Topics in Logic, Set Theory and Computability

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1 Topics in Logic, Set Theory and Computability Homework Set #3 Due Friday 4/6 at 3pm (by or in person at ) Exercises from Handouts 7-C-2 7-E-6 7-E-7(a) 8-A-4 8-A-9(a) 8-B-2 8-C-2(a,b,c) 8-D-4(a) 8-D-5(a) 8-E-1 8-E-2 Correction 8-C-2. Let f : X Y and B Y. (a) Prove that f (f (B)) B. (b) Provide a counterexample of B f (f (B)). (c) Provide a necessary and sufficient condition (or conditions) for f (f (B)) = B. (d) Prove your assertion in (c) is true.

2 Solutions to Homework #3 7-C-2. Let s be a relation from X to Y and suppose C X a nonempty collection of subsets of X. (a) Prove that s (»C)»{ s(a) : A C }. (b) Provide a counterexample of»{ s(a) : A C } s (»C). (c) Provide a necessary and sufficient condition (or conditions) for s (»C) =»{ s(a) : A C }. : (a) Let y s (»C). Then there exists x»c such that (x,y) s. This implies that x A for all A C. Hence for every A C, we have x A and (x,y) s. Thus y s(a) for every A C. This implies that y»{ s(a) : A C }. Therefore s (»C)»{ s(a) : A C }. (b) Take X = Y = 2 = { 0, 1 } with s = { (0,0), (1,0) } and C = { { 0 }, { 1 } }. The image of the sets in C are s({ 0 }) = s({ 1 }) = { 0 }. This implies that»{ s(a) : A C } = { 0 } whereas s (»C) = s ( ) =. Therefore, it is certainly not the case that»{ s(a) : A C } s (»C). (c) Equality comes from the condition given for the definition of an injection of a function. Using part (a), we only have to establish the following theorem. Theorem: Let s X Y. Then»{ s(a) : A C } s(»c) for every nonempty collection C X if and only if s ({y}) is a singleton or the empty set for every y Y. : Suppose x 1, x 2 s ({y}). [Now show x 1 = x 2.] Choose the collection to be C = { A, B } where A = { x 1 } and B = { x 2 }. Then s(a) = s(b) = { y }. This implies that»{ s(a) A C } = s(a) s(b) = { y } and s (»C) = s (A B). If x 1 x 2, then s(a B) = s( ) =, which would contradict the fact that»{ s(a) : A C } s(»c). Therefore, we must have that x 1 = x 2. : Let C be a nonempty collection of subsets of X. Suppose y»{ s(a) : A C }. [Now show y s(»c).] Then y s(a) for every A C. Since s ({y}) is a singleton, then there exists x X such that { x } = s ({y}). Then we must have that x A for every A C. Thus, x»c and this implies that y s(»c). (a) (b) (c) E-6. Let s be a relation on X. Prove that if s is transitive and reflexive, then s 2 = s. Is the converse true?

3 Suppose s is transitive and reflexive. Since s is transitive, then s 2 s by Exercise 7-E-1(g). [Show s s 2.] Suppose (x,y) s. [Show (x,y) s 2.] Since (x,y) s, then y X. Since s is reflexive, then (y,y) s. Since (x,y), (y,y) s, then (x,y) s s = s 2. Thus s s 2 which implies that s 2 = s. The converse is not true. Take X = { 0 } and s =. Then s 2 = s but s is not reflexive. Proof Converse E-7. Let r be a relation on X. Prove each of the following. (a) The relation t = r r 1 is symmetric and is the smallest symmetric relation containing r. [That is, if s is a symmetric relation such that r s t, then s = t.] Claim: The relation t = r r 1 is symmetric. Let (x,y) t = r r 1. Then either (x,y) r or (x,y) r 1. The former implies (y,x) r 1. The latter implies (y,x) r. So either (y,x) r 1 or (y,x) r. This implies (y,x) r r 1 = t. Thus t = r r 1 is symmetric. Ñ Claim: The relation t = r r 1 is the smallest symmetric relation containing r. Suppose s is symmetric with r s t. [Now show s = t (we need only show that t s).] Let (x,y) t = r r 1. This implies that (x,y) r or (x,y) r 1. If (x,y) r, then (x,y) s (since r s - and so we are done). If (x,y) r 1, then, (y,x) r. Since r s, then (y,x) s. Since s is symmetric, then (x,y) s. Ñ symmetric smallest A-4. Let f : X Y. Show that the relation f 1 is not necessarily a function. Let X = { a }, Y = { b, c }, and define f = { (a,b) }. Then f 1 Y X is the relation f 1 = { (b,a) } but f 1 is not a function since dom f 1 = { b } Y

4 8-A-9. Construct the sets (a) 2 3. Since 2 = { 0, 1 } and 3 = { 0, 1, 2 }, then 2 3 = { f 1, f 2, f 3, f 4, f 5, f 6, f 7, f 8 } where f 1 = { (0, 0), (1, 0), (2, 0) }, f 2 = { (0, 0), (1, 0), (2, 1) }, f 3 = { (0, 0), (1, 1), (2, 0) }, f 4 = { (0, 0), (1, 1), (2, 1) }, f 5 = { (0, 1), (1, 0), (2, 0) }, f 6 = { (0, 1), (1, 0), (2, 1) }, f 7 = { (0, 1), (1, 1), (2, 0) }, and f 8 = { (0, 1), (1, 1), (2, 1) }. Instructor s Note: We could have given a simplified answer using the characteristic function. It is 2 3 = { c 0, c 1, c {1}, c {2}, c 2, c {0,2}, c {1,2}, c 3 }. 8-B-2. Prove that X Y = iff X = and Y It would have been helpful to do some of Exercise 8-A-9 to know that 0 0 = 1, 1 0 = 1, and 0 1 = 0. : Proof of the contrapositive. Suppose it is not true that X = and Y. This implies that either X or Y =. Case 1: Suppose X. Then either Y = or Y. If Y =, then X Y = { }. If Y, then choose any a X. Define f X Y by f (y) = a for all y Y. Thus X Y. Case 2: Suppose Y =. Then X Y = { }. In both cases we have X Y. Thus X Y = implies X = and Y. : Suppose that X = and Y. Since X Y = (by Exercise 6-G), then there are no possible functions f that satisfy f X Y =. Thus, X Y =. Overall, X Y = iff X = and Y C-2. Let f : X Y and B Y. (a) Prove that f (f (B)) B. (b) Provide a counterexample of B f (f (B)). (c) Provide a necessary and sufficient condition (or conditions) for f (f (B)) = B. (a) Suppose y f (f (B)). Then there exists x f (B) such that f (x) = y. Since x f (B), then there exists z B such that f (x) = z. Since f is a function, then z = y and so y B.

5 (b) Let X = 1 = { 0 }, Y = 2 = { 0, 1 } and define f : X Y as f (0) = 0 with B = { 1 }. Since f (B) =, then f (f (B)) = and so it is not true that B f (f (B)). (c) f (f (B)) = B iff B ran f. (a) (b) (c) D-4. Let f :A B and g :B C. Prove each of the following. (a) If f and g are surjections, then g f is a surjection. Let f : A B and g : B C be surjections. To show g f : A C is a surjection, suppose z C. [Now show there is an element in the domain of g f that maps to z.] Since g is a surjection, then there exists y B such that g(y) = z. Since y B and f is a surjection, then there exists x A such that f (x) = y. Since g(y) = z and f (x) = y then g(f (x)) = z. This implies that (g f )(x) = z. Thus, g f is a surjection D-5. Let C X a nonempty collection of sets in X and let f : X Y be an injection. (a) Prove that»{ f (A) : A C } f (»C). Let y»{ f (A) : A C }. [Show y f (»C).] Then y f (A) for every A C. To each A C there is associated x A such that f (x) = y. Because f is an injection, there is only one preimage of y. Let s call this preimage x. Since x A for every A C, then we must have x»c. Since f (x ) = y, then y f (»C) E-1. Show that 2 3 ~ 6. Since 2 3 = { (0,0), (0,1), (0,2), (1,0), (1,1), (1,2) } and 6 = { 0, 1, 2, 3, 4, 5 }, then define a bijection f : by f = { ((0,0), 0), ((0,1), 1), ((0,2), 2), ((1,0), 3), ((1,1), 4), ((1,2), 5) }

6 8-E-2. Show 0 ~ 4. Since 0 = { 0, 1, {1}, 2 } by Exercise 5-H(d) and 4 = { 0, 1, 2, 3 }, then define a bijection f : 0 4 by f = { (0, 0), (1, 1), ({1}, 2), (2, 3) }