# 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!

• I agree with GFauxPas: what does $S=\{\emptyset, \{\emptyset\}, \{\emptyset \{\emptyset\}\}\}$ mean? Did you mean $S=\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$? – Rory Daulton Oct 26 '14 at 19:29
• @RoryDaulton Perhaps that is what is inferred by the question (and maybe why I'm confused), but the way I have written it is the way it was given to me. – hax0r_n_code Oct 26 '14 at 19:30
• Then the way it was given to you is nonsense, so we should just add the missing comma. The question then makes sense. – Rory Daulton Oct 26 '14 at 19:31
• @RoryDaulton I agree with you. I suppose it is missing the comma as that makes more sense. – hax0r_n_code Oct 26 '14 at 19:34

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.

• Thank you for this explanation. One other question, though, is how can $\emptyset$ be an element of the set? I think I understand how $\{\emptyset\}$ can be a member, but I don't know how just $\emptyset$ can be. Sorry if this is a confusing question. – hax0r_n_code Oct 26 '14 at 19:43
• $\emptyset$ cannot contain a member, but there is no problem with it being a member of another set such as $\{\emptyset\}$. Think of $\emptyset$ as an empty box. Nothing is inside that box, but that box can be inside another box. $\{\emptyset\}$ is a box that contains a box that contains nothing. And that whole complex could be inside another box. – Rory Daulton Oct 26 '14 at 19:48

$\{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}, \{\{\emptyset, \{\emptyset\}\}\}, \{\emptyset, \{\emptyset\}\}, \{ \emptyset, \{\emptyset, \{\emptyset\}\}\}, \{\{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}, \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\} \}$

No problem to have $\emptyset$ in another set. There is a problem if you have $x \in \emptyset$ though.

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 !

• "Every set has the empty set as an element" - definitely not true – aschepler Oct 26 '14 at 19:52
• Indeed. What about $\{\{\emptyset\}\}$ ? Perhaps you meant "as a subset" ? – DanielV Oct 26 '14 at 19:53
• @aschepler can you elaborate? I have always understood the empty set is an element of every set. Maybe subset is more correct. – Excalibur2000 Oct 26 '14 at 20:18
• @Excalibur2000: "Element" and "subset" are two entirely different concepts. The empty set is a subset of every set, but is is only an element of sets that explicitly, um, has it as an element. – hmakholm left over Monica Oct 26 '14 at 22:43