1
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ I agree with GFauxPas: what does $S=\{\emptyset, \{\emptyset\}, \{\emptyset \{\emptyset\}\}\}$ mean? Did you mean $S=\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$? $\endgroup$ – Rory Daulton Oct 26 '14 at 19:29
  • $\begingroup$ @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. $\endgroup$ – hax0r_n_code Oct 26 '14 at 19:30
  • 1
    $\begingroup$ Then the way it was given to you is nonsense, so we should just add the missing comma. The question then makes sense. $\endgroup$ – Rory Daulton Oct 26 '14 at 19:31
  • $\begingroup$ @RoryDaulton I agree with you. I suppose it is missing the comma as that makes more sense. $\endgroup$ – hax0r_n_code Oct 26 '14 at 19:34
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – hax0r_n_code Oct 26 '14 at 19:43
  • 2
    $\begingroup$ $\emptyset$ 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. $\endgroup$ – Rory Daulton Oct 26 '14 at 19:48
1
$\begingroup$

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

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

$\endgroup$
-1
$\begingroup$

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 !

$\endgroup$
  • 1
    $\begingroup$ "Every set has the empty set as an element" - definitely not true $\endgroup$ – aschepler Oct 26 '14 at 19:52
  • $\begingroup$ Indeed. What about $\{\{\emptyset\}\}$ ? Perhaps you meant "as a subset" ? $\endgroup$ – DanielV Oct 26 '14 at 19:53
  • $\begingroup$ @aschepler can you elaborate? I have always understood the empty set is an element of every set. Maybe subset is more correct. $\endgroup$ – Excalibur2000 Oct 26 '14 at 20:18
  • $\begingroup$ @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. $\endgroup$ – Henning Makholm Oct 26 '14 at 22:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.