nyuuzyou/wikis huggingface dataset:
Set of Condensation Points of Countable Set is Empty/Lemma

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be a subset of $S$.
Then:
:if $A$ is countable,
:then there exists no point $x$ of $S$ such that $x$ is a condensation point of $A$.
== Proof ==
Assume
:$A$ is countable.
 there exists a point $x$ of $S$ such that
: $x$ is a condensation point of $A$
By definition of topological space:
:$S \in \tau$
Then by definition of condensation point:
:$A \cap S$ is uncountable
By Intersection with Subset is Subset:
:$A \cap S = A$
$A$ is countable contradicts $A$ is uncountable.
Thus the result follows by Proof by Contradiction.
== Sources ==
* 
* 
Category:Condensation Points
Set of Condensation Points of Union is Union of Sets of Condensation Points

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $A, B$ be subsets of $S$.
Then:
:$\left({A \cup B}\right)^0 = A^0 \cup B^0$
=== Lemma ===
== Proof ==
=== Set of Condensation Points of Union Subset Union of Sets of Condensation Points ===
Let $x \in \left({A \cup B}\right)^0$.
By definition of set of condensation points:
:$x$ is condensation point of $A \cup B$
By Lemma:
:$x$ is condensation point of $A$ or $x$ is condensation point of $B$
By definition of set of condensation points:
:$x \in A^0$ or $x \in B^0$
Thus by definition of union:
:$x \in A^0 \cup B^0$
=== Union of Sets of Condensation Points Subset Set of Condensation Points of Union ===
By Set is Subset of Union:
:$A \subseteq A \cup B \land B \subseteq A \cup B$
By Set of Condensation Points is Monotone:
:$A^0 \subseteq \left({A \cup B}\right)^0 \land B^0 \subseteq \left({A \cup B}\right)^0$
Thus by Union is Smallest Superset:
:$A^0 \cup B^0 \subseteq \left({A \cup B}\right)^0$
Thus by definition of set equality:
:$\left({A \cup B}\right)^0 = A^0 \cup B^0$
== Sources ==
* 
* 
Category:Condensation Points
Set of Condensation Points of Union is Union of Sets of Condensation Points/Lemma

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A, B$ be subsets of $S$.
Let $x$ be a point of $S$.
Then:
:if $x$ is condensation point of $A \cup B$,
:then $x$ is condensation point of $A$ or $x$ is condensation point of $B$.
== Proof ==
Assume $x$ is condensation point of $A \cup B$.
:$x$ is not condensation point of $A$ and $x$ is not condensation point of $B$.
By definition of condensation point:
:$\exists U_1 \in \tau: x \in U_1 \land A \cap U_1$ is countable
By definition of condensation point:
:$\exists U_2 \in \tau: x \in U_2 \land B \cap U_2$ is countable
By definition of intersection:
:$x \in U_1 \cap U_2$
By definition of topological space:
:$U_1 \cap U_2 \in \tau$
By Intersection is Subset:
:$A \cap U_1 \cap U_2 \subseteq A \cap U_1 \land B \cap U_1 \cap U_2 \subseteq B \cap U_2$
By Set Union Preserves Subsets:
:$\left({A \cap U_1 \cap U_2}\right) \cup \left({B \cap U_1 \cap U_2}\right) \subseteq \left({A \cap U_1}\right) \cup \left({B \cap U_2}\right)$
By Countable Union of Countable Sets is Countable:
:$\left({A \cap U_1}\right) \cup \left({B \cap U_2}\right)$ is countable
Then by Subset of Countable Set is Countable:
:$(1): \quad \left({A \cap U_1 \cap U_2}\right) \cup \left({B \cap U_1 \cap U_2}\right)$ is countable
By Intersection Distributes over Union:
:$\left({A \cap U_1 \cap U_2}\right) \cup \left({B \cap U_1 \cap U_2}\right) = \left({A \cup B}\right) \cap \left({U_1 \cap U_2}\right)$
By definition of condensation point:
:$\left({A \cup B}\right) \cap \left({U_1 \cap U_2}\right)$ is uncountable
This contradicts $(1)$.
Thus the result fallows by Proof by Contradiction.
== Sources ==
* 
* 
Category:Condensation Points
Set of Congruence Relations on Algebraic Structure forms Complete Lattice

== Theorem ==
Let $\struct {S, \odot}$ be an algebraic structure.
Let $\map \RR \odot$ be the set of all congruence relations on $\struct {S, \odot}$.
Then $\struct {\map \RR \odot, \subseteq}$ is a complete lattice.
== Proof ==
Elements of $\map \RR \odot$ are subsets of $S \times S$.
First we consider the trivial relation $S \times S$ itself.
From Trivial Relation is Universally Congruent, $S \times S$ is a congruence relation on $\struct {S, \odot}$.
Let $\TT$ be a subset of $\map \RR \odot$.
Consider the intersection $\HH = \ds \bigcap \TT$.
Let $x_1, y_1, x_2, y_2 \in S$ be arbitrary, such that:
:$x_1 \mathrel \HH y_1$ and $x_2 \mathrel \HH y_2$
We have:
Hence $\HH = \ds \bigcap \TT$ is a congruence relation on $\struct {S, \odot}$.
That is:
:$\ds \bigcap \TT \in \map \RR \odot$
The appropriate conditions are fulfilled, and from Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice:
:$\struct {\map \RR \odot, \subseteq}$ is a complete lattice.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.14$
Category:Congruence Relations
Category:Examples of Complete Lattices
Set of Cuts forms Ordered Field

== Theorem ==
Let $\CC$ denote the set of cuts.
Let $\struct {\CC, +, \times, \le}$ denote the ordered structure formed from $\CC$ and:
:the operation $+$ of addition of cuts
:the operation $\times$ of multiplication of cuts
:the ordering $\le$ of cuts.
Then $\struct {\CC, +, \times, \le}$ is an ordered field.
== Proof ==
First we show that $\struct {\CC, +, \times}$ is a field by demonstrating that it fulfils the field axioms:
It has been established from Set of Cuts under Addition forms Abelian Group that $\struct {\CC, +}$ forms an abelian group.
Thus $\text A 0$ through to $\text A 4$ are fulfilled.
It remains to verify that $\struct {\CC, +, \times}$ fulfils the remaining field axioms.
=== ===
From Product of Cuts is Cut:
:$\forall \alpha, \beta \in \CC: \alpha \beta \in \CC$
Thus $\struct {\CC, \times}$ is closed.
=== ===
From Multiplication of Cuts is Associative:
:$\paren {\alpha \beta} \gamma = \alpha \paren {\beta \gamma}$
Thus the operation $\times$ of multiplication of cuts is associative on $\struct {\CC, \times}$.
=== ===
From Multiplication of Cuts is Commutative:
:$\alpha \beta = \beta \alpha$
Thus the operation $\times$ of multiplication of cuts is commutative on $\struct {\CC, \times}$.
=== ===
Consider the rational cut $1^*$ associated with the (rational) number $1$:
:$1^* = \set {r \in \Q: r < 1}$
From Cut Associated with 1 is Identity for Multiplication of Cuts:
:$\alpha \times 1^* = \alpha$
for all $\alpha \in \CC$.
From Multiplication of Cuts is Commutative it follows that:
:$1^* \times \alpha = \alpha$
That is, $1^*$ is the identity element of $\struct {\CC, \times}$.
=== ===
We have that $1^*$ is the identity element of $\struct {\CC, \times}$.
Let $\alpha \ne 0^*$.
From Existence of Unique Inverse Element for Multiplication of Cuts, there exists a unique cut $\dfrac 1 \alpha$ such that:
:$\alpha \times \dfrac 1 \alpha = 1^*$
From Multiplication of Cuts is Commutative it follows that:
:$\dfrac 1 \alpha \times \alpha = 1^*$
Thus every element $\alpha$ of $\struct {\CC, \times}$ such that $\alpha \ne 0^*$ has an inverse $\dfrac 1 \alpha$.
=== ===
From Multiplication of Cuts Distributes over Addition:
:$\alpha \paren {\beta + \gamma} = \alpha \beta + \alpha \gamma$
for all $\alpha$, $\beta$ and $\gamma$ in $\CC$.
All the field axioms are thus seen to be fulfilled, and so $\struct {\CC, +, \times}$ is a field.
Finally it is shown that $<$ is a total ordering on $\struct {\CC, +, \times}$:
This is demonstrated in Ordering on Cuts is Total.
From Properties of Ordered Field, this is all that is required to be shown.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Real Numbers
Category:Cuts
Category:Totally Ordered Fields
Set of Cuts under Addition forms Abelian Group

== Theorem ==
Let $\CC$ denote the set of cuts.
Let $\struct {\CC, +}$ denote the algebraic structure formed from $\CC$ and the operation $+$ of addition of cuts.
Then $\struct {\CC, +}$ forms an abelian group.
== Proof ==
In the below, $\alpha$, $\beta$ and $\gamma$ denote arbitrary cuts.
Taking the abelian group axioms in turn:
=== ===
From Sum of Cuts is Cut:
:$\forall \alpha, \beta \in \CC: \alpha + \beta \in \CC$
Thus $\struct {\CC, +}$ is closed.
=== ===
From Addition of Cuts is Associative:
:$\paren {\alpha + \beta} + \gamma = \alpha + \paren {\beta + \gamma}$
Thus the operation $+$ of addition of cuts is associative on $\struct {\CC, +}$.
=== $\text C$: Commutativity ===
From Addition of Cuts is Commutative:
:$\alpha + \beta = \beta + \alpha$
Thus the operation $+$ of addition of cuts is commutative on $\struct {\CC, +}$.
=== ===
Consider the rational cut $0^*$ associated with the (rational) number $0$:
:$0^* = \set {r \in \Q: r < 0}$
From Identity Element for Addition of Cuts:
:$\alpha + 0^* = \alpha$
for all $\alpha \in \CC$.
From Addition of Cuts is Commutative it follows that:
:$0^* + \alpha = \alpha$
That is, $0^*$ is the identity element of $\struct {\CC, +}$.
=== ===
We have that $0^*$ is the identity element of $\struct {\CC, +}$.
From Existence of Unique Inverse Element for Addition of Cuts, there exists a unique cut $-\alpha$ such that:
:$\alpha + \paren {-\alpha} = 0^*$
From Addition of Cuts is Commutative it follows that:
:$\paren {-\alpha} + \alpha = 0^*$
Thus every element of $\struct {\CC, +}$ has an inverse $-\alpha$.
All the abelian group axioms are thus seen to be fulfilled, and so $\struct {\CC, +}$ is a group.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.21$. Remark
Category:Cuts
Category:Examples of Abelian Groups
Set of Distributional Derivatives of Dirac Delta Distribution is Linearly Independent

== Theorem ==
Let $\delta \in \map {\DD'} \R$ be the Dirac delta distribution.
Then for any $n \in \N$ the set of distributional derivatives $\set {\delta, \delta' \ldots, \delta^{\paren n} }$ is linearly independent in $\map {\DD'} \R$.
== Proof ==
 there exist scalars $c_0, c_1, \ldots, c_n \in \R$ such that:
:$\ds \sum_{i \mathop = 0}^n c_i \delta^{\paren i} = \mathbf 0$
where $\mathbf 0 : \map \DD \R \to 0$ is the zero distribution.
Let $\phi \in \map \DD \R$ be a test function.
Let $\lambda \in \R_{\mathop > 0}$
Let $\map {\phi_\lambda} x := \map \phi {\lambda x}$.
We have that $\phi_\lambda$ is a test function.
Then:
This has to hold for any $\lambda > 0$.
Hence $\paren \star$:
:$c_0 \map \phi 0 = \ldots = c_n \map {\phi^{\paren n} } 0 = 0$
We have that:
:$\forall \phi \in \map \DD \R : \forall x \in \R : \forall n \in \N : x^n \phi \in \map \DD \R$
By Leibniz's Rule in One Variable:
:$\ds \map {\paren {x^n \phi}^{\paren n} } 0 = \paren {-1}^n \binom n n n! \map \phi 0 \ne 0$
Using $x^n \phi$ as our test function in $\paren \star$ yields:
:$c_0 = \ldots = c_n = 0$
and we have a contradiction.
== Sources ==
* : Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense
Category:Distributional Derivatives
Category:Dirac Delta Function
Set of Division Subrings forms Complete Lattice

== Theorem ==
Let $\struct {D, +, \circ}$ be a division ring.
Let $\mathbb K$ be the set of all division subrings of $K$.
Then $\struct {\mathbb K, \subseteq}$ is a complete lattice.
== Proof ==
Let $\O \subset \mathbb S \subseteq \mathbb K$.
By Intersection of Division Subrings is Largest Division Subring Contained in all Division Subrings:
:$\bigcap \mathbb S$ is the largest division subring of $K$ contained in each of the elements of $\mathbb S$.
By Intersection of Division Subrings Containing Subset is Smallest:
:The intersection of the set of all division subrings of $K$ containing $\bigcup \mathbb S$ is the smallest division 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 division subrings of $K$.
Therefore $\struct {\mathbb K, \subseteq}$ is a complete lattice.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.1$
Category:Complete Lattices
Category:Division Subrings
Set of Divisors of Integer

== Theorem ==
Let $n \in \Z_{>1}$.
Let $n$ be expressed in its prime decomposition:
:$n = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$
where $p_1 < p_2 < \dotsb < p_r$ are distinct primes and $k_1, k_2, \ldots, k_r$ are positive integers.
The set of divisors of $n$ is:
:$\set {p_1^{h_1} p_2^{h_2} \dotsm p_r^{h_r}: 0 \le h_i \le k_i, i = 1, 2, \ldots, r}$
== Proof ==
Each integer in the given set is a divisor of $n$ because:
:$(1): \quad \forall i: k_i - h_i \ge 0$
:$(2): \quad n = \paren {p_1^{h_1} p_2^{h_2} \dotsm p_r^{h_r} } p_1^{k_1 - h_1} p_2^{k_2 - h_2} \ldots p_r^{k_r - h_r}$
from Exponents of Primes in Prime Decomposition are Less iff Divisor.
By the Fundamental Theorem of Arithmetic, these integers are distinct.
It is necessary to show that the integers in this set are the only divisors of $n$.
Let $d > 1$ and let $p \in \mathbb P: p \divides d$.
It remains to be shown that:
:$\forall i: h_1 \le k_i$
First note that:
:$d \divides n \implies \forall i: p_i^{k_i} \divides n$
From above, all the primes $p_i$ are distinct.
Therefore by Prime not Divisor implies Coprime:
:$p_1 \nmid p_2^{k_2} p_3^{k_3} \dotsm p_r^{k_r} \implies \gcd \set {p_1, p_2^{k_2} p_3^{k_3} \ldots p_r^{k_r} } = 1$
So:
:$p_1^{h_1} \divides n \implies n = p_1^{k_1} \paren {p_2^{k_2} p_3^{k_3} \dotsm p_r^{k_r} }$
 
By Euclid's Lemma:
:$p_1^{h_1} \divides p_1^{k_1} \implies h_1 \le k_1$
and the same argument applies to each of the other prime factors of $n$.
The result follows.
Category:Number Theory
Set of Doubletons of Natural Numbers is Countable

== Theorem ==
Let $S$ be the set defined as:
:$S = \set {\set {n_1, n_2}: n_1, n_2 \in \N, n_1 \ne n_2}$
where $\N$ denotes the set of natural numbers.
Then $S$ is countably infinite.
== Proof ==
We can write the elements of $S \times T$ in the form of an infinite table:
:$\begin{array} {*{4}c}
 \tuple {1, 0} & & & \\
 \tuple {2, 0} & \tuple {2, 1} & & \\
 \tuple {3, 0} & \tuple {3, 1} & \tuple {3, 2} & \\
 \vdots & \vdots & \vdots & \ddots \\
\end{array}$
This table clearly contains all the elements of $S$.
Let $\set {n_1, n_2} \in S$ be written such that $n_1 > n_2$.
Then $\set {n_1, n_2}$ will be found in row $n_1$ and column $n_2 + 1$ of the table.
The index of the $n$th row is the $n$th triangular number.
Let $f: S \to \N$ be defined as:
:$\forall \set {n_1, n_2} \in S: \map f {\set {n_1, n_2} } = \dfrac {n_1 \paren {n_1 + 1} } 2 + n_2$
It is seen that $f$ is an injection.
With a little more work it can be shown to be a bijection as well, but that condition is not necessary.
The result follows.
== Sources ==
* : Chapter $1$: General Background: $\S 2$ Countable or uncountable?
Category:Cartesian Product of Countable Sets is Countable
Set of Elementary Events belonging to k Events is Event

== Theorem ==
Let $\EE$ be an experiment with a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $A_1, A_2, \ldots, A_m$ be events in the event space $\Sigma$ of $\EE$.
Let $S$ denote the set of all elementary events of $\EE$ which are elements of exactly $k$ of the events $A_1, A_2, \ldots, A_m$.
Then $S$ is an event of $\Sigma$.
== Proof ==
Let $r_1, r_2, \ldots r_k$ be a set of $k$ elements of $\set {1, 2, \ldots, m}$.
Then:
:$\paren {A_{r_1} \cap A_{r_2} \cap \cdots \cap A_{r_k} } \setminus \paren { A_{r_{k + 1} } \cup A_{r_{k + 2} } \cup \cdots \cup A_{r_m} }$
contains exactly those elements of $\Omega$ which are contained in exactly those events $A_{r_1}, A_{r_2}, \ldots, \cap A_{r_k}$.
By Elementary Properties of Event Space, that last set is an event of $\EE$.
By selecting all combinations of $\set {1, 2, \ldots, m}$, we can form the subsets of $\Sigma$ containing all those elementary events of exactly $k$ events.
Each one is an element of $\Sigma$.
We then form the union $S_k$ of each of those element of $\Sigma$.
By Elementary Properties of Event Space, $S_k$ is an event of $\EE$.
By construction, $S_k$ contains all and only the elementary events of $\EE$ which are elements of exactly $k$ of the events $A_1, A_2, \ldots, A_m$.
== Also see ==
* Elementary Properties of Event Space
== Sources ==
* : $1$: Events and probabilities: $1.2$: Outcomes and events: Exercise $4$
Category:Event Spaces
Set of Endomorphisms of Non-Abelian Group is not Ring

== Theorem ==
Let $\struct {G, \oplus}$ be a group which is non-abelian.
Let $\mathbb G$ be the set of all group endomorphisms of $\struct {G, \oplus}$.
Let $*: \mathbb G \times \mathbb G \to \mathbb G$ be the operation defined as:
:$\forall u, v \in \mathbb G: u * v = u \circ v$
where $u \circ v$ is defined as composition of mappings.
Then the algebraic structure $\struct {\mathbb G, \oplus, *}$ is not a ring.
== Proof ==
In order to be a ring, it is necessary that the additive operation $\oplus$ is commutative.
However, as $\struct {G, \oplus}$ is specifically defined as being non-abelian, a fortiori $\oplus$ is not commutative.
Hence the result.
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $10$
Category:Rings of Endomorphisms
Set of Endomorphisms on Entropic Structure is Closed in Induced Structure on Set of Self-Maps

== Theorem ==
Let $\struct {S, \odot}$ be a magma.
Let $\struct {S, \odot}$ be an entropic structure.
Let $S^S$ be the set of all mappings from $S$ to itself.
Let $\struct {S^S, \oplus}$ denote the algebraic structure on $S^S$ induced by $\odot$.
Let $T \subseteq S^S$ denote the set of endomorphisms on $\struct {S, \odot}$.
Then $\struct {T, \oplus_T}$ is closed in $\struct {S^S, \oplus}$.
=== Converse does not hold ===
== Proof ==
Recall the definition of algebraic structure on $S^S$ induced by $\odot$:
Let $f: S \to S$ and $g: S \to S$ be self-maps on $S$, and thus elements of $S^S$.
The pointwise operation on $S^S$ induced by $\odot$ is defined as:
:$\forall x \in S: \map {\paren {f \oplus g} } x = \map f x \odot \map g x$
Let $f, g \in T$ be arbitrary.
That is, let $f: S \to S$, $g: S \to S$ be endomorphisms on $\struct {S, \odot}$.
Let $x, y \in S$ be arbitrary.
Then:
demonstrating that $f \oplus g$ is a homomorphism from $S$ to itself.
Hence the result by definition of closed algebraic structure.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.12 \ \text{(g)}$
Category:Set of Endomorphisms on Entropic Structure is Closed in Induced Structure on Set of Self-Maps
Category:Entropic Structures
Category:Pointwise Operations
Category:Endomorphisms
Category:Self-Maps
Set of Endomorphisms on Entropic Structure is Closed in Induced Structure on Set of Self-Maps/Converse

== Theorem ==
Let $\struct {S, \odot}$ be a magma.
Let $S^S$ be the set of all mappings from $S$ to itself.
Let $\struct {S^S, \oplus}$ denote the algebraic structure on $S^S$ induced by $\odot$.
Let $T \subseteq S^S$ denote the set of endomorphisms on $\struct {S, \odot}$.
Let $\struct {T, \oplus_T}$ be closed in $\struct {S^S, \oplus}$.
Then it is not necessarily the case that $\struct {S, \odot}$ is an entropic structure.
== Proof ==
Proof by Counterexample:
Let $S = \set {a, b}$.
Let $\odot$ be the operation on $S$ defined by Cayley table as:
:$\begin{array} {c|cc} \odot & a & b \\
\hline
a & b & b \\
b & a & b
\end{array}$
First we note that:
demonstrating that $\struct {S, \odot}$ is specifically not an entropic structure.
Let $T \subseteq S^S$ denote the set of endomorphisms on $\struct {S, \odot}$.
Let $f, g \in T$ be arbitrary.
That is, let $f: S \to S$, $g: S \to S$ be endomorphisms on $\struct {S, \odot}$.
Let $x, y \in S$ be arbitrary.
We are to show that:
:$\map {\paren {f \oplus g} } {x \odot y} = \paren {\map {\paren {f \oplus g} } x} \odot \paren {\map {\paren {f \oplus g} } y}$
Let us investigate all the mappings from $S$ to $S$.
From Cardinality of Set of All Mappings, there are exactly $4$ of these, so inspecting them all is not onerous.
Let $I_S$ denote the identity mapping:
From Identity Mapping is Automorphism, $I_S$ is an automorphism.
Hence $I_S$ is an endomorphism on $\struct {S, \odot}$.
That is:
:$I_S \in T$
Let $f_a$ and $f_b$ denote the constant mappings:
We have that:
and so $f_a$ is not a homomorphism, and so not an endomorphism on $\struct {S, \odot}$.
Hence:
:$f_a \notin T$
We have that:
and so $f_b$ is a homomorphism, and so an endomorphism on $\struct {S, \odot}$.
Hence:
:$f_b \in T$
Now let $h \in S^S$ be defined as:
We have:
and so $h$ is not a homomorphism, and so not an endomorphism on $\struct {S, \odot}$.
Hence:
:$h \notin T$
So we are left with:
:$T = \set {I_S, f_b}$
From Composite of Endomorphisms is Endomorphism, each of $I_S \circ I_S$, $I_S \circ f_b$, $f_b \circ I_S$ and $f_b \circ f_b$ are endomorphisms.
By definition of identity mapping:
:$I_S \circ I_S = I_S$
:$I_S \circ f_b = f_b$
:$f_b \circ I_S = f_b$
and from Composite with Constant Mapping is Constant Mapping:
$f_b \circ f_b = f_b$
Hence we have that $\struct {T, \oplus}$ is closed in $\struct {S^S, \oplus}$.
But as we have seen, $\struct {S, \odot}$ is not an entropic structure.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.12 \ \text{(h)}$
Category:Set of Endomorphisms on Entropic Structure is Closed in Induced Structure on Set of Self-Maps
Category:Entropic Structures
Category:Pointwise Operations
Category:Endomorphisms
Set of Even Integers is Countably Infinite

== Theorem ==
Let $\Bbb E$ be the set of even integers.
Then $\Bbb E$ is countably infinite.
== Proof ==
Let $f: \Bbb E \to \Z$ be the mapping defined as:
:$\forall x \in \Bbb E: \map f x = \dfrac x 2$
$f$ is well-defined as $x$ is even and so $\dfrac x 2 \in \Z$.
Let $x, y \in \Bbb E$ such that $\map f x = \map f y$.
Then:
Thus $f$ is injective by definition.
Consider the inverse $f^{-1}$.
By inspection:
:$\forall x \in \Z: \map {f^{-1} } x = 2 x$
$f^{-1}$ is well-defined, and $2 x$ is even.
Thus $f^{-1}$ is a mapping from $\Z$ to $\Bbb E$.
Then:
Thus $f^{-1}$ is injective by definition.
It follows by the Cantor-Bernstein-Schröder Theorem that there exists a bijection between $\Z$ and $\Bbb E$.
== Sources ==
* : Introduction: Review of Algebra, Geometry, and Trigonometry: $\text{0-1}$: The Real Numbers
Category:Countable Sets
Category:Odd Integers
Set of Even Integers is Equivalent to Set of Integers

== Theorem ==
Let $\Z$ denote the set of integers.
Let $2 \Z$ denote the set of even integers.
Then:
:$2 \Z \sim \Z$
where $\sim$ denotes set equivalence.
== Proof ==
To demonstrate set equivalence, it is sufficient to construct a bijection between the two sets.
Let $f: \Z \to 2 \Z$ defined as:
:$\forall x \in \Z: \map f x = 2 x$
demonstrating injectivity.
demonstrating surjectivity.
Hence by definition $f$ is a bijection.
== Sources ==
* : Chapter $\text {I}$: Sets of Points: $1 \cdot 2$. Infinite sets
* : Chapter $3$. Mappings: Exercise $10$
Category:Set Equivalence
Category:Integers
Set of Even Numbers is Primitive Recursive

== Theorem ==
Let $E \subseteq \N$ be the set of all even natural numbers.
Then $E$ is primitive recursive.
== Proof ==
If $n \in E$ then $n$ is of the form $n = 2 k$ where $k \in \N$.
We have that:
* if the characteristic function $\chi_E \left({n}\right) = 1$ then $\chi_E \left({n + 1}\right) = 0$.
* if the characteristic function $\chi_E \left({n}\right) = 0$ then $\chi_E \left({n + 1}\right) = 1$.
So $\chi_E$ can be defined by:
:$\chi_E \left({n}\right) = \begin{cases}
1 & : n = 0 \\
\overline{\operatorname{sgn}} \left({\chi_E \left({n-1}\right)}\right) & : n > 0
\end{cases}$
So $\chi_E$ is obtained by primitive recursion from the constant $1$ and the primitive recursive function $\overline{\operatorname{sgn}}$.
Hence the result.
Category:Primitive Recursive Functions