Discrete
Math
Section 2.2: Set Operations
Dr. Jason Snyder
Collin College
Definition 1
Let π΄ and π΅ be sets. The union of the sets π΄ and π΅, denoted by π΄ ∪ π΅, is the set that
contains those elements that are either in π΄ or in π΅, or in both.
That is π΄ ∪ π΅ = {π₯|π₯ ∈ π΄ ∨ π₯ ∈ π΅}.
Example 1
Let π΄ = {1,2,3,5} and let π΅ = {3,4,5,6}. List the elements of π΄ ∪ π΅ using the roster
method.
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Definition 2
Let π΄ and π΅ be sets. The intersection of the sets π΄ and π΅, denoted by π΄ ∩ π΅, is the
set that contains those elements in both π΄ and π΅.
That is π΄ ∩ π΅ = {π₯|π₯ ∈ π΄ ∧ π₯ ∈ π΅}.
Example 2
Let π΄ = {1,2,3,5} and let π΅ = {3,4,5,6}. List the elements of π΄ ∩ π΅ using the roster
method.
Definition 3
Two sets are called disjoint if their intersection is the empty set.
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Definition 4
Let π΄ and π΅ be sets. The difference of π΄ and π΅, denoted by π΄ − π΅, is the set
containing those elements that are in π΄ but not in π΅. The difference of π΄ and π΅ is also
sometimes called the complement of π΅ in π΄.
That is π΄ − π΅ = {π₯|π₯ ∈ π΄ ∧ π₯ ∉ π΅}.
The difference of set π΄ and π΅ is sometimes denoted by π΄ β π΅.
Example 3
Let π΄ = {1,2,3,5} and let π΅ = {3,4,5,6}. List the elements of π΄ − π΅ and π΅ − π΄ using the
roster method.
Definition 5
Let π be a universal set. The complement of the set π΄, denoted by π΄, is the difference
of π and π΄. That is π΄ = π − π΄.
Example 4
For this problem, let β€ be the universal set. Let π΄ = {π₯ ∈ β€|π₯ ≥ 10}. What is the set π΄?
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Set Identities
Identity Laws
ο· π΄∩π =π΄
ο· π΄∪∅=π΄
Domination Laws
ο· π΄∪π =π
ο· π΄∩∅=∅
Idempotent Laws
ο· π΄∪π΄ =π΄
ο· π΄∩π΄ =π΄
Complementation Law
(π΄Μ
) = π΄
ο· Μ
Μ
Μ
Μ
Μ
Commutative Laws
ο· π΄∪π΅ =π΅∪π΄
ο· π΄∩π΅ =π΅∩π΄
Associative Laws
ο· π΄ ∪ (π΅ ∪ πΆ) = (π΄ ∪ π΅) ∪ πΆ
ο· π΄ ∩ (π΅ ∩ πΆ) = (π΄ ∩ π΅) ∩ πΆ
Distributive Laws
ο· π΄ ∪ (π΅ ∩ πΆ) = (π΄ ∪ π΅) ∩ (π΄ ∪ πΆ)
ο· π΄ ∩ (π΅ ∪ πΆ) = (π΄ ∩ π΅) ∪ (π΄ ∩ πΆ)
De Morgan’s Laws
ο· Μ
Μ
Μ
Μ
Μ
Μ
Μ
π΄ ∪ π΅ = π΄Μ
∩ π΅Μ
ο· Μ
Μ
Μ
Μ
Μ
Μ
Μ
π΄ ∩ π΅ = π΄Μ
∪ π΅Μ
Absorption Laws
ο· π΄ ∪ (π΄ ∩ π΅) = π΄
ο· π΄ ∩ (π΄ ∪ π΅) = π΄
Complement Laws
ο· π΄ ∪ π΄Μ
= π
ο· π΄ ∩ π΄Μ
= ∅
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Example 5
Prove that Μ
Μ
Μ
Μ
Μ
Μ
Μ
π΄ ∪ π΅ = π΄Μ
∩ π΅Μ
.
Example 6
Prove that π΄ ∪ (π΅ ∩ πΆ) = (π΄ ∪ π΅) ∩ (π΄ ∪ πΆ).
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Example 7
Prove that Μ
Μ
Μ
Μ
Μ
Μ
Μ
Μ
Μ
Μ
Μ
Μ
Μ
Μ
Μ
π΄ ∪ (π΅ ∩ πΆ) = (πΆΜ
∪ π΅Μ
) ∩ π΄Μ
.
Example 8
Let π΄ = {0,2,4,6,8}, π΅ = {0,1,2,3,4}, and πΆ = {0,3,6,9}. What are π΄ ∪ π΅ ∪ πΆ and
π΄ ∩ π΅ ∩ πΆ?
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Definition 6
Let π΄1 , π΄2 , …, π΄π be a collection of π sets. The union of these sets, π΄1 ∪ π΄2 ∪ β― ∪ π΄π is
the set containing those elements that belong to at least of the sets π΄1 , π΄2 , …, π΄π .
This union is denoted by
π
π΄1 ∪ π΄2 ∪ β― ∪ π΄π = β π΄π .
π=1
Definition 7
Let π΄1 , π΄2 , …, π΄π be a collection of π sets. The intersection of these sets,
π΄1 ∩ π΄2 ∩ β― ∩ π΄π is the set containing those elements that belong to all of the sets π΄1 ,
π΄2 , …, π΄π .
This union is denoted by
π
π΄1 ∩ π΄2 ∩ β― ∩ π΄π = β π΄π .
π=1
Example 9
Let π΄π = {π, π + 1, π + 2, … } for π = 1, 2, 3, 4, …
Find
π
β π΄π
π=1
π
and
β π΄π
π=1
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This can be extended to other, even infinite, collections of sets.
Example 10
Let π΄π = {π, π + 1, π + 2, … } and π΅π = {1,2,3, … , π} for π = 1, 2, 3, 4, …
Find
∞
β π΄π
∞
and
β π΄π .
π=1
π=1
∞
∞
Also find
β π΅π
π=1
and
β π΅π .
π=1
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