Sets

A set is a well-defined collection of distinct objects. "Well-defined" means that given any object, we can decide with certainty whether it belongs to the collection or not. The objects in a set are called its elements or members. Sets are usually denoted by capital letters (A, B, C, ...) and elements by small letters (a, b, x, ...).

If an object x is an element of a set A, we write x ∈ A (read "x belongs to A"). If x is not an element of A, we write x ∉ A. The number of distinct elements in a finite set A is called its cardinal number, written n(A).

There are two standard ways to describe a set:

• Roster form (tabular form): all elements are listed inside braces, separated by commas, each element written only once and the order not mattering. Example: the set of vowels in English is {a, e, i, o, u}.
• Set-builder form: the set is described by a common property satisfied by every element, written as A = {x : p(x)}, read "A is the set of all x such that x satisfies property p".

Worked Example: Write the set of natural numbers less than 6 in both roster form and set-builder form.

Solution: The natural numbers less than 6 are 1, 2, 3, 4, 5.

Roster form: A = {1, 2, 3, 4, 5}.

Set-builder form: A = {x : x ∈ N, x < 6}.

Here n(A) = 5, since A has 5 distinct elements.

A set having no element at all is called the empty set (or null set), denoted by φ or { }. Note that {0} and {φ} are not empty sets — each contains one element.

A set is called a finite set if the process of counting its elements terminates, i.e. it has a definite, countable number of elements. A set that is not finite, i.e. whose elements cannot be counted to an end, is called an infinite set. A set with exactly one element is called a singleton set.

Two sets A and B are said to be equal, written A = B, if they have exactly the same elements — that is, every element of A is in B and every element of B is in A. If A and B are not equal, we write A ≠ B.

Worked Example: Let A = {1, 2, 3} and B = {x : x ∈ N, x² < 12}. State whether A and B are finite or infinite, and check whether A = B.

Solution: A = {1, 2, 3} has exactly 3 elements, so the counting terminates and A is a finite set.

For B, we need natural numbers x with x² < 12. Testing x = 1, 2, 3, 4: 1²=1, 2²=4, 3²=9 (all < 12), but 4²=16 (not < 12). So B = {1, 2, 3}, which also has exactly 3 elements, so B is a finite set.

Since A = {1, 2, 3} and B = {1, 2, 3} contain exactly the same elements, A = B.

A set A is called a subset of a set B, written A ⊆ B, if every element of A is also an element of B, i.e. a ∈ A ⟹ a ∈ B. Every set is a subset of itself, and the empty set φ is a subset of every set. If A ⊆ B and A ≠ B, A is called a proper subset of B, written A ⊂ B, and B is called a superset of A.

In any discussion, all sets under consideration are usually subsets of one basic set called the universal set, denoted U. Intervals of real numbers, such as (a, b), [a, b], (a, b], [a, b), are also examples of subsets of R described using inequalities.

The collection of all possible subsets of a set A (including φ and A itself) is called the power set of A, denoted P(A). If a finite set A has n elements, then P(A) has exactly 2ⁿ elements.

Worked Example: Let A = {1, 2, 3}. Write the power set of A, and verify the formula n(P(A)) = 2ⁿ.

Solution: The subsets of A are: φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.

So P(A) = { φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} }, which has 8 elements.

Since n(A) = 3, the formula gives 2³ = 8, which matches. Verified.

A Venn diagram represents the universal set U by a rectangle, and its subsets by circles (or closed curves) drawn inside it. Venn diagrams make it easy to visualise relationships and operations between sets.

For two sets A and B:

• Union: A ∪ B = {x : x ∈ A or x ∈ B} — all elements that are in A, or in B, or in both.
• Intersection: A ∩ B = {x : x ∈ A and x ∈ B} — elements common to both A and B. If A ∩ B = φ, then A and B are called disjoint sets.
• Difference: A − B = {x : x ∈ A and x ∉ B} — elements of A that are not in B. In general, A − B ≠ B − A.
• Complement: A′ = U − A = {x ∈ U : x ∉ A} — elements of the universal set that are not in A.

Worked Example: Let U = {1, 2, 3, ..., 10}, A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}. Find A ∪ B, A ∩ B, A − B, B − A and A′.

Solution:

A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8} (all elements in A or B).

A ∩ B = {4, 5} (elements common to both).

A − B = {1, 2, 3} (elements of A not in B).

B − A = {6, 7, 8} (elements of B not in A) — note this is different from A − B.

A′ = U − A = {6, 7, 8, 9, 10} (elements of U not in A).

Set operations follow several standard laws, similar to the laws of arithmetic, that hold for all sets A, B, C drawn from a universal set U:

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

Worked Example: Using U = {1, 2, ..., 10}, A = {1, 3, 5, 7, 9}, B = {2, 3, 5, 7}, verify De Morgan's law (A ∪ B)′ = A′ ∩ B′.

Solution: A ∪ B = {1, 2, 3, 5, 7, 9}, so (A ∪ B)′ = U − (A ∪ B) = {4, 6, 8, 10}.

Now A′ = U − A = {2, 4, 6, 8, 10} and B′ = U − B = {1, 4, 6, 8, 9, 10}.

A′ ∩ B′ = {4, 6, 8, 10} (common elements of A′ and B′).

Since (A ∪ B)′ = {4, 6, 8, 10} = A′ ∩ B′, the law is verified.

For any two finite sets A and B, the number of elements in their union is given by the inclusion-exclusion principle: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). This is because elements common to both sets get counted twice when n(A) and n(B) are simply added, so n(A ∩ B) is subtracted once to correct the count. If A and B are disjoint (A ∩ B = φ), this simplifies to n(A ∪ B) = n(A) + n(B).

This idea extends to three finite sets A, B, C: n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C). Also, for the difference of two sets, n(A − B) = n(A) − n(A ∩ B).

In practical problems, it is often useful to first mark the "only A", "only B", "both", and "neither" regions on a Venn diagram before applying the formula.

Worked Example: In a class of 50 students, 20 take Physics, 25 take Chemistry, and 10 take both subjects. How many students take at least one of the two subjects, and how many take neither?

Solution: Let P = set of students taking Physics, n(P) = 20; C = set of students taking Chemistry, n(C) = 25; n(P ∩ C) = 10.

Number taking at least one subject: n(P ∪ C) = n(P) + n(C) − n(P ∩ C) = 20 + 25 − 10 = 35.

Number taking neither subject = total students − n(P ∪ C) = 50 − 35 = 15.