avewright/proofwiki-math huggingface dataset:
Theorem: Inverse of Transitive Relation is Transitive

Let $\RR$ be a relation on a set $S$.
Let $\RR$ be transitive.
Then its inverse $\RR^{-1}$ is also transitive.

Proof:
Let $\RR$ be transitive.
Then:
:$\tuple {x, y}, \tuple {y, z} \in \RR \implies \tuple {x, z} \in \RR$
Thus:
:$\tuple {y, x}, \tuple {z, y} \in \RR^{-1} \implies \tuple {z, x} \in \RR^{-1}$
and so $\RR^{-1}$ is transitive.
{{qed}}
Theorem: Inverse of Transitive Relation is Transitive

Let $\RR$ be a relation on a set $S$.
Let $\RR$ be transitive.
Then its inverse $\RR^{-1}$ is also transitive.

Proof:
Let $\RR$ be transitive.
Thus by definition:
:$\RR \circ \RR \subseteq \RR$
Thus:
{{begin-eqn}}
{{eqn | l = \RR^{-1} \circ \RR^{-1}
      | r = \paren {\RR \circ \RR}^{-1}
      | c = Inverse of Composite Relation
}}
{{eqn | o = \subseteq
      | r = \RR^{-1}
      | c = Inverse of Subset of Relation is Subset of Inverse
}}
{{end-eqn}}
{{qed}}
Hence the result by definition of transitive relation.
Theorem: Inverse of Antitransitive Relation is Antitransitive

Let $\RR$ be a relation on a set $S$.
If $\RR$ is antitransitive, then so is $\RR^{-1}$.

Proof:
Let $\RR$ be antitransitive.
Then:
:$\tuple {x, y}, \tuple {y, z} \in \RR \implies \tuple {x, z} \notin \RR$
Thus:
:$\tuple {y, x}, \tuple {z, y} \in \RR^{-1} \implies \tuple {z, x} \notin \RR^{-1}$
and so $\RR^{-1}$ is antitransitive.
{{qed}}
Theorem: Inverse of Non-Transitive Relation is Non-Transitive

Let $\RR$ be a relation on a set $S$.
If $\RR$ is non-transitive, then so is $\RR^{-1}$.

Proof:
Let $\RR$ be non-transitive.
Then:
:$\exists x_1, y_1, z_1 \in S: \tuple {x_1, y_1}, \tuple {y_1, z_1} \in \RR, \tuple {x_1, z_1} \in \RR$
:$\exists x_2, y_2, z_2 \in S: \tuple {x_2, y_2}, \tuple {y_2, z_2} \in \RR, \tuple {x_2, z_2} \notin \RR$
So:
:$\exists x_1, y_1, z_1 \in S: \tuple {y_1, x_1}, \tuple {z_1, y_1} \in \RR^{-1}, \tuple {z_1, x_1} \in \RR^{-1}$
:$\exists x_2, y_2, z_2 \in S: \tuple {y_2, x_2}, \tuple {z_2, y_2} \in \RR^{-1}, \tuple {z_2, x_2} \notin \RR^{-1}$
So $\RR^{-1}$ is non-transitive.
{{qed}}
Theorem: Inverse of Ordered Ring Isomorphism is Ordered Ring Isomorphism

Let $\struct {S, +, \circ, \preceq}$ and $\struct {T, \oplus, *, \preccurlyeq}$ be ordered rings.
Let $\phi: S \to T$ be an ordered ring isomorphism.
Then $\phi^{-1}: T \to S$ is also an ordered ring isomorphism.

Proof:
By definition, $\phi$ is a bijection.
By Bijection iff Inverse is Bijection, $\phi^{-1}$ is also a bijection.
By definition, an ordered ring isomorphism from $\phi: \struct {S, +, \circ, \preceq} \to \struct {T, \oplus, *, \preccurlyeq}$ is:
:an order isomorphism from the ordered set $\struct {S, \preceq}$ to the ordered set $\struct {T, \preccurlyeq}$
:a group isomorphism from the group $\struct {S, +}$ to the group $\struct {T, \oplus}$
:a semigroup isomorphism from the semigroup $\struct {S, \circ}$ to the semigroup $\struct {T, *}$.
From Inverse of Order Isomorphism is Order Isomorphism, $\phi^{-1}: \struct {T, \preccurlyeq} \to \struct {S, \preceq}$ is an order isomorphism.
From Inverse of Algebraic Structure Isomorphism is Isomorphism:
:$\phi^{-1}: \struct {T, \oplus} \to \struct {S, +}$ is a group isomorphism
:$\phi^{-1}: \struct {T, *} \to \struct {S, \circ}$ is a semigroup isomorphism.
From Isomorphism of Abelian Groups, $\phi^{-1}: \struct {T, \oplus} \to \struct {S, +}$ preserves the commutativity of $S$ and $T$.
Hence the result.
{{qed}}
Theorem: Orthogonal Group is Subgroup of General Linear Group

Let $k$ be a field.
Let $\map {\operatorname O} {n, k}$ be the $n$th orthogonal group on $k$.
Let $\GL {n, k}$ be the $n$th general linear group on $k$.
Then $\map {\operatorname O} {n, k}$ is a subgroup of $\GL {n, k}$.

Proof:
From Unit Matrix is Orthogonal, the unit matrix $\mathbf I_n$ is orthogonal.
Let $\mathbf A, \mathbf B \in \map {\operatorname O} {n, k}$.
Then, by definition, $\mathbf A$ and $\mathbf B$ are orthogonal.
Then by Inverse of Orthogonal Matrix is Orthogonal:
:$\mathbf B^{-1}$ is a orthogonal matrix.
By Product of Orthogonal Matrices is Orthogonal Matrix:
:$\mathbf A \mathbf B^{-1}$ is a orthogonal matrix.
Thus by definition of orthogonal group:
:$\mathbf A \mathbf B^{-1} \in \map {\operatorname O} {n, k}$
Hence the result by One-Step Subgroup Test.
{{qed}}
Category:Orthogonal Groups
Category:General Linear Group
fxd28dwjyo8rwky5erj948jjit849mq
Theorem: Orthogonal Group is Group

Let $k$ be a field.
The $n$th orthogonal group on $k$ is a group.

Proof:
A direct corollary of Orthogonal Group is Subgroup of General Linear Group.
{{qed}}
Theorem: Composite of Ordered Ring Isomorphisms is Ordered Ring Isomorphism

Let $\struct {S_1, +_1, \circ_1, \preccurlyeq_1}, \struct {S_2, +_2, \circ_2, \preccurlyeq_2}, \struct {S_3, +_3, \circ_3, \preccurlyeq_3}$ be ordered rings.
Let $\phi: S_1 \to S_2$ and $\psi: S_2 \to S_3$ be ordered ring isomorphisms.
Then the composite mapping $\psi \circ \phi: S_1 \to S_3$ is also an ordered ring isomorphism.

Proof:
From Composite of Order Isomorphisms is Order Isomorphism, $\psi \circ \phi: \struct {S_1, \preccurlyeq_1} \to \struct {S_3, \preccurlyeq_3}$ is an order isomorphism.
From Composite of Isomorphisms in Algebraic Structure is Isomorphism, $\psi \circ \phi$ is an algebraic structure isomorphism.
From Isomorphism Preserves Groups, it follows that $\psi \circ \phi$ is a group isomorphism from $\struct {S_1, +_1}$ to $\struct {S_3, +_3}$.
From Isomorphism Preserves Semigroups, it follows that $\psi \circ \phi$ is a semigroup isomorphism from $\struct {S_1, \circ_1}$ to $\struct {S_3, \circ_3}$.
Hence the result.
{{qed}}
Theorem: Composite of Ordered Ring Monomorphisms is Ordered Ring Monomorphism

Let $\struct {S_1, +_1, \circ_1, \preccurlyeq_1}, \struct {S_2, +_2, \circ_2, \preccurlyeq_2}, \struct {S_3, +_3, \circ_3, \preccurlyeq_3}$ be ordered rings.
Let $\phi: S_1 \to S_2$ and $\psi: S_2 \to S_3$ be ordered ring monomorphisms.
Then the composite mapping $\psi \circ \phi: S_1 \to S_3$ is also an ordered ring monomorphism.

Proof:
From Composite of Order Embeddings is Order Embedding, $\psi \circ \phi: \struct {S_1, \preceq_1} \to \struct {S_3, \preceq_3}$ is an order embedding.
From Composite of Monomorphisms is Monomorphism, $\psi \circ \phi$ is a monomorphism.
From Group Monomorphism preserves Group, it follows that $\psi \circ \phi$ is a group monomorphism from $\struct {S_1, +_1}$ to $\struct {S_3, +_3}$.
From Semigroup Monomorphism preserves Semigroup, it follows that $\psi \circ \phi$ is a semigroup monomorphism from $\struct {S_1, \circ_1}$ to $\struct {S_3, \circ_3}$.
Hence the result.
{{qed}}
Theorem: Rescaling is Linear Transformation

Let $\struct {R, +, \times}$ be a commutative ring.
Let $\struct {V, +, \ast}_R$ be an $R$-module.
Then for any $r \in R$, the rescaling operator:
:$m_r: V \to V, v \mapsto r \ast v$
is a linear transformation.

Proof:
Let $v \in V$ and $s \in R$.
Then:
{{begin-eqn}}
{{eqn | l = \map {m_r} {s \ast v}
      | r = r \ast \paren {s \ast v}
      | c = {{Defof|Rescaling Operator}}
}}
{{eqn | r = \paren {r \times s} \ast v
      | c = $V$ is an $R$-module
}}
{{eqn | r = \paren {s \times r} \ast v
      | c = $R$ is a commutative ring
}}
{{eqn | r = s \ast \paren {r \ast v}
      | c = $V$ is an $R$-module
}}
{{eqn | r = s \ast \map {m_r} v
      | c = {{Defof|Rescaling Operator}}
}}
{{end-eqn}}
Next, for $v, w \in V$:
{{begin-eqn}}
{{eqn | l = m_r \paren {v + w}
      | r = r \ast \paren {v + w}
      | c = {{Defof|Rescaling Operator}}
}}
{{eqn | r = r \ast v + r \ast w
      | c = $V$ is an $R$-module
}}
{{eqn | r = \map {m_r} v + \map {m_r} w
      | c = {{Defof|Rescaling Operator}}
}}
{{end-eqn}}
It follows that $m_r$ is a linear transformation.
{{qed}}
Category:Linear Transformations
ay8t4x4ybnvzbe8nhngyutqp0ddulw1
Theorem: Determinant of Rescaling Matrix

Let $R$ be a commutative ring.
Let $r \in R$.
Let $r \, \mathbf I_n$ be the square matrix of order $n$ defined by:
:$\sqbrk {r \, \mathbf I_n}_{i j} = \begin {cases} r & : i = j \\ 0 & : i \ne j \end {cases}$
Then:
:$\map \det {r \, \mathbf I_n} = r^n$
where $\det$ denotes determinant.

Proof:
From Determinant of Diagonal Matrix, it follows directly that:
:$\map \det {r \, \mathbf I_n} = \ds \prod_{i \mathop = 1}^n r = r^n$
{{qed}}
Theorem: Inverse of Rescaling Matrix

Let $R$ be a commutative ring with unity.
Let $r \in R$ be a unit in $R$.
Let $r \, \mathbf I_n$ be the $n \times n$ rescaling matrix of $r$.
Then:
:$\paren {r \, \mathbf I_n}^{-1} = r^{-1} \, \mathbf I_n$

Proof:
By definition, a rescaling matrix is also a diagonal matrix.
Hence Inverse of Diagonal Matrix applies, and since $r$ is a unit, it gives the desired result.
{{qed}}
Category:Matrix Algebra
fntdpuxeksnrvrzyedfpvwv41vz2cis
Theorem: Integral Multiple of Ring Element/General Result

:$\forall m, n \in \Z: \forall x \in R: \paren {m \cdot x} \circ \paren {n \cdot x} = \paren {m n} \cdot \paren {x \circ x}$.

Proof:
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\paren {m \cdot x} \circ \paren {n \cdot x} = \paren {m n} \cdot \paren {x \circ x}$
In what follows, we make extensive use of Integral Multiple of Ring Element:
:$\forall n \in \Z: \forall x \in R: \paren {m \cdot x} \circ x = m \cdot \paren {x \circ x} = x \circ \paren {m \cdot x}$
First we verify $\map P 0$.
When $n = 0$, we have:
{{begin-eqn}}
{{eqn | l = \paren {m \cdot x} \circ \paren {0 \cdot x}
      | r = \paren {m \cdot x} \circ 0_R
      | c = 
}}
{{eqn | r = 0_R
      | c = 
}}
{{eqn | r = 0 \cdot \paren {x \circ x}
      | c = 
}}
{{eqn | r = \paren {m 0} \cdot \paren {x \circ x}
      | c = 
}}
{{end-eqn}}
So $\map P 0$ holds.
Theorem: Characteristic of Division Ring is Zero or Prime

Let $\struct {D, +, \circ}$ be a division ring.
Let $\Char D$ be the characteristic of $D$.
Then $\Char D$ is either $0$ or a prime number.

Proof:
By definition, a division ring has no proper zero divisors.
If $\struct {D, +, \circ}$ is finite, then from Characteristic of Finite Ring with No Zero Divisors, $\Char D$ is prime.
On the other hand, suppose $\struct {D, +, \circ}$ is not finite.
Then there are no $x, y \in D, x \ne 0 \ne y$ such that $x + y = 0$.
Thus it follows that $\Char D$ is $0$.
{{qed}}
Theorem: Characteristic of Integral Domain is Zero or Prime

Let $\struct {D, +, \circ}$ be an integral domain.
Let $\Char D$ be the characteristic of $D$.
Then $\Char D$ is either $0$ or a prime number.

Proof:
By definition, an integral domain has no proper zero divisors.
If $\struct {D, +, \circ}$ is finite, then from Characteristic of Finite Ring with No Zero Divisors, $\Char D$ is prime.
On the other hand, suppose $\struct {D, +, \circ}$ is not finite.
Then there are no $x, y \in D, x \ne 0 \ne y$ such that $x + y = 0$.
Thus it follows that $\Char D$ is $0$.
{{qed}}
Theorem: Sigma-Algebra Generated by Complements of Generators

Let $\Sigma$ be a $\sigma$-algebra on a set $X$.
Let $\GG$ be a generator for $\Sigma$.
Then:
:$\GG' := \set {X \setminus G: G \in \GG}$
the set of relative complements of $\GG$, is also a generator for $\Sigma$.

Proof:
{{begin-eqn}}
{{eqn | q = \forall G \in \GG
      | l = G
      | o = \in
      | r = \Sigma
      | c = {{Defof|Sigma-Algebra Generated by Collection of Subsets/Generator|Generator of Sigma-Algebra}}
}}
{{eqn | ll= \leadsto
      | q = \forall G \in \GG
      | l = X \setminus G
      | o = \in
      | r = \Sigma
      | c = {{Defof|Sigma-Algebra}}: $\text{(SA3)}$
}}
{{eqn | ll= \leadsto
      | q = \forall G \in \GG'
      | l = G
      | o = \in
      | r = \Sigma
}}
{{eqn | ll= \leadsto
      | l = \map \sigma {\GG'}
      | o = \subseteq
      | r = \Sigma
      | c = {{Defof|Sigma-Algebra Generated by Collection of Subsets}}
}}
{{eqn | q = \forall G \in \GG'
      | l = G
      | o = \in
      | r = \map \sigma {\GG'}
      | c = {{Defof|Sigma-Algebra Generated by Collection of Subsets}}
}}
{{eqn | ll= \leadsto
      | q = \forall G \in \GG
      | l = X \setminus G
      | o = \in
      | r = \map \sigma {\GG'}
}}
{{eqn | ll= \leadsto
      | q = \forall G \in \GG
      | l = X \setminus \paren {X \setminus G}
      | o = \in
      | r = \map \sigma {\GG'}
      | c = {{Defof|Sigma-Algebra}}: $\text{(SA3)}$
}}
{{eqn | ll= \leadsto
      | q = \forall G \in \GG
      | l = X \cap G
      | o = \in
      | r = \map \sigma {\GG'}
      | c = Set Difference with Set Difference
}}
{{eqn | ll= \leadsto
      | q = \forall G \in \GG
      | l = G
      | o = \in
      | r = \map \sigma {\GG'}
      | c = Intersection with Subset is Subset
}}
{{eqn | ll= \leadsto
      | l = \Sigma
      | o = \subseteq
      | r = \map \sigma {\GG'}
      | c = {{Defof|Sigma-Algebra Generated by Collection of Subsets/Generator|Generator of Sigma-Algebra}}
}}
{{eqn | ll= \leadsto
      | l = \Sigma
      | r = \map \sigma {\GG'}
      | c = {{Defof|Set Equality|index = 2}}
}}
{{end-eqn}}
Hence the result.
{{qed}}
Category:Sigma-Algebras Generated by Collection of Subsets
42a2gkr2ld2zq5bbknfpjpebkw7n4ye
Theorem: Scalar Product with Inverse Unity

:$\paren {-1_R} \circ x = - x$

Proof:
Follows directly from Scalar Product with Inverse.
{{qed}}
Theorem: Scalar Product with Multiple of Unity

:$\paren {n \cdot 1_R} \circ x = n \cdot x$
that is:
:$\paren {\map {\paren {+_R}^n} {1_R} } \circ x = \map {\paren {+_G}^n} x$

Proof:
Follows directly from Scalar Product with Product.
{{qed}}
Theorem: Subring Module is Module/Special Case

Let $S$ be a subring of the ring $\struct {R, +, \circ}$.
Let $\circ_S$ be the restriction of $\circ$ to $S \times R$.
Then $\struct {R, +, \circ_S}_S$ is an $S$-module.

Proof:
From Ring is Module over Itself, it follows that:
:$\struct {R, +, \circ}_R$ is an $R$-module.
The result follows directly from Subring Module is Module.
{{qed}}
Theorem: Trivial Module is Module

Let $\struct {G, +_G}$ be an abelian group whose identity is $e_G$.
Let $\struct {R, +_R, \circ_R}$ be a ring.
Let $\struct {G, +_G, \ast}_R$ be the trivial $R$-module, such that:
:$\forall \lambda \in R: \forall x \in G: \lambda \ast x = e_G$
Then $\struct {G, +_G, \ast}_R$ is a module.

Proof:
Checking the module axioms in turn:
:{{Module-axiom|1}}: $\quad \lambda \circ \paren {x +_G y} = e_G = e_G +_G e_G = \paren {\lambda \circ x} +_G \paren {\lambda \circ y}$
:{{Module-axiom|2}}: $\quad \paren {\lambda +_R \mu} \circ x = e_G = e_G +_G e_G = \paren {\lambda \circ x} +_G \paren {\mu \circ x}$
:{{Module-axiom|3}}: $\quad \paren {\lambda \times_R \mu} \circ x = e_G = \lambda \circ e_G = \lambda \circ \paren {\mu \circ x}$
Thus the trivial module is indeed a module.
{{qed}}
Theorem: Trivial Module is Not Unitary

Let $\struct {G, +_G}$ be an abelian group whose identity is $e_G$.
Let $\struct {R, +_R, \circ_R}$ be a ring.
Let $\struct {G, +_G, \circ}_R$ be the trivial $R$-module, such that:
:$\forall \lambda \in R: \forall x \in G: \lambda \circ x = e_G$
Then unless $R$ is a ring with unity and $G$ contains only one element, this is ''not'' a unitary module.

Proof:
By definition, for a trivial module to be unitary, $R$ needs to be a ring with unity.
For {{Module-axiom|4}} to apply, we require that:
:$\forall x \in G: 1_R \circ x = x$
But for the trivial module:
:$\forall x \in G: 1_R \circ x = e_G$
So {{Module-axiom|4}} can apply only when:
:$\forall x \in G: x = e_G$
Thus for the trivial module to be unitary, it is necessary that $G$ be the trivial group, and thus to contain one element.
{{qed}}
Theorem: Null Module is Module

Let $\struct {R, +_R, \circ_R}$ be a ring.
Let $G$ be the trivial group.
Let $\struct {G, +_G, \ast}_R$ be the null module.
Then $\struct {G, +_G, \ast}_R$ is a module.

Proof:
Follows from the fact that $\struct {G, +_G, \ast}_R$ has to be, by definition, a trivial module:
$\ast$ can only be defined as:
:$\forall \lambda \in R: \forall x \in G: \lambda \ast x = e_G$
The result then follows from Trivial Module is Module.
{{qed}}
Category:Module Theory
0tlsqipf6e4cnq5oqkgv0z0ay7pjtty
Theorem: Condition on Equality of Generated Sigma-Algebras

Let $X$ be a set.
Let $\GG$, $\HH$ be sets of subsets of $X$.
Suppose that:
:$\GG \subseteq \HH \subseteq \map \sigma \GG$
where $\sigma$ denotes generated $\sigma$-algebra.
Then:
:$\map \sigma \GG = \map \sigma \HH$

Proof:
From Generated Sigma-Algebra Preserves Subset, it follows that:
:$\map \sigma \GG \subseteq \map \sigma \HH$
Since $\map \sigma \GG$ is a $\sigma$-algebra containing $\HH$:
:$\map \sigma \HH \subseteq \map \sigma \GG$
from the definition of generated $\sigma$-algebra.
Hence the result, from the definition of set equality.
{{qed}}
Category:Sigma-Algebras Generated by Collection of Subsets
jfx0ajlffio339ib8y9gni31up3t6yw
Theorem: Sigma-Algebra is Monotone Class

Let $\Sigma$ be a $\sigma$-algebra on a set $X$.
Then $\Sigma$ is also a monotone class.

Proof:
By definition, $\Sigma$, being a $\sigma$-algebra, is closed under countable unions.
From Sigma-Algebra Closed under Countable Intersection, it is also closed under countable intersections.
Thence, by definition, $\Sigma$ is a monotone class.
{{qed}}
Category:Sigma-Algebras
Category:Monotone Classes
fzkm77bth2qkus1e99usmrmrjf455ur
Theorem: Linear Combination of Sequence is Linear Combination of Set

Let $G$ be an $R$-module.
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of elements of $G$.
Let $b$ be an element of $G$.
Then:
:$b$ is a linear combination of the sequence $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$
{{iff}}:
:$b$ is a linear combination of the set $\set {a_k: 1 \mathop \le k \mathop \le n}$

Proof:
=== Necessary Condition ===
By definition of linear combination of subset:
:Every linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is a linear combination of $\set {a_k: 1 \mathop \le k \mathop \le n}$.
{{qed|lemma}}
Theorem: Trace Sigma-Algebra of Generated Sigma-Algebra

Let $X$ be a Set, and let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Let $A \subseteq X$ be a subset of $X$.
Then the following equality holds:
:$A \cap \map \sigma \GG = \map \sigma {A \cap \GG}$
where
:$\map \sigma \GG$ denotes the smallest $\sigma$-algebra on $X$ that contains $\GG$
:$\map \sigma {A \cap \GG}$ denotes the smallest $\sigma$-algebra on $A$ that contains ${A \cap \GG}$
:$A \cap \map \sigma \GG$ denotes the trace $\sigma$-algebra on $A$
:$A \cap \GG$ is a shorthand for $\set {A \cap G: G \in \GG}$

Proof:
By definition of generated $\sigma$-algebra:
:$\GG \subseteq \map \sigma \GG$
whence from Set Intersection Preserves Subsets:
:$A \cap \GG \subseteq A \cap \map \sigma \GG$
and therefore, by definition of generated $\sigma$-algebra:
:$\map \sigma {A \cap \GG} \subseteq A \cap \map \sigma \GG$
For the reverse inclusion, define $\Sigma$ by:
:$\Sigma := \set {E \subseteq X: A \cap E \in \map \sigma {A \cap \GG} }$
We will show that $\Sigma$ is a $\sigma$-algebra on $X$.
Since $A \in \map \sigma {A \cap \GG}$:
:$A \cap X = A \in \map \sigma {A \cap \GG}$
and therefore $X \in \Sigma$.
Suppose that $E \in \Sigma$.
Then by Set Intersection Distributes over Set Difference and Intersection with Subset is Subset:
:$\paren {X \setminus E} \cap A = \paren {X \cap A} \setminus \paren {E \cap A} = A \setminus \paren {E \cap A}$
Since $E \cap A \in \map \sigma {A \cap \GG}$ and this is a $\sigma$-algebra on $A$:
:$A \setminus \paren {E \cap A} \in \map \sigma {A \cap \GG}$
Finally, let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence in $\Sigma$.
Then by Intersection Distributes over Union:
:$\ds \paren {\bigcup_{n \mathop \in \N} E_n} \cap A = \bigcup_{n \mathop \in \N} \paren {E_n \cap A}$
The latter expression is a countable union of elements of $\map \sigma {A \cap \GG}$, hence again in $\map \sigma {A \cap \GG}$.
Therefore, $\Sigma$ is a $\sigma$-algebra.
It is also apparent that $\GG \subseteq \Sigma$ since:
:$A \cap \GG \subseteq \map \sigma {A \cap \GG}$
by definition of generated $\sigma$-algebra.
Thus, as $\Sigma$ is a $\sigma$-algebra:
:$\map \sigma \GG \subseteq \Sigma$
and therefore:
:$A \cap \map \sigma \GG \subseteq \map \sigma {A \cap \GG}$
Hence the result, by definition of set equality.
{{qed}}
Category:Trace Sigma-Algebras
Category:Sigma-Algebras Generated by Collection of Subsets
mbx1au39vn8gad07p8fvc6sj1xrzb0p
Theorem: Intersection is Subset/General Result

Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\mathbb S \subseteq \powerset S$.
Then:
:$\ds \forall T \in \mathbb S: \bigcap \mathbb S \subseteq T$
=== Family of Sets ===
In the context of a family of sets, the result can be presented as follows:
{{:Intersection is Subset/Family of Sets}}

Proof:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = \bigcap \mathbb S
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall T \in \mathbb S
      | l = x
      | o = \in
      | r = T
      | c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
      | q = \forall T \in \mathbb S
      | l = \bigcap \mathbb S
      | o = \subseteq
      | r = T
      | c = {{Defof|Subset}}
}}
{{end-eqn}}
{{qed}}
Category:Set Intersection
Category:Subsets
j3fcbz868axxqufbjjn04beiw2p6ara
Theorem: Union of Relative Complements of Nested Subsets

Let $R \subseteq S \subseteq T$ be sets with the indicated inclusions.
Then:
:$\relcomp T S \cup \relcomp S R = \relcomp T R$
where $\complement$ denotes relative complement.
Phrased via Set Difference as Intersection with Relative Complement:
:$\paren {T \setminus S} \cup \paren {S \setminus R} = T \setminus R$
where $\setminus$ denotes set difference.

Proof:
From Union with Set Difference:
:$T = T \setminus S \cup S$
and therefore by Set Difference is Right Distributive over Union:
:$T \setminus R = \paren {\paren {T \setminus S} \setminus R} \cup \paren {S \setminus R}$
Now, by Set Difference with Union and Union with Superset is Superset:
:$\paren {T \setminus S} \setminus R = T \setminus \paren {S \cup R} = T \setminus S$
Combining the above yields:
:$T \setminus R = \paren {T \setminus S} \cup \paren {S \setminus R}$
{{qed}}
Theorem: Homomorphic Image of R-Module is R-Module

Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ_G}_R$ be an $R$-module.
Let $\struct {H, +_H, \circ_H}_R$ be an $R$-algebraic structure.
Let $\phi: G \to H$ be a homomorphism.
Then the homomorphic image of $\phi$ is an $R$-module.

Proof:
Let us write $\phi \sqbrk G$ to denote the homomorphic image of $\phi$.
From Image of Group Homomorphism is Subgroup, $\phi \sqbrk G$ is a subgroup of $\struct {H, +_H}$.
For any $\map \phi g$ and $\map \phi {g'}$ in $\phi \sqbrk G$, we have:
{{begin-eqn}}
{{eqn | l = \map \phi g +_H \map \phi {g'}
      | r = \map \phi {g +_G g'}
      | c = {{Defof|Morphism Property}}
}}
{{eqn | r = \map \phi {g' +_G g'}
      | c = {{Defof|Abelian Group}}
}}
{{eqn | r = \map \phi {g'} +_H \map \phi g
      | c = {{Defof|Morphism Property}}
}}
{{end-eqn}}
hence $\phi \sqbrk G$ is an abelian group.
Now we can turn to showing that $\phi \sqbrk G$ is an $R$-module.
To do this, we take the module axioms in turn.