let’s take a Set.
S= {a, b}
then P(S) = { { } , {a} , {b} , {a , b} }
To make subsequent expansion less I am going to use variables to represent these subsets, i.e. P (S) = {W, X, Y, Z}
P (P (S)) = { {},
{W},
{X},
{Y},
{Z},
{W, X},
{W, Y},
{W, Z},
{X, Y},
{X, Z},
{Y, Z},
{W, X, Y},
{W, X, Z},
{W, Y, Z},
{X, Y, Z},
{W, X, Y, Z}
}
We can substitute as needed:
W -> {}
X -> {a}
Y -> {b}
Z -> {a, b}
P (P (S)) = { {}, and P(S) = { {} , {a} , {b} , {a , b} }
{{}},
{{a}},
{{b}},
{{a, b}},
{{}, {a}},
{{}, {b}},
{{}, {a, b}},
{{a},{b}},
{{a}, {a, b}},
{{b}, {a, b}},
{{}, {a}, {b}},
{{}, {a}, {a, b}},
{{}, {b}, {a, b}},
{{a}, {b}, {a, b}},
{{}, {a}, {b}, {a, b}}
}
A. P(P(S)) =P(S)
It is completely wrong statement.Both can't be equal to each other.
B. P(S) ∩ P(P(S)) = { Ø }
This is statement seems to be true because
There is only one element is common and that is { { } } ={ Ø }
But it is False
CounterExample : Let S={∅}
P(S)={∅, {∅} }
P(P(S))={ ∅, {∅} , { {∅} } , {∅, {∅} } }
P(S) ∩ P(P(S)) = { Ø ,{Ø} }
C. P(S) ∩ S= { {} , {a} , {b} , {a , b} } ∩ {a, b}
There is nothing is common so
P(S) ∩ S = Ø
Note: In P(S) {a, b} is a single element and in S ‘a’ and ‘b’ 2 different element so nothing is common.
D. S ∉ P(S)
It is false statement.for S ∈ P(S) it can be true.
Hence All options are Wrong.