Homework 3 Solutions Problem 7.8. Proposition. Let X = {0, 1}, and let B be the set of all countable subsets of X ω . Then |B| = |X ω |. Proof. Let M be the set of all functions Z+ × Z+ → X, i.e. the set of all infinite matrices x11 x12 x13 x 21 x22 x23 {xij } = x 31 x32 x33 . .. .. .. . . ··· ··· ··· ... where each xij ∈ X. Note that such a matrix has countably many entries (since Z+ × Z+ is countable), so |M| = |X ω |. Now, consider the function f : M → B that maps each matrix to its set of rows, i.e. f ({xij }) = (x11 , x12 , x13 , . . .), (x21 , x22 , x23 , . . .), (x31 , x32 , x33 , . . .), . . . The image of f is B − {∅}. Indeed, since a matrix can have repeated rows, each element of B − {∅} has infinitely many preimages. If we modify the function f slightly to map one of the matrices to ∅, we obtain a surjection f : M → B, and therefore |M| ≥ |B|. Finally, we can define an injection g : X ω → B by g(x) = {x}, so |X ω | ≤ |B|. Then |X ω | ≤ |B| ≤ |M| = |X ω |, which proves that |X ω | = |B|. 1 Problem 10.4. Let Z− denote the set of negative integers in the usual order. Proposition. A simply ordered set A fails to be well-ordered if and only if it has a subset with the same order type as Z− . Proof. The backward direction is obvious, since a subset with the order type of Z− does not have a minimum element. For the forward direction, suppose that A is not well-ordered. Then there must exist a nonempty subset S of A that has no minimum element. Fix an element b1 ∈ S. Since b1 is not the minimum element of S, there must exist a b2 ∈ S for which b2 < b1 . Since b2 is not the minimum element of S, there must exist a b3 ∈ S for which b3 < b2 . Continuing in this way, we can choose an infinite decreasing sequence b1 > b2 > b3 > · · · of elements of S. Then {. . . , b3 , b2 , b1 } is a subset of A with the same order type as Z− . Proposition. Let A be a simply ordered set. If every countable subset of A is well-ordered, then A is well-ordered. Proof. By the previous proposition, if A is not well-ordered then A has a countable subset that is not well-ordered. 2 Problem 10.8. Proposition. Let A1 and A2 be disjoint sets, well-ordered by <1 and <2 , respectively. Define an order relation < on A1 ∪ A2 by letting a < b if either a, b ∈ A1 and a <1 b, or if a, b ∈ A2 and a <2 b, or if a ∈ A1 and b ∈ A2 . Then < is a well-ordering. Proof. Let S ⊂ A1 ∪ A2 be nonempty. If S has any elements from A1 , then the minimum element of S ∩ A1 is a minimum element for S. If S has no elements from A1 , then S ⊂ A2 , which implies that S has a minimum element. Proposition. Let {Aα }α∈I be a disjoint family of sets, well-ordered by relations {<α }α∈I , and let <I be a well-ordering of the index set I. Define an order relation < on S α∈I Aα by letting a < b if either a, b ∈ Aα and a <α b for some α ∈ I, or if a ∈ Aα and b ∈ Aβ for some α, β ∈ I with α <I β. Then < is a well-ordering. Proof. Let S ⊂ S α∈I Aα be nonempty. Since <I is a well-ordering, there exists a minimum element α ∈ I for which S ∩ Aα is nonempty. Then the minimum element of S ∩ Aα under <α will also be the minimum element of S under <. 3
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