Solution Sangchul Lee February 19, 2018 1 Solutions of Homework 1 Problem 1.2 Let A and B be nonempty subsets of R + :: {x R : x > 0} which are bounded above. Let us define C = {xy : x A and y B} Show that supc = (supa)(supb). Proof. We show that one bounds the other. Proof of supc (supa)(supb). Let z C be arbitrary. Then there exists x A and y B for which z = xy holds. Since 0 < x supa and 0 < y supb, we have z (supa)(supb). This shows that (supa)(supb) is an upper bound of C and hence supc (sup A)(sup B). Proof of supc (supa)(supb). This direction is much trickier. We present two proofs. 1 st way. We temporarily fix y B. Then for all x A, we have xy supc and hence x (supc)/y for all x A. So (supc)/y is an upper bound of A. Then it follows that supa (supc)/y, or equivalently, y (supc)/(supa). (This is possible because supa > 0.) Since this is true for all y B, we have supb (supc)/(supa) by the same argument. Therefore (sup A)(sup B) supc as required. 2 nd way. Assume otherwise that supc < (supa)(supb). Let ε = (supa)(supb) supc > 0 and notice that (supa)(supb) + supc 0 < 2supB (supa)(supb) + supc 0 < 2supA So there exists x A and y B such that supa ( supc xy > supa which is a contradiction. ε 2supB 2supB ε )( supb = supa ε 2supB < supa, = supb ε 2supA < supb. < x and supb ε ε 2supA 2supA < y. Then with these x and y, ) > (supa)(supb) ε = supc, 2 Homework 2 (This part is skipped.) 1
3 Homework 3 We review some facts which seems intuitively clear but nevertheless deserve some proofs. Lemma (Well-ordering principle). Let S be a non-empty subset of N = {1,2, }. Then S has the smallest element. Proof of Lemma. We prove by contradiction. Let S be a non-empty subset of N which has no smallest element. We seek to obtain a contradiction. Define T = {n N : {1,,n} S = }. We claim that T = N. 1 / S, for otherwise 1 would have been the smallest element of S. So 1 T. Assume that n T. If n+1 S, then every element x of S must satisfy x > n, which leads to the contradiction that n + 1 is the smallest element of S. So n + 1 / S and thus {1,,n + 1} S =, implying that n + 1 T. So by the mathematical induction, we must have T = N. This implies that S =, hence the desired contradiction is obtained. //// Lemma (Pigeonhole principle). Let J 0 = and J n = {1,,n} for n 1. Then for non-negative integers m,n 0, (1) There exists an injection J m J n if and only if m n. (2) There exists a surjection J m J n if and only if m n. (3) There exists a bijection J m J n if and only if m = n. Proof. (1) We first give a sketch of proof and then fill the necessary detail. When m n one can simply consider the function k k from J m to J n, which is obviously an injection. For the other direction, if f : J m J n is an injection, by permuting the set J n we may assume that f (k) = k for k J m. This proves that m = f (m) n. Now we formalize this idea. Assume that f : J m J n is an injection. For each pair of elements i, j J n, not necessarily distinct, we define the function σ i, j : J n J n by j, if k = i σ i, j (k) = i, if k = j k, otherwise. That is, σ i, j permutes i and j but fixed other elements of J n. Then each σ i, j is a bijection on J n. Now we define f 0,, f m by the following recursion f 0 = 0 and f k = σ fk 1 (k),k f k 1 for k = 1,,m. (Those who are familiar with algorithm will recognize this as selection sort.) Then we claim that Claim. For each k J m, f k is injective and f k (i) = i for each i J k. Indeed, when k = 1 this follows from f 1 (1) = σ f (1),1 ( f (1)) = 1. Now suppose that k 2 and the claim holds for f k 1. Since f k 1 (i) = i for all i J k 1, injectivity of f k 1 tells that f k 1 (k) k. So σ fk 1 (k),k fixes the set J k 1 and we have f k (i) = i for i J k 1. Also f k (k) = σ fk 1 (k),k( f k 1 (k)) = k, hence f k fixes J k. Finally, injectivity of f k follows from the injectivity of σ fk 1 (k),k and f k 1. Therefore the claim follows by inducting on k. By the claim, we have m = f m (m) J n. Therefore m n as required. (2) If m n, then the function k min{k,n} is a surjection from J m to J n. Conversely, if f : J m J n is a surjection, we define g : J n J m by g(k) = min f 1 ({k}). This is well-defined by the well-ordering principle. Then g is an injection and hence n m by the previous part. (3) This is a direct consequence of part (1) and (2). 2
Problem 3.1 Let E be a nonempty set. Show that E is infinite if and only if E has the same cardinality with at least one of its proper subset. Proof. (= ) : Let E be infinite. We first establish the following lemma: Lemma. If E is infinite, then E has a countable subset. Proof of Lemma. Assume that we have distinct elements x 0,,x n E. Since the function k x k from {0,,n} to E is injective and E is infinite, this map cannot be surjective and thus we can pick an element x n+1 from E \ {x 0,,x n }. This recursively defines a sequence {x n } n=0 whose terms are all distinct, and therefore {x n : n 0} is a countable subset of E. //// So it boils down to proving the claim for conuntably infinite sets. Let S be countably infinite and enumerate this by S = {x 0,x 1, }. If we let S = {x 1,x 2, }, then S is a proper subset of S and has a bijection between S. Indeed, one such bijection can be constructed by writing f : {0,1, } S for the function f (n) = x n and defining g : S S by g(x) = f ( f 1 (x) + 1). Now for arbitrary infinite set E, use Lemma to find a countable subset S. Now find bijection g : S S to a proper subset S of S using the previous step, and extend g to a function h : E S (E \ S) by setting { g(x), x S h(x) = x, x E \ S Then h is a bijection between E and its proper subset E := S (E \ S). ( =) : Assume that E is finite. This amounts to saying that there exits a bijection f : E J N for some N 0, where we recall that J 0 = and J n = {1,,n} for n 1. If N = 0, then there is no proper subset of E =, so we may assume that N 1. Let f : E J N be a bijection and let F E be a proper subset of E. Then we can pick a point x N E \ F. Now by permuting the order, we may assume that f (x N ) = N. Then it follows that the restriction f F : F J N 1 is an injection. So if M denotes the cardinality of F, i.e. if F J M, then by the pigeonhole principle we have M N 1. Then by the pigeonhole principle again, there is no injection from F J M to E J N. Remark. In this proof we used a seemingly innocuous claim that we can pick an element of a set. However, this turns out to be highly non-trivial in the realm of infinite sets, and is in fact has to be taken as an axiom which is called the axiom of choice (AC). This tells that for any collection of sets {E α } there exists a function f : {E α } α E α such that f (E α ) E α. (Such a function is called a choice function.) Applying this claim to the power set {S : S E} shows that for each subset S of E we can pick an element of S. Problem 3.2 (a) Let X be countable, and let Y be an infinite subset of X. Show that Y is countable. You can follow the proof in the book if you would like to, the point is to make you understand the proof. (b) Using (a), show that if {X n } n=1 is a family of countable sets, then n N X n is countable. Explain where we need (a). Proof. (a) We first prove the claim when X = N. We recursively define {n k } k=1 by the following procedure: Let n 1 = miny, and if n 1,,n k are chosen, let n k+1 = min(y \ {n 1,,n k }). We inductively check that Y \ {n 1,,n k } is 3
non-empty for each k, hence this prcedure is well-defined. Now we prove that k n k is a bijection from N to Y. To this end, we claim that Claim. For each k N, we have (1) n 1 < n 2 < < n k, (2) k n k, and (3) Y {1,,n k } = {n 1,,n k }. Indeed this is an easy exercise on mathematical induction. When k = 1, both (1) and (2) are obvious and (3) is immediate from the definition n 1 = miny. Assume that the claim is true for k. Then every n Y \ {n 1,,n k } must satisfy n > n k and hence n k+1 > n k, proving (1) for k + 1. Then n k+1 n k + 1 k + 1 and hence (2) holds for k + 1. Finally, if n Y satisfies n n k+1 then either n n k so that n {n 1,,n k } by the induction hypothesis, or n k < n n k+1 so that n = n k+1 in view of the definition of n k+1. This proves (3) for k + 1 As an immediate consequence, (1) implies that k n k is injective. To show that this map is surjective, assume otherwise. Then k 0 := min(y \ {n k : k N}) exists. But this implies that k 0 Y {1,,k 0 } (2) Y {1,,n k0 } (3) = {n 1,,n k0 }, a contradiction. Therefore k n k is surjective and hence bijective. For a general countable set X and its infinite subset Y, consider a bijection φ : X N and notice that this induces a bijection from Y to φ(y ). Since φ(y ) is an infinite subset of N, the previous step tells that φ(y ) is countable, hence the same holds for Y. (b) Writing S = n N X n for the union, we find that S is infinite. For each n N choose a bijection f n : N X n. Then this induces a map f : N N S given by (n,k) f n (k). Also pick a bijection g : N N N and consider the composition φ = f g : N S. We begin by noting that f is surjective. Indeed, for each x S there exists n N such that x X n, and hence with k = fn 1 (x) we have f (n,k) = f n (k) = x. Then φ is also surjective. So we can define ψ : S N by ψ(x) = minφ 1 ({x}). Since each set φ 1 ({x}) is non-empty, φ is indeed well-defined. It is injective, since for diffferent x,y the corresponding sets φ 1 ({x}) and φ 1 ({y}) are disjoint. So ψ induces a bijection between S and its image T := ψ(s). Since T is an infinite subset of N, it is countable by part (a) and therefore S is countable. Problem 3.3 Let us denote (0,1) = {x R : 0 < x < 1}. For x (0,1), let f (x) = (a 1,a 2,a 3, ) {0,1,,9} N where the a k {0,1,,9} are chosen uniquely by the decimal expansion constructed in class. Recall that in particular we showed that r n := 0.a 1 a 2 a n satisfies r n x < r n + 1/10 n. (a) Show that f : (0,1) {0,1,,9} N is a one-to-one map. (b) Show that the set S := { f (x) : x (0,1)} is uncountable. (c) Conclude that (0,1) is uncountable, and thus so is R. Proof. (a) Suppose that x,y (0,1) are distinct. We may assume that x < y. By the Archimedean property, there exists n 1 such that n(y x) > 1. Now recall that 10 n n. (This can be easily proved by invoking induction. 10 1 = 10 1, 4
and if 10 n n then 10 n+1 = 10 10 n 10n = n + 9n n + 1.) This gives y > x + 10 n. So if f (x) = (a 1,a 2, ) and f (y) = (b 1,b 2, ), then 0.a 1 a n x < y 10 n < 0.b 1 b n and hence we must have (a 1,,a n ) (b 1,,b n ). This tells that f (x) f (y) as required. (b) For an economic proof, it suffices to show that S contains an uncountable subset. (We already proved that, if S is countable, then every infinite subset is also countable. We are utilizing the contrapositive of this statement.) Consider the set {1,2} N. For each (a 1,a 2, ) {1,2} N we associate the number x = sup{0.a 1 a n : n 1}. If we write r n = 0.a 1 a n, then for any m,n 1 we have If n m then r n r m r( m + 10 m. ) If n > m, then r n r m + 2 + + 2 10 m+1 10 n = r m + 2 ( 1 9 10 m 10 1 ) n rm + 2 9 1 10 m. So x is indeed a real number and satisfies r m x r m + (2/9)10 m < r m + 10 m for all m 1. This tells that f (x) = (a 1,a 2, ) and hence {1,2} N S. Finally we claim that {1,2} N is uncountable. To this end, we invoke the famous Cantor s diagonal argument. Let φ : N {1,2} N be any function. Then we construct an element b = (b 1,b 2, ) of {1,2} N as follows: { 1, if a n = 2 where φ(n) = (a 1,a 2, ) b n = 2, if a n = 1 where φ(n) = (a 1,a 2, ).. By the construction, we have b φ(n) for all n. So it follows that φ(n) does not contain b. Since this is true for all functions f : N {1,2} N, it follows that there is no surjection from N to {1,2} N and in particular {1,2} N is uncountable. (c) Since (0,1) S and S is uncountable, so is (0,1). Now since R contains (0,1) as subset, the same is true for R. Problem 3.4 Find a 1 1, onto map between [0,1] and (0,1). Proof. Write S = (0,1) \ { 1 n : n 1}. Then both (0,1) \ S = { 1 n : n 2} and [0,1] \ S = {0} { 1 n : n 1} are countable sets, and hence there is a bijection f 0 : (0,1) \ S [0,1] \ S. Now extend f 0 to a function f : (0,1) [0,1] by letting { f 0 (x), x (0,1) \ S f (x) = x, x S. This function is a bijection from (0,1) to [0,1]. 4 Homework 4 Problem 4.1 Let E be a non-empty subset of R that is both open and closed. Show that E = R. 5
Proof. Assume otherwise that E R. Then both E and E c are non-empty and both open and closed. Interchanging the role of E and E c if necessary, we can pick a E and b E c such that a < b. Then consider γ = sup(e [a,b]). We will show that γ E and γ E c, thereby yielding a contradiction. Notice that E [a,b] is a non-empty, bounded and closed subset of R. So γ = sup(e [a,b]) E [a,b] follows from Theorem 2.28. Since γ [a,b] and γ / E c, we must have γ < b. Then (γ,b] E c. Since E c is closed, we must have [γ,b] = (γ,b] E c as well. This implies γ E c, a contradiction! Therefore we must have E = R. Problem 4.2 Show that any open subset of R is a countable or finite union of disjoint open intervals. Proof. Let U be an opne subset of R. For each r U, we define as the union of all open intervals in U that contain r. I(r) = {(a,b) : a < r < b + and (a,b) U} We begin by making some easy obsevations. Since each r U is an interior point of U, we have N ε (r) I(r) for some ε > 0. In particular, I(r) is non-empty. Also, since it is an arbitrary union of open subsets of U, I(r) is also an open subset of U. Next, we claim that I(r) is an open interval. We consider different cases. Case 1. If I(r) is bounded below, then α := infi(r) is a real number. Since N ε (r) I(r) for some ε > 0, we know that α r ε < r. We show that I(r) (,r] = (α,r]. Since α is the infimum of I(r), no number less than α lies in I(r). Also, since α cannot be an interior point of I(r), we must have α / I(r). So it suffices to show that every x (α,r) lies in I(r). Indeed, let x (α,r). Then there exists x I(r) such that x (α,x). Then there exists an open interval (a,b) U such that x,r (a,b). This implies that x [x,r] (a,b) I(r) as desired. Case 2. If I(r) is not bounded below, then we show that I(r) (,r] = (,r]. The proof is almost similar to the previous case. Let x (,r). Since x is not a lower bound of I(r), there exists x I(r) such that x < r. Then there exists an open interval (a,b) U such that x,r (a,b). So x [x,r] (a,b) I(r). Same argument applies to I(r) [r, ), showing that we must have I(r) [r, ) = [r,β) for some β with r < β +. Combining together, we have I(r) = (α, β) for some α < r < β +. This proves the claim. Next, we prove that for r,s U we have either I(r) = I(s) or I(r) I(s) =. There is nothing to prove if either r = s or I(r) I(s) =. So it suffices to consider the case where r s and I(r) I(s). In this case, I(r) I(s) is again an open interval that contains both r and s. By the definition, this implies that I(r) I(s) I(r) and I(r) I(s) I(s), hence we have I(r) = I(s). Finally, for each x U there exists r Q U such that r I(x), which then implies that I(r) I(x) and hence x I(r). Therefore we have U I(r). r Q U As we already observed, I(r) are subsets of U. So the inclusion r Q U I(r) U is clear. Since Q U is at most countable, this completes the proof. 6
Problem 4.3 For the space of polynomials let us define { X := P(x) = n k=0 d(p,q) := a k x k : n N and a k R ( m k=0 a k b k 2 ) 1/2 }, for two polynomials P(x) = n k=0 a kx k and Q(x) = m k=0 b kx k in X, with m n: we define a k = 0 if k > n. (a) Show that d is a metric on X. (b) Let 0 be the constant zero polynomial. Show that the set B := {P(x) X : d(p,0) 1} is closed and bounded but not compact. Proof. (a) We check that d satisfies all the defining property of a metric. We will write P(x) = k a k x k, Q(x) = k b k x k and R(x) = k c k x k for generic polynomials P,Q,R X. If P Q, then there exists k 0 such that a k b k. Then we have d(p,q) a k b k > 0. That d(p,p) = 0 is obvious. Again, d(p,q) = d(q,p) is clear from definition. Let m 0 be the maximum degree of P,Q,R. Then for p = (a 0,,a m ), q = (b 0,,b m ) and r = (c 0,,c m ) in R m+1, notice that d(p,q) = p q where the right-hand side is the Euclidean distance between p and q. Of course, similar statement holds for d(q,r) and d(p,r) as well. So by borrowing the fact that Euclidean distance in R m+1 is a metric, we have Therefore d is a metric on X. d(p,r) = p r p q + q r = d(p,q) + d(q,r). (b) B is bounded since B N 2 (0). To show that B is closed, notice that we have the following very general statement: Lemma. Let (X,d) be any metric space. Then for any p X and r > 0, the set D r (p) = {q X : d(p,q) r} is closed in X. To this end, we show that D r (p) c is open. Let q D r (p) c. Then d(p,q) > r. Set ε = d(p,q) r > 0. Then for each q N ε (q), we have d(p,q ) + d(q,q) d(p,q) d(p,q ) d(p,q) d(q,q) > d(p,q) ε = r and hence q D r (p) c. This tells that N ε (q) D r (p) c. Therefore every point of D r (p) c is an interior point of D r (p) c and it is open in X as required. //// Finally, we prove that B is not compact. To this end, we consider the subset E = {x n : n 0} of B. Notice that d(0,x n ) = 1 for each n, so we indeed have E B. Next we claim that E is closed in X. If P(x) = m k=0 a kx k is not in E, then we must have d(p,x n ) > 0 for all n. But if n > m then d(p,x n ) 1. So inf{d(p,x n ) : n 0} min({1} {d(p,x n ) : 0 n m}) > 0 This tells that N ε (P) E = for a sufficiently small ε > 0. Also, if n l then d(x l,x n ) = 2. So N 2 (xn ) E = {x n }. Combining altogether, we find that E has no limit point, hence Ē = E and E is closed. We are ready to conclude our proof. If B is compact, then E is also compact by Theorem 2.35. But this contradicts the fact that {N 2 (xn ) : n 0} is an open cover of E which has no finite subcover. Therefore B cannot be compact. 7
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