in here P(A) is power set of(A) 1.Prove that A B iff P (A) P (B). 2. Prove that for any set A, A = P (A). 3. Suppose {Ai | i I} and {Bi | i I} are indexed families of sets. (a) Prove that iI (Ai Bi)(iIAi)(iIBi). (b) Find an example for which iI (Ai Bi ) (iI Ai ) (iI Bi ). 4. Prove that for every integer x, x 2 + x is even. 5. Prove that if x P(x) x Q(x) then x(P(x) Q(x)). (Hint: Re member that P Q is equivalent to P Q). 6. Let U be any set. (a) Prove that there is a unique AP(U) such that for every B P (U), A B =B. (b) Prove that there is a unique AP(U) such that for every B P(U), A B = A. 7. Suppose A is a set, and for every family of sets F, if F = A then A F. Prove that A has exactly one element. (Hint: For both the existence and uniqueness parts of the proof, try proof by contradiction.) 8. Prove that if lim xc f(x)= L then lim xc 7 f(x)=7L. 9. Suppose {Ai | i I} and {Bi | i I} are indexed families of sets. (a) Prove that iI (Ai Bi)(iIAi)(iIBi). (b) For each (i, j) I I let C(i,j) = Ai Bj, and let P = I I. Prove that pP Cp =(iI Ai)(iI Bi). 10. Let L and E be the relations defined in parts 4 and 5 of Example 4.2.2. Describe the following relations: (b) E(L1L). 11. Suppose R is a relation from A to B and S is a relation from B to C. (a) Prove that Dom(S R) Dom(R). (b) Prove that if Ran(R) Dom(S) then Dom(S R) = Dom(R). (c) Formulate and prove similar theorems about Ran(S R). 12. Suppose R is a relation from A to B and S and T are relations from B to C. (a) Prove that (SR)\\(T R)(S\\T)R. (b) What's wrong with the following proof that (S \\ T) R (S R) \\ (T R)? Proof. Suppose (a, c) (S \\ T ) R. Then we can choose some b B such that (a, b) R and (b, c) S \\ T , so (b, c) S and (b,c) T. Since (a,b)R and (b,c)S, (a,c)SR. Similarly, since (a,b) R and (b,c) T, (a,c) T R. Therefore (a, c) (S R) \\ (T R). Since (a, c) was arbitrary, this shows that (S \\ T ) R (S R) \\ (T R). (c) Must it be true that (S\\T)R(SR)\\(TR)? Justify your answer with either a proof or a counterexample. in here P(A) is power set of(A) 1.Prove that A B iff P (A) P (B). ()Suppose A is a subset of B. Let x be an arbitrary element of P(A), Since x is in P(A) it follows x is a subset of A. Since A is a subset of B, so x is a subset of B also. Therefore x is in P(B). Since x is arbitrary, we can conclude P(A) is a subset of P(B). ()Suppose P(A) is a subset of P(B). Let x be an arbitrary element of A, so there should exist at least one set y in P(A) such that x is in y. Since P(A) is a subset of P(B), so y is in P(B). It means y is a subset of B and x is in y, therefore x is in B also. Since x is arbitrary, we can conclude A is a subset of B. 2. Prove that for any set A, A = P (A). We will prove that A is a subset of UP(A) and also UP(A) is a subset of A. Let us prove A is a subset of UP(A). Let x be an arbitrary element of A. Since A is in P(A), so by definition of UP(A), x should be an element of UP(A) also. Since x is arbitrary, A is a subset of UP(A). Let us prove UP(A) is a subset of A. Let x be an arbitrary element of UP(A), so there exists an a in P(A) such that x is in a. Since a is in P(A), therefore a is a subset of A and hence x is in A also. Since x is arbitrary we can conclude that UP(A) is a subset of A. Therefore A = P(A). 3. Suppose {Ai | i I} and {Bi | i I} are indexed families of sets. (a) Prove that iI (Ai Bi)(iIAi)(iIBi). Let x be an arbitrary element of Ui in I(AiBi). It follows that there exists a i in I such that x in Ai and x in Bi. Since we have a particular i such that x in Ai and i in I, therefore x in Ui in I Ai. By similar argument x in Ui in I Bi as well. Thus x in (Ui in I Ai) (x in Ui in I Bi) . Since x is arbitrary, we can conclude that Ui in I(AiBi) is a subset of (Ui in I Ai) (Ui in I Bi) (b) Find an example for which iI (Ai Bi ) (iI Ai ) (iI Bi ). A1 = {1,3}, A2 = {4} B1 = {1}, B2 = {3,4} A1B1 = {1} A2B2 = {4} Ui in I(AiBi) = {1,4} ...(i) Ui in I Ai = {1,3,4} Ui in I Bi = {1,3,4} (Ui in I Ai) (Ui in I Bi) = {1,3,4} ...(ii) We can see that (i) and (ii) are not same. 4. Prove that for every integer x, x 2 + x is even. Considering two cases. Case1: x is even Then there exists an integer k such that x = 2k x2 = 4k2 x2+x = 4k2+2k x2+x = 2k(2k+1) k(k+1) is an integer, so x2+x is even Case2: x is odd Then there exists an integer k such that x = 2k + 1 x2 = (2k+1) 2 x2 = 4k2+1+4k x2+x = (4k2+1+4k)+(2k+1) x2+x = 4k2+6k+2 x2+x = 2(2k2+3k+1) Since (2k2+3k+1) is an integer, x2+x is even 5. Prove that if x P(x) x Q(x) then x(P(x) Q(x)). (Hint: Re member that P Q is equivalent to P Q). Suppose, For all x P(x) there exists x Q(x) ( For all x P(x)) or (there exists x Q(x)) (there exists xP(x)) or (there exists xQ(x)) there exists x(P(x) or Q(x)) there exists x(P(x)Q(x)) Thus if for all xP(x) there exists xQ(x) then there exists x(P(x)Q(x)) 6. Let U be any set. (a) Prove that there is a unique AP(U) such that for every B P (U), A B =B. We prove existence, try A = U. For any B in P(U), B is a subset of U and hence UB=B We prove uniqueness, assume there is another set C in P(U) such that for every B in P(U)(CB=B). We can put B = U in this equation, then CU=U. Therefore C=U=A (b) Prove that there is a unique AP(U) such that for every B P(U), A B = A. We prove existence, try A = . For any set B in P(U), clearly B= We prove uniqueness, assume there is another set C in P(U) such that for every B in P(U)(CB=C). We can put B= in this equation, then C=C. Therefore C==A 7. Suppose A is a set, and for every family of sets F, if F = A then A F. Prove that A has exactly one element. (Hint: For both the existence and uniqueness parts of the proof, try proof by contradiction.) We first prove that A and then establish that A can not have more than one element and that proves that A has only 1 element. Proof for A: We use proof by contradiction. Suppose A=. For F= in particular, UF==A but clearly A. So A Proof for A not having more than 1 element: Again we use proof by contradiction. Suppose A has more than one element. Consider an element x in A and Ax as A has more than one element by assumption. Now consider the particular family of sets F = {{x}, A\\{x}} clearly, UF=A, but AF. So A can not have more than one element. 8. Prove that if lim xc f(x)= L then lim xc 7 f(x)=7L. Suppose limxc f(x)=L. Let be an arbitrary positive number. It follows that /7 is also a positive number. So there exists >0 such that If 0<|x-c|< then |f(x)-L|