nyuuzyou/wikis huggingface dataset:
Set Difference with Union/Venn Diagram

== Theorem ==
== Proof ==
Demonstration by Venn diagram:
: 
Consider the diagram on the .
The red area forms $R \setminus \paren {S \cup T}$.
Consider the diagram in the middle.
The red and orange areas together form $R \setminus S$.
The red area alone forms $\paren {R \setminus S} \setminus T$.
Consider the diagram on the .
The red and orange areas together form $R \setminus T$.
The red area alone forms $\paren {R \setminus T} \setminus S$.
It is seen that the red areas are the same on all diagrams.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.4 \ \text{(c)}$
Category:Set Difference with Union
Category:Examples of Venn Diagrams
Set Difference with Union is Set Difference

== Theorem ==
The set difference between two sets is the same as the set difference between their union and the second of the two sets:
Let $S, T$ be sets.
Then:
:$\paren {S \cup T} \setminus T = S \setminus T$
== Proof ==
Consider $S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universal set.
Category:Set Difference
Category:Set Union
Set Equality is Equivalence Relation

== Theorem ==
Let $S$ be a set.
Set equality is an equivalence relation on the power set $\powerset S$ of $S$.
== Proof ==
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
Let $A \in \powerset S$.
From Set Equals Itself:
:$A = A$
So set equality has been shown to be reflexive on $\powerset S$.
=== Symmetry ===
Let $A, B \in \powerset S$.
Let $A = B$.
Then by definition of set equality:
:$A \subseteq B$
:$B \subseteq A$
from which it follows by definition of set equality that $B = A$.
So set equality has been shown to be symmetric on $\powerset S$.
=== Transitivity ===
Let $A, B, C \in \powerset S$.
Let $A = B$ and $B = C$.
Then by definition of set equality:
:$(1): \quad A \subseteq B$
:$(2): \quad B \subseteq C$
:$(3): \quad C \subseteq B$
:$(4): \quad B \subseteq A$
From $(1)$ and $(2)$ and Subset Relation is Transitive:
:$A \subseteq C$
From $(3)$ and $(4)$ and Subset Relation is Transitive:
:$C \subseteq A$
from which it follows by definition of set equality that $A = C$.
So set equality has been shown to be transitive on $\powerset S$.
Set equality has been shown to be reflexive, symmetric and transitive on $\powerset S$.
Hence by definition it is an equivalence relation on $\powerset S$.
== Sources ==
* : Exercise $1.2: 4$
Category:Set Theory
Category:Examples of Equivalence Relations
Set Equation: Intersection

== Theorem ==
Let $A$ and $B$ be sets.
Consider the set equation:
:$A \cap X = B$
The solution set of this is:
:$\O$ if $B \nsubseteq A$
:$\set {B \cup Y: A \nsubseteq Y}$ otherwise.
== Proof ==
Category:Set Theory
Category:Set Intersection
Set Equivalence Less One Element

== Theorem ==
Let $S$ and $T$ be sets which are equivalent:
:$S \sim T$
Let $a \in S$ and $b \in T$.
Then:
:$S \setminus \set a \sim T \setminus \set b$
where $\setminus$ denotes set difference.
== Proof ==
As $S \sim T$, there exists a bijection $f: S \to T$.
Let $g: \paren {S \setminus \set a} \to \paren {T \setminus \set b}$ be the mapping defined as follows:
$\forall x \in S \setminus \set a: \map g x =
\begin{cases}
\map f x: \map f x \ne b \\
\map f a: \map f x = b
\end{cases}$
First it is shown that $g$ is an injection.
Let $x, y \in S \setminus \set a$ and $x \ne y$.
The following cases apply:
$(1): \quad \map f x \ne b$ and $\map f y \ne b$
By definition of $g$:
:$\map f x = \map g x$
and:
:$\map f y = \map g y$
It follows from the injectivity of $f$ that:
:$\map g x \ne \map g y$
$(2): \quad \map f x = b$
By definition of $g$:
:$\map g x = \map f a$
We have that:
:$y \in S \setminus \set a$
Therefore by definition of set difference:
:$y \ne a$
So by the injectivity of $f$:
:$\map f y \ne \map f a = \map g x$
Also by the injectivity of $f$:
:$\map f y \ne b$
Thus, by the definition of $g$:
:$\map g y = \map f y \ne \map f a = \map g x$
$(2): \quad \map f y = b$
The proof is the same as that of case $2$, with the roles of $x$ and $y$ reversed.
Next it is shown that $g$ is a surjection.
Let $y \in T \setminus \set b$.
By definition of surjectivity of $f$:
:$\exists x \in S: \map f x = y$
It is to be shown that:
:$\exists x' \in S, x \ne a: \map f {x'} = y$
$(1): \quad \map f a = b$
By hypothesis:
:$\map f x = y \ne b$
Then by the rule of transposition:
:$x \ne a$
so let:
:$x' = x$
$(2): \quad \map f a \ne b$
By the rule of transposition:
:$\map {f^{-1} } b \ne a$
where $\map {f^{-1} } b$ denotes the preimage of $b$ under $f$.
:Case $(2.1): \quad \map f a = y$
By the definition of $g$:
:$\map g {\map {f^{-1} } b} = y$
so let:
:$x' = \map {f^{-1} } b$
:Case $(2.2): \quad \map f a \ne y$
By hypothesis:
:$\map f x = y \ne \map f a$
Then by the rule of transposition:
:$x \ne a$
so let:
:$x' = x$
Thus:
:$\forall y \in T \setminus \set b: \exists x \in S \setminus \set a: \map g x = y$
and so $g$ is a surjection.
Thus $g$ is both an injection and a surjection, and is by definition a bijection.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 17$: Finite Sets: Theorem $17.2$
* : $1$: Set Theory and Logic: $\S 6$: Finite Sets: Lemma $6.1$
Category:Equivalence Relations
Set Equivalence behaves like Equivalence Relation

== Theorem ==
Set equivalence behaves like an equivalence relation.
That is:
where $S, T, S_1, S_2, S_3$ are sets.
== Proof ==
For two sets to be equivalent, there needs to exist a bijection between them.
In the following, let $\phi$, $\phi_1$ and $\phi_2$ be understood to be bijections.
=== Reflexive ===
Thus $\sim$ is seen to behave like a reflexive relation.
=== Symmetric ===
Thus $\sim$ is seen to behave like a symmetric relation.
=== Transitive ===
Thus $\sim$ is seen to behave like a transitive relation.
== Warning ==
== Also see ==
The definition of a cardinal of a set as the equivalence class of that set under set equivalence.
== Sources ==
* : $\S 13$: Arithmetic
* : Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability
* : $\S 3.7$. Similar sets
* : Chapter $\text {III}$: The Natural Numbers: $\S 17$: Finite Sets: Theorem $17.1$
* : $\S 2.3$: Equivalence of sets (footnote $6$)
* : Appendix $\text{A}.4$: Functions: Problem Set $\text{A}.4$: $26$
* : $2$. Definition of Equivalence. The Concept of Cardinality. The Axiom of Choice: Theorem $2.1$
* : $\S 10.2$
Category:Examples of Equivalence Relations
Category:Set Equivalence
Category:Set Equivalence behaves like Equivalence Relation
Set Equivalence behaves like Equivalence Relation/Symmetric

== Theorem ==
Set equivalence behaves like a symmetric relation:
:$S \sim T \implies T \sim S$
== Proof ==
== Sources ==
* : Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability
* : $\S 3.7$. Similar sets
* : Chapter $\text {III}$: The Natural Numbers: $\S 17$: Finite Sets: Theorem $17.1$
* : Part $1$: Not Entirely Naive Set Theory: Chapter $3$: Cardinality: Exercise $1 \ \text{(b)}$
* : $2$. Definition of Equivalence. The Concept of Cardinality. The Axiom of Choice: Theorem $2.1$
Category:Set Equivalence behaves like Equivalence Relation
Category:Symmetric Relations
Set Equivalence behaves like Equivalence Relation/Transitive

== Theorem ==
Set equivalence behaves like a transitive relation:
:$S_1 \sim S_2 \land S_2 \sim S_3 \implies S_1 \sim S_3$
== Proof ==
== Sources ==
* : Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability
* : $\S 3.7$. Similar sets
* : Chapter $\text {III}$: The Natural Numbers: $\S 17$: Finite Sets: Theorem $17.1$
* : Part $1$: Not Entirely Naive Set Theory: Chapter $3$: Cardinality: Exercise $1 \ \text{(c)}$
* : $2$. Definition of Equivalence. The Concept of Cardinality. The Axiom of Choice: Theorem $2.1$
Category:Set Equivalence behaves like Equivalence Relation
Category:Transitive Relations
Set Equivalence behaves like Equivalence Relation/Warning

== Set Equivalence behaves like Equivalence Relation: Warning ==
It has been shown that set equivalence exhibits the same properties as an equivalence relation.
However, it is important to note that set equivalence is not strictly speaking a relation.
This is because the collection of all sets is itself specifically not a set, but a class.
Hence it is incorrect to refer to $\sim$ as an equivalence relation, although it is useful to be able to consider it as behaving like an equivalence relation.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $3$: Cardinality: Exercise $1 \ \text{(c)}$
Category:Set Equivalence behaves like Equivalence Relation
Set Equivalence of Regular Representations

== Theorem ==
If $S$ is a finite subset of a group $G$, then:
:$\card {a \circ S} = \card S = \left|{S \circ a}\right|$
That is, $a \circ S$, $S$ and $S \circ a$ are equivalent: $a \circ S \sim S \sim S \circ a$.
== Proof ==
Follows immediately from the fact that both the left and right regular representation are permutations, and therefore bijections.
== Sources ==
* : $\S 41.2$ Multiplication of subsets of a group
Category:Group Theory
Category:Regular Representations
Set Finite iff Injection to Initial Segment of Natural Numbers

== Theorem ==
Let $S$ be a set.
Then $S$ is finite for some $n \in \N$ there exists an injection $f: S \to \N_{< n}$.
Here, $\N_{< n}$ denotes an initial segment of $\N$.
== Proof ==
=== Necessary Condition ===
Suppose that $S$ is finite.
By definition, this means there exists a bijection $f: S \to \N_{< n}$.
Then $f$ is a fortiori also the sought injection.
=== Sufficient Condition ===
Let $f: S \to \N_{< n}$ be an injection.
Then by Injection to Image is Bijection, $S$ is equivalent to a subset of $\N_{<n}$.
By Subset of Finite Set is Finite, it follows that $S$ is finite.
== Sources ==
* : $1$: Set Theory and Logic: $\S 6$: Finite Sets: Corollary $6.7$
Category:Set Theory
Category:Injections
Set Finite iff Surjection from Initial Segment of Natural Numbers

== Theorem ==
Let $S$ be a set.
Then $S$ is finite for some $n \in \N$ there exists a surjection $f: \N_{<n} \to S$.
Here, $\N_{<n}$ denotes an initial segment of $\N$.
== Proof ==
=== Necessary Condition ===
Suppose that $S$ is finite.
By definition, this means there exists a bijection $f: \N_{<n} \to S$.
Then $f$ is a fortiori also the sought surjection.
=== Sufficient Condition ===
Let $f: \N_{<n} \to S$ be a surjection.
Define $g: S \to \N_{<n}$ by:
:$\map g s := \min \map {f^{-1} } s$
where $\map {f^{-1} } s$ is the preimage of $s$ under $f$.
Note that $\map {f^{-1} } s$ is not empty because $f$ is a surjection.
By the Well-Ordering Principle, $\map {f^{-1} } s \subseteq \N$ has a smallest element.
Hence $g$ is well-defined.
Next we show that $g$ is injective.
So suppose that $\map g s = \map g {s'}$ for some $s, s' \in S$:
Hence $g$ is injective.
Then by Injection to Image is Bijection, $S$ is equivalent to a subset of $\N_{<n}$.
By Subset of Finite Set is Finite, it follows that $S$ is finite.
== Sources ==
* : $1$: Set Theory and Logic: $\S 6$: Finite Sets: Corollary $6.7$
Category:Set Theory
Category:Surjections
Set Intersection/Examples

== Examples of Set Intersection ==
=== Example: $2$ Arbitrarily Chosen Sets: $1$ ===
=== Example: $2$ Arbitrarily Chosen Sets: $2$ ===
=== Example: $2$ Arbitrarily Chosen Sets of Complex Numbers: $1$ ===
=== Example: $2$ Arbitrarily Chosen Sets of Complex Numbers: $2$ ===
=== Example: $3$ Arbitrarily Chosen Sets ===
=== Example: $3$ Arbitrarily Chosen Sets of Complex Numbers ===
=== Example: $4$ Arbitrarily Chosen Sets of Complex Numbers ===
=== Example: Blue-Eyed British People ===
=== Example: Overlapping Integer Sets ===
=== Example: $2$ Circles in Complex Plane ===
=== Example: $3$ Circles in Complex Plane ===
=== Example: Arbitrary Integer Sets ===
=== Example: Intersection with Power Set ===
=== Arbitrary Example $1$ ===
Category:Examples of Set Intersection
Set Intersection/Examples/2 Circles in Complex Plane

== Example of Set Intersection ==
Let $A$ and $B$ be sets defined by circles embedded in the complex plane as follows:
Then $A \cap B$ can be illustrated graphically as:
:
where the intersection is depicted in yellow.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Miscellaneous Problems: $49 \ \text{(a)}$
Category:Examples of Set Intersection
Set Intersection/Examples/3 Arbitrarily Chosen Sets of Complex Numbers

== Example of Set Intersection ==
Let:
Then:
:$B \cap C = \O$
and so it follows that:
:$A \cap \paren {B \cap C} = \O$
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Point Sets: $46 \ \text {(f)}$
Category:Examples of Set Intersection
Set Intersection/Examples/3 Circles in Complex Plane

== Example of Set Intersection ==
Let $A$, $B$ and $C$ be sets defined by circles embedded in the complex plane as follows:
Then $A \cap B \cap C$ can be illustrated graphically as:
:
where the intersection is depicted in yellow.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Point Sets: $123 \ \text{(a)}$
Category:Examples of Set Intersection
Set Intersection/Examples/4 Arbitrarily Chosen Sets of Complex Numbers

== Example of Set Intersection ==
Let:
Then:
:$\paren {A \cup C} \cap \paren {B \cup D} = \set {1, -i}$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Point Sets: $121 \ \text{(c)}$
Category:Examples of Set Union
Category:Examples of Set Intersection
Set Intersection/Examples/Finite Subfamily of Unbounded Above Open Real Intervals

== Example of Intersection of Set of Sets ==
Let $\R$ denote the set of real numbers.
For a given $a \in \R$, let $S_a$ denote the (real) interval:
:$S_a = \openint a \to = \set {x \in \R: x > a}$
Let $\SS$ denote the family of sets indexed by $\R$:
:$\SS := \family {S_a}_{a \mathop \in \R}$
Let $\TT$ be a finite subfamily of $\SS$.
Then:
:$\bigcap \TT$ is not empty.
== Proof ==
According to the definition of subfamily, let $\TT$ be the family of sets indexed by $U$, where $U \subseteq \R$ is a finite subset of $\R$.
By construction it is seen that $\TT \subseteq \R$
By definition, $\R$ is totally ordered by $>$.
By Subset of Toset is Toset, $U$ is also totally ordered by $>$.
From Finite Totally Ordered Set is Well-Ordered, $U$ has a greatest element $M$, say.
Thus:
:$\TT = \set {x \in \R: x > M}$
Let $N \in \R$ such that $N > M$.
Then it is seen that $N \in \TT$ and the result follows.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Unions and Intersections
Category:Examples of Set Intersection
Set Intersection/Examples/Intersection with Power Set

== Example of Set Intersection ==
Let $S$ be the set defined as:
:$S = \set {1, 2, \set {1, 2} }$
Then the power set of $S$ is:
:$\powerset S = \set {\O, \set 1, \set 2, \set {\set {1, 2} }, \set {1, 2}, \set {1, \set {1, 2} }, \set {2, \set {1, 2} }, \set {1, 2, \set {1, 2} } }$
and the intersection of $S$ with $\powerset S$ is:
:$S \cap \powerset S = \set {\set {1, 2} }$
== Sources ==
* : $\S 2$. Sets of sets: Exercise $4$
Category:Examples of Set Intersection
Set Intersection/Examples/Set of Initial Segments

== Example of Intersection of Set of Sets ==
Let $\Z$ denote the set of integers.
Let $\map \Z n$ denote the initial segment of $\Z_{> 0}$:
:$\map \Z n = \set {1, 2, \ldots, n}$
Let $\mathscr S := \set {\map \Z n: n \in \Z_{> 0} }$
That is, $\mathscr S$ is the set of all initial segments of $\Z_{> 0}$.
Then:
:$\bigcap \mathscr S = \set 1$
== Sources ==
* : $\S 1.8$. Sets of sets: Example $28$
Category:Examples of Set Intersection
Set Intersection/Examples/Set of Unbounded Above Open Real Intervals

== Example of Intersection of Set of Sets ==
Let $\R$ denote the set of real numbers.
For a given $a \in \R$, let $S_a$ denote the (real) interval:
:$S_a = \openint a \to = \set {x \in \R: x > a}$
Let $\SS$ denote the family of sets indexed by $\R$:
:$\SS := \family {S_a}_{a \mathop \in \R}$
Then:
:$\bigcap \SS = \O$
== Proof ==
 there exists $x \in \R$ such that $x \in \SS$.
Then by definition of set intersection:
:$\forall a \in \R: x \in S_a$
That is:
:$\forall a \in \R: x > a$
But then we have:
:$x + 1 \in \R$
leading to:
:$x > x + 1$
which contradicts the properties of real numbers, which lead to:
:$x < x + 1$
Hence the result.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Unions and Intersections
Category:Examples of Set Intersection
Set Intersection Distributes over Set Difference

== Theorem ==
Set intersection is distributive over set difference.
Let $R, S, T$ be sets.
Then:
:$\paren {R \setminus S} \cap T = \paren {R \cap T} \setminus \paren {S \cap T}$
:$R \cap \paren {S \setminus T} = \paren {R \cap S} \setminus \paren {R \cap T}$
where:
:$R \setminus S$ denotes set difference
:$R \cap T$ denotes set intersection.
== Proof ==
Then:
== Sources ==
* : $\S 5$: Complements and Powers
* : $\S 1$: Sets and Functions: Problem $4 \ \text{(a)}$
* : $1$: Set Theory and Logic: $\S 1$: Fundamental Concepts: Exercise $1.2 \ \text{(g)}$
Category:Set Difference
Category:Set Intersection
Category:Examples of Distributive Operations
Set Intersection Not Cancellable

== Theorem ==
Let $S$ be a set and let $\powerset S$ be the power set of $S$.
Let $S_1, S_2, T \in \powerset S$.
Suppose that $S_1 \cap T = S_2 \cap T$.
Then it is not necessarily the case that $S_1 = S_2$.
== Proof ==
Proof by Counterexample:
Let $S = \set {1, 2, 3}$.
Let $T = \set 3$.
Let $S_1 = \set {1, 3}, S_2 = \set {2, 3}$
Then $S_1 \cap T = S_2 \cap T = \set 3$ but $S_1 \ne S_2$.
Category:Set Intersection
Set Intersection Preserves Subsets

== Theorem ==
Let $A, B, S, T$ be sets.
Then:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$
=== Corollary ===
=== Families of Sets ===
== Proof ==
Let $A \subseteq B$ and $S \subseteq T$.
Then:
 
Now we invoke the Praeclarum Theorema of propositional logic:
:$\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s}$
applying it as:
:$\paren {x \in A \implies x \in B, \ x \in S \implies x \in T} \leadsto \paren {x \in A \land x \in S \implies x \in B \land x \in T}$
The result follows directly from the definition of set intersection:
:$\paren {x \in A \implies x \in B, \ x \in S \implies x \in T} \leadsto \paren {x \in A \cap S \implies x \in B \cap T}$
and from the definition of subset:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$
Category:Set Intersection
Category:Subsets
Category:Set Intersection Preserves Subsets
Set Intersection Preserves Subsets/Corollary

== Theorem ==
Let $A, B, S$ be sets.
Then:
:$A \subseteq B \implies A \cap S \subseteq B \cap S$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.6$: Set Identities and Other Set Relations: Exercise $1 \ \text{(a)}$
Category:Set Intersection Preserves Subsets
Set Intersection Preserves Subsets/Corollary/Proof 1

== Theorem ==
Let $A, B, S$ be sets.
Then:
:$A \subseteq B \implies A \cap S \subseteq B \cap S$
== Proof ==
Let $A \subseteq B$, and let $S$ be any set.
From Set Intersection Preserves Subsets:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$
for arbitrary sets $S$ and $T$.
Substituting $S$ for $T$:
:$A \subseteq B, \ S \subseteq S \implies A \cap S \subseteq B \cap S$
From Set is Subset of Itself, $S \subseteq S$ for all sets $S$.
Hence the result:
:$A \subseteq B \implies A \cap S \subseteq B \cap S$
Category:Set Intersection Preserves Subsets
Set Intersection Preserves Subsets/Corollary/Proof 2

== Corollary to Set Intersection Preserves Subsets ==
Let $A, B, S$ be sets.
Then:
:$A \subseteq B \implies A \cap S \subseteq B \cap S$
== Proof ==
Recall the Factor Principles, themselves a corollary of the Praeclarum Theorema:
:$\paren {p \implies q} \vdash \paren {p \land r} \implies \paren {q \land r}$
This is applied as:
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: $\text{(e)}$
Category:Set Intersection Preserves Subsets
Set Intersection Preserves Subsets/Families of Sets

== Theorem ==
Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be indexed families of subsets of a set $S$.
Let:
:$\forall \beta \in I: A_\beta \subseteq B_\beta$
Then:
:$\ds \bigcap_{\alpha \mathop \in I} A_\alpha \subseteq \bigcap_{\alpha \mathop \in I} B_\alpha$
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
By definition of subset:
:$\ds \bigcap_{\alpha \mathop \in I} A_\alpha \subseteq \bigcap_{\alpha \mathop \in I} B_\alpha$
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $1 \ \text{(e)}$
Category:Set Intersection Preserves Subsets
Category:Indexed Families
Set Intersection Preserves Subsets/Families of Sets/Corollary

== Theorem ==
Let $I$ be an indexing set.
Let $\family {B_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of a set $S$.
Let $A$ be a set such that $A \subseteq B_\alpha$ for all $\alpha \in I$.
Then:
:$\ds A \subseteq \bigcap_{\alpha \mathop \in I} B_\alpha$
== Proof ==
For each $\alpha \in I$, define $A_\alpha := A$.
Then by Set Intersection is Idempotent, it follows that:
:$\ds \bigcap_{\alpha \mathop \in I} A_\alpha = A$
Since $A \subseteq B_\alpha$ for all $\alpha \in I$, the premises of Set Intersection Preserves Subsets are satisfied.
Applying this theorem gives:
:$\ds A = \bigcap_{\alpha \mathop \in I} A_\alpha \subseteq \bigcap_{\alpha \mathop \in I} B_\alpha$
which is precisely the desired result.
Category:Set Intersection Preserves Subsets
Category:Indexed Families
Set Intersection Preserves Subsets/Families of Sets/Intersection is Empty Implies Intersection of Subsets is Empty

== Theorem ==
Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be indexed families of subsets of a set $S$.
Let:
:$\forall \beta \in I: A_\beta \subseteq B_\beta$
Then:
:$\ds \bigcap_{\alpha \mathop \in I} B_\alpha = \O \implies \bigcap_{\alpha \mathop \in I} A_\alpha = \O$
== Proof ==
Let $\ds \bigcap_{\alpha \mathop \in I} B_\alpha = \O$.
From Set Intersection Preserves Subsets/Families of Sets:
:$\ds \bigcap_{\alpha \mathop \in I} A_\alpha \subseteq \bigcap_{\alpha \mathop \in I} B_\alpha = \O$
From Subset of Empty Set:
:$\ds \bigcap_{\alpha \mathop \in I} A_\alpha = \O$
Category:Set Intersection Preserves Subsets
Category:Indexed Families
Set Intersection expressed as Intersection Complement

== Theorem ==
Let $A$ and $B$ be subsets of a universal set $\Bbb U$.
Let $\uparrow$ denote the operation on $A$ and $B$ defined as:
:$\paren {A \uparrow B} \iff \paren {\relcomp {\Bbb U} {A \cap B} }$
where $\relcomp {\Bbb U} A$ denotes the complement of $A$ in $\Bbb U$.
Then:
:$A \cap B = \paren {A \uparrow B} \uparrow \paren {A \uparrow B}$
== Proof ==
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $16$
Category:Set Intersection
Category:Set Complement
Set Intersection is Idempotent

== Theorem ==
Set intersection is idempotent:
:$S \cap S = S$
=== Indexed Family ===
== Proof ==
== Also see ==
* Set Union is Idempotent
== Sources ==
* : $\S 4$: Unions and Intersections
* : Exercise $1.1: \ 8 \ \text{(a)}$
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : Chapter $1$: Set Theory: $1.3$: Set operations
* : $\S 1.5$. The algebra of sets
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B i}$
* : $\text I$: Random Events: $1$. Relations among Random Events: Problem $1$
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 5 \ \text{(c)}$
* : $\S 1.6$: Set Identities and Other Set Relations: Exercise $2 \ \text{(c)}$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: $\text{(a)}$
* : Chapter $1$: Sets and mappings: $\S 1.2$: Sets: Exercise $1 \ \text{(i)}$
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.1 \ \text{(i)}$
* : $\S 2$
* : $\text {(v)}$
* : $\text {(i)}$
Category:Set Intersection
Category:Examples of Idempotence‎