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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)

My approach:- S = {1},

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

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

So with this approach I can eliminate option a and c.Option d is also false as set always an element of its power set.I think b is also false ,but i am getting confused in b option.

Generally we say that intersection of two sets is a set.And intersection of two disjoint sets is Ø (empty set).

Now P(S) ∩ P(P(S)) has {} in common ,which is same as Ø. But what is confusing me is we say intersection is a set having common elements,Now as my common elements are Ø so my intersection result is a set having Ø i.e {Ø}.These are not disjoint sets as they have something in common that is Ø .So is the representation of { Ø } is correct to represent the intersection as the Ø is common or will it be Ø.

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  • $\begingroup$ Yes, the intersection would be $\{\emptyset\}$ and not $\emptyset$, since $\emptyset$ is an element of both sets in question. $\endgroup$ – Chris Jun 16 '17 at 18:59
  • $\begingroup$ Also, "intersection-theory" is NOT appropriate here, OP.... $\endgroup$ – Chris Jun 16 '17 at 19:00
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The assertion of $(b)$ is also false, unless some restrictions are imposed on $S$. If not, we can take $S = \{a, \{a\}\}$; then, \begin{eqnarray} \mathcal{P}(S) &&=&& \{\emptyset, \{a\}, \{\{a\}\}, \{a, \{a\}\}\} \\ \mathcal{P}(\mathcal{P}(S)) &&=&& \{\emptyset, \{\{a\}\}, {}\dots \} \end{eqnarray}

and so $\mathcal{P}(S) \cap \mathcal{P}(\mathcal{P}(S)) \supset \{\emptyset, \{\{a\}\}\}$.

After looking over the question you showed us, it seems to me either that the answer is "none," or else that the question as written is ill-posed.

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  • $\begingroup$ Thank you chris The answer must be none of these. $\endgroup$ – rahul sharma Jun 16 '17 at 21:33
  • $\begingroup$ You're welcome, OP. If you're satisfied with this answer, you can select the "checkmark" to indicate that this answers your question. $\endgroup$ – Chris Jun 16 '17 at 21:44
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Your intuition is absolutely correct!

We get the intersection of two disjoint sets as $\phi$ that is empty set. But, just think about this: is $\left\lbrace \phi \right\rbrace$ really empty?

The answer is No! It is a set with empty set as its element. Therefore that set is non - empty. It has one element which is the empty set.

Therefore the answer you gave is absolutely correct and thus we may say that $P(S)$ and $P(P(S))$ are NOT disjoint.

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  • $\begingroup$ Thanks aniruddha.I understood it:) $\endgroup$ – rahul sharma Jun 16 '17 at 21:33

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