If $P(S)$ denotes power set of a set $S$ then is $P(S) \cap P(P(S)) = \{\emptyset\}$?
My book says its true.
I considered a simple set $S=\{1,2\}$
Now $P(S) = \{\emptyset, \{1\}, \{2\}, \{1,2\}\}$
$$
P(P(S))= \{\emptyset,\{\emptyset\}, \{\{1\}\}, \{\{2\}\},\{\{1,2\}\}, \{\{1\}\,\{2\}\}, \{\{1\},\{1,2\}\}, \{\{2\},\{1,2\}\} , \ldots\}
$$
Now $P(S) \cap P(P(S))=\emptyset$ and not $\{\emptyset\}$.
So I think that above claim is wrong.
If I go by theorem of power set intersection which says $P(A) \cap P(B)=P(A\cap B)$ and put $A=P(S)$ and $B=P(S)$, then I get
$$
P(S) \cap P(P(S)) = P(S \cap P(S)) = P(\emptyset)={\emptyset}.
$$
So if I go by theorem, then the claim seems to be correct.
Where am I making a mistake?
Best Answer
If $S=\{\emptyset, \{\emptyset\}\}$, then $P(S) = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}, \{\{\emptyset\}\}\}$ and $P(P(S))$ will contain the whole of $P(S)$ as a subset.
So the statement as such does not generally follow. Basically whenever a set contains an element as well as a set containing only that element, its powerset will contain the set containing that element as well (obviously).