Explain the correctness of the statement

Question

Question : 

Explain the correctness of the mathematical statement  :
Statement 1 : P(S)  $\cap$ S = S  ?
Statement 2 :  P(S)  $\cap$ P(P(S)) = $\phi$
My Approach :

Staement 1

Let S= { 1,2}

P(S) = {   $\phi$ , {1} ,{2}, {1,2} }

P(S)  $\cap$  S = $\phi$
{ 1,2}   $\cap$ {{ 1,2}} =  $\phi$  

Statement 2

Let S={1}

P(S) ={$\phi$,{1} }

P(P(S)) = {$\phi$ , {$\phi$} , {$\phi$,1}}

P(S)  $\cap$ P(P(S)) = { $\phi$ }

Is My answers are correct ??

Comments

check anurag's  answer below.

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@Leen. He has taken wrong power set of an empty set. See my commenton his answer.

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@Leen. OK, Anurag is right. Powerset of empty set contains only 1 element.

Answer 1

Statement 1 is contingent and statement 2 is false

Testing the statements on just a particular instance of S might not give the proper insight about the correctness or incorrectness of any mathematical statement.

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Statement 1 : $P\left ( S \right ) \cap S = S$

For all those sets $S$ whose power set contains every element of $S$, Statement 1 would be true. 

The power set of any set always contains the empty set, so if we choose our $S$ to be the empty set then Statement 1 would be true.

That is, if $S = \Phi$, then the power set $P\left ( S \right ) = P\left ( \Phi \right ) = \left \{ \Phi \right \}$.

Thus $P\left ( S \right ) \cap S =\left \{ \Phi \right \} \cap \Phi = \Phi =S$, where the second equality follows from the fact that intersection of any set with the empty set equals the empty set. 

Similarly, there are infinitely many possible ways to choose the set $S$ such that statement 1 holds, e.g. $\left \{ \Phi , \left \{ \Phi \right \} \right \}, \left \{ \Phi , \left \{ \Phi \right \}, \left \{ \left \{ \Phi \right \} \right \} \right \}$ and so on.

On the contrary, there also exist infinitely possible ways to choose the set $S$ such that statement 1 does not holds, e.g. $\left \{ 1 \right \}, \left \{ 1 , \left \{ 1 \right \} \right \}, \left \{ 1 , \left \{ 1 \right \}, \left \{ \left \{ 1 \right \} \right \} \right \}$ and so on.

So statement 1 is a contingent statement.

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 Statement 2 : $P\left ( S \right ) \cap P\left ( P\left ( S \right ) \right ) = \Phi$

Both $P\left ( S \right )$ and $P\left ( P\left ( S \right ) \right )$ are power sets, so for sure both of these sets will contain the empty set.We need not to worry about rest of the elements of $P\left ( S \right )$ and $P\left ( P\left ( S \right ) \right )$ for proving statement 2 incorrect, which can be done as follows:

$P\left ( S \right ) \cap P\left ( P\left ( S \right ) \right ) = \left \{ \Phi , ... \right \} \cap \left \{ \Phi , ... \right \} = \left \{ \Phi , ... \right \}\neq \Phi$

That is, the intersection of $P\left ( S \right )$ and $P\left ( P\left ( S \right ) \right )$ would for sure contain the empty set (although, it may contain other elements also apart from the empty set), so in the worst case even if the intersection of $P\left ( S \right )$ and $P\left ( P\left ( S \right ) \right )$ contains only the empty set then also the set containining the empty set is not equal to the empty set i.e. ($\left \{ \Phi \right \} \neq \Phi$).

So statement 2 is false.

    

Answer 2

These 2 rules are very important (Derived from the exercises of Kenneth Rosen).

1)  if   $a \epsilon Set$, it means that 'a' is an element of the set {x1, x2, ........., xn}   i.e a=xi  

2) if $A\sqsubseteq B$ , then Vx( x$\epsilon$A    $\rightarrow$  x$\epsilon$B  )

   i.e whenever we are comparing sets, we should compare their elements.

Now, considering first question.

S={} then P(S) = { $\phi$ }. So, S$\bigcap$P(S) = $\phi$ ={}=S as there are no elements common between them.

S={1} then P(S) = { $\phi$, {1} } .So, S$\bigcap$P(S) = $\phi$ ={} $\neq$ S.

Thus, answer to first question is contingency as Anurag has pointed.

Now, considering second question.

S = {1}

P(S) = { $\phi$, {1} }

P(P(S)) = { $\phi$, {$\phi$}, {{1}}, { $\phi$, {1} }  }

Thus, P(S)$\bigcap$P(P(S)) = {$\phi$}   (since $\phi$ is the common element between both).

Thus, answer to 2nd question is false.