nyuuzyou/wikis huggingface dataset:
Set of Linear Transformations over Commutative Ring forms Submodule of Module of All Mappings/Unitary

== Theorem ==
Let $R$ be a commutative ring with unity whose unity is $1_R$.
Let $\struct {G, +_G, \circ}_R$ and $\struct {H, +_H, \circ}_R$ be $R$-modules.
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $\struct {H, +_H, \circ}_R$ be a unitary module.
Then $\map {\LL_R} {G, H}$ is also a unitary module.
== Proof ==
From Set of Linear Transformations over Commutative Ring forms Submodule of Module of All Mappings, $\map {\LL_R} {G, H}$ is a module.
It remains to be shown that $\map {\LL_R} {G, H}$ is a unitary module, that is:
:$\forall \phi \in \map {\LL_R} {G, H}: 1_R \circ \phi = \phi$
So, let $\struct {H, +_H, \circ}_R$ be a unitary $R$-module.
Then, by :
:$\forall x \in H: 1_R \circ x = x$
Thus:
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Linear Transformations
Category:Unitary Modules
Set of Linear Transformations under Pointwise Addition forms Abelian Group

== Theorem ==
Let $\struct {G, +_G}$ and $\struct {H, +_H}$ be abelian groups.
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ}_R$ and $\struct {H, +_H, \circ}_R$ be $R$-modules.
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $\oplus_H$ be defined as pointwise addition on $\map {\LL_R} {G, H}$:
:$\forall u, v \in \map {\LL_R} {G, H}: \forall x \in G: \map {\paren {u \oplus_H v} } x := \map u x +_H \map v x$
Then $\struct {\map {\LL_R} {G, H}, \oplus_H}$ is an abelian group.
== Proof ==
From Structure Induced by Group Operation is Group, $\struct {H^G, \oplus_H}$ is a group
Let $\phi, \psi \in \map {\LL_R} {G, H}$.
From Addition of Linear Transformations:
:$\phi \oplus_H \psi \in \map {\LL_R} {G, H}$
From Negative Linear Transformation:
:$-\phi \in \map {\LL_R} {G, H}$
Thus, from the Two-Step Subgroup Test:
:$\struct {\map {\LL_R} {G, H}, \oplus_H}$ is a subgroup of $\struct {H^G, \oplus_H}$.
It remains to be shown that $\struct {\map {\LL_R} {G, H}, \oplus_H}$ is abelian.
Let $u$ and $v$ be arbitrary elements of $\map {\LL_R} {G, H}$.
Indeed, we have that:
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Linear Transformations
Category:Abelian Groups
Set of Liouville Numbers is Uncountable

== Theorem ==
The set of Liouville numbers is uncountable.
== Proof ==
By Corollary to Liouville's Constant is Transcendental, all numbers of the form:
where
:$a_1, a_2, a_3, \ldots \in \set {1, 2, \ldots, 9}$
are Liouville numbers.
Therefore each sequence in $\set {1, 2, \ldots, 9}$ defines a unique Liouville number.
By Set of Infinite Sequences is Uncountable, there are uncountable sequences in $\set {1, 2, \ldots, 9}$.
As the set of Liouville numbers has an uncountable subset, it is also uncountable by Sufficient Conditions for Uncountability.
Category:Transcendental Numbers
Set of Literals Satisfiable iff No Complementary Pairs

== Theorem ==
Let $S$ be a set of literals.
Then $S$ is satisfiable it contains no complementary pairs.
== Proof ==
=== Necessary Condition ===
Suppose $S$ contained the complementary pair $\set {p, \neg p}$.
By the Principle of Non-Contradiction, $\set {p, \neg p}$ is unsatisfiable.
By Superset of Unsatisfiable Set is Unsatisfiable, so is $S$.
Hence if $S$ is satisfiable, it cannot contain any complementary pairs.
=== Sufficient Condition ===
Suppose $S$ contains no complementary pair.
Define the boolean interpretation $v: \PP_0 \to \set {\T, \F}$ by:
$\quad \map v p = \begin{cases}
\T & \text{if } p \in S \\
\F & \text{otherwise}
\end{cases}$
Let $\mathbf A \in S$ be an arbitary literal.
If $\mathbf A = p$ is a positive literal, then $\map v {\mathbf A} = \T$ is true by the first clause of the definition.
Now suppose $\mathbf A = \neg p$ is a negative literal.
Then $p \notin S$, for $S$ contains no complementary pairs.
Therefore $\map v p = \F$, and hence:
:$\map v {\neg p} = \map v {\mathbf A} = \T$
Since $v$ has been demonstrated to be a model of $S$, $S$ is satisfiable.
== Sources ==
* : $\S 2.6.1$: Theorem $2.60$
Category:Boolean Interpretations
Category:Propositional Logic
Set of Local Minimum is Countable

== Theorem ==
Let $X$ be a subset of $\R$.
The set:
:$\leftset {x \in X: x}$ is local minimum in $\rightset X$
is countable.
== Proof ==
Define:
:$Y := \leftset {x \in X: x}$ is local minimum in $\rightset X$
By definition of $Y$ and definition of local minimum in set:
:$\forall x \in Y: \exists y \in \R: y < x \land \openint y x \cap X = \O$
By the Axiom of Choice, define a mapping $f: Y \to \powerset \R$ as:
:$\forall x \in Y: \exists y \in \R: \map f x = \openint y x \land y < x \land \map f x \cap X = \O$
We will prove that $f$ is an injection by definition:
Let $x_1, x_2 \in Y$ such that
:$\map f {x_1} = \map f {x_2}$
By definition of $f$:
:$\exists y_1 \in \R: \map f {x_1} = \openint {y_1} {x_1} \land y_1 < x_1 \land \map f {x_1} \cap X = \O$
and:
:$\exists y_2 \in \R: \map f {x_2} = \openint {y_2} {x_2} \land y_2 < x_2 \land \map f {x_2} \cap X = \O$
Then:
:$\openint {y_1} {x_1} = \openint {y_2} {x_2}$
Thus $x_1 = x_2$.
This ends the proof of injection.
By Cardinality of Image of Injection:
:$(1): \quad \card Y = \card {\map {f^\to} Y} = \card {\Img f}$
where
:$\card Y$ denotes the cardinality of $Y$,
:$\map {f^\to} Y$ denotes the image of $Y$ under $f$,
:$\Img f$ denotes the image of $f$.
We will prove that $\Img f$ is pairwise disjoint by definition.
Let $A, B \in \Img f$ such that
:$A \ne B$.
Then by definition of image:
:$\exists x_1 \in Y: \map f {x_1} = A$
and
:$\exists x_2 \in Y: \map f {x_2} = B$.
By difference of $A$ and $B$:
:$x_1 \ne x_2$
By definition of $f$:
:$\exists y_1 \in \R: \map f {x_1} = \openint {y_1} {x_1} \land y_1 < x_1 \land \map f {x_1} \cap X = \O$
and:
:$\exists y_2 \in \R: \map f {x_2} = \openint {y_2} {x_2} \land y_2 < x_2 \land \map f {x_2} \cap X = \O$
Aiming at contradiction suppose that
:$A \cap B \ne \O$.
$x_1 < x_2$ or $x_1 > x_2$.
In case when $x_1 < x_2$, $x_1 \in \map f {x_2}$ what contradicts with $\map f {x_2} \cap X = \O$.
In case when $x_1 > x_2$, analogically.
This ends the proof that $\Img f$ is pairwise disjoint.
By Set of Pairwise Disjoint Intervals is Countable:
:$\Img f$ is countable.
Thus by $(1)$ and Set is Countable if Cardinality equals Cardinality of Countable Set the result:
:$Y$ is countable.
== Sources ==
* 
Category:Real Analysis
Category:Countable Sets
Set of Logical Formulas is Inconsistent iff it has Finite Inconsistent Subset

== Theorem ==
Let $\FF$ be a collection of logical formulas.
Then:
:$\FF$ be inconsistent
:
there exists a finite subset of $\FF$ which it itself inconsistent.
== Proof ==
=== Sufficient Condition ===
Let $\FF$ be inconsistent.
Then it is possible to assemble a proof in a finite set of statements of a contradiction.
This finite set of statements uses within it a finite subset $\GG \subseteq \FF$ of the logical formulas of $\FF$.
Hence $\GG$ is that inconsistent finite subset of $\FF$ whose existence is proposed.
=== Necessary Condition ===
Let $\FF$ have a finite subset $\GG$ which is inconsistent.
Then it is possible to assemble a proof of a contradiction using logical formulas of $\GG$.
But those logical formulas are also logical formulas of $\FF$.
Then by definition $\FF$ is likewise inconsistent.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles
Category:Logical Consistency
Set of Mappings can be Ordered by Subset Relation

== Theorem ==
Let $S \times T$ be the product of two sets.
Let $\FF$ be a set of mappings on $S \times T$.
Then $\FF$ can be ordered by the subset relation.
== Proof ==
By the definition of mapping, a mapping is a specific type of relation.
The result then follows from Set of Relations can be Ordered by Subset Relation.
Category:Order Theory
Category:Subset Relation
Set of Mappings from Integers to Boolean Set is Uncountable

== Example of Use of Cantor's Diagonal Argument ==
Let $S$ be the Boolean set defined as:
:$S = \set {0, 1}$
Let $\mathbb G$ be the set of all mappings from the integers $\Z$ to $S$:
:$\mathbb G = \set {f: \Z \to S}$
Then $\mathbb G$ is uncountably infinite.
== Proof ==
This is an instance of the corollary to Cantor's Diagonal Argument.
== Sources ==
* : Chapter $1$: Preliminaries: $1.4$ Set Notation: Infinite sets
Category:Examples of Use of Cantor's Diagonal Argument
Set of Mappings which map to Same Element induces Equivalence Relation

== Theorem ==
Let $X$ and $Y$ be sets.
Let $E$ be the set of all mappings from $X$ to $Y$.
Let $b \in X$.
Let $\RR \subseteq E \times E$ be the relation on $E$ defined as:
:$\RR := \set {\tuple {f, g} \in \RR: \map f b = \map g b}$
Then $\RR$ is an equivalence relation.
== Proof ==
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
:$\forall f \in E: \map f b = \map f b$
That is:
:$\forall f \in E: \tuple {f, f} \in \RR$
Thus $\RR$ is seen to be reflexive.
=== Symmetry ===
Thus $\RR$ is seen to be symmetric.
=== Transitivity ===
Let $f, g, h \in E$.
Thus $\RR$ is seen to be transitive.
$\RR$ has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 7$: Relations: Exercise $5$
Category:Mapping Theory
Category:Equivalence Relations
Set of Matrices under Entrywise Addition forms Vector Space

== Theorem ==
Let $m, n \in \Z_{>0}$ be (strictly) positive integers.
Let $\map {\MM_\GF} {m, n}$ be the $m \times n$ matrix space over a field $\GF$.
Let $\struct {\map {\MM_\GF} {m, n}, +}$ denote the abelian group formed from $\map {\MM_\GF} {m, n}$ under matrix entrywise addition.
Let $\struct {\map {\MM_\GF} {m, n}, +, \times}_\GF$ denote the unitary module over $\GF$ where $\times$ denotes the matrix scalar product.
Then $\struct {\map {\MM_\GF} {m, n}, +, \times}_\GF$ forms a vector space.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
It is to be demonstrated that $\map {\MM_\GF} {m, n}$ satisfies all the vector space axioms.
First we note that from Matrix Entrywise Addition forms Abelian Group, $\struct {\map {\MM_\GF} {m, n}, +}$ is indeed an abelian group.
Hence vector space axioms $(\text V 0)$ to $(\text V 4)$ are satisfied.
=== Proof of ===
From Matrix Scalar Product Distributes over Number Addition:
For all $\mathbf A$ in $\map {\MM_\GF} {m, n}$ and $\lambda, \mu \in \GF$:
:$\paren {\lambda + \mu} \mathbf A = \lambda \mathbf A + \mu \mathbf A$
So holds for $\struct {\map {\MM_\GF} {m, n}, +, \times}_\GF$.
=== Proof of ===
From Matrix Scalar Product Distributes over Matrix Entrywise Addition:
For all $\mathbf A, \mathbf B$ in $\map {\MM_\GF} {m, n}$ and $\lambda \in \GF$:
:$\lambda \paren {\mathbf A + \mathbf B} = \lambda \mathbf A + \lambda \mathbf B$
So holds for $\struct {\map {\MM_\GF} {m, n}, +, \times}_\GF$.
=== Proof of ===
From Matrix Scalar Product is Associative:
For all $\mathbf A$ in $\map {\MM_\GF} {m, n}$ and $\lambda, \mu \in \GF$:
:$\lambda \paren {\mu \mathbf A} = \paren {\lambda \mu} \mathbf A$
So holds for $\struct {\map {\MM_\GF} {m, n}, +, \times}_\GF$.
=== Proof of ===
Let $1_\GF$ denote the multiplicative identity of $\GF$.
We have:
:$\forall i \in \set {1, 2, \ldots, m}$ and $\forall j \in \set {1, 2, \ldots, n}$:
::$1_\GF \sqbrk a_{i j} = \sqbrk {1_\GF a}_{i j}$
where $\sqbrk a_{i j}$ denotes an arbitrary element of $\mathbf A \in \map {\MM_\GF} {m, n}$.
So holds for $\struct {\map {\MM_\GF} {m, n}, +, \times}_\GF$.
Thus $\struct {\map {\MM_\GF} {m, n}, +, \times}_\GF$ satisfies all the vector space axioms and so is a vector space.
== Sources ==
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $4$. LINEAR VECTOR SPACE: Example $5$
Category:Set of Matrices under Entrywise Addition forms Vector Space
Category:Vector Spaces of Matrices
Category:Examples of Vector Spaces
Category:Matrix Entrywise Addition
Set of Meet Irreducible Elements Excluded Top is Order Generating

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a continuous complete lattice.
Let $X = \map {\mathit {IRR} } L \setminus \set \top$
where:
:$\map {\mathit {IRR} } L$ denotes the set of all meet irreducible element of $S$
:$\top$ denotes the top of $L$.
Then $X$ is order generating.
== Proof ==
We will prove that:
:$\forall x, y \in S: \paren {y \npreceq x \implies \exists p \in X: x \preceq p \land y \npreceq p}$
Let $x, y \in S$ such that:
:$y \npreceq x$
By Not Preceding implies There Exists Meet Irreducible Element Not Preceding:
:$\exists p \in S: p$ is meet irreducible and $x \preceq p$ and $y \npreceq p$
By definition of greatest element:
:$p \ne \top$ and $p \in \map {\mathit {IRR} } L$
By definitions of difference and singleton:
:$p \in X$
Thus
:$\exists p \in X: x \preceq p \land y \npreceq p$
Hence by Order Generating iff Not Preceding implies There Exists Element Preceding and Not Preceding:
:$X$ is order generating.
== Sources ==
* 
* 
Category:Continuous Lattices
Category:Order Generating Subsets
Category:Meet Irreducible Elements
Set of Monomials is Closed Under Multiplication

== Theorem ==
Let $M$ be the set of all monomials on the set $\set {X_j: j \in J}$, with multiplication $\circ$ defined by:
:$\ds \paren {\prod_{j \mathop \in J} X_j^{k_j} } \circ \paren {\prod_{j \mathop \in J} X_j^{k_j'} } = \paren {\prod_{j \mathop \in J} X_j^{k_j + k_j'} }$
Then $M$ is closed under $\circ$.
== Proof ==
Let $\ds m_1 = \prod_{j \mathop \in J} X_j^{k_j}, m_2 = \prod_{j \mathop \in J} X_j^{k_j'}$ be two monomials.
Their product is:
:$\ds m_1 \circ m_2 = \paren {\prod_{j \mathop \in J} X_j^{k_j + k_j'} }$
If $k_j + k_j' \ne 0$ then either $k_j \ne 0$ or $k_j' \ne 0$ (or both are nonzero).
Therefore if $k_j + k_j' \ne 0$ for infinitely many $j$, then either $m_1$ or $m_2$ is not a monomial.
Category:Monomials
Set of Natural Numbers Equals Union of its Successor

== Theorem ==
Let $\omega$ denote the set of natural numbers as defined by the von Neumann construction on a Zermelo universe $V$.
Then:
:$\bigcup \omega^+ = \omega$
== Proof ==
We have that:
:$\omega^+ = \omega \cup \set \omega$
and so:
:$\omega \subseteq \omega^+$
By definition:
:$\bigcup \omega^+ = \set {x: \exists y \in \omega^+: x \in y}$
Thus:
:$x \in \bigcup \omega^+ \implies x \in \omega$
So by definition of set equality:
:$\bigcup \omega^+ = \omega$
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 3$ Derivation of the Peano postulates and other results: Exercise $3.3 \ \text {(b)}$
Category:Natural Numbers
Set of Natural Numbers Equals its Union

== Theorem ==
Let $\omega$ denote the set of natural numbers as defined by the von Neumann construction on a Zermelo universe $V$.
Then:
:$\bigcup \omega = \omega$
== Proof ==
We have that:
:$\omega \in V$
where $V$ is a Zermelo universe.
By the we have that $\omega$ is a set.
By the , $\omega$ is transitive.
Hence by Class is Transitive iff Union is Subclass:
:$\bigcup \omega \subseteq \omega$
Let $n \in \omega$.
Then by definition of the von Neumann construction:
:$n + 1 = n \cup \set n$
That is:
:$\exists X \in \omega: n \in X$
where in this case $X = n + 1$.
Thus we have that:
:$n \in \set {x: \exists X \in \omega: x \in X}$
and so by definition of union of class:
:$n \in \bigcup \omega$
Thus we have that:
:$\omega \subseteq \bigcup \omega$
which, together with:
:$\bigcup \omega \subseteq \omega$
gives us, by set equality:
:$\bigcup \omega = \omega$
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 3$ Derivation of the Peano postulates and other results: Exercise $3.3 \ \text {(a)}$
Category:Natural Numbers
Set of Natural Numbers can be Derived using Axiom of Abstraction

== Theorem ==
Let $\N$ denote the set of natural numbers.
By application of the , $\N$ can be derived as a valid object in Frege set theory.
== Proof ==
== Sources ==
* : Chapter $1$: General Background: $\S 7$ Frege set theory
Category:Natural Numbers
Category:Axiom of Abstraction
Set of Natural Numbers is Ordinal

== Theorem ==
The set of natural numbers $\N$ is an ordinal.
== Proof ==
From Natural Number is Ordinal, every element of $\N$ is an ordinal.
From Union of Set of Ordinals is Ordinal, $\bigcup \N$ is therefore itself an ordinal.
From Set of Natural Numbers Equals its Union:
:$\bigcup \N = \N$
Hence the result.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 3$ Some ordinals
Category:Natural Numbers
Category:Ordinals
Set of Natural Numbers is Primitive Recursive

== Theorem ==
The set of natural numbers $\N$ is primitive recursive.
== Proof ==
The characteristic function $\chi_\N: \N \to \N$ is defined as:
:$\forall n \in \N: \chi_\N \left({n}\right) = 1$.
So:
: $\chi_\N \left({n}\right) = f^1_1 \left({n}\right)$
The constant function $f^1_1$ is primitive recursive.
Hence the result.
Category:Primitive Recursive Functions
Category:Natural Numbers
Set of Natural Numbers is Smallest Ordinal Greater than All Natural Numbers

== Theorem ==
The set of natural numbers $\N$ is the smallest ordinal which is greater than all natural numbers.
== Proof ==
From Set of Natural Numbers is Ordinal, $\N$ is an ordinal.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 3$ Some ordinals
Category:Natural Numbers
Category:Ordinals
Set of Non-Negative Real Numbers is not Well-Ordered by Usual Ordering

== Theorem ==
The set of non-negative real numbers $\R_{\ge 0}$ is not well-ordered under the usual ordering $\le$.
== Proof ==
 $\R_{\ge 0}$ is a well-ordered set.
Then by definition, all subsets of $\R_{\ge 0}$ has a smallest element.
Take the subset $\R_{> 0}$:
:$\R_{> 0} = \left\{ {x \in \R_{\ge 0}: x > 0}\right\} = \R_{\ge 0} \setminus \left\{ {0}\right\}$
Suppose $x \in \R_{> 0}$ is a smallest element.
Then $\dfrac x 2 \in \R_{> 0}$.
But $\dfrac x 2 < x$, which contradicts the supposition that $x \in \R_{> 0}$ is a smallest element.
Hence there can be no such smallest element.
So by Proof by Contradiction, $\R_{\ge 0}$ is not well-ordered by $\le$.
== Sources ==
* : $\S 1.2.1$: Mathematical Induction: Exercise $15 \ \text{(c)}$
Category:Real Numbers
Category:Well-Orderings
Set of Non-Zero Natural Numbers is Primitive Recursive

== Theorem ==
Let $\N^*$ be defined as $\N^* = \N \setminus \set 0$.
The subset $\N^* \subset \N$ is primitive recursive.
== Proof ==
We have that the characteristic function $\chi_{\set 0}$ of $\set 0$ is primitive recursive.
We note that:
:If $n = 0$ then $\map {\chi_{\set 0} } n = 1$ therefore $\map {\chi_{\set 0} } {\map {\chi_{\set 0} } n} = 0$.
:If $n > 0$ then $\map {\chi_{\set 0} } n = 0$ therefore $\map {\chi_{\set 0} } {\map {\chi_{\set 0} } n} = 1$.
Thus $\map {\chi_{\set 0} } {\map {\chi_{\set 0} } n} = \map {\chi_{\N^*} } n$.
So $\chi_{\N^*}$ is obtained by substitution from the primitive recursive function $\chi_{\set 0}$.
Hence the result.
Category:Primitive Recursive Functions
Set of Normal Subgroups of Group is Subsemigroup of Power Set Semigroup

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\circ_\PP$ be the operation induced by $\circ$ on $\powerset G$, the power set of $G$.
Let $\HH$ be the set of all normal subgroups of $\struct {G, \circ}$.
Then the algebraic structure $\struct {\HH, \circ_\PP}$ is a subsemigroup of the algebraic structure $\struct {\powerset G, \circ_\PP}$.
== Proof ==
From Power Structure of Group is Semigroup, we have that $\struct {\powerset G, \circ_\PP}$ is a semigroup.
Note that $\HH \subseteq \powerset G$.
From Subset Product of Normal Subgroups is Normal, $\struct {\HH, \circ_\PP}$ is closed.
By Subsemigroup Closure Test, $\struct {\HH, \circ_\PP}$ is a subsemigroup of $\struct {\powerset G, \circ_\PP}$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.11$
Category:Normal Subgroups
Category:Subset Products
Category:Semigroups
Set of Normal Subgroups of Group is Subsemigroup of Power Set under Intersection

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\HH$ be the set of all normal subgroups of $\struct {G, \circ}$.
Then the algebraic structure $\struct {\HH, \cap}$ is a subsemigroup of the algebraic structure $\struct {\powerset G, \cap}$.
== Proof ==
From Power Set with Intersection is Commutative Monoid, we have that $\struct {\powerset G, \cap}$ is a fortiori a semigroup.
Note that $\HH \subseteq \powerset G$.
Let $H_1$ and $H_2$ be normal subgroups of $\struct {G, \circ}$.
From Intersection of Normal Subgroups is Normal, $H_1 \cap H_2$ is also a normal subgroup of $\struct {G, \circ}$.
Hence $\struct {\HH, \cap}$ is closed.
By Subsemigroup Closure Test, $\struct {\HH, \cap}$ is a subsemigroup of $\struct {\powerset G, \cap}$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.11$
Category:Normal Subgroups
Category:Set Intersection
Category:Power Set
Category:Semigroups
Set of Null Strings forms Formal Language

== Example of Formal Language ==
The set consisting of the null string $\epsilon$ forms a formal language.
== Proof ==
$\set \epsilon$ vacuously forms a set of symbols.
The formal grammar is vacuous.
The result follows by definition of formal language.
== Also see ==
* Empty Set forms Formal Language: a different formal language from this one
== Sources ==
* : Chapter $1$: Preliminaries: $1.1$ Strings, Alphabets and Languages
Category:Examples of Formal Languages
Set of Numbers of form n - 1 over n is Bounded Above

== Theorem ==
Let $S$ be the subset of the set of real numbers $\R$ defined as:
:$S = \set {\dfrac {n - 1} n: n \in \Z_{>0} }$
$S$ is bounded above with supremum $1$.
$S$ has no greatest element.
== Proof ==
We have that:
:$\dfrac {n - 1} n = 1 - \dfrac 1 n$
As $n > 0$ it follows from Reciprocal of Strictly Positive Real Number is Strictly Positive that $\dfrac 1 n > 0$.
Thus $1 - \dfrac 1 n < 1$ and so $S$ is bounded above by $1$.
Next it is to be shown that $1$ is the supremum of $S$.
Suppose $x$ is the supremum of $S$ such that $x < 1$.
Then:
:$1 - x = \epsilon$
where $\epsilon \in \R_{>0}$.
By the Axiom of Archimedes we have that:
:$\exists m \in \Z_{>0}: n > \dfrac 1 \epsilon$
and so from Reciprocal Function is Strictly Decreasing:
:$\exists m \in \Z_{>0}: \dfrac 1 n < \epsilon$
Thus:
:$1 - \dfrac 1 m > 1 - \epsilon = x$
and so $x$ is not the supremum of $S$ after all.
Thus $\sup S$ cannot be less than $1$.
It follows that:
:$\sup S = 1$
Next it is noted that:
:$\forall n \in \Z_{>0}: \dfrac 1 n > 0$
and so:
:$\forall x \in S: x < 1$
Thus as $\sup S \notin S$ it follows from Greatest Element is Supremum that $S$ has no greatest element.
== Sources ==
* : $\S 3$: Natural Numbers: Exercise $\S 3.6 \ (1)$
Category:Suprema
Set of Odd Integers is Countably Infinite

== Theorem ==
Let $\Bbb O$ be the set of odd integers.
Then $\Bbb O$ is countably infinite.
== Proof ==
Let $f: \Bbb O \to \Z$ be the mapping defined as:
:$\forall x \in \Bbb O: \map f x = \dfrac {x + 1} 2$
$f$ is well-defined as $x + 1$ is even and so $\dfrac {x + 1} 2 \in \Z$.
Let $x, y \in \Bbb O$ such that $\map f x = \map f y$.
Then:
Thus $f$ is injective by definition.
Consider the inverse $f^{-1}$.
By inspection:
:$\forall x \in \Z: \map {f^{-1} } x = 2 x - 1$
$f^{-1}$ is well-defined, and $2 x - 1$ is odd.
Thus $f^{-1}$ is a mapping from $\Z$ to $\Bbb O$.
Then:
Thus $f^{-1}$ is injective by definition.
It follows by the Cantor-Bernstein-Schröder Theorem that there exists a bijection between $\Z$ and $\Bbb O$.
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.1$: Functions: Example $2.1.6$
Category:Countable Sets
Category:Odd Integers
Set of Orbits forms Partition

== Theorem ==
Let $G$ be a group.
Let $X$ be a set.
Let $G$ act on $X$.
Then the set of orbits of the group action forms a partition of $X$.
== Proof ==
Follows from the Fundamental Theorem on Equivalence Relations.
== Also see ==
* Group Action Induces Equivalence Relation
Category:Group Actions