Find the power set $P(S)$ for $S=\{\emptyset, \{\emptyset\}, \{\emptyset \{\emptyset\}\}\}$ Find the power set $P(S)$ for $S=\{\emptyset, \{\emptyset\}, \{\emptyset \{\emptyset\}\}\}$
OK this problem confuses me for many reasons, but here is what I know.  The cardinality of a set is $2^n$ where $n$ is the number of elements in the set.  In this problem, however, how can the empty set be an element?  
If I were just going to say this set has $2^n$ elements that would mean the set has $2^3$ or 8 elements, but I don't know what those elements would be other than empty sets.
Any help in understanding this problem is greatly appreciated!
 A: Let's assume the actual question is to find the power set of $S=\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$. Using Von Neumann's definition of the natural numbers, this is equivalent to finding the power set of $S=\{0,1,2\}$, where $0=\emptyset$, $1=\{0\}$, and $2=\{0,1\}$. The power set is then
$$\{\emptyset,\,\{0\},\,\{1\},\,\{2\},\,\{0,1\},\,\{0,2\},\,\{1,2\},\,\{0,1,2\}\}$$
Just change the $2$'s to $\{0,1\}$, then the $1$'s to $\{0\}$, then the $0$'s to $\emptyset$--and you are done!
You will end up with a royal mess, of course. I found it difficult enough to format the easier form of the power set above.
A: $\{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}, \{\{\emptyset, \{\emptyset\}\}\}, \{\emptyset, \{\emptyset\}\}, \{ \emptyset, \{\emptyset, \{\emptyset\}\}\}, \{\{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}, \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\} \}$
No problem to have $\emptyset$ in another set. There is a problem if you have $x \in \emptyset$ though.
A: Every set has the empty set as an element. However #(S) should be zero where #(S) is the cardinality of set S. 
The power set of the empty set or sets of empty sets should be simply the empty set.
Remember however that the power set of a set is not a cardinality. It is a set constructor that builds a set from the elements of set S. So technically your answer should be the set which is the factorial combination of each element ie if set S = { a, b, c} then P(S) = {a, b, c, {a,b}, {a,c}, {b, c}, {a, b, c} }
IMHO of course !
