# Unit 2: THE TEACHING OF SETS (ALGEBRA OF SETS)

##### This unit describes sets and operations on sets. It also explains how sets and operations on sets are taught.

1
THE TEACHING OF SETS (ALGEBRA OF SETS)
The Idea of Sets:
The concept of set occurs at all levels of mathematics, from the nursery stage to the University.
Before children enter school, they have experienced the collection of objects. They have seen a
congregation of people in the church, bundles of firewood, flocks of sheep, etc. These collections
give the idea of sets.
To introduce set to children, we use familiar physical objects such as seeds, stones, trees, shells,
pencils, houses etc. From the concrete object we use pictures of objects and geometrical shapes
like in fig 1.

Symbols must not be used in the early stages of the development of sets.
A set is a collection of things that are considered together in some way. We can speak of the set
of books in the school library, the set of people in a church, the set of JHS 1 students in a
particular school, the set of counting numbers less than 10.
A set made up of a book, stone, pencil, chalk, is an example of a set with many kinds of objects
but the set made up of boys in a class is a set of similar objects. You must not confused your
pupils with a set of objects and a group of objects. A set of objects can have one or no member
but a group of objects denotes more than one member. The members of a set need not to be of
the same kind i.e a set can have similar things or can contain any variety of objects but the
members of a group are of the same kind.
Set of triangles Set of bowls Set of flowers
Fig 1. Pictures of objects
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Naming Sets.
In writing down sets, we come across notations which are used everywhere. A set is denoted by a
capital letter. For example, we may talk of set A, set B and set C etc. The set is made up of a
number of objects. If an object belongs to a set it is called a member or an element of the set.
Suppose our set F is made up of Kofi, Kwame, Yaw, then we say that Kwame is a member of the
set F. 4 is in the set A, of counting numbers less than 10, therefore 4 is an element of the set A.
The symbol ∈ is used to indicate membership of a set. If we write 4 ∈ A, it implies 4 is an
element of set ‘A’ or 4 is contained in set ‘A’ or 4 belongs to set ‘A’.
However if we write 20 ∉ A, it implies 20 is not an element of set A. For example if
B = {a, b, c, d, e} then d ∈ B but f ∉ B.
Methods of describing a Set
A set may be described in two distinct ways as:- (a) Tabulation method (b) The rule method
Tabulation Method: In the tabulation method we indicate a set by listing the elements and
enclosing them in braces. The braces should be a curly bracket { }
Rule Method: In the rule method, we describe the set with a descriptive phrase which qualifies
only those objects belonging to the set. Usually we put the descriptive phrase under bracket.
Consider the following examples.
(i) A is the set of counting numbers less than 10
In the tabulation method, we have A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
In the rule method, we will write A = {all counting numbers less than 10}
(ii) V is the set of vowels in the English alphabet
In the tabulation method, we have V = {a, e, i, o, u}
In the rule method we have V = {all vowels in the English alphabet}
Set builder Notation: In describing a set, we very often use set builder notations to make our
description short.
For example the set A = {all counting numbers less than 10} may be written using set builder
notation as A = {x: x ∈ N, x < 10} or A= {x I x ∈ N, x < 10}.
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Finite and Infinite Set
A set is called finite, if it has a finite number of elements, i.e if we can enumerate the elements of
the set in some order and then count the elements until a last one is reached.
eg. Set A = {1, 2, 3, 4, 5, 6, 7, 8, 9} is a finite set.
If we cannot count all the elements in the set, then the set is said to be infinite. The set of
counting numbers is infinite set since we cannot find the last counting number. In the tabulation
form we indicate an infinite set by listing some of the elements and then use dots to indicate that
there are other elements. eg. Set A = {1, 2, 3, 4……………} indicates the set of counting
numbers which is infinite.
Equivalent Set
Two sets are said to be equivalent if they have the same number of elements. In such situation it
is possible to associate the elements of one set with the elements of the other in such a way
elements of each set is associated with exactly one element of the other.
Let us consider set A = {1, 2, 3, 4, 5, 6} and set B = {a, b, c, d, e, f}
Set A = {1 2 3 4 5 6 }
↓ ↓ ↓ ↓ ↓ ↓
B= {a b c d e f }
Set A here is equivalent to set B since they have the same number of elements which is 6. To
introduce equivalent sets to children we use matching. Consider the sets below in fig 2
(i) (ii)
Fig 2: Matching of set
In fig 2 (i), the sets match exactly i.e. the two sets are in a one – to – one correspondence and this
indicates equivalent sets. In fig 2 (ii), the sets do not match exactly, there is no one to one
4
correspondence between the sets. We see that the first set has more members than the second set
hence the two set are not equivalent.
With the activities such as (i) matching of sets (ii) making equivalent sets (ii) adding more
objects to a set etc. teachers may help children to recognize sets that are equivalent.
Empty Set (Null Set)
Empty set sometimes called a null set is the set that has no element eg. The set of all human
beings with eight legs is an empty set. The empty set is indicated by { } and denoted by Ø. ie Ø
= { }. Note that {Ø} is not an empty set, it is a set that contains one element.
To develop the idea of an empty set; teachers must help children to make empty set in the class
room. In making the empty set, the teacher could say make a set of bottle tops, make a set of
books, make a set of children with three legs in the classroom. See fig 14.3
The children will realize that the set of boys with three legs
contains no member. This indicates empty set. Pictures { }
such as those in fig 4 can help children understand the empty set.
(i) (ii)
Subset and power set
The set A is a subset of the set B, if every element of A is an element of B. Consider the set
P = {1, 2, 3, 4} and the set Q = {1, 2, 3, 4, 5, 6, 7, 8}. We see that every element of P is an
element in Q. This means that P is a subset of Q. The symbol for subset is ⊂ i. e P ⊂ Q implies P
is a subset of Q.
Fig 4
Set of bottle tops Set of books Set of boys with three legs
Fig 3
5
Note that (i) Every set is a subset of itself i.e A ⊂ A and B ⊂ B. (ii) The empty set is a
subset of any set i.e { } ⊂ A and { } ⊂ B
Generating subsets and power set of a given set
If set A = {a}, the possible subsets are:
Let the pupils write (i) the empty set { } since the empty set is a subset of any set and (ii) {𝑎},
the set it itself since every set is a subset of itself. The subset of A are { } and {𝑎}
The number of element in set A is 1 and the number of subsets are 2=2
1
The power set of a given set is the set of all the possible subsets of the given set. The power
set of A written as P(A) is { { }, {𝑎}}. The cardinality of P(A) is 21 = 2
If set B = {a, b} the possible subset are:
Let the pupils write (i) the empty set { } since the empty set is a subset of any set and (ii) {𝑎, 𝑏},
the set it itself since every set is a subset of itself. (iii) Other subset of B are {a} and {b}. All the
possible subset of B are { } {𝑎}, {b} and {𝑎, 𝑏}. The number of element in set B is 2 and the
number of subsets are 4=22
. The power set of B written as P(B) is {{ } {𝑎}, {b},{𝑎, 𝑏}}. The
cardinality of P(B) is 22 =4
If set C = {a, b, c} the possible subset are:
Let the pupils write
(i) the empty set { } since the empty set is a subset of any set;
(ii) the set it itself {𝑎, 𝑏, 𝑐}, since every set is a subset of itself;
(iii) Other subset of C are {a}, {b}, {c}, {a, b}, {a, c}, {b, c}
All the possible subset of C are { } {𝑎}, {b} and {𝑎, 𝑏}. The number of element in set C is3 and
the number of subsets are 8=23
. The power set of C written as P(C) is {{ } {𝑎}, {b}, {c}, {𝑎, 𝑏},
{a, c}, {b, c},{𝑎, 𝑏, 𝑐} } . The cardinality of P(C) is 23 =8
For an empty set the number element is zero. The only subset of the empty set is the empty set
itself. Therefore the empty set has one subset and one power set. The number of subset or the
cardinality of the power set is 20 = 1
We will see that for one member we have two subsets, for two elements we have 4 subsets and
for three members we have eight subsets. This means that 1 →2 = 21
, 2 → 4 = 22
, 3 → 8 = 23
6
Therefore for four members we will have 24 = 16 subsets and for five members we will have25 =
32 subset. In general if we have n members or elements in a set, number of subsets or the
cardinality of the power set will be given by 2n
.
Equal Set
Two sets A and B are equal if they have
(i) The same number of elements (ii) Similar elements.
In such a case, A ⊂ B and B ⊂ A. Consider A = {1, 2, 3, 4, 5} and B = {1, 2, 3, 4, 5, 6} C = {5,
4, 3, 2, 1}, D = {a, b, c, d, e}. We can see that A = C, A ≠ D but A is equivalent to D.
Disjoint Sets
Two sets A and B are disjoint if and only if A and B have no element in common. For example
Set A = {1, 2, 3, 4} and set B = {5, 6, 7, 8, 9} are disjoint since they do not have any element in
common. However, the set P = {1, 2, 3, 4, 5, 6} and the set Q = {5, 7, 8, 9, 10} are not disjoint
since they have a common element 5.
Universal Set (U)
The universal set is the overall set of elements with which we are concerned throughout our
discussion. The universal set in any discussion is therefore the totality of members under
consideration. The universal set can be the set of counting numbers. In other times, the universal
set can be counting numbers less than 11 i.e {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Operation on Sets
The important operations on sets are
(a) The complement (b) The union (c) the intersection (d) the difference
The Complement
The complement of a set A is the set of elements of the universal set that do not belong to A. We
may denote the complement of set A by A’. For example if our universal set N = {1, 2, 3, 4, 5, 6,
7, 8, 9, 10}, and the set A = {1, 3, 5, 7, 10} then A’ = {2, 4, 6, 8, 9} is complement of the set A.
7
The Union of Sets
To help pupils get the real meaning of the operation of the Union of sets, go through the
following activities with the pupils.
Consider the chart in fig 5 which shows the students in a JHS class who passed mathematics,
English and science.
Let M be the set of students who passed mathematics.
E be the set of students who passed English
S be the set of students who passed science
NAME MATHEMATICS ENGLISH SCIENCE
Mary P F P
John P F P
Kofi P P P
Peter F P F
Mensah P P F
Paul F F F
Akua P P P
Fig. 5
Note: P = Pass F= Failed
Question: List the members in (i) Set M (ii) Set E (iii) Set S
(ii) E = {Kofi, Peter Mensah, Akua}
(iii) S= {Mary, John, Kofi, Akua}
Question
1. List the members of the set A who passed mathematics or English or both
2. List the members of the set B who passes mathematics or science or both.
1. A = {Mary, John, Kofi, Peter, Adwoa, Mensah, Akua}
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2. B= {Mary, John, Kofi, Adowa, Mensah Akua}
The set A was obtained by writing down the names of all the students who are in the set M or set
E or both set M and E and the set B was from the names of all the students in the set M or set S
or both. We call the set A the union of set M and E and set B the union of set M and S. The
symbol ∪ is used to represent union and we can represent the statement that A is the union of the
set M and E as A = M ∪ E and the statement B is the union of set M and S as B = M ∪ S.
The union of two sets P and Q is the set of elements that are in P or Q or both i.e the elements in
the union belongs to at least one of the sets P and Q. As seen above, we denote union by the
symbol ∪.
P ∪ Q ⟹ P Union Q. So P ∪ Q = {x: x ∈ P or x ∈ Q}. For example.
If P = {a, b, c, d, e} Q = {b. d. f. g. h} then P ∪ Q = {a, b, c, d, e, f, g, h}
Example
If X = {prime factors of 56}, Y = {prime factors of 210} Find the elements of X ∪ Y.
Solution
Guide the students to find the factors of 56 and 210 using the factor tree and write them as set X
and Y respectively.
56 210

∴ X ∪ Y = {2, 3, 5, 7}
X = {2, 7}, Y = {2, 3, 5, 7}
Assist them to select the elements that are contained in X or Y or both as the set X∪Y
2 28
2 14
2 7
2 105
3
35
5 7
9
The Intersection of Sets
From the activities on the charts in fig 5, we may look for the students who passed in both
mathematics and science, the students who passed in both mathematics and English. If C is the
students who passed in both mathematics and English and D the set of students who passed in
mathematics and science then let children list the members in:
(i) Set C and (ii) Set D
Answer: C = {Kofi, Mensah, Akua} D = {Mary, John, Kofi, Akua}
The set C is the intersection of the set M and E and the set D is the intersection of the set M and
S. The intersection of two sets P and Q, is the set of elements that is contained in both P and Q.
We denote intersection by the symbol ∩. P ∩ Q ⟹ P intersection Q.
∴ P ∩ Q = {x: x ∈ P and x ∈ Q}
For example, if A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6, 7, 8}, C = {6, 7, 8, 9}, then
A ∩ B = {3, 4, 5} A ∩ C = { } B ∩ C = {6, 7, 8}
Example
List the members of each of the sets.
B = {whole numbers from 20 to 30}
D = {factors of 63}
List the members of (i) B ∩ D (ii) B ∪ D
Solution
Guide the students to write/list the whole numbers as set B and factors of 63 as set D.
B = {20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30} D = {1, 3, 7, 9, 21, 63}
For (i) assist them to select the elements that are contained in both B and D as the set B∩D
B ∩ D = {21}
For (ii) assist them to select the elements that are contained in B or D or both as the set B∪D
B ∪ D = {1, 3, 7, 9, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 63}
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The difference of Sets
Another operation on any two sets A and B is the difference “A minus B”, which “subtracts”
from set A all elements which are in set B. for example, if set A = {1, 2, 3, 4, 5, 6} and the set
B= {3, 5, 6}, then, “A minus B” is given by set A – B = {1, 2, 4}
Definition
Suppose A and B are two sets, then the difference, A minus B, denoted by A – B is the set of
elements that are found in A but not in B. that is, A – B = {x: x ∈ A and x ∉ 𝐵}.
Note that A – B is also called the relative complement of B to A.
Venn Diagram
Venn diagrams are pictorial representation of the relations among subsets of a set. Very often, in
a Venn diagram, we represent the universal set by a rectangular region and the subsets of the
universal set by closed circular/oval region within the rectangle.
If P is subset of the universal set ∪ and P’ the complement of P then we can illustrate this
information in fig 6 (i)
3 5
1 7 9
(i) (ii)
Fig 6 Venn diagram showing complement
For example if ∪ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10) and P = {1, 3, 5, 7, 9} the information can be
represented on the Venn diagram as in fig 6 (ii) above
Consider the Venn diagrams below.
P
P’
2
P
P’
4
10 6 8
11
(i) A ⊂ B (ii) A ∩ B = { } (iii) A ∩ B is the shaded region
A∩B = A, A∪B = B
Diagram (i) indicates that set A is contained in set B, this implies that set A is a subset of set B.
Diagram (ii) shows that set A and B are disjoint. This implies that A ∩ B = { }. Diagram (iii)
indicates that set A and set B have some elements in common. Their common set is given by the
Properties of operations on Sets
Commutative property
Given the sets A = {2, 3, 4} and B = {1, 3, 5, 7}. Find:
(i) A ∩ B
(ii) B ∩ A
(iii) A∪ B
(iv) B ∪ A
(v) what can you say about A ∩ B and B ∩ A
(vi) what can you say about A∪ B and B ∪ A
Solution
(i) Ask the students to write down the intersecting sets
A ∩ B = {2, 3, 4}∩{1, 3, 5, 7}
Assist them to identify and write the set of all the common elements that are found in
set A and B as A intersection B. ie A ∩ B = {3}
(ii) Ask the students to write down the intersecting sets
B ∩ A = {1, 3, 5, 7} ∩{2, 3, 4}
U
B
A
B
A
U U
A B
12
Assist them to identify and write the set of all the elements that are found in both sets
as B intersection A. ie B ∩ A = {3}
(v) The sets A ∩ B and B ∩ A are the same. Therefore A ∩ B = B ∩ A, the intersection
of sets is commutative.
(iii) Ask the students to write down the two sets as
A∪ B = {2, 3, 4}∪{1, 3, 5, 7}
Assist them to identify and write the set of all the elements that are found in A or B or
both sets as A union B. ie A∪ B = {1, 2, 3, 4, 5, 7}
(iv) Ask the students to write down the two sets as
B ∪ A = {1, 3, 5, 7} ∪{2, 3, 4}
Assist them to identify and write the set of all the elements that are found in A or B or
both sets as A union B. ie A∪ B = {1, 2, 3, 4, 5, 7}
The sets A∪ B and B ∪ A are the same. Therefore A ∪ B = B ∪ A, the union of sets
is commutative.
Associative property
Given the three sets A = {2, 3, 4},
B = {1, 3, 5, 7} and C = {3, 4, 5, 6}
i) Find (A ∪ B) ∪ C and A ∪ (B ⋃ C)
ii) What can you say about (A ∪ B)∪C and A ⋃ (B ∪ C)?
iii) Find (A ∩ B) ∩ C and A ∩ (B ∩ C)
iv) What can you say about (A ∩ B) ∩ C and A ∩ (B ∩ C)?
Solution
i) For (A ∪ B) ∪ C, guide the students/pupils to find (A ∪ B) first, ie, (A ∪ B) = {1, 2,
3, 4, 5, 7}. The second step is to guide the students to find the union of the resulting
set and C ie (A ∪ B) ∪ C = {1, 2, 3, 4, 5, 6, 7}
Similarly for A ∪ (B ⋃ C), guide the students/pupils to first find (B ⋃ C), ie (B ⋃ C)
= {1, 3, 4, 5, 6, 7}. The second step is to guide the students to find the union of the A
and resulting set ie A ∪ (B ⋃ C) = {1, 2, 3, 4, 5, 6, 7}
ii) In conclusion, from the results in i) above, it can be seen that (A ∪ B) ∪ C = A ∪ (B
∪ C) = {1, 2, 3, 4, 5, 6, 7}. This shows that the union of sets is associative.
13
iii) For (A ∩ B) ∩ C, guide the students/pupils to find (A ∩ B) first, ie, (A ∩ B)= {3}.
The second step is to guide the students to find the intersection of the resulting set
and C ie (A ∩ B) ∩ C = {3} ∩ {3, 4, 5, 6} = {3}
Similarly for A ∩ (B ∩ C), guide the students/pupils to first find (B ∩ C), ie (B ∩ C)
= {3, 5}. The second step is to guide the students to find the intersection of the A and
resulting set, ie, A ∩ (B ∩ C) = {2, 3, 4} ∩ {3, 5} = {3}
iv) In conclusion, from the results in iii) above, it can be seen that (A ∩ B) ∩ C = A ∩ (B
∩ C)= {3}. Therefore intersection of sets is associative.
Distributive property
Let A = {2, 3, 4}, B = {1, 3, 5, 7} and C = {3, 4, 5, 6}
i) Find A ∩ (B ∪ C) and (A ∩ B) ∪ (A ∩ C)
ii) What can you say about A ∩ (B ∪ 𝐶) and (A ∩ B) ∪ (A ∩ C)?
iii) Find A ∪ (B ∩ C) and (A ∪ B) ∩ (A ∪ C)
iv) What can you say about A ∪ ((B ∩ C) and (A ∪ B) ∩ (A ∪ C)?
Solution
i) For A ∩ (B ∪ C), guide the students/pupils to find (B ∪ C) first, ie, (B ∪ C)= {1, 3, 5,
7}∪ {3, 4, 5, 6} = {1, 3, 4, 5, 6, 7}. The second step is to guide the students to find
the intersection of set A and the resulting set ie A ∩ (B ∪ C)= {2, 3, 4}∩ {1, 3, 4, 5,
6, 7} = {3, 4}
Similarly for (A ∩ B) ∪ (A ∩ C), guide the students/pupils to first find (A ∩ B) and
(A ∩ C) ie (A ∩ B) = {3} and (A ∩ C) = {3, 4}. The second step is to guide the
students to find the union of the (A ∩ B) and (A ∩ C) ie (A ∩ B) ∪ (A ∩ C), = {3} ∪
{3, 4} ={3, 4}.
ii) In conclusion, from the results in i) above, it can be seen that A ∩ (B ∪ C) = (A ∩ B)
∪ (A ∩ C) ={3, 4}. This shows that the intersection of sets is distributive over the
union of sets.
iii) For A ∪ (B ∩ C), guide the students/pupils to find (B ∩ C) first, ie, (B ∩ C) = {3, 5}.
The second step is to guide the students to find the union of A and the resulting set ie
A ∪ (B ∩ C)= {2, 3, 4}∪ {3, 5}= {2, 3, 4, 5}
14
Similarly for (A ∪ B) ∩ (A ∪ C), guide the students/pupils to first find (A ∪ B) and
(A ∪ C), ie (A ∪ B) = {1, 2, 3, 4, 5, 7} and (A ∪ C) = {2, 3, 4, 5, 6}. The second step
is to guide the students to find the intersection of the (A ∪ B) and (A ∪ C), ie, (A ∪
B) ∩ (A ∪ C)= {1, 2, 3, 4, 5, 7}∩ {2, 3, 4, 5, 6}= {2, 3, 4, 5}.
iv) In conclusion, from the results in iii) above, it can be seen that A ∪ (B ∩ C) = (A ∪
B) ∩ (A ∪ C). Therefore union (∪) of sets is distributive over intersection (∩) of sets.

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