0 votes 0 votes Wanted asked Jan 12, 2017 Wanted 516 views answer comment Share Follow See all 9 Comments See all 9 9 Comments reply Sushant Gokhale commented Jan 13, 2017 reply Follow Share The operation defined is: (A-B) $\cup$ (B-A) = A.B' + A'.B = A XOR B Now, XOR is associative and commmutative. But, I think its not closed. So, option D is answer, right? 0 votes 0 votes Wanted commented Jan 13, 2017 reply Follow Share counter example of not closed...? i think it is close in powerset . 1 votes 1 votes Sushant Gokhale commented Jan 13, 2017 reply Follow Share Consider power set of {A, B}. Does it contain set corresponding to set difference of A and B? No 0 votes 0 votes Wanted commented Jan 13, 2017 reply Follow Share what will b its powerset ? first tell that . 0 votes 0 votes Sushant Gokhale commented Jan 13, 2017 reply Follow Share Powerset will be { $\Phi$, {A}, {B}, {A, B} } 0 votes 0 votes Hradesh patel commented Jan 17, 2017 reply Follow Share @sushant here (A- B)U (B-A) = A*B is apply on power set element { Φ, {A}, {B}, {A, B} } then why its not closed. 0 votes 0 votes Sushant Gokhale commented Jan 17, 2017 reply Follow Share @Hradesh. If the sets A,B are not disjoint, then does the set (A-B)$\cup$(B-A) belong to { Φ, {A}, {B}, {A, B} } ? 0 votes 0 votes Hradesh patel commented Jan 17, 2017 reply Follow Share we can i say that power set is not closed always for disjoint set. means its not groupoid 0 votes 0 votes Sushant Gokhale commented Jan 17, 2017 reply Follow Share If its not closed for some, means its not closed :) 0 votes 0 votes Please log in or register to add a comment.