Power sets in set theory What is the set $\mathcal{P}(\mathcal{P}(\mathcal{P}( \emptyset )))$? 
Would it be $\{\{\{ \emptyset\}\}\}$?
I understand that $\mathcal{P}(\{a,b,c\}) = \{ \{ \}, \{a\}, \{b\}, \{c\}, \{a,b\}, \{a,c\}, \{b,c\}, \{a,b,c\} \}$.
 A: I would take this in steps. Remember that $P(A)$ is the set of all subsets of a set $A$. Recall also that the empty set is a subset of all sets.
$P(\emptyset) = \{\emptyset\}$ since the only subset of $\emptyset$ is $\emptyset$.
Then $P(P(\emptyset)) = P(\{\emptyset\}) = \{\emptyset, \{\emptyset\}\}$.
And so ...
A: Let's denest the braces one-by-one:
$$A=2^\varnothing=\{\varnothing\}$$
$$B=2^A=\{\varnothing,\{\varnothing\}\}$$
$$C=2^B=\{\varnothing,\{\varnothing\},\{\{\varnothing\}\},\{\varnothing,\{\varnothing\}\}\}$$
Therefore
$$\mathcal{P(P(P(}\varnothing)))=\{\varnothing,\{\varnothing\},\{\{\varnothing\}\},\{\varnothing,\{\varnothing\}\}\}$$
As you can see, computing power sets by hand becomes mentally unwieldy quickly.
A: For every set $A$ we have $A\subseteq A$ so that $A\in\wp(A)$. So also for $\varnothing$. If $B\neq\varnothing$ then some $b$ exists with $b\in B$. So in that case $B\subseteq\varnothing$ would lead to $b\in\varnothing$, wich is evidently not true. We conclude now that $\varnothing$ is the only subset of $\varnothing$. That is:$$\wp(\varnothing)=\{\varnothing\}$$
Looking at $\{\varnothing\}$ we find exactly two subsets: $\varnothing$ (which is a subset of every set) and $\{\varnothing\}$. This leads to:$$\wp(\wp(\varnothing))=\wp(\{\varnothing\})=\{\varnothing,\{\varnothing\}\}$$
This set has $4$ subsets (find them!). I leave the rest to you. 
