nyuuzyou/wikis huggingface dataset:
Equivalence of Mappings between Finite Sets of Same Cardinality

== Theorem ==
Let $S$ and $T$ be finite sets such that $\card S = \card T$.
Let $f: S \to T$ be a mapping.
Then the following statements are equivalent:
:$(1): \quad f$ is bijective
:$(2): \quad f$ is injective
:$(3): \quad f$ is surjective.
== Proof ==
$(2)$ implies $(3)$:
Let $f$ be an injection.
Then by Cardinality of Image of Injection:
:$\card S = \card {f \sqbrk S}$
where $f \sqbrk S$ denotes the image of $S$ under $f$.
Therefore the subset $f \sqbrk S$ of $T$ has the same number of elements as $T$.
Therefore:
:$f \sqbrk S = T$
and so $f$ is a surjection.
$(3)$ implies $(1)$:
Let $f$ be surjective.
We have that $\card S = \card T$.
By Cardinality of Codomain of Surjection, $f$ is bijective.
$(1)$ implies $(2)$:
Let $f$ be a bijection.
By definition of bijection, $f$ is also an injection.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 17$: Finite Sets: Theorem $17.7$
* : $\S 22.4$: Injections; bijections; inverse of a bijection
* : Appendix $\text{A}.10$: Finite Sequences: Proposition $\text{A}.10.1: \ 3$
* : Chapter $2$: 'And you do addition?': $\S 2.1$: Functions: Ponderable $2.1.6$
Category:Injections
Category:Surjections
Category:Bijections
Equivalence of Metrics is not Defined/Mistake

== Source Work ==
:
:Part $\text I$: Basic Definitions
::Section $5$. Metric Spaces
:::Complete Metric Spaces
== Mistake ==
:The concept of equivalence of metrics is not defined, although the concept is mentioned and used in the context of complete metric spaces.
=== Equivalent Metrics ===
The definition is as follows:
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
Category:Mistakes/Counterexamples in Topology
Equivalence of Semantic Consequence and Logical Implication

== Theorem ==
Let $U = \set {\phi_1, \phi_2, \ldots, \phi_m, \ldots}$ be a countable set of propositional formulas.
Let $\psi$ be a propositional formula.
Then $U \models \psi$ $U \vdash \psi$.
That is, semantic consequence is equivalent to provable consequence.
== Direct Proof ==
=== Necessary Condition ===
This is a statement of the Extended Soundness Theorem for Propositional Tableaux and Boolean Interpretations:
Let $\mathbf H$ be a countable set of propositional formulas.
Let $\mathbf A$ be a propositional formula.
If $\mathbf H \vdash \mathbf A$, then $\mathbf H \models \mathbf A$.
=== Sufficient Condition ===
This is the statement of the Extended Completeness Theorem for Propositional Tableaux and Boolean Interpretations.
== Comment ==
There are two things being proved here:
:$(1): \quad$ Suppose we have a sequent $\phi_1, \phi_2, \ldots, \phi_m, \ldots \vdash \psi$, the validity of which has been established, for example, by a tableau proof.
The result:
:if $\phi_1, \phi_2, \ldots, \phi_m, \ldots \vdash \psi$ then $\set {\phi_1, \phi_2, \ldots, \phi_m, \ldots} \models \psi$
establishes that if all the propositions $\phi_1, \phi_2, \ldots, \phi_m, \ldots$ evaluate to true, then so does $\psi$.
This establishes that propositional logic is sound.
:$(2): \quad$ Suppose we have determined that $\set {\phi_1, \phi_2, \ldots, \phi_m, \ldots} \models \psi$.
The result:
:if $\set {\phi_1, \phi_2, \ldots, \phi_m, \ldots} \models \psi$ then $\phi_1, \phi_2, \ldots, \phi_m, \ldots \vdash \psi$
establishes that if we can show that there is a model for a proposition, then we will be able to find a tableau proof for it.
This establishes that propositional logic is complete.
== Sources ==
* : $\S 1.11$: Compactness: Corollary $1.11.6$
Category:Propositional Logic
Equivalence of Subobjects is Equivalence

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $C$ be an object of $\mathbf C$.
Let $\map {\mathbf {Sub}_{\mathbf C} } C$ be the category of subobjects of $C$.
The relation $\sim$ on $\map {\mathbf {Sub}_{\mathbf C} } C$ defined by:
:$m \sim m'$ $m$ and $m'$ are equivalent
is an equivalence.
== Proof ==
By Inclusion Relation on Subobjects is Preordering, $\subseteq$ is a preordering.
From Preorder Induces Equivalence, we see that $\sim$ is the equivalence induced by $\subseteq$.
In particular, $\sim$ is an equivalence.
Category:Subobjects
Category:Examples of Equivalence Relations
Equivalence of Well-Ordering Principle and Induction

== Theorem ==
The Well-Ordering Principle, the Principle of Finite Induction and the Principle of Complete Finite Induction are logically equivalent.
That is:
:Principle of Finite Induction: Given a subset $S \subseteq \N$ of the natural numbers which has these properties:
:: $0 \in S$
:: $n \in S \implies n + 1 \in S$
:then $S = \N$.
:
:Principle of Complete Finite Induction: Given a subset $S \subseteq \N$ of the natural numbers which has these properties:
:: $0 \in S$
:: $\set {0, 1, \ldots, n} \subseteq S \implies n + 1 \in S$
:then $S = \N$.
:
:Well-Ordering Principle: Every non-empty subset of $\N$ has a minimal element.
== Proof ==
To save space, we will refer to:
: The Well-Ordering Principle as WOP
: The Principle of Finite Induction as PFI
: The Principle of Complete Finite Induction as PCI.
=== PFI implies PCI ===
=== PCI implies WOP ===
=== WOP implies PFI ===
=== Final assembly ===
So, we have that:
* PFI implies PCI: The Principle of Mathematical Induction implies the Principle of Complete Induction
* PCI implies WOP: The Principle of Complete Induction implies the Well-Ordering Principle
* WOP implies PFI: The Well-Ordering Principle implies the Principle of Mathematical Induction.
This completes the result.
== Sources ==
* : $\S 2.3$: Theorem $2.17$
: which demonstrates that PFI implies WOP
* : $\S 2.3$: Theorem $2.18$
: which demonstrates that WOP implies PCI
* : $\S 1.17$: Finite Induction and Well-Ordering for Positive Integers: Theorem $17.3$
: which demonstrates that PFI and WOP are equivalent
* : $\S 10$: The well-ordering principle:
: Students familiar with the principle of induction can gain some sense of perspective in these matters by thinking out how the well-ordering principle can be proved from the principle of mathematical induction and vice versa.
* : $\S 2.1$: The integers
* University of Western Ontario: Mathematics 310a: The Principle of Mathematical Induction and the Well Ordering Principle
Category:Proofs by Induction
Category:Well-Ordering Principle
Category:Equivalence of Well-Ordering Principle and Induction
Equivalence of Well-Ordering Principle and Induction/Proof/PCI implies WOP

== Theorem ==
The Principle of Complete Induction implies the Well-Ordering Principle.
That is:
:Principle of Complete Induction: Given a subset $S \subseteq \N$ of the natural numbers which has these properties:
::$0 \in S$
::$\set {0, 1, \ldots, n} \subseteq S \implies n + 1 \in S$
:then $S = \N$.
implies:
:Well-Ordering Principle: Every nonempty subset of $\N$ has a minimal element.
== Proof ==
To save space, we will refer to:
:The Principle of Complete Induction as PCI
:The Well-Ordering Principle as WOP.
Let us assume that the PCI is true.
Let $\O \subsetneqq S \subseteq \N$.
We need to show that $S$ has a minimal element, and so demonstrate that the WOP holds.
 that:
:$(C): \quad S$ has no minimal element.
Let $\map P n$ be the propositional function:
:$n \notin S$
Suppose $0 \in S$.
We have that $0$ is a lower bound for $\N$.
Hence by Lower Bound for Subset, $0$ is also a lower bound for $S$.
$0 \notin S$, otherwise $0$ would be the minimal element of $S$.
This contradicts our supposition $(C)$, namely, that $S$ does not have a minimal element.
So $0 \notin S$ and so $\map P 0$ holds.
Suppose $\map P j$ for $0 \le j \le k$.
That is:
:$\forall j \in \closedint 0 k: j \notin S$
where $\closedint 0 k$ denotes the closed interval between $0$ and $k$.
Now if $k + 1 \in S$ it follows that $k + 1$ would then be the minimal element of $S$.
So then $k + 1 \notin S$ and so $\map P {k + 1}$.
Thus we have proved that:
:$(1): \quad \map P 0$ holds
:$(2): \quad \paren {\forall j \in \closedint 0 k: \map P j} \implies \map P {k + 1}$
So we see that PCI implies that $\map P n$ holds for all $n \in \N$.
But this means that $S = \O$, which is a contradiction of the fact that $S$ is non-empty.
So, by Proof by Contradiction, $S$ must have a minimal element.
That is, $\N$ satisfies the Well-Ordering Principle.
Thus PCI implies WOP.
== Sources ==
* : Introduction $\S 4$: The natural numbers
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers
* University of Western Ontario: Mathematics 310a: The Principle of Mathematical Induction and the Well Ordering Principle
Category:Equivalence of Well-Ordering Principle and Induction
Equivalence of Well-Ordering Principle and Induction/Proof/PFI implies PCI

== Theorem ==
The Principle of Finite Induction implies the Principle of Complete Finite Induction.
That is:
:Principle of Finite Induction: Given a subset $S \subseteq \N$ of the natural numbers which has these properties:
::$0 \in S$
::$n \in S \implies n + 1 \in S$
:then $S = \N$.
implies:
:Principle of Complete Finite Induction: Given a subset $S \subseteq \N$ of the natural numbers which has these properties:
::$0 \in S$
::$\set {0, 1, \ldots, n} \subseteq S \implies n + 1 \in S$
:then $S = \N$.
== Proof ==
To save space, we will refer to:
:The Principle of Finite Induction as PFI
:The Principle of Complete Finite Induction as PCI.
Let us assume that the PFI is true.
Let $S \subseteq \N$ which satisfy:
:$(A): \quad 0 \in S$
:$(B): \quad \set {0, 1, \ldots, n} \subseteq S \implies n + 1 \in S$.
We want to show that $S = \N$, that is, the PCI is true.
Let $P \paren n$ be the propositional function:
:$P \paren n \iff \set {0, 1, \ldots, n} \subseteq S$
We define the set $S'$ as:
:$S' = \set {n \in \N: P \paren n \text { is true} }$
$P \paren 0$ is true by $(A)$, so $0 \in S'$.
Assume $P \paren k$ is true where $k \ge 0$.
So $k \in S'$, and by hypothesis:
:$\set {0, 1, \ldots, k} \subseteq S$
So by $(B)$:
:$k + 1 \in S$
Thus:
:$\set {0, 1, \ldots, k, k + 1} \subseteq S$.
That last statement means $P \paren {k + 1}$ is true.
This means $k + 1 \in S'$.
Thus we have satisfied the conditions:
:$0 \in S'$
:$n \in S' \implies n + 1 \in S'$
That is, $S' = \N$, and $P \paren n$ holds for all $n \in \N$.
Hence, by definition:
:$S = \N$
So PFI gives that $S = \N$.
Therefore PFI implies PCI.
== Sources ==
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers
* University of Western Ontario: Mathematics 310a: The Principle of Mathematical Induction and the Well Ordering Principle
Category:Equivalence of Well-Ordering Principle and Induction
Equivalence of Well-Ordering Principle and Induction/Proof/WOP implies PFI

== Theorem ==
The Well-Ordering Principle implies the Principle of Finite Induction.
That is:
:Well-Ordering Principle: Every non-empty subset of $\N$ has a minimal element
implies:
:Principle of Finite Induction: Given a subset $S \subseteq \N$ of the natural numbers which has these properties:
::$0 \in S$
::$n \in S \implies n + 1 \in S$
:then $S = \N$.
== Proof ==
To save space, we will refer to:
:The Well-Ordering Principle as WOP
:The Principle of Finite Induction as PFI.
We assume the truth of WOP.
Let $S \subseteq \N$ which satisfy:
:$(D): \quad 0 \in S$
:$(E): \quad n \in S \implies n+1 \in S$.
We want to show that $S = \N$, that is, the PFI is true.
 that:
:$S \ne \N$
Consider $S' = \N \setminus S$, where $\setminus$ denotes set difference.
From Set Difference is Subset, $S' \subseteq \N$.
So from WOP, $S'$ has a minimal element.
A lower bound of $\N$ is $0$.
By Lower Bound for Subset, $0$ is also a lower bound for $S'$.
By hypothesis, $0 \in S$.
From the definition of set difference, $0 \notin S'$.
So this minimal element of $S'$ has the form $k + 1$ where $k \in \N$.
We can consider the von Neumann construction of the natural numbers.
By definition of natural number addition, it is noted that $k + 1 \in \N$ is the immediate successor element of $k \in \N$.
Thus $k \in S$ but $k + 1 \notin S$.
From $(E)$, this contradicts the definition of $S$.
Thus if $S' \ne \O$, it has no minimal element.
This contradicts the Well-Ordering Principle, and so $S' = \O$.
So $S = N$.
Thus we have proved that WOP implies PFI.
== Sources ==
* : $\S 2.1$: Chapter $2$: Integers and natural numbers: The integers
* University of Western Ontario: Mathematics 310a: The Principle of Mathematical Induction and the Well Ordering Principle
Category:Equivalence of Well-Ordering Principle and Induction
Equivalences are Interderivable

== Theorem ==
If two propositional formulas are interderivable, they are equivalent:
:$\paren {p \dashv \vdash q} \dashv \vdash \paren {p \iff q}$
=== Forward Implication ===
=== Reverse Implication ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Exercises, Group $\text{III}$
Category:Biconditional
Category:Equivalences are Interderivable
Equivalences are Interderivable/Proof 1

== Theorem ==
== Proof ==
The result follows directly from the truth table for the biconditional:
$\begin{array}{|cc||ccc|} \hline
p & q & p & \iff & q \\
\hline
\F & \F & \F & \T & \F \\
\F & \T & \F & \F & \T \\
\T & \F & \T & \F & \F \\
\T & \T & \T & \T & \T \\
\hline
\end{array}$
By inspection, it is seen that $\map \MM {p \iff q} = \T$ $\map \MM p = \map \MM q$.
== Sources ==
* : $\S \text{II}$: The Logic of Statements $(2): \ 6$: Using logical equivalences
Category:Equivalences are Interderivable
Category:Truth Table Proofs
Equivalences are Interderivable/Proof 2

== Theorem ==
== Proof ==
Let $v$ be an arbitrary interpretation.
Then by definition of interderivable:
:$\map v {p \iff q}$ $\map v p = \map v q$
Since $v$ is arbitrary, $\map v p = \map v q$ holds in all interpretations.
That is:
:$p \dashv \vdash q$
== Sources ==
* : Chapter $2$: Propositional Calculus: $\S 2.4$: Logical equivalence and substitution: Theorem $2.4.4$
Category:Equivalences are Interderivable
Equivalent Cauchy Sequences have Equal Limits of Norm Sequences

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be Cauchy sequences in $R$.
Let $\ds \lim_{n \mathop \to \infty} {x_n - y_n} = 0$.
Then:
:$\ds \lim_{n \mathop \to \infty} \norm {x_n} = \lim_{n \mathop \to \infty} \norm {y_n}$
== Proof ==
Let:
:$l = \ds \lim_{n \mathop \to \infty} \norm {x_n}$
and:
:$m = \ds \lim_{n \mathop \to \infty} \norm {y_n}$
By Norm Sequence of Cauchy Sequence has Limit, both of these limits exist.
Then:
By Reverse Triangle Inequality on Normed Division Ring, for $n \in \N$:
:$l = \size {\norm {x_n} - \norm {y_n} } \le \norm {x_n - y_n} = \to 0$
as $n \to \infty$.
By the Squeeze Theorem:
:$\ds \lim_{n \mathop \to \infty} \paren {\size {\norm {x_n} - \norm {y_n} } } = 0$
So:
:$\size {l - m} = 0$
and therefore:
$l = m$
Category:Cauchy Sequences
Category:Normed Division Rings
Equivalent Characterisations of Irrational Periodic Continued Fraction

== Theorem ==
Let $x \in \R \setminus \Q$ be an irrational number.
Let $\sequence {a_n}_{n \mathop \ge 0}$ be its continued fraction.
Let $N \ge 0$ be a natural number.
:$(1):\quad$ The sequence of partial denominators $\sequence {a_n}_{n \mathop \ge 0}$ is periodic for $n \ge N$.
:$(2):\quad$ The sequence of complete quotients $\sequence {x_n}_{n \mathop \ge 0}$ is periodic for $n \ge N$.
:$(3):\quad$ There exists $M > N$ such that $x_M = x_N$.
== Proof ==
Category:Continued Fractions
Equivalent Characterizations of Abelian Group

== Theorem ==
Let $G$ be a group.
== Proof ==
=== $(1)$ iff $(2)$ ===
See Inversion Mapping is Automorphism iff Group is Abelian.
=== $(1)$ iff $(3)$ ===
See Group is Abelian iff it has Cross Cancellation Property.
=== $(1)$ iff $(4)$ ===
See Group is Abelian iff it has Middle Cancellation Property.
=== $(1)$ iff $(5)$ ===
See Group is Abelian iff Opposite Group is Itself.
Hence all $5$ statements are logically equivalent.
Category:Abelian Groups
Equivalent Characterizations of Finer Equivalence Relation

== Theorem ==
Let $X$ be a set.
Let $\equiv$ and $\sim$ be equivalence relations on $X$.
# $\equiv$ is finer than $\sim$:
#:$\forall x, y \in X : x \equiv y \implies x \sim y$
#The graph of $\equiv$ is contained in the graph of $\sim$.
#Every $\equiv$-equivalence class is contained in a $\sim$-equivalence class.
#Every $\sim$-equivalence class is saturated under $\equiv$.
== Proof ==
=== 1 implies 2 ===
By definition of subset, $\map \TT \equiv \subseteq \map \TT \sim$.
=== 2 implies 1 ===
=== 1 implies 3 ===
Let $x \in X$.
Then:
By definition of subset, $\eqclass x \equiv \subseteq \eqclass x \sim$.
=== 3 implies 1 ===
For this proof and the next, we use this result implied by $3$:
Let $y \in X$.
$3$ implies:
:$\exists z \in X: \eqclass y \equiv \subseteq \eqclass z \sim$
By Element in its own Equivalence Class:
:$y \in \eqclass y \equiv \subseteq \eqclass z \sim$
:$y \in \eqclass y \sim$
By contrapositive of Equivalence Classes are Disjoint:
:$\eqclass y \sim = \eqclass z \sim$
Hence $\eqclass y \equiv \subseteq \eqclass y \sim$.
Thus:
=== 3 implies 4 ===
Let $x \in X$.
so $\eqclass x \sim$ is a union of equivalence classes under $\equiv$.
Hence $\eqclass x \sim$ is saturated under $\equiv$.
=== 4 implies 3 ===
Let $x \in X$.
Then $\eqclass x \sim$ is saturated under $\equiv$:
:$\ds \exists S \subseteq X: \bigcup_{y \mathop \in S} \eqclass y \equiv = \eqclass x \sim$
By Element in its own Equivalence Class:
:$x \in \eqclass x \sim$
:$x \in \eqclass x \equiv$
Hence by definition of Union of Family:
:$\exists z \in S: x \in \eqclass z \equiv$
By contrapositive of Equivalence Classes are Disjoint:
== Sources ==
* Chapter $\text I$: Description of Formal Mathematics: $\S 6$ Equivalence relations $7$. Quotients of equivalence relations
Category:Equivalence Relations
Equivalent Conditions for Cover by Collection of Subsets

== Theorem ==
Let $X$ be a set.
Then the following conditions are equivalent for a subset $\CC \subseteq \powerset X$ of the power set of $X$:
:$(1): \quad \CC$ is a cover for $X$.
:$(2): \quad \ds X = \bigcup \CC$.
:$(3): \quad \ds \exists \SS \subseteq \CC: X = \bigcup \SS$.
== Proof ==
=== $(1)$ implies $(2)$ ===
By definition, $\CC$ covers $X$ $X \subseteq \ds \bigcup \CC$.
By Union of Subsets is Subset, we have that:
:$\ds \bigcup \CC \subseteq X$
since $\CC \subseteq \powerset X$.
By definition of set equality, it follows that $X = \ds \bigcup \CC$.
=== $(2)$ implies $(3)$ ===
Since $\CC \subseteq \CC$, we can take $\SS = \CC$.
Hence $(3)$ is immediate from $(2)$.
=== $(3)$ implies $(1)$ ===
By assumption, $X = \ds \bigcup \SS$ for some $\SS \subseteq \CC$.
By Union is Increasing, we have that:
:$\ds \bigcup \SS \subseteq \bigcup \CC$
Hence, $X \subseteq \ds \bigcup \CC$, that is $\CC$ is a cover for $X$.
Category:Covers
Equivalent Conditions for Dedekind-Infinite Set

== Theorem ==
For a set $S$, the following conditions are equivalent:
:$(1): \quad$ $S$ is Dedekind-infinite.
:$(2): \quad$ $S$ has a countably infinite subset.
The above equivalence can be proven in Zermelo-Fraenkel set theory.
If the axiom of countable choice is accepted, then it can be proven that the following condition is also equivalent to the above two:
:$(3): \quad$ $S$ is infinite.
== Proof ==
=== (1) implies (2) ===
Let S be a Dedekind-infinite set.
This means that S is equilvant to a proper subset of itself.
That means there exists a map F from S to a proper subset of itself.
Let A be an element of S not in the image of F.
Also let A_n be a sequence such that A_0 = A and A_(n+1) = F(A_n).
The elements of A_n form a countably infinite set.
=== (2) implies (1) ===
Let S be a set with a countably infinite subset A.
We can define F such that if x is an element of A, then f(x) = f(A_n) = f(A_(n+1)) (and f(x) = x otherwise)
Thus F is a bijection between S and a subset of S (S minus A_0).
This means it is equilvant to that subset.
Category:Infinite Sets
Equivalent Conditions for Element is Loop

== Theorem ==
Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $\sigma$ denote the closure operator on $M$.
Let $\rho$ denote the rank function of $M$.
Let $\mathscr B$ denote the set of all bases of $M$.
Let $x \in S$.
:$(1)\quad x$ is a loop
:$(2)\quad x \in \map \sigma \O$ 
:$(3)\quad \map \rho {\set x} = 0$
:$(4)\quad \set x$ is a circuit
:$(5)\quad x$ is not an element of any $B \in \mathscr B$
== Proof ==
=== Condition $(1)$ iff Condition $(2)$ ===
Follows immediately from Element is Loop iff Member of Closure of Empty Set.
 
=== Condition $(1)$ iff Condition $(3)$ ===
Follows immediately from Element is Loop iff Rank is Zero.
 
=== Condition $(1)$ iff Condition $(4)$ ===
Follows immediately from Element is Loop iff Singleton is Circuit.
 
=== Condition $(1)$ iff Condition $(5)$ ===
Follows immediately from the contrapositive statement of Element is Member of Base iff Not Loop.
Category:Matroid Loops
Equivalent Conditions for Entropic Structure

== Theorem ==
Let $\struct {S, \odot}$ be an algebraic structure.
This page gathers together some conditions on $\struct {S, \odot}$ which are logical equivalences to $\struct {S, \odot}$ being an entropic structure.
=== Pointwise Operation is Homomorphism ===
=== Pointwise Operation of Homomorphisms from External Direct Product is Homomorphism ===
=== Mapping from External Direct Product is Homomorphism ===
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.13$
Category:Equivalent Conditions for Entropic Structure
Category:Entropic Structures
Equivalent Conditions for Entropic Structure/Mapping from External Direct Product is Homomorphism

== Theorem ==
Let $\struct {S, \odot}$ be an algebraic structure.
Let $\struct {S \times S, \otimes}$ denote the external direct product of $\struct {S, \odot}$ with itself:
:$\forall \tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S: \tuple {x_1, y_1} \otimes \tuple {x_2, y_2} = \tuple {x_1 \odot x_2, y_1 \odot y_2}$
Consider the operation $\odot$ as a mapping from $S \times S$ to $S$.
That is:
:$\forall a, b \in S: \map \odot {a, b} = a \odot b$
Then:
:$\odot: S \times S \to S$ is a homomorphism from $\struct {S \times S, \otimes}$ to $\struct {S, \odot}$
:
:$\struct {S, \odot}$ is an entropic structure.
== Proof ==
=== Sufficient Condition ===
Let $\struct {S, \odot}$ be such that $\odot: S \times S \to S$ is a homomorphism.
Let $\tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S$ be arbitrary.
We have:
and it is seen $\struct {S, \odot}$ is an entropic structure by definition.
=== Necessary Condition ===
Let $\struct {S, \odot}$ be an entropic structure.
Then:
Hence by definition $\odot: S \times S \to S$ is a homomorphism.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.13$
Category:Equivalent Conditions for Entropic Structure
Category:Homomorphisms (Abstract Algebra)
Category:External Direct Products