MAT1000 ASSIGNMENT 1. a k 3 k. x =
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1 MAT1000 ASSIGNMENT 1 VITALY KUZNETSOV Question 1 (Exercise 2 on page 37). Tne 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. We also note that this decomposition is not unique since, for example 1/3 = 2/3k. We show that x C if and only if x has a representation as above where every a k is 0 or 2. Proof. Recall that be definition the Cantor set C = k=0 C k, where C 0 = [0, 1] and C k+1 is obtained from C k be removing the open middle-third subinterval of each of the disjoint intervals that constitute C k. First we use induction to show that for all non-negative integers k, if a is such that [a, b] is one of the disjoint intervals constituting C k, then a can be written as n=1 a n3 n, where a n {0, 2} for 0 < n k and a n = 0 for n > k. If k = 0 then C 0 = [0, 1] and 0 = n=1 0/3k as needed. Now we assume that the statement holds for k N and we show that it is also true for k + 1. Assume that s is such that [s, t] is one of the disjoint intervals constituting C k+1. If s is also an endpoint of a disjoint interval in C k then we know that s = n=1 a n3 n, where a n {0, 2} for 0 < n k and a n = 0 for n > k. Hence, a n {0, 2} for 0 < n k + 1 and a n = 0 for n > k + 1 as well. Now if s is not an endpoint of the interval in C k then by construction there is an interval [a, t] in C k such that [s, t] [a, t]. Furthermore, s a = 2(t a)/3 by construction of C. Since length of [a, t] is 3 k (on each step we reduce the length of disjoint intervals by a factor of 3), we have that s = a + 2/3 k+1. Using induction hypothesis, s = k n=1 a n3 n + 2/3 k+1. That is, s = n=1 a n3 n such that a n {0, 2} for 0 < n k + 1 and a n = 0 for n > k + 1 as claimed. Now we make an observation that C k consists of exactly 2 k disjoint intervals, which is exactly the number of points in [0, 1] of the form s = n=1 a n3 n, where a n {0, 2} for 0 < n k and a n = 0 for n > k. Therefore, this observation combined with our previous result implies that each such number is an endpoint of one of the disjoint intervals constituting C k. We also note that if s is an endpoint of an interval [s, t] in C k then t = s + 1/3 k = s + n=k+1 2/3n. That is, t = same as for s and the rest are equal to 2. n=1 a n3 n such that the first k of a n s are the Now if x = n=1 a n3 n such that a n {0, 2} for all n, then we have that s k = k n=1 a n3 n x = n=1 a n3 n k n=1 a n3 n + n=k+1 2/3n = t k for all k. That, is x [s k, t k ] C k for all k, which shows that x C. Date: September 29,
2 2 VITALY KUZNETSOV Conversely, assume that x C. Then for each k, there is an interval [x k, y k ] in C k containing x. First observe that the sequence x k converges to x since x x k < 2/3 k for each k N. Next we note that the previous results imply that for i < j, we can write x j = x i + j n=i+1 a n3 n with a n {0, 2}. Thus, x = lim k x k = lim k k n=1 a n3 n = n=1 a n3 n with a n {0, 2} for all n. (b) The Cantor-Lebesgue function is defined on C by F (x) = where x = a k3 k with a k {0, 2} and b k = a k 2. We show that F is well-defined and continuous on C, and moreover F (0) = 0 as well as F (1) = 1. Proof. To show that F is well-defined it suffices to show that a ternary representation of x = a k3 k with a k {0, 2} is unique. Assume x C and x = a k3 k = a k 3 k with a k, a k {0, 2}. For a contradiction, let a n a n for some n (and by well-ordering principle of natural numbers we can choose the smallest such n). Then by our result from part (a), we see that x lies in an interval [s, t] C n with s = n 1 a k3 k. At the same time it must also lie in the interval [a, b] C n with a = s + 2/3 n. We know that such intervals are disjoint which leads to contradiction. Thus, a k = a k for all k and F is well-defined. Next we show that F is continuous. It is enough to show that if x n is a sequence of points in C which converges to x, then F (x n ) converges to F (x). Fix ɛ > 0. Then there is n 0 N such that 1/2 n 0 < ɛ. Since x n converges to x there is n 1 > n 0 such that for all n > n 1, x x n < 1/3 n 1. Then x and x n must lie in the same interval in C n for all n > n 1. Therefore, ternary expansion in terms of 0 and 2 of x and x n must agree for the first n 1 digits. Thus, F (x) F (x n ) < k=n k = 2 n 1 < 2 n 0 < ɛ for all n > n 1, which implies that F (x n ) converges to F (x). This proves that F is continuous. Finally, using geometric series we have that 0 = 0/3k and 1 = F (0) = 0/2k = 0 and F (1) = 1/2k = 1. b k 2 k 2/3k, thus (c) We prove that F : C [0, 1] is surjective, that is, for every y [0, 1] there is x C such that F (x) = y. Proof. Let y [0, 1]. Then y has a binary expansion, that is y = b k2 k such that b k {0, 1}. We let a k = 2b k {0, 2}. By part (a), x = a k3 k C. Now we observe that by construction, F (x) = b k2 k = y, which shows that F is indeed surjective. (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 compliment of C, then F (a) = F (b). Hence we may define F to have the constant value F (a) in that interval. Proof. First we show that if (a, b) is an open interval of the compliment of C, then F (a) = F (b). If (a, b) is a subinterval of the compliment of C with endpoints in C, then this interval was removed from [0, 1] during construction of C on stage n + 1, for some n 0. In fact it was removed from some interval [s, t] C n. By part (a), we know that s = n a k3 k. Since the length of [s, t] is 1/3 n, a = s + 1/3 n+1 = n a k3 k + k=n+2 2/3k and b = n a k3 k + 2/3 n+1.
3 We compute that F (a) = n b k2 k + k=n+2 2 k required. MAT1000 ASSIGNMENT 1 3 = n b k2 k + 2 n 1 = F (b) as Next we show that this extension of F is in fact continuous. It is immediate that F is continuous at x C, since there is a neighbourhood of x on which F is constant. Now assume x C and we wish to show that F is continuous at x. As before, it enough to show that if x n is a sequence that converges to x, then F (x n ) converges to F (x). If x n C, then let y n = x n. If x n C, then it lies in (s n, t n ) C c. In that case, let y n = s n if x < x n and y n = t n if x n < x. By construction, y n C and y n x x n x for all n. Therefore, y n is a sequence of points in C that converge to x. Hence, by continuity of F on C we must have that F (y n ) converges to F (x). We note that by construction we have that F (x n ) = F (y n ) for all n, which shows that F (x n ) converges to F (x) as required. Thus F is indeed continuous on [0, 1]. Question 2 (Exercise 4 on page 38). We construct a closed set Ĉ so that starting with [0, 1] at the k th stage of the construction one removes 2 k 1 centrally situated open intervals each of length l k with l 1 + 2l k 1 l k < 1. (a) We show that if 2 k 1 l k < 1 then m(ĉ) = 1 2 k 1 l k > 0. Proof. We consider a compliment of Ĉ in [0, 1], that is the set S = [0, 1] Ĉ. We know that the set S is a union of the collection {I α } α A of open intervals removed from [0, 1] during the construction of Ĉ. Each interval in this collection is disjoint from all the other intervals in the collection since each time we remove centrally situated intervals. Furthermore, this collection of open intervals is countable since on each stage of the construction we remove 2 k 1 intervals. Therefore, S is measurable as countable union of measurable sets and by countable additivity we have m(s) = α m(i α). Since each interval removed on the k th stage of the construction is of length l k this sum is equal to 2k 1 l k. Now we use the fact that [0, 1] is a disjoint union of Ĉ and S and countable additivity to conclude that 1 = m[0, 1] = m(ĉ)+m(s) = m(ĉ)+ 2k 1 l k (note that we implicitly use the fact that Ĉ is measurable as a compliment of measurable set). Finally, we write m(ĉ) = 1 2k 1 l k > 0 and the proof is complete. (b) We show that if x Ĉ then there exists a sequence of points {x n} n=1 such that x n Ĉ, yet x n x and x I n, where I n is a sub-interval in the compliment of Ĉ with I n 0. Proof. Let x Ĉ and define Ĉn to be the set containing all points of [0, 1] which has not been removed during the first n stages of the construction of Ĉ (this set is analogous to the set C n in the construction of the Cantor set). By construction, Ĉ n is a disjoint union of 2 n closed intervals. Note that there are exactly 2 n closed intervals in Ĉn since each interval in Ĉ n 1 gives a rise to two new intervals in Ĉn, that is on each step the number of intervals is twice the number of intervals on the previous step. Let [a, b] be one of these intervals in Ĉn such that x [a, b]. On the next stage of the construction an open interval (s, t) of length l n is removed from [a, b]. We take x n to be the midpoint of (s, t) and we let I n = (s, t). Note that the distance from x to x n is less than the length of the interval [a, b] since both of these points lie in it. On the other hand, by construction each of the intervals in Ĉn has equal length and using the fact that there are 2 n such intervals, we conclude that the
4 4 VITALY KUZNETSOV length of [a, b] is bounded above by 2 n. Therefore, we obtain a sequence {x n } n=1 such that x n x < 2 n, x n Ĉ and x n I n [0, 1] Ĉn for all n. It follows that this sequence converges to x and it remains to verify that I n vanishes as n. We know that the series n=1 2n 1 l n converge and hence it must be the case that lim n 2 n 1 l n = 0. We observe that 0 l n 2 n 1 l n and by Squeeze theorem we conclude that I n = l n 0 as n as desired. Remark: The sequence of intervals I n constructed in the proof above has the following property. For all n, if J is an open interval which contains x n and is a subset of the compliment of Ĉ, then it must be the case J I n = (s, t). This is the case, since we always remove centrally situated open intervals when constructing Ĉ and this implies that endpoints s and t will never be removed and will remain in Ĉ. This observation will be used in the next proof. (c) We show that as consequence of (b), Ĉ is perfect and it contains no open interval. Proof. We have seen earlier that the compliment S of Ĉ in [0, 1] is a union of open intervals and hence it is open. Therefore, Ĉ is closed in [0, 1] and it implies that it is also closed in R. Therefore, to show that Ĉ is perfect we only need to verify that it has no isolated points. For a contradiction, assume that x is an isolated point of Ĉ. Then there exist r > 0 such that (x r, x + r) Ĉ = {x}. Now let {x n} n=1 be a sequence of points in S converging to x satisfying properties in (b). Since x n x as n, there exists n 0 N such that for all n n 0, x x n < r. That is, for all n n 0, x n lies either in (x r, x) or (x, x + r). Hence, by the remark we have made earlier, for all n n 0, I n contains either of these two intervals which implies that the length of each I n with n n 0 is at least r > 0. However, this contradicts the fact that I n 0 as n. Therefore, we conclude that Ĉ has no isolated points and is perfect. Now we prove that Ĉ contains no open interval. We assume otherwise and derive a contradiction. Let (a, b) Ĉ and consider the midpoint x of (a, b). Since x Ĉ, by part (b) we know that there exist a sequence {x n } n=1 such that x n Ĉ and x n x. Then there exists n N such that x x n < b a which implies that x 2 n (a, b) Ĉ. This is a contradiction since x n Ĉ by assumption. Thus, we conclude that Ĉ contains no open interval. (d) We show that Ĉ is uncountable.1 Proof. To show that Ĉ is uncountable we will construct a surjection f : Ĉ [0, 1] and since [0, 1] is uncountable it will complete the proof. Let x Ĉ and [a, b] be one of the disjoint intervals in Ĉn 1, n > 1 such that x [a, b]. On the next stage of the construction of Ĉ an open interval (s, t) is removed from [a, b], resulting in disjoint intervals [a, s] and [s, t] lying in Ĉn. We let In 0 = [a, s] and In 1 = [s, t]. Since x Ĉ Ĉn it must lie in one of these intervals. We define a n = 0 if x In 0 and a n = 1 otherwise. Finally, we set f(x) = n=1 a n2 n (Note that this series converge since n=1 a n2 n n=1 2 n = 1 and f is well-defined). 1 Note that my original argument was only valid in the case when Ĉ has positive measure and was based on a simple fact that any set of positive measure is uncountable. I would like to thank Mustazee Rahman for this correction.
5 MAT1000 ASSIGNMENT 1 5 Now we show that f is surjective. Let y [0, 1] be arbitrary. Then y has a binary expansion n=1 a n2 n where a n {0, 1}. Note that the sequence {a n } n=1 induces a nested sequence of intervals In an constructed as above starting with I 0 = [0, 1]. By Nested Intervals Theorem, n=1in an is non-empty. Furthermore, since each In an is one of the disjoint intervals in Ĉn, we know from part (b) that In an < 2 n and hence In an 0 as n. Therefore, Nested Intervals Theorem tells us that in this case n=1in an = {x} for some unique real number x. Note that x Ĉ since x Ĉn for all n. Moreover, by definition of f we have that f(x) = y and this proves that f is surjective. Question 3 (Exercise 5 on page 39). Suppose E is a given set and O n is the open set O n = {x : d(x, E) < 1/n} (a) If E is compact, then m(e) = lim n m(o n ). Proof. We start with an observation that O n+1 O n since if x O n+1 then d(x, E) < 1 < n+1 1 and hence x O n n. Also, each O n is measurable since each O n is an open set. Thus, O n s form a decreasing sequence of measurable sets. Therefore, if we show that E = n=1 O n and m(o 1 ) < then the desired conclusion will follow from downward measure continuity. We first show that E = n=1 O n. To simplify the notation we will write O = n=1 O n. If x E then d(x, E) = inf y E x y = 0 < 1/n for all n N. Thus, x O n for all n N and hence x O, which shows that E O. On the other hand, if x O n then d(x, E) = inf y E x y < 1/n and hence there exists y n E such that x y n < 1/n. Therefore, if x O then for all n N there exists y n E such that x y n < 1/n. Note that the sequence y n converges to x by construction and this implies that x E since E is compact (and hence it is closed). Thus we conclude that O E and it follows that O = E. It remains to show that m(o 1 ) <. We will do so by showing that O 1 is bounded and the conclusion will follow since every bounded measurable set has a finite measure. Since E is compact it is bounded and thus it is contained in some ball B r where r is the radius of the ball. Consider now B r+2 which is a ball of radius r + 2 centred at the same point as B r. We claim that O 1 B r+2, that is no point of O 1 lies outside of B r+2. For a contradiction, assume x Br+2 c O 1. In this case we must have that d(x, B r ) 2 since d(b r, Br+2) c = inf z y = 2 and infinum is taken over all y B r, x Br+2. c However, since E B r and x O 1 we have that 1 > d(x, E) d(x, B r ) 2 which is a contradiction. Therefore, O 1 is bounded and the proof is complete. (b) We give an example when conclusion in (a) fails for E which is closed and unbounded or for E which is open and bounded. Example 1. We first give an example when conclusion in (a) fails when E is closed and unbounded. First consider E = N. Note that E is closed in R since its compliment (, 0) k=0 (k, k + 1) which is open (as a union of open sets). Furthermore, the set E is unbounded, since any ball (interval) in R contains finitely many natural numbers and hence can not contain the whole E. We notice that for an arbitrary positive ɛ we can cover E with cubes [k ɛ, k + ɛ ]. That is, E 2 k+1 2 k+1 k=0 [k ɛ, k + ɛ ] for any ɛ > 0. 2 k+1 2 k+1 Therefore, by monotonicity and subaddivity we have m(e) k=0 m([k ɛ, k + ɛ ]) 2 k+1 2 k+1 k=0 ɛ2 k = 2ɛ and since ɛ was arbitrary we conclude that m(e) = 0. On the other hand, we have O n = {x : d(x, E) < 1/n} = k=0 (k 1/n, k + 1/n). For all n > 1 this
6 6 VITALY KUZNETSOV union is disjoint and we can use additivity to write m(o n ) = k=0 2/n >. Therefore, m(e) = 0 lim n m(o n ) as desired. Example 2. Now we give an example when conclusion in (a) fails with E being open and bounded. Since Q is countable let q 1, q 2, q 3,... be an enumeration of Q [0, 1]. We define E = (q k 1, q 2 k+2 k + 1 ). We note that E is open as a union of open intervals 2 k+2 and it is bounded since E [ 1, 2]. Furthermore, by subadditivity m(e) m(q k 1, q 2 k+2 k + 1 ) = 2 k+2 2 k 1 = 1/2. On the other hand, since Q [0, 1] is dense in [0, 1], we know that for any x [0, 1] there is q Q [0, 1] such that x q < 1/n. In other words, d(x, E) < 1/n for any x [0, 1] which implies that [0, 1] O n for all n. Therefore, by monotonicity 1 = m([0, 1]) m(o n ) for all n, which implies that 1 lim n m(o n ). Thus, m(e) 1/2 < 1 lim n m(o n ) as desired.
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