Math 300: Final Exam Practice Solutions

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We construct out of this a number in (0, 1) as follows: start with 0.x 11 , then append the digits x 21 and x 12 , then x 31 , x 22 , and x 13 , and so on following the successive diagonals moving from lower-left to upper-right. Define the function f : (0, 1) ∞ → (0, 1) by sending the sequence (x 1 , x 2 , x 3 , . . .) to this number: f(x 1 , x 2 , x 3 , . . .) = 0.x 11 x 21 x 12 x 31 x 22 x 13 · · · . This function is bijective since it is invertible with inverse f −1 : (0, 1) → (0, 1) ∞ defined as follows. Take y ∈ (0, 1), write it according to its decimal expansion y = 0.y 1 y 2 y 3 . . ., and then “unwind” the process from above to create a sequence of numbers: x 1 := 0.y 1 y 3 y 6 y 10 . . . x 2 := 0.y 2 y 5 y 9 . . . x 3 := 0.y 4 y 8 . . . x 4 := 0.y 7 . . . where we list y 1 in the upper-left corner, then y 2 and y 3 along the next diagonal down, then y 4 , y 5 , y 6 along the next diagonal down, and so on. The inverse of f is f −1 (y) = (x 1 , x 2 , x 3 , . . .). Since (0, 1) and R have the same cardinality, there exists a bijection g : (0, 1) → R. Also, as a result of Exercise 6.1.9 in the book (or a generalization of this with a similar proof), (0, 1) ∞ and R ∞ have the same cardinality so there exists a bijection h : R ∞ → (0, 1) ∞ . The composition gfh is then a bijection R ∞ → R, so R ∞ and R have the same cardinality as claimed. (b) Take any list of elements of Z ∞ : (n 11 , n 12 , n 13 , . . .) (n 21 , n 22 , n 23 , . . .) (n 31 , n 32 , n 33 , . . .) . Define an element (m 1 , m 2 , m 3 , . . .) of Z ∞ simply by choosing for m i an integer different from n ii . This element is then not in the list above since it differs from the kth element of this list in the kth location. Hence any listing of elements of Z ∞ cannot possibly contain all elements of Z ∞ , so Z ∞ is uncountable. 3. Consider the relation ∼ between sets defined by: A ∼ B if A and B have the same cardinality. (a) Show that ∼ is an equivalence relation. (b) Determine which of the following sets are equivalent to which. No justification is needed. P(Z), Q × Q, P(P({∅})), F (R), P(Z × Z × Z), (0, 1) ∩ Q, P(P(Z + )), R 2 , F ({2, 3}) Remark. This is a mixture of older material and newer material. In part (a) we must as usual show that R is reflexive, symmetric, and transitive. This amounts to using properties of bijective functions. In part (b) we use facts about cardinality we’ve already built up, and is a good example of how to compare cardinalities. Solution. (a) For any set A, the identity function i A : A → A sending anything in A to itself is bijective so A has the same cardinality as A. Thus ARA so R is reflexive. Now suppose that ARB. Then A and B have the same cardinality, meaning there exists a bijection f : A → B. Then f −1 : B → A is also bijective, so B and A have the same cardinality. Hence BRA so R is symmetric. 2
Finally, suppose that ARB and BRC. Then A and B have the same cardinality so there exists a bijection f : A → B, and B and C have the same cardinality so there exists a bijection g : B → C. The composition gf : A → C is then also bijective, so A and C have the same cardinality. Thus ARC so R is transitive and we conclude that R is an equivalence relation. (b) No justification is needed, but we give some anyway. Two sets are in the same equivalence classes when they have the same cardinality. • P(P({∅})) and F ({2, 3}) both have 4 elements and thus have the same cardinality. These are the only finite sets in this list. • Q × Q and (0, 1) ∩ Q are both countably infinite, and so have the same cardinality. • Since Z × Z × Z and Z have the same cardinality, their power sets have the same cardinality, which is the same as the cardinality of R. Since R 2 and R also have the same cardinality, we get that P(Z), P(Z × Z × Z), and R 2 all have the same cardinality. • Finally, an example on the “Worked Examples” handout shows that F (R) and P(R) have the same cardinality. Since R and P(Z + ) have the same cardinality, their power sets do as well so we get that F (R) and P(P(Z + )) have the same cardinality. Notice that in none of these did we construct any explicit bijections, but rather used cardinality facts we’ve previously built up. 4. Show that Z × Z + and E × O have the same cardinality, where E is the set of even integers and O the set of odd integers, by constructing an explicit bijection between them. Remark. Since an explicit bijection is what is being asked for, an explicit bijection is what we should give. This will require giving a candidate for the bijection f : Z → Z + → E × O and then showing that f is indeed bijective. To do all this we use previous bijections we’ve seen on homework or in class involving integers, even integers, and odd integers. Also, using the same kind of idea as in Exercise 6.1.9 in the book, we should really be looking for bijections Z → E and Z + → O separately. Proof. Define f : Z × Z + → E × O by f(m, n) = { (2m, n − 1) if n is even (2m, −n) if n is odd. We claim that f is bijective. To check injectivity, suppose that f(m, n) = f(a, b). Since whether or not the second component of f(m, n) = f(a, b) is positive or negative depends on whether n, b are even or odd, these can be equal only when n and b are both even or both odd. If they are both even, then f(m, n) = f(a, b) becomes (2m, n − 1) = (2a, b − 1), so 2m = 2a and n − 1 = b − 1, giving m = n and a = b. If n, b are both odd, then f(m, n) = f(a, b) becomes (2m, −n) = (2a, −b), so 2m = 2a and −n = −b, again giving m = n and a = b. Thus f(m, n) = f(a, b) implies (m, n) = (a, b) in either case, so f is injective. 3
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Math 300: Final Exam Practice Solutions

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