nyuuzyou/wikis huggingface dataset:
Riesz-Markov-Kakutani Representation Theorem/Notation

== Riesz-Markov-Kakutani Representation Theorem ==
== Notation used ==
The purpose of this page is to document the notation used for the proof of the Riesz-Markov-Kakutani Representation Theorem.
For an open set $V \in \tau$ and a mapping $f \in \map {C_c} X$:
:$f \prec V \iff \supp f \subset V$
where $\supp f$ denotes the support of $f$.
For a compact set $K \subset X$ and a mapping $f \in \map {C_c} X$:
:$K \prec f \iff \forall x \in K: \map f x = 1$
Category:Riesz-Markov-Kakutani Representation Theorem
Riesz Representation Theorem (Hilbert Spaces)

== Theorem ==
Let $H$ be a Hilbert space.
Let $L$ be a bounded linear functional on $H$.
Then there is a unique $h_0 \in H$ such that:
:$\forall h \in H: L h = \innerprod h {h_0}$
== Proof ==
If $L \equiv 0$ identically, then:
:$L h = 0 = \innerprod h 0$
and the theorem holds.
By Kernel of Bounded Linear Transformation is Closed Linear Subspace:
:$M := \ker L$ is a closed linear subspace of $H$.
Then we can decompose $H$ as a direct sum:
:$H \cong M \oplus M^\perp$
where $M^\perp$ denotes the orthocomplement of $M$.
From:
:$L \not \equiv 0$
we have:
:$M^\perp \ne \set 0$
Choose a $z \in M^\perp$ with norm $1$.
By linearity of $L$, for any $h \in H$:
So:
:$z L h - h L z \in \ker L = M$
Then:
Thus:
:$L h = \innerprod h {h_0}$
for $h_0 = z \paren {L z}^*$.
To show uniqueness, assume $h_0$ and $h_1$ both satisfy the above equation for all $h \in H$:
The result follows from setting $h = h_0 - h_1$ and invoking the positive definiteness of the inner product.
== Examples ==
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 3.$ The Riesz Representation Theorem: $3.4$ The Riesz Representation Theorem
Category:Hilbert Spaces
Category:Bounded Linear Functionals
Category:Representation Theorems
Riesz Representation Theorem (Hilbert Spaces)/Examples/L2 Space

== Example of Use of Riesz Representation Theorem (Hilbert Spaces) ==
Let $\struct{ X, \Sigma, \mu }$ be a measure space.
Let $\map {L^2} \mu$ be the associated $L^2$ space.
Let $F: \map {L^2} \mu \to \GF$ be a bounded linear functional.
Then there exists a unique $f_0 \in \map {L^2} \mu$ such that:
:$\ds \forall f \in \map {L^2} \mu: \map F f = \int f \overline{f_0} \rd \mu$
== Proof ==
By $L^2$ Space forms Hilbert Space, $\map {L^2} \mu$ is a Hilbert space with the $L^2$ inner product $\innerprod \cdot \cdot$.
Hence, the Riesz Representation Theorem (Hilbert Spaces) applies, so that there exists a unique $f_0 \in \map {L^2} \mu$ such that:
:$\forall f \in \map {L^2} \mu: \map F f = \innerprod f {f_0}$
Unpacking the definition of the $L^2$ inner product, the result follows.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 3.$ The Riesz Representation Theorem: Corollary $3.5$
Category:Examples of Use of Riesz Representation Theorem (Hilbert Spaces)
Category:Lp Spaces
Riesz Representation Theorem (Hilbert Spaces)/Examples/Space of Square Summable Mappings

== Example of Use of Riesz Representation Theorem (Hilbert Spaces) ==
Let $\map {\ell^2} \N$ be the space of square summable mappings on $\N$.
Let $N \in \N$.
Let $\GF$ be a subfield of $\C$.
Let $L_N: \map {\ell^2} \N \to \GF$ be defined by:
:$\map {L_N} {\sequence{ a_n } } := a_N$
Let $\delta_N \in \map {\ell^2} \N$ be given by:
:$\forall n \in \N: \paren{ \delta_N }_n = \begin{cases}
1 & n = N \\
0 & n \ne N
\end{cases}$
Then:
:$\forall a \in \map {\ell^2} \N: \map {L_N} a = \innerprod a {\delta_N}$
where $\innerprod \cdot \cdot: \map {\ell^2} \N \times \map {\ell^2} \N \to \GF$ denotes the inner product on $\map {\ell^2} \N$.
== Proof ==
By Space of Square Summable Mappings is Hilbert Space, $\map {\ell^2} \N$ is a Hilbert space.
Since for all $a \in \map {\ell^2} \N$:
it follows that $L_N$ is a bounded linear functional.
Therefore, the Riesz Representation Theorem (Hilbert Spaces) applies.
Hence there exists a unique $b \in \map {\ell^2} \N$ such that:
:$\forall a \in \map {\ell^2} \N: \map {L_N} a = \innerprod a b$
Let us check that $\delta_N$ fulfils the claim:
The result follows.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 3.$ The Riesz Representation Theorem: Exercise $2$
Category:Examples of Use of Riesz Representation Theorem (Hilbert Spaces)
Right-Hand Derivative not Limit of Derivative from Right

== Theorem ==
Let $f$ be a real function.
Let the right-hand derivative $f'_+$ of $f$ exist.
Then it is not necessarily the case that:
:$\map {f'_+} a$
is the same thing as:
:$\map {f'} {a^+}$
== Proof ==
By definition:
:$\map {f'_+} a := \ds \lim_{h \mathop \to 0} \dfrac {\map f {a + h} - \map f a} h$
while:
:$\map {f'} {a^+} := \ds \lim_{x \mathop \to a^+} \map {f'} x$
Let:
:$\map f x = \begin {cases} x^2 \sin \dfrac 1 x & : x \ne 0 \\ 0 & : x = 0 \end {cases}$.
Then:
but:
== Also see ==
* Extendability Theorem for Derivatives Continuous on Open Intervals
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.2$ Derivatives: Example $\text E$
Category:Differential Calculus
Right-Hand Differentiable Function is Right-Continuous

== Theorem ==
Let $f$ be a real function defined on an interval $I$.
Let $a$ be a point in $I$ where $f$ is right-hand differentiable.
Then $f$ is right-continuous at $a$.
== Proof ==
By hypothesis, $\map {f'_+} a$ exists.
First we note that $a$ cannot be the right hand end point of $I$ because values in $I$ greater than $a$ need to exist for $\map {f'_+} a$ to exist.
We form the following expression:
:$\ds \lim_{x \mathop \to a^+} \paren {\map f x - \map f a}$
We need to show that it is defined and to find its value.
We find:
Note that this proves that $\ds \lim_{x \mathop \to a^+} \paren {\map f x - \map f a}$ exists.
We continue by manipulating the result above:
	
which means that $f$ is right-continuous at $a$.
Category:Continuous Real Functions
Category:Differentiable Real Functions
Right-Limits of Filtration of Sigma-Algebra form Filtration

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space. 
Let $\sequence {\FF_t}_{t \ge 0}$ be a continuous-time filtration of $\Sigma$. 
For each $t \in \hointr 0 \infty$, let $\FF_{t^+}$ be the right-limit of $\sequence {\FF_t}_{t \ge 0}$ at $t$. 
Then $\sequence {\FF_{t^+} }_{t \ge 0}$ is a filtration of $\Sigma$. 
== Proof ==
We need to show that $\FF_{t^+}$ is a sub-$\sigma$-algebra of $\Sigma$ for each $t \in \hointr 0 \infty$, and:
:$\FF_{t^+} \subseteq \FF_{q^+}$
for $t, q \in \hointr 0 \infty$ with $t \le q$.
From the definition of the right-limit, we have: 
:$\ds \FF_{t^+} = \bigcap_{s > t} \FF_s$
for each $t \in \hointr 0 \infty$.
From Intersection of Sigma-Algebras, we have that $\FF_{t^+}$ is a sub-$\sigma$-algebra of $\Sigma$ for each $t \in \hointr 0 \infty$.
Let $t, q \in \hointr 0 \infty$ with $t \le q$. 
Then, we have: 
:$\set {s \in \R : s > q} \subseteq \set {s \in \R : s > t}$
and so: 
:$\ds \bigcap_{s > t} \FF_s \subseteq \bigcap_{s > q} \FF_s$
so that:
:$\FF_{t^+} \subseteq \FF_{q^+}$
Category:Filtrations of Sigma-Algebras
Right-Truncatable Prime/Sequence

== Sequence ==
The sequence of right-truncatable primes begins:
:$2, 3, 5, 7, 23, 29, 31, 37, 53, 59, 71, 73, 79, 233, 239, 293, 311, 313, 317, 373, 379, \ldots$
Category:Right-Truncatable Primes
Right Angle to Tangent of Circle goes through Center

== Theorem ==
== Proof ==
:
Let $DE$ touch the circle $ABC$, and let $AC$ be drawn at right angles to $DE$.
Suppose the center is not on $AC$.
Let $F$ be the center of $ABC$, and let $FC$ be joined from there to the point of contact.
By Radius at Right Angle to Tangent, $FC$ is perpendicular to $DE$.
Therefore $\angle FCE$ is a right angle.
But $\angle ACE$ is also a right angle.
Thus $\angle FCE = \angle ACE$, which is impossible unless $F$ lies on $AC$.
Therefore the center of $ABC$ lies on $AC$.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Tangents to Circles
Right Cancellable Commutative Operation is Left Cancellable

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be right cancellable and also commutative.
Then $\circ$ is also left cancellable.
== Proof ==
Let $\circ$ be both right cancellable and commutative on a set $S$.
Then:
== Also see ==
* Left Cancellable Commutative Operation is Right Cancellable
Category:Commutativity
Category:Cancellability
Right Cancellable Element is Right Cancellable in Subset

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, \circ} \subseteq \struct {S, \circ}$.
Let $x \in T$ be right cancellable in $S$.
Then $x$ is also right cancellable in $T$.
== Proof ==
Let $x \in T$ be right cancellable in $S$.
That is:
:$\forall a, b \in S: a \circ x = b \circ x \implies a = b$
Therefore:
:$\forall c, d \in T: c \circ x = d \circ x \implies c = d$
Thus $x$ is right cancellable in $T$.
Category:Abstract Algebra
Right Cancellable Elements of Semigroup form Subsemigroup

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $C_\rho$ be the set of right cancellable elements of $\struct {S, \circ}$.
Then $\struct {C_\rho, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
== Proof ==
Let $C_\rho$ be the set of right cancellable elements of $\struct {S, \circ}$:
:$C_\rho = \set {x \in S: \forall a, b \in S: a \circ x = b \circ x \implies a = b}$
Let $x, y \in C_\rho$.
Then:
Thus $\struct {C_\rho, \circ}$ is closed.
Therefore by the Subsemigroup Closure Test $\struct {C_\rho, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
Category:Subsemigroups
Category:Cancellability
Right Cancellable iff Right Regular Representation Injective

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Then $a \in S$ is right cancellable the right regular representation $\map {\rho_a} x$ is injective.
== Proof ==
Suppose $a \in S$ is right cancellable.
Then:
:$\forall x, y \in S: x \circ a = y \circ a \implies x = y$
From the definition of the right regular representation:
:$\map {\rho_a} x = x \circ a$
Thus:
:$\forall x, y \in S: \map {\rho_a} x = \map {\rho_a} y \implies x = y$
and so the right regular representation is injective.
Suppose $\map {\rho_a} x$ is injective.
Then:
:$\forall x, y \in S: \map {\rho_a} x = \map {\rho_a} y \implies x = y$
From the definition of the right regular representation:
:$\map {\rho_a} x = x \circ a$
Thus:
:$\forall x, y \in S: x \circ a = y \circ a \implies x = y$
and so $a$ is right cancellable.
== Also see ==
* Left Cancellable iff Left Regular Representation Injective
* Cancellable iff Regular Representations Injective
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.6$
Category:Regular Representations
Category:Cancellability
Right Composition of Compact Linear Transformation with Bounded Linear Transformation is Compact

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,}_X}$, $\struct {Y, \norm {\, \cdot \,}_Y}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces over $\GF$.
Let $S : X \to Y$ be a bounded linear transformation.
Let $T : Y \to Z$ be a compact linear transformation.
Then $T S : X \to Z$ is a compact linear transformation.
== Proof == 
Let $\sequence {x_n}_{n \mathop \in \N}$ be a bounded sequence.
So there exists $K \ge 0$ such that $\norm {x_n}_X \le K$ for each $n \in \N$. 
Since $S$ is bounded, there exists $M > 0$ such that: 
:$\norm {S x}_Y \le M \norm x_X$
for each $x \in X$.
Then:
:$\norm {S x_n}_Y \le M \norm {x_n}_X \le M K$
So $\sequence {S x_n}_{n \mathop \in \N}$ is bounded sequence.
So since $T$ is compact linear transformation.
So there exists a subsequence $\sequence {S x_{n_j} }_{j \mathop \in \N}$ such that $\sequence {T S x_{n_j} }_{j \mathop \in \N}$ converges.
So we have shown that for each bounded sequence $\sequence {x_n}_{n \mathop \in \N}$ in $X$, there exists a subsequence such that $\sequence {T S x_{n_j} }_{j \mathop \in \N}$ converges.
So $TS$ is compact. 
Category:Compact Linear Transformations
Right Congruence Class Modulo Subgroup is Right Coset

== Theorem ==
Let $G$ be a group, and let $H \le G$ be a subgroup.
Let $\RR^r_H$ be the equivalence defined as right congruence modulo $H$.
The equivalence class $\eqclass g {\RR^r_H}$ of an element $g \in G$ is the right coset $H g$.
This is known as the right congruence class of $g \bmod H$.
== Proof ==
Let $x \in \eqclass g {\RR^r_H}$.
Then:
Now let $x \in g H$.
Then:
Thus:
:$\eqclass g {\RR^r_H} = H g$
That is, the equivalence class $\eqclass g {\RR^r_H}$ of an element $g \in G$ equals the right coset $g H$.
== Also see ==
* Left Congruence Class Modulo Subgroup is Left Coset
* Right Coset Space forms Partition
* Element of Group is in Unique Right Coset of Subgroup
== Sources ==
* : $\S 6.1$. The quotient sets of a subgroup
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Theorem $11.1$
* : $\S 1.9$: Subgroups: Theorem $11$
* : $\S 42.3$ Another approach to cosets
Category:Cosets
Right Congruence Modulo Subgroup is Equivalence Relation

== Theorem ==
Let $G$ be a group, and let $H$ be a subgroup of $G$.
Let $x, y \in G$.
Let $x \equiv^r y \pmod H$ denote the relation that $x$ is right congruent modulo $H$ to $y$
Then the relation $\equiv^r$ is an equivalence relation.
== Proof ==
Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
For clarity of expression, we will use the notation:
:$\tuple {x, y} \in \RR^r_H$
for:
:$x \equiv^r y \pmod H$
From the definition of right congruence modulo a subgroup, we have:
:$\RR^r_H = \set {\tuple {x, y} \in G \times G: x y^{-1} \in H}$
We show that $\RR^r_H$ is an equivalence:
=== Reflexive ===
We have that $H$ is a subgroup of $G$.
From Identity of Subgroup:
:$e \in H$
Hence:
:$\forall x \in G: x x^{-1} = e \in H \implies \tuple {x, x} \in \RR^r_H$
and so $\RR^r_H$ is reflexive.
=== Symmetric ===
But then:
:$\tuple {x y^{-1} }^{-1} = y x^{-1} \implies \tuple {y, x} \in \RR^r_H$
Thus $\RR^r_H$ is symmetric.
=== Transitive ===
Thus $\RR^r_H$ is transitive.
So $\RR^r_H$ is an equivalence relation.
== Also see ==
* Definition:Right Coset
* Definition:Right Coset Space
* Left Congruence Modulo Subgroup is Equivalence Relation
== Sources ==
* : $\S 6.1$. The quotient sets of a subgroup: Lemma $\text {(ii)}$
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Theorem $11.1$
* : $\S 1.9$: Subgroups: Theorem $11$
* : $\S 42.1$ Another approach to cosets
* : Chapter $5$: Cosets and Lagrange's Theorem
Category:Congruence Modulo Subgroup
Category:Examples of Equivalence Relations
Right Coset Equals Subgroup iff Element in Subgroup

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
Let $x \in G$.
Let $H x$ denote the right coset of $H$ by $x$.
Then:
:$H x = H \iff x \in H$
== Proof ==
== Also see ==
* Left Coset Equals Subgroup iff Element in Subgroup
== Sources ==
* : $\S 6.1$. The quotient sets of a subgroup: Example $112$
* : $\S 42.6 \ \text {(3R)}$ Another approach to cosets
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $7 \ \text{(iii)}$
Category:Cosets
Right Coset Space forms Partition

== Theorem ==
Let $G$ be a group, and let $H \le G$ be a subgroup.
The right coset space of $H$ forms a partition of its group $G$:
 
== Proof ==
Follows directly from:
:Right Congruence Modulo Subgroup is Equivalence Relation
:Relation Partitions Set iff Equivalence.
== Also see ==
* Left Coset Space forms Partition
== Sources ==
* : $\S 6.1$. The quotient sets of a subgroup: Theorem
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Theorem $11.1$
* : $0$: Some Conventions and some Basic Facts
* : $\S 42.4$ Another approach to cosets
* : Chapter $5$: Cosets and Lagrange's Theorem
Category:Coset Space forms Partition
Right Coset by Identity

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
Then:
: $H = H e$
where $H e$ is the right coset of $H$ by $e$.
== Proof ==
We have:
So $H = H e$.
== Also see ==
This is consistent with the definition of the concept of coset by means of the subset product:
: $H e = H \set e$
* Left Coset by Identity
== Sources ==
* : $\S 6.1$. The quotient sets of a subgroup: Example $112$
* : $\S 1.9$: Example $30$
* : $\S 3.3$: Group actions and coset decompositions
Category:Coset by Identity
Right Cosets are Equal iff Element in Other Right Coset

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $x, y \in G$.
Let $H x$ denote the right coset of $H$ by $x$.
Then:
:$H x = H y \iff x \in H y$
== Proof ==
== Also see ==
* Left Cosets are Equal iff Element in Other Left Coset
== Sources ==
* : Chapter $5$: Cosets and Lagrange's Theorem: Exercise $3$
Category:Cosets
Right Cosets are Equal iff Left Cosets by Inverse are Equal

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
Let $g_1, g_2 \in G$.
Then:
:$H g_1 = H g_2 \iff {g_1}^{-1} H = {g_2}^{-1} H$
where:
:${g_1}^{-1}$ and ${g_2}^{-1}$ denote the inverses of $g_1$ and $g_2$ in $G$
:$H g_1$ and $H g_2$ denote the right cosets of $H$ by $g_1$ and $g_2$ respectively
:${g_1}^{-1} H$ and ${g_2}^{-1} H$ denote the left cosets of $H$ by ${g_1}^{-1}$ and ${g_2}^{-1}$ respectively.
== Proof ==
== Sources ==
* : $0$: Some Conventions and some Basic Facts: Exercise $10$
Category:Cosets
Right Cosets are Equal iff Product with Inverse in Subgroup

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $x, y \in G$.
Let $H x$ denote the right coset of $H$ by $x$.
Then:
:$H x = H y \iff x y^{-1} \in H$
== Proof ==
== Also see ==
* Left Cosets are Equal iff Product with Inverse in Subgroup
== Sources ==
* : $\S 6.2$. Mappings of quotient sets: $\text{(i)}$
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Theorem $11.1$
* : $0$: Some Conventions and some Basic Facts
* : $\S 42.6 \ \text {(2R)}$ Another approach to cosets
* : Chapter $5$: Cosets and Lagrange's Theorem: Exercise $4$
Category:Cosets
Right Distributive Law for Natural Numbers

== Theorem ==
The operation of multiplication is right distributive over addition on the set of natural numbers $\N_{> 0}$:
:$\forall x, y, n \in \N_{> 0}: \paren {x + y} \times n = \paren {x \times n} + \paren {y \times n}$
== Proof ==
Using the axiomatization:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\forall a, b \in \N_{> 0}: \paren {a + b} \times n = \paren {a \times n} + \paren {b \times n}$
=== Basis for the Induction ===
$\map P 1$ is the case:
and so $\map P 1$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\forall a, b \in \N_{> 0}: \paren {a + b} \times k = \paren {a \times k} + \paren {b \times k}$
Then we need to show:
:$\forall a, b \in \N_{> 0}: \paren {a + b} \times \paren {k + 1} = \paren {a \times \paren {k + 1} } + \paren {b \times \paren {k + 1} }$
=== Induction Step ===
This is our induction step:
The result follows by the Principle of Mathematical Induction.
== Also see ==
* Left Distributive Law for Natural Numbers
== Sources ==
* : $\S 2.1$: Theorem $2.5$
Category:Natural Numbers/1-Based
Right Ideal is Right Module over Ring

== Theorem ==
Let $\struct {R, +, \times}$ be a ring.
Let $J \subseteq R$ be a right ideal of $R$.
Let $\circ : J \times R \to J$ be the restriction of $\times$ to $J \times R$.
Then $\struct {J, +, \circ}$ is a right module over $\struct {R, +, \times}$.
=== Corollary ===
== Proof ==
By definition of a right ideal then $\circ$ is well-defined.
=== ===
Follows directly from .
=== ===
Follows directly from .
=== ===
Follows directly from 
== Also see ==
* Left Ideal is Left Module over Ring
== Sources ==
* : Chapter $4$: Rings and Modules: $\S 4.1$: The Definitions Recalled
Category:Module Theory
Category:Ideal Theory
Category:Right Ideal is Right Module over Ring
Right Ideal is Right Module over Ring/Ring is Right Module over Ring

== Theorem ==
Let $\struct {R, +, \times}$ be a ring.
Then $\struct {R, +, \times}$ is a right module over $\struct {R, +, \times}$.
== Proof ==
From Ring is Ideal of Itself, $R$ is a right ideal.
From Right Ideal is Right Module over Ring, $\struct {R, +, \times}$ is a right module over $\struct {R, +, \times}$.
== Also see ==
* Ring is Left Module over Ring
== Sources ==
* : Chapter $4$: Rings and Modules: $\S 4.1$: The Definitions Recalled
Category:Right Ideal is Right Module over Ring
Right Identity Element is Idempotent

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $e_R \in S$ be a right identity with respect to $\circ$.
Then $e_R$ is idempotent under $\circ$.
== Proof ==
By the definition of a right identity:
:$\forall x \in S: x \circ e_R = x$
Thus in particular:
:$e_R \circ e_R = e_R$
Therefore $e_R$ is idempotent under $\circ$.
== Also see ==
* Left Identity Element is Idempotent
* Identity Element is Idempotent
== Sources ==
* : $\S 1.2$
Category:Identities are Idempotent
Right Identity in Semigroup may not be Unique

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $e_R$ be a right identity of $\struct {S, \circ}$.
Then it is not necessarily the case that $e_R$ is unique.
== Proof ==
;Proof by Counterexample
Let $\struct {S, \gets}$ be an algebraic structure in which the operation $\gets$ is the left operation.
From Structure under Left Operation is Semigroup, $\struct {S, \gets}$ is a semigroup.
From Element under Left Operation is Right Identity, every element of $\struct {S, \gets}$ is a right identity.
The result follows.
Category:Identity Elements
Category:Semigroups
Right Identity while exists Right Inverse for All is Identity

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup with a right identity $e_R$ such that:
:$\forall x \in S: \exists x_R: x \circ x_R = e_R$
That is, every element of $S$ has a right inverse with respect to the right identity.
Then $e_R$ is also a left identity, that is, is an identity.
== Proof ==
Let $x \in S$ be any element of $S$.
From Right Inverse for All is Left Inverse we have that $x_R \circ x = e_R$.
Then:
So $e_R$ behaves as a left identity as well as a right identity.
That is, by definition, $e_R$ is an identity element.
== Also see ==
* Left Identity while exists Left Inverse for All is Identity
== Sources ==
* : $\S 1.4$: Lemma $4$
Category:Semigroups
Right Inverse Mapping/Examples/Real Part of Complex Number

== Example of Right Inverse Mapping ==
Let $f: \C \to \R$ be the mapping:
:$\forall z \in \C: \map f z = \map \Re z$
From Real Part as Mapping is Surjection, $f$ is a surjection.
Hence it has a right inverse $g: \R \to \C$ which, for example, can be defined as:
:$\forall x \in \R: \map g x = x + i$
This right inverse is not unique.
For example, the mapping $h: \R \to \C$ defined as:
:$\forall x \in \R: \map h x = x - i$
is also a right inverse.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{Q}$
Category:Examples of Right Inverse Mappings
Category:Real Parts
Right Inverse Mapping/Examples/Real Square Function to Non-Negative Reals

== Example of Right Inverse Mapping ==
Let $f: \R \to \R_{\ge 0}$ be the real square function whose codomain is the set of non-negative reals:
:$\forall x \in \R: \map f x = x^2$
From Real Square Function to $\R_{\ge 0}$, $f$ is a surjection.
Hence it has a right inverse $g: \R_{\ge 0} \to \R$ which, for example, can be defined as:
:$\forall x \in \R_{\ge 0}: \map g x = +\sqrt x$
This right inverse is not unique.
For example, the mapping $h: \R_{\ge 0} \to \R$ defined as:
:$\forall x \in \R_{\ge 0}: \map h x = -\sqrt x$
is also a right inverse, as is the arbitrarily defined mapping $j: \R_{\ge 0} \to \R$ defined as:
:$\forall x \in \R_{\ge 0}: \map j x = \begin {cases} \sqrt x & : x \le 5 \\ -\sqrt x & : x > 5 \end {cases}$
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Composition of Functions
Category:Examples of Right Inverse Mappings
Category:Square Function