GATE CSE 2000 | Question: 2.6

Question

Let $P(S)$ denotes the power set of set $S.$ Which of the following is always true?

• A. $P(P(S)) = P(S)$
• B. $P(S) ∩ P(P(S)) = \{ Ø \}$
• C. $P(S) ∩ S = P(S)$
• D. $S ∉ P(S)$

Comments

one more option should be there.

E. None of these

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https://math.stackexchange.com/questions/2836572/can-a-set-and-its-powerset-have-anything-in-common?fbclid=IwAR2cWb5g3nNf3hY-0zD8bvdK1M7tKOsHleiRmxlhmW8Bv7Fd0VVMl1fYNRQ

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Option A,C,D are $\color{red}{\text{NEVER true}}$ for any set $S.$

There is NO set $S(finite \,\,or\,\,\,infinite)$, for which statements in Option A,C,D are true. 

Statement in option B is NOT always true for every set $S,$ but it is true if set $S$ contains simple elements i.e. elements which are not set.

$\color{red}{\text{Find Video Solution Below:}}$

Detailed Video Solution

Answer 1

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

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

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

• $(A)\; P(P(S)) = P(S)$ - 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.

Comments

I would like to provide 1 more eg

S={ a,  {a} }

P(S) ={ phi, {a} , {{a} }  ,  { a,  {a} } }

Now,  S ^ P(S)  = {{a} }

P(S) ^ P(P(S) ) = { phi , {{a} } }

Here the theorem given by Sachin sir above is  valid

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Do you really need that long explanation for part b?

any set and power set will not have any common element

hence  set ∩ Power(Set)  = {} or  ϕ

{ϕ} means set contains one element , hence answer is straight forward false

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This can help

https://math.stackexchange.com/questions/2836572/can-a-set-and-its-powerset-have-anything-in-common

Answer 2

S belongs to P(S) is true so d is false

intersection of powerset and set is phi so c is false

powerset and powerset of powerset will only have phi as common element so ans is b

Comments

@Pooja Can you correct this?

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@pooja palod i think your statement(intersection of powerset and set is set itself) is wrong. 

intersection of powerset and set should be  ∅ not set itself.

Answer 3

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

1. P(P(S)) = P(S)
   False
    
2. P(S) ∩ P(P(S)) = { Ø }
   False
    
3. P(S) ∩ S = P(S)
   False
    
4. S ∉ P(S)
   False

There has been a confusion with what $\phi$ is and what $\{\}$ is.
$\phi = \{\} = \text{empty set}$

Comments

P(S) will not contain {∅}, it will contain ∅ only

∅ != {∅}

so option (b) turns out to be ∅ not {∅} , so false(S)P(P(S))={0}={{0},{ϕ}}={{{0}},{{ϕ

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I acknowledge that, Here, A, C, D cannot be true by any way. best option we are left with is option B,

Answer 4

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.  

 

 

Comments

@Pragy I guess the question assumes all set elements are simple. Do you agree?

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if we add p(s) intersection s= s option this will be always true hoga i think

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@Aryan Asrafi 1

As Arjun sir said to take the simple set

Let S = { 1,2 }

P(S) = { $\phi$,{1},{2},{1,2} } 

Now (P(S) $\cap$ S) = { } = $\phi$

Nothing is common in S and P(S).