nyuuzyou/wikis huggingface dataset:
Algebraic Structures formed by Left and Right Operations are not Isomorphic for Cardinality Greater than 1

== Theorem ==
Let $S$ be a set.
Let $\gets$ and $\to$ denote the left operation and right operation respectively.
Let $\card S > 1$.
The algebraic structures $\struct {S, \gets}$ and $\struct {S, \to}$ are not isomorphic.
== Proof ==
 there exists an isomorphism $\phi$ from $\struct {S, \gets}$ to $\struct {S, \to}$.
Because $\card S > 1$ we have that there exist at least $2$ distinct elements of $S$.
Let these be $x$ and $y$.
Hence:
Hence $\map \phi x = \map \phi y$ and so $\phi$ is not an injection.
Hence $\phi$ is not a bijection and so not an isomorphism.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.11 \ \text {(b)}$
Category:Left and Right Operations
Category:Examples of Isomorphisms (Abstract Algebra)
Algebraic Variety/Examples/Circle

== Example of Algebraic Variety ==
Consider the circle described by Equation of Circle in Cartesian Plane:
:$(1): \quad {x_1}^2 + {x_2}^2 - r^2 = 0$
whose radius is $r$.
Then the circle is the solution set of $(1)$.
== Sources ==
* 
* 
Category:Examples of Algebraic Varieties
Algebraically Closed Field is Infinite

== Theorem ==
Let $F$ be an algebraically closed field.
Then $F$ is infinite.
== Proof ==
We prove the contrapositive: that a Galois field cannot be algebraically closed.
Let $F$ be Galois.
Define the polynomial:
:$\ds \map f x = 1 + \prod_{a \mathop \in F} \paren {x - a}$
By definition, a field is a ring.
Thus by Ring Product with Zero:
:$\ds \forall x \in F: \prod_{a \mathop \in F} \paren {x - a} = 0$
But:
:$\forall a \in F: \map f a = 1 + 0 \ne 0$
Therefore $\map f x$ cannot have a root in $F$.
Thus $F$ is not algebraically closed.
The result follows by the Rule of Transposition.
Category:Field Theory
Algebraically Closed Field is Perfect

== Theorem ==
Let $F$ be an algebraically closed field.
Then $F$ is perfect.
== Proof ==
Let $E / F$ be any algebraic extension.
Since $F$ is an algebraically closed field, $E = F$.
By Field is Separable over itself, $E$ is separable over $F$.
Hence $F$ is perfect.
Category:Field Extensions
Category:Perfect Fields
Algorithmic Complexity/Examples/Multiplication of Square Matrices

== Example of Algorithmic Complexity ==
Consider two square matrices $\mathbf A$ and $\mathbf B$ of order $n$.
Consider the matrix multiplication operation:
:$\ds \mathbf A \times \mathbf B := \forall i j \in \closedint 1 n: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$
This takes $2 n^3$ steps when using the usual method.
However, using the Strassen algorithm, this takes a number of steps proportional to $n^{\lg 7}$, where $\lg$ denotes logarithm base $2$.
As $\lg 7 < 3$, it follows that the Strassen algorithm uses fewer steps than the usual method, for sufficiently large $n$.
== Proof ==
== Sources ==
* 
* 
Category:Examples of Algorithmic Complexity
Aliquot Sequence/Examples/138

== Example of Aliquot Sequence ==
The aliquot sequence that starts at $138$ reaches as high as $179 \, 931 \, 895 \, 322$ after $117$ steps, then terminates at $1$ after a total of $177$ steps.
== Historical Note ==
== Sources ==
* : $276$
* 
* : $276$
Category:Aliquot Sequences
Category:138
Aliquot Sequence/Examples/276

== Example of Aliquot Sequence ==
The eventual fate of the aliquot sequence that starts at $276$ is unknown.
By step $469$ it is $45$ digits long:
:$149 \, 384 \, 846 \, 598 \, 254 \, 844 \, 243 \, 905 \, 695 \, 992 \, 651 \, 412 \, 919 \, 855 \, 640$
== Historical Note ==
== Sources ==
* : $276$
* : $276$
Category:Aliquot Sequences
Category:276
Aliquot Sequence/Examples/276/Historical Note

== Historical Note on Aliquot Sequence starting from $276$ ==
The fate of the aliquot sequence that starts at $276$ was explored by and others.
However far the calculation is pushed, it still does not approach termination.
 reported in of $1986$ that after $469$ steps, the $45$-digit number had been reached:
:$149 \, 384 \, 846 \, 598 \, 254 \, 844 \, 243 \, 905 \, 695 \, 992 \, 651 \, 412 \, 919 \, 855 \, 640$
By the time of of $1997$, he was able to report that the calculation had been pushed to $487$ steps.
== Sources ==
* : $276$
* : $276$
Category:Historical Notes
All Bases of Matroid have same Cardinality

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\rho: \powerset S \to \Z$ be the rank function of $M$.
Let $B$ be a base of $M$.
Then:
:$\size B = \map \rho S$
That is, all bases of $M$ have the same cardinality, which is the rank of $M$.
=== Corollary ===
== Proof ==
By definition of the rank function:
:$\map \rho S = \max \set{\size X : X \subseteq S, X \in \mathscr I}$
Let $B_1$ be an independent subset such that:
:$\size {B_1} = \map \rho S$
It is shown that $B_1$ is a base.
Let $X$ be an independent superset of $B_1$.
From Cardinality of Subset of Finite Set:
:$\size {B_1} \le \size X$
By choice of $B_1$:
:$\size X \le \size {B_1}$
It follows that:
:$\size X = \size {B_1}$
From Cardinality of Proper Subset of Finite Set:
:$X = B_1$
It follows that $B_1$ is a maximal independent subset.
That is, $B_1$ is a base.
It is now shown that any other base has the same cardinality as $B_1$.
Let $B_2$ be any other base.
By choice of $B_1$:
:$\size {B_2} \le \size {B_1}$
 $\size {B_2} < \size {B_1}$.
From Independent Set can be Augmented by Larger Independent Set there exists $Z \subseteq B_1 \setminus B_2$ such that:
:$B_2 \cup Z \in \mathscr I$
:$\size{B_2 \cup Z} = \size{B_1}$
This contradicts the maximality of $B_2$ in $\mathscr I$.
Then:
:$\size {B_2} = \size {B_1} = \map \rho S$ 
== Sources ==
* Chapter $1.$ $\S 5.$ Properties of independent sets and bases
Category:Matroid Bases
All Bases of Matroid have same Cardinality/Corollary

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $B \subseteq S$ be a base of $M$.
Let $X \subseteq S$ be any independent subset of $M$.
Then:
:$\card X \le \card B$
== Proof ==
From Independent Subset is Contained in Maximal Independent Subset :
:$\exists B' \subseteq S : X \subseteq B'$ and $B'$ is a maximal independent subset of $S$
By definition of a base:
:$B'$ is a base of $M$
From Cardinality of Subset of Finite Set:
:$\card X \le \card {B'}$
From All Bases of Matroid have same Cardinality:
:$\card{B'} = \card B$
Hence:
:$\card X \le \card B$
Category:Matroid Bases
All Elements Self-Inverse then Abelian

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Suppose that every element of $G$ is self-inverse.
Then $G$ is abelian.
== Proof ==
Every element of $G$ is self-inverse, that is:
:$\forall x \in G: x \circ x = e$
In particular, for all $x, y \in G$:
:$\paren {x \circ y} \circ \paren {x \circ y} = e$
that is, $x \circ y$ is also self-inverse.
From Self-Inverse Elements Commute iff Product is Self-Inverse, it follows that:
:$\forall x, y \in G: y \circ x = x \circ y$
that is: $x$ and $y$ commute for all $x, y \in G$.
Hence $G$ is an abelian group.
== Sources ==
* : $\S 1.4$: Exercise $2.2$
* : Chapter $\text{II}$: Groups: Exercise $\text{T}$
* : Chapter $2$: The Definition of Group Structure: $\S 29 \delta$
Category:Abelian Groups
All Elements of Left Operation are Left Zeroes

== Theorem ==
Let $\struct {S, \leftarrow}$ be an algebraic structure in which the operation $\leftarrow$ is the left operation.
Then no matter what $S$ is, $\struct {S, \leftarrow}$ is a semigroup all of whose elements are left zeroes.
Thus it can be seen that any left zero in a semigroup is not necessarily unique.
== Proof ==
It is established in Structure under Left Operation is Semigroup that $\struct {S, \leftarrow}$ is a semigroup.
From the definition of left operation:
:$\forall x, y \in S: x \leftarrow y = x$
from which it can immediately be seen that all elements of $S$ are indeed left zeroes.
From More than One Right Zero then No Left Zero, it also follows that there is no right zero.
== Also see ==
* Element under Left Operation is Right Identity
* All Elements of Right Operation are Right Zeroes
== Sources ==
* : Chapter $4$. Groups: Exercise $6$
Category:Left Operation
Category:Zero Elements
All Elements of Right Operation are Right Zeroes

== Theorem ==
Let $\struct {S, \to}$ be an algebraic structure in which the operation $\to$ is the right operation.
Then no matter what $S$ is, $\struct {S, \to}$ is a semigroup all of whose elements are right zeroes.
Thus it can be seen that any right zero in a semigroup is not necessarily unique.
== Proof ==
It is established in Structure under Right Operation is Semigroup that $\struct {S, \to}$ is a semigroup.
From the definition of right operation:
:$\forall x, y \in S: x \to y = y$
from which it can immediately be seen that all elements of $S$ are indeed right zeroes.
From More than One Right Zero then No Left Zero, it also follows that there is no left zero.
== Also see ==
* Element under Right Operation is Left Identity
* All Elements of Left Operation are Left Zeroes
== Sources ==
* : Chapter $4$. Groups: Exercise $6$
Category:Right Operation
Category:Zero Elements
All Factors Divide Integer then Whole Divides Integer

== Theorem ==
Let $S = \set {a_1, a_2, \ldots, a_r} \subseteq \Z$ be a finite subset of the integers.
Let $S$ be pairwise coprime.
Let:
:$\forall j \in \set {1, 2, \ldots, r}: a_r \divides b$
where $\divides$ denotes divisibility.
Then:
:$\ds \prod_{j \mathop = 1}^r a_j \divides b$
== Proof ==
Proof by induction:
In the following, it is assumed at all times that $S = \set {a_1, a_2, \ldots, a_r} \subseteq \Z$ is pairwise coprime.
For all $r \in \N_{> 1}$, let $\map P r$ be the proposition:
:$\ds \prod_{j \mathop = 1}^r a_j \divides b$
=== Basis for the Induction ===
$\map P 2$ is the case:
:$a_1 a_2 \divides b$
By definition of divisibility:
:$b = a_1 q_1 = a_2 q_2$
where $q_1, q_2 \in \Z$.
By Integer Combination of Coprime Integers:
:$\exists x, y \in \Z: a_1 x + a_2 y = 1$
So:
and so $a_1 a_2 \divides b$.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \prod_{j \mathop = 1}^k a_j \divides b$
Then we need to show:
:$\ds \prod_{j \mathop = 1}^{k + 1} a_j \divides b$
=== Induction Step ===
This is our induction step:
By hypothesis:
:$a_{k + 1} \divides b$
By Integer Coprime to all Factors is Coprime to Whole:
:$\ds a_{k + 1} \perp \prod_{j \mathop = 1}^k a_j$
By the induction hypothesis:
:$\ds \prod_{j \mathop = 1}^k a_j \divides b$
So by the basis for the induction:
:$\ds \prod_{j \mathop = 1}^{k + 1} a_j \divides b$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
$\ds r \in \N_{> 1}: \prod_{j \mathop = 1}^r a_j \divides b$
== Sources ==
* : Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Proposition $3 \ \text{(i)}$
Category:Coprime Integers
All Horses are the Same Colour/Historical Note

== Historical Note on All Horses are the Same Colour ==
The paradox All Horses are the Same Colour was originally raised by , and appears in his $1954$ work Induction and Analogy in Mathematics.
It arose from the popular (at the time: the $1930$s) colloquialism:
:That's a horse of a different color!
meaning:
:That's unexpectedly new and different.
Category:Historical Notes
All Infima Preserving Mapping is Upper Adjoint of Galois Connection

== Theorem ==
Let $\struct {S, \preceq}$ be a complete lattice.
Let $\struct {T, \precsim}$ be an ordered set.
Let $g: S \to T$ be an all infima preserving mapping.
Then there exists a mapping $d: T \to S$ such that $\struct {g, d}$ is Galois connection and:
:$\forall t \in T: \map d t = \map \min {g^{-1} \sqbrk {t^\succsim} }$
where:
:$\min$ denotes the minimum
:$g^{-1} \sqbrk {t^\succsim}$ denotes the image of $t^\succsim$ under relation $g^{-1}$
:$t^\succsim$ denotes the upper closure of $t$
== Proof ==
=== Lemma 1 ===
Let us define a mapping $d: T \to S$ as:
:$\forall t \in T: \map d t := \map \inf {g^{-1} \sqbrk {t^\succsim} }$
=== Lemma 2 ===
Thus by Galois Connection is Expressed by Minimum:
:$\struct {g, d}$ is a Galois connection.
Thus by lemma $2$:
:$\forall t \in T: \map d t = \map \min {g^{-1} \sqbrk {t^\succsim} }$
== Sources ==
* 
* 
Category:Galois Connections
Category:All Infima Preserving Mapping is Upper Adjoint of Galois Connection
All Infima Preserving Mapping is Upper Adjoint of Galois Connection/Lemma 1

== Theorem ==
Let $\struct {S, \preceq}$ be a complete lattice.
Let $\struct {T, \precsim}$ be an ordered set.
Let $g: S \to T$ be an all infima preserving mapping.
Then:
:$g$ is an increasing mapping.
== Proof ==
Define a mapping $d: T \to S$:
:$\forall t \in T: \map d t := \map \inf {g^{-1} \sqbrk {t^\succsim} }$
Let $x, y \in S$ such that
:$x \preceq y$
By Upper Closure is Decreasing:
:$y^\succeq \subseteq x^\succeq$
By Infimum of Upper Closure of Element:
:$\map \inf {x^\succeq} = x$ and $\map \inf {y^\succeq} = y$
By definition of all infima preserving mapping:
:$g$ preserves the infimum on $x^\succeq$
and:
:$g$ preserves the infimum on $y^\succeq$
By definition of infimum on subset preserving mapping:
:$\map \inf {\map {g^\to} {x^\succeq} } = \map g x$ and $\map \inf {\map {g^\to} {y^\succeq} } = \map g y$
By Image of Subset under Mapping is Subset of Image:
:$\map {g^\to} {y^\succeq} \subseteq \map {g^\to} {x^\succeq}$
Thus by Infimum of Subset:
:$\map g x \precsim \map g y$
== Sources ==
* 
* 
Category:All Infima Preserving Mapping is Upper Adjoint of Galois Connection
All Infima Preserving Mapping is Upper Adjoint of Galois Connection/Lemma 2

== Theorem ==
Let $\struct {S, \preceq}$ be a complete lattice.
Let $\struct {T, \precsim}$ be an ordered set.
Let $g: S \to T$ be an all infima preserving mapping.
Let us define a mapping $d: T \to S$ as:
:$\forall t \in T: \map d t := \map \inf {g^{-1} \sqbrk {t^\succsim} }$
Then:
:$\forall t \in T: \map d t = \map \min {g^{-1} \sqbrk {t^\succsim} }$
== Proof ==
:$\forall t \in T: \map d t = \map \min {g^{-1} \sqbrk {t^\succsim} }$
Let $t \in T$.
By definition of $d$:
:$\map d t = \map \inf {g^{-1} \sqbrk {t^\succsim} }$
By Image of Inverse Image:
:$g \sqbrk {g^{-1} \sqbrk {t^\succsim} } \subseteq t^\succsim$
By Infimum of Subset and Infimum of Upper Closure of Element:
:$t = \map \inf {t^\succsim} \precsim \map \inf {g \sqbrk {g^{-1} \sqbrk {t^\succsim} } }$
By definition of upper closure of element:
:$\map \inf {g \sqbrk {g^{-1} \sqbrk {t^\succsim} } } \in t^\succsim$
By definition of complete lattice:
:$g^{-1} \sqbrk {t^\succsim}$ admits an infimum
By definitions mapping preserves the infimum:
:$\map \inf {g \sqbrk {g^{-1} \sqbrk {t^\succsim} } } = \map g {\map d t}$
Thus by definition of image of set:
:$\map d t \in g^{-1} \sqbrk {t^\succsim}$
Thus
:$\map d t = \map \min {g^{-1} \sqbrk {t^\succsim} }$
== Sources ==
* 
* 
Category:All Infima Preserving Mapping is Upper Adjoint of Galois Connection
All Normal Vectors of Simple Closed Contour Cannot Point into Interior/Lemma 1

== Lemma ==
Let $C$ be a simple closed contour in the complex plane $\C$ with parameterization $\gamma: \closedint a b \to \C$.
Let $t \in \openint a b$ such that $\gamma$ is complex-differentiable at $t$.
Let $r_0, R_0 \in \R_{>0}$ such that:
:for all $s \in \openint {t - R_0} {t + R_0}$ and for all $\epsilon \in \openint 0 {r_0}$: $\map \gamma s + \epsilon i \map {\gamma'} s \notin \Img C$, and $\map \gamma s - \epsilon i \map {\gamma'} s \notin \Img C$.
There exist $r, R \in \R_{>0}$ with $r \le r_0 , R \le R_0$ such that:
:$V_{r,R} = \set {\map \gamma s + \epsilon i \map {\gamma'} s : s \in \openint {t - R} {t + R}, \epsilon \in \openint {-r} r}$
is an open set in $\C$.
== Proof ==
This proof assumes that $\gamma '$ is continuously differentiable.
Let $x, y : \openint a b \to \R$ be the real functions defined by:
:$\map \gamma t = \map x t + i \map y t$
Let $f: \openint {t - R_0} {t + R_0} \times \openint {-r_0} {r_0} \to \R^2$ be defined by $\map f {s, \epsilon} = \tuple {\map x s - \epsilon \map {y'} s, \map y s + \epsilon \map {x'} s}$.
The Jacobian matrix of $f$ at $\tuple {s, \epsilon}$ is:
:$\mathbf J_f = \begin {pmatrix}
\map {x'} s - \epsilon \map {y' '} s & - \map {y'} s \\
\map {y'} s + \epsilon \map {x' '} s & \map {x'} s
\end {pmatrix}$
and the Jacobian determinant of $f$ evaluated at $\tuple {t, 0}$ is:
:$\map {\map \det {\mathbf J_f} } {t, 0} = \map {x'} t^2 + \map {y'} t^2$
As $\map {\gamma'} t \ne 0$ by definition of smooth path, it follows that $\map {\map \det {\mathbf J_f} } {t, 0} \ne 0$.
From Matrix is Nonsingular iff Determinant has Multiplicative Inverse, it follows that the Jacobian matrix of $f$ evaluated at $\tuple {t, 0}$ is nonsingular.
From Inverse Function Theorem for Real Functions, it follows that there exist open sets:
:$U \subseteq \openint {t - R_0} {t + R_0} \times \openint {-r_0} {r_0}, V \subseteq \R^2$ 
such that the restriction of $f$ to $U \times V$ is bijective, and the inverse $f^{-1}$ is continuous.
Equip $\R^2$ with the direct product norm.
Norms on Finite-Dimensional Real Vector Space are Equivalent shows that the direct product norm is equivalent to the Euclidean norm on $\R^2$.
From Open Sets in Vector Spaces with Equivalent Norms Coincide, it follows that $U$ is open in $\R^2$ equipped with the direct product norm.
Then there exist an open ball $\map {B_r} t \subseteq U$ with center $t$ of the form:
:$\map {B_r} t = \openint {t - r} {t + r} \times \openint {-r} r$
From Bijection is Open iff Inverse is Continuous, it follows that $f \sqbrk {\map {B_r} t}$ is open in $\R^2$.
Complex Plane is Homeomorphic to Real Plane shows that we can identify the complex plane $\C$ with the real plane $\R^2$ by the homeomorphism $\map \phi {x, y} = x + i y$
We identify $\map f {s, \epsilon}$ with $\map \phi {\map f {s, \epsilon} } = \map \gamma s + \epsilon i \map {\gamma '} s$.
Then $V_{r, r} := \phi \sqbrk {f \sqbrk {\map {B_r} t} }$ fulfils the claim of the theorem.
Category:Orientation of Complex Contour
All Suprema Preserving Mapping is Lower Adjoint of Galois Connection

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $\struct {T, \precsim}$ be a complete lattice.
Let $d: T \to S$ be all suprema preserving mapping.
Then there exists a mapping $g: S \to T$ such that $\tuple {g, d}$ is Galois connection and:
:$\forall s \in S: \map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$
where
:$\min$ denotes the maximum
:$d^{-1} \sqbrk {s^\preceq}$ denotes the image of $s^\preceq$ under relation $d^{-1}$
:$s^\preceq$ denotes the lower closure of $t$
== Proof ==
Define a mapping $d: T \to S$:
:$\forall s \in S: \map g s := \map \sup {d^{-1} \sqbrk {s^\preceq} }$
We will prove as lemma 1 that:
:$d$ is an increasing mapping.
Let $x, y \in T$ such that
:$x \precsim y$
By Lower Closure is Increasing:
:$x^\precsim \subseteq y^\precsim$
By Supremum of Lower Closure of Element:
:$\map \sup {x^\precsim} = x$ and $\map \sup {y^\precsim} = y$
By definition of mapping preserves all suprema:
:$d$ preserves the supremum on $x^\precsim$ and $d$ preserves the supremum on $y^\precsim$
By definition of mapping preserves the supremum on set:
:$\map \sup {\map {d^\to} {x^\precsim} } = \map d x$ and $\map \sup {\map {d^\to} {y^\precsim} } = \map d y$
By Image of Subset under Mapping is Subset of Image:
:$\map {d^\to} {x^\precsim} \subseteq \map {d^\to} {y^\precsim}$
Thus by Supremum of Subset:
:$\map d x \preceq \map d y$
This ends the proof of lemma 1.
We will prove as lemma 2 that
:$\forall s \in S: \map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$
Let $s \in S$.
By definition of $g$:
:$\map g s = \map \sup {d^{-1} \sqbrk {s^\preceq} }$
By Image of Inverse Image:
:$d \sqbrk {d^{-1} \sqbrk {s^\preceq} } \subseteq s^\preceq$
By Supremum of Subset and Supremum of Lower Closure of Element:
:$\map \sup {d \sqbrk {d^{-1} \sqbrk {s^\preceq} } } \preceq \map \sup {s^\preceq} = s$
By definition of lower closure of element:
:$\map \sup {d \sqbrk {d^{-1} \sqbrk {s^\preceq} } } \in s^\preceq$
By definition of complete lattice:
:$d^{-1} \sqbrk {s^\preceq}$ admits a supremum
By definition of mapping preserves the supremum:
:$\map \sup {d \sqbrk {d^{-1} \sqbrk {s^\preceq} } } = \map d {\map g s}$
Thus by definition of image of set:
:$\map g s \in d^{-1} \sqbrk {s^\preceq}$
Thus
:$\map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$
This ends the proof of lemma 2.
Thus by Galois Connection is Expressed by Maximum:
:$\tuple {g, d}$ is a Galois connection.
Thus by lemma 2:
:$\forall s \in S: \map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$
== Sources ==
* 
* 
Category:Galois Connections
Almost-Everywhere Equality Relation for Lebesgue Space is Equivalence Relation

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space, and let $p \in \closedint 1 \infty$.
Let $\map {\mathcal L^p} {X, \Sigma, \mu}$ be the Lebesgue $p$-space of $\struct {X, \Sigma, \mu}$.
We define the $\mu$-almost-everywhere equality relation $\sim_\mu$ on $\map {\mathcal L^p} {X, \Sigma, \mu}$ by: 
:$\forall f, g \in \mathcal L^p: f \sim_\mu g$ $\norm {f - g}_p = 0$
where $\norm {\, \cdot \,}_p$ is the $p$-seminorm.
Then $\sim_\mu$ is an equivalence relation.
== Proof 1 ==
== Proof 2 ==
Category:Definitions/Functional Analysis
Category:Definitions/Measure Theory
Category:Definitions/Lebesgue Spaces
Category:Definitions/Almost-Everywhere Equality Relation
Almost-Everywhere Equality Relation for Lebesgue Space is Equivalence Relation/Proof 1

== Theorem ==
== Proof ==
Let $f, g, h \in \LL^p$.
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
From P-Seminorm of Function Zero iff A.E. Zero, we have:
:$\norm {f - f}_p = 0$
Therefore:
:$f \sim_\mu f$
Hence $\sim_\mu$ is a reflexive relation.
=== Symmetry ===
Suppose:
:$f \sim_\mu g$
Then:
:$\norm {f - g}_p = 0$
Then:
:$\norm {g - f}_p = 0$
Therefore:
:$g \sim_\mu f$
Hence $\sim_\mu$ is a symmetric relation.
=== Transitivity ===
Suppose:
:$f \sim_\mu g$
and:
:$g \sim_\mu h$
Then:
:$\norm {f - g}_p = 0$
and:
:$\norm {g - h}_p = 0$
Then:
:$\norm {f - h}_p = 0$
Hence $\sim_\mu$ is a transitive relation.
All the criteria are therefore seen to hold for $\sim_\mu$ to be an equivalence relation.
== Sources ==
* : $12.5 \ \text {(ii)}$
Category:Definitions/Functional Analysis
Category:Definitions/Measure Theory
Category:Definitions/Lebesgue Spaces
Category:Definitions/Almost-Everywhere Equality Relation
Almost-Everywhere Equality Relation for Lebesgue Space is Equivalence Relation/Proof 2

== Theorem ==
== Proof ==
Let $f, g, h \in \LL^p$.
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
By Equality is Reflexive, we have:
:$f \sim_\mu f$ $\map f x = \map f x$ for $\mu$-almost all $x \in X$.
Therefore:
:$f \sim_\mu f$
Hence $\sim_\mu$ is a reflexive relation.
=== Symmetry ===
Suppose:
:$f \sim_\mu g$
Then:
:$\map f x = \map g x$ for $\mu$-almost all $x \in X$.
Then by Equality is Symmetric
:$\map g x = \map f x$ for $\mu$-almost all $x \in X$.
Therefore:
:$g \sim_\mu f$
Hence $\sim_\mu$ is a symmetric relation.
=== Transitivity ===
Suppose:
:$f \sim_\mu g$
and:
:$g \sim_\mu h$
Then:
:$\map f x = \map g x$ for $\mu$-almost all $x \in X$.
and:
:$\map g x = \map h x$ for $\mu$-almost all $x \in X$.
Then by Equality is Transitive:
:$\map f x = \map h x$ for $\mu$-almost all $x \in X$.
Hence $\sim_\mu$ is a transitive relation.
All the criteria are therefore seen to hold for $\sim_\mu$ to be an equivalence relation.
Category:Definitions/Functional Analysis
Category:Definitions/Measure Theory
Category:Definitions/Lebesgue Spaces
Category:Definitions/Almost-Everywhere Equality Relation
Almost-Everywhere Equality Relation for Measurable Sets is Equivalence Relation

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
We define the $\mu$-almost-everywhere equality relation $\sim_\mu$ on $\Sigma$ by: 
:$A \sim_\mu B$ $\map \mu {A \symdif B} = 0$
where $\symdif$ denotes set symmetric difference.
Then:
:$\sim_\mu$ is an equivalence relation.
== Proof ==
== Sources ==
* Chapter $2$: Isomorphism, Conjugacy, and Spectral Isomorphism
Category:Definitions/Almost-Everywhere Equality Relation