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Unformatted text preview: SOLUTIONS FOR HOMEWORK 2 1.3.8. For k ∈ N let Ak = {1, 2, . . . , k }. Then Ak  = k , yet ∪∞ Ak = N. k =1 1.3.12. Let Fm be the family of subsets of N with precisely m elements. Let Nm = N × . . . × N (m times) be the mfold Cartesian product of N with itself, that is, the set of all ordered mtuples whose elements are positive integers. Use induction on m to prove that Nm is countable. The case of m = 1 is clear. The case of m = 2 was proved in class: there exists a bijection from N to N × N. This forms the base of induction. Now, suppose Nm is countable, and prove the same for Nm+1 . By the induction hypothesis, there exists an injection fm : Nm → N. For a = (a1 , . . . , am , am+1 ) ∈ Nm+1 , let a′ = (a1 , . . . , am ). Deﬁne fm+1 by setting fm+1 (a) = f2 (fm (a′ , am+1 )). You can check that fm+1 is injective, hence Nm is countable. Next we deﬁne a map g : Fm → Nm , taking a set of m elements to the mtuple, listing them in the increasing order. This is an injection, hence so is fm+1 ◦ g . Therefore, Fm is countable. Finally, F = ∅ ∪ (∪∞=1 Fm ), hence countable (as a countable union of countable sets. m 2.1.1. (b) First make an important observation: the negative element is unique. More precisely: Lemma. If a, b ∈ R, and a + b = 0, then b = −a. Proof. If a + b = 0, then b + a = 0 (commutativity of addition), hence −a = 0 + (−a) = (b + a) + (−a) = b + (a + (−a)) = b + 0 = b, which is what we need. Thus, for b ∈ R, −b is the unique c ∈ R satisfying b + c = 0. In particular, −(−a) is the unique real number with the property that (−a) + (−(−a)) = 0. But (−a) + a = 0, hence a = −(−a). 2.1.2. (b) As in the previous problem, start with Lemma. If a, b ∈ R, then (−a)b = −ab. Proof. By the distributive law, ab + (−a)b = (a + (−a))b = 0 · b = 0, hence, by the previous lemma, (−a)b = −ab. Therefore, (−a)(−b) = −a(−b) = −((−b)a) = −(−ba) = ba = ab. Problem A: Denote by S the set of all subsets A ⊂ N for which both A and N\A are inﬁnite. Prove that S is uncountable. We can wrote P (N) (the power set of N) as a union of three disjoint sets: S , F , and F ′, the latter being the set of all those subsets of N whose complements belong to F .
1 2 SOLUTIONS FOR HOMEWORK 2 Above, we established that F is countable. The operation A → N\A is a bijection from F to F ′ , hence F ′ is also countable. If S is countable, then so is P (N), which is impossible by Cantor’s theorem (1.3.13). Problem B: Suppose A is an inﬁnite set, and B is a countable set. Prove that the sets A and A ∪ B are equipollent. Note that, by considering B \A instead of B , we can assume that A and B are disjoint. We consider the case of B being countably inﬁnite (the ﬁnite case is handled in a similar fashion). There exists a bijection f : N → B . Moreover, as shown in the notes, A ≥ N, hence there exists an injection g : N → A. Now deﬁne h : A ∪ B → A by setting x ∈ A\g (N) x g (2k ) x = g (k ) for some k ∈ N . h(x) = g (2k − 1) x = f (k ) for some k ∈ N Clearly, h is a bijection. Problem C (a bonus problem – very little partial credit is given): For a set A, denote by AN the set of all inﬁnite sequences of elements of A. Show that RN (the set of all sequences of real numbers) is equipollent with R. You can use the fact that R is equipollent with the power set of N, and identify the latter with the set of all 0 − 1 sequences. By CantorBernsteinSchroeder Theorem, we have to prove the existence of injections f : R → RN , and g : RN → R. f is easy to construct: deﬁne f (x) = (x, x, . . .). To construct g , recall that there exists a bijection φ : R → {0, 1}N . Denote by φi (x) the ith term of the sequence φ(x). There exists a bijection ψ : N → N × N. Deﬁne π and σ by setting ψ (a) = (π (a), σ (a)) (π and σ map N to itself). For x = (x1 , x2 , . . .) ∈ RN , deﬁne g (x) as a 0 − 1 sequence, whose mth element equals ψπ(m) (xσ(m) ). Check that g is an injection! Back to: the syllabus, the main page of the course, the assignment. ...
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This note was uploaded on 09/30/2009 for the course MATH 444 taught by Professor Junge during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 JUNGE
 Sets

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