24 votes 24 votes The number of elements in the power set $P(S)$ of the set $S=\{\{\emptyset\}, 1, \{2, 3\}\}$ is: $2$ $4$ $8$ None of the above Set Theory & Algebra gate1995 set-theory&algebra normal set-theory + – Kathleen asked Oct 8, 2014 recategorized Apr 25, 2021 by Lakshman Bhaiya Kathleen 16.2k views answer comment Share Follow See all 7 Comments See all 7 7 Comments reply Swati Rauniyar commented Oct 18, 2017 reply Follow Share Would we not consider that an element itself has two elements (2,3) which itself have its own power set of size = 22 ? 0 votes 0 votes Dileep kumar M 6 commented Nov 19, 2017 reply Follow Share NO. {2,3} is considered as one element in set S. if you were confused replaced that with x. 0 votes 0 votes BharathiCH commented Dec 20, 2018 reply Follow Share 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 votes 0 votes Abhisek Tiwari 4 commented Dec 21, 2018 reply Follow Share @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 ∅. 1 votes 1 votes BharathiCH commented Dec 22, 2018 reply Follow Share @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? 0 votes 0 votes Abhisek Tiwari 4 commented Dec 22, 2018 reply Follow Share @BharathiCH Yes {1} and { {∅},1} different. [Note it should be {{∅},1} instead of {∅,1} 2 votes 2 votes endurance1 commented Mar 3, 2021 reply Follow Share It is explicitly said in the question that what ever S is, it a set according to paper setter, so we just need to calculate the cardinality and apply the formula. 0 votes 0 votes Please log in or register to add a comment.
Best answer 34 votes 34 votes Number of elements in power set $= 2^{\text{(no of elements in the set)}} = 2^{3} = 8$ Elements are $\{∅,\{\{∅\}\},\{1\},\{\{2,3\}\},\{\{∅\},1\},\{1,\{2,3\}\},\{\{∅\},\{2,3\}\},\{\{∅\},1,\{2,3\}\}\}$ Hence, Option is $(C): 8.$ jayendra answered Dec 27, 2014 edited May 10, 2019 by Arjun jayendra comment Share Follow See 1 comment See all 1 1 comment reply Deepak Poonia commented Apr 23, 2022 reply Follow Share Video Solution 2 votes 2 votes Please log in or register to add a comment.
5 votes 5 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. Warrior answered Jul 29, 2017 Warrior comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Since there are three elements in the set S , so the power of this set P( S)= 2 ^3= 8. Hence the option C is correct. DIBAKAR MAJEE answered Apr 26, 2020 DIBAKAR MAJEE comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes S = {(φ), 1, (2, 3)} P(S) = {φ, {{φ}}, {1}, {{2, 3}}, {{φ}, 1}, {1, {2, 3}}, {{φ}, 1, {2, 3}}} In P(S) it contains 8 elements. varunrajarathnam answered Oct 3, 2020 varunrajarathnam comment Share Follow See all 0 reply Please log in or register to add a comment.