Operations on Sets — Union, Intersection and Difference

Sets become powerful when you combine them. Four operations do the work: union, intersection, difference and complement — and two elegant laws (De Morgan's) tie them together.

The four operations

• Union A ∪ B: elements in A or B (or both) = {x : x ∈ A or x ∈ B}.
• Intersection A ∩ B: elements in both = {x : x ∈ A and x ∈ B}.
• Difference A − B: in A but not in B.
• Complement A′ = U − A: everything in the universal set outside A.

Disjoint sets: A ∩ B = ∅ (no common element).

Key algebraic laws

• Commutative: A ∪ B = B ∪ A; A ∩ B = B ∩ A.
• Associative: (A ∪ B) ∪ C = A ∪ (B ∪ C).
• Distributive: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
• Identity: A ∪ ∅ = A; A ∩ U = A.
• De Morgan's laws: (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′.

Counting formula

For finite sets: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). For three sets add back the triple overlap: n(A∪B∪C) = n(A)+n(B)+n(C) − n(A∩B) − n(B∩C) − n(A∩C) + n(A∩B∩C).

Exam Tricks & Tips

• 🎯 De Morgan flips the operation: complement of a union is intersection of complements, and vice-versa.
• 🎯 A − B = A ∩ B′ — rewrite differences as intersections to simplify proofs.
• 🎯 The formula n(A∪B) = n(A)+n(B)−n(A∩B) is the workhorse of every counting word problem.
• 🎯 A ∪ A′ = U and A ∩ A′ = ∅ — a set and its complement partition U.
• 🎯 Disjoint means A ∩ B = ∅, so then n(A∪B) = n(A)+n(B) exactly.
• ❌ Common mistake: forgetting to subtract n(A ∩ B) once, which double-counts the overlap and inflates n(A ∪ B).

Expected exam pattern

Simplify set expressions using the laws, verify De Morgan's laws, or apply the union counting formula in survey-style word problems (1–4 marks).

Cover the answer, recall, then check. 12 cards on set operations and laws.

Q1. Define A ∪ B.
A1. {x : x ∈ A or x ∈ B} — elements in A or B or both.

Q2. Define A ∩ B.
A2. {x : x ∈ A and x ∈ B} — elements in both sets.

Q3. What does A − B mean?
A3. Elements in A but not in B.

Q4. Rewrite A − B using intersection and complement.
A4. A − B = A ∩ B′.

Q5. What is the complement A′?
A5. U − A, all elements of the universal set not in A.

Q6. State De Morgan's laws.
A6. (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′.

Q7. When are sets disjoint?
A7. When A ∩ B = ∅ (no common element).

Q8. State the two-set union counting formula.
A8. n(A ∪ B) = n(A) + n(B) − n(A ∩ B).

Q9. What is A ∪ A′ and A ∩ A′?
A9. A ∪ A′ = U; A ∩ A′ = ∅.

Q10. Is A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)? Name the law.
A10. Yes — the distributive law.

Q11. If n(A)=20, n(B)=15, n(A∩B)=5, find n(A∪B).
A11. 20 + 15 − 5 = 30.

Q12. State the three-set union formula.
A12. n(A∪B∪C) = n(A)+n(B)+n(C) − n(A∩B) − n(B∩C) − n(A∩C) + n(A∩B∩C).

Combining sets like numbers

Just as numbers can be added and subtracted, sets can be combined. The three core operations — union, intersection, difference — plus the complement, are the algebra you'll use in probability, functions and beyond.

The four operations

Beginner: what each does

• Union A ∪ B = elements in A or B (or both). Think everything thrown together.
  A ∪ B = {x : x ∈ A or x ∈ B}.
• Intersection A ∩ B = elements in both A and B (the overlap).
  A ∩ B = {x : x ∈ A and x ∈ B}.
• Difference A − B = in A but not in B.
  A − B = {x : x ∈ A and x ∉ B}.
• Complement A′ = U − A, everything in the universal set outside A.

Disjoint sets: A ∩ B = ∅ (no common elements).

Intermediate: the laws that save time

• Commutative: A ∪ B = B ∪ A; A ∩ B = B ∩ A.
• Associative: (A ∪ B) ∪ C = A ∪ (B ∪ C).
• Distributive: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
• Identity: A ∪ ∅ = A; A ∩ U = A.
• Idempotent: A ∪ A = A; A ∩ A = A.

Advanced: De Morgan's Laws

The most-tested identities in the chapter:

(A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′

In words: the complement of a union is the intersection of complements (and vice-versa). The operation flips and the bar distributes.

Worked example

Q. U = {1,…,10}, A = {2,4,6,8}, B = {4,8,10}. Find A ∪ B, A ∩ B, A − B, A′, and verify (A ∪ B)′ = A′ ∩ B′.
Solution.

• A ∪ B = {2,4,6,8,10}
• A ∩ B = {4,8}
• A − B = {2,6}
• A′ = {1,3,5,7,9,10}; B′ = {1,2,3,5,6,7,9}
• (A ∪ B)′ = {1,3,5,7,9}; A′ ∩ B′ = {1,3,5,7,9} ✓ — De Morgan holds.

Where it shows up

Search engines: "cats AND dogs" = intersection, "cats OR dogs" = union, "cats NOT kittens" = difference. Set operations are literally how filters work.

Exam tricks & shortcuts

• Union = OR = ∪ looks like a cup catching everything. Intersection = AND = ∩ is the cap of overlap. ("U for Union, and ∪ scoops it all Up.")
• To find A − B fast, delete from A everything that also sits in B.
• De Morgan mnemonic: "Break the bar, change the sign" — the long bar splits and ∪ ↔ ∩.

Key formulas

• Union: A ∪ B = {x : x ∈ A or x ∈ B}.
• Intersection: A ∩ B = {x : x ∈ A and x ∈ B}.
• Difference: A − B = {x : x ∈ A and x ∉ B}.
• Complement: A′ = U − A.
• De Morgan's laws: (A ∪ B)′ = A′ ∩ B′; (A ∩ B)′ = A′ ∪ B′.
• Distributive: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).

Disjoint sets have A ∩ B = ∅; the complement is taken with respect to the universal set U.

Worked Example

Problem: Given A = {1, 2, 3, 4} and B = {3, 4, 5, 6}, find A ∪ B, A ∩ B, A − B, and B − A.

Solution:
Step 1 — Union (A ∪ B): all elements in A or B (or both), listed once:
A ∪ B = {1, 2, 3, 4, 5, 6}.

Step 2 — Intersection (A ∩ B): elements common to both sets:
A ∩ B = {3, 4}.

Step 3 — Difference (A − B): elements in A but not in B:
A − B = {1, 2}.

Step 4 — Difference (B − A): elements in B but not in A:
B − A = {5, 6}.

Step 5 — Cross-check: (A − B), (A ∩ B), and (B − A) are disjoint and together give A ∪ B: {1,2} ∪ {3,4} ∪ {5,6} = {1,2,3,4,5,6}. ✓

Answer: A ∪ B = {1,2,3,4,5,6}; A ∩ B = {3,4}; A − B = {1,2}; B − A = {5,6}.