All notesMathTalk
CAMBRIDGE O LEVEL · 2025–2027

Making sense
of sets.

From a collection of elements to a clear visual language. Learn to read, draw and reason with sets.

Syllabus 1.2Visual notesWorked examples
BY THE END

Use set notation · describe relationships · interpret two- and three-set Venn diagrams · solve counting problems

01

Sets are collections with a rule

Think of a box containing red pencils. A set is the mathematical version of a collection: we need a clear rule for deciding which objects belong. Each object in the collection is called an element.

The rule must be definite. “Students older than 14” can be checked; “interesting students” cannot be decided in the same precise way. We usually give a set a capital letter and list its elements inside curly brackets. The brackets describe membership, not an addition calculation.

A set is a well-defined collection of distinct objects called elements. A Venn diagram makes the relationships between collections visible.

List the elements

A = {2, 4, 6, 8}

Use curly brackets. Order does not matter, and repeating an element does not add a new member.

{2, 4, 4, 6, 8} = {8, 6, 4, 2}

Count the elements

n(A) = 4

The set contains four different elements. n(A) counts them; it does not add their values.

4 ∈ A reads “4 is an element of A.”

Pause and explainDoes repeating an element make a set larger?

No. A set records which distinct elements belong, so {2, 2, 4} and {2, 4} describe the same set.

02

A small toolkit of symbols

Set symbols are abbreviations for ordinary sentences. Before learning their shapes, say what you want to describe: an object belongs, two collections overlap, or one collection sits entirely inside another.

An element and a set play different roles. The number 3 is an element; {3} is a set containing one element. Keep that distinction in mind as you read the notation table. Translate each symbolic statement into words before deciding whether it is true.

Read the symbol as a phrase first. Then apply it to the elements.

n(A)

Number of elements

If A = {2, 4, 6}, n(A) = 3.

∈ / ∉

Element / not an element

4 ∈ A, but 5 ∉ A.

A′

Complement

Everything in the universal set that is not in A.

∅

Empty set

A set with no elements: ∅ = { }, n(∅) = 0.

ξ

Universal set

All the elements under consideration; sometimes written U.

A ⊆ B

Subset

Every element of A is in B. Equality is allowed.

A ⊈ B

Not a subset

At least one element of A is not in B.

A ∪ B

Union

In A or B or both.

A ∩ B

Intersection

In both A and B.

Pause and explainWhat is the difference between ∅ and {0}?

∅ contains no elements. The set {0} contains one element, the number zero.

03

See what the symbols do

Imagine sorting students by whether they play football and whether they swim. Some do both, so two separate lists would hide an important overlap. A Venn diagram gives each possible membership a region.

Place someone who does both activities in the overlap, not once in each outer region. The union includes everyone in either circle; the intersection includes only the shared part. When adding circle totals, notice that an overlap member appears in both totals even though it is only one person.

The rectangle is the universal set. Each circle is a set. An element in the overlap belongs to both sets.

VENN LAB · CHOOSE AN OPERATION
ξAB2 · 48 · 10612391 · 5 · 7 · 11
ξ = {1, 2, 3, …, 12} · A = even numbers · B = multiples of 3

A ∪ B

Shade everything in A or B, including their overlap.

{2, 3, 4, 6, 8, 9, 10, 12}

n(A ∪ B) = 8

The lime region shows exactly which elements satisfy the operation.
WORKED EXAMPLE

Count a union without counting the overlap twice.

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

In the diagram, n(A) = 6, n(B) = 4 and n(A ∩ B) = 2.

So n(A ∪ B) = 6 + 4 − 2 = 8. The elements 6 and 12 were each counted twice when we added 6 + 4, so subtract one copy of the overlap.

Pause and explainDoes “A or B” include an element in both sets?

Yes. The union A ∪ B includes the overlap as well as the two single-set regions.

04

Three sets. One region at a time.

With three activities, a student can belong to none, one, two or all three. The central region belongs to every circle, so it affects each pairwise overlap as well as each individual total.

Fill the smallest, most specific region first: all three sets. Next find the pair-only regions by subtracting the centre from the stated pair totals. Only then calculate the single-set regions. This order prevents a student being counted twice within one circle.

In a survey of 24 students, 11 study Art (A), 9 study Biology (B), and 10 study Computing (C). Also, n(A ∩ B) = 5, n(A ∩ C) = 4, n(B ∩ C) = 3, and 2 study all three.

WORKED EXAMPLE

Complete the Venn diagram and find how many study none.

ξABC43532124
Numbers inside the regions are counts of students, not individual elements.

Fill from the centre out

  1. Put 2 in the all-three overlap.
  2. n(A ∩ B) = 5 includes those 2, so the A-and-B-only region is 5 − 2 = 3.
  3. Similarly, A-and-C-only = 4 − 2 = 2; B-and-C-only = 3 − 2 = 1.
  4. A only = 11 − 3 − 2 − 2 = 4. B only = 9 − 3 − 1 − 2 = 3. C only = 10 − 2 − 1 − 2 = 5.

Exactly one subject

4 + 3 + 5 = 12

Use only the three outer regions.

Exactly two subjects

3 + 2 + 1 = 6

Use the pair-only regions, excluding the centre.

Pause and explainIf 5 belong to A and B and 2 belong to all three, how many belong to A and B only?

Three. The pair total includes the two in the centre, so subtract 2 from 5.

05

Write a set. Describe a relationship.

A long list is not always the easiest way to describe a set. Instead, we can give a condition that every member must satisfy. This is called set-builder notation, and it should be read as a sentence.

The chosen number domain matters. The integers between 2 and 5 form a short list, but the real numbers between them also include values such as 2.1 and 3.75. When comparing sets, a subset statement asks whether every element of the first is also in the second.

Set-builder notation reads like a sentence

{x: …}

“The set of x such that …”

A = {x: x is a natural number} = {1, 2, 3, …}. The colon means “such that.”

{(x, y): …}

A set of ordered pairs

B = {(x, y): y = mx + c} describes points on a straight line. If m = 2 and c = 1, (0, 1) and (2, 5) belong to B.

a ≤ x ≤ b

An interval, including both endpoints

C = {x: a ≤ x ≤ b} contains all real x from a to b, unless a different number domain is given. For integer x with 2 ≤ x ≤ 5, C = {2, 3, 4, 5}.

{a, b, c, …}

A list with a continuing pattern

D = {a, b, c, …} lists elements directly. Make the continuation clear from the question’s context.

WORKED EXAMPLE

Decide whether one set is a subset of another.

P = {2, 4} and Q = {2, 4, 6}. Every element of P is in Q, so P ⊆ Q. But 6 ∈ Q and 6 ∉ P, so Q ⊈ P.

A subset circle sits entirely inside the other set. Every set is a subset of itself, and ∅ is a subset of every set.

Pause and explainIf P = {2, 4} and Q = {2, 4, 6}, why is Q not a subset of P?

The element 6 belongs to Q but not to P. One counterexample is enough to disprove the subset statement.

06

Your turn. Read, shade, count.

Use a small sketch while practising, even if the question only contains symbols. Write the universal set first, because “not in A” has no fixed meaning until we know the collection we are choosing from.

Read each word carefully: “only”, “both”, “exactly” and “at least” select different regions. After filling a counting diagram, add every separate region, including the outside, and check that the result equals the total population.

Sketch a Venn diagram on paper where useful. Label the universal set before you place any elements.

1

Let ξ = {1, 2, …, 10}, A = {2, 4, 6, 8, 10} and B = {2, 3, 5, 7}. Find A ∩ B, A ∪ B and A′.

Reveal worked answer
A ∩ B = {2}. A ∪ B = {2, 3, 4, 5, 6, 7, 8, 10}. A′ = {1, 3, 5, 7, 9}. Include shared elements only once in the union.
2

In a class of 30, 18 play football, 14 play cricket and 6 play both. How many play only football? How many play neither?

Reveal worked answer
Only football = 18 − 6 = 12. At least one sport = 18 + 14 − 6 = 26. Neither = 30 − 26 = 4.
3

Using the three-subject diagram, find n((A ∪ B) ∩ C).

Reveal worked answer
(A ∪ B) ∩ C means in C and also in A or B. Add A-and-C-only, B-and-C-only, and the centre: 2 + 1 + 2 = 5.
4

List E = {x: x is an integer and −2 ≤ x < 3}. State n(E).

Reveal worked answer
E = {−2, −1, 0, 1, 2}, so n(E) = 5. Include −2 because of ≤; exclude 3 because of <.
5

True or false: ∅ = {0}; {3} ⊆ {1, 3, 5}; 3 ∈ {1, 3, 5}.

Reveal worked answer
False; true; true. The empty set has zero elements, whereas {0} has one. {3} is a set, so use ⊆; 3 is an element, so use ∈.