Question

If $P$ denotes the power set of $A$ and $A$ is the void set, then what is the number of elements in $P\{P\{P\{P(A)\}\}\}\,$?
A. $0$
B. $1$
C. $4$
D. $16$

Answer 1

Hint: To solve this question we will use the formula of calculating the number of elements in power set ${{2}^{n}}$. We will first calculate the number of elements of power set by taking the value of $n=0$ as $A$ is the void set. Then we will calculate the number of elements in power set by increasing the value of $n$ accordingly and hence determining the value of $P\{P\{P\{P(A)\}\}\}\,$.

Formula Used: Number of power set : ${{2}^{n}}$ where $n$ is the number of elements.

Complete step by step solution: We have given a power set $P(A)$ where$A$ is the void set and we have to find the number of elements in $P\{P\{P\{P(A)\}\}\}\,$.
We know that number of set can be calculated with the formula ${{2}^{n}}$ where $n$ is the total number of element of the set. And void set means that there are no elements present in that set which means $0$ element.
So the number of elements in the power set of $A$ will be,
$\begin{align}
  & P(A)={{2}^{n}} \\
 & ={{2}^{0}} \\
 & =1
\end{align}$
When $P(A)$ contains one element ,
$\begin{align}
  & P\left\{ P(A) \right\}={{2}^{n}} \\
 & ={{2}^{1}} \\
 & =2
\end{align}$
When $P\left\{ P(A) \right\}$ contains two elements,
$\begin{align}
  & P\left\{ P\left\{ P(A) \right\} \right\}={{2}^{n}} \\
 & ={{2}^{2}} \\
 & =4
\end{align}$
When $P\left\{ P\left\{ P(A) \right\} \right\}$ contains four elements,
$\begin{align}
  & P\left\{ P\left\{ P\left\{ P(A) \right\} \right\} \right\}={{2}^{n}} \\
 & ={{2}^{4}} \\
 & =16
\end{align}$
The number of elements in $P\{P\{P\{P(A)\}\}\}\,$is $16$ where $P$ denotes the power set of $A$ and $A$ is the void set

Option ‘D’ is correct

Note: A power set can be defined as the set which includes all the subsets of a particular set including the void set. It is denoted as $P(S)$ where $S$ is any set. As a set has ${{2}^{n}}$ subsets in total therefore a power set has ${{2}^{n}}$ elements in it.