nyuuzyou/wikis huggingface dataset:
Set of all Self-Maps under Composition forms Semigroup

== 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 semigroup.
== Proof ==
Let $f, g \in S^S$.
As the domain of $g$ and codomain of $f$ are the same, the composition $f \circ g$ is defined.
By the definition of composition, $f \circ g$ is a mapping from the domain of $g$ to the codomain of $f$.
Thus $f \circ g: S \to S$, so $f \circ g \in S^S$.
Since this holds for all $f, g \in S^S$, $\struct {S^S, \circ}$ is closed.
By Composition of Mappings is Associative, $\circ$ is associative.
Since $\struct {S^S, \circ}$ is closed and $\circ$ is associative:
:$\struct {S^S, \circ}$ is a semigroup.
== Sources ==
* : Chapter $\text{I}$: Semi-Groups and Groups: $1$: Definition and examples of semigroups
* : Chapter $4$. Groups: Exercise $7$
* : $\S 1$: Some examples of groups: Theorem $1.12$
* : Chapter $2$: The Definition of Group Structure: $\S 26 \alpha$
* : $\S 29$. Semigroups: definition and examples: $(2)$
* : Chapter $3$: Elementary consequences of the definitions: Remark
Category:Composite Mappings
Category:Examples of Semigroups
Set together with Condensation Points is not necessarily Closed

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $\CC$ denote the set of condensation points of $H$.
Then it is not necessarily the case that $H \cup \CC$ is a closed set of $T$.
== Proof ==
Proof by Counterexample:
Let $T = \struct {\R, \tau}$ denote the right order topology on $\R$.
Let $H \subseteq \R$ be a non-empty finite subset of $\R$.
Let $\CC$ denote the set of condensation points of $H$.
Since $H$ is finite, it has no condensation points, that is, $\CC = \O$.
Therefore $H \cup \CC = H$ is finite.
From Closed Sets of Right Order Space on Real Numbers, closed sets of $T$ are either infinite or is the empty set.
Since $H \cup \CC$ is finite, it is not a closed set of $T$.
== Also see ==
* Definition:Closure (Topology)
* Set together with Omega-Accumulation Points is not necessarily Closed
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Set Closures
Category:Condensation Points
Set together with Omega-Accumulation Points is not necessarily Closed

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $\Omega$ denote the set of $\omega$-accumulation points of $H$.
Then it is not necessarily the case that $H \cup \Omega$ is a closed set of $T$.
== Proof ==
Proof by Counterexample:
Let $T = \struct {\R, \tau}$ denote the right order topology on $\R$.
Let $H \subseteq \R$ be a non-empty finite subset of $\R$.
Let $\Omega$ denote the set of $\omega$-accumulation points of $H$.
Since $H$ is finite, it has no $\omega$-accumulation points, that is, $\Omega = \O$.
Therefore $H \cup \Omega = H$ is finite.
From Closed Sets of Right Order Space on Real Numbers, closed sets of $T$ are either infinite or is the empty set.
Since $H \cup \Omega$ is finite, it is not a closed set of $T$.
== Also see ==
* Definition:Closure (Topology)
* Set together with Condensation Points is not necessarily Closed
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Set Closures
Category:Omega-Accumulation Points
Set which is Equivalent to Countable Set is Countable

== Lemma ==
Let $S$ be countable set.
Let $T$ be a set.
Let $T$ be equivalent to $S$.
Then $T$ is countable.
== Proof ==
By definition of set equivalence:
:$S$ is in one-to-one correspondence with $T$.
We have that $S$ is countable.
By definition of countable set:
:$S$ is in one-to-one correspondence with a subset of the natural numbers.
$T$ is in one-to-one correspondence with $S$.
By Composite of Bijections is Bijection:
:$T$ is in one-to-one correspondence with a subset of the natural numbers.
Hence by definition of countable set:
$T$ is countable.
Category:Countable Sets
Category:Set Equivalence
Set which is Superinductive under Progressing Mapping has Fixed Point

== Theorem ==
Let $S$ be a set.
Let $g: S \to S$ be a progressing mapping on $S$.
Let $S$ be superinductive under $g$.
Then there exists $x \in S$ such that $x$ is a fixed point of $g$.
=== Corollary ===
== Proof ==
Let us assume the hypothesis.
Let $M$ be the intersection of all subsets of $S$ that are superinductive under $g$.
From:
:Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping
:Intersection of Set whose Every Element is Closed under Chain Unions is also Closed under Chain Unions
it follows by definition that $M$ is minimally superinductive under $g$.
Hence by definition $M$ is a $g$-Tower.
By Intersection of Non-Empty Class is Set, $M$ is a set.
Thus by Union of $g$-Tower is Greatest Element and Unique Fixed Point:
:$\ds \bigcup M$ is a fixed point of $g$
:$\ds \bigcup M \in M$
Because $M \subseteq S$ it follows that:
:$\ds \bigcup M \in S$
and so $\ds \bigcup M$ is the $x$ that is being proved to exist.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 2$ Superinduction and double superinduction: Theorem $2.8$
Category:Superinductive Classes
Category:Progressing Mappings
Set which is Superinductive under Progressing Mapping has Fixed Point/Corollary

== Corollary to Set which is Superinductive under Progressing Mapping has Fixed Point ==
Let $S$ be a non-empty set of sets.
Let $g: S \to S$ be a progressing mapping on $S$ such that:
:$(1): \quad S$ is closed under $g$
:$(2): \quad S$ is closed under chain unions.
Let $b \in S$.
Then there exists $x \in S$ such that:
:$b \subseteq x$
and:
:$\map g x = x$
== Proof ==
Let us assume the hypothesis.
;Case $b = \O$
This case reduces to Set which is Superinductive under Progressing Mapping has Fixed Point.
Hence the result follows for $b = \O$.
;Case $b \ne \O$
Let $S_b$ be the set of all elements $x$ of $S$ such that $b \subseteq x$, together with $\O$:
:$S_b = \set {x \in S: b \subseteq x} \cup \set \O$
It is clear by inspection that $S_b$ is closed under chain unions.
Let us define the mapping $g': S_b \to S_b$ as follows:
:$\forall x \in S_b: \map {g'} x = \begin{cases} b & : x = \O \\ \map g x & : x \ne \O \end{cases}$
Then:
:$g'$ is a progressing mapping on $S_b$
:$S_b$ is closed under $g'$
:$\O \in S_b$
So $S_b$ is superinductive under $g'$.
Hence by Set which is Superinductive under Progressing Mapping has Fixed Point:
:$\exists x \in S_b: x = \map {g'} x$
For such an $x$ we have that $x \ne \O$ because $\map {g'} \O = b$.
Hence:
:$b \subseteq x$
Also because $x \ne \O$:
:$\map {g'} x = \map g x$
and so:
:$\map g x = x$
Hence:
:$b \subseteq x$
and:
:$\map g x = x$
and the proof is complete.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 2$ Superinduction and double superinduction: Theorem $2.10$
Category:Superinductive Classes
Category:Progressing Mappings
Set whose every Finite Subset is Nest is also Nest

== Theorem ==
Let $x$ be a set of sets with the property that:
:every finite subset of $x$ is a nest.
Then $x$ is a nest.
== Proof ==
Let $x$ be a set of sets with the given property.
Let $a, b \in x$ be arbitrary.
Then:
:$\set {a, b} \subseteq x$
and so $\set {a, b}$ is a nest.
That is:
:$a \subseteq b$ or $b \subseteq a$
As $a$ and $b$ are arbitrary, it follows that:
:$\forall a, b \in x: a \subseteq b$ or $b \subseteq a$
The result follows from Subset Relation is Ordering.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles
Category:Nests
Set with Choice Function is Well-Orderable

== Theorem ==
Let $S$ be a set.
Let there exist a choice function for $S$.
Then $S$ is well-orderable.
== Proof ==
Let us define the notation:
:$\map {\PP_{\ne \O} } S := \powerset S \setminus \set \O$
for the power set of $S$ without the empty set.
Let $C: \map {\PP_{\ne \O} } S \to S$ be a choice function for $S$.
Let $x \subseteq S$ be a subset of $S$.
From Subset is Element of Power Set we have:
:$x \subseteq S \iff x \in \powerset S$
Let us define $g: \powerset S \to S$ as follows:
:$\forall x \in \powerset S: \map g x := \begin {cases} x \cup \paren {\map C {S \setminus x} } & : x \ne S \\ S & : x = S \end {cases}$
where $S \setminus x$ denotes the difference between $S$ and $x$.
We note that when $x \ne S \iff S \setminus x \ne \O$, hence $\map C {S \setminus x}$ is always defined in the above.
Thus:
:$\map g S = S$
and for an arbitrary proper subset $x$ of $S$, $\map g x$ adds to $x$ exactly $1$ element of $S$ which is not in $x$.
That is:
:$g$ is a slowly progressing mapping
and:
:$S$ is the unique fixed point of $g$.
Hence $S$ fulfils the assumptions of Set with Slowly Progressing Mapping on Power Set with Self as Fixed Point is Well-Orderable.
The result follows.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 4$ Well ordering and choice: Theorem $4.7$
Category:Choice Functions
Category:Well-Orderings
Set with Complement forms Partition

== Theorem ==
Let $\O \subset S \subset \mathbb U$.
Then $S$ and its complement $\map \complement S$ form a partition of the universal set $\mathbb U$.
== Proof ==
Follows directly from Set with Relative Complement forms Partition:
If $\O \subset T \subset S$, then $\set {T, \relcomp S T}$ is a partition of $S$.
Category:Set Complement
Category:Universal Set
Category:Set Partitions
Set with Dispersion Point is Biconnected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be a connected set in $T$.
Let $p \in H$ be a dispersion point of $H$.
Then $H$ is biconnected.
== Proof ==
 $H$ is not biconnected.
Then by definition there exist disjoint non-degenerate connected sets $U, V$ such that $H = U \cup V$.
, let $p \in U$.
Then $V \subset H \setminus \set p$.
As $p$ is a dispersion point of $H$, $H \setminus \set p$ is totally disconnected.
Thus, by definition, $H \setminus \set p$ contains no non-degenerate connected sets.
But this contradicts our definition of $V$, as $V \subset H \setminus \set p$.
It follows by Proof by Contradiction that $H$ is biconnected.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Biconnectedness and Continua
Category:Biconnected Sets
Category:Dispersion Points
Set with Slowly Progressing Mapping on Power Set with Self as Fixed Point is Well-Orderable

== Theorem ==
Let $S$ be a set of sets.
Let there exist a slowly progressing mapping $g: \powerset S \to \powerset S$, where $\powerset S$ denotes the power set of $S$.
Let $g$ have exactly one fixed point $S$.
Then $S$ is well-orderable.
== Proof ==
Let the hypothesis be assumed.
We have that:
:$\O \in \powerset S$
:$\powerset S$ is closed under $g$
:$\powerset S$ is closed under chain unions.
Hence $\powerset S$ is superinductive under $g$.
Let $M$ be the intersection of all subsets of $S$ that are superinductive under $g$.
From:
:Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping
:Intersection of Set whose Every Element is Closed under Chain Unions is also Closed under Chain Unions
it follows by definition that $M$ is minimally superinductive under $g$.
Hence by definition $M$ is a slow $g$-tower.
By Union of $g$-Tower is Greatest Element and Unique Fixed Point:
:$\ds \bigcup M \in M$ and $\ds \bigcup M$ is a fixed point of $g$.
But $S$ is the only fixed point of $g$.
Hence:
:$\ds \bigcup M = S$
From Union of Slow $g$-Tower is Well-Orderable, it follows that $S$ is well-orderable.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 4$ Well ordering and choice: Theorem $4.6$
Category:Slowly Progressing Mappings
Category:Well-Orderings
Set with Two Parallel Elements is Dependent

== Theorem ==
Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $A \subseteq S$.
Let $x, y \in S$.
Let $x, y$ be parallel elements.
If $x, y \in A$ then $A$ is dependent.
== Proof ==
Let $x, y \in A$.
 
From Doubleton of Elements is Subset:
:$\set{x, y} \subseteq A$
By the definition of parallel elements:
:$\set {x, y}$ is dependent
From Superset of Dependent Set is Dependent:
:$A$ is dependent
== Sources ==
* Chapter $1.$ $\S 4.$ Loops and parallel elements
Category:Matroid Parallelism
Category:Matroid Dependent Subsets
Sets in Modified Fort Space are Disconnected

== Theorem ==
Let $T = \struct {S, \tau_{a, b}}$ be a modified Fort space.
Let $H$ be a subset of $S$ with more than one point.
Then $H$ is disconnected.
== Proof ==
By Isolated Points in Subsets of Modified Fort Space:
:$\exists x \in H: x$ is isolated
By Point in Topological Space is Open iff Isolated, $\set x$ is open in $T$.
By Modified Fort Space is $T_1$ and definition of $T_1$ space, $\set x$ is closed in $T$.
Therefore $\relcomp S {\set x}$ is open in $T$.
Then we have:
This shows that $H$ is disconnected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $27$. Modified Fort Space: $4$
Category:Disconnected Sets
Category:Modified Fort Space
Sets in Modified Fort Space are Separated/Mistake

== Source Work ==
:
:Part $\text {II}$: Counterexamples
::Section $27$: Modified Fort Space
:::Item $4$
== Mistake ==
:Every point is a component since every set containing more than one point is separated, ...
Separated is the wrong word to use here, as that refers to the relationship between two subsets of a space.
The correct word is disconnected.
== Also see ==
* Modified Fort Space is Totally Disconnected
== Sources ==
* : Part $\text {II}$: Counterexamples: $27$. Modified Fort Space: $4$
Category:Mistakes/Counterexamples in Topology
Sets of 4 Integers a, b, c, d for which Every Integer is in form ax^2 + by^2 + cz^2 + du^2

== Theorem ==
There are exactly $55$ sets of $4$ integers $\left\{ {a, b, c, d}\right\}$ such that all integers can be written in the form:
:$n = a x^2 + b y^2 + c z^2 + d w^2$
for integer $x, y, z, w$.
== Proof ==
== Sources ==
* 
* : $55$
* : $55$
Category:Number Theory
Category:55
Sets of 4 Prime Quadruples

== Sequence ==
The following are the $38$ sets of prime quadruples up to $100 \, 000$:
:$5, 7, 11, 13$
:$11, 13, 17, 19$
:$101, 103, 107, 109$
:$191, 193, 197, 199$
:$821, 823, 827, 829$
:$1481, 1483, 1487, 1489$
:$1871, 1873, 1877, 1879$
:$2081, 2083, 2087, 2089$
:$3251, 3253, 3257, 3259$
:$3461, 3463, 3467, 3469$
:$5651, 5653, 5657, 5659$
:$9431, 9433, 9437, 9439$
:$13001, 13003, 13007, 13009$
:$15641, 15643, 15647, 15649$
:$15731, 15733, 15737, 15739$
:$16061, 16063, 16067, 16069$
:$18041, 18043, 18047, 18049$
:$18911, 18913, 18917, 18919$
:$19421, 19423, 19427, 19429$
:$21011, 21013, 21017, 21019$
:$22271, 22273, 22277, 22279$
:$25301, 25303, 25307, 25309$
:$31721, 31723, 31727, 31729$
:$34841, 34843, 34847, 34849$
:$43781, 43783, 43787, 43789$
:$51341, 51343, 51347, 51349$
:$55331, 55333, 55337, 55339$
:$62981, 62983, 62987, 62989$
:$67211, 67213, 67217, 67219$
:$69491, 69493, 69497, 69499$
:$72221, 72223, 72227, 72229$
:$77261, 77263, 77267, 77269$
:$79691, 79693, 79697, 79699$
:$81041, 81043, 81047, 81049$
:$82721, 82723, 82727, 82729$
:$88811, 88813, 88817, 88819$
:$97841, 97843, 97847, 97849$
:$99131, 99133, 99137, 99139$
== Sources ==
* 
* : $99,131$
Category:Prime Numbers
Sets of 4 Prime Quadruples/Mistake

== Source Work ==
:
:The Dictionary
::$99,131$
== Mistake ==
:There are $35$ sets of $4$ consecutive prime numbers of the form $10 n + 1$, $10 n + 3$, $10 n + 7$, $10 n + 9$, below $100,000$.
There are in fact $37$.
== Sources ==
* : $99,131$
Category:Mistakes/Curious and Interesting Numbers
Sets of Operations on Set of 3 Elements

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\AA$, $\BB$, $\CC_1$, $\CC_2$, $\CC_3$ and $\DD$ be respectively the set of all operations $\circ$ on $S$ such that the groups of automorphisms of $\struct {S, \circ}$ are as follows:
Let $\CC := \CC_1 \cup \CC_2 \cup \CC_3$.
The following results can be deduced:
=== Automorphism Group of $\AA$ ===
=== Automorphism Group of $\BB$ ===
=== Automorphism Group of $\CC_n$ ===
=== Automorphism Group of $\DD$ ===
=== Isomorphism Classes ===
=== Operations with Identity ===
=== Commutative Operations ===
== Also see ==
* Count of Binary Operations on Set
* Count of Binary Operations with Identity
* Count of Commutative Binary Operations on Set
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14$
Category:Sets of Operations on Set of 3 Elements
Category:Symmetric Group on 3 Letters
Sets of Operations on Set of 3 Elements/Automorphism Group of A

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\AA$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ is the symmetric group on $S$, that is, $\map \Gamma S$.
Then:
:$\AA$ has $3$ elements.
=== Cayley Table ===
=== Isomorphism Classes ===
=== Operations with Identity ===
=== Commutative Operations ===
== Proof ==
Recall the definition of (group) automorphism:
:$\phi$ is an automorphism on $\struct {S, \circ}$ :
::$\phi$ is a permutation of $S$
::$\phi$ is a homomorphism on $\struct {S, \circ}$: $\forall a, b \in S: \map \phi {a \circ b} = \map \phi a \circ \map \phi b$
Hence $\AA$ can be defined as the set of operations $\circ$ on $S$ such that every permutation on $S$ is an automorphism of $\struct {S, \circ}$.
The set $\map \Gamma S$ of all permutations on $S = \set {a, b, c}$ has $6$ elements.
From Identity Mapping is Group Automorphism, $I_S$ is always an automorphism on $\struct {S, \circ}$.
Hence it is not necessary to analyse the effect of $I_S$ on the various elements of $S$.
Let us denote each of the remaining permutations on $S$ as follows:
So, let $\circ \in \AA$.
From Permutation of Set is Automorphism of Set under Left Operation and Permutation of Set is Automorphism of Set under Right Operation:
:${\to} \in \AA$
:${\gets} \in \AA$
where $\to$ and $\gets$ are the right operation and left operation respectively.
Next we note that from Structure of Cardinality 3+ where Every Permutation is Automorphism is Idempotent:
:$\forall x \in S: x \circ x = x$
Let us explore various options for $a \circ b$.
Let $a \circ b = a$.
Then by definition of the above mappings $p$, $q$, $r$, $s$ and $t$, and the definition of a homomorphism:
This is none other than the left operation, which has already been counted.
Let $a \circ b = b$.
 
Then by definition of the above mappings $p$, $q$, $r$, $s$ and $t$, and the definition of a homomorphism:
This is none other than the right operation, which has already been counted.
Let $a \circ b = c$.
 
Then by definition of the above mappings $p$, $q$, $r$, $s$ and $t$, and the definition of a homomorphism:
Thus we have identified another operation on $S$ for which all permutations on $S$ are automorphisms of $\struct {S, \circ}$.
We now note that once the product element $a \circ b$ has been selected to be either $a$, $b$ or $c$, the full structure of $\struct {S, \circ}$ is forced.
Hence there are no other operations $\circ$ on $S$ but these ones we have counted.
That is, there is a total of $3$ such operations.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(a)}$
Category:Sets of Operations on Set of 3 Elements
Sets of Operations on Set of 3 Elements/Automorphism Group of A/Cayley Table

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\AA$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ is the symmetric group on $S$, that is, $\map \Gamma S$.
From Automorphism Group of $\AA$, there are $3$ such operations $\circ$ on $S$.
One is the right operation, one is the left operation, and the third is neither.
The Cayley table of the operation $\circ$ on $S$ such that:
:every permutation of $S$ is an automorphism on $\struct {S, \circ}$
:$\circ$ is neither the right operation nor the left operation
can be presented as follows:
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a & c & b \\
b & c & b & a \\
c & b & a & c \\
\end {array}$
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(a)}$
Category:Sets of Operations on Set of 3 Elements
Sets of Operations on Set of 3 Elements/Automorphism Group of A/Commutative Operations

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\AA$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ is the symmetric group on $S$, that is, $\map \Gamma S$.
Then:
:Exactly $1$ of the operations of $\AA$ is commutative.
== Proof ==
Recall from Automorphism Group of $\AA$ the elements of $\AA$, expressed in Cayley table form:
:$\begin {array} {c|ccc}
\to & a & b & c \\
\hline
a & a & b & c \\
b & a & b & c \\
c & a & b & c \\
\end {array}
\qquad
\begin {array} {c|ccc}
\gets & a & b & c \\
\hline
a & a & a & a \\
b & b & b & b \\
c & c & c & c \\
\end {array}
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a & c & b \\
b & c & b & a \\
c & b & a & c \\
\end {array}$
From Cayley Table for Commutative Operation is Symmetrical about Main Diagonal, it is apparent by inspection that of the above:
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a & c & b \\
b & c & b & a \\
c & b & a & c \\
\end {array}$
is the only Cayley table of a commutative operation.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(d)}$
Category:Sets of Operations on Set of 3 Elements
Sets of Operations on Set of 3 Elements/Automorphism Group of A/Isomorphism Classes

== Theorem ==
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Let $\AA$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ is the symmetric group on $S$, that is, $\map \Gamma S$.
Then:
:The elements of $\AA$ are each in its own isomorphism class.
== Proof ==
Recall from Automorphism Group of $\AA$ the elements of $\AA$, expressed in Cayley table form:
:$\begin {array} {c|ccc}
\to & a & b & c \\
\hline
a & a & b & c \\
b & a & b & c \\
c & a & b & c \\
\end {array}
\qquad
\begin {array} {c|ccc}
\gets & a & b & c \\
\hline
a & a & a & a \\
b & b & b & b \\
c & c & c & c \\
\end {array}
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a & c & b \\
b & c & b & a \\
c & b & a & c \\
\end {array}$
We have from Algebraic Structures formed by Left and Right Operations are not Isomorphic for Cardinality Greater than 1 that $\struct {S, \to}$ and $\struct {S, \gets}$ are not isomorphic.
Observe from the Cayley table $\circ$ that:
:$(1): \quad \forall x, y \in S: x \ne y \implies x \circ y \ne x \land x \circ y \ne y$
 there exists an isomorphism $\phi$ from $\struct {S, \to}$ to $\struct {S, \circ}$.
We have:
Then:
Hence:
:$\map \phi c = \map \phi b$
which means $\phi$ is not an injection.
That is, $\phi$ is not a bijection and hence not an isomorphism.
Hence by Proof by Contradiction there exists no isomorphism from $\struct {S, \to}$ to $\struct {S, \circ}$.
In a similar way:
 there exists an isomorphism $\phi$ from $\struct {S, \gets}$ to $\struct {S, \circ}$.
We have:
Then:
Hence:
:$\map \phi c = \map \phi a$
which means $\phi$ is not an injection.
That is, $\phi$ is not a bijection and hence not an isomorphism.
Hence by Proof by Contradiction there exists no isomorphism from $\struct {S, \gets}$ to $\struct {S, \circ}$.
It has been shown that none of the elements of $\AA$ is isomorphism to any of the others.
The result follows by definition of isomorphism class.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.14 \ \text{(b)}$
Category:Sets of Operations on Set of 3 Elements