E-Math notes · Chapter 8

Set Language and Notation: O-Level E-Math Notes

Sets give you a precise language for groups of things, and Venn diagrams let you see them. In this chapter you learn the notation, combine sets with union, intersection and complement, and fill and shade Venn diagrams.

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Chapter
8

Figure 1 ·

A \cup B

3 outcomes · 9 worked examples · every step shown ✦

Set language and notation

Learn the symbols as a language: each one is a short way to say something you already understand.

Union, intersection and complement

$n(A \cup B) = n(A) + n(B) - n(A \cap B)$

Venn diagrams

Every element of $\mathscr{E}$ lives in exactly one of four regions: $A$ only, both, $B$ only, or outside both.

By the end of this chapter

Use set language and the notation for elements, subsets, the empty set, the universal set, number of elements and complement.
Find the union and intersection of two sets and count the union without counting the overlap twice.
Draw, fill and shade Venn diagrams with a test letter in every region.

Start here

Maths In Real Life

Three real situations where sets and Venn diagrams count people without counting anyone twice. The pictures are drawn first, then the Venn diagram is placed on top by script so the numbers match the problem exactly.

$The after-school clubs: two overlapping circles drawn on top of the scene with the counts in each region$

Maths In Real Life 1

The after-school clubs

A Venn diagram counts people once, even when they belong to two groups.

In a class of $36$ students, $20$ are in the choir ( $A$ ), $15$ are in the badminton team ( $B$ ) and $8$ are in both.

(a)Find how many are in at least one of the two groups.
(b)Find how many are in neither group.
(c)Find how many are in the choir group only.

The four regions of the survey.

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Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(a)

\displaystyle n(A \cup B) = n(A) + n(B) - n(A \cap B)

The counting formula: the people in BOTH sets were counted twice, so subtract them once.

\displaystyle n(A \cup B) = 20 + 15 - 8 = 27

Substitute: choir is

A

, badminton is

B

(b)

\displaystyle n(\mathscr{E}) - n(A \cup B) = 36 - 27 = 9

People in NEITHER set: everyone minus those in at least one set.

(c)

\displaystyle n(A \text{ only}) = 20 - 8 = 12

Take the overlap out of set

A

Final answer

(a)27 are in at least one group.

(b)9 are in neither group.

$The lunch-time survey: two overlapping circles drawn on top of the scene with the counts in each region$

Maths In Real Life 2

The lunch-time survey

Surveys use overlapping sets to avoid counting the same person twice.

At a hawker centre, $50$ customers are surveyed. $31$ ordered noodles ( $A$ ), $26$ ordered soup ( $B$ ) and $14$ ordered both.

(a)Find how many are in at least one of the two groups.
(b)Find how many are in neither group.
(c)Find how many are in the noodles group only.

The four regions of the survey.

Show full solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(a)

\displaystyle n(A \cup B) = n(A) + n(B) - n(A \cap B)

The counting formula: the people in BOTH sets were counted twice, so subtract them once.

\displaystyle n(A \cup B) = 31 + 26 - 14 = 43

Substitute: noodles is

A

, soup is

B

(b)

\displaystyle n(\mathscr{E}) - n(A \cup B) = 50 - 43 = 7

People in NEITHER set: everyone minus those in at least one set.

(c)

\displaystyle n(A \text{ only}) = 31 - 14 = 17

Take the overlap out of set

A

Final answer

(a)43 are in at least one group.

(b)7 are in neither group.

$The apps teenagers use: two overlapping circles drawn on top of the scene with the counts in each region$

Maths In Real Life 3

The apps teenagers use

The overlap of two groups is where both memberships are true.

Of $40$ teenagers surveyed, $28$ use a messaging app ( $A$ ), $22$ use a video app ( $B$ ) and $15$ use both.

(a)Find how many are in at least one of the two groups.
(b)Find how many are in neither group.
(c)Find how many are in the messaging group only.

The four regions of the survey.

Show full solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(a)

\displaystyle n(A \cup B) = n(A) + n(B) - n(A \cap B)

The counting formula: the people in BOTH sets were counted twice, so subtract them once.

\displaystyle n(A \cup B) = 28 + 22 - 15 = 35

Substitute: messaging is

A

, video is

B

(b)

\displaystyle n(\mathscr{E}) - n(A \cup B) = 40 - 35 = 5

People in NEITHER set: everyone minus those in at least one set.

(c)

\displaystyle n(A \text{ only}) = 28 - 15 = 13

Take the overlap out of set

A

Final answer

(a)35 are in at least one group.

(b)5 are in neither group.

Supports N8.1

Set language and notation

A set is a collection of things, called its elements. We write a set inside curly brackets, for example $\{a, b, c\}$ . The order does not matter and an element is never written twice.

Links To

For a quick refresher on sets, see the free E-Math revision page on sets and matrices

The Core Idea

Learn the symbols as a language: each one is a short way to say something you already understand.

$\in$ means "is an element of", $n(A)$ counts the elements, $A'$ is everything NOT in $A$ , and $\mathscr{E}$ is the universal set: everything we are talking about.

Case table 1: the notation of sets
Symbol	Read as	Meaning
$\mathscr{E}$	the universal set	everything under discussion
$\varnothing$ or $\{\ \}$	the empty set	a set with no elements
$x \in A$	$x$ is an element of $A$	$3 \in \{1, 2, 3\}$
$x \notin A$	$x$ is not an element of $A$	$5 \notin \{1, 2, 3\}$
$n(A)$	the number of elements in $A$	$n(\{4, 6, 8\}) = 3$
$A'$	the complement of $A$	all elements of $\mathscr{E}$ not in $A$
$A \cup B$	the union of $A$ and $B$	in $A$ or $B$ (or both)
$A \cap B$	the intersection of $A$ and $B$	in both $A$ and $B$
$A \subseteq B$	$A$ is a subset of $B$	every element of $A$ is in $B$
$A \nsubseteq B$	$A$ is not a subset of $B$	some element of $A$ is not in $B$
$A \subset B$	$A$ is a proper subset of $B$	a subset of $B$ that is not $B$ itself
$A \not\subset B$	$A$ is not a proper subset of $B$	either not a subset, or equal to $B$

SGSchoolKaki Tip

Say it out loud: " $A \cap B$ " is " $A$ and $B$ ", " $A \cup B$ " is " $A$ or $B$ ". The cap looks like an upside-down cup: the cup holds everything, the cap keeps only what is common.

Example 1.1

Listing a set and counting it

List the elements of the following set and write down $n(A)$ , the number of elements: $A = \{x : x \text{ is an integer},\ \allowbreak 3 < x \le 9\}.$

Show solutionHide solution

The set

A

is described by a rule. List every value that satisfies it.

Read the rule slowly, then test each candidate.

\displaystyle A = \{x : x \text{ is an integer},\ \allowbreak 3 < x \le 9\}

The rule.

\displaystyle A = \{4,\ \allowbreak 5,\ \allowbreak 6,\ \allowbreak 7,\ \allowbreak 8,\ \allowbreak 9\}

List the elements in order, once each, inside curly brackets.

\displaystyle n(A) = 6

n(A)

is the number of elements in the set: count them.

Final answer

A = \{4,\ \allowbreak 5,\ \allowbreak 6,\ \allowbreak 7,\ \allowbreak 8,\ \allowbreak 9\}

and

n(A) = 6

Your Turn 1.1

List the elements of the following set and write down $n(A)$ , the number of elements: $A = \{x : x \text{ is a positive} \allowbreak \text{ multiple of } 4,\ \allowbreak x \le 30\}.$

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The set

A

is described by a rule. List every value that satisfies it.

Read the rule slowly, then test each candidate.

\displaystyle A = \{x : x \text{ is a positive} \allowbreak \text{ multiple of } 4,\ \allowbreak x \le 30\}

The rule.

\displaystyle A = \{4,\ \allowbreak 8,\ \allowbreak 12,\ \allowbreak 16,\ \allowbreak 20,\ \allowbreak 24,\ \allowbreak 28\}

List the elements in order, once each, inside curly brackets.

\displaystyle n(A) = 7

n(A)

is the number of elements in the set: count them.

Final answer

A = \{4,\ \allowbreak 8,\ \allowbreak 12,\ \allowbreak 16,\ \allowbreak 20,\ \allowbreak 24,\ \allowbreak 28\}

and

n(A) = 7

Example 1.2

Is it an element or not?

Here is a set described by a rule: $B = \{x : x \text{ is a prime number},\ \allowbreak x < 20\}.$ Write each of $7 \in B, 9 \in B, 13 \in B$ and decide whether it is true or false.

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\displaystyle B = \{x : x \text{ is a prime number},\ \allowbreak x < 20\}

The rule for the set.

\displaystyle B = \{2,\ \allowbreak 3,\ \allowbreak 5,\ \allowbreak 7,\ \allowbreak 11,\ \allowbreak 13,\ \allowbreak 17,\ \allowbreak 19\}

List its elements first.

\displaystyle 7 \in B

7

is in the list, so

7 \in B

(true).

\displaystyle 9 \notin B

9

is NOT in the list, so the statement

9 \in B

is false.

\displaystyle 13 \in B

13

is in the list, so

13 \in B

(true).

Final answer

7 \in B

: true;

9 \notin B

: true, so the statement

9 \in B

is false;

13 \in B

: true

Your Turn 1.2

Here is a set described by a rule: $B = \{x : x \text{ is a positive} \allowbreak \text{ multiple of } 6,\ \allowbreak x \le 40\}.$ Write each of $18 \in B, 20 \in B, 36 \in B$ and decide whether it is true or false.

Show worked solutionHide worked solution

\displaystyle B = \{x : x \text{ is a positive} \allowbreak \text{ multiple of } 6,\ \allowbreak x \le 40\}

The rule for the set.

\displaystyle B = \{6,\ \allowbreak 12,\ \allowbreak 18,\ \allowbreak 24,\ \allowbreak 30,\ \allowbreak 36\}

List its elements first.

\displaystyle 18 \in B

18

is in the list, so

18 \in B

(true).

\displaystyle 20 \notin B

20

is NOT in the list, so the statement

20 \in B

is false.

\displaystyle 36 \in B

36

is in the list, so

36 \in B

(true).

Final answer

18 \in B

: true;

20 \notin B

: true, so the statement

20 \in B

is false;

36 \in B

: true

Think About It

Is the empty set

\varnothing

a subset of

\{a, b\}

? Is

\{a, b\}

a proper subset of itself?

Show the worked answerHide the worked answer

\displaystyle S = \{\text{a},\ \allowbreak \text{b}\}

The set.

A set with

2

elements has

2^{2}

subsets, because each element is either IN a subset or OUT of it.

Two choices for each element.

\displaystyle \varnothing \subseteq S

The empty set has NO element that could be missing from

S

: every element of

\varnothing

is in

S

\displaystyle \{\text{a},\ \allowbreak \text{b}\} \not\subset S

S

is not a PROPER subset of itself:

\{\text{a},\ \allowbreak \text{b}\}

S

itself, so it is not a PROPER subset.

Final answerYes to the first: to fail,

\varnothing

would need an element that is not in the set, and it has none. No to the second: a proper subset must be smaller than the set.

Example 1.3

Subsets and proper subsets

$S = \{\text{a},\ \allowbreak \text{b},\ \allowbreak \text{c}\}$ . For each statement, decide whether it is true: $\{\text{a},\ \allowbreak \text{b}\} \subset S$ , $\{\text{a},\ \allowbreak \text{b},\ \allowbreak \text{c}\} \subset S$ , $\varnothing \subseteq S$ , $\{\text{a},\ \allowbreak \text{d}\} \subseteq S$ . How many subsets does $S$ have?

Show solutionHide solution

\displaystyle S = \{\text{a},\ \allowbreak \text{b},\ \allowbreak \text{c}\}

The set.

A set with

3

elements has

2^{3}

subsets, because each element is either IN a subset or OUT of it.

Two choices for each element.

\displaystyle \{\text{a},\ \allowbreak \text{b}\} \subset S

Two of the three elements: a subset of

S

that is not

S

itself.

\displaystyle \{\text{a},\ \allowbreak \text{b},\ \allowbreak \text{c}\} \not\subset S

All of the elements:

\{\text{a},\ \allowbreak \text{b},\ \allowbreak \text{c}\}

S

itself, so it is not a PROPER subset.

\displaystyle \varnothing \subseteq S

No elements: every element of

\varnothing

is in

S

\displaystyle \{\text{a},\ \allowbreak \text{d}\} \nsubseteq S

It contains

d

\{\text{a},\ \allowbreak \text{d}\}

has an element that is not in

S

Final answer

\{\text{a},\ \allowbreak \text{b}\} \subset S

;

\{\text{a},\ \allowbreak \text{b},\ \allowbreak \text{c}\} \not\subset S

;

\varnothing \subseteq S

;

\{\text{a},\ \allowbreak \text{d}\} \nsubseteq S

; 8 subsets in all

Your Turn 1.3

$S = \{\text{p},\ \allowbreak \text{q},\ \allowbreak \text{r},\ \allowbreak \text{s}\}$ . For each statement, decide whether it is true: $\{\text{p},\ \allowbreak \text{q},\ \allowbreak \text{r}\} \subset S$ , $\{\text{p},\ \allowbreak \text{q},\ \allowbreak \text{r},\ \allowbreak \text{s}\} \subset S$ , $\varnothing \subset S$ , $\{\text{p},\ \allowbreak \text{t}\} \subseteq S$ . How many subsets does $S$ have?

Show worked solutionHide worked solution

\displaystyle S = \{\text{p},\ \allowbreak \text{q},\ \allowbreak \text{r},\ \allowbreak \text{s}\}

The set.

A set with

4

elements has

2^{4}

subsets, because each element is either IN a subset or OUT of it.

Two choices for each element.

\displaystyle \{\text{p},\ \allowbreak \text{q},\ \allowbreak \text{r}\} \subset S

Three of the four elements: a subset of

S

that is not

S

itself.

\displaystyle \{\text{p},\ \allowbreak \text{q},\ \allowbreak \text{r},\ \allowbreak \text{s}\} \not\subset S

All of the elements:

\{\text{p},\ \allowbreak \text{q},\ \allowbreak \text{r},\ \allowbreak \text{s}\}

S

itself, so it is not a PROPER subset.

\displaystyle \varnothing \subset S

No elements: a subset of

S

that is not

S

itself.

\displaystyle \{\text{p},\ \allowbreak \text{t}\} \nsubseteq S

It contains

t

\{\text{p},\ \allowbreak \text{t}\}

has an element that is not in

S

Final answer

\{\text{p},\ \allowbreak \text{q},\ \allowbreak \text{r}\} \subset S

;

\{\text{p},\ \allowbreak \text{q},\ \allowbreak \text{r},\ \allowbreak \text{s}\} \not\subset S

;

\varnothing \subset S

;

\{\text{p},\ \allowbreak \text{t}\} \nsubseteq S

; 16 subsets in all

SGSchoolKaki Tip

Do not mix up $\in$ and $\subseteq$ . $3 \in A$ says the element $3$ is in the set; $\{3\} \subseteq A$ says the set $\{3\}$ is inside $A$ .

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Notes version 1.0 · Last updated 3 Nov 2026 · Chapter 8 of 19