nyuuzyou/wikis huggingface dataset:
Set of Subrings forms Complete Lattice

== Theorem ==
Let $\struct {K, +, \circ}$ be a ring.
Let $\mathbb K$ be the set of all subrings of $K$.
Then $\struct {\mathbb K, \subseteq}$ is a complete lattice.
== Proof ==
Let $\O \subset \mathbb S \subseteq \mathbb K$.
By Intersection of Subrings is Largest Subring Contained in all Subrings:
:$\bigcap \mathbb S$ is the largest subring of $K$ contained in each of the elements of $\mathbb S$.
By Intersection of Subrings Containing Subset is Smallest:
:The intersection of the set of all subrings of $K$ containing $\bigcup \mathbb S$ is the smallest subring of $K$ containing $\bigcup \mathbb S$.
Thus:
:Not only is $\bigcap \mathbb S$ a lower bound of $\mathbb S$, but also the largest, and therefore an infimum.
:The supremum of $\mathbb S$ is the intersection of the set of all subrings of $K$.
Therefore $\struct {\mathbb K, \subseteq}$ is a complete lattice.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.4$: Corollary
Category:Complete Lattices
Category:Subrings
Set of Subsemigroups forms Complete Lattice

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $\mathbb S$ be the set of all subsemigroups of $S$.
Then:
:$\struct {\mathbb S, \subseteq}$ is a complete lattice.
where for every set $\mathbb H$ of subsemigroups of $S$:
:the infimum of $\mathbb H$ necessarily admitted by $\mathbb H$ is $\ds \bigcap \mathbb H$.
== Proof ==
From Semigroup is Subsemigroup of Itself:
:$\struct {S, \circ} \in \mathbb S$
Let $\mathbb H$ be a non-empty subset of $\mathbb S$.
Let $T = \bigcap \mathbb H$.
Then:
Hence, from Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice:
:$\struct {\mathbb S, \subseteq}$ is a complete lattice
where $\ds \bigcap \mathbb H$ is the infimum of $\mathbb H$.
Category:Examples of Complete Lattices
Category:Subsemigroups
Set of Subsemigroups of Commutative Semigroup form Subsemigroup of Power Structure

== Theorem ==
Let $\struct {S, \circ}$ be a commutative semigroup.
Let $\struct {\powerset S, \circ_\PP}$ denote the power structure of $\struct {S, \circ}$.
Let $\TT$ be the set of all subsemigroups of $S$.
Then $\struct {\TT, \circ_\PP}$ is a subsemigroup of $\struct {\powerset S, \circ_\PP}$.
== Proof ==
First we establish that from Power Structure of Semigroup is Semigroup:
:$\struct {\powerset S, \circ_\PP}$ is a semigroup.
From Subset Product within Commutative Structure is Commutative:
:$\struct {\powerset S, \circ_\PP}$ is a commutative semigroup.
Let $A$ and $B$ be arbitrary subsemigroups of $S$.
As $A$ and $B$ are subsemigroups of $S$, they themselves are closed for $\circ$.
That is:
:$\forall x, y \in A: x \circ y \in A$
and:
:$\forall x, y \in B: x \circ y \in B$
By definition of operation induced on $\powerset S$:
:$A \circ_\PP B = \set {a \circ b: a \in A, b \in B}$
We are to show that:
:$\forall x, y \in A \circ_\PP B: x \circ y \in A \circ_\PP B$
Let $x, y \in A \circ_\PP B$ such that $x = a_x \circ b_x$, $y = a_y \circ b_y$.
We have:
That is:
:$x, y \in A \circ_\PP B \implies x \circ y \in A \circ_\PP B$
and $A \circ_\PP B$ is seen to be $\struct {\TT, \circ_\PP}$.
Hence the result.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.9$
Category:Commutative Semigroups
Category:Power Structures
Category:Subsemigroups
Set of Subset of Reals with Cardinality less than Continuum has not Interval in Union Closure

== Theorem ==
Let $\BB$ be a set of subsets of $\R$, the set of all real numbers.
Let:
:$\card \BB < \mathfrak c$
where
:$\card \BB$ denotes the cardinality of $\BB$
:$\mathfrak c = \card \R$ denotes continuum.
Let $\FF = \set {\bigcup \GG: \GG \subseteq \BB}$.
Then:
:$\exists x, y \in \R: x < y \land \hointr x y \notin \FF$
== Proof ==
Define:
:$ Z = \leftset {x \in \R: \exists U \in \FF: x}$ is local minimum in $\rightset U$
By Set of Subsets of Reals with Cardinality less than Continuum Cardinality of Local Minimums of Union Closure less than Continuum:
:$\card Z < \mathfrak c$
Then by Cardinalities form Inequality implies Difference is Nonempty:
:$\R \setminus Z \ne \O$
Hence by definition of empty set:
:$\exists z: z \in \R \setminus Z$
By definition of difference:
:$z \in \R \land z \notin Z$
Thus $z < z + 1$.
We will show that $z$ is local minimum in $\hointr z {z + 1}$.
Thus:
:$z \in \hointr z {z + 1}$
Hence:
:$z - 1 < z$
Thus:
:$\openint {z - 1} z \cap \hointr z {z + 1} = \O$
Then by definition $z$ is a local minimum in $\hointr z {z + 1}$.
Because $z \notin Z$:
:$\hointr z {z + 1} \notin \FF$
== Sources ==
* 
Category:Infinite Sets
Set of Subsets is Cover iff Set of Complements is Free

== Theorem ==
Let $S$ be a set.
Let $\CC$ be a set of sets.
Then $\CC$ is a cover for $S$ $\set {\relcomp S X: X \in \CC}$ is free.
== Proof ==
Let $S$ be a set.
Let us recall the definition of free:
=== Necessary Condition ===
Let $\CC$ be a cover for $S$:
:$S \subseteq \bigcup \CC$
Suppose:
:$\set {\relcomp S X: X \in \CC}$
is not free.
Then there exists an $x \in S$ such that:
:$\forall X \in \CC: x \in \relcomp S X$
But by the definition of (relative) complement:
:$\forall X \in \CC: x \notin X$
That is:
:$x \notin \bigcup \CC$
which contradicts:
:$S \subseteq \bigcup \CC$
=== Sufficient Condition ===
Let:
:$\set {\relcomp S X: X \in \CC}$
be free.
Suppose:
:$\bigcup \CC$
is not a cover for $S$.
Then there exists some $x \in S$ such that:
:$\forall X \in \CC: x \notin X$
Hence:
:$\forall \relcomp S X \in \CC: x \in X$
which contradicts the supposition that: 
:$\set {\relcomp S X: X \in \CC}$
is free.
Category:Set Theory
Set of Subsets of Element of Minimally Inductive Class under Progressing Mapping is Finite

== Theorem ==
Let $M$ be a class which is minimally inductive under a progressing mapping $g$.
Let $x \in M$.
Let $S$ be the set of all $y \in M$ such that $y \subseteq x$.
Then $S$ is finite.
== Proof ==
The proof proceeds by general induction.
For all $x \in M$, let $\map P x$ be the proposition:
:$\set {y \in M: y \subseteq x}$ is finite.
=== Basis for the Induction ===
We have that:
:$\set {y \in M: y \subseteq \O} = \O$
From Cardinality of Empty Set, $\O$ is finite.
Thus $\map P \O$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P x$ is true, where $x \in M$, then it logically follows that $\map P {\map g x}$ is true.
So this is the induction hypothesis:
:$\set {y \in M: y \subseteq x}$ is finite.
from which it is to be shown that:
:$\set {y \in M: y \subseteq \map g x}$ is finite.
=== Induction Step ===
This is the induction step:
Let $S = \set {y \in M: y \subseteq x}$.
By the induction hypothesis, $S$ is finite.
Consider the set:
:$S' = \set {y \in M: y \subseteq \map g x}$
By definition of progressing mapping:
:$x \subseteq \map g x$
Hence it follows that:
:$S \subseteq S'$
Let $y \in S' \setminus S$.
Then:
:$y \notin S$
and so:
:$y \nsubseteq x$
From Minimally Inductive Class under Progressing Mapping induces Nest:
:$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$
As it is not the case that $y \subseteq x$, it must be that:
:$\map g x \subseteq y$
From sandwich principle:
:$x \subseteq y \subseteq \map g x \implies x = y \lor y = \map g x$
As $x \ne y$ (or then $y \subseteq x$), it must be the case that:
:$y = \map g x$
So:
:$S' \setminus S = \map g x$
and so is finite, containing $1$ element.
We have that:
:$S' = S \cup \paren {S' \setminus S}$
and it follows from Union of Finite Sets is Finite that $S'$ is finite.
So $\map P x \implies \map P {\map g x}$ and the result follows by the Principle of General Induction.
Therefore:
:$\forall x \in M: \set {y \in M: y \subseteq x}$ is finite.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 6$ Finite Sets: Exercise $6.4$
Category:Minimally Inductive Classes
Category:Progressing Mappings
Category:Finite Sets
Set of Subsets of Finite Character of Countable Set is of Type M

== Theorem ==
Let $D$ be a countable set.
Let $S$ be a set of subsets of $D$ such that $S$ is of finite character.
Then $S$ is of type $M$, that is:
:every element of $S$ is a subset of a maximal element of $S$ under the subset relation.
== Proof ==
We have that $\ds \bigcup S \subseteq D$.
Hence by Subset of Countable Set is Countable, $\ds \bigcup S$ is itself countable.
Hence from Countable Set has Choice Function, $S$ has a choice function.
The result follows from Set of Finite Character with Choice Function is of Type $M$.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles
Category:Finite Character
Category:Choice Functions
Category:Type M Sets
Category:Countable Sets
Set of Subsets of Reals with Cardinality less than Continuum Cardinality of Local Minimums of Union Closure less than Continuum

== Theorem ==
Let $\BB$ be a set of subsets of $\R$.
Let:
:$\size \BB < \mathfrak c$
where
:$\size \BB$ denotes the cardinality of $\BB$
:$\mathfrak c = \size \R$ denotes the cardinality of the continuum.
Let
:$X = \leftset {x \in \R: \exists U \in \set {\bigcup \GG: \GG \subseteq \BB}: x}$ is a local minimum in $\rightset U$
Then:
:$\size X < \mathfrak c$
== Proof ==
We will prove that:
:$(1): \quad \size \BB \aleph_0 < \mathfrak c$
where $\aleph_0 = \size \N$ by Aleph Zero equals Cardinality of Naturals.
In the case when $\size \BB = \mathbf 0$ we have by Zero of Cardinal Product is Zero:
:$\size \BB \aleph_0 = \mathbf 0 < \mathfrak c$
In the case when $\mathbf 0 < \size \BB < \aleph_0$:
In the case when $\size \BB \ge \aleph_0$ we have:
Define:
:$Y = \leftset {x \in \R: \exists U \in \BB: x}$ is a local minimum in $\rightset U$
We will show that $X \subseteq Y$ by definition of subset.
Let $x \in X$.
By definition of $X$:
:$\exists U \in \leftset {\bigcup \GG: \GG \subseteq \BB}: x$ is local minimum in $\rightset U$
:$\exists \GG \subseteq \BB: U = \bigcup \GG$
By definition of local minimum:
:$x \in U$
By definition of union:
:$\exists V \in \GG: x \in V$
By definition of subset:
:$V \in \BB$
By definition of local minimum:
:$\exists y \in \R: y < x \land \openint y x \cap U = \O$
By Set is Subset of Union:
:$V \subseteq U$
Then:
:$\exists y \in \R: y < x \land \openint y x \cap V = \O$
By definition:
:$x$ is local minimum in $V$
Thus by definition of $Y$
:$x \in Y$
So
:$(2): \quad X \subseteq Y$
Define $\family {Z_A}_{A \mathop \in \BB}$ as:
:$Z_A = \leftset {x \in \R: x}$ is local minimum in $\rightset A$
We will prove that:
:$(3): \quad Y \subseteq \ds \bigcup_{A \mathop \in \BB} Z_A$
Let $x \in Y$.
By definition of $Y$:
:$\exists U \in \BB: x$ is local minimum in $U$
By definition of $Z_U$:
:$x \in Z_U$
Thus by definition of union:
:$x \in \ds \bigcup_{A \mathop \in \BB} Z_A$
This ends the proof of inclusion.
By Set of Local Minimum is Countable:
:$\forall A \in \BB: Z_A$ is countable
By Countable iff Cardinality not greater Aleph Zero:
:$\forall A \in \BB: \size {Z_A} \le \aleph_0$
By Cardinality of Union not greater than Product:
:$(4): \quad \ds \size {\bigcup_{A \mathop \in \BB} Z_A} \le \size \BB \aleph_0$
Thus:
== Sources ==
* 
Category:Infinite Sets
Category:Cardinality of Continuum
Category:Real Numbers
Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice

== Theorem ==
Let $A$ be a set.
Let $\SS$ be a set of subsets of $A$ such that:
:$A \in \SS$
:for every non-empty subset $\TT$ of $\SS$, $\ds \bigcap \TT \in \SS$, where $\ds \bigcap \TT$ denotes the intersection of $\TT$.
Then:
:the ordered set $\struct {\SS, \subseteq}$ is a complete lattice
where:
:$\ds \bigcap \TT$ is the infimum necessarily admitted by $\TT$.
== Proof ==
From Subset Relation is Ordering, $\struct {\SS, \subseteq}$ is indeed an ordered set.
We have by definition of $\SS$ that:
:$\forall H \in \SS: H \subseteq A$
Hence $A$ is the greatest element of $\SS$ with respect to the ordered set $\struct {A, \subseteq}$.
Then from Intersection is Largest Subset:
:$\forall K \in \TT: \ds \bigcap \TT \subseteq K$
and so $\TT$ admits an infimum, which is $\ds \bigcap \TT$.
Hence the conditions of Ordered Set with Greatest Element whose Subsets have Infimum is Complete Lattice.
The result follows.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.11 \ \text{(b)}$
Category:Complete Lattices
Category:Subsets
Set of Successive Numbers contains Unique Multiple

== Theorem ==
Let $m \in \Z_{\ge 1}$.
Then $\set {m, m + 1, \ldots, m + n - 1}$ contains a unique integer that is a multiple of $n$.
That is, in any set containing $n$ successive integers, $n$ divides exactly one of those integers.
== Proof ==
Let $S_m = \set {m, m + 1, \ldots, m + n - 1}$ be a set containing $n$ successive integers.
The proof proceeds by induction on $m$, the smallest number in $S_m$.
=== Basis for the Induction ===
The case $m = 1$ is verified as follows:
:$S_1 = \set {1, 2, \ldots, n}$
This contains a multiple of $n$, namely, $n$ itself.
If $S_1$ is a singleton then this element is trivially unique.
If there are any other integers in $S_1$, they are less than $n$ and more than $0$, and so cannot be a multiple of $n$.
This is the basis for the induction.
=== Induction Hypothesis ===
Fix $m \in \N$ with $m \ge 1$.
Assume that:
:$S_m = \set {m, m + 1, \ldots, m + n - 1}$
contains a unique multiple of $n$.
This is our induction hypothesis.
=== Induction Step ===
This is our induction step:
Consider the set:
:$S_{m + 1} = \set {m + 1, m + 2, \ldots, m + n - 1, m + n}$
As $n \equiv 0 \pmod n$, $m + n$ is a multiple of $n$ $m$ is a multiple of $n$.
That means that we can replace $m + n$ with $m$ and instead analyze:
:${S_{m + 1} }' = \set {m + 1, m + 2, \ldots, m + n - 1, m}$
But ${S_{m + 1} }' = S_m$, which contains a unique multiple of $n$ by the induction hypothesis.
The result follows by the Principle of Mathematical Induction.
Category:Number Theory
Set of Total Functions is Not Recursive

== Theorem ==
The set $\operatorname{Tot}$ of natural numbers which code URM programs which compute total functions of one variable is not recursive.
== Proof ==
We perform a proof by Cantor's Diagonal Argument.
Suppose the contrary, that $\operatorname{Tot}$ is a recursive set.
First we define a recursive function $h$ which enumerates the code numbers of URM programs which compute total functions of one variable.
The program of this sort with the smallest code is:
Line		Command
$1$ 		$\map Z n$
which computes the zero function.
Its code number is $2^3 = 8$.
So we can define the recursive function $h$ as:
:$\map h n = \begin{cases}
8 & : n = 0 \\
\map {\mu z} {z \in \operatorname{Tot} \text { and } \map h {n - 1} < z} & : n > 0
\end{cases}$
where $\mu z$ is the minimization operation.
As there are infinitely many URM programs which compute a total function of one variable, $\map h n$ is always defined.
More importantly, if $f: \N \to \N$ is a total recursive function, there is some $m \in \N$ such that the URM program with code number $\map h n$ computes $f$.
Now we define the function $\Phi: \N^2 \to \N$ as:
:$\map \Phi {m, n} = \map {\Phi_1} {\map h m, n}$
where $\Phi_1$ is the universal URM computable function of one variable.
Then $\Phi$ is a total recursive function, because:
:$\Phi_1$ and $h$ are both recursive
:The program coded by $m$ halts for all inputs $n$.
It has the property that if $f: \N \to \N$ is a total recursive function, there is some $m \in \N$ such that $\map \Phi {m, n} = \map f n$ for all $n \in \N$.
Now we use Cantor's Diagonal Argument.
It follows immediately that the function $g: \N \to \N$ given by:
:$\map g n = \map \Phi {n, n} + 1$
is also recursive.
Hence there is some number $m_0$ such that:
:$\map g n = \map \Phi {m_0, n}$
for all $n \in \N$.
In particular:
:$\map g {m_0} = \map \Phi {m_0, m_0}$
But from the definition of $g$:
:$\map g {m_0} = \map \Phi {m_0, m_0} + 1$
This contradiction arose because we assumed that $\operatorname{Tot}$ is recursive.
Hence the result.
Category:URM Programs
Category:Recursion Theory
Set of Transpositions is not Subgroup of Symmetric Group

== Theorem ==
Let $S$ be a finite set with $n$ elements such that $n > 2$.
Let $G = \struct {\map \Gamma S, \circ}$ denote the symmetric group on $S$.
Let $H \subseteq G$ denote the set of all transpositions of $S$ along with the identity mapping which moves no elements of $S$.
Then $H$ does not form a subgroup of $G$.
== Proof ==
First it is noted that $H \subseteq G$, and that the identity mapping is an element of $H$.
Hence to demonstrate that $H$ is a subgroup of $G$, one may use the Two-Step Subgroup Test.
Let $\phi \in H$ be a transposition.
Then from Transposition is Self-Inverse, $\phi^{-1} \in H$.
So $H$ is closed under inversions.
Let $a, b, c \in S$.
Let $\pi, \phi \in H$ such that:
:$\pi = \tuple {a \ b}$
:$\phi = \tuple {b \ c}$
Then by inspection:
:$\phi \circ \pi = \tuple {a \ c \ b}$
Thus $\phi \circ \pi$ is a $3$-cycle and so not a transposition.
Hence $H$ is not closed under composition.
So, by the Two-Step Subgroup Test, $H$ is not a subgroup of $G$.
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{E iv}$
Category:Symmetric Groups
Set of Upper Closures of Compact Elements is Basis implies Complete Scott Topological Lattice is Algebraic

== Theorem ==
Let $L = \struct {S, \preceq, \tau}$ be a complete Scott Definition:Topological Lattice.
Let $\BB = \set {x^\succeq: x \in \map K L}$ be a basis of $L$, where:
:$x^\succeq$ denotes the upper closure of $x$,
:$\map K L$ denotes the compact subset of $L$.
Then $L$ is algebraic.
== Proof ==
Thus by Compact Closure is Directed:
:$\forall x \in S:x^{\mathrm {compact} }$ is directed
where $x^{\mathrm {compact} }$ denotes the compact closure of $x$.
Thus by definition of complete lattice:
:$L$ is up-complete.
Let $x \in S$.
By definition of lower closure of element:
:$x$ is upper closure for $x^\preceq$
By definition of supremum:
:$\map \sup {x^\preceq} \preceq x$
By Compact Closure is Intersection of Lower Closure and Compact Subset:
:$x^{\mathrm {compact} } = x^\preceq \cap \map K L$
By Intersection is Subset:
:$x^{\mathrm {compact} } \subseteq x^\preceq$
By Supremum of Subset:
:$\map \sup {x^{\mathrm {compact} } } \preceq \map \sup {x^\preceq}$
By definition of transitivity:
:$\map \sup {x^{\mathrm {compact} } } \preceq x$
:$x \ne \map \sup {x^{\mathrm {compact} } }$
We will prove that:
:$x \notin \paren {x^{\mathrm {compact} } }^\preceq$
:
:$x \in \paren {\map \sup {x^{\mathrm {compact} } } }^\preceq$
By definition of lower closure of element:
:$x \preceq \map \sup {x^{\mathrm {compact} } }$
Thus by definition of antisymmetry:
:this contradicts $x \ne \map \sup {x^{\mathrm {compact} } }$
By definition of relative complement:
:$x \in \relcomp S {\paren {\map \sup {x^{\mathrm {compact} } } }^\preceq}$
By Complement of Lower Closure of Element is Open in Scott Topological Ordered Set:
:$\relcomp S {\paren {\map \sup {x^{\mathrm {compact} } } }^\preceq}$ is open.
By definition of basis:
:$\relcomp S {\paren {\map \sup {x^{\mathrm {compact} } } }^\preceq} = \bigcup \set {G \in \BB: G \subseteq \relcomp S {\paren {\map \sup {x^{\mathrm {compact} } } }^\preceq} }$
By definition of union:
:$\exists X \in \set {G \in \BB: G \subseteq \relcomp S {\paren {\map \sup {x^{\mathrm {compact} } } }^\preceq} }: x \in X$
Then:
:$X \in \BB \land X \subseteq \relcomp S {\paren {\map \sup {x^{\mathrm {compact} } } }^\preceq}$
By definition of $\BB$:
:$\exists k \in \map K L: X = k^\succeq$
By definition of compact subset:
:$k$ is compact.
By definition of upper closure of element:
:$k \preceq x$
By definition of compact closure:
:$k \in x^{\mathrm {compact} }$
By definitions of supremum and upper bound:
:$k \preceq \map \sup {x^{\mathrm {compact} } }$
By definition of upper closure of element:
:$\map \sup {x^{\mathrm {compact} } } \in X$
By gefinition of subset:
:$\map \sup {x^{\mathrm {compact} } } \in \relcomp S {\paren {\map \sup {x^{\mathrm {compact} } } }^\preceq}$
By definition of difference:
:$\map \sup {x^{\mathrm {compact} } } \notin \paren {\map \sup {x^{\mathrm {compact} } } }^\preceq$
Thus by definition of lower closure of element:
:this contradicts $\map \sup {x^{\mathrm {compact} } } \in \paren {\map \sup {x^{\mathrm {compact} } } }^\preceq$
== Sources ==
* 
* 
Category:Continuous Lattices
Category:Topological Order Theory
Set of Vectors defined by Directed Line Segments in Space forms Vector Space

== Theorem ==
Let $\R^3$ be a real cartesian space of $3$ dimensions.
Consider the set $S$ of directed line segments in $\R^3$.
Let the equivalence relation $\sim$ be applied to $\R^3$ such that:
:$\forall L_1, L_2 \in \R^3: L_1 \sim L_2$ there exists a translation $T$ such that $\map T {L_1} = L_2$
Let $\mathbb V$ denote the set of equivalence classes of $\sim$ on $S$.
The elements of $\mathbb V$ form a vector space where:
:$\forall \mathbf v_1, \mathbf v_2 \in \mathbb V: \mathbf v_1 + \mathbf v_2$ denotes the element of $\mathbb V$ exemplified by the result of the operation of joining of a representative element of element of $\mathbf v_2$ to the end of a similarly representative element of element of $\mathbf v_1$.
:$\forall \mathbf v \in \mathbb V, \lambda \in \R: \lambda \mathbf v$ denotes the element of $\mathbb V$ exemplified by the result of the operation of mulitplying a representative element of element of $\mathbf v$ by the scale factor $\lambda$.
== Proof ==
== Sources ==
* : $\S 1$: $n$-Dimensional vector spaces: Definition $1$
Category:Vector Spaces
Set of Words Generates Group

== Theorem ==
Let $S \subseteq G$ where $G$ is a group.
Let $\hat S$ be defined as $S \cup S^{-1}$, where $S^{-1}$ is the set of all the inverses of all the elements of $S$.
Then $\gen S = \map W {\hat S}$, where $\map W {\hat S}$ is the set of words of $\hat S$.
=== Corollary ===
== Proof ==
Let $H = \gen S$ where $S \subseteq G$.
$H$ must certainly include $\hat S$, because any group containing $s \in S$ must also contain $s^{-1}$.
Thus $\hat S \subseteq H$.
By , $H$ must also contain all products of a finite number of elements of $\hat S$.
Thus $\map W {\hat S} \subseteq H$.
Now we prove that $\map W {\hat S} \le G$.
By the Two-Step Subgroup Test:
Let $x, y \in \map W {\hat S}$.
	
As $x$ and $y$ are both products of a finite number of elements of $\hat S$, it follows that so is their product $x y$.
Thus $x y \in \map W {\hat S}$ and is satisfied.
Let $x = s_1 s_2 \ldots s_n \in \map W {\hat S}$.
Then $x^{-1} = s_n^{-1} \ldots s_2^{-1} s_1^{-1} \in \map W {\hat S}$.
Thus the conditions of the Two-Step Subgroup Test are fulfilled, and $\map W {\hat S} \le G$.
Thus $\map W {\hat S}$ is the subgroup of $G$ generated by $S$.
== Sources ==
* : $\S 5.3$. Subgroup generated by a subset
* : $\S 1.9$: Theorem $20$
* : Chapter $\text{II}$: Groups: Exercise $\text{K}$
* : $\S 1.2$
Category:Group Theory
Category:Generated Subgroups
Category:Set of Words Generates Group
Set of Words Generates Group/Corollary

== Corollary to Set of Words Generates Group ==
Let $G$ be a group.
Let $T \subseteq G$.
Let $\map W T$ be the set of words of $T$.
If $T$ is closed under taking inverses, then $\map W T$ is a subgroup of $G$.
== Proof ==
This follows directly from Set of Words Generates Group and the fact that $T$ has the same properties as $\hat S$ in that result.
Category:Set of Words Generates Group
Set of all Self-Maps under Composition forms Monoid

== Theorem ==
Let $S$ be a set.
Let $S^S$ be the set of all mappings from $S$ to itself.
Let the operation $\circ$ represent composition of mappings.
Then the algebraic structure $\struct {S^S, \circ}$ is a monoid whose identity element is the identity mapping on $S$.
== Proof ==
By Set of all Self-Maps under Composition forms Semigroup, $\struct {S^S, \circ}$ is a semigroup.
By Identity Mapping is Left Identity and Identity Mapping is Right Identity the identity mapping on $S$ is the identity element of $\struct {S^S, \circ}$.
Since $\struct {S^S, \circ}$ is a semigroup with an identity element, it is a monoid by definition.
== Sources ==
* : $\S 4.3$. Units and zeros: Example $73$
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions
* : $\S 3.1$: Monoids: Exercise $(2)$
Category:Composite Mappings
Category:Examples of Monoids
Category:Self-Maps