Intersection of Power set and Power set of power set. 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?
 A: In your first argument, you're claiming that $P(S)$ and $P(P(S))$ have empty intersection, but they have one element in common, namely $\emptyset$, so their intersection is $\{\emptyset\}$, as claimed.
A: Note that $\emptyset \in P(S)$ and $\emptyset \in P(P(S))$ so $\{\emptyset\}\subseteq P(S) \cap P(P(S))$
A: 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).
A: Here is a necessary and sufficient condition for this to hold:
$$S \cap P(S) = \emptyset \Leftrightarrow P(S) \cap P(P(S)) = \{\emptyset\}\;.$$
Clearly, any powerset contains $\emptyset$ as element.  All other elements of $P(S)$ are non-empty subsets of $S$, all other elements of $P(P(S))$ are non-empty subsets of $P(S)$.  Any common element among those must thus be made from shared elements between $S$ and $P(S)$, and if there is a shared element $x$ between $S$ and $P(S)$, $\{x\}$ is a shared element between $P(S)$ and  $P(P(S))$.
An easy way to satisfy this is to have no sets as elements of $S$ since $P(S)$ has only sets as elements.
