TITLE: Power set cardinality exercise, where am I wrong?
QUESTION [5 upvotes]: Exercise. Find the cardinality.
$|P(P(P(A\times\varnothing)))|$
My solution 1:
$|P(P(P(A\times\varnothing)))|=2^{|P(P(A\times\varnothing))|}=2^{2^{|P(A\times \varnothing)|}}=2^{2^{2^{|A\times\varnothing|}}}=2^{2^{2^{|\varnothing|}}}= 2^{2^{2^0}}=2^2=4$
My solution 2:
$|P(P(P(A\times\varnothing)))|=|P(P(P(\varnothing)))|=|P(P(\{\varnothing \}))|=|P(\{\varnothing,\{\varnothing\} \})|=|\{\varnothing, \{\varnothing\},\{\{\varnothing\} \}, \{\varnothing, \{\varnothing\} \}    \}|=4$
But I got it wrong in my homework. Where am I wrong?

REPLY [2 votes]: Your solution is correct. I suspect that in the question "$\emptyset$" may have been a typo (or miscopied) for "$\{\emptyset\}$," in which case the answer would be $2^{2^{2^{\vert A\vert}}}$ instead of $4$. (There's also the sad possibility that the claimed correct answer is $0$, which is what one gets if one forgets that $\mathcal{P}(\emptyset)=\{\emptyset\}\not=\emptyset$.)