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+11 votes
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The number of elements in the power set $P(S)$ of the set $S=\{\{\phi\}, 1, \{2, 3\}\}$ is:

  1. $2$
  2. $4$
  3. $8$
  4. None of the above
asked in Set Theory & Algebra by Veteran (59.8k points)
edited by | 1.5k views
0

Would we not consider that an element itself has two elements (2,3) which itself have its own power set of size = 2?

0
NO. {2,3} is considered as one element in set S. if you were confused replaced that with x.
0
What will be no. of powersets if it is {∅,1,{2,3}}? Will it be 4 as cardinality of ∅ is 0 where as cardinality of {∅}=1 ??
0

@BharathiCH what do u mean by no of power set? no of element in power set is 2^(set size)

in {∅,1,{2,3}} there are 3 element i.e phi,1,{2,3} [Note 2,3 single element] now power set 2^3=8

Note ∅ and {∅} not same. first means no element whereas second means one element which is ∅.

0

@Abhisek Tiwari 4

Then the elements will be as

{∅,{∅},{1},{{2,3}},{∅,1},{1,{2,3}},{∅,{2,3}},{∅,1,{2,3}}}..

Then {∅,1} and {1} are they not equal?

+1

@BharathiCH 

Yes {1} and { {∅},1} different. [Note it should be {{∅},1} instead of {∅,1}

2 Answers

+18 votes
Best answer

no of elements in power set is = $2$(no of elements in the set) = $2^{3}$ = $8$

Elements are {∅,{{∅}},{1},{{2,3}},{{∅},1},{1,{2,3}},{{∅},{2,3}},{{∅},1,{2,3}}}

Hence, Option is $(C)$ 8.

answered by Loyal (8.1k points)
edited by
+3 votes

S={{ϕ},1,{2,3}}

cardinality of S= 3

then cardinality of Power set of S = 2^3 =8

Hence ,Option (C)8  is the correct answer.

answered by Loyal (7.3k points)

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