nyuuzyou/wikis huggingface dataset:
Idempotent Semigroup/Examples/Relation induced by Inverse Element/Properties/6

== Example of Idempotent Semigroup ==
Let $\struct {S, \circ}$ be an idempotent semigroup.
Let $\RR$ be the relation on $S$ defined as:
:$\forall a, b \in S: a \mathrel \RR b \iff \paren {a \circ b \circ a = a \land b \circ a \circ b = b}$
That is, such that $a$ is the inverse of $b$ and $b$ is the inverse of $a$.
The quotient structure $\struct {S / \RR, \circ_\RR}$ is a commutative idempotent semigroup.
The equivalence classes under $\RR$ are anticommutative subsemigroups of $\struct {S, \circ}$.
== Proof ==
From Idempotent Semigroup: Relation induced by Inverse Element: $5$:
:$\RR$ is a congruence relation on $\struct {S, \circ}$.
Hence $\struct {S / \RR, \circ_\RR}$ is indeed a quotient structure, and Quotient Structure is Well-Defined applies.
From Quotient Structure of Semigroup is Semigroup, we have that $\struct {S / \RR, \circ_\RR}$ is a semigroup.
Then we have:
Thus $\struct {S / \RR, \circ_\RR}$ is an idempotent semigroup.
To prove commutativity, it is sufficient to prove that $\paren {a \circ b} \mathrel \RR \paren {b \circ a}$.
We have:
and:
That is:
:$\paren {a \circ b} \mathrel \RR \paren {b \circ a}$
Hence:
Thus we have shown that $\struct {S / \RR, \circ_\RR}$ is a commutative idempotent semigroup.
Let $\eqclass x \RR$ be an arbitrary element of $S / \RR$.
We have that $\eqclass x \RR$ is a subset of $\struct {S, \circ}$.
We now show that $\struct {\eqclass x \RR, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
From Subsemigroup Closure Test, it is sufficient to show that $\struct {\eqclass x \RR, \circ}$ is closed.
Hence, let $a, b \in \eqclass x \RR$.
Then:
:$a \mathrel \RR b$
and so:
We have:
and:
That is:
:$a \mathrel \RR a \circ b$
and so:
:$a, b \in \eqclass x \RR \implies a \circ b \in \eqclass x \RR$
Thus is satisfied.
Hence from Subsemigroup Closure Test $\eqclass x \RR$ is a subsemigroup of $\struct {S, \circ}$
It remains to be shown that $\struct {\eqclass x \RR, \circ}$ is anticommutative.
Suppose $a, b \in \eqclass x \RR$ such that $a \circ b = b \circ a$.
We have:
Then:
That is:
:$a \circ b = b \circ a \implies a = b$
and so $\struct {\eqclass x \RR, \circ}$ is an anticommutative subsemigroup of $\struct {S, \circ}$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.19 \ \text {(h)}$
* 
Category:Idempotent Semigroups
Identification Mapping is Continuous

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ be a topological space.
Let $S_2$ be a set.
Let $f: S_1 \to S_2$ be a mapping.
Let $\tau_2$ be the identification topology on $S_2$ with respect to $f$ and $\struct {S_1, \tau_1}$.
Then the identification mapping $f$ is continuous.
== Proof ==
By definition of identification topology:
:$\tau_2 = \set {V \in \powerset {S_2}: f^{-1} \sqbrk V \in \tau_1}$
That is:
:$V \in \tau_2 \implies f^{-1} \sqbrk V \in \tau_1$
This is precisely the definition of a continuous mapping.
Category:Identification Topology
Identification Topology equals Quotient Topology on Induced Equivalence

== Theorem ==
Let $\struct {S_1, \tau_1}$ be a topological space.
Let $S_2$ be a set.
Let $f: S_1 \to S_2$ be a surjective mapping.
Let $\tau_2$ be the identification topology on $S_2$ $f$ and $\struct {S_1, \tau_1}$:
:$\tau_2 = \set {V \subseteq S_2 : f^{-1} \sqbrk V \in \tau_1}$
Let $\RR_f \subseteq S_1 \times S_1$ be the equivalence on $S_1$ induced by $f$:
:$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$
Let $q_{\RR_f}$ be the quotient mapping induced by $\RR_f$.
Let $\tau_{\RR_f}$ be the quotient topology on $S_1 / \RR_f$ by $q_{\RR_f}$:
:$\tau_{\RR_f} := \set {U \subseteq S_1 / \RR_f: q_{\RR_f}^{-1} \sqbrk U \in \tau_1}$
Then:
:$\struct{ S_2, \tau_2 }$ is homeomorphic to the quotient space $\struct{ S_1 / \RR_f, \tau_{\RR_f} }$.
== Proof ==
Define a mapping $\tilde f: S_1 / \RR_f \to S_2$ by:
:$\map {\tilde f}{\eqclass s { \RR_f} } = \map f s$
:$\begin{xy} \xymatrix@L+2mu@+1em{
 S_1 \ar[r]^*{f}
 \ar[rd]_*{ q_{\RR_f} }
&
 S_2 
\\ &
 S_1 / \RR_f \ar@{-->}[u]^*{\tilde f}
}\end{xy}$
Then $\tilde f$ is well-defined, as for all $s' \in \eqclass s { \RR_f}$, we have:
:$\map f {s'} = \map f s$
For any $s_1 , s_2 \in S_1$ with $\eqclass {s_1}{\RR_f} \ne \eqclass {s_2}{\RR_f}$, we have:
:$\map f {s_1} \ne \map f {s_2}$
so:
:$\map {\tilde f}{\eqclass {s_1}{ \RR_f} } \ne \map {\tilde f}{\eqclass {s_2}{ \RR_f} }$
It follows that $\tilde f$ is injective.
For all $t \in S_2$, we can find $s \in S_1$ such that $\map f s = t$, as $f$ is surjective.
Then $\map {\tilde f}{\eqclass s { \RR_f} } = t$, so $\tilde f$ is surjective.
Let $V \in \tau_2$.
As Identification Mapping is Continuous, it follows that:
:$ f^{-1} \sqbrk V \in \tau_1$
Then:
By definition of quotient topology, we have:
:$\set {\eqclass s {\RR_f} \in S_1 / \RR_f : \map f s \in V} \in \tau_{\RR_f}$
By definition of continuity, $\tilde f$ is continuous.
Let $U \in \tau_{\RR_f}$.
By definition of quotient topology, we have:
:$q_{\RR_f}^{-1} \sqbrk U \in \tau_1$
By definition of identification topology, we have:
:$f \sqbrk {q_{\RR_f}^{-1} \sqbrk U } \in \tau_2$
Then:
By definition of continuity, $\tilde f^{-1}$ is continuous.
By definition of homeomorphism, $\tilde f$ is a homeomorphism between $\struct {S_1 / \RR_f, \tau_{\RR_f} }$ and $\struct {S_2, \tau_2}$.
== Sources ==
* : $2$: Topological Spaces and Continuous Functions: $\S 22$: The Quotient Topology
Category:Identification Topology
Category:Quotient Topology
Identification Topology is Finest Topology for Mapping to be Continuous

== Theorem ==
Let $T_1 := \struct {S_1, \tau_1}$ be a topological space.
Let $S_2$ be a set.
Let $f: S_1 \to S_2$ be a mapping.
Let $\tau_2$ be the identification topology on $S_2$ with respect to $f$ and $\struct {S_1, \tau_1}$.
Let $T_2 := \struct {S_2, \tau_2}$ be the resulting topological space.
Then $\tau_2$ is the finest topology on $S_2$ such that $f: T_1 \to T_2$ is continuous.
== Proof ==
It is established in Identification Mapping is Continuous that $f$ is continuous.
Let $\tau \subseteq \powerset {S_2}$ be a topology on $S_2$ for which $f$ is $\struct {\tau_1, \tau}$-continuous.
Let $U \in \tau$.
Then by definition of $\struct {\tau_1, \tau}$-continuous:
:$f^{-1} \sqbrk U \in \tau_1$
By definition of $\tau_2$ it then follows that $U \in \tau_2$.
By definition of subset it follows that $\tau \subseteq \tau_2$.
Thus it has been shown that $\tau_2$ is finer than any topology on $S_2$ for which $f$ is $\struct {\tau_1, \tau}$-continuous.
Hence the result, by definition of finest topology.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Functions
Category:Identification Topology
Identification Topology is Topology

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ be a topological space.
Let $S_2$ be a set.
Let $f: S_1 \to S_2$ be a mapping.
Let $\tau_2$ be the identification topology on $S_2$ with respect to $f$ and $\struct {S_1, \tau_1}$.
Then $\tau_2$ is a topology on $S_2$.
== Proof ==
By definition:
:$\tau_2 = \set {V \in \powerset {S_2}: f^{-1} \sqbrk V \in \tau_1}$
We examine each of the open set axioms in turn:
=== ===
Let $\family {U_i}_{i \mathop \in I}$ be an indexed family of elements of $\tau_2$.
Let $\ds V = \bigcup_{i \mathop \in I} U_i$ be the union of $\family {U_i}_{i \mathop \in I}$.
From Preimage of Union under Mapping: Family of Sets:
:$\ds f^{-1} \sqbrk {\bigcup_{i \mathop \in I} U_i} = \bigcup_{i \mathop \in I} f^{-1} \sqbrk {U_i}$
We have $f^{-1} \sqbrk {U_i} \in \tau_1$.
As $\tau_1$ is a topology:
:$\ds \bigcup_{i \mathop \in I} f^{-1} \sqbrk {U_i} \in \tau_1$
So $f^{-1} \sqbrk V \in \tau_1$.
Thus $V \in \tau_2$ and so $V$ is open by definition.
=== ===
Let $U$ and $V$ be elements of $\tau_2$.
From Preimage of Intersection under Mapping:
:$f^{-1} \sqbrk {U \cap V} = f^{-1} \sqbrk U \cap f^{-1} \sqbrk V$
As $\tau_1$ is a topology:
:$f^{-1} \sqbrk U \cap f^{-1} \sqbrk V \in \tau_1$
Hence $U \cap V$ is open by definition.
=== ===
As $f$ is a mapping its domain is $S_1$.
That is:
:$f^{-1} \sqbrk {S_2} = S_1$
By definition of a topology, $S_1$ is open in $T_1$.
Thus by definition $S_2 \in \tau_2$.
All the open set axioms are fulfilled, and the result follows.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 39$
Category:Identification Topology
Identities of Boolean Algebra are also Zeroes

== Theorem ==
Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra, defined as in Definition 1.
Let the identity for $\vee$ be $\bot$ and the identity for $\wedge$ be $\top$.
Then:
That is:
:$\bot$ is a zero element for $\wedge$
:$\top$ is a zero element for $\vee$.
== Proof ==
Let $x \in S$.
Then:
So $x \vee \top = \top$.
The result $x \wedge \bot = \bot$ follows from the Duality Principle.
== Also known as ==
== Sources ==
* : $\S 1.5$: Theorem $1.15, \ 1.15 \ \text{(b)}$
* 
* : $\S 2$: Exercise $2$
* 
Category:Identities of Boolean Algebra are also Zeroes
Category:Identity Elements
Category:Zero Elements
Category:Boolean Algebras
Identity (Equation)/Examples/Difference of Two Squares

== Example of Identity in context of Equations ==
The result Difference of Two Squares:
:$\forall x, y \in \R: x^2 - y^2 \equiv \paren {x + y} \paren {x - y}$
is an example of an identity.
== Sources ==
* 
* 
Category:Examples of Identities (Equations)
Identity Element for Addition of Cuts

== Theorem ==
Let $\alpha$ be a cut.
Let $0^*$ be the rational cut associated with the (rational) number $0$:
:$0^* = \set {r \in \Q: r < 0}$
Then:
:$\alpha + 0^* = \alpha$
where $+$ denotes the operation of addition of cuts.
== Proof ==
Let $r \in \alpha + 0^*$.
$\alpha + 0^*$ is the set of all rational numbers of the form $p + q$ such that $p \in \alpha$, $q \in 0^*$, that is, $q < 0$.
It follows that:
:$p + q < p$
and so:
:$p + q \in \alpha$
that is:
:$r \in \alpha$
Let $r \in \alpha$.
Let $s \in \Q$ be a rational number such that $s > r$ and $s \in \alpha$.
This is possible because $r$ is not the greatest element of $\alpha$ by definition of a cut.
Let $q = r - s$.
Then $q < 0$ and so $q \in 0^*$.
Thus we have:
:$r = s + q$
and so:
:$r \in \alpha + 0^*$
We have that:
:$r \in \alpha \implies r \in \alpha + 0^*$
and:
:$r \in \alpha + 0^* \implies r \in \alpha$
and so by definition of set equality:
:$\alpha + 0^* = \alpha$
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.14$. Theorem $\text {(c)}$
Category:Cuts
Category:Addition
Identity Element for Dirichlet Convolution

== Theorem ==
Let $f$ be an arithmetic function.
Let $*$ denote Dirichlet convolution.
Let $\iota$ be the identity arithmetic function.
Then:
:$\iota * f = f$
== Proof ==
We have:
Hence the result.
== Also see ==
* Properties of Dirichlet Convolution
Category:Dirichlet Convolution
Identity Element for Power Structure

== Theorem ==
Let $\struct {S, \circ}$ be a magma whose underlying set $S$ is non-empty.
Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $\struct {S, \circ}$.
Then:
:a subset $J$ of $S$ is an identity element of $\struct {\powerset S, \circ_\PP}$
:
:there exists an identity element $e$ of $\struct {S, \circ}$, such that $J = \set e$.
== Proof ==
=== Sufficient Condition ===
Let $J \subseteq S$ such that $J$ is an identity element of $\struct {\powerset S, \circ_\PP}$.
We have:
That is, the elements of $J$ are all left identities for $\circ$.
Similarly:
That is, the elements of $J$ are all right identities for $\circ$.
So we have established that $J$ contains at least one element which is both a left identity and a right identity.
From Left and Right Identity are the Same, there is only one element of $J$, and it is the necessarily unique identity element for $\circ$.
That is:
:there exists an identity element $e$ of $\struct {S, \circ}$, such that $J = \set e$.
=== Necessary Condition ===
Let $\struct {S, \circ}$ have an identity element $e$.
Let $J = \set e$.
Then we have:
and similarly:
So:
:$\forall A \in \powerset S: A \circ_\PP J = A = J \circ_\PP A$
and it is seen that $J$ is an identity element for $\circ_\PP$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.6$
Category:Power Structures
Category:Identity Elements
Identity Element in Unital *-Algebra is Hermitian

== Theorem ==
Let $\struct {A, \ast}$ be a unital $\ast$-algebra.
Let ${\mathbf 1}_A$ be the identity element of $A$. 
Then we have:
:${\mathbf 1}_A^\ast = {\mathbf 1}_A$
== Proof ==
From Product of Element in *-Star Algebra with its Star is Hermitian, we have:
:$\paren { {\mathbf 1}_A {\mathbf 1}_A^\ast}^\ast = {\mathbf 1}_A {\mathbf 1}_A^\ast$
We have, since $\ast$ is an involution: 
:$\paren { {\mathbf 1}_A {\mathbf 1}_A^\ast} = {\mathbf 1}_A^{\ast \ast} = {\mathbf 1}_A$
and:
:${\mathbf 1}_A {\mathbf 1}_A^\ast = {\mathbf 1}_A^\ast$
Hence we conclude:
:${\mathbf 1}_A^\ast = {\mathbf 1}_A$
== Sources ==
* : $2.1$: $C^\ast$-Algebras
Category:Hermitian Elements of *-Algebras
Identity Element is Idempotent

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $e \in S$ be an identity with respect to $\circ$.
Then $e$ is idempotent under $\circ$.
== Proof 1 ==
By the definition of an identity element:
:$\forall x \in S: e \circ x = x$
Thus in particular:
:$e \circ e = e$
Therefore $e$ is idempotent under $\circ$.
== Proof 2 ==
Follows from Left Identity Element is Idempotent, Right Identity Element is Idempotent and definition of Two-Sided Identity.
== Also see ==
* Left Identity Element is Idempotent
* Right Identity Element is Idempotent
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory: $(1.8)$
Category:Identities are Idempotent
Identity Element is Order Unit on Set of Hermitian Elements of Unital C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $\le$ be the canonical preordering on $A$.
Let $A_{\mathbf {SA}}$ be the set of Hermitian elements of $A$.
Then ${\mathbf 1}_A$ is an order unit for $\tuple {A_{\mathbf{SA}}, \le}$. 
== Proof ==
Fix $a \in A$ Hermitian.
Then from Bounds on Hermitian Element of Unital C*-Algebra in terms of Bounds on Spectrum:
:$-\norm a {\mathbf 1}_A \le_A a \le \norm a {\mathbf 1}_A$
Take $N \in \N$ such that $N > \norm a$. 
From Non-Negative Multiple of Positive Element of C*-Algebra is Positive: Corollary, we have:
:$-N {\mathbf 1}_A \le_A -\norm a {\mathbf 1}_A$
and:
:$\norm a {\mathbf 1}_A \le_A N {\mathbf 1}_A$
Since $\le_A$ is a preordering, we have:
:$-N {\mathbf 1}_A \le a \le_A N {\mathbf 1}_A$
Since $a \in A$ was arbitrary Hermitian, ${\mathbf 1}_A$ is an order unit.
Category:C*-Algebras
Category:Preordered Vector Spaces
Identity Element of Addition on Numbers

== Theorem ==
On all the number systems:
* natural numbers $\N$
* integers $\Z$
* rational numbers $\Q$
* real numbers $\R$
* complex numbers $\C$
the identity element of addition is zero ($0$).
== Proof ==
This is demonstrated in these pages:
* Identity Element of Natural Number Addition is Zero
* Integer Addition Identity is Zero
* Rational Addition Identity is Zero
* Real Addition Identity is Zero
* Complex Addition Identity is Zero
== Also see ==
* Identity Element of Multiplication on Numbers
== Sources ==
* : $\S 4.3$. Units and zeros: Example $71$
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses
* : Part $\text I$: Prologue: Chapter $1$: Basic Properties of Numbers: $(\text P 2)$
* 
* 
* 
Category:Examples of Identity Elements
Category:Addition
Category:Numbers
Identity Element of Multiplication on Numbers

== Theorem ==
On all the number systems:
* natural numbers $\N$
* integers $\Z$
* rational numbers $\Q$
* real numbers $\R$
* complex numbers $\C$
the identity element of multiplication is one ($1$).
== Proof ==
This is demonstrated in these pages:
* Identity Element of Natural Number Multiplication is One
* Integer Multiplication Identity is One
* Rational Multiplication Identity is One
* Real Multiplication Identity is One
* Complex Multiplication Identity is One
== Also see ==
* Identity Element of Addition on Numbers
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 1$: Introduction: $(1.3)$
* : $\S 4.3$. Units and zeros: Example $71$
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses
* : Part $\text I$: Prologue: Chapter $1$: Basic Properties of Numbers: $(\text P 6)$
* : $1$
* 
* 
* : Chapter $\S 2.4$: Composition of 
Category:Examples of Identity Elements
Category:Multiplication
Category:Numbers
Identity Element of Natural Number Addition is Zero

== Theorem ==
The identity element for the natural numbers under addition is zero ($0$):
:$\forall n \in \N: 0 + n = n$
== Proof ==
Take the definition of natural numbers $\N$ as a naturally ordered semigroup $\struct {\N, +, \le}$.
The result follows from Zero is Identity in Naturally Ordered Semigroup.
== Sources ==
* : $\S 1$: Numbers: $1.1$ Natural Numbers and Integers: Examples $1.1 \ \text {(b)}$
Category:Natural Number Addition
Category:Examples of Identity Elements
Identity Element of Natural Number Multiplication is One

== Theorem ==
Let $\N$ be the natural numbers.
Let $1$ be the element one of $\N$.
Then $1$ is the identity element of multiplication:
:$\forall n \in \N: n \times 1 = n = 1 \times n$
== Proof ==
Firstly, by definition of multiplication:
Next, recall that multiplication is recursively defined as:
:$\forall m, n \in \N: \begin{cases}
m \times 0 & = 0 \\
m \times \paren {n + 1} & = m \times n + m
\end{cases}$
From the Principle of Recursive Definition, there is only one mapping $f$ satisfying this definition for $m = 1$; that is, such that:
:$\forall n \in \N: \begin {cases}
\map f 0 = 0 \\
\map f {n + 1} = \map f n + 1
\end{cases}$
Consider now $f'$ defined as $\map {f'} n = n$.
Then evidently $\map {f'} 0 = 0$.
Also:
showing that $f'$ also satisfies the definition for $1 \times n$.
Hence $n \times 1 = 1 \times n = n$ for all $n \in \N$.
That is, $1$ is the identity element for multiplication.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.10$
* : $\S 1$: Numbers: $1.1$ Natural Numbers and Integers: Examples $1.1 \ \text {(g)}$
Category:Examples of Identity Elements
Category:Natural Number Multiplication
Identity Elements occupy Diagonal of Cayley Table in Inverse Row Form

== Theorem ==
Let $\struct {G, \circ}$ be a finite group.
Let $\CC$ be a Cayley table for $\struct {G, \circ}$ presented in inverse row form.
Then all the entries in the main diagonal of $\CC$
are instances of the identity element.
== Proof ==
By definition of inverse row form, the rows of $\CC$ are headed by the inverse elements of the elements which head the corresponding columns.
The entries in the main diagonal of $\CC$ have the same column number as row number.
Let $\sqbrk c_{k k}$ denote the entry of $\CC$ corresponding to the element where the $k$th row intersects the $k$th column.
Let $a$ be the element which heads column $k$.
Then, by definition, $a^{-1}$ is the element which heads row $k$.
Thus the element which occupies entry $\struct c_{k k}$ is $a^{-1} \circ a = e$.
Hence the result.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 5$: The Multiplication Table
Category:Cayley Tables
Identity Function is Completely Multiplicative

== Theorem ==
Let $I_{\Z_{>0}}: \Z_{>0} \to \Z_{>0}$ be the identity function:
:$\forall n \in \Z_{>0}: I_{\Z_{>0}} \left({n}\right) = n$
Then $I_{\Z_{>0}}$ is completely multiplicative.
== Proof ==
:$\forall m, n \in \Z_{>0}: I_{\Z_{>0}} \left({m n}\right) = m n = I_{\Z_{>0}} \left({m}\right) I_{\Z_{>0}} \left({n}\right)$
Category:Completely Multiplicative Functions
Category:Identity Mappings
Identity Function is Computably Uniformly Continuous Real Function

== Theorem ==
Let $I_\R : \R \to \R$ denote the identity function on $\R$.
Then, $I_\R$ is a computably uniformly continuous real function.
== Proof ==
Let $d : \N \to \N$ be defined as:
:$\map d n = n$
As $d$ is precisely the projection function $\pr_1^1$, it is primitive recursive.
By Primitive Recursive Function is Total Recursive Function, $d$ is a total recursive function.
Let $n \in \N$ and $x, y \in \R$ satisfy:
:$\size {x - y} < \dfrac 1 {\map d n + 1}$
By definition of identity function:
:$\map {I_\R} x = x$
:$\map {I_\R} y = y$
By definition of $d$:
:$\map d n = n$
Thus:
:$\size {\map {I_\R} x - \map {I_\R} y} < \dfrac 1 {n + 1}$
As $n$, $x$, and $y$ were arbitrary, it follows that $I_\R$ is computably uniformly continuous by definition.
Category:Computability Theory
Identity Function is Sequentially Computable Real Function

== Theorem ==
Let $I_\R : \R \to \R$ denote the identity function on $\R$.
Then $I_\R$ is a sequentially computable real function.
== Proof ==
Let $\sequence {x_n}$ be a computable real sequence.
By definition of identity function:
:$\map {I_\R} x = x$
Therefore:
:$\sequence {\map {I_\R} {x_n}} = \sequence {x_n}$
and is thus computable.
As $\sequence {x_n}$ was arbitrary, it follows that $I_\R$ is sequentially computable by definition.
Category:Computability Theory
Identity Functor is Functor

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $\operatorname{id}_{\mathbf C}: \mathbf C \to \mathbf C$ be the identity functor on $\mathbf C$.
Then $\operatorname{id}_{\mathbf C}$ is a functor.
== Proof ==
Let $f, g$ be morphisms of $\mathbf C$ such that $g \circ f$ is defined.
Then:
Also, for any object $C$ of $\mathbf C$:
Hence $\operatorname{id}_{\mathbf C}$ is shown to be a functor.
Category:Category Theory
Identity Functor is Left Identity

== Theorem ==
Let $\mathbf C$ and $\mathbf D$ be metacategories.
Let $F: \mathbf C \to \mathbf D$ be a functor, and let $\operatorname{id}_{\mathbf D}$ be the identity functor on $\mathbf D$.
Then the composite functor $\operatorname{id}_{\mathbf D} F$ satisfies:
:$\operatorname{id}_{\mathbf D} F = F$
== Proof ==
We have, for all objects $C$ of $\mathbf C$:
:$\operatorname{id}_{\mathbf D}F C = \operatorname{id}_{\mathbf D} \left({F C}\right) = F C$
by definition of composition of functors and of identity functor.
Similarly, we have, for a morphism $f$ of $\mathbf C$:
:$\operatorname{id}_{\mathbf D}F f = \operatorname{id}_{\mathbf D} \left({F f}\right) = F f$
Hence the result.
Category:Category Theory
Identity Functor is Right Identity

== Theorem ==
Let $\mathbf C$ and $\mathbf D$ be metacategories.
Let $F: \mathbf C \to \mathbf D$ be a functor, and let $\operatorname{id}_{\mathbf C}$ be the identity functor on $\mathbf C$.
Then the composite functor $F \operatorname{id}_{\mathbf C}$ satisfies:
:$F \operatorname{id}_{\mathbf C} = F$
== Proof ==
We have, for all objects $C$ of $\mathbf C$:
:$F \operatorname{id}_{\mathbf C} C = F \left({\operatorname{id}_{\mathbf C} C}\right) = F C$
by definition of composition of functors and of identity functor.
Similarly, we have, for a morphism $f$ of $\mathbf C$:
:$F \operatorname{id}_{\mathbf C} f = F \left({\operatorname{id}_{\mathbf C} f}\right) = F f$
Hence the result.
Category:Category Theory
Identity Mapping between Metrics separated by Scale Factor is Continuous

== Theorem ==
Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
Let $d_1$ and $d_2$ be such that:
:$\forall x, y \in A: \map {d_2} {x, y} \le K \map {d_2} {x, y}$
Let $I_A: A \to A$ be the identity mapping on $A$.
Then $I_A$ is continuous from $M_1$ to $M_2$.
== Proof ==
Let $\epsilon \in \R_{>0}$.
Let $a \in A$.
Set $\delta = \dfrac \epsilon K$.
Then:
Hence by definition $I_A$ is continuous at $a$.
As $a$ is arbitrary, it follows that this is true for all $a \in A$.
Thus $I_A$ is continuous on the whole of $M_1$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces: Lemma $7.8$
Category:Identity Mappings
Category:Continuous Mappings on Metric Spaces
Identity Mapping forms Galois Connection

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered sets.
Let $\operatorname{id}_S : S \to S$ be the identity mapping on $S$.
Then:
:$\tuple {\operatorname{id}_S, \operatorname{id}_S}$ is a Galois connection.
== Proof ==
We have
The result follows.
Category:Galois Connections