Find the Power sets of P[P{P(Φ)}]

Question

My approach : Null set contains itself only so number of elements is 1. Or we can apply the direct formula of 2^n [where n is the number of elelments] Now i got P{P(1)} Applying the formula again It becomes, P{2}=2²=4 But it was by the formula, Now i tried to write the sets and got like this, Let P(Φ)={Φ} Then P[P(Φ)]= {Φ,{Φ}} Here i got tge two sets but I can't figure out what will be the last 4 sets. Any way I convinced myself , P[P{P(Φ)}]={Φ,{Φ},{{Φ}}, {Φ,{Φ},{{Φ}} } } I thought as each set contains itself But suddenly I relized that if it is so then P[P(Φ)]= {Φ,{Φ}, {Φ,{Φ}} } Then there will be three Power sets.

Answer 1

$\emptyset$ has no elements and only itself, i.e. $\emptyset$, as subset. Hence $$P(\emptyset)=\{\emptyset\}.$$

$\{\emptyset\}$ has one element $\emptyset$, hence two subsets $\emptyset$ and $\{\emptyset\}$ itself. Hence $$P(P(\emptyset))=\{\emptyset,\{\emptyset\}\}.$$

$\{\emptyset,\{\emptyset\}\}$ has two elements $\emptyset$ and $\{\emptyset\}$, hence has four subsets $\emptyset$ and $\{\emptyset\}$ and $\{\{\emptyset\}\}$ and $\{\emptyset,\{\emptyset\}\}$ itself. Hence $$ P(P(P(\emptyset)))=\{\emptyset,\{\emptyset\},\{\{\emptyset\}\},\{\emptyset,\{\emptyset\}\}\}.$$ This gets funny if one uses $\{\}$ as notation for the emptyset: $ \{\{\},\{\{\}\},\{\{\{\}\}\},\{\{\},\{\{\}\}\}\}$.

Answer 2

Let us save some braces

$$P(\phi)=\{\phi\}=A$$

$$ P(P(\phi))=P(A)=\{\phi, A\}=B$$

$$ P(P(P(\phi))) =P(B)=\{\phi, \{A\} , \{\phi\},B\}$$