Question

If $A = \left\{ {1,3,5,7} \right\}$, then what is the cardinality of the power set $P\left( A \right)$?
A.8
B.15
C.16
D.17

Answer 1

Hint: We can find the number of elements in A by counting. Then we can find the number of elements in the power set of A using the equation $n\left( {P\left( A \right)} \right) = {2^n}$ where n is the number of elements in A. we can obtain the cardinality of the power set by substituting the cardinality of A in the equation.

Complete step-by-step answer:
We are given the set $A = \left\{ {1,3,5,7} \right\}$. By counting we get the cardinality of set A is 4.
We know that power set of a set is the collection of all the subsets of a set. The number of elements or cardinality is given by the equation $n\left( {P\left( A \right)} \right) = {2^n}$.
On substituting the value of n in the equation, we get,
$ \Rightarrow n\left( {P\left( A \right)} \right) = {2^4}$
As \[{2^4} = 16\], using this we get,
$ \Rightarrow n\left( {P\left( A \right)} \right) = 16$
Therefore, the cardinality of the power set of A is 16.
So the correct answer is option C.

Note: Alternate solution to this problem is given by,
We know that power set of a set A is the set of all the possible subsets of the set A. So, the cardinality of the power set will be number of subsets of A
We have $A = \left\{ {1,3,5,7} \right\}$
So, we can write all its subsets. We know that the set itself and the null set are the subsets of a set.
$\left\{ {1,3,5,7} \right\},\left\{ {} \right\}$
Now we can take 1 element at a time. There will be 4 such subsets.
$\left\{ 1 \right\},\left\{ 3 \right\},\left\{ 5 \right\},\left\{ 7 \right\}$
Now we can take 2 elements at a time and there will be 6 such subsets.
$\left\{ {1,3} \right\},\left\{ {1,5} \right\},\left\{ {1,7} \right\},\left\{ {3,5} \right\},\left\{ {3,7} \right\},\left\{ {5,7} \right\}$
Now we can take 3 elements at a time. There will be 4 such subsets.
$\left\{ {1,3,5} \right\},\left\{ {1,3,7} \right\},\left\{ {1,5,7} \right\},\left\{ {3,5,7} \right\}$
Then the power set is given by,
$P\left( A \right) = \left\{ {\left\{ {1,3,5,7} \right\},\left\{ {} \right\},\left\{ 1 \right\},\left\{ 3 \right\},\left\{ 5 \right\},\left\{ 7 \right\},\left\{ {1,3} \right\},\left\{ {1,5} \right\},\left\{ {1,7} \right\},\left\{ {3,5} \right\},\left\{ {3,7} \right\},\left\{ {5,7} \right\},\left\{ {1,3,5} \right\},\left\{ {1,3,7} \right\},\left\{ {1,5,7} \right\},\left\{ {3,5,7} \right\}} \right\}$
By counting, we can say that there are 16 elements in the powerset. So the cardinality of the power set $P\left( A \right)$ is 16.