
The cardinality of the set $P\{P[P(\phi )]\}$ is
A. \[0\]
B. \[1\]
C. \[2\]
D. \[4\]
Answer
162.6k+ views
Hint: To solve this question, we will first calculate the power set of $\phi $ that is $P(\phi )$which is termed as the empty or null set and it contains only one element $\phi $. We will then calculate the power set of $P[P(\phi )]$ and then again calculate the power set of $P[P(\phi )]$ that is $P\{P[P(\phi )]\}$ and determine the number of elements it will contain. We will then count the total number of elements in set $P\{P[P(\phi )]\}$ and determine its cardinality.
Complete step by step solution:We are given a set $P\{P[P(\phi )]\}$ and we have to find its cardinality. We know that cardinality of a power set is the total number of elements which are present in the set. We know that a power set is the set of all its subsets including the void set and is denoted with the symbol $P(S)$ .
We will first find the value of $P(\phi )$.
We know that power set of $\phi $ that is $P(\phi )$contains only one element which is $\phi $. So,
$P(\phi )=\phi $
Let us consider $P(\phi )=\phi $ as $A$ that is $P(\phi )=\phi =A$.
Now power set of $P(\phi )$ that is $P[P(\phi )]$ will be,
$\begin{align}
& P[P(\phi )]=P[A] \\
& =\{\phi ,A\} \\
& =\{\phi ,\{\phi \}\}
\end{align}$
We will now calculate the power set of $P[P(\phi )]$ that is $P\{P[P(\phi )]\}$.
$\begin{align}
& P\{P[P(\phi )]\}=P\{\phi ,\{\phi \}\} \\
& =\phi ,\{\phi \},\{\{\phi \}\},\{\phi ,\{\phi \}\}
\end{align}$
There are four elements in the set of $P\{P[P(\phi )]\}$ so its cardinality will be \[4\].
Option ‘D’ is correct
Note:We can also calculate cardinality with the help of the formula ${{2}^{n}}$ where $n$ is the total number of elements in the set. As $P(\phi )$ is a null set so its cardinality will be ,
$\begin{align}
& P(\phi )={{2}^{n}} \\
& ={{2}^{0}} \\
& =1
\end{align}$
Cardinality of $P[P(\phi )]$ will be,
$\begin{align}
& P[P(\phi )]={{2}^{1}} \\
& =2
\end{align}$
Cardinality of $P\{P[P(\phi )]\}$ will be,
$\begin{align}
& P\{P[P(\phi )]\}={{2}^{2}} \\
& =4
\end{align}$
Complete step by step solution:We are given a set $P\{P[P(\phi )]\}$ and we have to find its cardinality. We know that cardinality of a power set is the total number of elements which are present in the set. We know that a power set is the set of all its subsets including the void set and is denoted with the symbol $P(S)$ .
We will first find the value of $P(\phi )$.
We know that power set of $\phi $ that is $P(\phi )$contains only one element which is $\phi $. So,
$P(\phi )=\phi $
Let us consider $P(\phi )=\phi $ as $A$ that is $P(\phi )=\phi =A$.
Now power set of $P(\phi )$ that is $P[P(\phi )]$ will be,
$\begin{align}
& P[P(\phi )]=P[A] \\
& =\{\phi ,A\} \\
& =\{\phi ,\{\phi \}\}
\end{align}$
We will now calculate the power set of $P[P(\phi )]$ that is $P\{P[P(\phi )]\}$.
$\begin{align}
& P\{P[P(\phi )]\}=P\{\phi ,\{\phi \}\} \\
& =\phi ,\{\phi \},\{\{\phi \}\},\{\phi ,\{\phi \}\}
\end{align}$
There are four elements in the set of $P\{P[P(\phi )]\}$ so its cardinality will be \[4\].
Option ‘D’ is correct
Note:We can also calculate cardinality with the help of the formula ${{2}^{n}}$ where $n$ is the total number of elements in the set. As $P(\phi )$ is a null set so its cardinality will be ,
$\begin{align}
& P(\phi )={{2}^{n}} \\
& ={{2}^{0}} \\
& =1
\end{align}$
Cardinality of $P[P(\phi )]$ will be,
$\begin{align}
& P[P(\phi )]={{2}^{1}} \\
& =2
\end{align}$
Cardinality of $P\{P[P(\phi )]\}$ will be,
$\begin{align}
& P\{P[P(\phi )]\}={{2}^{2}} \\
& =4
\end{align}$
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