When is was making some exercises I encountered the following exercise:


Let $P(A)$ denote the set of all subsets of an arbitrary set $A$.

List first the elements of $P(\emptyset)$, then the elements of $P(P(\emptyset))$.

Finally, check in two steps whether you have listed the correct number of elements.

I wasn't quite sure how to handle this exercise.

My idea was that $P(\emptyset)$ = {$\emptyset$}, thus $P(P(\emptyset))$ = {$\emptyset$}

I am not sure if this is correct and I don't know how to check if they are the correct number of elements.

Maybe with $2^n$ because this counts the number of all subsets of a set $A$?

I hope someone can correct me and help me out.


$P(\{x\}) = \{\emptyset,\{x\}\}$ for any $x$. Then take $x=\emptyset$. Then note that $\emptyset \neq\{\emptyset\}$.

So $P(\emptyset) = \{\emptyset\}$. And $P(P(\emptyset))=\{\emptyset,\{\emptyset\}\}$.

  • $\begingroup$ Many thanks Sigur and Thomas, and how do I check if I got the correct number of elements? $\endgroup$ – Nedellyzer Jul 3 '13 at 18:52
  • $\begingroup$ Yes, for a set $A$ of size $n$, the size of $P(A)$ is $2^n$. $\endgroup$ – Thomas Andrews Jul 3 '13 at 18:59

Hint: for every set $A$, $P(A)$ always contains among its elements the empty set $\emptyset$ and also the total set $A$, since both are always subsets of $A$.

So, if $A\neq \emptyset$ then you can count at least two elements in $P(A)$.

  • $\begingroup$ @ Sigur But A = $\emptyset$ in this case, so then it is correct what I've written down? $\endgroup$ – Nedellyzer Jul 3 '13 at 18:45
  • $\begingroup$ Note that if $A=\emptyset$ then $B=P(\emptyset)=\{\emptyset\}$ and $P(B)=P(P(\emptyset))=\{\emptyset,\{\emptyset\}\}$. $\endgroup$ – Sigur Jul 3 '13 at 18:49

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