nyuuzyou/wikis huggingface dataset:
Left Inverse and Right Inverse is Inverse

== Theorem ==
Let $\struct {S, \circ}$ be a monoid with identity element $e_S$.
Let $x \in S$ such that $x$ has both a left inverse and a right inverse. That is:
:$\exists x_L \in S: x_L \circ x = e_S$
:$\exists x_R \in S: x \circ x_R = e_S$
Then $x_L = x_R$, that is, $x$ has an inverse.
Furthermore, that element is the only inverse (both right and left) for $x$
== Proof ==
We note that as $\struct {S, \circ}$ is a monoid, $\circ$ is associative by definition.
Its uniqueness comes from Inverse in Monoid is Unique.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Exercise $4.9 \ \text{(a)}$
Category:Monoids
Category:Inverse Elements
Left Inverse for All is Right Inverse

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup with a left identity $e_L$ such that:
:$\forall x \in S: \exists x_L: x_L \circ x = e_L$
That is, every element of $S$ has a left inverse with respect to the left identity.
Then $x \circ x_L = e_L$, that is, $x_L$ is also a right inverse with respect to the left identity.
== Proof ==
Let $y = x \circ x_L$. Then:
So $x \circ x_L = e_L$, and $x_L$ behaves as a right inverse as well as a left inverse with respect to the left identity.
== Also see ==
* Right Inverse for All is Left Inverse
* Left Identity while exists Left Inverse for All is Identity
* Right Identity while exists Right Inverse for All is Identity
== Sources ==
* : $\S 1.4$
* : Chapter $5$: Semigroups: Exercise $5$
Category:Semigroups
Category:Inverse Elements
Left Limit of Distribution Function of Finite Borel Measure

== Theorem ==
Let $\mu$ be a finite Borel measure.
Let $F_\mu$ be the distribution function of $\mu$.
Then we have:
:$\ds \lim_{y \mathop \to x^-} \map {F_\mu} y = \map \mu {\openint {-\infty} x}$ 
== Proof ==
From Monotonic Sequence Characterization of Left Limit of Function, the claim is equivalent to:
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n < x$ for each $n$, that converge to $x$ we have:
::$\map {F_\mu} {x_n} \to \map \mu {\openint {-\infty} x}$
That is:
:$\map \mu {\hointl {-\infty} {x_n} } \to \map \mu {\openint {-\infty} x}$
Let $\sequence {x_n}_{n \mathop \in \N}$ be a monotone sequence with $x_n < x$ for each $n \in \N$, satisfying $x_n \to x$. 
From Limit of Countable Union of Unbounded Below Closed Intervals, we have:
:$\hointl {-\infty} {x_n} \uparrow \openint {-\infty} x$
where $\uparrow$ denotes the limit of an increasing sequence of sets.
From Measure of Limit of Increasing Sequence of Measurable Sets, we therefore have:
:$\map \mu {\hointl {-\infty} {x_n} } \to \map \mu {\openint {-\infty} x}$
Invoking Monotonic Sequence Characterization of Left Limit of Function, we obtain:
:$\ds \lim_{y \mathop \to x^-} \map {F_\mu} y = \map \mu {\openint {-\infty} x}$
Category:Distribution Function of Finite Borel Measure
Left Module Does Not Necessarily Induce Right Module over Ring

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct{G, +_G, \circ}$ be a left module over $\struct {R, +_R, \times_R}$.
Let $\circ' : G \times R \to G$ be the binary operation defined by:
:$\forall \lambda \in R: \forall x \in G: x \circ' \lambda = \lambda \circ x$
Then $\struct{G, +_G, \circ'}$ is not necessarily a right module over $\struct {R, +_R, \times_R}$
== Proof ==
Proof by Counterexample:
Let $\struct {S, +, \times}$ be a ring with unity
Let $\struct {\map {\MM_S} 2, +, \times}$ denote the ring of square matrices of order $2$ over $S$.
From Ring of Square Matrices over Ring is Ring, $\struct {\map {\MM_S} 2, +, \times}$ is a ring.
Let $G = \set {\begin{bmatrix}
x & 0 \\
y & 0
\end{bmatrix} : x, y \in S }$
=== Lemma ===
From Left Ideal is Left Module over Ring, $\struct {G, +, \times}$ is a left module over $\struct {\map {\MM_S} 2, +, \times}$.
Let $R = \map {\MM_S} 2$
Let $\mathbf A = \begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}, 
\mathbf B = \begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix}, 
\mathbf C = \begin{bmatrix}
1 & 0 \\
1 & 0
\end{bmatrix}$
Then:
and:
Hence for $\mathbf A, \mathbf B \in R, \mathbf C \in G$:
:$ \paren {\mathbf A \times \mathbf B} \times \mathbf C \neq \paren {\mathbf B \times \mathbf A} \times \mathbf C$
Let $\circ : G \times R \to R$ be the binary operation defined by:
:$\forall \mathbf \Lambda \in R: \forall \mathbf X \in G: \mathbf X \circ \mathbf \Lambda = \mathbf \Lambda \times \mathbf X$
From Left Module induces Right Module over same Ring iff Actions are Commutative, $\struct {G, +, \circ}$ is not a right module over $\struct {R, +, \times}$
== Also see ==
* Right Module Does Not Necessarily Induce Left Module over Ring
Category:Module Theory
Category:Ring Theory
Category:Left Module Does Not Necessarily Induce Right Module over Ring
Left Module Does Not Necessarily Induce Right Module over Ring/Lemma

== Theorem ==
Let $\struct {S, +, \times}$ be a ring with unity.
Let $\struct {\map {\MM_S} 2, +, \times}$ denote the ring of square matrices of order $2$ over $S$.
Let:
:$G = \set {\begin{bmatrix}
x & 0 \\
y & 0
\end{bmatrix} : x, y \in S }$
Then:
:$G$ is a left ideal of $\struct {\map {\MM_S} 2, +, \times}$.
== Proof ==
From Test for Left Ideal, the following need to be proved:
:$(1): \quad G \ne \O$
:$(2): \quad \forall \mathop {\mathbf X}, \mathop {\mathbf Y} \in G: \mathbf X + \paren {-\mathbf Y} \in G$
:$(3): \quad \forall \mathop{\mathbf J} \in G, \mathop {\mathbf R} \in \map {\MM_S} 2: \mathbf R \times \mathbf J \in G$
=== Condition $(1): \quad G \ne \O$ ===
By definition of $G$:
:$\quad \begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix} \in G$
=== Condition $(2): \quad \forall \mathop {\mathbf X}, \mathop{\mathbf Y} \in G: \mathbf X + \paren {-\mathbf Y} \in G$ ===
Let:
:$\quad \mathbf X = \begin{bmatrix}
x_1 & 0 \\
x_2 & 0
\end{bmatrix}, \quad \mathbf Y = \begin{bmatrix}
y_1 & 0 \\
y_2 & 0
\end{bmatrix} \in G$
Then:
:$\quad \mathbf X - \mathbf Y = \begin{bmatrix}
x_1 - y_1 & 0 \\
x_2 - y_2 & 0
\end{bmatrix} \in G$
=== Condition $(3): \quad \forall \mathop{\mathbf J} \in G, \mathop{\mathbf R} \in \map {\MM_S} 2: \mathbf R \times \mathbf J \in G$ ===
Let:
:$\quad \mathbf J = \begin{bmatrix}
j_1 & 0 \\
j_2 & 0
\end{bmatrix} \in G, \quad \mathbf R = \begin{bmatrix}
r_{1 1} & r_{2 1} \\
r_{1 2} & r_{2 2}
\end{bmatrix} \in \map {\MM_S} 2$
Then:
:$\quad \mathbf R \times \mathbf J = \begin{bmatrix}
r_{1 1} \times j_1 + r_{2 1} \times j_2 & 0 \\
r_{1 2} \times j_1 + r_{2 2} \times j_2 & 0
\end{bmatrix} \in G$
Category:Left Module Does Not Necessarily Induce Right Module over Ring
Left Module induces Right Module over same Ring iff Actions are Commutative

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ}$ be a left module over $\struct {R, +_R, \times_R}$.
Let $\circ': G \times R \to G$ be the binary operation defined by:
:$\forall \lambda \in R: \forall x \in G: x \circ' \lambda = \lambda \circ x$
Then $\struct {G, +_G, \circ'}$ is a right module over $\struct {R, +_R, \times_R}$ :
:$\forall \lambda, \mu \in R: \forall x \in G: \paren {\lambda \times_R \mu} \circ x = \paren {\mu \times_R \lambda} \circ x$ 
== Proof ==
=== Necessary Condition ===
Let $\struct {G, +_G, \circ'}$ be a right module over $\struct {R, +_R, \times_R}$.
Then:
=== Sufficient Condition ===
Let the scalar multiplication $\circ$ satisfy:
:$\forall \lambda, \mu \in R: \forall x \in G: \paren {\lambda \times_R \mu} \circ x = \paren {\mu \times_R \lambda} \circ x$
It needs to be shown that $\struct {G, +_G, \circ'}$ satisfies the right module axioms.
==== ====
Let $\lambda, \mu \in R, x \in G$.
Then:
==== ====
Let $\lambda \in R, x, y \in G$.
Then:
==== ====
Let $\lambda, \mu \in R, x \in G$.
Then:
== Also see ==
* Right Module induces Left Module over same Ring iff Actions are Commutative
Category:Module Theory
Category:Ring Theory
Left Module over Commutative Ring induces Bimodule

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a commutative ring.
Let $\struct{G, +_G, \circ}$ be a left module over $\struct {R, +_R, \times_R}$.
Let $\circ' : G \times R \to G$ be the binary operation defined by:
:$\forall \lambda \in R: \forall x \in G: x \circ' \lambda = \lambda \circ x$
Then $\struct{G, +_G, \circ, \circ'}$ is a bimodule over $\struct {R, +_R, \times_R}$.
== Proof ==
From Left Module over Commutative Ring induces Right Module, $\struct{G, +_G, \circ'}$ is a right module.
Let $\lambda, \mu \in R$ and $x \in G$.
Then:
Hence $\struct {G, +_G, \circ, \circ'}$ is a bimodule over $\struct {R, +_R, \times_R}$ by definition.
== Also see ==
* Right Module over Commutative Ring induces Bimodule
== Sources ==
* : Chapter $4$: Rings and Modules: $\S 4.1$: The Definitions Recalled
Category:Module Theory
Category:Ring Theory
Left Module over Commutative Ring induces Right Module

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a commutative ring.
Let $\struct {G, +_G, \circ}$ be a left module over $\struct {R, +_R, \times_R}$.
Let $\circ' : G \times R \to G$ be the binary operation defined by:
:$\forall \lambda \in R: \forall x \in G: x \circ' \lambda = \lambda \circ x$
Then $\struct {G, +_G, \circ'}$ is a right module over $\struct {R, +_R, \times_R}$.
== Proof ==
From Ring is Commutative iff Opposite Ring is Itself, $\struct {R, +_R, \times_R}$ is its own opposite ring.
From Left Module over Ring Induces Right Module over Opposite Ring, $\struct {G, +_G, \circ'}$ is a right module over $\struct {R, +_R, \times_R}$.
== Also see ==
* Right Module over Commutative Ring induces Left Module
== Sources ==
* : Chapter $4$: Rings and Modules: $\S 4.1$: The Definitions Recalled
Category:Module Theory
Category:Ring Theory
Left Module over Ring Induces Right Module over Opposite Ring

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {R, +_R, *_R}$ be the opposite ring of $\struct {R, +_R, \times_R}$.
Let $\struct{G, +_G, \circ}$ be a left module over $\struct {R, +_R, \times_R}$.
Let $\circ' : G \times R \to G$ be the binary operation defined by:
:$\forall \lambda \in R: \forall x \in G: x \circ’ \lambda = \lambda \circ x$
Then $\struct{G, +_G, \circ'}$ is a right module over $\struct {R, +_R, *_R}$.
== Proof ==
It is shown that $\struct{G, +_G, \circ'}$ satisfies the right module axioms.
By definition of the opposite ring:
:$\forall x, y \in R: x *_R y = y \times_R x$
=== ===
Let $\lambda \in R$ and $x, y \in G$.
=== ===
Let $\lambda, \mu \in R$ and $x \in G$.
=== ===
Let $\lambda, \mu \in S$ and $x \in G$.
== Also see ==
* Right Module over Ring Induces Left Module over Opposite Ring
== Sources ==
* : Chapter $4$: Rings and Modules: $\S 4.1$: The Definitions Recalled
Category:Module Theory
Category:Ring Theory
Left Operation has no Left Identities

== Theorem ==
Let $S$ be a set with more than $1$ element.
Let $\struct {S, \gets}$ be an algebraic structure in which the operation $\gets$ is the left operation.
Then $\struct {S, \gets}$ has no left identities.
== Proof ==
From Element under Left Operation is Right Identity, every element of $\struct {S, \gets}$ is a right identity.
Because there are at least $2$ elements in $\struct {S, \gets}$, it follows that $\struct {S, \gets}$ has more than one right identity.
From More than one Right Identity then no Left Identity, it follows that $\struct {S, \gets}$ has no left identity.
== Also see ==
* Right Operation has no Right Identities
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Exercise $4.3 \ \text{(b)}$
Category:Identity Elements
Category:Left Operation
Left Operation is Anticommutative

== Theorem ==
The left operation is anticommutative:
:$\forall x, y: x \leftarrow y = y \leftarrow x \iff x = y$
== Proof ==
Immediate from the definition of the left operation.
== Also see ==
* Right Operation is Anticommutative
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.17 \ \text{(a)}$
Category:Left Operation
Category:Examples of Anticommutativity
Left Operation is Associative

== Theorem ==
The left operation is associative:
:$\forall x, y, z: \paren {x \gets y} \gets z = x \gets \paren {y \gets z}$
== Proof ==
== Also see ==
* Right Operation is Associative
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.4$
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.9$
Category:Left Operation
Category:Examples of Associative Operations
Left Operation is Closed for All Subsets

== Theorem ==
Let $S$ be a set.
Let $\leftarrow$ be the left operation on $S$.
That is:
:$\forall x, y \in S: x \leftarrow y = x$
Let $\powerset S$ be the power set of $S$.
Then for all $T \in \powerset S$, $\leftarrow$ is closed on $T$.
That is, for all $T \in \powerset S$:
:$\struct {T, \leftarrow}$ is a subsemigroup of $\struct {S, \leftarrow}$.
== Proof ==
From Structure under Left Operation is Semigroup we have that $\struct {S, \leftarrow}$ is a semigroup, whatever the nature of $S$.
Let $T \in \powerset S$.
Then:
:From Structure under Left Operation is Semigroup, $\struct {T, \leftarrow}$ is a semigroup, and therefore a subsemigroup of $\struct {S, \leftarrow}$.
This applies whatever $S$ is and whatever the subset $T$ is.
== Also see ==
* Right Operation is Closed for All Subsets
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.1$
Category:Left Operation
Left Operation is Distributive over Idempotent Operation

== Theorem ==
Let $\struct {S, \circ, \leftarrow}$ be an algebraic structure where:
:$\leftarrow$ is the left operation
:$\circ$ is any arbitrary binary operation.
Then:
:$\leftarrow$ is distributive over $\circ$
:
:$\circ$ is idempotent.
== Proof ==
From Left Operation is Right Distributive over All Operations:
:$\forall a, b, c \in S: \paren {a \circ b} \leftarrow c = \paren {a \leftarrow c} \circ \paren {b \leftarrow c}$
for all binary operations $\circ$.
It remains to show that $\leftarrow$ is left distributive over $\circ$ $\circ$ is idempotent.
=== Necessary Condition ===
Let $\circ$ be idempotent.
Then:
Thus $\leftarrow$ is left distributive over $\circ$.
=== Sufficient Condition ===
Let $\leftarrow$ be left distributive over $\circ$.
Let $a \in S$ be arbitrary.
Then:
Hence $\circ$ is idempotent.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.23 \ \text{(b)}$
Category:Left Operation
Category:Distributive Operations
Category:Idempotence
Left Operation is Entropic

== Theorem ==
The left operation is entropic:
:$\forall a, b, c, d: \paren {a \gets b} \gets \paren {c \gets d} = \paren {a \gets c} \gets \paren {b \gets d}$
== Proof ==
== Also see ==
* Right Operation is Entropic
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.12 \ \text{(c)}$
Category:Left Operation
Category:Examples of Entropic Operations
Left Operation is Idempotent

== Theorem ==
The left operation is idempotent:
:$\forall x: x \leftarrow x = x$
== Proof ==
Immediate from the definition of the left operation.
== Also see ==
* Right Operation is Idempotent
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.17 \ \text{(a)}$
Category:Left Operation
Category:Examples of Idempotence
Left Operation is Right Distributive over All Operations

== Theorem ==
Let $\struct {S, \circ, \leftarrow}$ be an algebraic structure where:
:$\leftarrow$ is the left operation
:$\circ$ is any arbitrary binary operation.
Then $\leftarrow$ is right distributive over $\circ$.
== Proof ==
The result follows by definition of right distributivity.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.23 \ \text{(a)}$
Category:Left Operation
Left Operation is not Commutative

== Theorem ==
Let $S$ be a finite set.
Let $\leftarrow$ denote the left operation on $S$.
Then $\leftarrow$ is not commutative on $S$ unless $S$ is a singleton.
== Proof ==
Let $S$ be a singleton, $S = \set s$, say.
Then:
:$s \leftarrow s = s$
and so $\leftarrow$ is trivially commutative on $S$
Otherwise, $\exists s, t \in S$ such that $s \ne t$.
Then:
:$s \leftarrow t = s$
but:
:$t \leftarrow s = t$
and the result follows by definition of commutative operation.
== Also see ==
* Left Operation is Anticommutative
* Right Operation is not Commutative
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.4$
Category:Left Operation
Category:Examples of Commutative Operations
Left Regular Representation is Group Action

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $*: G \times G \to G$ be the operation:
:$\forall g, h \in G: g * h = \map {\lambda_g} h$
where $\lambda_g$ is the left regular representation of $G$ with respect to $g$.
Then $*$ is a group action.
== Proof ==
The group action axioms are investigated in turn.
Let $g, h, a \in G$.
Thus:
demonstrating that holds.
Then:
demonstrating that holds.
== Also defined as ==
Some sources denote a mapping by placing the symbol defining that mapping on the left of its operand.
Thus under such a convention:
: $\lambda_g \left({a}\right)$ is written $a \lambda_g$
and:
: $g * a$ is written $a * g$ (or even $a g$ in sources which do not place a high regard on clarity).
Thus for the left regular representation to be defined as being a group action there exists the need to use it on the inverse:
Such a treatment can be found in .
As it runs contrary to the conventions used on , beyond its mention here it will not be used.
However, this example indicates how the arbitrary nature of notational conventions can cause the details of results to be equally arbitrarily dependent upon the convention used.
== Also see ==
* Right Regular Representation by Inverse is Group Action
* Group Acts on Itself
== Sources ==
* : $\S 3.3$: Group actions and coset decompositions: Examples of group actions: $\text{(iv)}$
Category:Group Actions
Left Regular Representation is Transitive Group Action

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $*: G \times G \to G$ be the group action:
:$\forall g, h \in G: g * h = \map {\lambda_g} h$
where $\lambda_g$ is the left regular representation of $G$ with respect to $g$.
Then $*$ is a transitive group action.
== Proof ==
Let $g, h \in G$.
Then:
$h$ is arbitrary, therefore the above holds for all $h \in G$.
By definition of orbit:
:$\Orb h = \set {t \in G: \exists g \in G: g * h = t}$
That is:
:$\Orb h = G$
Hence the result by definition of transitive group action.
== Also see ==
* Right Regular Representation by Inverse is Transitive Group Action
== Sources ==
* : $\S 3.3$: Group actions and coset decompositions: Examples of group actions: $\text{(iv)}$
Category:Transitive Group Actions
Category:Regular Representations
Left Regular Representation of Subset Product

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $T \subseteq S$ be a subset of $S$.
Let $\lambda_a: S \to S$ be the left regular representation of $S$ with respect to $a$.
Then:
:$\lambda_a \sqbrk T = \set a \circ T = a \circ T$
where $a \circ T$ denotes subset product with a singleton.
== Proof ==
== Sources ==
* : $\S 41$. Multiplication of subsets of a group
Category:Regular Representations
Left Regular Representation wrt Left Cancellable Element on Finite Semigroup is Bijection

== Theorem ==
Let $\struct {S, \circ}$ be a finite semigroup.
Let $a \in S$ be left cancellable.
Then the left regular representation $\lambda_a$ of $\struct {S, \circ}$ with respect to $a$ is a bijection.
== Proof ==
By Left Cancellable iff Left Regular Representation Injective, $\lambda_a$ is an injection.
By hypothesis, $S$ is finite.
From Injection from Finite Set to Itself is Surjection, $\lambda_a$ is a surjection.
Thus $\lambda_a$ is injective and surjective, and therefore a bijection.
== Also see ==
* Right Regular Representation wrt Right Cancellable Element on Finite Semigroup is Bijection
Category:Semigroups
Category:Regular Representations
Category:Cancellability
Left Self-Distributive Operation with Right Identity is Idempotent

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be left self-distributive.
Let $\struct {S, \circ}$ have a right identity.
Then $\circ$ is an idempotent operation.
== Proof ==
Let the right identity of $\struct {S, \circ}$ be $e_R$.
We have:
The result follows by definition of idempotent operation.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.23 \ \text{(c)}$
Category:Idempotence
Category:Self-Distributive Operations
Category:Identity Elements
Left Shift Operator is Linear Mapping

== Theorem ==
Let $X = Y = \ell^2$ be 2-sequence spaces over real numbers.
Let $L : X \to Y$ be the left shift operator.
Then $L$ is a linear mapping.
== Proof ==
Let $x = \tuple {x_1, x_2,x_3, \ldots}, y = \tuple {y_1, y_2, y_3, \ldots} \in \ell^2$
Let $\alpha \in \R$.
=== Distributivity ===
=== Positive homogenity ===
== Sources ==
* : Chapter $\S 2.1$: Continuous and linear maps. Linear transformations
Category:Linear Transformations
Left Shift Operator on 2-Sequence Space is Continuous

== Theorem ==
Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the $2$-sequence space with $2$-norm.
Let $L : \ell^2 \to \ell^2$ be the left shift operator.
Then $L$ is continuous on $\struct {\ell^2, \norm {\, \cdot \,}_2}$.
== Proof ==
Let $\sequence {a_n}_{n \mathop \in \N} = \tuple {a_1, a_2, a_3, \ldots}$ be a $2$-sequence.
By Continuity of Linear Transformation between Normed Vector Spaces, $L$ is continuous in $\struct {\ell^2, \norm {\, \cdot \,}_2}$.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$
Category:Operator Theory
Category:Continuous Mappings
Left Zero Divisor of Commutative Ring is Right Zero Divisor

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring.
Let $x \in R$ be a left zero divisor of $R$.
Then $x$ is also a right zero divisor of $R$.
== Proof ==
Let $x \in R$ be a left zero divisor of $R$.
Let $0_R$ denote the zero of $R$.
Then:
Hence $x$ is a right zero divisor of $R$ by definition.
== Sources ==
* 
* 
Category:Zero Divisors
Left and Right Coset Spaces are Equivalent

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$.
Let:
:$x H$ denote the left coset of $H$ by $x$
:$H y$ denote the right coset of $H$ by $y$.
Then:
:$\order {\set {x H: x \in G} } = \order {\set {H y: y \in G} }$
That is:
:The number of right cosets is the same as the number of left cosets of $G$ with respect to $H$.
:The left and right coset spaces are equivalent.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Index of Subgroup, which is the number of left (or right) cosets of a subgroup.
* Subgroup is Normal iff Left Cosets are Right Cosets
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 37 \beta$
Category:Cosets
Category:Left and Right Coset Spaces are Equivalent
Left and Right Coset Spaces are Equivalent/Proof 1

== Theorem ==
== Proof ==
Let there be exactly $r$ different left cosets of $H$ in $G$.
Let a complete repetition-free list of these left cosets be:
:$a_1 H, a_2 H, a_3 H \ldots, a_r H: a_1, a_2, \ldots, a_r \in G$
From Left Coset Space forms Partition, every element of $G$ is contained in exactly one of the left cosets.
Let $x \in G$.
Then, for $1 \le i \le r$:
Since $x^{-1} \in a_i H$ is true for precisely one value of $i$, it follows that $x \in H a_i^{-1}$ is also true for precisely that value of $i$.
So there are exactly $r$ different right cosets of $H$ in $G$, and a complete repetition-free list of these is:
:$H a_1^{-1}, H a_2^{-1}, H a_3^{-1}, \ldots, H a_r^{-1}$
The result follows.
== Sources ==
* : $\S 43.2$: Lagrange's theorem
Category:Left and Right Coset Spaces are Equivalent