nyuuzyou/wikis huggingface dataset:
Intersection is Largest Subset/Family of Sets

== Theorem ==
Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Then for all sets $X$:
:$\ds \paren {\forall i \in I: X \subseteq S_i} \iff X \subseteq \bigcap_{i \mathop \in I} S_i$
where $\ds \bigcap_{i \mathop \in I} S_i$ is the intersection of $\family {S_i}$.
== Proof ==
Let $X \subseteq S_i$ for all $i \in I$.
Then from Set is Subset of Intersection of Supersets: General Result:
:$\ds X \subseteq \bigcap_{i \mathop \in I} S_i$
Now suppose that $\ds X \subseteq \bigcap_{i \mathop \in I} S_i$.
From Intersection is Subset: Family of Sets we have:
:$\ds \forall i \in I: \bigcap_{j \mathop \in I} S_j \subseteq S_i$
So from Subset Relation is Transitive, it follows that:
:$\ds \forall i \in I: X \subseteq \bigcap_{j \mathop \in I} S_j \subseteq S_i$
So it follows that $\forall i \in I: X \subseteq S_i$.
So:
:$\ds X \subseteq \bigcap_{i \mathop \in I} S_i \implies \paren {\forall i \in I: X \subseteq S_i}$
Hence:
:$\ds \paren {\forall i \in I: X \subseteq S_i} \iff X \subseteq \bigcap_{i \mathop \in I} S_i$
== Also see ==
* Union is Smallest Superset: Family of Sets
== Sources ==
* : $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.1 \ (1)$
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $4 \ \text{(a)}$
Category:Set Intersection
Category:Subsets
Category:Indexed Families
Intersection is Largest Subset/General Result

== Theorem ==
Let $T$ be a set.
Let $\powerset T$ be the power set of $T$.
Let $\mathbb T$ be a subset of $\powerset T$.
Then:
:$\paren {\forall X \in \mathbb T: S \subseteq X} \iff S \subseteq \bigcap \mathbb T$
=== Family of Sets ===
In the context of a family of sets, the result can be presented as follows:
== Proof ==
Let $\mathbb T \subseteq \powerset T$.
Suppose that $\forall X \in \mathbb T: S \subseteq X$.
Consider any $x \in S$.
By definition of subset, it follows that:
:$\forall X \in \mathbb T: x \in X$
Thus it follows from definition of set intersection that:
:$x \in \bigcap \mathbb T$
Thus by definition of subset, it follows that:
:$S \subseteq \bigcap \mathbb T$
So:
:$\paren {\forall X \in \mathbb T: S \subseteq X} \implies S \subseteq \bigcap \mathbb T$
Now suppose that $S \subseteq \bigcap \mathbb T$.
From Intersection is Subset: General Result we have:
:$\forall X \in \mathbb T: \bigcap \mathbb T \subseteq X$
So from Subset Relation is Transitive, it follows that:
:$\forall X \in \mathbb T: S \subseteq \bigcap \mathbb T \subseteq X$
So it follows that $\forall X \in \mathbb T: S \subseteq X$.
So:
:$S \subseteq \bigcap \mathbb T \implies \paren {\forall X \in \mathbb T: S \subseteq X}$
Hence:
:$\paren {\forall X \in \mathbb T: S \subseteq X} \iff S \subseteq \bigcap \mathbb T$
== Caution ==
Be careful of the way this result is stated:
:$\paren {\forall X \in \mathbb T: S \subseteq X} \iff S \subseteq \bigcap \mathbb T$
Without the brackets, this would read:
:$\forall X \in \mathbb T: S \subseteq X \iff S \subseteq \bigcap \mathbb T$
That is:
:For all $X \in \mathbb T$, it is the case that $S \subseteq X$ $S \subseteq \bigcap \mathbb T$
This is not the same thing at all.
For example, if $\mathbb T = \powerset T$ (as it well might), then $T \in \mathbb T$, and $\bigcap \mathbb T = \O$.
This would imply that:
:$T \subseteq \bigcap \mathbb T$
That is:
:$T \subseteq \O$ 
which is obviously rubbish.
== Also see ==
* Union is Smallest Superset: General Result
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.6 \ \text{(e)}$
Category:Set Intersection
Category:Subsets
Intersection is Subset

== Theorem ==
The intersection of two sets is a subset of each:
:$S \cap T \subseteq S$
:$S \cap T \subseteq T$
=== General Result ===
== Proof ==
Similarly for $T$.
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection
* : Exercise $1.1: \ 8 \ \text{(f)}$
* : $\S 1.3$. Intersection: Example $13$
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 5$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems
* : $\S 7$: Unions and Intersections
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.1 \ \text{(ii)}$
Category:Set Intersection
Category:Subsets
Intersection is Subset/Family of Sets

== Theorem ==
Let $\family {S_\alpha}_{\alpha \mathop \in I}$ be a family of sets indexed by $I$.
Then:
:$\ds \forall \beta \in I: \bigcap_{\alpha \mathop \in I} S_\alpha \subseteq S_\beta$
where $\ds \bigcap_{\alpha \mathop \in I} S_\alpha$ is the intersection of $\family {S_\alpha}_{\alpha \mathop \in I}$.
== Proof ==
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $1 \ \text{(b)}$
Category:Set Intersection
Category:Subsets
Category:Indexed Families
Intersection is Subset/General Result

== Theorem ==
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:
== Proof ==
Category:Set Intersection
Category:Subsets
Intersection is Subset of Union of Intersections with Complements

== Theorem ==
Let $R, S, T$ be sets.
Then:
:$S \cap T \subseteq \paren {R \cap S} \cup \paren {\overline R \cap T}$
where $\overline R$ denotes the complement of $R$.
== Proof ==
Let $x \in S \cap T$.
Then by definition of set intersection, $x \in S \land x \in T$.
From Conjunction implies Disjunction of Conjunctions with Complements, it follows that:
:$\paren {x \in S \land \psi} \lor \paren {x \in T \land \neg \psi}$
where $\psi$ is any arbitrary statement.
Let $\psi$ be the statement $x \in R$.
Thus:
:$\paren {x \in S \land x \in T} \implies \paren {x \in S \land x \in R} \lor \paren {x \in T \land x \notin R}$
By definition of subset, set intersection, set union and set complement, it follows that:
:$S \cap T \subseteq \paren {R \cap S} \cup \paren {\overline R \cap T}$
== Illustration by Venn Diagram ==
A Venn diagram illustrating this result is given below:
:
The red field marks $R \cap S$.
The blue field marks $\overline R \cap T$.
The purple field marks $S \cap T$, where both the red and blue are seen to intersect with the purple field.
== Sources ==
* : $\S 5$: Complements and Powers
Category:Set Union
Category:Set Intersection
Category:Set Complement
Intersection of *-Subalgebras is *-Subalgebra

== Theorem ==
Let $\tuple {A, \ast}$ be a $\ast$-algebra over $\C$.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ be an $I$-indexed family of $\ast$-subalgebras of $A$.
Let:
:$\ds B = \bigcap_{\alpha \mathop \in I} A_\alpha$
Then $B$ is a $\ast$-subalgebra of $A$. 
== Proof ==
From Intersection of Subalgebras is Subalgebra, $B$ is a subalgebra of $A$.
It remains to show that for each $x \in B$ we have $x^\ast \in B$.
Let $x \in B$.
Then $x \in A_\alpha$ for each $\alpha \in I$. 
Since each $A_\alpha$ is a $\ast$-subalgebra, we have $x^\ast \in A_\alpha$ for each $\alpha \in I$. 
Hence we have:
:$\ds x^\ast \in \bigcap_{\alpha \mathop \in I} A_\alpha$
So $x^\ast \in B$.
Category:*-Algebras
Intersection of Abelian Subgroups is Normal Subgroup of Subgroup Generated by those Subgroups

== Theorem ==
Let $G$ be a group.
Let $L$ and $M$ be abelian subgroups of $G$.
Let $H = \gen {L, M}$ be the subgroup of $G$ generated by $L$ and $M$.
Then $L \cap M$ is a normal subgroup of $H$.
== Proof ==
From 
We have that $L$ and $M$ are abelian.
From Intersection of Subgroups is Subgroup, we have that $L \cap M$ is a subgroup of both $L$ and $M$.
From Subgroup of Abelian Group is Normal, we also have that $L \cap M$ is a normal subgroup of both $L$ and $M$.
Thus from Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup:
:$L \subseteq \map N {L \cap M}$
and:
:$M \subseteq \map N {L \cap M}$
Hence:
:$L \cup M \subseteq \map N {L \cap M}$
But from Normalizer is Subgroup, $\map N {L \cap M}$ is a subgroup of $G$.
So it follows that:
:$H \subseteq \map N {L \cap M}$
Hence the result by Subgroup is Normal Subgroup of Normalizer.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $12 \ \text{(ii)}$
Category:Normal Subgroups
Intersection of Additive Groups of Integer Multiples

== Theorem ==
Let $m, n \in \Z_{> 0}$ be (strictly) positive integers.
Let $\struct {m \Z, +}$ and $\struct {n \Z, +}$ be the corresponding additive groups of integer multiples.
Then:
:$\struct {m \Z, +} \cap \struct {n \Z, +} = \struct {\lcm \set {m, n} \Z, +}$
== Proof ==
By definition:
:$m \Z = \set {x \in \Z: m \divides x}$
Thus:
Hence the result.
== Sources ==
* : Chapter $5$: Subgroups: Exercise $5$
* : Chapter $2$: Subgroups and Cosets: $\S 36 \alpha$
Category:Additive Groups of Integer Multiples
Intersection of Algebra Ideals is Ideal

== Theorem ==
Let $K$ be a field.
Let $A$ be an algebra over $K$.
Let $\family {I_\alpha}_{\alpha \mathop \in I}$ be an $I$-indexed family of ideals of $A$.
Let:
:$\ds I = \bigcap_{\alpha \mathop \in I} I_\alpha$
Then $I$ is an ideal of $A$.
== Proof ==
From Set of Linear Subspaces is Closed under Intersection, $I$ is a vector subspace of $A$.
Now, let $a, b \in I$.
Then we have $a, b \in I_\alpha$ for all $\alpha \in I$. 
Since $I_\alpha$ is an ideal for each $\alpha \in I$, we have $a b \in I_\alpha$ and $b a \in I_\alpha$ for each $\alpha \in I$.
Hence we have $a b \in I$ and $b a \in I$. 
So $I$ is an ideal of $A$.
Category:Ideals of Algebras
Intersection of All Division Subrings is Prime Subfield

== Theorem ==
Let $\struct {K, +, \circ}$ be a division ring.
Let $P$ be the intersection of the set of all division subrings of $K$.
Then $P$ is the prime subfield of $K$.
== Proof ==
By Intersection of Division Subrings is Division Subring, the intersection $P$ of the set of all division subrings of $K$ is a division ring.
Let $\map Z K$ be the center of $K$.
From Center of Ring is Commutative Subring, $\map Z K$ is a commutative subring of $K$.
Therefore $\map Z K$ is a commutative division ring
Thus $\map Z K$ is a subfield of $K$.
But as $P$ is contained in $\map Z K$, it is itself commutative.
By its definition, $P$ contains no proper subfield and hence is a prime field.
Also, $P$ is contained in every other subfield of $K$.
Therefore $P$ is the only prime subfield of $K$.
Category:Subfields
Category:Division Subrings
Intersection of All Ring Ideals Containing Subset is Smallest

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring
Let $S \subseteq R$ be a subset of $R$.
Let $L$ be the intersection of the set of all ideals of $R$ containing $S$.
Then $L$ is the smallest ideal of $R$ containing $S$. 
== Proof ==
From Intersection of All Subrings Containing Subset is Smallest, $L$ is the smallest subring of $R$ containing $S$.
From Intersection of Ring Ideals is Ideal‎, $L$ is an ideal of $R$.
As $L$ is the smallest subring of $R$ containing $S$, and it is an ideal of $R$, there can be no smaller ideal of $R$ containing $S$ as it would then not be a subring.
So the intersection of the set of all ideals of $R$ containing $S$ is the smallest ideals of $R$ containing $S$. 
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.4$
Category:Set Intersection
Category:Ideal Theory
Intersection of All Subrings Containing Subset is Smallest

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $S \subseteq R$ be a subset of $R$.
Let $L$ be the intersection of the set of all subrings of $R$ containing $S$.
Then $L$ is the smallest subring of $R$ containing $S$. 
== Proof ==
From Intersection of Subrings is Subring, $L$ is indeed a subring of $R$.
Let $T$ be a subring of $R$ containing $S$.
Let $x, y \in L$.
By the Subring Test, we have that:
By Intersection is Largest Subset, it follows that $x, y \in T$.
But $T$ is also a subring of $R$.
So, by the Subring Test again, we have that:
So by definition of subset, $L \subseteq T$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.4$
Category:Set Intersection
Category:Subrings
Intersection of Antisymmetric Relations is Antisymmetric

== Theorem ==
The intersection of two antisymmetric relations is also an antisymmetric relation.
== Proof ==
Let $\RR_1$ and $\RR_2$ be antisymmetric relations on a set $S$.
Let $\RR_3 = \RR_1 \cap \RR_2$.
Hence we have:
 $\RR_3$ is not antisymmetric.
Then:
Hence we have:
That is, neither $\RR_1$ nor $\RR_2$ are antisymmetric.
From this contradiction it follows that $\RR_1 \cap \RR_2$ is an antisymmetric relation.
Category:Antisymmetric Relations
Category:Set Intersection
Intersection of Applications of Down Mappings at Element equals Way Below Closure of Element

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below meet-continuous lattice.
Let $\operatorname {Ids}$ be the set of all ideals in $L$.
Let for all $I \in \operatorname {Ids}$: $m_I: S \to \operatorname {Ids}$ be a mapping:
:$\forall x \in S: x \preceq \sup I \implies \map {m_I} x = \set {x \wedge i: i \in I}$
and
:$\forall x \in S: x \npreceq \sup I \implies \map {m_I} x = x^\preceq$
where $x^\preceq$ denotes the lower closure of $x$.
Let $x \in S$.
Then
:$\ds \bigcap \set {\map {m_I} x: I \in \operatorname {Ids} } = x^\ll$
where $x^\ll$ denotes the way below closure of $x$.
== Proof ==
Thus
== Sources ==
* 
* 
Category:Way Below Relation
Category:Meet-Continuous Lattices
Intersection of Auxiliary Relations is Auxiliary Relation

== Theorem ==
Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $\FF$ be a non-empty set of auxiliary relations on $S$.
Then
:$\ds \bigcap \FF$ is auxiliary relation.
== Proof ==
By definition of non-empty set:
:$\exists \RR: \RR \in \FF$
We will prove that:
:$\ds \forall x, y \in S: \tuple {x, y} \in \bigcap \FF \implies x \preceq y$
Let $x, y \in S$ such that:
:$\ds \tuple {x, y} \in \bigcap \FF$
By definition of intersection:
:$\tuple {x, y} \in \RR$
Thus by definition of auxiliary relation:
:$x \preceq y$
We will prove that:
:$\ds \forall x, y, z, u \in S: x \preceq y \land \tuple {y, z} \in \bigcap \FF \land z \preceq u \implies \tuple {x, u} \in \bigcap \FF$
Let $x, y, z, u \in S$ such that:
:$\ds x \preceq y \land \tuple {y, z} \in \bigcap \FF \land z \preceq u$
By definition of intersection:
:$\forall r \in \FF: \tuple {y, z} \in r$
By definition of auxiliary relation:
:$\forall r \in \FF: \tuple {x, u} \in r$
Thus by definition of intersection:
:$\ds \tuple {x, u} \in \bigcap \FF$
We will prove that:
:$\ds \forall x, y, z \in S: \tuple {x, z} \in \bigcap \FF \land \tuple {y, z} \in \bigcap \FF \implies \tuple {x \vee y, z} \in \bigcap \FF$
Let $x, y, z \in S$ such that:
:$\ds \tuple {x, z} \in \bigcap \FF \land \tuple {y, z} \in \bigcap \FF$
By definition of intersection:
:$\forall r \in \FF: \tuple {x, z} \in r \land \tuple {y, z} \in r$
By definition of auxiliary relation:
:$\forall r \in \FF: \tuple {x \vee y, z} \in r$
Thus by definition of intersection:
:$\tuple {x \vee y, z} \in \bigcap \FF$
By definition of auxiliary relation:
:$\forall x \in S: \forall r \in \FF: \tuple {\bot, x} \in r$
Thus by definition of intersection:
:$\ds \forall x \in S: \tuple {\bot, x} \in \bigcap \FF$
Thus by definition:
:$\ds \bigcap \FF$ is auxiliary relation.
== Sources ==
* 
* 
Category:Auxiliary Relations
Intersection of Balanced Sets in Vector Space is Balanced

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$. 
Let $\sequence {E_\alpha}_{\alpha \mathop \in I}$ be an $I$-indexed family of balanced subsets of $X$.
Then:
:$\ds E = \bigcap_{\alpha \mathop \in I} E_\alpha$ is balanced.
== Proof ==
Let $x \in E$.
Then $x \in E_\alpha$ for all $\alpha \in I$. 
Let $\lambda \in \Bbb F$ have $\cmod \lambda \le 1$.
Since $E_\alpha$ is balanced for each $\alpha \in I$, we have $\lambda x \in E_\alpha$ for each $\alpha \in I$.
So $\lambda x \in E$.
Category:Balanced Sets
Category:Set Intersection
Intersection of C*-Subalgebras is C*-Subalgebra

== Theorem ==
Let $\tuple {A, \norm {\, \cdot \,}, \ast}$ be a $\text C^\ast$-algebra.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ be an $I$-indexed family of $\text C^\ast$-subalgebras of $A$.
Let:
:$\ds B = \bigcap_{\alpha \mathop \in I} A_\alpha$
Then $B$ is a $\text C^\ast$-subalgebra of $A$. 
== Proof ==
From Intersection of *-Subalgebras is *-Subalgebra, $B$ is a $\ast$-subalgebra of $A$. 
By definition, $A_\alpha$ is a complete metric subspace of $A$ for each $\alpha \in I$.
Hence from Subspace of Complete Metric Space is Closed iff Complete, $A_\alpha$ is closed in $A$.
Hence as the intersection of closed sets, $B$ is closed in $A$.
Hence from Closed *-Subalgebra of C*-Algebra is C*-Algebra, $B$ is a $\text C^\ast$-subalgebra of $A$.
Category:C*-Algebras
Intersection of Chain of Prime Ideals of Commutative Ring is Prime Ideal

== Theorem ==
Let $R$ be a commutative ring.
Let $\Spec R$ be the spectrum of $R$, ordered by the subset relation.
Let $\set {P_\alpha}_{\alpha \mathop \in A}$ be a non-empty chain of prime ideals of $\Spec R$.
Let $\ds P = \bigcap_{\alpha \mathop \in A} P_\alpha$ be their intersection.
Then $P$ is a prime ideal of $R$.
== Proof ==
By Intersection of Ring Ideals is Ideal, $P$ is an ideal of $R$.
We show that $P$ is prime.
Let $a, b\in R$ with $a, b \notin P$.
We show that $a b \notin P$.
Because $a \notin P$, there exists $\alpha \in A$ with $a \notin P_\alpha$.
Because $b \notin P$, there exists $\beta \in A$ with $b \notin P_\beta$.
Because $\set {P_\gamma}_{\gamma \mathop \in A}$ is totally ordered, $P_\alpha \subseteq P_\beta$ or $P_\beta \subseteq P_\alpha$.
, we can assume $P_\alpha \subseteq P_\beta$, which gives us $a, b \notin P_\alpha$.
Since $P_\alpha$ is an prime ideal, $a b \notin P_\alpha$.
Thus $a b \notin P$.
== Also see ==
* Union of Chain of Prime Ideals of Commutative Ring with Unity is Prime Ideal
Category:Commutative Rings
Category:Prime Ideals of Rings
Intersection of Class Exists and is Unique

== Theorem ==
Let $V$ be a basic universe.
Let $A \subseteq V$ be a class.
Let $\bigcap A$ denote the intersection of $A$.
Then $\bigcap A$ is guaranteed to exist and is unique.
== Proof ==
By the Axiom of Specification we can create the subclass of $V$:
:$\bigcap A = \set {x \in V: \forall y \in A: x \in y}$
Hence $\bigcap A$ exists.
Suppose $\QQ \subseteq V$ such that $\QQ$ and $\bigcap A$ are both the intersection of $A$.
Then:
:$\QQ = \set {x \in V: \forall y \in A: x \in y}$
Thus:
:$x \in \QQ \implies x \in \bigcap A$
and:
:$x \in \bigcap A \implies x \in \QQ$
Hence by the Axiom of Extension:
:$\QQ = \bigcap A$
and uniqueness has been demonstrated.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom
Category:Class Intersection
Intersection of Class and Set is Set

== Theorem ==
Let $C$ be the class:
:$C = \set { u : \map \phi {u, p_1, \ldots, p_n} }$
Then for all sets $X$, $C \cap X$ is a set.
== Proof ==
By the definition of class intersection:
:$a \in C \cap X \implies a \in C \land a \in X$
Thus:
:$a \in C \cap X \implies a \in X$
The subclass definition gives:
:$C \cap X \subseteq X$
By Subclass of Set is Set, $C \cap X$ is a set.
== Sources ==
* : Chapter $1$: Separation Schema
Category:Gödel-Bernays Class Theory
Category:Class Intersection
Category:Subclasses
Intersection of Class is Subset of Intersection of Subclass

== Theorem ==
Let $V$ be a basic universe.
Let $A$ and $B$ be classes of $V$:
:$A \subseteq V, B \subseteq V$
such that it is not the case that $A = B = \O$.
Let $\bigcap A$ and $\bigcap B$ denote the intersection of $A$ and intersection of $B$ respectively.
Let $A$ be a subclass of $B$:
:$A \subseteq B$
Then $\bigcap B$ is a subset of $\bigcap A$:
:$\bigcap B \subseteq \bigcap A$
== Proof ==
First we consider the degenerate case where $A = B = \O$.
By Intersection of Empty Class we have that:
:$\bigcap \O = V$
Thus we have:
:$\bigcap B = \bigcap A = V$
and neither $\bigcap B$ nor $\bigcap A$ are in fact sets.
So while in this case $\bigcap B \subseteq \bigcap A$, $\bigcap B$ is a subclass of $\bigcap A$ and not actually a subset.
Next we cover the case where $A = \O$ is the empty class.
It is noted that Empty Class is Subclass of All Classes, and so $A \subseteq B$.
By Intersection of Empty Class we have that:
:$\bigcap A = V$
and so by definition of basic universe $\bigcap B \subseteq V$.
By Intersection of Non-Empty Class is Set, $\bigcap B$ is a set.
By definition of basic universe:
:$\bigcap B \in V$
By the , $V$ is transitive.
Hence as $\bigcap B \in V$ it follows that:
:$\bigcap B \subseteq V$
and the result holds.
By Intersection of Non-Empty Class is Set, both $\bigcap A$ and $\bigcap B$ are sets.
Let $x \in \bigcap B$.
Then:
:$\forall y \in B: x \in y$
That is:
:$y \in B \implies x \in y$
But as $A \subseteq B$, we have that:
:$y \in A \implies y \in B$
and therefore:
:$y \in A \implies x \in y$
That is:
:$\forall y \in A: x \in y$
and so by definition of intersection:
:$x \in \bigcap A$
Hence the result by definition of subset.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.2. \ \text {(b)}$
Category:Class Intersection
Intersection of Closed Intervals of Positive Reals is Zero

== Theorem ==
Let $\R_{> 0}$ be the set of strictly positive real numbers.
For all $x \in \R_{> 0}$, let $B_x$ be the closed real interval $\closedint 0 x$.
Then:
:$\ds \bigcap_{x \mathop \in \R_{> 0} } B_x = \set 0$
== Proof ==
Let $\ds B = \bigcap_{x \mathop \in \R_{> 0} } B_x$.
We have that:
:$\forall x \in \R_{> 0}: 0 \in \closedint 0 x$
So by definition of intersection:
:$0 \in B$
and so by Singleton of Element is Subset:
:$\ds \set 0 \subseteq \bigcap_{x \mathop \in \R_{> 0} } B_x$
 $\exists y \in \R_{> 0}: y \in B$.
By definition of $B_x$:
:$y \notin \closedint 0 {\dfrac y 2} = B_{y/2}$
and so by definition of intersection of family:
:$y \notin B$
From this contradiction it follows that there can be no elements in $B$ apart from $0$.
That is:
:$\ds \bigcap_{x \mathop \in \R_{> 0} } A_x \subseteq \set 0$
By definition of set equality:
:$\ds \bigcap_{x \mathop \in \R_{> 0} } B_x = \set 0$
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $5$
Category:Real Intervals
Intersection of Closed Set with Compact Subspace is Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be closed in $T$.
Let $K \subseteq S$ be compact in $T$.
Then $H \cap K$ is compact in $T$.
== Proof 1 ==
== Proof 2 ==
Category:Compact Spaces
Category:Closed Sets
Category:Intersection of Closed Set with Compact Subspace is Compact
Intersection of Closed Set with Compact Subspace is Compact/Proof 1

== Theorem ==
== Proof ==
Let $\tau_K$ be the subspace topology on $K$.
Let $T_K = \left({K, \tau_K}\right)$ be the topological subspace determined by $K$.
By Closed Set in Topological Subspace, $H \cap K$ is closed in $T_K$.
By Closed Subspace of Compact Space is Compact, $H \cap K$ is compact in $T_K$.
By Compact in Subspace is Compact in Topological Space, $H \cap K$ is compact in $T$.
Category:Intersection of Closed Set with Compact Subspace is Compact
Intersection of Closed Set with Compact Subspace is Compact/Proof 2

== Theorem ==
== Proof ==
Let $\family {U_\alpha}$ be an open cover of $H \cap K$:
:$\ds H \cap K \subseteq \bigcup_\alpha U_\alpha$
Then:
:$\ds K \subseteq \bigcup_\alpha U_\alpha \cup \paren {S \setminus H}$
Since $H$ is closed in $T$, $\paren {S \setminus H}$ is open in $T$.
Hence $\family {U_\alpha} \cup S \setminus H$ is an open cover of $K$.
We have that $K$ is compact in $T$.
It follows by definition that a finite subcover:
:$\set {U_{\alpha_1}, U_{\alpha_2}, \ldots, U_{\alpha_n}, S \setminus H}$
of $K$ exists.
Thus:
:$H \cap K \subseteq \set {U_{\alpha_1}, U_{\alpha_2}, \ldots, U_{\alpha_n} }$
and $H \cap K$ is compact in $T$.
Category:Intersection of Closed Set with Compact Subspace is Compact