The four fundamental operations on sets are:

• Union (∪): A ∪ B = {x: x ∈ A or x ∈ B}
• Intersection (∩): A ∩ B = {x: x ∈ A and x ∈ B}
• Complement (A'): A' = {x ∈ U: x ∉ A}
• Difference (A − B): A − B = {x: x ∈ A and x ∉ B}

Set expressions can be simplified using the laws of algebra of sets. These laws allow us to transform complex expressions into simpler equivalent forms without changing their meaning.

Key Laws of Algebra of Sets

Worked Example: Simplifying a Set Expression

Simplify: (A ∪ B)' ∩ A

Solution:

Step 1: Apply De Morgan's law to (A ∪ B)' $(A \cup B)' = A' \cap B'$

So the expression becomes: $(A' \cap B') \cap A$

Step 2: Use the commutative law: $A \cap (A' \cap B')$

Step 3: Use the associative law: $(A \cap A') \cap B'$

Step 4: Apply the complement law (A ∩ A' = ∅): $\emptyset \cap B'$

Step 5: Apply the identity law (∅ ∩ B' = ∅): $\boxed{\emptyset}$

The cardinality of a set A, written n(A), is the number of elements in A.

Cardinality of Union of Two Sets

For any two finite sets A and B: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$

This formula accounts for elements counted twice in the intersection.

Cardinality of Union of Three Sets

For three finite sets A, B, and C: $n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)$

Worked Example: Cardinality of Two Sets

Given: n(A) = 28, n(B) = 10, n(A ∪ B) = 33

Find: n(A ∩ B)

Solution:

Using the formula: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$

Rearranging: $n(A \cap B) = n(A) + n(B) - n(A \cup B)$ $n(A \cap B) = 28 + 10 - 33$ $n(A \cap B) = 5$

Therefore, there are 5 elements in the intersection of A and B.

Worked Example: Cardinality of Three Sets

Given:

• n(A) = 10, n(B) = 7, n(C) = 9
• n(A ∩ B) = 4, n(B ∩ C) = 3, n(A ∩ C) = 3
• n(A ∩ B ∩ C) = 1

Find: n(A ∪ B ∪ C)

Solution:

Apply the three-set formula: $n(A \cup B \cup C) = 10 + 7 + 9 - 4 - 3 - 3 + 1$ $n(A \cup B \cup C) = 27 - 10 + 1$ $n(A \cup B \cup C) = 17$

Problem: In a class of 45 students, 21 play football, 17 play basketball, and 12 play volleyball. Also, 8 play football and basketball only, 5 play basketball and volleyball only, 3 play football and volleyball only, and 2 play all three games. How many students play at least one game?

Solution: Let F, B, and V represent the sets of students playing football, basketball, and volleyball respectively.

Using the formula with the given information:

• n(F) = 21
• n(B) = 17
• n(V) = 12
• n(F ∩ B) = 8 + 2 = 10 (including those who play all three)
• n(F ∩ V) = 3 + 2 = 5
• n(B ∩ V) = 5 + 2 = 7
• n(F ∩ B ∩ V) = 2

$n(F \cup B \cup V) = 21 + 17 + 12 - 10 - 5 - 7 + 2 = 30$

Therefore, 30 students play at least one game, meaning 15 students play none.

In Tanzania, set theory and cardinality are used in market research and business surveys. For example, a shop owner in Arusha conducting a survey about preferred mobile network providers (Airtel, Vodacom, Tigo) can use set operations to find how many customers use at least one network, how many use exactly two networks, or how many use none of them. This helps in planning marketing strategies and stock management for items like SIM cards and airtime bundles worth millions of Tanzanian shillings.