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Let U be a set, and let A CU. Recall the indicator function XA: U + Z, defined by XA(x) = ſi, rEA 0, A. Now, let A, B CU and

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Since, A, B CU and consider the symmetric difference of A and B is defined as AAB = (A - B)U(B – A). (a) To Show :- AAB CU. SCase-III Assume x EU but I e B and x A. Then I E AAB. So, XB(x) = 1, XA(X) = 0) and XAaB(x) =1. Hence, XAAB(x) = XA(2) + XB(x

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