Integrals of Integrable Functions",
"Product Measure"
] | [
"Definition:Sigma-Finite Measure Space",
"Definition:Product Measure Space",
"Definition:Integrable Function/Measure Space",
"Definition:Real-Valued Function",
"Definition:Real-Valued Function",
"Definition:Integrable Function/Measure Space",
"Definition:Integrable Function/Measure Space"
] | [
"Definition:Given",
"Definition:Integrable Function/Measure Space",
"Fubini's Theorem/Lemma",
"Tonelli's Theorem",
"Positive Part of Vertical Section of Function is Vertical Section of Positive Part",
"Tonelli's Theorem",
"Negative Part of Vertical Section of Function is Vertical Section of Negative Par... |
proofwiki-5683 | Product Sigma-Algebra Generated by Projections | Let $\struct {X, \Sigma_1}$ and $\struct {Y, \Sigma_2}$ be measurable spaces.
Let $\Sigma_1 \otimes \Sigma_2$ be the product $\sigma$-algebra on $X \times Y$.
Let $\pr_1: X \times Y \to X$ and $\pr_2: X \times Y \to Y$ be the first and second projections, respectively.
Then:
:$\Sigma_1 \otimes \Sigma_2 = \map \sigma {\... | {{Proofread}}
{{tidy}}
Let $R = \set {A \times B: A \in \Sigma_1, B \in \Sigma_2}$ be a generator for the product $\sigma$-algebra $\Sigma_1 \otimes \Sigma_2$.
Note that by Preimage of Element under Projection:
:$\inv {\pr_1} A = A \times Y \in R$
and:
:$\inv {\pr_2} B = X \times B \in R$
for any $A \in \Sigma_1$ and... | Let $\struct {X, \Sigma_1}$ and $\struct {Y, \Sigma_2}$ be [[Definition:Measurable Space|measurable spaces]].
Let $\Sigma_1 \otimes \Sigma_2$ be the [[Definition:Product Sigma-Algebra|product $\sigma$-algebra]] on $X \times Y$.
Let $\pr_1: X \times Y \to X$ and $\pr_2: X \times Y \to Y$ be the [[Definition:First Proj... | {{Proofread}}
{{tidy}}
Let $R = \set {A \times B: A \in \Sigma_1, B \in \Sigma_2}$ be a generator for the [[Definition:Product Sigma-Algebra|product $\sigma$-algebra]] $\Sigma_1 \otimes \Sigma_2$.
Note that by [[Preimage of Element under Projection]]:
:$\inv {\pr_1} A = A \times Y \in R$
and:
:$\inv {\pr_2} B = X \t... | Product Sigma-Algebra Generated by Projections | https://proofwiki.org/wiki/Product_Sigma-Algebra_Generated_by_Projections | https://proofwiki.org/wiki/Product_Sigma-Algebra_Generated_by_Projections | [
"Product Sigma-Algebras"
] | [
"Definition:Measurable Space",
"Definition:Product Sigma-Algebra",
"Definition:Projection (Mapping Theory)/First Projection",
"Definition:Projection (Mapping Theory)/Second Projection",
"Definition:Sigma-Algebra Generated by Collection of Mappings"
] | [
"Definition:Product Sigma-Algebra",
"Preimage of Element under Projection",
"Generated Sigma-Algebra Preserves Subset"
] |
proofwiki-5684 | Cardinal Zero is Less than Cardinal One | The zero cardinal $0$ is less than one:
:$0 < 1$ | We have that the Cardinals are Totally Ordered.
Let $\RR \subseteq \O \times \set \O$ be any arbitrary relation between $\O$ and $\set \O$.
We have that $\RR$ is vacuously many-to-one.
Also vacuously, $\RR$ is left-total.
Thus by definition, $\RR$ is in fact a mapping.
From Empty Mapping is Unique, this relation $\RR$ ... | The [[Definition:Zero (Cardinal)|zero cardinal]] $0$ is less than [[Definition:One (Cardinal)|one]]:
:$0 < 1$ | We have that the [[Cardinals are Totally Ordered]].
Let $\RR \subseteq \O \times \set \O$ be any arbitrary [[Definition:Relation|relation]] between $\O$ and $\set \O$.
We have that $\RR$ is [[Definition:Vacuous Set|vacuously]] [[Definition:Many-to-One Relation|many-to-one]].
Also [[Definition:Vacuous Set|vacuously]]... | Cardinal Zero is Less than Cardinal One | https://proofwiki.org/wiki/Cardinal_Zero_is_Less_than_Cardinal_One | https://proofwiki.org/wiki/Cardinal_Zero_is_Less_than_Cardinal_One | [
"Cardinals"
] | [
"Definition:Zero (Cardinal)",
"Definition:One (Cardinal)"
] | [
"Cardinals are Totally Ordered",
"Definition:Relation",
"Definition:Empty Set",
"Definition:Many-to-One Relation",
"Definition:Empty Set",
"Definition:Left-Total Relation",
"Definition:Mapping",
"Empty Mapping is Unique",
"Definition:Mapping",
"Definition:Empty Set",
"Definition:Injection",
"D... |
proofwiki-5685 | Right Quasigroup if (1-3) Parastrophe of Magma is Magma | Let $\struct {S, \circ}$ be a magma.
Let the $(1-3)$ parastrophe of $\struct {S, \circ}$ be a magma.
Then $\struct {S, \circ}$ is a right quasigroup. | By the definition of a right quasigroup it must be shown that:
:$\forall a, b \in S: \exists ! x \in S: x \circ a = b$
{{AimForCont}} there exists $a, b \in S$ such that $x \circ a = b$ does not have a unique solution for $x$.
Then in the $(1-3)$ parastrophe of $\struct {S, \circ}$ we see that $\circ$ as a mapping eith... | Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]].
Let the [[Definition:(1-3) Parastrophe|$(1-3)$ parastrophe]] of $\struct {S, \circ}$ be a [[Definition:Magma|magma]].
Then $\struct {S, \circ}$ is a [[Definition:Right Quasigroup|right quasigroup]]. | By the definition of a [[Definition:Right Quasigroup|right quasigroup]] it must be shown that:
:$\forall a, b \in S: \exists ! x \in S: x \circ a = b$
{{AimForCont}} there exists $a, b \in S$ such that $x \circ a = b$ does not have a unique solution for $x$.
Then in the [[Definition:(1-3) Parastrophe|$(1-3)$ parastr... | Right Quasigroup if (1-3) Parastrophe of Magma is Magma | https://proofwiki.org/wiki/Right_Quasigroup_if_(1-3)_Parastrophe_of_Magma_is_Magma | https://proofwiki.org/wiki/Right_Quasigroup_if_(1-3)_Parastrophe_of_Magma_is_Magma | [
"Parastrophes",
"Quasigroups",
"Magmas"
] | [
"Definition:Magma",
"Definition:(1-3) Parastrophe",
"Definition:Magma",
"Definition:Quasigroup/Right Quasigroup"
] | [
"Definition:Quasigroup/Right Quasigroup",
"Definition:(1-3) Parastrophe",
"Definition:Mapping",
"Definition:Left-Total Relation",
"Definition:Many-to-One Relation",
"Definition:Magma",
"Proof by Contradiction",
"Definition:Assumption",
"Category:Parastrophes",
"Category:Quasigroups",
"Category:M... |
proofwiki-5686 | Idempotent Non-Trivial Quasigroup is Not a Loop | Let $\struct {S, \circ}$ be an idempotent quasigroup whose underlying set $S$ comprises more than one element.
Then $\struct {S, \circ}$ is not an algebra loop, that is, it has no identity element. | {{AimForCont}} $\struct {S, \circ}$ has an identity element $e$.
Then by Identity Element is Idempotent:
:$e \circ e = e$
Consider $e' \in S$ where $e' \ne e$.
Since $e$ is an identity element:
:$e' \circ e = e'$
Also, by assumption, $\circ$ is idempotent, so:
:$e' \circ e' = e'$
Then by the definition of a quasigroup,... | Let $\struct {S, \circ}$ be an [[Definition:Idempotent Algebraic Structure|idempotent]] [[Definition:Quasigroup|quasigroup]] whose [[Definition:Underlying Set of Structure|underlying set]] $S$ comprises more than one [[Definition:Element|element]].
Then $\struct {S, \circ}$ is not an [[Definition:Algebra Loop|algebra... | {{AimForCont}} $\struct {S, \circ}$ has an [[Definition:Identity Element|identity element]] $e$.
Then by [[Identity Element is Idempotent]]:
:$e \circ e = e$
Consider $e' \in S$ where $e' \ne e$.
Since $e$ is an [[Definition:Identity Element|identity element]]:
:$e' \circ e = e'$
Also, by [[Definition:Assumption|... | Idempotent Non-Trivial Quasigroup is Not a Loop | https://proofwiki.org/wiki/Idempotent_Non-Trivial_Quasigroup_is_Not_a_Loop | https://proofwiki.org/wiki/Idempotent_Non-Trivial_Quasigroup_is_Not_a_Loop | [
"Quasigroups",
"Algebra Loops",
"Idempotence"
] | [
"Definition:Idempotence/Algebraic Structure",
"Definition:Quasigroup",
"Definition:Underlying Set/Abstract Algebra",
"Definition:Element",
"Definition:Algebra Loop",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Identity Element is Idempotent",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Assumption",
"Definition:Idempotence/Operation",
"Definition:Quasigroup",
"Definition:Regular Representations/Left Regular Representatio... |
proofwiki-5687 | B-Algebra is Right Cancellable | Let $\struct {X, \circ}$ be a $B$-algebra.
Then $\circ$ is right-cancellable for $X$. That is:
:$\forall x, y, z \in X: x \circ z = y \circ z \implies x = y$ | Let $x, y \in X$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {x \circ y} \circ \paren {0 \circ y}
| r = x \circ \paren {\paren {0 \circ y} \circ \paren {0 \circ y} }
| c = {{B-algebra-axiom|3}}
}}
{{eqn | r = x \circ 0
| c = {{B-algebra-axiom|1}}
}}
{{eqn | r = x
| c = {{B-algebra-axiom|2}}
}}
{{end... | Let $\struct {X, \circ}$ be a [[Definition:B-Algebra|$B$-algebra]].
Then $\circ$ is [[Definition:Right Cancellable Operation|right-cancellable]] for $X$. That is:
:$\forall x, y, z \in X: x \circ z = y \circ z \implies x = y$ | Let $x, y \in X$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {x \circ y} \circ \paren {0 \circ y}
| r = x \circ \paren {\paren {0 \circ y} \circ \paren {0 \circ y} }
| c = {{B-algebra-axiom|3}}
}}
{{eqn | r = x \circ 0
| c = {{B-algebra-axiom|1}}
}}
{{eqn | r = x
| c = {{B-algebra-axiom|2}}
}}
{{e... | B-Algebra is Right Cancellable | https://proofwiki.org/wiki/B-Algebra_is_Right_Cancellable | https://proofwiki.org/wiki/B-Algebra_is_Right_Cancellable | [
"B-Algebras"
] | [
"Definition:B-Algebra",
"Definition:Right Cancellable Operation"
] | [
"Category:B-Algebras"
] |
proofwiki-5688 | Right Regular Representation of 0 is Bijection in B-Algebra | Let $\struct {X, \circ}$ be a $B$-algebra.
Then the right regular representation of $\struct {X, \circ}$ {{WRT}} $0$ is a bijection. | {{B-algebra-axiom|2}} states:
:$\forall x \in X: x \circ 0 = x$
and so, for all $x \in X$:
:$\map {\rho_0} x = x$
That is:
:$\rho_0 = I_X$
which is the identity mapping on $X$.
The result follows from Identity Mapping is Bijection.
{{qed}}
Category:B-Algebras
Category:Regular Representations
7qdxvqnkmbjcb8n4ss82iikpiuf... | Let $\struct {X, \circ}$ be a [[Definition:B-Algebra|$B$-algebra]].
Then the [[Definition:Right Regular Representation|right regular representation]] of $\struct {X, \circ}$ {{WRT}} $0$ is a [[Definition:Bijection|bijection]]. | {{B-algebra-axiom|2}} states:
:$\forall x \in X: x \circ 0 = x$
and so, for all $x \in X$:
:$\map {\rho_0} x = x$
That is:
:$\rho_0 = I_X$
which is the [[Definition:Identity Mapping|identity mapping]] on $X$.
The result follows from [[Identity Mapping is Bijection]].
{{qed}}
[[Category:B-Algebras]]
[[Category:Re... | Right Regular Representation of 0 is Bijection in B-Algebra | https://proofwiki.org/wiki/Right_Regular_Representation_of_0_is_Bijection_in_B-Algebra | https://proofwiki.org/wiki/Right_Regular_Representation_of_0_is_Bijection_in_B-Algebra | [
"B-Algebras",
"Regular Representations"
] | [
"Definition:B-Algebra",
"Definition:Regular Representations/Right Regular Representation",
"Definition:Bijection"
] | [
"Definition:Identity Mapping",
"Identity Mapping is Bijection",
"Category:B-Algebras",
"Category:Regular Representations"
] |
proofwiki-5689 | Product of Cardinals is Commutative | Let $\mathbf a$ and $\mathbf b$ be cardinals.
Then:
: $\mathbf a \mathbf b = \mathbf b \mathbf a$
where $\mathbf a \mathbf b$ denotes the product of $\mathbf a$ and $\mathbf b$. | Let $\mathbf a = \card A$ and $\mathbf b = \card B$ for some sets $A$ and $B$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf a \mathbf b
| r = \card {A \times B}
| c = {{Defof|Product of Cardinals}}
}}
{{eqn | r = \card {B \times A}
| c = Cardinality of Cartesian Product of Finite Sets/Corollary
}}
{{eqn | ... | Let $\mathbf a$ and $\mathbf b$ be [[Definition:Cardinal|cardinals]].
Then:
: $\mathbf a \mathbf b = \mathbf b \mathbf a$
where $\mathbf a \mathbf b$ denotes the [[Definition:Product of Cardinals|product]] of $\mathbf a$ and $\mathbf b$. | Let $\mathbf a = \card A$ and $\mathbf b = \card B$ for some [[Definition:Set|sets]] $A$ and $B$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf a \mathbf b
| r = \card {A \times B}
| c = {{Defof|Product of Cardinals}}
}}
{{eqn | r = \card {B \times A}
| c = [[Cardinality of Cartesian Product of Finite Se... | Product of Cardinals is Commutative | https://proofwiki.org/wiki/Product_of_Cardinals_is_Commutative | https://proofwiki.org/wiki/Product_of_Cardinals_is_Commutative | [
"Cardinals"
] | [
"Definition:Cardinal",
"Definition:Product of Cardinals"
] | [
"Definition:Set",
"Cardinality of Cartesian Product of Finite Sets/Corollary"
] |
proofwiki-5690 | Product of Cardinals is Associative | Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals.
Then:
: $\mathbf a \paren {\mathbf {b c} } = \paren {\mathbf {a b} } \mathbf c$
where $\mathbf {a b}$ denotes the product of $\mathbf a$ and $\mathbf b$. | Let $\mathbf a = \card A$, $\mathbf b = \card B$ and $\mathbf c = \card C$ for some sets $A$, $B$ and $C$.
By definition of product of cardinals:
:$\mathbf a \paren {\mathbf {b c} }$ is the cardinal associated with $A \times \paren {B \times C}$.
Consider the mapping $f: A \times \paren {B \times C} \to \paren {A \time... | Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be [[Definition:Cardinal|cardinals]].
Then:
: $\mathbf a \paren {\mathbf {b c} } = \paren {\mathbf {a b} } \mathbf c$
where $\mathbf {a b}$ denotes the [[Definition:Product of Cardinals|product]] of $\mathbf a$ and $\mathbf b$. | Let $\mathbf a = \card A$, $\mathbf b = \card B$ and $\mathbf c = \card C$ for some [[Definition:Set|sets]] $A$, $B$ and $C$.
By definition of [[Definition:Product of Cardinals|product of cardinals]]:
:$\mathbf a \paren {\mathbf {b c} }$ is the [[Definition:Cardinal|cardinal associated with $A \times \paren {B \times... | Product of Cardinals is Associative | https://proofwiki.org/wiki/Product_of_Cardinals_is_Associative | https://proofwiki.org/wiki/Product_of_Cardinals_is_Associative | [
"Cardinals"
] | [
"Definition:Cardinal",
"Definition:Product of Cardinals"
] | [
"Definition:Set",
"Definition:Product of Cardinals",
"Definition:Cardinal",
"Definition:Mapping",
"Equality of Ordered Tuples",
"Equality of Ordered Tuples",
"Equality of Ordered Tuples",
"Equality of Ordered Tuples",
"Definition:Injection",
"Definition:Surjection",
"Definition:Injection",
"De... |
proofwiki-5691 | B-Algebra Identity: xy=x(0(0y)) | Let $\struct {X, \circ}$ be a $B$-algebra.
Then:
:$\forall x, y \in X: x \circ y = x \circ \paren {0 \circ \paren {0 \circ y} }$ | Let $x, y \in X$.
Then:
{{begin-eqn}}
{{eqn | l = x \circ y
| r = \paren {x \circ y} \circ 0
| c = {{B-algebra-axiom|2}}
}}
{{eqn | r = x \circ \paren {0 \circ \paren {0 \circ y} }
| c = {{B-algebra-axiom|3}}
}}
{{end-eqn}}
Hence the result.
{{qed}}
Category:B-Algebras
pximfi3644a38aax08lelqdnv4cyajl | Let $\struct {X, \circ}$ be a [[Definition:B-Algebra|$B$-algebra]].
Then:
:$\forall x, y \in X: x \circ y = x \circ \paren {0 \circ \paren {0 \circ y} }$ | Let $x, y \in X$.
Then:
{{begin-eqn}}
{{eqn | l = x \circ y
| r = \paren {x \circ y} \circ 0
| c = {{B-algebra-axiom|2}}
}}
{{eqn | r = x \circ \paren {0 \circ \paren {0 \circ y} }
| c = {{B-algebra-axiom|3}}
}}
{{end-eqn}}
Hence the result.
{{qed}}
[[Category:B-Algebras]]
pximfi3644a38aax08lelqd... | B-Algebra Identity: xy=x(0(0y)) | https://proofwiki.org/wiki/B-Algebra_Identity:_xy=x(0(0y)) | https://proofwiki.org/wiki/B-Algebra_Identity:_xy=x(0(0y)) | [
"B-Algebras"
] | [
"Definition:B-Algebra"
] | [
"Category:B-Algebras"
] |
proofwiki-5692 | B-Algebra Identity: x (y z) = (x (0 z)) y | Let $\struct {X, \circ}$ be a $B$-algebra.
Then:
:$\forall x, y, z \in X: x \circ \paren {y \circ z} = \paren {x \circ \paren {0 \circ z} } \circ y$ | Let $x, y, z \in X$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {x \circ \paren {0 \circ z} } \circ y
| r = x \circ \paren {y \circ \paren {0 \circ \paren {0 \circ z} } }
| c = {{B-algebra-axiom|2}}
}}
{{eqn | r = x \circ \paren {y \circ z}
| c = Identity: $y \circ \paren {0 \circ \paren {0 \circ z} } = y... | Let $\struct {X, \circ}$ be a [[Definition:B-Algebra|$B$-algebra]].
Then:
:$\forall x, y, z \in X: x \circ \paren {y \circ z} = \paren {x \circ \paren {0 \circ z} } \circ y$ | Let $x, y, z \in X$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {x \circ \paren {0 \circ z} } \circ y
| r = x \circ \paren {y \circ \paren {0 \circ \paren {0 \circ z} } }
| c = {{B-algebra-axiom|2}}
}}
{{eqn | r = x \circ \paren {y \circ z}
| c = [[B-Algebra Identity: xy=x(0(0y))|Identity: $y \circ \par... | B-Algebra Identity: x (y z) = (x (0 z)) y | https://proofwiki.org/wiki/B-Algebra_Identity:_x_(y_z)_=_(x_(0_z))_y | https://proofwiki.org/wiki/B-Algebra_Identity:_x_(y_z)_=_(x_(0_z))_y | [
"B-Algebras"
] | [
"Definition:B-Algebra"
] | [
"B-Algebra Identity: xy=x(0(0y))"
] |
proofwiki-5693 | B-Algebra Identity: xy = 0 iff x = y | Let $\struct {X, \circ}$ be a $B$-algebra.
Then:
:$\forall x, y \in X: x \circ y = 0 \iff x = y$ | === Sufficient Condition ===
Suppose that $x = y$.
Then by {{B-algebra-axiom|1}}:
:$x \circ y = x \circ x = 0$
{{qed|lemma}} | Let $\struct {X, \circ}$ be a [[Definition:B-Algebra|$B$-algebra]].
Then:
:$\forall x, y \in X: x \circ y = 0 \iff x = y$ | === Sufficient Condition ===
Suppose that $x = y$.
Then by {{B-algebra-axiom|1}}:
:$x \circ y = x \circ x = 0$
{{qed|lemma}} | B-Algebra Identity: xy = 0 iff x = y | https://proofwiki.org/wiki/B-Algebra_Identity:_xy_=_0_iff_x_=_y | https://proofwiki.org/wiki/B-Algebra_Identity:_xy_=_0_iff_x_=_y | [
"B-Algebras"
] | [
"Definition:B-Algebra"
] | [] |
proofwiki-5694 | First Power of Element in B-Algebra | Let $\left({X, \circ}\right)$ be a $B$-algebra.
Then:
:$\forall x \in X: x^1 = x$
where $x^k$ for $k \in \N$ denotes the $k$th power of the element $x$. | {{begin-eqn}}
{{eqn | l = x^1
| r = x^0 \circ \left({0 \circ x}\right)
| c = {{Defof|B-Algebra Power of Element}}
}}
{{eqn | r = 0 \circ \left({0 \circ x}\right)
| c = {{Defof|B-Algebra Power of Element}}
}}
{{eqn | r = 0 \circ x
| c = $0$ in $B$-Algebra is Left Cancellable Element
}}
{{eqn | r ... | Let $\left({X, \circ}\right)$ be a [[Definition:B-Algebra|$B$-algebra]].
Then:
:$\forall x \in X: x^1 = x$
where $x^k$ for $k \in \N$ denotes the [[Definition:B-Algebra Power of Element|$k$th power of the element $x$]]. | {{begin-eqn}}
{{eqn | l = x^1
| r = x^0 \circ \left({0 \circ x}\right)
| c = {{Defof|B-Algebra Power of Element}}
}}
{{eqn | r = 0 \circ \left({0 \circ x}\right)
| c = {{Defof|B-Algebra Power of Element}}
}}
{{eqn | r = 0 \circ x
| c = [[0 in B-Algebra is Left Cancellable Element|$0$ in $B$-Alge... | First Power of Element in B-Algebra | https://proofwiki.org/wiki/First_Power_of_Element_in_B-Algebra | https://proofwiki.org/wiki/First_Power_of_Element_in_B-Algebra | [
"B-Algebras"
] | [
"Definition:B-Algebra",
"Definition:Power (B-Algebra)"
] | [
"0 in B-Algebra is Left Cancellable Element",
"0 in B-Algebra is Left Cancellable Element",
"Category:B-Algebras"
] |
proofwiki-5695 | B-Algebra Power Law | Let $\struct {X, \circ}$ be a $B$-algebra.
Let $n, m \in \N$ such that $n \ge m$.
Then:
:$\forall x \in X: x^n \circ x^m = x^{n - m}$
where $x^k$ for $k \in \N$ denotes the $k$th power of the element $x$. | Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\forall m \in \N_{> 0}, m \le n: \forall x \in X: x^n \circ x^m = x^{n - m}$ | Let $\struct {X, \circ}$ be a [[Definition:B-Algebra|$B$-algebra]].
Let $n, m \in \N$ such that $n \ge m$.
Then:
:$\forall x \in X: x^n \circ x^m = x^{n - m}$
where $x^k$ for $k \in \N$ denotes the [[Definition:B-Algebra Power of Element|$k$th power of the element $x$]]. | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\forall m \in \N_{> 0}, m \le n: \forall x \in X: x^n \circ x^m = x^{n - m}$ | B-Algebra Power Law | https://proofwiki.org/wiki/B-Algebra_Power_Law | https://proofwiki.org/wiki/B-Algebra_Power_Law | [
"B-Algebras"
] | [
"Definition:B-Algebra",
"Definition:Power (B-Algebra)"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-5696 | B-Algebra is Left Cancellable | Let $\struct {X, \circ}$ be a $B$-algebra.
Then $\circ$ is a left cancellable operation. | Let $x, y, z \in X$:
{{begin-eqn}}
{{eqn | l = x \circ y
| r = x \circ z
}}
{{eqn | ll= \leadstoandfrom
| l = 0 \circ \paren {x \circ y}
| r = 0 \circ \paren {x \circ z}
| c = $0$ in $B$-Algebra is Left Cancellable Element
}}
{{eqn | ll= \leadstoandfrom
| l = \paren {0 \circ \paren {0 \cir... | Let $\struct {X, \circ}$ be a [[Definition:B-Algebra|$B$-algebra]].
Then $\circ$ is a [[Definition:Left Cancellable Operation|left cancellable operation]]. | Let $x, y, z \in X$:
{{begin-eqn}}
{{eqn | l = x \circ y
| r = x \circ z
}}
{{eqn | ll= \leadstoandfrom
| l = 0 \circ \paren {x \circ y}
| r = 0 \circ \paren {x \circ z}
| c = [[0 in B-Algebra is Left Cancellable Element|$0$ in $B$-Algebra is Left Cancellable Element]]
}}
{{eqn | ll= \leadstoan... | B-Algebra is Left Cancellable | https://proofwiki.org/wiki/B-Algebra_is_Left_Cancellable | https://proofwiki.org/wiki/B-Algebra_is_Left_Cancellable | [
"B-Algebras"
] | [
"Definition:B-Algebra",
"Definition:Left Cancellable Operation"
] | [
"0 in B-Algebra is Left Cancellable Element",
"B-Algebra Identity: x (y z) = (x (0 z)) y",
"0 in B-Algebra is Left Cancellable Element",
"0 in B-Algebra is Left Cancellable Element",
"B-Algebra is Right Cancellable",
"Category:B-Algebras"
] |
proofwiki-5697 | B-Algebra is Quasigroup | Let $\struct {X, \circ}$ be a $B$-algebra.
Then $\struct {X, \circ}$ is a quasigroup. | By the definition of a quasigroup it must be shown that $\forall x \in X$ the left and right regular representations $\lambda_x$ and $\rho_x$ are permutations on $X$.
As $\struct {X, \circ}$ is a magma:
:$\forall x \in X$ the codomain of $\lambda_x$ and $\rho_x$ is $X$.
Hence it is sufficient to prove that $\lambda_x$ ... | Let $\struct {X, \circ}$ be a [[Definition:B-Algebra|$B$-algebra]].
Then $\struct {X, \circ}$ is a [[Definition:Quasigroup|quasigroup]]. | By the definition of a [[Definition:Quasigroup|quasigroup]] it must be shown that $\forall x \in X$ the [[Definition:Regular Representations|left and right regular representations]] $\lambda_x$ and $\rho_x$ are [[Definition:Permutation|permutations]] on $X$.
As $\struct {X, \circ}$ is a [[Definition:Magma|magma]]:
:$... | B-Algebra is Quasigroup | https://proofwiki.org/wiki/B-Algebra_is_Quasigroup | https://proofwiki.org/wiki/B-Algebra_is_Quasigroup | [
"B-Algebras",
"Examples of Quasigroups"
] | [
"Definition:B-Algebra",
"Definition:Quasigroup"
] | [
"Definition:Quasigroup",
"Definition:Regular Representations",
"Definition:Permutation",
"Definition:Magma",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Bijection",
"B-Algebra is Left Cancellable",
"B-Algebra is Right Cancellable",
"Cancellable iff Regular Representations Injective",
"... |
proofwiki-5698 | Quasigroup is not necessarily B-Algebra | Let $\struct {S, \circ}$ be a quasigroup.
Then $\struct {S, \circ}$ is not necessarily a $B$-algebra. | ;Proof by Counterexample
From Group is Quasigroup we take an arbitrary small group.
Consider the Cayley table of the group of order $3$:
:$\begin{array}{c|cccccc}
& 0 & 1 & 2 \\
\hline
0 & 0 & 1 & 2 \\
1 & 1 & 2 & 0 \\
2 & 2 & 0 & 1 \\
\end{array}$
By inspection we see that {{B-algebra-axiom|2}} does not hold as $1 \... | Let $\struct {S, \circ}$ be a [[Definition:Quasigroup|quasigroup]].
Then $\struct {S, \circ}$ is not necessarily a [[Definition:B-Algebra|$B$-algebra]]. | ;[[Proof by Counterexample]]
From [[Group is Quasigroup]] we take an arbitrary small [[Definition:Group|group]].
Consider the [[Definition:Cayley Table|Cayley table]] of the [[Definition:Group|group]] of [[Definition:Order of Structure|order]] $3$:
:$\begin{array}{c|cccccc}
& 0 & 1 & 2 \\
\hline
0 & 0 & 1 & 2 \\
1... | Quasigroup is not necessarily B-Algebra | https://proofwiki.org/wiki/Quasigroup_is_not_necessarily_B-Algebra | https://proofwiki.org/wiki/Quasigroup_is_not_necessarily_B-Algebra | [
"B-Algebras",
"Examples of Quasigroups"
] | [
"Definition:Quasigroup",
"Definition:B-Algebra"
] | [
"Proof by Counterexample",
"Group is Quasigroup",
"Definition:Group",
"Definition:Cayley Table",
"Definition:Group",
"Definition:Order of Structure"
] |
proofwiki-5699 | B-Algebra Power Law with Zero | :$\forall x \in X: n, m \in \N_{>0} \implies x^m \circ x^n = 0 \circ x^{n - m}$ | First we show that:
:$\forall x \in X: x \circ x^2 = 0 \circ x$
{{begin-eqn}}
{{eqn | l = x \circ x^2
| r = x \circ \paren {x^1 \circ \paren {0 \circ x} }
| c = {{Defof|Power (B-Algebra)|Power ($B$-Algebra)}}
}}
{{eqn | r = x \circ \paren {x \circ \paren {0 \circ x} }
| c = First Power of Element in $... | :$\forall x \in X: n, m \in \N_{>0} \implies x^m \circ x^n = 0 \circ x^{n - m}$ | First we show that:
:$\forall x \in X: x \circ x^2 = 0 \circ x$
{{begin-eqn}}
{{eqn | l = x \circ x^2
| r = x \circ \paren {x^1 \circ \paren {0 \circ x} }
| c = {{Defof|Power (B-Algebra)|Power ($B$-Algebra)}}
}}
{{eqn | r = x \circ \paren {x \circ \paren {0 \circ x} }
| c = [[First Power of Element i... | B-Algebra Power Law with Zero | https://proofwiki.org/wiki/B-Algebra_Power_Law_with_Zero | https://proofwiki.org/wiki/B-Algebra_Power_Law_with_Zero | [
"B-Algebras"
] | [] | [
"First Power of Element in B-Algebra",
"Definition:Proposition",
"Principle of Mathematical Induction",
"Definition:Lemma",
"B-Algebra Identity: x (y z) = (x (0 z)) y",
"B-Algebra Power Law",
"Category:B-Algebras"
] |
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