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!
Best Answer
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.