Math 172 HW 1 Solutions

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1 Math 172 HW 1 Solutions Joey Zou April 15, 2017 Problem 1: Prove that the Cantor set C constructed in the text is totally disconnected and perfect. In other words, given two distinct points x, y C, there is a point z C that lies between x and y, and yet C has no isolated points. Solution: To show that C is totally disconnected, let x, y C with x y. We need to show the existence of z C lying between x and y. Following the hint, since x y, we have x y > 0 = x y > 1/3 k for some integer k, and hence x and y belong to two different intervals in C k, as each interval in C k has length 1/3 k. Between any two distinct intervals in C k, there is a gap consisting of points not belonging to C k (and hence not belonging to C); hence, there exist points between x and y not belonging to C. To show that C has no isolated points, given x C and any ɛ > 0, we need to show there exists y C with 0 < x y < ɛ. Choose k large enough so that 1/3 k < ɛ. For x C, we have x C k, and since C k consists of a disjoint union of intervals of length 1/3 k, each of the endpoints of the interval in C k containing x is at most 1/3 k (and hence at most ɛ) distance away from x, and at least one such endpoint is not x. So let y be such an endpoint. By construction, the endpoints of the intervals all belong to C, so y C. Thus we have found y C (not equal to x) such that 0 < x y < 1/3 k < ɛ, as desired. Problem 2 The Cantor set C can also be described in terms of ternary expansions. (a) Every number in [0, 1] has a ternary expansion x = a k 3 k, where a k = 0, 1, or 2. Note that this decomposition is not unique since, for example, 1/3 = 2/3 k. Prove that x C if and only if x has a representation as above where every a k is either 0 or 2. Solution: Notice that we can write { } C k = a i 3 i a i = 0 or 2 for all 1 i k. i=1 It follows that C, as the intersection of the C k s, consists exactly of the numbers where a i is 0 or 2 for all i. k=2 1

2 (b) The Cantor-Lebesgue function is defined on C by b k F (x) = 2 k if x = a k 3 k, where b k = a k /2. In this definition, we choose the expansion of x in which a k = 0 or 2. Show that F is well-defined and continuous on C, and moreover F (0) = 0 as well as F (1) = 1. Solution: To show F is well-defined, it is enough to show that if x C (which by part (a) is equivalent to x having some ternary expansion where all the digits are 0 or 2), then x has a unique ternary expansion with all digits 0 or 2. Thus, consider the sums a k 3 k and a k 3 k, where the sequences {a k } and {a k } consist of all 0 s and 2 s and are not equal. Suppose K is the smallest k for which a k a k. Then a k a k = 0 for k < K, and a K a K = 2. It follows that a k 3 k a k3 k = (a K a K)3 K + (a k a k)3 k and since a k a k 2 for each k as a k, a k {0, 2}, we have (a k a k)3 k 2 3 k = 2 3 (K+1) 1 1/3 = 3 K and thus In particular, a k 3 k a k3 k = (a K a K)3 K + a K a K 3 K a k 3 k 2 3 K 3 K = 3 K. (a k a k)3 k (a k a k)3 k a k 3 k if {a k } {a k }. Thus, every element of C has a unique ternary expansion where all the digits are 0 or 2. Notice the above calculation also shows that if a k 3 k a k 3 k < 3 K, then in fact a k = a k for all k K. For such sequences, if b k = a k /2 and b k = a k /2, then b k = b k for k K, b k b k 1 for k > K, and b k 2 k b k2 k = (b k b k)2 k b k b k 2 k 2 k = 2 K. Thus, if x, y C satisfy x y < 3 K, then F (x) F (y) 2 K. This shows that F is continuous on C (in fact uniformly continuous): indeed, for any ɛ > 0, if we take 2 K < ɛ, then for δ = 3 K we have x y < δ = 3 K = F (x) F (y) 2 K < ɛ. Finally, it is clear that F (0) = 0 and F (1) = 1. 2

3 (c) Prove that F : C [0, 1] is surjective, that is, for every y [0, 1] there exists x C such that F (x) = y. Solution: This follows immediately from the fact that every number in [0, 1] has a binary expansion of the form a k {0, 2}. b k 2 K, with b k {0, 1}, i.e. each b k is of the form a k /2 with (d) One can also extend F to be a continuous function on [0, 1] as follows. Note that if (a, b) is an open interval of the complement of C, then F (a) = F (b). Hence we may define F to have the constant value F (a) in that interval. Solution: The only thing to check is that the extension of F from C to [0, 1] (which I will denote F : [0, 1] [0, 1]) is indeed continuous. In part (b) we showed that F was uniformly continuous on C (even if you showed just continuity and not uniform continuity, the fact that C is compact would imply uniform continuity anyway). That is, for any ɛ > 0, there exists δ > 0 such that x, y C, x y < δ = F (x) F (y) < ɛ. I claim that this choice of δ works for F as well; i.e. for ɛ > 0, if δ > 0 is as in the F case, then x y < δ = F (x) F (y) < ɛ. Indeed, suppose x, y [0, 1] satisfied x y < δ. We aim to show F (x) F (y) < ɛ. If x and y belong to the same interval in the complement of C, then F (x) = F (y) since F is constant on that interval, and we are done. Otherwise, there exists a point in C between x and y (including possibly x and y themselves), so let x = inf{z C x z y} and y = sup{z C x z y}, and notice that the inf and sup are actually attained since C is closed, i.e. x, y C. Furthermore, if x = x, i.e. x C, then F (x) = F (x ), while if x > x, i.e. x C, then z C for all x z < x by minimality of x, and hence x is the right endpoint of the interval in the complement of C to which x belongs, so F (x) = F (x ) as well. Similarly, F (y) = F (y ). Finally, since x and y are between x and y, we have x y x y < δ, and hence F (x ) F (y ) < ɛ (as x, y C). Thus, we have F (x) F (y) = F (x ) F (y ) < ɛ as desired. Problem 3 Consider the unit interval [0, 1], and let ξ be a fixed real number with 0 < ξ < 1 (the case ξ = 1/3 corresponds to the Cantor set C in the text). In stage 1 of the construction, remove the centrally situated open interval in [0, 1] length ξ. In stage 2, remove two central intervals each of relative length ξ, one in each of the remaining intervals after stage 1, and so on. Let C ξ denote the set which remains after applying the above procedure indefinitely. (a) Prove that the complement of C ξ in [0, 1] is the union of open intervals of total length equal to 1. Solution: Let C k be the set left over after k steps in the construction, similar to the C k in the standard Cantor set construction. The complement Ck C of C k for any (finite) k is certainly a union of disjoint open intervals in [0, 1], and since C k+1 is constructed from C k by removing open intervals which were originally contained in the interior of C k and hence do not intersect Ck C, it follows that CC k+1 consists of the union of C k and these new intervals, all of which are disjoint from the intervals in C k. Thus, C k+1 just adds more disjoint intervals to the disjoint union of intervals in C k. Since C ξ = C k, it follows that Cξ C = Ck C, and since {CC k } is nested (i.e. C C k CC k+1 ), with each CC k+1 adding more disjoint intervals to the disjoint union, it 3

4 follows that the union of all Ck C is a collection of disjoint open intervals as well. (A more direct way to see this is to note that C ξ contains 0 and 1 and is closed, since it is the intersection of closed sets, and hence Cξ C is an open subset of [0, 1] not containing 0 nor 1, and hence an open subset of (0, 1), i.e. an open subset of R, which can always be expressed as a union of countably many disjoint open intervals.) To find the sum of the length of the intervals in Cξ C, we note that there are countably many such intervals, and since the lengths are nonnegative, we may rearrange and group the lengths as needed to calculate the sum. Notice that each interval is a subset of some Ck C. So let J be the collection of intervals of Cξ C}, and let J k be the collection of intervals which belong to Ck C but not to Ck 1 C, so that the J k s are disjoint, and J = J k. Then I = I = lim I J K K I. K Now, I is the sum of the length of all intervals belonging to J k for some k K, which is precisely the collection of all intervals of CK C. At step k of the construction, the interval [0, 1] is divided into the finitely many intervals belonging to C k and Ck C, and hence the sum of the lengths of the intervals of C k and Ck C is 1. The sum of the lengths of the intervals of C k is (1 ξ) k, since at each step each interval has an interval of relative length ξ removed, and so the length of the remaining intervals is decreased by a factor of (1 ξ) at each step. It follows that K I = 1 (1 ξ) K K 1 and hence the sum of the lengths of the intervals in C C ξ (b) Show directly that m (C ξ ) = 0. is 1. Solution: Since C ξ is covered by C k for every k, and the sum of the lengths of the intervals of C k is (1 ξ) k, it follows that m (C ξ ) (1 ξ) k for every k, so taking k yields m (C ξ ) 0, i.e. m (C ξ ) = 0. Problem 14 The outer Jordan content J (E) of a set E in R is defined by J (E) = inf N I j, where the inf is taken over every finite covering E N I j, by intervals I j. (a) Prove that J (E) = J (E) for every set E (here E denotes the closure of E). Solution: In the definition of J, we can take the intervals to be closed, since any covering by finitely many (say N) closed intervals can be extended to a covering by open intervals of total length at most ɛ more by extending each closed interval to an open interval of length at 4

5 most ɛ/n more, and vice versa. Using this definition, we see that if E N I j, then certainly E N I j, while if E N I j, then taking closures yields E N I j = N I j = N I j (notice that closure distributes over finite unions, and I j = I j by assumption that the intervals are closed). It follows that a finite covering for E is also a finite covering for E, and vice versa, so by definition it follows that J (E) = J (E). (b) Exhibit a countable subset E [0, 1] such that J (E) = 1 while m (E) = 0. Solution: Take E = Q [0, 1]. Then E = [0, 1], so J (E) = J (E) = J ([0, 1]) = 1. However, m (E) = 0 since E is countable. Problem 22 Let χ [0,1] be the characteristic function of [0, 1]. Show that there is no everywhere continuous function f on R such that f(x) = χ [0,1] (x) almost everywhere. Solution: Let f be any function on R such that f(x) = χ [0,1] (x) almost everywhere. This precisely means that if we let E = {x R f(x) χ[0,1] (x)}, then E has exterior measure zero. Now, for any δ > 0, we have that E cannot contain the interval (0, δ), since otherwise by monotonicity we would have that the measure of E would be greater than the measure of (0, δ), whose measure is already positive. This means that for every δ > 0 there exists x (0, δ) such that x E, i.e. there exists x (0, δ) with f(x) = χ [0,1] (x). Hence, if we let δ n 0 be a sequence of positive reals decreasing to 0 (e.g. δ n = 1/n), then for each n there exists x n (0, δ n ) with f(x n ) = χ [0,1] (x n ), and since χ [0,1] (x) = 1 for all 0 < x < 1, it follows that f(x n ) = 1 for all sufficiently large n, with x n converging to 0. Similarly, for every δ > 0 we have that E cannot contain ( δ, 0), and with δ n as before we can thus find y n ( δ n, 0) (in particular, y n 0 as δ n 0) satisfying f(y n ) = χ [0,1] (y n ) = 0 (as y n < 0). Thus, there exist two sequences x n, y n both converging to zero, with f(x n ) 1, but f(y n ) 0. This implies that f cannot be continuous at 0. In particular, no continuous function can equal χ [0,1] almost everywhere. 5