nyuuzyou/wikis huggingface dataset:
Intersection of Closed Sets is Closed/Closure Operator

== Theorem ==
Let $S$ be a set.
Let $f: \powerset S \to \powerset S$ be a closure operator on $S$.
Let $\CC$ be the set of all subsets of $S$ that are closed with respect to $f$.
Let $\AA \subseteq \CC$.
Then $\bigcap \AA \in \CC$.
== Proof ==
Let $Q = \bigcap \AA$.
By the definition of closure operator, $f$ is inflationary, order-preserving, and idempotent.
Let $A \in \AA$.
By Intersection is Largest Subset, $Q \subseteq A$.
Since $f$ is order-preserving, $\map f Q \subseteq \map f A$.
By the definition of closed set, $\map f A = A$
Thus $\map f Q \subseteq A$.
This holds for all $A \in \AA$.
Thus by Intersection is Largest Subset:
:$\map f Q \subseteq \bigcap \AA$
Since $\bigcap \AA = Q$:
:$\map f Q \subseteq Q$
Since $f$ is inflationary:
:$Q \subseteq \map f Q$
Thus by definition of set equality:
:$Q = \map f Q$
Therefore $Q$ is closed with respect to $f$.
Category:Closure Operators
Category:Set Intersection
Intersection of Closed Sets is Closed/Normed Vector Space

== Theorem ==
Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Then the intersection of an arbitrary number of closed sets of $M$ (either finitely or infinitely many) is itself closed.
== Proof ==
Let $I$ be an indexing set (either finite or infinite).
Let $\ds \bigcap_{i \mathop \in I} V_i$ be the intersection of a indexed family of closed sets of $M$ indexed by $I$.
By definition of closed set, each of $X \setminus V_i$ are by definition open in $M$.
From De Morgan's laws: Difference with Intersection:
:$\ds X \setminus \bigcap_{i \mathop \in I} V_i = \bigcup_{i \mathop \in I} \paren {X \setminus V_i}$
We have that $\ds \bigcup_{i \mathop \in I} \paren {X \setminus V_i}$ is the union of a indexed family of open sets of $M$ indexed by $I$.
By Union of Open Sets of Normed Vector Space is Open, $\ds \bigcup_{i \mathop \in I} \paren {X \setminus V_i} = X \setminus \bigcap_{i \mathop \in I} V_i$ is likewise open in $M$.
Then by definition of closed set, $\ds \bigcap_{i \mathop \in I} V_i$ is closed in $M$.
== Sources ==
* : $\S 1.3$: Normed and Banach spaces. Topology of normed spaces
Category:Intersection of Closed Sets is Closed
Category:Closed Sets (Normed Vector Spaces)
Intersection of Closed Sets is Closed/Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then the intersection of an arbitrary number of closed sets of $T$ (either finitely or infinitely many) is itself closed.
== Proof ==
Let $I$ be an indexing set (either finite or infinite).
Let $\ds \bigcap_{i \mathop \in I} V_i$ be the intersection of a indexed family of closed sets of $T$ indexed by $I$.
Then from De Morgan's laws: Difference with Intersection:
:$\ds S \setminus \bigcap_{i \mathop \in I} V_i = \bigcup_{i \mathop \in I} \paren {S \setminus V_i}$
By definition of closed set, each of $S \setminus V_i$ are by definition open in $T$.
We have that $\ds \bigcup_{i \mathop \in I} \paren {S \setminus V_i}$ is the union of a indexed family of open sets of $T$ indexed by $I$.
Therefore, by definition of a topology:
:$\ds \bigcup_{i \mathop \in I} \paren {S \setminus V_i} = S \setminus \bigcap_{i \mathop \in I} V_i$ 
is likewise open in $T$.
Then by definition of closed set, $\ds \bigcap_{i \mathop \in I} V_i$ is closed in $T$.
== Sources ==
* : $2.24 b$
* : Chapter $\text {I}$: Topological Spaces: $1$. Open Sets and Closed Sets: Theorem $1$
* : Chapter $3$: Topological Spaces: $\S 2$: Topological Spaces: Exercise $4$
Category:Intersection of Closed Sets is Closed
Intersection of Closures of Rationals and Irrationals is Reals

== Theorem ==
Let $\struct {\R, \tau}$ be the real number line with the usual (Euclidean) topology.
Let $\Q$ be the set of rational numbers.
Then:
:$\Q^- \cap \paren {\R \setminus \Q}^- = \R$
where:
:$\R \setminus \Q$ denotes the set of irrational numbers
:$\Q^-$ denotes the closure of $\Q$.
== Proof ==
From Closure of Rational Numbers is Real Numbers:
:$\Q^- = \R$
From Closure of Irrational Numbers is Real Numbers:
:$\paren {\R \setminus \Q}^- = \R$
The result follows from Set Intersection is Idempotent.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30 \text { - } 31$. The Rational and Irrational Numbers: $3$
Category:Rational Numbers
Category:Irrational Numbers
Category:Real Number Line with Euclidean Topology
Intersection of Compact and Closed Subsets of Normed Finite-Dimensional Real Vector Space with Euclidean Norm is Compact

== Theorem ==
Let $\struct {\R^d, \norm {\, \cdot \,}_2}$ be the normed finite-dimensional real vector space with Euclidean norm.
Let $K$ be a compact subset of $\struct {\R^d, \norm {\, \cdot \,}_2}$.
Let $F$ be a closed subset of $\struct {\R^d, \norm {\, \cdot \,}_2}$.
Then $F \cap K$ is compact in $\struct {\R^d, \norm {\, \cdot \,}_2}$.
== Proof ==
By assumption, $K$ is compact.
We have that a compact subset of normed vector space is closed and bounded.
Hence, $K$ is closed and bounded.
Since $K$ is bounded:
:$\exists C \in \R_{> 0} : \forall \mathbf x \in K : \norm {\mathbf x}_2 \le C$.
Then:
:$\forall \mathbf x \in K \cap F : \norm {\mathbf x}_2 \le C$
Hence, $K \cap F$ is bounded.
By assumption, $K$ and $F$ are closed.
We have that intersection of closed sets is closed in normed vector space.
Therefore, $K \cap F$ is closed.
By Heine-Borel theorem, $K \cap F$ is compact.
== Sources ==
* : Chapter $\S 1.5$: Normed and Banach spaces. Compact sets
Category:Closed Sets
Category:Compact Spaces
Intersection of Complete Meet Subsemilattices induces Closure Operator

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $I$ be an indexing set.
Let $\family {f_i}_{i \mathop \in I}$ be an indexed family of closure operators on $S$.
Let $C_i = \map {f_i} S$ be the set of closed elements with respect to $f_i$ for each $i \in I$.
Suppose that for each $i \in I$, $C_i$ is a complete meet subsemilattice of $S$ in the following sense:
:For each $D \subseteq C_i$, $D$ has an infimum in $S$ such that $\inf D \in C_i$.
Then $C = \ds \bigcap_{i \mathop \in I} C_i$ induces a closure operator on $S$.
== Proof ==
=== Lemma ===
By the lemma, $C$ is a complete meet semilattice.
Let $x \in S$.
Then $C \cap x^\succcurlyeq$ has an infimum in $S$ which lies in $C$, where $x^\succcurlyeq$ is the upper closure of $x$.
By the definition of infimum:
:$x \preceq \inf \struct {C \cap x^\succcurlyeq}$
so this infimum is in fact the smallest element of $C \cap x^\succcurlyeq$.
Thus $C$ induces a closure operator on $S$ by Closure Operator from Closed Elements.
Category:Closure Operators
Category:Intersection of Complete Meet Subsemilattices induces Closure Operator
Intersection of Complete Meet Subsemilattices induces Closure Operator/Lemma

== Lemma for Intersection of Complete Meet Subsemilattices induces Closure Operator ==
Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $\family {f_i}_{i \mathop \in I}$ be an indexed family of closure operators on $S$.
Let $C_i = \map {f_i} S$ be the set of closed elements with respect to $f_i$ for each $i \in I$.
Let $\family {C_i}_{i \mathop \in I}$ be an indexed family of complete meet subsemilattices of $S$.
Then $C = \ds \bigcap_{i \mathop \in I} C_i$ is also a complete meet subsemilattice.
=== Proof ===
Let $D \subseteq C$.
By Intersection is Largest Subset, $D \subseteq C_i$ for each $i \in I$.
Thus $D$ has an infimum in $S$ and $\inf D \in C_i$ for each $i \in I$.
By the definition of intersection, $\inf D \in C$.
Category:Intersection of Complete Meet Subsemilattices induces Closure Operator
Intersection of Congruence Classes

== Theorem ==
Let $\RR_m$ denote congruence modulo $m$ on the set of integers $\Z$.
Then:
:$\RR_m \cap \RR_n = \RR_{\lcm \set {m, n} }$
where $\lcm \set {m, n}$ is the lowest common multiple of $m$ and $n$.
In the language of modulo arithmetic, this is equivalent to:
:$a \equiv b \pmod m, a \equiv b \pmod n \implies a \equiv b \pmod {\lcm \set {m, n} }$
=== Corollary ===
== Proof ==
Let $\tuple {a, b} \in \RR_m \cap \RR_n$.
That is, let $\tuple {a, b} \in \RR_m$ and $\tuple {a, b} \in \RR_n$.
That means, by definition of congruence:
:$a \equiv b \pmod m$
:$a \equiv b \pmod n$
Thus by definition of congruence:
:$\exists r, s \in \Z: a - b = r m, a - b = s n$
Let $d = \gcd \set {m, n}$ so that $m = d m', n = d n', m' \perp n'$.
Substituting for $m$ and $n$:
:$r d m' = s d n'$ and so $r m' = s n'$.
So $n' \divides r m'$ and $m' \perp n'$ so by Euclid's Lemma $n' \divides r$.
So we can put $r = k n'$ and get:
:$a - b = r m = k m n' = k m \dfrac n d = k \dfrac {m n} d$
But:
:$\dfrac {m n} d = \dfrac {m n} {\gcd \set {m, n} }$
So by Product of GCD and LCM:
:$a - b = k \lcm \set {m, n}$
So:
:$a \equiv b \pmod {\lcm \set {m, n} }$
and hence the result.
== Examples ==
== Sources ==
* : Chapter $1$: Equivalence Relations: $\S 18 \beta$
* : Chapter $1$: Properties of the Natural Numbers: $\S 23 \delta$
Category:Modulo Arithmetic
Category:Set Intersection
Category:Lowest Common Multiple
Category:Intersection of Congruence Classes
Intersection of Congruence Classes/Corollary

== Corollary to Intersection of Congruence Classes ==
Let $\RR_m$ denote congruence modulo $m$ on the set of integers $\Z$.
If $m \perp n$ then $\RR_m \cap \RR_n = \RR_{m n}$.
== Proof ==
By Intersection of Congruence Classes:
:$\RR_m \cap \RR_n = \RR_{\lcm \set {m, n} }$
$m \perp n$ means $\gcd \set {m, n} = 1$.
From Product of GCD and LCM it follows that $\lcm \set {m, n} = m n$.
Hence the result.
Category:Intersection of Congruence Classes
Intersection of Congruence Classes/Examples/Intersection of 1 mod 3 with -1 mod 4

== Examples of Use of Intersection of Congruence Classes ==
:$\eqclass 1 3 \cap \eqclass {-1} 4 = \eqclass 7 {12}$
== Proof ==
From Intersection of Congruence Classes:
:$\eqclass 1 3 \cap \eqclass {-1} 4 = \eqclass x {12}$
for some $x$ which is to be found.
We have that:
and that:
Hence the result by taking the intersection.
== Sources ==
* : Chapter $2$. Equivalence Relations: Exercise $8$
Category:Intersection of Congruence Classes
Intersection of Convex Sets is Convex Set (Order Theory)

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $\CC$ be a set of convex sets in $S$.
Then $\ds \bigcap \CC$ is convex.
== Proof ==
Let $a, b, c \in S$.
Let $a, c \in \ds \bigcap \CC$.
Let $a \prec b \prec c$.
By the definition of intersection:
:$\forall T \in \CC$: $a, c \in T$
Since each $T \in \CC$ is convex:
:$\forall T \in \CC$: $b \in T$
Thus by the definition of intersection:
:$b \in \ds \bigcap \CC$
Thus $\ds \bigcap \CC$ is convex.
Category:Convex Sets (Order Theory)
Intersection of Convex Sets is Convex Set (Vector Spaces)

== Theorem ==
Let $V$ be a vector space over $\R$ or $\C$.
Let $\CC$ be a family of convex subsets of $V$.
Then the intersection $\ds \bigcap \CC$ is also a convex subset of $V$.
== Proof ==
Let $x, y \in \ds \bigcap \CC$.
Then by definition of set intersection, $\forall C \in \CC: x, y \in C$.
The convexity of each $C$ yields:
:$\forall t \in \closedint 0 1: t x + \paren {1 - t} y \in C$
Therefore, these elements are also in $\ds \bigcap \CC$, by definition of set intersection.
Hence $\ds \bigcap \CC$ is also convex.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality
Category:Vector Spaces
Category:Convex Sets (Vector Spaces)
Intersection of Coprime Cyclic Subgroups is Trivial

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $x, y \in G$ such that:
:$\order x \perp \order y$
where:
:$\order x, \order y$ denotes the orders of $x$ and $y$ in $G$ respectively
:$\perp$ denotes the coprimality relation.
Then:
:$\gen x \cap \gen y = \set e$
where $\gen x, \gen y$ denotes the subgroup of $G$ generated by $x$ and $y$ in $G$ respectively.
== Proof ==
From Order of Cyclic Group equals Order of Generator:
:$\order x = \order {\gen x}$
and:
:$\order y = \order {\gen y}$
where $\order {\gen x}, \order {\gen y}$ denote the orders of $\gen x$ and $\gen y$ respectively.
From Intersection of Subgroups is Subgroup:
:$\gen x \cap \gen y$ is a subgroup of both $\gen x$ and $\gen y$.
From Lagrange's Theorem:
:$\order {\gen x \cap \gen y} \divides \order {\gen x}$
and:
:$\order {\gen x \cap \gen y} \divides \order {\gen y}$
But as $\order {\gen x} \perp \order {\gen y}$:
:$\order {\gen x \cap \gen y} = 1$
Hence the result by definition of trivial subgroup.
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $17$
Category:Cyclic Groups
Intersection of Division Subrings Containing Subset is Smallest

== Theorem ==
Let $\struct {D, +, \circ}$ be a division ring.
Let $S \subseteq D$ be a subset of $D$.
Let $L$ be the intersection of the set of all division subrings of $D$ containing $S$.
Then $L$ is the smallest division subring of $D$ containing $S$. 
== Proof ==
From Intersection of Division Subrings is Division Subring, $L$ is indeed a division subring of $D$.
Let $T$ be a division subring of $D$ containing $S$.
Let $x, y \in L$.
By the Division Subring Test, we have that:
By Intersection is Largest Subset, it follows that $x, y \in T$.
But $T$ is also a division subring of $D$.
So, by the Division Subring Test again, we have that:
So by definition of subset, $L \subseteq T$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.1$
Category:Division Subrings
Intersection of Division Subrings is Division Subring

== Theorem ==
Let $\struct {D, +, \circ}$ be a division ring.
Let $\mathbb K$ be a non-empty set of division subrings of $D$.
Then the intersection $\ds \bigcap \mathbb K$ of the members of $\mathbb K$ is itself a division subring of $D$.
== Proof ==
Let $\ds L = \bigcap \mathbb K$.
Let $0$ be the zero of $\struct {D, +, \circ}$.
By Intersection of Subgroups is Subgroup: General Result, $\struct {L, +}$ is a subgroup of $\struct {D, +}$.
By the One-Step Subgroup Test:
:$\forall x, y \in L: x + \paren {-y} \in L$
By Non-Zero Elements of Division Ring form Group:
:$\struct {D \setminus \set 0, \circ}$ is a group
:$\struct {K \setminus \set 0, \circ}$ is a group for each $K \in \mathbb K$
By Set Difference over Subset, $\struct {K \setminus \set 0, \circ}$ is a subgroup of $\struct {D \setminus \set 0, \circ}$ for each $K \in \mathbb K$.
By Set Difference is Right Distributive over Set Intersection:
:$\ds L \setminus \set 0 = \bigcap_{K \mathop \in \mathbb K} \paren {K \setminus \set 0}$
By Intersection of Subgroups is Subgroup: General Result, $\struct {L \setminus \set 0, \circ}$ is a subgroup of $\struct {D \setminus \set 0, \circ}$.
By the Two-Step Subgroup Test and Ring Product with Zero:
By the Division Subring Test it follows that $\struct {L, +, \circ}$ is a division subring of $\struct {D, +, \circ}$.
Category:Division Subrings
Intersection of Division Subrings is Largest Division Subring Contained in all Division Subrings

== Theorem ==
Let $\struct {D, +, \circ}$ be a division ring.
Let $\mathbb K$ be a non-empty set of division subrings of $D$.
Let $\ds \bigcap \mathbb K$ be the intersection of the elements of $\mathbb K$.
Then $\ds \bigcap \mathbb K$ is the largest division subring of $D$ contained in each element of $\mathbb K$.
== Proof ==
Let $\ds L = \bigcap \mathbb K$.
Let $0$ be the zero of $\struct {D, +, \circ}$.
From Intersection of Division Subrings is Division Subring, $\struct {L, +, \circ}$ is a division subring of $\struct {D, +, \circ}$.
By Intersection of Subgroups is Subgroup, $\struct {L, +}$ is the largest subgroup of $\struct {D, +}$ contained in each element of $\mathbb K$.
By Intersection of Subgroups is Subgroup, $\struct {L \setminus \set 0, \circ}$ is the largest subgroup of $\struct {D \setminus \set 0, \circ}$ contained in each element of $\mathbb K$.
Let $S$ be a subring of $D$ such that:
:$\forall K \in \mathbb K: S \subseteq K$
By Intersection is Largest Subset, $S \subseteq L$.
Thus $L$ is the largest division subring of $D$ contained in each element of $\mathbb L$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.1$
Category:Division Subrings
Intersection of Doubleton

== Theorem ==
Let $x$ and $y$ be sets.
Let $\set {x, y}$ be a doubleton.
Then $\bigcap \set {x, y}$ is a set such that:
:$\bigcap \set {x, y} = x \cap y$
== Proof ==
From Intersection of Non-Empty Class is Set it follows that $x \cap y$ is a set.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.4. \ \text {(b)}$
Category:Class Intersection
Category:Doubleton Classes
Intersection of Elements of Power Set

== Theorem ==
Let $S$ be a set.
Let:
:$\ds \mathbb S = \bigcap_{X \mathop \in \powerset S} X$
where $\powerset S$ is the power set of $S$.
Then $\mathbb S = \O$.
== Proof ==
By Intersection is Subset:
:$\ds \forall X \in \powerset S: \bigcap_{X \mathop \in \powerset S} X \subseteq X$
From Empty Set is Element of Power Set:
:$\O \in \powerset S$
So:
:$\ds \bigcap_{X \mathop \in \powerset S} X \subseteq \O$
From Empty Set is Subset of All Sets:
:$\ds \O \subseteq \bigcap_{X \mathop \in \powerset S} X$
So by definition of set equality:
:$\ds \bigcap_{X \mathop \in \powerset S} X = \O$
== Sources ==
* : $\S 5$: Complements and Powers
Category:Power Set
Category:Set Intersection
Category:Empty Set
Intersection of Empty Set

== Theorem ==
Consider the set of sets $\mathbb S$ such that $\mathbb S$ is the empty set $\O$.
Then the intersection of $\mathbb S$ is $\mathbb U$:
:$\mathbb S = \O \implies \bigcap \mathbb S = \mathbb U$
where $\mathbb U$ is the universal set.
A paradoxical result.
=== Intersection of Empty Class ===
In the context of class theory the result is in the same form:
== Proof ==
Let $\mathbb S = \O$.
Then from the definition:
:$ \bigcap \mathbb S = \set {x: \forall X \in \mathbb S: x \in X}$
Consider any $x \in \mathbb U$.
Then as $\mathbb S = \O$, it follows that:
:$\forall X \in \mathbb S: x \in X$
from the definition of vacuous truth.
It follows directly that:
:$\bigcap \mathbb S = \set {x: x \in \mathbb U}$
That is:
:$\bigcap \mathbb S = \mathbb U$
== Also presented as ==
Using the terminology of indexed families, this can also be written as:
:$\ds \bigcap_{i \mathop \in \O} S_i = \set {x: \forall i \in \O: x \in S_i} = \mathbb U$
where it is important to specify the nature of $\mathbb U$.
== Paradoxical Implications ==
== Also see ==
* Union of Empty Set
== Sources ==
* : $\S 5$: Complements and Powers
* : $1$. Preliminary ideas; sets, vector spaces: $1.1$. Sets
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.7$
* : Chapter $\text{I}$: Sets and Functions: Unions and Intersections
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets
* : $\S 1$: Naive Set Theory: $\S 1.4$: Sets of Sets: Exercise $1.4.2: \ \text{(ii)}$
* : $1$. Notation, Conventions: $12$
* : $1$: Set Theory and Logic: $\S 1$: Fundamental Concepts
* : Appendix $\text{A}$: Set Theory: Families
Category:Set Intersection
Category:Veridical Paradoxes
Category:Intersection of Empty Set
Intersection of Empty Set/Class Theory

== Theorem ==
Let $V$ be a basic universe.
Let $\O$ denote the empty class.
Then the intersection of $\O$ is $V$:
:$\bigcap \O = V$
== Proof ==
By definition of $V$, we have:
:$\bigcap \O \subseteq V$
By definition of empty class, there exists no set $x \in V$ which is an element of $\O$.
Hence it is vacuously true that every element of $V$ is an element of every element of $\O$.
Therefore every $x \in V$ is an element of $\bigcap \O$.
Therefore:
:$V \subseteq \bigcap \O$
It follows by definition of class equality that:
:$\bigcap \O = V$
== Paradoxical Implications ==
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Theorem $5.1 \ (2)$
Category:Class Intersection
Category:Intersection of Empty Set
Intersection of Empty Set/Paradoxical Implications

== Theorem ==
Consider the Intersection of Empty Set:
:$\mathbb S = \O \implies \bigcap \mathbb S = \mathbb U$
Although it appears counter-intuitive, the reasoning is sound.
This result is therefore classed as a veridical paradox.
 declares, in Section $5$ of his of $1960$ that:
:There is no profound problem here; it is merely a nuisance to be forced always to be making qualifications and exceptions just because some set somewhere along some construction might turn out to be empty. There is nothing to be done about this; it is just a fact of life.
However, later in that same work (in Section $9$, in the context of indexed families of sets) he says that:
:... an empty intersection does not make sense.
a sentiment which is repeated in the $2008$ collaboration with , , p. $457$.
In , the author recognizes the result, but does not adopt it:
:If one has a given large set $X$ that is specified at the outset of the discussion to be one's "universe of discourse," and one considers only subsets of $X$ throughout, it is reasonable to let $\ds \bigcap_{A \mathop \in \AA} A = X$ when $\AA$ is empty. Not all mathematicians follow this convention, however. To avoid difficulty, we shall not define the intersection when $\AA$ is empty.
 accepts this result, but cautiously:
:A natural mnemonic for these extreme cases is that $\bigcap \SS$ "grows larger" as $\SS$ "grows smaller", and $\bigcup \SS$ grows smaller as $\SS$ grows smaller. No other convention is possible, but the case $\SS = \O$ will often be treated redundantly by itself in definitions and proofs, as a reminder of the null case.
 dismiss it casually:
:As usual, we adopt the convention that in case $A = \O$ the expression $\bigcap A$ is defined only in case we work with the subsets of an underlying set $X$. In this case we put $\bigcap A = X$.
and do not even acknowledge that there may be a problem in the first place:
:In the case where $I = \O$, we have
::$\ds \bigcap_{i \mathop \in I} A_i = X$;''
:this moreover is the only case where $X$ plays a role; in fact, if $I$ is not empty, clearly we have:
::$\ds \bigcap_{i \mathop \in I} A_i = \set {x \mid x \in A_i \text { for every } i \in I}$.
== Sources ==
* : $\S 5$: Complements and Powers
* : $\S 9$: Families
* : $1$. Preliminary ideas; sets, vector spaces: $1.1$. Sets
* : Chapter $\text{I}$: Sets and Functions: Unions and Intersections
* : $1$. Notation, Conventions: $12$
* : $1$: Set Theory and Logic: $\S 1$: Fundamental Concepts
* : Appendix $\text{A}$: Set Theory: Families
Category:Intersection of Empty Set
Intersection of Equivalences

== Theorem ==
The intersection of two equivalence relations is itself an equivalence relation.
== Proof ==
Let $\RR_1$ and $\RR_2$ be equivalence relations on $S$.
Let $\RR_3 = \RR_1 \cap \RR_2$.
Checking in turn each of the criteria for equivalence:
=== Reflexive ===
Equivalence relations are by definition reflexive.
So, by Intersection of Reflexive Relations is Reflexive, so is $\RR_3$.
=== Symmetric ===
Equivalence relations are by definition symmetric.
So, by Intersection of Symmetric Relations is Symmetric, so is $\RR_3$.
=== Transitive ===
Equivalence relations are by definition transitive.
So, by Intersection of Transitive Relations is Transitive, so is $\RR_3$.
Thus $\RR_3$ is an equivalence.
== Sources ==
* : Chapter $1$: Equivalence Relations: $\S 17 \beta$
Category:Equivalence Relations
Category:Set Intersection
Intersection of Events/Examples/Both Prime and Even

== Examples of Intersections of Events ==
Consider the experiment $\EE$ such that $2$ (positive) integers are drawn at random from a table of random numbers.
Let $A$ be the event that at least $1$ of these integers is prime.
Let $B$ be the event that at least $1$ of these integers is even.
Then their intersections $A \cap B$ means:
:at least one of the $2$ integers is even and at least one of the $2$ integers is prime.
That is, either:
:one of the $2$ integers is $2$ (two)
or:
:one of the $2$ integers is even and the other is prime.
== Sources ==
* : $\text I$: Random Events: $1$. Relations among Random Events: Example $1.2$
Category:Examples of Intersections of Events