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Unformatted text preview: SOLUTIONS TO EXERCISES FOR MATHEMATICS 205A Fall 2014 General Remarks The main objective of the course is to present basic graduate level material, but an important secondary objective of many point set topology courses is to is to build the students’ skills in writing proofs and communicating them to others. Higher skill levels are needed because proofs in graduate level mathematics courses are frequently longer, more abstract and less straightforward than their counterparts in undergraduate level courses. The files with solutions to exercises are named solutions*.pdf, where ∗ is some number, perhaps followed by a letter to indicate supplementary content like drawings to accompany arguments. Exercises marked with one or two asterisks should be viewed as having lower priorities unless their solutions are specifically assigned as readings for the course. These solutions are posted mainly for students to compare with their own efforts and to determine whether their solutions are correct or can be improved upon (compensating in part for the lack of homework grading personnel), but in some cases the solutions might also be useful if and when a student reaches an impasse. At some points, solutions to some of the more complicated exercises may be parts of reading assignments. However, as a general rule the solutions should not be viewed as an excuse for not trying to work the exercises, especially those that are specifically assigned in course announcements (i.e., the aabUpdate∗ files in the course directory). Problems at the level of those in the assigned exercises will appear on both course and qualifying examinations, so it is important for students to be able to work out solutions on their own. Suggestions for working exercises The directory file polya.pdf contains a systematic and general list of suggestions for approaching and solving mathematical problems, and the file mathproofs.pdf — which was written for undergraduate level courses but is still relevant at higher levels — discusses the more formal aspects of mathematical proofs. Here are a few additional comments: 1. Many exercises can be solved by imitating arguments in the course text or notes, so it is usually worthwhile to see if an exercise can be analyzed by modifying a previously seen argument. The following quotation from the Poetics of Aristotle (384–322 B.C.E.; see Section I, Part IV) seems appropriate: The instinct of imitation is implanted in man from childhood, one difference between him and other animals being that he is the most imitative of living creatures, and through imitation learns his earliest lessons. 1 The differences between human and animal behavior in some species might not be as vast as they seemed to the ancient Greeks, but the passage still reflects the importance of imitation in human thought and action. 2. The first efforts (and in many cases subsequent efforts!) at solving exercises will not necessarily be as clear or polished as some proofs and solutions in textbooks or the files posted to the course directory. Often the first attempts to find solutions are at least somewhat messy, and they usually get better as a result of increased experience, skills, and trial and error. The following quotation due to A. S. Besicovitch (1891–1970) summarizes everything in a somewhat ironic manner: A mathematician’s reputation rests on the number of bad proofs he has given. 3. It is usually good to try anything that might work rather than not getting started with work on a problem. This advice reflects a frequently repeated quotation due to L. R. (“Yogi”) Berra (1925–): When you come to a fork in the road ... Take it. Likewise, if a solution to a problem is not apparent after some thought, it is often worthwhile to be systematic and look at everything that can be said about the given situation, no matter how insignificant it might seem at first. This is summarized in a quotation from one of the Sherlock Holmes stories by A. C. Doyle (1859–1930): You know my method. It is founded upon the observation of trifles. 4. Although rigorous mathematical proofs must be expressed in words and symbols rather than pictures, in many cases good (or even not so good) drawings are extremely helpful — sometimes even indispensable — sources for insights which can suggest an approach to proving a mathematical statement. This is particularly true for courses with substantial amounts of geometric content like Mathematics 205A. The following two quotations reflect this point: Geometry is the science of correct reasoning [based] on incorrect figures. (1887–1985) — G. Poly´a For me, following a geometrical argument purely logically, without a picture for it constantly in front of me, is impossible. — F. Klein (1849–1925) 5. The first step in putting together a logical argument is to come up with something that appears to be correct, but one important test of an argument’s clarity and validity is to refine the ideas so that they become equally clear and convincing to someone else who is reasonably well-informed about the subject. This generally requires healthy doses of skepticism and self-criticism; these can certainly be overdone, but initially most of us need to be more concerned about not going far enough. The next page contains two well-known satirical cartoons by Sidney Harris (1933–) on convincing arguments. We should also note that that there are several anthologies of Harris’ excellent cartoons on humorous aspects of scientists and their work, and they are definitely worth reading and viewing. 2 ;ljljklkl;ljkljk SOLUTIONS TO EXERCISES FOR MATHEMATICS 205A — Part 1 Fall 2014 I. Foundational material I.1 : Basic set theory Problems from Munkres, § 9, p. 64 2. (a) − (c) For each of the first three parts, choose a 1–1 correspondence between the integers or the rationals and the positive integers, and consider the well-orderings that the latter inherit from these maps. For each nonempty subset, define the choice function to be the first element of that subset with respect to the given ordering. TECHNICAL FOOTNOTE. (This uses material from a graduate level measure theory course.) In Part (d) of the preceding problem, one cannot find a choice function without assuming something like the Axiom of Choice. The following explanation goes beyond the content of this course but is hopefully illuminating. The first step involves the results from Section I.3 which show that the set of all functions from {0, 1} to the nonnegative integers is in 1–1 correspondence with the real numbers. If one could construct a choice function over all nonempty subsets of the real numbers, then among other things one can prove that that there is a subset of the reals which is not Lebesgue measurable without using the Axiom of Choice (see any graduate level book on Lebesgue integration; for example, Section 3.4 of Royden). On the other hand, there are models for set theory in which every subset of the real numbers is Lebesgue measurable (see R. Solovay, A model of set theory in which every set of reals is Lebesgue measurable, Annals of Math. (2) 92 (1970), pp. 1–56). — It follows that one cannot expect to have a choice function for arbitrary families of nonempty subsets of the reals unless one makes some extra assumption related to the Axiom of Choice. 5. (a) For each b ∈ B pick h(b) ∈ A such that f (a) = b; we can find these elements by applying the Axiom of Choice to the family of subsets f −1 [{b}] because surjectivity implies each of these subsets is nonempty. It follows immediately that b = f ( h(b) ). (b) For each x ∈ A define a function gx : B → A whose graph consists of all points of the form f (a), a  ∈ B×A togther with all points of the form (b, x) if b does not lie in the image of A. The injectivity of f implies that this subset is the graph of some function g x , and by construction we have gx o f (a) = a for all a ∈ A. This does NOT require the Axiom of Choice; for each x ∈ A we have constructed an EXPLICIT left inverse to f . — On the other hand, if we had simply said that one should pick some element of A for each element of B − f [A], then we WOULD have been using the Axiom of Choice. 1 Additional exercise 1. The commutativity law for ⊕ holds because B ⊕ A = (B − A) ∪ (A − B) by definition and the commutativity of the set-theoretic union operation. The identity A ⊕ A = ∅ follows because A ⊕ A = (A − A) ∪ (A − A) = ∅ and A ⊕ ∅ = A because A ⊕ ∅ = (A − ∅) ∪ (∅ − A) = A ∪ ∅ = A . In order to handle the remaining associative and distributive identities it is necessary to write things out explicitly, using the fact that every Boolean expression involving a finite list of subsets can be written as a union of intersections of subsets from the list. It will be useful to introduce some algebraic notation in order to make the necessary manipulations more transparent. Let X ⊃ A ∪ B ∪ C denote the complement of Y ⊂ X by Yb (or by Y b if Y is some compound algebraic expression), and write P ∩ Q simply as P Q. Then the symmetric difference can be b ∪ (B A). b It then follows that rewritten in the form (AB)     b ∪ BA b C b ∪ C AB b ∪ BA b b= (A ⊕ B) ⊕ C = AB     bC b ∪ BA bC b ∪ C (A b ∪ B)(B b ∪ A) = AB bC b ∪ BA bC b∪C A bB b ∪ AB = AB Similarly, we have bC b ∪ AB b C b∪A bBC b ∪ ABC . AB     b ∪ CB b b ∪ BC b ∪ CB b A b= A ⊕ (B ⊕ C) = A B C   b ∪ C)(C b ∪ B) ∪ B C bA b ∪ CB bA b = A(B bC b ∪ BC) ∪ B C bA b ∪ CB bA b= A (B bC b ∪ AB b C b∪A bBC b ∪ ABC . AB This proves the associativity of ⊕ because both expressions are equal to the last expression displayed above. The proof for distributivity is similar but shorter (the left side of the desired equation has only one ⊕ rather than two, and we only need to deal with monomials of degree 2 rather than 3): b ∪ C B) b = AB C b ∪ ABC b A (B ⊕ C) = A(B C AB ⊕ AC = (AB)(AC)b ∪ (AC)(AB)b =     b ∪ C) b ∪ AC(A b ∪ B) b = AB C b ∪ ABC b AB(A Thus we have shown that both of the terms in the distributive law are equal to the same set. 2 I.2 : Products, relations and functions Problem from Munkres, § 4, p. 44 4. (a) This is outlined in the course notes. Additional exercises 1. (i) Suppose that (x, y) lies in the left hand side. Then x ∈ A and y ∈ B ∩ D. Since the latter means y ∈ B and y ∈ D, this means that (x, y) ∈ (A × B) ∩ (A × D) . Now suppose that (x, y) lies in the set displayed on the previous line. Since (x, y) ∈ A × B we have x ∈ A and y ∈ B, and similarly since (x, y) ∈ A × D we have x ∈ A and y ∈ D. Therefore we have x ∈ A and y ∈ B ∩ D, so that (x, y) ∈ A × (B ∩ D). Thus every member of the the first set is a member of the second set and vice versa, and therefore the two sets are equal. (ii) Suppose that (x, y) lies in the left hand side. Then x ∈ A and y ∈ B ∪ D. If y ∈ B then (x, y) ∈ A × B, and if y ∈ D then (x, y) ∈ A × D; in either case we have (x, y) ∈ (A × B) ∪ (A × D) . Now suppose that (x, y) lies in the set displayed on the previous line. If (x, y) ∈ A × B then x ∈ A and y ∈ B, while if (x, y) ∈ A × D then x ∈ A and y ∈ D. In either case we have x ∈ A and y ∈ B ∪ D, so that (x, y) ∈ A × (B ∩ D). Thus every member of the the first set is a member of the second set and vice versa, and therefore the two sets are equal. (iii) Suppose that (x, y) lies in the left hand side. Then x ∈ A and y ∈ Y − D. Since y ∈ Y we have (x, y) ∈ A × Y , and since y 6∈ D we have (x, y) 6∈ A × D. Therefore we have A × (Y − D) ⊂ (A × Y ) − (A × D) . Suppose now that (x, y) ∈ (A × Y )−(A × D). These imply that x ∈ A and y ∈ Y but (x, y) 6∈ A×D; since x ∈ A the latter can only be true if y 6∈ D. Therefore we have that x ∈ A and y ∈ Y − D, so that A × (Y − D) ⊃ (A × Y ) − (A × D) . This proves that the two sets are equal. (iv) Suppose that (x, y) lies in the left hand side. Then we have x ∈ A and y ∈ B, and we also have x ∈ C and y ∈ D. The first and third of these imply x ∈ A ∩ C, while the second and fourth imply y ∈ B ∩ D. Therefore (x, y) ∈ (A ∩ C) × (B ∩ D) so that (A × B) ∩ (C × D) ⊂ (A ∩ C) × (B ∩ D) . Suppose now that (x, y) lies in the set on the right hand side of the displayed equation. Then x ∈ A ∩ C and y ∈ B ∩ D. Since x ∈ A and y ∈ B we have (x, y) ∈ A × B, and likewise since x ∈ C and y ∈ D we have (x, y) ∈ C × D, so that (A × B) ∩ (C × D) ⊃ (A ∩ C) × (B ∩ D) . 3 Therefore the two sets under consideration are equal. (v) Suppose that (x, y) lies in the left hand side. Then either we have x ∈ A and y ∈ B, or else we have x ∈ C and y ∈ D. The first and third of these imply x ∈ A ∪ C, while the second and fourth imply y ∈ B ∪ D. Therefore (x, y) is a member of (A ∪ C) × (B ∪ D) so that (A × B) ∪ (C × D) ⊂ (A ∪ C) × (B ∪ D) . Supplementary note: To see that the sets are not necessarily equal, consider what happens if A ∩ C = B ∩ D = ∅ but all of the four sets A, B, C, D are nonempty. Try drawing a picture in the plane to visualize this. (vi) Suppose that (x, y) lies in the left hand side. Then x ∈ X and y ∈ Y but (x, y) 6∈ A × B. The latter means that the statement x ∈ A and y ∈ B is false, which is logically equivalent to the statement either x 6∈ A or y 6∈ B .  If x 6∈ A, then it follows that (x, y) ∈ (X − A) × Y , while if y 6∈ B then it follows that (x, y) ∈ X × (Y − B) . Therefore we have   X ×Y − A×B ⊂   X × (Y − B) ∪ (X − A) × Y . Suppose now that (x, y) lies in the set on the right hand side of the containment relation on the displayed line. Then we have (x, y) ∈ X × Y and also either x 6∈ A or y 6∈ B . The latter is logically equivalent to x ∈ A and y ∈ B and this in turn means that (x, y) 6∈ A × B and hence proves the reverse inclusion of sets. 2. Since the object of this exercise is to address ambiguities in notation, the key to solving this exercise is to find unambiguous ways of distinguishing between the two interpretations of the symbolism f −1 [C]. The formulation in the exercise goes part way towards doing this, but one must still be careful. We shall systematically use the inverse function identities g o f = idA and f o g = idB which relate a 1–1 onto function f : A → B with its inverse function g : B → A. As in the statement of the exercise, let C ⊂ B. Suppose that a ∈ IM AGE g[C]. Then a = g(b) for some b ∈ C, which implies that f (a) = f (g(b)) = b, so that f (a) ∈ C and hence a ∈ IN V ERSE IM AGE f −1 [C]. Conversely, suppose that a ∈ IN V ERSE IM AGE f −1 [C]. Then f (a) ∈ C, so that g(f (a)) = a ∈ C, which means that a ∈ IM AGE g[C]. Therefore each of the sets IM AGE g[C] and IN V ERSE IM AGE f −1 [C] is contained in the other, so the two sets are equal. 3. We shall first prove that R# is an equivalence relation. The relation is reflexive. The definition of R # stipulates that x R# x. The relation is symmetric. By the preceding step we need only consider the case where x 6= y. If we are given a finite sequence {v 0 · · · , vm } as described in the definition such 4 that v0 = x and vm = y, then the reverse sequence with w j = vm−j satisfies the criterion in the definition implies with w0 = y and wm = x. The relation is transitive. Suppose that we have a finite sequence {v 0 · · · , vm } as described in the definition such that v 0 = x and vm = y, and another finite sequence {u0 · · · , un } as described in the definition such that u 0 = y and un = z, then we can concatenate (string together) the original sequences into a new sequence {t 0 · · · , tm+n } such that tj = vj if j ≤ m and tj = uj−m if j ≥ m (the formulas are consistent at the overlapping value m, for which tm = y). This new sequence still satisfies the criterion in the definition with t0 = x and tm+n = z. To complete the proof, we need to verify that if S is an equivalence relation such that x S y whenever x R y, then u R# v implies that u S v. If u = v then the latter follows because we are working with equivalence relations, which are reflexive. If u 6= v and u S v, let {t 0 · · · , tm } be a sequence starting with u and ending with v such that the for each i we have either t i R ti+1 or ti+1 R ti . Then by the hypotheses on S we know that t i S ti+1 for all i, and therefore by repeated application of the transitivity property we have u S v. 4. We shall refer to the file math205Asolutions01a.pdf for drawings which may help explain the underlying ideas; as usual, the proof must be written so that it does not formally depend upon these drawings. The first step is to show that if (i, j) ∈ E, then every point of the form (i + t, j + t) in B — where t runs through all admissible integers such that the point in question belongs to B — also lies in E. In other words, if i0 − j 0 = i − j, then (i0 , j 0 ) E (i, j). For points in B the difference values i − j are the 15 integers between ± 7, so this shows that there are at most 15 equivalence classes (in the first drawing, the squares with i − j = CONSTANT are on the diagonal lines and have the same color). To prove the assertion in the first sentence, observe that (i, j) R (i + ε, j + ε) for ε = ±1 by definition, and by definition of E this yields (i, j) E (i + ε, j + ε). The statement for general values of t now follows by repeated application of the final assertion in the previous sentence and the transitivity of E. Next, let F be the binary relation with (i 0 , j 0 ) F (i, j) if i0 + j 0 and i + j are both even or both odd. This is an equivalence relation by one of the exercises in Sutherland and the fact that two ordered pairs are F are related if and only if they have the same values under the function ϕ : B → {EVEN, ODD} whose value is determined by whether i+j is even or odd. The definition of R implies that if (i, j) R (p, q) then both i+j and p+q are even or odd, and therefore (i, j) F (p, q) whenever (i, j) R (p, q). By Exercise 0, it follows that (i, j) F (p, q) whenever (i, j) E (p, q), and since F has two equivalence classes the equivalence relation E must also have at least two equivalence classes. Finally, we need to show that E has exactly two equivalence classes. The idea is similar to that of the first step; namely, if (i, j) ∈ E, then every point of the form (i + t, j − t) in B — where t runs through all admissible integers such that the point in question belongs to B — also lies in E. The main difference in the argument is the need to observe that we also have (i, j) R (i + ε, j − ε) for ε = ±1 by the definition of R. By the same reasoning as in the first step, this implies that if i0 + j 0 = i + j, then (i0 , j 0 ) E (i, j). — To conclude the argument, it suffices to observe that the set of all (i, j) ∈ B with i + j = 9 the difference i − j takes all odd values between −7 and +7, while the set of all (i, j) with i + j = 8 takes all even values between −6 and +6 (in the second drawing, observe how the two lines with slope −1 cut through all the lines with slope +1). This proves that there are at most two equivalence classes for E, and by the preceding paragraph there must be precisely two equivalence classes. 5 5. It turns out that, in order to make things less repetitive, the best place to start is by observing that if [x] = [y] then x S y. This follows from the reflexive property of the equivalence relation R 2 . Note that this also yields the reflexive property for S. Suppose now that x S y, so that [x] R2 [y]. Since R2 is an equivalence relation, this means that [y] R2 [x], which in turn implies that y S x. Finally, suppose that x S y and y S z, so that [x] R2 [y] and [y] R2 [z]. Since R2 is transitive we have [x] R2 [z], and this yields x S z, so that S is an equivalence relation on X. I.3 : Cardinal numbers Problem from Munkres, § 7, p. 51 4. (a) Let Q[t] denote the ring of polynomials with rational coefficients, and for each integer d > 0 let Q[t]d denote the set of polynomials with degree equal to d. There is a natural identification of Q[t]d with the subset of Qd+1 consisting of n-tuples whose last coordinate is nonzero, and therefore Q[t]d is countable. Since a countable union of countable sets is countable (Munkres, Theorem 7.5, pp. 48–49), it follows that Q[t] is...
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