nyuuzyou/wikis huggingface dataset:
Equivalence iff Diagonal and Inverse Composite

== Theorem ==
Let $\RR$ be a relation on $S$.
Then $\RR$ is an equivalence relation on $S$ :
:$\Delta_S \subseteq \RR$
and:
:$\RR = \RR \circ \RR^{-1}$
== Proof ==
=== Necessary Condition ===
Let $\RR$ be an equivalence relation.
Then by definition, it is reflexive, symmetric and transitive.
As $\RR$ is reflexive, we have $\Delta_S \subseteq \RR$ from Relation Contains Diagonal Relation iff Reflexive.
As $\RR$ is transitive:
:$\RR \circ \RR \subseteq \RR$
from Relation contains Composite with Self iff Transitive.
But as $\RR$ is symmetric:
:$\RR = \RR^{-1}$
from Relation equals Inverse iff Symmetric.
Thus:
:$\RR \circ \RR^{-1} \subseteq \RR$
Now we need to show that $\RR \subseteq \RR \circ \RR^{-1}$:
Let $\tuple {x, y} \in \RR$.
Then as $\RR$ is reflexive:
:$\tuple {x, y} \in \RR \land \tuple {y, y} \in \RR$
and so:
:$\tuple {x, y} \in \RR \circ \RR$
As $\RR = \RR^{-1}$, it follows that:
:$\RR \subseteq \RR \circ \RR^{-1}$
So it has been shown:
:$\Delta_S \subseteq \RR$ and $\RR = \RR \circ \RR^{-1}$
=== Sufficient Condition ===
Now, let $\Delta_S \subseteq \RR$ and $\RR = \RR \circ \RR^{-1}$.
From Relation Contains Diagonal Relation iff Reflexive, $\RR$ is reflexive.
Let $\tuple {x, y} \in \RR$.
Then:
:$\tuple {x, y} \in \RR \circ \RR^{-1}$
So:
:$\exists z \in S: \tuple {x, z} \in \RR, \tuple {z, y} \in \RR^{-1}$
and so:
:$\tuple {y, z} \in \RR$
That is, by definition, $\RR$ is transitive.
As $\tuple {x, z} \in \RR$:
:$\tuple {z, x} \in \RR^{-1}$
and so:
:$\tuple {y, x} \in \RR$
So it follows that $\RR$ is symmetric.
Thus $\RR$ is reflexive, symmetric and transitive.
Therefore, by definition, $\RR$ is an equivalence relation.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 10$: Equivalence Relations: Exercises $10.7$
Category:Equivalence Relations
Equivalence of Almost Equal Locally Integrable Functions and their Distributions

== Theorem ==
Let $f, g \in \map {L^1_{loc}} \R$ be locally integrable functions.
Let $T_f, T_g$ be the distributions associated with $f$ and $g$ respectively.
Then the following statements are equivalent:
:$T_f = T_g$
:For almost all $\mathbf x \in \R^d$ we have $\map f {\mathbf x} = \map g {\mathbf x}$.
== Proof ==
== Sources ==
* : Chapter $\S 6.1$: A glimpse of distribution theory. Test functions, distributions, and examples
Category:Distributions
Category:Locally Integrable Functions
Equivalence of Axiom Schemata for Finite Group

== Theorem ==
The following axiom schemata for the definition of a finite group are logically equivalent:
=== Finite Group Axioms ===
An algebraic structure which fulfils the finite group axioms:
=== Group Axioms ===
An algebraic structure of finite order which fulfils the group axioms:
== Proof ==
Let $S$ be an algebraic structure of finite order which fulfils the group axioms.
It is to be shown that $S$ also fulfils all the finite group axioms.
We have that $\text {FG} 0$ and $\text {FG} 1$ are the same as $\text G 0$ and $\text G 1$ and so $\text {FG} 0$ and $\text {FG} 1$ are fulfilled.
By hypothesis $S$ is of finite order.
That is, there exists some (strictly) positive integer $n$ such that $s$ has exactly $n$ elements.
That is, $\text {FG} 2$ is fulfilled.
As $S$ fulfils the group axioms, it is by definition a group.
Therefore the Cancellation Laws hold:
:$b a = c a \implies b = c$
:$a b = a c \implies b = c$
That is, $\text {FG} 3$ is fulfilled.
Thus, $S$ fulfils all the finite group axioms.
Let $S$ be an algebraic structure which fulfils the finite group axioms.
It is to be shown that $S$ also fulfils all the group axioms.
We have that $\text G 0$ and $\text G 1$ are the same as $\text {FG} 0$ and $\text {FG} 1$ and so $\text G 0$ and $\text G 1$ are fulfilled.
By definition, an algebraic structure which fulfils group axioms $\text G 0$ and $\text G 1$ is a semigroup.
From Cancellable Finite Semigroup is Group, it then follows that $S$ is a group.
That is, $S$ fulfils all the group axioms $\text G 0$, $\text G 1$, $\text G 2$ and $\text G 3$.
Hence the result.
Category:Finite Groups
Equivalence of Axiom Schemata for Groups

== Theorem ==
In the definition of a group, the axioms for the existence of an identity element and for closure under taking inverses can be replaced by the following two axioms:
:Given a group $G$, there exists at least one element $e \in G$ such that $e$ is a left identity;
:For any element $g$ in a group $G$, there exists at least one left inverse of $g$.
Alternatively, we can also replace the aforementioned axioms with the following two:
:Given a group $G$, there exists at least one element $e \in G$ such that $e$ is a right identity;
:For any element $g$ in a group $G$, there exists at least one right inverse of $g$.
Thus we can formulate the group axioms as either of the following:
=== Group Axioms (Left) ===
=== Group Axioms (Right) ===
== Proof ==
Suppose we define a group $G$ in the usual way, but make the first pair of axiom replacements listed above:
:the existence of a left identity
:every element has a left inverse.
Let $e \in G$ be a left identity and $g \in G$.
Then, from Left Inverse for All is Right Inverse, each left inverse is also a right inverse with respect to the left identity.
Also from Left Identity while exists Left Inverse for All is Identity we have that the left identity is also a right identity.
Also we have that such an Identity is Unique, so this element can rightly be called the identity.
So we have that:
:$G$ has an identity;
:each element of $G$ has an element that is both a left inverse and a right inverse with respect to this identity.
Therefore, the validity of the two axiom replacements is proved.
The proof of the alternate pair of replacements (existence of a right identity and closure under taking right inverses) is similar.
== Warning ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.12$
* : Chapter $2$: The Definition of Group Structure: $\S 26 \gamma$
* : $\text{I}$: Groups: $\S 1$ Semigroups, Monoids and Groups: Proposition $1.3$
* : $\S 3.2$: Groups; the axioms: Theorem $1$
Category:Group Theory
Equivalence of Axiom Schemata for Groups/Warning

== Theorem ==
Suppose we build an algebraic structure with the following axioms:
Then this does not (necessarily) define a group (although clearly a group fulfils those axioms).
== Proof ==
Let $\struct {S, \circ}$ be the algebraic structure defined as:
:$\forall x, y \in S: x \circ y = x$
That is, $\circ$ is the left operation.
From Element under Left Operation is Right Identity, every element serves as a right identity.
Then given any $a \in S$, we have that $x \circ a = x$ and as $x$ is an identity, axiom $(3)$ is fulfilled as well.
But from More than one Right Identity then no Left Identity, there is no left identity and therefore no identity element.
Hence $\struct {S, \circ}$ is not a group.
== Sources ==
* : $\S 1$: Some examples of groups
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Exercise $3$
Category:Group Theory
Equivalence of Complete Semilattice and Complete Lattice

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
:
:$(\text 1) \quad \struct {S, \preceq}$ is a complete join semilattice
:$(\text 2) \quad \struct {S, \preceq}$ is a complete meet semilattice
:$(\text 3) \quad \struct {S, \preceq}$ is a complete lattice
== Proof ==
=== Statement $(\text 1)$ implies Statement $(\text 2)$ ===
Let $\struct {S, \preceq}$ be a complete join semilattice.
Let $A \subseteq S$.
Let $T = \leftset{s \in S : s}$ is a lower bound for $\rightset{A}$
By definition of complete join semilattice:
:$\sup T$ exists in $\struct {S, \preceq}$
Let $a = \sup T$.
We have:
:$\forall s \in A : s$ is an upper bound for $T$
By definition of supremum:
:$\forall s \in A: a \preceq s$
By definition of lower bound:
:$a$ is a lower bound for $A$
Since $a$ is the supremum of all lower bounds of $A$:
:$a = \inf A$
Since $A$ was arbitrary:
:$\forall A \subseteq S : \inf A \in S$
Hence $\struct {S, \preceq}$ is a complete meet semilattice by definition.
=== Statement $(\text 2)$ implies Statement $(\text 3)$ ===
We have:
:Statement $(\text 2)$ implies Statement $(\text 1)$
is the dual statement of:
:Statement $(\text 1)$ implies Statement $(\text 2)$
by Dual Pairs (Order Theory).
We have:
:Statement $(\text 1)$ implies Statement $(\text 2)$
So: 
:Statement $(\text 2)$ implies Statement $(\text 1)$
follows from the Duality Principle.
The result now follows directly from:
:* Definition:Complete Meet Semilattice
:* Definition:Complete Join Semilattice
:* Definition:Complete Lattice
=== Statement $(\text 3)$ implies Statement $(\text 1)$ ===
The result follows immediately from:
:* Definition:Complete Lattice
:* Definition:Complete Join Semilattice
== Sources ==
* : Chapter $\text I$: Preliminaries, $\S4.3$ Proposition
Category:Complete Lattices
Equivalence of Definitions for Alternating Bilinear Mapping on Ring of Characteristic Not 2

== Theorem ==
Let $R$ be a commutative ring.
Let $\struct {A_R, \oplus}$ be an algebra over $R$.
Let $R$ have a characteristic not equal to $2$.
Then the following definitions for alternating bilinear mappings are equivalent:
:$\oplus$ is an alternating bilinear mapping for all $a \in A_R$, $a \oplus a = 0$
:$\oplus$ is an alternating bilinear mapping for all $a, b \in A_R$, $a \oplus b + b \oplus a = 0$
== Proof ==
Let $\oplus$ be a bilinear mapping with the property that:
:$\forall a \in A_R: a \oplus a = 0$
Then for all $u, v \in A_R$:
Let $\oplus$ be a bilinear mapping with the property that:
:$\forall a, b \in A_R: a \oplus b + b \oplus a = 0$
Let $u \in A_R$.
Then:
 
Because the characteristic of $R$ is not $2$:
:$u \oplus u = 0$
Category:Alternating Bilinear Mappings
Equivalence of Definitions of Abelian Category

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $C$ be an abelian category by definition $1$.
Thus $C$ is an abelian category by definition $2$.
=== $(2)$ implies $(1)$ ===
Let $C$ be an abelian category by definition $2$.
Thus $C$ is an abelian category by definition $1$.
Category:Abelian Categories
Equivalence of Definitions of Abelian Group

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $G$ be an abelian group by definition 1.
Then by definition:
:$\forall a \in G: \forall x \in G: a x = x a$
Thus:
:$\forall a \in G: a \in \map Z G$
By definition of subset:
:$G \subseteq \map Z G$
By definition of center:
:$\map Z G \subseteq G$
So by definition of set equality:
:$G = \map Z G$
Thus $G$ is an abelian group by definition 2.
=== Definition 2 implies Definition 1 ===
Let $G$ be an abelian group by definition 2.
Then by definition:
:$G = \map Z G$
So by the definition of center:
:$\forall a \in G: \forall x \in G: a x = x a$
Thus $G$ is an abelian group by definition 1.
== Also see ==
* Definition:Center of Group
* Center of Group is Abelian Subgroup
== Sources ==
* : $\S 37.3$ Some important general examples of subgroups
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(vii)}$
* 
Category:Abelian Groups
Equivalence of Definitions of Absolute Convergence of Product

== Theorem ==
Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.
Let $\sequence {a_n}$ be a sequence in $\mathbb K$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== 1 implies 2 ===
By the Monotone Convergence Theorem, it suffices to show that the partial sums of $\ds \sum_{n \mathop = 1}^\infty a_n$ are bounded.
Because $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \norm {a_n} }$ converges, its partial products are bounded.
By Bounds for Finite Product of Real Numbers:
:$\ds \sum_{n \mathop = 1}^N \norm {a_n} \le \prod_{n \mathop = 1}^N \paren {1 + \norm {a_n} }$
=== 2 implies 1 ===
==== Proof 1 ====
By the Monotone Convergence Theorem, it suffices to show that the partial products of $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \norm {a_n} }$ are bounded.
By Bounds for Finite Product of Real Numbers:
:$\ds \prod_{n \mathop = 1}^N \paren {1 + \norm {a_n} } \le \map \exp {\sum_{n \mathop = 1}^N \norm {a_n} }$
Because $\ds \sum_{n \mathop = 1}^\infty \norm {a_n}$ converges, its partial sums are bounded.
==== Proof 2 ====
By the Monotone Convergence Theorem, it suffices to show that the partial products of $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \norm {a_n} }$ are bounded.
By the AM-GM Inequality:
:$\ds \prod_{n \mathop = 1}^N \paren {1 + \norm {a_n} } \le \paren {\frac {N + \sum_{n \mathop = 1}^N \norm {a_n} } N }^N \le \paren {1 + \frac M N}^N$
where $M>0$ is such that $\ds \sum_{n \mathop = 1}^N \norm {a_n} \le M$ for all $N$.
By definition of the real exponential, $\paren {1 + \dfrac M N}^N \to \map \exp M$ as $N \to \infty$.
By Convergent Sequence in Metric Space is Bounded, $\paren {1 + \dfrac M N}^N$ is bounded.
== Also see ==
* Equivalence of Definitions of Absolute Convergence of Product of Complex Numbers
Category:Infinite Products
Equivalence of Definitions of Absolute Convergence of Product of Complex Numbers

== Theorem ==
Let $\sequence {a_n}$ be a sequence of complex numbers.
Let $\log$ denote the complex logarithm.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== 1 iff 2 ===
Follows directly from Equivalence of Definitions of Absolute Convergence of Product.
=== 2 implies 3 ===
By Terms in Convergent Series Converge to Zero, there exists $n_0 \in \N$ such that $\size {a_n} \le \dfrac 1 2$ for $n > n_0$.
Thus $a_n \ne -1$ for $n > n_0$.
By Bounds for Complex Logarithm:
:$\size {\map \log {1 + a_n} } \le \dfrac 3 2 \size {a_n}$
for $n > n_0$.
By the Comparison Test, $\ds \sum_{n \mathop = n_0 + 1}^\infty \map \log {1 + a_n}$ is absolutely convergent.
=== 3 implies 2 ===
By Terms in Convergent Series Converge to Zero, $\map \log {1 + a_n} \to 0$.
By Complex Exponential is Continuous, $1 + a_n \to 1$.
That is, $a_n \to 0$.
Let $n_1 \in \N$ be such that $\size {a_n} \le \dfrac 1 2$ for $n > n_1$.
By Bounds for Complex Logarithm, $\dfrac 1 2 \size {a_n} \le \size {\map \log {1 + a_n} }$ for $n > \map \max {n_0, n_1}$.
By the Comparison Test, $\ds \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent.
== Also see ==
* Logarithm of Infinite Product of Complex Numbers for a refined relation between the series $\ds \sum_{n \mathop = 1}^\infty \map \log {1 + a_n}$ and the product $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$
== Sources ==
* $VII$: Compact and Convergence in the Space of Analytic Functions: $\S 5$: Weierstrass Factorization Theorem: Proposition $5.4$
Category:Complex Analysis
Category:Infinite Products
Equivalence of Definitions of Absolute Value Function

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $f: \R \to \R$ be the function defined as:
:$\forall x \in \R: \map f x = \begin{cases}
x & : x > 0 \\
0 & : x = 0 \\
-x & : x < 0
\end{cases}$
Let $x > 0$.
We have that:
Thus $\map f x = +\sqrt {x^2}$
=== Definition 2 implies Definition 1 ===
Let $f: \R \to \R$ be the function defined as:
:$\forall x \in \R: \map f x = +\sqrt {x^2}$
Let $x > 0$.
Then by definition of square and square root:
:$+\sqrt {x^2} = x$
Let $x = 0$.
Then $+\sqrt {0^2} = 0$.
Let $x < 0$.
Thus let $x = -y$.
Then:
:$+\sqrt {\paren {-y}^2} = y = -x$
== Sources ==
* : $\S 2.2$: Functions: Exercise $\text{B} \ 1 \ \text{(d)}$
* : $\S 1$: Real Numbers: $\S 1.14$: Modulus
Category:Absolute Value Function
Equivalence of Definitions of Abundant Number

== Theorem ==
The following definitions of a abundant number are equivalent:
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
By definition of abundance:
:$\map A n = \map {\sigma_1} n - 2 n$
By definition of divisor sum function:
:$\map {\sigma_1} n$ is the sum of all the divisors of $n$.
Thus $\map {\sigma_1} n - n$ is the aliquot sum of $n$.
The result follows.
Category:Abundant Numbers
Equivalence of Definitions of Adherent Point

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
=== Definition by Open Neighborhood ===
=== Definition by Closure ===
=== Definition by Neighborhood ===
== Proof ==
=== Definition by Open Neighborhood is equivalent to Definition by Closure ===
This is shown in Condition for Point being in Closure.
=== Definition by Open Neighborhood is equivalent to Definition by Neighborhood ===
Category:Adherent Points
Category:Equivalence of Definitions of Adherent Point
Equivalence of Definitions of Adherent Point/Definition by Open Neighborhood iff Definition by Neighborhood

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
=== Definition by Open Neighborhood ===
=== Definition by Neighborhood ===
== Proof ==
=== Necessary Condition ===
Let every open neighborhood $U$ of $x$ satisfy:
:$H \cap U \ne \O$
Let $N$ be an arbitrary neighborhood of $x$.
By definition of a neighborhood:
:$\exists V \in \tau: x \in V \subseteq N$
From Set is Open iff Neighborhood of all its Points, $V$ is an open neighborhood of $x$.
Thus:
:$H \cap V \ne \O$
By the contrapositive statement of Subsets of Disjoint Sets are Disjoint:
:$H \cap N \ne \O$
Because $N$ was arbitrary, it follows that every neighborhood $N$ of $x$ satisfies:
:$H \cap N \ne \O$
=== Sufficient Condition ===
Let every neighborhood $N$ of $x$ satisfy:
:$H \cap N \ne \O$
By definition, every open neighborhood $U$ of $x$ is a neighborhood of $x$.
So every open neighborhood $U$ of $x$ satisfies:
:$H \cap U \ne \O$
Category:Equivalence of Definitions of Adherent Point
Equivalence of Definitions of Affine Hull

== Theorem ==
Let $S$ be a set.
=== Definition $1$ ===
=== Definition $2$ ===
== Proof ==
=== Definition $(1)$ implies Definition $(2)$ ===
Let $\HH$ be the affine hull of $S$ by definition $1$.
Thus $\HH$ is the affine hull of $S$ by definition $2$.
=== Definition $(2)$ implies Definition $(1)$ ===
Let $\HH$ be the affine hull of $S$ by definition $2$.
Thus $\HH$ is the affine hull of $S$ by definition $1$.
== Sources ==
* 
* 
Category:Affine Hulls
Equivalence of Definitions of Affine Space

== Theorem ==
=== Associativity Axioms ===
=== Group Action ===
=== Weyl's Axioms ===
== Proof ==
=== Associativity Axioms implies Weyl's Axioms ===
Assume the axioms $(A1)$, $(A2)$, $(A3)$.
Then for any $p, q \in \EE$ we have:
Therefore by Identity is Unique applied to the vector space $V$ we have:
Now let $p \in \EE$, $v \in V$ as in $(W1)$.
We must show there exists a unique $q \in \EE$ such that:
:$v = q - p$
Let:
:$q = p + v$
Then:
Now let $r \in \EE$ be arbitrary such that:
:$v = r - p$
Then:
This shows that $q$ is unique and establishes $(W1)$.
Now let:
:$p, q, r \in \EE$ 
as in $(W2)$.
Then:
This establishes $(W2)$.
=== Weyl's Axioms implies Group Action ===
Assume the axioms $(W1)$, $(W2)$.
Let $\phi: \EE \times V \to \EE$ be the group action defined by:
:$\forall \tuple {p, v} \in \EE \times V: p + v := \map \phi {p, v} = q$
where $q \in \EE$ is the unique point such that:
:$v = q - p$ 
given by $(W1)$.
We must verify:
To establish $(RGA1)$ let $p \in \EE$ and $u, v \in V$.
Then by $(W1)$:
Then we have:
Therefore by uniqueness in $(W1)$ we must have:
:$r = s$
Therefore:
Now to establish $(RGA2)$ let $p \in \EE$ and choose any other point $q \in \EE$.
Then by $(W2)$:
:$q - p = \paren {q - p} +_V \paren {p - p}$
So:
:$\paren {p - p} = 0_V$
or:
:$p + 0_V = p$
which establishes $(RGA2)$.
Next we must show that the action is free, that is:
:$\forall v \in V: \forall p \in \EE: p + v = p \implies v = 0_V$
Let $v \in V$ be any vector such that:
:$p + v = p$
that is:
:$p - p = v$
We have shown for $(RGA2)$ that:
:$p - p = 0_V$
and $-$ is a mapping which associates to any $p, q \in \EE$ a unique point in $q - p \in V$.
It follows that:
:$v = 0_V$
i.e. the action $+$ is free.
Finally we show that the action is transitive, that is:
:$\forall p, q \in \EE: \exists v \in V: p + v = q$
For any $p, q \in \EE$ we let:
:$v = q - p$
By the definition of the action $+$ this means that $p + v = q$, which shows that the action is transitive.
=== Group Action implies Associativity Axioms ===
Let $\phi: \EE \times V \to \EE$ be a free and transitive group action of $\struct {V, +_V}$ on $\EE$.
For $\tuple {p, v} \in \EE \times V$ write $p + v = \map \phi {p, v}$.
For any points $p, q \in \EE$, by the definition of a transitive group action there exists $v \in V$ such that:
:$p + v = q$
Now let us show that the vector $v$ with this property is unique.
If:
:$p + v_1 = p + v_2$ 
then:
Now by the definition of a free group action $p + \paren {v_1 - v_2} = 0$ implies that $v_1 - v_2 = 0$.
That is:
:$v_1 = v_2$
which shows that there is a unique vector $v$ such that:
:$p + v = q$
Therefore we can define a mapping:
:$- : \EE \times \EE \to V$ 
that associates to $\tuple {p, q} \in \EE \times \EE$ the unique vector: 
:$v = q - p \in V$ 
such that:
:$p + v = q$
Now that the mappings $+$ and $-$ are defined, we verify $(A1)$, $(A2)$ and $(A3)$.
First:
This establishes $(A1)$.
Now $(A2)$ is:
:$p + \paren {u + v} = \paren {p + u} + v$
But this is simply the statement $(RGA1)$ of a group action.
Finally for $(A3)$, let $p, q \in \EE$ and $v \in V$.
Then:
Category:Affine Geometry
Equivalence of Definitions of Algebra of Sets

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\RR$ be a system of sets such that $\forall A, B \in \RR$:
:$(1): \quad A \cup B \in \RR$
:$(2): \quad \relcomp X A \in \RR$
Let $A, B \in \RR$.
From the definition:
:$\forall A \in \RR: A \subseteq X$.
Hence from Intersection with Subset is Subset:
:$\forall A \in \RR: A \cap X = A$
Hence $X$ is the unit of $\RR$.
From Properties of Algebras of Sets, we have that $\RR$ is closed under set intersection.
From the definition of symmetric difference:
:$A \symdif B = \paren {A \setminus B} \cup \paren {B \setminus A}$
Since both set union and set difference are closed operations, it follows that symmetric difference is also closed.
So by definition 1 of ring of sets, it follows that $\RR$ is indeed a ring of sets.
=== Definition 2 implies Definition 1 ===
Let $\RR$ be a ring of sets with a unit $X$.
By definition, $X \in \RR$.
From definition 2 of ring of sets, $\RR$ is:
:$(1) \quad$ closed under set union
:$(2) \quad$ closed under set difference.
From Unit of System of Sets is Unique, we have that:
:$\forall A \in \RR: A \subseteq X$
from which we have that $X \setminus A = \relcomp X A$.
So $\RR$ is an algebra of sets by definition 1.
== Sources ==
* : $\S 1.1$: Rings and Algebras
Category:Algebras of Sets