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+31 votes
Let P(S) denotes the power set of set S. Which of the following is always true?

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

P(S) ∩ P(P(S)) = { Ø }

P(S) ∩ S = P(S)

S ∉ P(S)
asked in Set Theory & Algebra by Veteran (69k points)
edited by | 2.2k views
if we add p(s) intersection s= s option this will be always true hoga i think
no. $S \in P(S)$.

7 Answers

+27 votes
Best answer
S = {1},

P(S) = { {}, {1}},

P(P(S)) = { {} , {{}} , {{1}} , {{},{1}} }

A) P(P(S)) = P(S) ==> This is false. Counterexample given above.

C) P(S) ∩ S = P(S) , This is false. This intersection is usually Empty set.

D) S ∉ P(S). This is false. S belongs to P(S).


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

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\}
So Answer is none of the above, all options are false.

But if we consider Simple sets(Except Empty Set) only then best Suitable  Option is B among the given Options.
answered by Veteran (49.5k points)
edited by
Being objective question I would suggest to assume that every set is "simple" in this case as none of these is not an option.
if S= {}
then P(s) should be { {} } which is also supported by the fact that P(s) should have 2^n elements which is 1 if n=0
then P(P(s)) = { {}, { {} } }

Hence b option will be wrong for s= { {} }

The empty set is a set with no elements. We can use braces to show the empty set: { }. Alternatively, this symbol, Ø, is often used to show the empty set.

Your explanation is correct the given question " Which of the following is always true ?  "

Option B is always true

There is a theorem-
$ \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 $\mathcal{P}(S \cap P(S))=\left \{ \phi \right \}$
For option b, P(S) ∩ P(P(S)) = { Ø }, the intersection will be Ø and not  { Ø }.

{} 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?

Notice the empty set contains 0 elements, while its power set contains 2^0 = 1 elements.

Hence P(∅)= { ∅, {∅}} is incorrect.

Correct me if i am wrong.
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
+6 votes
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
answered by Veteran (34.3k points)
edited by
@Pooja Can you correct this?

@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.

+4 votes

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

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

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

answered by Veteran (31.1k points)
edited by

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

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

+2 votes
Let S = {1,2}

P(S) = {{},(1,1),(1,2),(2,1),(2,2)}

P(P(S)) = {{} , (1,1,1,1) , (1,1,1,2) , (1,1,2,1) , (1,1,2,2)...................................... }

So , intersection of P(S) and P(P(S)) = phi


Please correct me , if my understanding is wrong.
answered by Boss (5.9k points)
yes u r correct

Your answer is correct but you explanation is wrong.let s{1,2} then p(s)={{},{1},{2},{1,2}} including {phi} there will 4 element(2^2).

+2 votes

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},      

              {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}, {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.  



answered by Veteran (40.2k points)
edited by
@arjun sir if anything is wrong then please correct me.

Let $S = \{\emptyset\}$

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


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\}


@pragy Agrawal I am not saying P(P(S)) $\bigcap$ P(S) = ∅ is exactly Right.It can be true but all other options is completely false. what you think all options are wrong?

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.

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

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

0 votes
Answer: B
answered by Veteran (36.4k points)
–2 votes

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

answered by (15 points)

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