nyuuzyou/wikis huggingface dataset:
Intersection Condition for Direct Sum of Subspaces

== Theorem ==
Let $U$ and $W$ be subspaces of a vector space $V$.
Then $U + W$ is a direct sum $U \cap W = 0$. 
== Proof ==
We must first prove that if $U + W$ is a direct sum, then:
:$U \cap W = 0$
Let $U + W$ be a direct sum.
Let $\mathbf v \in U \cap W$ be an arbitrary vector in $U \cap W$.
Then:
:$\mathbf 0 = \mathbf v + \paren {-\mathbf v}$
where:
:$\mathbf v \in U$
:$-\mathbf v \in W$
== Also see ==
* Two-Step Vector Subspace Test
* Null Space is Subspace
== Sources ==
* 
Category:Linear Algebra
Intersection Distributes over Symmetric Difference

== Theorem ==
Intersection is distributive over symmetric difference:
:$\paren {R \symdif S} \cap T = \paren {R \cap T} \symdif \paren {S \cap T}$
:$T \cap \paren {R \symdif S} = \paren {T \cap R} \symdif \paren {T \cap S}$
== Proof ==
From Set Intersection Distributes over Set Difference, we have:
:$\paren {R \setminus S} \cap T = \paren {R \cap T} \setminus \paren {S \cap T}$
So:
The second part of the proof is a direct consequence of the fact that Intersection is Commutative.
== Sources ==
* : Chapter $1$. Sets: Exercise $7 \ \text{(iv)}$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $14$
* : $\text {(v)}$
Category:Set Intersection
Category:Symmetric Difference
Category:Examples of Distributive Operations
Intersection Distributes over Union

== Theorem ==
Set intersection is distributive over set union:
:$R \cap \paren {S \cup T} = \paren {R \cap S} \cup \paren {R \cap T}$
=== Family of Sets ===
=== General Result ===
== Proof ==
== Demonstration by Venn Diagram ==
== Examples ==
== Also known as ==
== Also see ==
* Union Distributes over Intersection
* Class Intersection Distributes over Class Union
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $2 \ \text{(d)}$
* : $\S 4$: Unions and Intersections
* : A Note on Set-Theoretic Concepts: $(3)$
* : Exercise $1.1: \ 8 \ \text{(e)}$
* : Chapter $1$: Set Theory: $1.3$: Set operations
* : $\S 1.5$. The algebra of sets: Example $17$
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Theorem $3.1$
* : $1$ Set Theory: $1$. Sets and Functions: $1.2$: Operations on sets: $(1)$
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 7 \ \text{(b)}$
* : $\S 1.6$: Set Identities and Other Set Relations: Theorem $6.2$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: $\text{(d)}$
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $2$
* : $\S 7.4 \ \text{(i)}$: Unions and Intersections
* : $1$: Complex Numbers: Supplementary Problems: Point Sets: $122 \ \text{(e)}$
* : Chapter $1$: Sets and mappings: $\S 1.2$: Sets: Exercise $1 \ \text{(vii)}$
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.1 \ \text{(vi)}$
* : $1$: Set Theory and Logic: $\S 1$: Fundamental Concepts
* : $\S 2$
* : $\S 2$
* 
* : Appendix $\text{A}.2$: Theorem $\text{A}.10$
* : $\text {(vi)}$
Category:Intersection Distributes over Union
Category:Distributive Laws of Set Theory
Category:Set Intersection
Category:Set Union
Category:Examples of Distributive Operations
Intersection Distributes over Union/Also known as

== Intersection Distributes over Union: Also known as == 
The result Intersection Distributes over Union, along with Union Distributes over Intersection, are together known as:
:the distributive laws (of set theory)
:the distributive properties (of set theory).
== Sources ==
* : $1$: Set Theory and Logic: $\S 1$: Fundamental Concepts
Category:Intersection Distributes over Union
Intersection Distributes over Union/Examples/3 Arbitrarily Chosen Sets

== Examples of Use of Intersection Distributes over Union ==
Let:
==== Intersection with Union ====
==== Union of Intersections ====
Thus it is seen that:
:$A \cap \paren {B \cup C} = \paren {A \cap B} \cup \paren {A \cap C}$
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Point Sets: $46$
Category:Examples of Use of Intersection Distributes over Union
Intersection Distributes over Union/Examples/Arbitrary Integer Sets

== Example of Use of Intersection Distributes over Union ==
Let:
Then:
:$\paren {A \cup B} \cap C = \set {1, 2, 3, 4} = \paren {A \cap C} \cup \paren {B \cap C}$
== Proof ==
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $1$
Category:Examples of Use of Intersection Distributes over Union
Intersection Distributes over Union/Family of Sets

== Theorem ==
Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ be a indexed family of subsets of a set $S$.
Let $B \subseteq S$.
Then:
:$\ds \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cap B} = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B$
where $\ds \bigcup_{\alpha \mathop \in I} A_\alpha$ denotes the union of $\family {A_\alpha}_{\alpha \mathop \in I}$.
=== Corollary ===
== Proof ==
By definition of subset:
:$\ds \bigcup_{\alpha \mathop \in I} \paren {A_\alpha \cap B} \subseteq \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B$
By definition of subset:
:$\ds \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B \subseteq \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cap B}$
By definition of set equality:
:$\ds\map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cap B} = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B$
== Also see ==
* Union Distributes over Intersection of Family
== Sources ==
* : $\S 9$: Families
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $1 \ \text{(f)}$
* : $\S 1$: Naive Set Theory: $\S 1.4$: Sets of Sets: Exercise $1.4.4 \ \text{(v)}$
Category:Intersection Distributes over Union
Category:Indexed Families
Intersection Distributes over Union/Family of Sets/Corollary

== Theorem ==
Let $I$ and $J$ be indexing sets.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\beta}_{\beta \mathop \in J}$ be indexed families of subsets of a set $S$.
Then:
:$\ds \bigcup_{\tuple {\alpha, \beta} \mathop \in I \times J} \paren {A_\alpha \cap B_\beta} = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap \paren {\bigcup_{\beta \mathop \in J} B_\beta}$
where $\ds \bigcup_{\alpha \mathop \in I} A_\alpha$ denotes the union of $\family {A_\alpha}_{\alpha \mathop \in I}$.
== Proof ==
== Sources ==
* : $\S 9$: Families
Category:Intersection Distributes over Union
Intersection Distributes over Union/General Result

== Theorem ==
Set intersection is distributive over set union:
Let $S$ and $T$ be sets.
Let $\powerset T$ be the power set of $T$.
Let $\mathbb T$ be a subset of $\powerset T$.
Then:
:$\ds S \cap \bigcup \mathbb T = \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$
== Proof ==
=== Intersection Subset of Union ===
Let $\ds x \in S \cap \bigcup \mathbb T$.
We need to show that $\ds x \in \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$ and then by definition of subset we will have shown that $\ds S \cap \bigcup \mathbb T \subseteq \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$.
So, we have that $\ds x \in S \cap \bigcup \mathbb T$.
By definition of set intersection, $x \in S$ and $\ds x \in \bigcup \mathbb T$.
From $\ds x \in \bigcup \mathbb T$ we know that:
:$\exists X \in \mathbb T: x \in X$
and so:
:$\ds \exists X \in \mathbb T: x \in S \cap X$
So by definition of set union:
:$\ds x \in \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$
So:
:$\ds S \cap \bigcup \mathbb T \subseteq \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$
=== Union Subset of Intersection ===
Let $\ds x \in \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$.
We need to show that $\ds x \in S \cap \bigcup \mathbb T$ and then by definition of subset we will have shown that $\ds \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X} \subseteq S \cap \bigcup \mathbb T$.
So, we have that $\ds x \in \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$.
By definition of set union:
:$\exists X \in \mathbb T: x \in S \cap X$
By definition of set intersection, we have that $x \in S$ and $x \in X$.
By definition of set union:
:$\ds x \in \bigcup \mathbb T$
So by definition of set intersection, we have that:
:$\ds x \in S \cap \bigcup \mathbb T$
So:
:$\ds \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X} \subseteq S \cap \bigcup \mathbb T$
So we have that:
:$\ds S \cap \bigcup \mathbb T \subseteq \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$
and:
:$\ds \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X} \subseteq S \cap \bigcup \mathbb T$
and so by definition of set equality:
:$\ds S \cap \bigcup \mathbb T = \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$
== Also see ==
* Union Distributes over Intersection/General Result
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.6 \ \text{(a)}$
* : Chapter $\text{I}$: Sets and Functions: Unions and Intersections: Theorem $2 \ \text{(i)}$
Category:Intersection Distributes over Union
Intersection Distributes over Union/Venn Diagram

== Theorem ==
== Proof ==
Demonstration by Venn diagram:
: 
In the left hand diagram, $R$ is depicted in yellow and $S \cup T$ is depicted in blue.
Their intersection $R \cap \paren {S \cup T}$ where they overlap is depicted in green.
In the right hand diagram, $\paren {R \cap S}$ is depicted in yellow and $\paren {R \cap T}$ is depicted in blue.
Their intersection, where they overlap, is depicted in green.
Their union is the total shaded area: yellow, blue and green.
As can be seen by inspection, the areas are the same.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets
* : Chapter $1$: Elementary, my dear Watson: $\S 1.2$: Elements, my dear Watson: Ponderable $1.2.1 \ \text{(a)}$
Category:Intersection Distributes over Union
Category:Examples of Venn Diagrams
Intersection Measure is Measure

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $F \in \Sigma$ be a measurable set.
Then the intersection measure $\mu_F$ is a measure on the measurable space $\struct {X, \Sigma}$.
== Proof ==
Verify the axioms for a measure in turn for $\mu_F$:
=== Axiom $(1)$ ===
The statement of axiom $(1)$ for $\mu_F$ is:
:$\forall E \in \Sigma: \map {\mu_F} E \ge 0$
For every $E \in \Sigma$ have:
=== Axiom $(2)$ ===
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$.
The statement of axiom $(2)$ for $\mu_F$ is:
:$\ds \map {\mu_F} {\bigcup_{n \mathop \in \N} E_n} = \sum_{n \mathop \in \N} \map {\mu_F} {E_n}$
This is verified by the following computation:
=== Axiom $(3')$ ===
The statement of axiom $(3')$ for $\mu_F$ is:
:$\map {\mu_F} \O = 0$
We have:
Having verified a suitable set of axioms, it follows that $\mu_F$ is a measure.
== Sources ==
* : $\S 4$: Problem $7$
Category:Intersection Measures
Category:Measures
Intersection Measure of Finite Measure is Finite Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a finite measure on $\struct {X, \Sigma}$. 
Let $A \in \Sigma$. 
Let $\mu_A$ be the intersection measure of $\mu$ by $A$.
Then $\mu_A$ is a finite measure.
== Proof ==
From Intersection Measure is Measure, $\mu_A$ is a measure.
Since $\mu$ is a finite measure, we have: 
:$\map \mu X < \infty$ 
Then, we have: 
So $\mu_A$ is a finite measure.
Category:Finite Measures
Category:Intersection Measures
Intersection Measures preserve Absolute Continuity

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $A \in \Sigma$.
Let $\mu$ and $\nu$ be measures such that:
:$\mu$ is absolutely continuous with respect to $\nu$.
Let $\mu_A$ be the intersection measure of $\mu$ with respect to $A$.
Then $\mu_A$ is absolutely continuous with respect to $\nu$.
== Proof ==
Let $B \in \Sigma$ be such that:
:$\map \nu B = 0$
From Null Sets Closed under Subset, we have that $\map \nu {A \cap B} = 0$.
Hence since $\mu$ is absolutely continuous with respect to $\nu$, we have $\map \mu {A \cap B} = 0$.
That is, $\map {\mu_A} B = 0$. 
Hence whenever $B \in \Sigma$ is a $\nu$-null set, it is a $\mu_A$-null set.
Hence $\mu_A$ is absolutely continuous with respect to $\nu$.
Category:Absolutely Continuous Measures
Intersection Measures preserve Mutual Singularity

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $A \in \Sigma$.
Let $\mu$ and $\nu$ be measures such that:
:$\mu$ is mutually singular with respect to $\nu$.
Let $\mu_A$ be the intersection measure of $\mu$ with respect to $A$.
Then $\mu_A$ is mutually singular with respect to $\nu$.
== Proof ==
Since $\mu$ is mutually singular with respect to $\nu$, there exists a $\nu$-null set $N$ such that $\map \mu {N^c} = 0$.
From Null Sets Closed under Subset, we have $\map \mu {A \cap N^c} = 0$.
Hence $\map {\mu_A} {N^c} = 0$. 
Hence $N$ is a $\nu$-null set such that $\map {\mu_A} {N^c} = 0$.
Hence $\mu_A$ is mutually singular with respect to $\nu$.
Category:Mutually Singular Measures
Intersection Operation on Supersets of Subset is Closed

== Theorem ==
Let $S$ be a set.
Let $T \subseteq S$ be a given subset of $S$.
Let $\powerset S$ denote the power set of $S$
Let $\mathscr S$ be the subset of $\powerset S$ defined as:
:$\mathscr S = \set {Y \in \powerset S: T \subseteq Y}$
Then the algebraic structure $\struct {\mathscr S, \cap}$ is closed.
== Proof ==
Let $A, B \in \mathscr S$.
We have that:
and:
Thus we have:
Hence the result by definition of closed algebraic structure.
== Also see ==
* Union Operation on Supersets of Subset is Closed
== Sources ==
* : Chapter $5$: Subgroups: Exercise $1$
Category:Set Union
Intersection Signed Measure is Signed Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$. 
Let $F \in \Sigma$.
Let $\mu_F$ be the intersection signed measure of $\mu$ by $F$.
Then $\mu_F$ is a signed measure.
== Proof ==
Since $\mu$ is a signed measure it takes values in either $\overline \R \setminus \set \infty$ or $\overline \R \setminus \set {-\infty}$.
That is: 
:$\map \mu E \in \overline \R \setminus \set \infty$ for each $E \in \Sigma$
or:
:$\map \mu E \in \overline \R \setminus \set {-\infty}$ for each $E \in \Sigma$.
In particular: 
:$\map {\mu_F} E = \map \mu {E \cap F} \in \overline \R \setminus \set \infty$ for each $E \in \Sigma$
or:
:$\map {\mu_F} E = \map \mu {E \cap F} \in \overline \R \setminus \set {-\infty}$ for each $E \in \Sigma$.
Now we verify the two conditions required of a signed measure.
We have: 
Now let $\sequence {S_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$.
Then, we have: 
So $\mu$ is a signed measure.
Category:Signed Measures
Category:Intersection Signed Measures
Intersection is Associative

== Theorem ==
Set intersection is associative:
:$A \cap \paren {B \cap C} = \paren {A \cap B} \cap C$
=== Family of Sets ===
== Proof ==
Therefore:
:$x \in A \cap \paren {B \cap C}$ $x \in \paren {A \cap B} \cap C$
Thus it has been shown that:
:$A \cap \paren {B \cap C}\ = \paren {A \cap B} \cap C$
== Also see ==
* Union is Associative
* Set Difference is not Associative
* Symmetric Difference is Associative
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $2 \ \text{(c)}$
* : $\S 4$: Unions and Intersections
* : A Note on Set-Theoretic Concepts: $(2)$
* : Exercise $1.1: \ 8 \ \text{(d)}$
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : Chapter $1$: Set Theory: $1.3$: Set operations
* : $\S 1.8$. Sets of sets: Example $27$
* : $\S 4.2$. Commutative and associative operations: Example $66$
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Theorem $3.1$
* : Chapter $\text{I}$: Sets and Functions: Unions and Intersections: $1$
* : $1$ Set Theory: $1$. Sets and Functions: $1.2$: Operations on sets
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 5 \ \text{(a)}$
* : $\S 1.6$: Set Identities and Other Set Relations: Exercise $2 \ \text{(f)}$
* : $\S 2$: Sets and functions: Operations
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: $\text{(c)}$
* : $1$: Complex Numbers: Supplementary Problems: Point Sets: $122 \ \text{(d)}$
* : Chapter $1$: Sets and mappings: $\S 1.2$: Sets: Exercise $1 \ \text{(v)}$
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets
* : $\S 2$
* : Appendix $\text{A}.2$: Theorem $\text{A}.10$
* : $\text {(i)}$
* : $\text {(iii)}$
Category:Direct Proofs
Category:Set Intersection
Category:Examples of Associative Operations‎
Category:Intersection is Associative
Intersection is Associative/Family of Sets

== Theorem ==
Let $\family {S_i}_{i \mathop \in I}$ and $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$ be indexed families of sets.
Let $\ds I = \bigcap_{\lambda \mathop \in \Lambda} I_\lambda$.
Then:
:$\ds \bigcap_{i \mathop \in I} S_i = \bigcap_{\lambda \mathop \in \Lambda} \paren {\bigcap_{i \mathop \in I_\lambda} S_i}$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* General Associativity of Set Union
Category:Intersection is Associative
Intersection is Associative/Family of Sets/Proof 1

== Theorem ==
== Proof ==
For every $\lambda \in \Lambda$, let $\ds T_\lambda = \bigcap_{i \mathop \in I_\lambda} S_i$.
Then:
Thus:
:$\ds \bigcap_{i \mathop \in I} S_i = \bigcap_{\lambda \mathop \in \Lambda} T_\lambda = \bigcap_{\lambda \mathop \in \Lambda} \paren {\bigcap_{i \mathop \in I_\lambda} S_i}$
== Sources ==
* : $\S 9$: Families
* : $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.2$
Category:Intersection is Associative
Intersection is Commutative

== Theorem ==
Set intersection is commutative:
:$S \cap T = T \cap S$
=== Family of Sets ===
== Proof ==
== Also see ==
* Union is Commutative
* Set Difference is Anticommutative
* Symmetric Difference is Commutative
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $2 \ \text{(b)}$
* : $\S 4$: Unions and Intersections
* : A Note on Set-Theoretic Concepts: $(1)$
* : Exercise $1.1: \ 8 \ \text{(c)}$
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : $\S 2.1$: Sets: Exercise $\text{B} \ 6$
* : Chapter $1$: Set Theory: $1.3$: Set operations
* : $\S 1.3$. Intersection: Example $13$
* : $\S 4.2$. Commutative and associative operations: Example $63$
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Theorem $3.1$
* : Chapter $\text{I}$: Sets and Functions: Unions and Intersections: $1$
* : $1$ Set Theory: $1$. Sets and Functions: $1.2$: Operations on sets
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 5 \ \text{(b)}$
* : $\S 2$: Sets and functions: Operations
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: $\text{(b)}$
* : $1$: Complex Numbers: Supplementary Problems: Point Sets: $122 \ \text{(b)}$
* : Chapter $1$: Sets and mappings: $\S 1.2$: Sets: Exercise $1 \ \text{(iii)}$
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets
* : $\S 2$
* : Appendix $\text{A}.2$: Theorem $\text{A}.10$
* : $\text {(ii)}$
* : $\text {(ii)}$
Category:Set Intersection
Category:Examples of Commutative Operations
Category:Intersection is Commutative
Intersection is Commutative/Family of Sets

== Theorem ==
Let $\family {S_i}_{i \mathop \in I}$ be an indexed family of sets.
Let $\ds I = \bigcap_{i \mathop \in I} S_i$ denote the intersection of $\family {S_i}_{i \mathop \in I}$.
Let $J \subseteq I$ be a subset of $I$.
Then:
:$\ds \bigcap_{i \mathop \in I} S_i = \bigcap_{j \mathop \in J} S_j \cap \bigcap_{k \mathop \in \relcomp I J} S_k = \bigcap_{k \mathop \in \relcomp I J} S_k \cap \bigcap_{j \mathop \in J} S_j$
where $\relcomp I J$ denotes the complement of $J$ relative to $I$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* General Commutativity of Set Union
Category:Intersection is Commutative
Intersection is Commutative/Family of Sets/Proof 1

== Theorem ==
== Proof ==
We have that both $\ds \bigcap_{j \mathop \in J} S_j$ and $\ds \bigcap_{k \mathop \in \relcomp I J} S_k$ are sets.
Hence by Intersection is Commutative we have:
:$\ds \bigcap_{j \mathop \in J} S_j \cap \bigcap_{k \mathop \in \relcomp I J} S_k = \bigcap_{k \mathop \in \relcomp I J} S_k \cap \bigcap_{j \mathop \in J} S_j$
It remains to be demonstrated that $\ds \bigcap_{i \mathop \in I} S_i = \bigcap_{j \mathop \in J} S_j \cap \bigcap_{k \mathop \in \relcomp I J} S_k$.
So:
That is:
:$\ds x \in \bigcap_{i \mathop \in I} S_i \iff x \in \bigcap_{j \mathop \in J} S_j \cap \bigcap_{k \mathop \in \relcomp I J} S_k$
The result follows by definition of set equality.
== Sources ==
* : $\S 9$: Families
Category:Intersection is Commutative
Intersection is Decreasing

== Theorem ==
Let $U$ be a set.
Let $\FF$ and $\GG$ be sets of subsets of $U$.
Then $\FF \subseteq \GG \implies \bigcap \GG \subseteq \bigcap \FF$, where by convention $\bigcap \O = U$.
That is, $\bigcap$ is a decreasing mapping from $\struct {\powerset {\powerset U}, \subseteq}$ to $\struct {\powerset U, \subseteq}$, where $\powerset U$ is the power set of $U$.
== Proof ==
Let $\FF \subseteq \GG$.
Let $x \in \bigcap \GG$.
Then for each $S \in \FF$, $S \in \GG$.
By the definition of intersection, $x \in S$.
Since this holds for all $S \in \FF$, $x \in \bigcap \FF$.
Since this holds for all $ x \in \bigcap \GG$:
:$\bigcap \GG \subseteq \bigcap \FF$
Category:Set Intersection
Intersection is Empty and Union is Universe if Sets are Complementary

== Theorem ==
Let $A$ and $B$ be subsets of a universe $\Bbb U$.
Then:
:$A \cap B = \O$ and $A \cup B = \Bbb U$
:
:$B = \relcomp {\Bbb U} A$
where $\relcomp {\Bbb U} A$ denotes the complement of $A$ with respect to $\Bbb U$.
== Proof ==
From Complement Union with Superset is Universe: Corollary:
:$A \cup B = \mathbb U \iff \relcomp {\Bbb U} A \subseteq B$
and from Empty Intersection iff Subset of Complement:
:$A \cap B = \O \iff B \subseteq \relcomp {\Bbb U} A$
The result follows by definition of set equality.
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $11 \ \text{(c)}$
Category:Set Intersection
Category:Set Union
Category:Set Complement
Category:Empty Set
Category:Universal Set
Intersection is Largest Subset

== Theorem ==
Let $T_1$ and $T_2$ be sets.
Then $T_1 \cap T_2$ is the largest set contained in both $T_1$ and $T_2$.
That is:
:$S \subseteq T_1 \land S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$
=== Set of Sets ===
=== General Result ===
== Proof ==
=== Sufficient Condition ===
From Set is Subset of Intersection of Supersets we have that:
:$S \subseteq T_1 \land S \subseteq T_2 \implies S \subseteq T_1 \cap T_2$
=== Necessary Condition ===
Let:
:$S \subseteq T_1 \cap T_2$
From Intersection is Subset we have $T_1 \cap T_2 \subseteq T_1$ and $T_1 \cap T_2\subseteq T_2$.
From Subset Relation is Transitive, it follows directly that $S \subseteq T_1$ and $S \subseteq T_2$.
So $S \subseteq T_1 \cap T_2 \implies S \subseteq T_1 \land S \subseteq T_2$.
From the above, we have:
:$S \subseteq T_1 \land S \subseteq T_2 \implies S \subseteq T_1 \cap T_2$
:$S \subseteq T_1 \cap T_2 \implies S \subseteq T_1 \land S \subseteq T_2$
Thus $S \subseteq T_1 \land S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$ from the definition of equivalence.
== Also see ==
* Union is Smallest Superset
== Sources ==
* : $1$: Set Theory and Logic: $\S 1$: Fundamental Concepts: Exercise $1.2 \ \text{(c)}$
Category:Set Intersection
Category:Subsets