Set theory

Introduction to Set Theory

Basic Definitions

Set, Element, Subset

Example:
Let A={1,2,3} and B={1,2,3,4,5}. Since every element of A is also in B, we say A is a subset of B: A⊆B.

Empty Set

∅={}

Universal Set


Set Notation

Roster Notation

Set-Builder Notation


Membership and Inclusion

Membership

a∈A

If a does not belong to A, we write:

a∉A

Subset and Inclusion

A⊆B

Example:
If A={1,2} and B={1,2,3}, then A⊆B because all elements of A are in B. However, A≠B, so A⊂B (proper subset).


Venn Diagrams

Common Venn Diagram Operations:

Example:
If A={1,2,3} and B={3,4,5}:

Venn diagrams provide an intuitive way to visualize these set operations.

Operations on Sets

Union ( ∪ )

A∪B={x∣x∈A or x∈B} A∪B={1,2,3,4,5}

Intersection ( ∩ )

A∩B={x∣x∈A and x∈B} A∩B={3}

Difference ( A \ B )

A∖B={x∣x∈A and x∉B} A∖B={1,2}

Complement ( A' )

A′={x∣x∈U and x∉A} A′={3,4,5}

Symmetric Difference

A△B=(A∖B)∪(B∖A) A△B={1,2,4,5}

Cartesian Product

A×B={(a,b)∣a∈A and b∈B} A×B={(1,x),(1,y),(2,x),(2,y)}

Pairs and Tuples

Example of Pairs and Tuples:

Basic Set Identities

Commutative, Associative, and Distributive Laws

Commutative Laws:

A∪B=B∪AA∩B=B∩A

Associative Laws:

(A∪B)∪C=A∪(B∪C)(A∩B)∩C=A∩(B∩C)

Distributive Laws:

A∩(B∪C)=(A∩B)∪(A∩C)A∪(B∩C)=(A∪B)∩(A∪C)

De Morgan’s Laws

(A∪B)′=A′∩B′(A∩B)′=A′∪B′

Example:
If A={1,2,3} and B={3,4,5} with universal set U={1,2,3,4,5,6}, then:


Absorption and Idempotent Laws

Absorption Laws:

A∪(A∩B)=AA∩(A∪B)=A

Idempotent Laws:

A∪A=AA∩A=A

Double Complement Law

(A′)′=A

Example:
If A={1,2,3} and the universal set U={1,2,3,4,5}, then:


Proofs Using Venn Diagrams and Algebraic Techniques

Proof with Venn Diagrams:

Example: Prove De Morgan’s Law (A∪B)′=A′∩B′ using a Venn diagram:

  1. Draw two overlapping circles representing sets A and B.
  2. Shade the area outside the union of A and B to represent (A∪B)′.
  3. Then, separately shade the intersection of A′ and B′ (the regions outside A and B).
  4. You will see that the shaded regions for both expressions are identical, proving the identity.

Proof with Algebraic Techniques:

Example: Prove the absorption law A∪(A∩B)=A:

  1. Start with A∪(A∩B).
  2. By the distributive law, we have:
A∪(A∩B)=(A∪A)∩(A∪B)
  1. By the idempotent law, A∪A=A, so:
A∪(A∩B)=A∩(A∪B)
  1. By the absorption law again, A∩(A∪B)=A. Therefore:
A∪(A∩B)=A

Example of De Morgan’s Law Using Algebraic Techniques:

  1. (A∪B)′ means all elements that are not in either A or B.
  2. A′∩B′ means all elements that are not in A and not in B.
  3. Since both describe the same set (elements not in A or B), they are equal, and thus:
(A∪B)′=A′∩B′

Relations

Definitions

Ordered Pairs and Cartesian Products

A×B={(a,b)∣a∈A and b∈B} A×B={(1,x),(1,y),(2,x),(2,y)}

Binary Relations


Properties of Relations

Reflexive Relations

∀a∈A,(a,a)∈R

Symmetric Relations

∀a,b∈A,(a,b)∈R⟹(b,a)∈R

Antisymmetric Relations

∀a,b∈A,(a,b)∈R and (b,a)∈R⟹a=b

Transitive Relations

∀a,b,c∈A,(a,b)∈R and (b,c)∈R⟹(a,c)∈R

Equivalence Relations

Equivalence Classes

[a]={b∈A∣aRb}

Partition of a Set


Partial and Total Orders

Partial Orders

Total Orders


Hasse Diagrams

Example:
For the set A={1,2,3,4} with the partial order R=≤, the Hasse diagram would look like this: