## 5 Answers

$S=\left \{ 1 \right \}$

$P(S)=\left \{ \left \{ \right \},\left \{ 1 \right \} \right \}$

$P(P(S))=\left \{ \left \{ \right \},\left \{ \left \{ \right \} \right \},\left \{ \left \{ 1 \right \} \right \},\left \{ \left \{ \right \},\left \{ 1 \right \} \right \} \right \}$

- $(A)\; P(P(S)) = P(S)$ - This is false. Counterexample given above.
- $(C)\; P(S) \cap S = P(S)$ - This is false. This intersection is usually Empty set.
- $(D)\; S \notin P(S)$ - This is false. $S$ belongs to $P(S)$.

Edit:-

$B.$ It seems like $B$ is true, but there is a counter-example for $B$ too. (Given By @Pragy Below)

$S = \{\emptyset\}$

$P(S) = \left \{\emptyset, \{\emptyset\} \right \}$

$P(P(S)) = \left \{ \emptyset, \{\emptyset\}, \left \{ \{ \emptyset \}\right \}, \left \{ \emptyset, \{\emptyset\} \right \} \right \}$

$P(S) \cap P(P(S)) = \left \{ \emptyset, \{\emptyset \}\right \} \qquad \neq \{\emptyset\}$

So, the answer is none of the above, all options are false.

But if we consider Simple sets(Except Empty Set) only then the best option is B among the given options.

### 9 Comments

$ \mathcal{P}(A) \cap \mathcal{P}(B)=\mathcal{P}(A \cap B) $

here in option B, $A=S$ and $B=\mathcal{P}(S)$. It becomes

$ \mathcal{P}(S) \cap \mathcal{P}(\mathcal{P}(S))=\mathcal{P}(S \cap P(S)) $

Now option B become- $\mathcal{P}(S \cap P(S))$

Can a set and its power set has anything is common ?

They can not have anything in common therefore - $S \cap P(S)=\phi$

and $\mathcal{P}(S \cap P(S))=\left \{ \phi \right \}$

https://proofwiki.org/wiki/Intersection_of_Power_Sets

{} is represented by Ø ,where as { Ø } is a set with single element that is Ø called as singleton set.So how is b true in case of non empty set?Even in given example {} is the intersection which is equivalent of Ø,where as { Ø } is different thing.Please check someone?

$\begin{align*}S &= \{0\}\\ P(S) &= \big\{ \phi, \{0\} \big \}\\ P\big(P(S)\big) &= \Big\{ \phi, \big \{ \phi \big \}, \big \{\! \{0\} \! \big \}, \big \{\! \phi,\{0\}\! \big \} \Big \}\end{align*}$

- P(P(S)) = P(S)

False

- P(S) ∩ P(P(S)) = { Ø }

False

- P(S) ∩ S = P(S)

False

- S ∉ P(S)

False

There has been a confusion with what $\phi$ is and what $\{\}$ is.

$\phi = \{\} = \text{empty set}$

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. **

### 7 Comments

Let $S = \{\emptyset\}$

What is $P(S) \cap P(P(S))$?

Answer:

$\begin{align}

S &= \{\emptyset\}\\[1em]

P(S) &= \Bigl \{\emptyset, \{\emptyset\} \Bigr \}\\[1em]

P(P(S)) &= \biggl \{ \emptyset, \{\emptyset\}, \Bigl \{ \{ \emptyset \}\Bigr \}, \Bigl \{ \emptyset, \{\emptyset\} \Bigr \} \biggr \}\\[1em]

P(S) \cap P(P(S)) &= \Bigl \{ \emptyset, \{\emptyset \}\Bigr \}\\[1em]

& \neq \{\emptyset\}

\end{align}$

..

I am not saying P(P(S)) ⋂ P(S) = ∅ is exactly

The question specifically mentions the words "always true"

what you think all options are wrong

Yes, all options are wrong.

but all other options is completely false

Things are either True, or False. If two things are False, both of them are False. It is not the case that one false thing can be more falsey than the other false thing.