Question

If $A$ is the void set $\phi$, then $P\left(A\right)$ has just one element $\phi$ i.e. $P[\phi] = \left\{\phi\right\}$. So number of elements of $P\left[P\left(P\left(\phi\right)\right)\right]$ is

• A. $4$
• B. $3$
• C. $5$
• D. $2^4$

Answer 1

Answer: (A)

We have, $P(\phi) = \left\{\phi\right\}$
$\therefore P\left(P\left(\phi\right)\right) = \left\{\phi, \left\{\phi\right\}\right\}$
$\Rightarrow P\left[P\left(P\left(\phi\right)\right)\right] = \left\{\phi, \left\{\phi\right\}, \left\{\left\{\phi\right\}\right\}, \left\{\phi, \left\{\phi\right\}\right\}\right\}$
Hence, $P\left[P\left(P\left(\phi\right)\right)\right]$ consists of $4$ elements i.e. $n\left[P\left[P\left(P\left(\phi\right)\right)\right]\right] = 4$