
The cardinality of the Power Set of \[\left\{ {\phi ,\left\{ \phi \right\},\left\{ {\phi ,\left\{ \phi \right\}} \right\}} \right\}\]
A. Cannot be less than 8
B. is equal to 6
C. is equal to 8
D. can be less than 8
Answer
570.9k+ views
Hint: Recall from the definition of the power set of a set. The power set is the set of all the subsets of a set. First of all, write all the possible subsets of the given set in a set. Count all the elements of the power set. The number of elements of the set gives the cardinality of the set.
Complete step-by-step answer:
We have to find the cardinality of the Power Set \[\left\{ {\phi ,\left\{ \phi \right\},\left\{ {\phi ,\left\{ \phi \right\}} \right\}} \right\}\]
We know that the power set is the set of all the sub-sets of a set.
Let $A$ be the original set.
Then we have $A = \left\{ {\phi ,\left\{ \phi \right\},\left\{ {\phi ,\left\{ \phi \right\}} \right\}} \right\}$
Also, every set and the null-set is the power set of the set $A$.
Then, \[P\left( A \right) = \left\{ {\phi ,\left\{ \phi \right\},\left\{ {\phi ,\left\{ \phi \right\}} \right\},\left\{ {\phi ,\left\{ {\phi ,\left\{ \phi \right\}} \right\}} \right\},\left\{ {\left\{ \phi \right\},\left\{ {\phi ,\left\{ \phi \right\}} \right\}} \right\},\left\{ {\left\{ \phi \right\}} \right\},\left\{ {\left\{ {\phi ,\left\{ \phi \right\}} \right\}} \right\},\left\{ {\left\{ {\phi ,\left\{ \phi \right\},\left\{ {\phi ,\left\{ \phi \right\}} \right\}} \right\}} \right\}} \right\}\]
Now, the cardinality refers to the number of elements in the set.
Here, we can see that there are 8 elements in the power set of the given set.
Thus, the cardinality of the power set of \[\left\{ {\phi ,\left\{ \phi \right\},\left\{ {\phi ,\left\{ \phi \right\}} \right\}} \right\}\] is equal to 8.
Hence, option C is correct.
Note: Write the elements of the power set carefully without missing any element otherwise it will lead to wrong answers. If a set has $n$ elements, then the number of subsets it can have is ${2^n}$. This also implies that the cardinality of the power set is ${2^n}$. So, alternatively, we can directly calculate the number of elements in the power set of \[\left\{ {\phi ,\left\{ \phi \right\},\left\{ {\phi ,\left\{ \phi \right\}} \right\}} \right\}\], which is equal to ${2^3} = 8$ as there are 3 elements in the given set.
Complete step-by-step answer:
We have to find the cardinality of the Power Set \[\left\{ {\phi ,\left\{ \phi \right\},\left\{ {\phi ,\left\{ \phi \right\}} \right\}} \right\}\]
We know that the power set is the set of all the sub-sets of a set.
Let $A$ be the original set.
Then we have $A = \left\{ {\phi ,\left\{ \phi \right\},\left\{ {\phi ,\left\{ \phi \right\}} \right\}} \right\}$
Also, every set and the null-set is the power set of the set $A$.
Then, \[P\left( A \right) = \left\{ {\phi ,\left\{ \phi \right\},\left\{ {\phi ,\left\{ \phi \right\}} \right\},\left\{ {\phi ,\left\{ {\phi ,\left\{ \phi \right\}} \right\}} \right\},\left\{ {\left\{ \phi \right\},\left\{ {\phi ,\left\{ \phi \right\}} \right\}} \right\},\left\{ {\left\{ \phi \right\}} \right\},\left\{ {\left\{ {\phi ,\left\{ \phi \right\}} \right\}} \right\},\left\{ {\left\{ {\phi ,\left\{ \phi \right\},\left\{ {\phi ,\left\{ \phi \right\}} \right\}} \right\}} \right\}} \right\}\]
Now, the cardinality refers to the number of elements in the set.
Here, we can see that there are 8 elements in the power set of the given set.
Thus, the cardinality of the power set of \[\left\{ {\phi ,\left\{ \phi \right\},\left\{ {\phi ,\left\{ \phi \right\}} \right\}} \right\}\] is equal to 8.
Hence, option C is correct.
Note: Write the elements of the power set carefully without missing any element otherwise it will lead to wrong answers. If a set has $n$ elements, then the number of subsets it can have is ${2^n}$. This also implies that the cardinality of the power set is ${2^n}$. So, alternatively, we can directly calculate the number of elements in the power set of \[\left\{ {\phi ,\left\{ \phi \right\},\left\{ {\phi ,\left\{ \phi \right\}} \right\}} \right\}\], which is equal to ${2^3} = 8$ as there are 3 elements in the given set.
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