4. Subset

A set \(A\) is said to be a subset of a set \(B\) if every element of \(A\) is also an element of \(B\) .

It is denoted by
\(A \subset B\) if whenever \(a \in A\) , then \(a \in B\)

If \(A \subset B\) and \(B \subset A\) , then \(A = B\).
 Every set is subset of itself \(A \subset A\)
 Empty set is subset of every set \(\phi \subset A\)
 if $ A \subset B$ and $ B \subset C$ ,then $ A \subset C$
Example
1. A ={1,2,3,4,5}
B= {1,2,3,4,5,6,7,8}
Now we can see all the elements of A are in set B
So , $A \subset B $
2. A ={a,b,c}
B= {a,b,c}
Now we can see all the elements of A are in set B
So , $A \subset B $
Now we also see that all the elements of B are in set A
$B \subset A $
3. A ={a,b,c}
B= {1,2,3}
Now we can see none of the element A are present in B
So $ A \not\subseteq B$
5. Proper Subset and Superset
 If \(A \subset B\) and \(A \ne B\) , then \(A\) is proper subset of \(B\).
 We may use symbol $\subseteq$ to define general subset and $\subset $ to define proper subset
 In case of proper subset, \(B\) is called superset of set \(A\)

 it is represented as $ B \supset A$
Example
1. A ={1,3,7}
B= {1,3,5,7,8}
Now we can see all the elements of A are in set B and $A \ne B$
So A is proper subset of B
$A \subset B $
and B is the superset of A
$ B \supset A$
2. A = ${x : x^2 1=0}$
B= {1}
Now here A can be represented in set builder form as
A= {1,1}
We can see all the elements belongs to elements in A
So B is proper subset of A
$B \subset A $
and A is the superset of B
$ A \supset B$
5. Subset of set of the real numbers
N : the set of all natural numbers
Z : the set of all integers
Q : the set of all rational numbers
R : the set of real numbers
Z+ : the set of positive integers
Q+ : the set of positive rational numbers, and
R+ : the set of positive real numbers
\(T = \left\{ {x:x \in R{\rm{ \; and \;}}x \notin Q} \right\}\), i.e., all real numbers that are not rational
\(N \subset Z \subset Q,{\rm{ }}Q \subset R,{\rm{ }}T \subset R,{\rm{ }}N \not\subset T\)
6. Interval as subset of R Real Number
\(a,b \in R,b > a\)
\((a,b)\) 
It is the open interval set between point and b such that All the points
between a and b belong to the open interval (a, b) but a, b themselves do not belong to this interval 
\(\{ y:a < y < b\} \) 
\([a,b]\) 
It is the closed interval set between point and b such that All the points
between a and b belong to the open interval (a, b) including a, b

\(\{ x:a \le x \le b\} \) 
\([a,b)\) 
It is the open interval set between point and b such that All the points
between a and b belong to the open interval (a, b) including a, but not b

\(\{ x:a \le x < b\} \) 
\((a,b)\) 
It is the open interval set between point and b such that All the points
between a and b belong to the open interval (a, b) including b, but not a

\(\{ x:a < x \le b\} \) 
Power Set

The collection of all subsets of a set \(X\) is called the power set of \(X\). It is denoted by \(P(X)\). In \(P(X)\), every element is a set.

if \(X = \left\{ {1,2,3} \right\}\), then
\(X\left( A \right) = \left\{ {\phi ,\left\{ 1 \right\},\left\{ 2 \right\},\left\{ 3 \right\},\left\{ {1,2} \right\},\left\{ {1,3} \right\},\left\{ {2,3} \right\},\left\{ {1,2,3} \right\}} \right\}\)
Also, note that \(n\left[ {X\left( A \right)} \right] = 8 = {2^3}\)

In general, if \(X\) is a set with \(n\left( X \right) = m\), then it can be shown that
\(n\left[ {P\left( A \right)} \right] = {2^m}\)
Example
List the subsets of {a,0,a}
Solution
A={1,0, 1 }.
The subset of A having no element is the empty set φ
 The subsets of A having one element are { –a }, { 0 }, { a }.
The subsets of A having two elements are {–a, 0}, {–a, a} ,{0, a}.
The subset of A having three elements of A is A itself.
So, all the subsets of A are φ, {–a}, {0}, {a}, {–a, 0}, {–a, a},{0, a} and {–a, 0, a}.
Also Read
link to this page by copying the following text