Question

Verify whether the given statement is true or false.
(i) Every set has a proper subset.

Answer 1

Hint: First we should know the correct definition of the proper subset which is given as: First set should have all the elements of the second set included and the second set should have some extra elements not present in the first set. Then the first set is a proper subset of the second set. From this, we can take an example and try to find out whether the sentence given is true or false.

Complete step-by-step answer:
Here, we should know the definition of proper subset.
Proper Subset: Consider set A and set B, if all the elements of set A are members of set B then set A is called proper subset of B. In mathematical form, it is written as $\left( A\subset B \right)$ . Only this condition is satisfied, if there is at least one element of set B that is not a member of set A i.e. $\left( A\ne B \right)$ .
Now, will see by taking an example:

Here, we have universal set U $=\left\{ 0,1,2,3,..... \right\}$ , set A $=\left\{ 1,2,3 \right\}$ , set B $=\left\{ 1,2,3 \right\}$ , set C $=\left\{ 5,6,7 \right\}$ and set D $=\left\{ 0,1,2,3,4,5 \right\}$
Now, we can see that members of set A are present in set D, so it is said that set A is a subset of set D which is denoted as $\left( A\subseteq D \right)$ . This means that $n\left( A \right)\le n\left( D \right)$ as we can see that set A has 3 members and set D has 6 members.
Here, we can say that A is the proper subset of D because all the elements of D are not present in set A i.e. represented as $\left( A\subset D \right)$ .
Now, we can see that set C contains three members out of which only one member i.e. number 5 is present in set D. So, here C is not a proper subset of D i.e. represented as $\left( C\not\subset D \right)$ .
Now, we can see that set A and set B have same element set i.e. $\left( A=B \right)$ so, it is said that A is subset of B $\left( A\subseteq B \right)$ but in set B there are no such elements which are not present in set A. So, here set A is not proper subset of set B.
Thus, the given statement every set has a proper subset is not always true.
Hence, the answer is false.


Note: Sometimes students make mistakes in considering the proper subset. For example if set C $=\left\{ 5,6,7 \right\}$ and set D $=\left\{ 0,1,2,3,4,5 \right\}$ then here only member element of set C is there in set D. So, do not take it as proper set because for proper subset all the element of set C should be present in set D included that set D should also have extra elements not present in set C. Along with this, remember some points of subset i.e. given as,
Any set is a subset of itself.
No set is a proper subset of itself.
The empty set is a subset of every set.
The empty set is a proper subset of every set except for the empty set.