# Set elements || Understand the elements of set with examples

In this post we will learn about elements or objects of sets with solved examples.

## What are set elements?

The objects or members present inside the set is called set elements.

Set elements are present inside the curly bracket that represent any particular set.

### Examples of set elements

(i) E = { 2, 4, 6, 8, 10 }

Here;
E is the set name.
2, 4, 6, 8 and 10 are the set elements.

(ii) Y = { “Red”, “Orange”, “Yellow”, “Green”}

Here,
Y is the set name.
Red, Orange, Yellow and Green are the set elements.

(iii) A = { -3, 0, 9, 15}

Here A is the set name.
-3, 0, 9 and 5 are the set elements.

## Representing set elements

We use the symbol epsilon ” \mathtt{\ \epsilon \ } ” to show that the given element is the part of the set.

For Example, consider the below set.

X = { 9, 19, 25, 30 }

Here;
X is the name of the set.
9, 19, 25 and 30 are the elements of set.

We can say that;
9 \mathtt{\ \epsilon \ } X, It means element 9 belongs to X.

Similarly we can write this for all the elements of X.
19 \mathtt{\ \epsilon \ } X ;
25 \mathtt{\ \epsilon \ } X ;
30 \mathtt{\ \epsilon \ } X ;

Remember the significance of above symbol as it is directly asked in the examinations.

### What does symbol \mathtt{\notin } mean in set theory?

The symbol \mathtt{\notin } represents that the element doesn’t belongs to a particular set.

For Example, consider the below set.
X = { 5, 19, 25, 54}

We know that number 60 doesn’t belongs to the set X.
This statement can be expressed as: 60 \mathtt{\notin } X.

## What is size of a set?

The number of elements present in a given set is known as size of a set.

For Example, consider the below set;
A = { 19, 23, 51, 64, 71 }

Note that there are 5 elements in the set; 19, 23, 51, 64 and 71.
Hence, the size of the set is 5.

Similarly, consider the below set Y’
Y = { 2, 4, 6, 8, 10, 12 . . . . . 2n}

Here Y is the never ending set of even numbers.
The number of elements in set Y is infinite.
Hence, the size of set is also infinite.

## Solved examples of elements of Set

(01) Given below is the set A.
A = { 99, -33, 21, 15, -73, 21 }
List down all the elements/members of the set.

Solution
The elements are set are: 99, -33, 21, 15, -73 and 21.

(02) List down all the elements of the below set X
X = { A, P, L, E }

Solution
The elements of the set are: A, P, L, E

(03) Given below are statements regarding set A. Check is the statement is True or False.

A = { -2, 6, -5, 10, -15, 20 }

(i) 6 \mathtt{\ \epsilon \ } A
(ii) 5 \mathtt{\ \epsilon \ } A
(iii) 20 \mathtt{\ \epsilon \ } A
(iv) -15 \mathtt{\ \epsilon \ } A
(v) 7 \mathtt{\ \epsilon \ } A

Solution

(i) 6 \mathtt{\ \epsilon \ } A
The statement is correct as 6 is an element of set A.

(ii) 5 \mathtt{\ \epsilon \ } A
The statement is incorrect as 5 is not an element of A.

(iii) 20 \mathtt{\ \epsilon \ } A
It’s a correct statement as element 20 belongs to set A.

(iv) -15 \mathtt{\ \epsilon \ } A
It’s a correct statement.

(v) 7 \mathtt{\ \epsilon \ } A
This statement is False.

(04) Given below is the set Y.
Y = { \mathtt{\frac{1}{2} ,\ \frac{1}{6} ,\ \frac{1}{18} ,\ \frac{1}{36}} }

Check if the below statements are correct or not.

(i) \mathtt{\frac{1}{2} \ \notin \ Y}
(ii) \mathtt{\frac{1}{3} \ \epsilon \ Y}
(iii) \mathtt{\frac{1}{18} \ \epsilon \ Y}
(iv) \mathtt{\frac{1}{7} \ \notin \ Y}
(v) \mathtt{\frac{1}{36} \ \epsilon \ Y}

Solution

(i) \mathtt{\frac{1}{2} \ \notin \ Y}

This is incorrect statement.
1/2 is an elements belongs to set A.
So the correct expression is; \mathtt{\frac{1}{2} \ \epsilon \ Y}

(ii) \mathtt{\frac{1}{3} \ \epsilon \ Y}
This is an incorrect statement.
1/3 is not the member of set Y.

(iii) \mathtt{\frac{1}{18} \ \epsilon \ Y} .
This is a correct expression.
Element 1/3 belongs to set Y.

(iv) \mathtt{\frac{1}{7} \ \notin \ Y}
This is a correct expression.
1/7 is not part of set Y.

(v) \mathtt{\frac{1}{36} \ \epsilon \ Y}
This is correct statement.
Element 1/36 belongs to set Y.

(05) Find the size of the below sets.

(i) A = { 2, 9, -13, 7, 13}
(ii) B = { 23, 19, 7 }
(iii) X = { A, M, U, V }
(iv) Y = { 0 }
(v) Z = { }

Solution
(i) A = { 2, 9, -13, 7, 13}
There are 5 elements in the given set, namely 2, 9, -13, 7, 13.
Hence, the size of set is 5.

(ii) B = { 23, 19, 7 }
There are 3 elements in the set.
Hence, the size of set is 3.

(iii) X = { A, M, U, V }
There are 4 elements in set, namely A, M, U , V.
Hence, the size of set is 4.

(iv) Y = { 0 }
There is only one element in set, namely 0
Hence, the size of set is 1.

(v) Z = { }
There is no element is set Z.
Hence, it is a null set.