Find Powerset of powerset of the empty set The powerset of the empty set is {$\emptyset$}. Attempt:
The power set of the powerset of the empty set is { { {$\emptyset$} } }? 
Next I want to find the powerset of the power set of the powerset of the empty set P(P(P($\emptyset$))). 
 A: Here's a visualization of $\emptyset$ (called $V_0$ here), $\mathcal P(\emptyset)$ (called $V_1$ here), $\mathcal P(\mathcal P(\emptyset))$ (called $V_2$ here), etc. An empty box represents the empty set. A box containing only an empty box represents the set containing only the empty set, and so forth.

Thus:


*

*$V_0=\{\}$

*$V_1=\{\{\}\}$

*$V_2=\{\{\},\{\{\}\}\}$ (this is the powerset of the powerset)

*$V_3=\{\{\},\{\{\}\},\{\{\{\}\}\},\{\{\},\{\{\}\}\}$

*etc.


$V_5$, not pictured, has $2^{2^{\large 2^{2}}}=65,\!536$ elements.
EDIT:
To be explicit, the definition is $V_0=\emptyset$ and $V_{n+1}=\mathcal P(V_n)$. There are two properties of these you should know:


*

*Every "hereditarily finite set" (that is, a set you could write with a finite number of curly braces) is contained in one of those $V$  sets.

*$V_n\subset V_{n+1}$ for all $n$. That is, each is a subset of the next one.


Can you see why these are true?
A: If it makes things easier you can use labels.  The "whooo, trippy, power set of a  power set of an empty set" nature shouldn't distract us from what is actually straightforward.
$A = \emptyset = \{\}$ is a set with zero elements.
$B= P(A) = \{A\} = \{\emptyset\} = \{\{\}\}$ is a set with $2^0 = 1$ element. (Note: for any set $K$, two subsets that will always exist are the two non-proper subsets of $K$ and $\emptyset$.  In this case $A$ = \emptyset$)
$B= P(A) = \{\emptyset\}$ is, in essence, nothing more or less than a set with one element.  That the one element just happens to be the emptyset ("whooo, freaky") doesn't make any difference at all.
$C=P(B) = \{B, \emptyset\}= \{\{A\}, \emptyset\}= \{\{\emptyset\}, \emptyset\}$
Or $C=P(B) = P(P(\emptyset) = \{\{\emptyset\}, \emptyset\}$.  This is a set with $2^1 =2$.  For each of the one element, $A=\emptyset$, in $B = P(\emptyset)$ the two subsets are i) The set that contains $A = \emptyset$, that is, the set $\{A\} = \{\emptyset\}$ and ii) the set that does not contain $A = \emptyset$, that is, ... the emptyset. 
