Solutions to Homework Assignment 2

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1 Solutions to Homework Assignment Real Analysis I February, 03 Notes: (a) Be aware that there maybe some typos in the solutions. If you find any, please let me know. (b) As is usual in proofs, most problems can be proved in many different ways. These solutions contain only one of these multiple ways, not the only way. Section.4 Problem.4.7 Assume B is a countable set. Thus, there exists f : N B, which is and onto. Let A B be an infinite subset of B. Show that A is countable. Proof. We will define a and onto function g from N to A by induction on k: Base Case: When k =, let n = min{n f(n) A} and set g() = f(n ). Induction Step: Assume that g(k) has been defined and we need to define g(k + ). Let n k+ = min{n f(n) A and n n,..., n k } and set g(k + ) = f(n k+ ). Observe that n < n <... and since A is an infinite set, the above definition defines g on all natural numbers. Claim. g is a function. Proof of Claim. Towards contradiction assume that g is not. I.e. there are k, k N such that k < k and g(k ) = g(k ). Equivalently f(n k ) = f(n k ). Since f is, it must be that n k = n k. But, n k = min{n f(n) A and n n,..., n k }. Since k < k, n k must be different than n k. Contradiction. Claim. g is an onto function.

2 Proof of Claim. Let a A. Since f is onto, there exists some n N such that f(m) = a. Let k be the smallest numbers such that n k > m. I.e. n k > m and n < n <... < n k m. If n k = m, then g(k ) = f(n k ) = f(m) = a and a is in the range of g. If n k < m, then m {n f(n) A and n n,..., n k }. So, m min{n f(n) A and n n,..., n k } = n k, which is a contradiction since m < n k. So, this case can not happen and we conclude that a is in the range of g. Equivalently, g is onto. Since g is both and onto, A is countable. Problem.4.8 (a) If A, A are countable, then A A is also countable. Generalize for unions of finitely many A,..., A n. (b) Explain why induction can not be used for part (b) of Theorem.4.3. (c) Prove that if A n is countable for all n N, then n A n is also countable. Proof. (a) If A and A are not disjoint, i.e. A A, then let B = A \ A. B is a subset of A, B A, and B is disjoint from A, B A =. B A implies that A B A A. On the other hand, if x A A, then x A B. This is obvious if x A. If x / A, then x must be in A. So, x A \ A, which is equivalent to x B. In either case, x belongs to A B, which implies that A A A B. Altogether, A A = A B. By problem.4.7, B is either finite or countable. Case I (a bit easier): B is finite. Then there exists a - and onto function f : N A and some natural number N (the size of B ) and a - and onto function g : {,,..., N} B. Consider the function h : N A B defined by: { g(n), if n {,,..., N} h(n) = f(n N), if n > N It is immediate that h is defined for all numbers n N. If x A, then by virtue of f being onto, there exists some n N such that f(n) = x. Then h(n + N) = f(n) = x, i.e. x is in the range of h. If x B, then by virtue of g being onto, there exists some n {,,..., N} such that g(n) = x. Then h(n) = g(n) = x, i.e. x is in the range of h. Overall, range(h) = A B and h is an onto function. Moreover, h is also a - function. Let n, n N be such that h(n ) = h(n ). Since h(n ), h(n ) are equal, they are both in A, or both in B (A is disjoint from B ). If they are both in B, then g(n ) = h(n ) = h(n ) = g(n ) and since g is a - function, n = n. If both h(n ), h(n ) are in A, then f(n N) = h(n ) = h(n ) = f(n N) and since f is a - function, n N = n N and n = n. So, in either case, n = n which proves that h is a - function.

3 Case II (a bit harder): B is countable. Then there exist - and onto functions f : N A and g : N B. Define h : N A B by: { g( n h(n) = ), if n is even f( n+ ), if n is odd Again, h is defined for all natural numbers n N and working as before we can prove that h is - and onto. Now, we can generalize the result to the unions of arbitrary A, A,..., A n by induction on n: n=. We just proved that. n >. Assume that for A,..., A n countable, n i= A i is countable. We need to prove that for A,..., A n, A n+ countable, n+ i= A i is countable. Using the induction hypothesis, n i= A i is countable. Then ( n+ n ) A i = A i A n+, i= i= and the union of two countable sets is countable. (b) Induction is good for proving statements that hold for all n N. Part (ii) of Theorem.4.3, states that n A n is countable. Therefore, induction is not good for proving part (ii) of Theorem.4.3. (c) The proof here uses a diagonal argument similar to the argument in the proof that Q is countable. We describe the idea: First we can assume that all A n s are disjoint. If this is not the case, then we replace A by B = A \ A, A 3 by B 3 = A 3 \ (A A ), etc. We can prove as in part (a) that n A n = A n B n and that A, B, B 3,... are all disjoint sets. Thus, without loss of generality, we will assume that all A n s are disjoint. Now, for each n N, A n is countable. I.e. there exists a - and onto function f n : N A n. Using these functions f n, we write: A = {f (), f (),...} A = {f (), f (),...}... A n = {f n (), f n (),...} The idea is to define a function h : N n A n that is - and onto, tracing all the values f n (m), n, m N through the diagonals. More specifically, we follow the following pattern:

4 This means that h() = f (), h() = f (), h(3) = f (), h(4) = f 3 (), h(5) = f (), h(6) = f (3) etc. It follows that h is onto (we trace all the values) and that h is - (no values are repeated). We could actually write a formula for h and then use it to prove that it is - and onto, but it becomes unnecessarily complicated. Problem.4.0 Show that the set of finite sequences of natural numbers is countable. Proof. Let S be the set of finite sequences of natural numbers. E.g. (0,, 43, ) S. The idea is to associate to each such sequence a natural number that serves as code for the sequence. Given the code we must be able to tell how long the sequence is and what the numbers in the sequence are. We need some number theory: A natural number p is prime if its only divisors are and p itself. E.g., 3, 5, 7 are primes, but 4, 6, 8 are not primes. For every natural number n, we can write n uniquely as the product of prime numbers p, p,..., p k : n = p e pe... pe k k, where the exponents e, e,..., e k are natural numbers. Let s see how we can use this theorem to code finite sequences. Map the sequence (0,, 43, ) to the number = We use here, 3, 5, 7 because they are the first four prime numbers. Reservely, given the number we can determine that the length of the corresponding sequence to this number is 4, since the first four prime numbers are being used, and we can determine the numbers in the sequence by subtracting from each exponent. I.e. (0,, 43, ). More generally now, map every finite sequence s = (n, n,..., n k ) S to f(s) = p n+ p n+... p n k+ k N, where p, p,..., p k are the first k prime numbers. Using the fact that every natural numbers can be written uniquely as the product of prime numbers, if f(s ) = f(s ), then s must be equal to s. Thus, f is a - function, but it is not onto N. Let B = range(f). Then B is an infinite subset of the natural numbers. By problem.4.7, B is countable. Therefore, there exists some - and onto function g : B N. Then the composition g f : S N is a - and onto function from S to N, which proves that S is countable. This problem is the hardest in this assignment. We use here without proof the fact the composition of two - and onto functions is - and onto. 4

5 Section.5 Problem.5.9 Proof. (a) A function from {0, } to N is just a pair of natural numbers. So, the set of all functions from {0, } to N is the same as the set of all pairs of natural numbers. Associate to each such pair (n, m) a rational number, e.g. n 3 m. The rational numbers are countable. So, the same is true for the set of all functions from {0, } to N. (b) The situation is different for all functions from N to {0, }. Every such function defines an infinite sequence of 0 s and s. By Exercise.5.4 there are uncountably many such sequences. (c) P (N) does contain an uncountable antichain. Section. Problem.. Proof. (a) For every 0, we need to find some N N such that for all n N, 6n + 0 < ɛ. Let N be bigger than (we found this expression by solving for n the inequality 6n + 6ɛ < ɛ). Then, N > 6ɛ N > 6ɛ 6N 6N + 6n, for all n N. + (b) Let 0 and we need some N N such that for all n N, 3n + n < ɛ. 5

6 Equivalently, 3n + n < ɛ (3n + ) 3(n + 5) (n + 5) < ɛ 3 < ɛ (n + 5) Let N be a natural number bigger than 3 4ɛ. Then, N > 3 4ɛ 3 4N 3 4N + 0 3, for all n N. (n + 5) 3n + n + 5 3, for all n N. (c) Let 0 and we need some N N such that for all n N, 0 n + 3 < ɛ. Equivalently, 0 n + 3 < ɛ ɛ < n + 3 ɛ < n + 3 Let N be a natural number bigger than ɛ. Then, N > ɛ N + 3 > ɛ n + 3 >, for all n N ɛ 0 n + 3 < ɛ, for all n N 6

7 Problem.. Proof. Consider the sequence (0,, 0,, 0,,...). It is divergent. We will prove that it is vercongent to 0. Let ɛ = > 0. Then for all n N, x n 0 < ɛ =. The strange definition of vercongence describes sequences that are bounded. Problem..4 Proof. Let ɛ =.5. We will prove that for all N there exists some n N such that x n 0 > ɛ (that s the negation of the definition of the limit). Since the sequence contains an infinitely many number of 0 s and s, given N find some n N such that x n =. Then x n 0 = 0 > ɛ. For, we can find some N. For ɛ, we can not find any N. Problem..6 Proof. (a) In the definition of the limit, if we found a value N that works for 0, then any larger value also works for the same ɛ. (b) This same N also works for any larger value of ɛ. Problem..8 Proof. (a) The sequence ( ) n is frequently in the set {}. (b) Eventually implies frequently. I.e. eventually is stronger. (c) A sequence (a n ) converges to a iff for every 0 the sequence (a n ) is eventually in the neighborhood V ɛ (a). Eventually is the term we need. (d) The sequence (,,,,,,...) takes the value infinitely many times but is not eventually in the interval (.9,.). It is frequently in this interval. Section.3 Problem.3. Let x n 0, for all n N. (a) Show that if (x n ) 0, then ( x n ) 0. (b) Show that if (x n ) x, then ( x n ) x. Proof. (a) We have that 0 N N n N x n 0 < ɛ. () Equivalently, x n < ɛ. We need that 0 N N n N, x n 0 < ɛ. Equivalently, x n < ɛ. Let ɛ be a positive number and use property () for ɛ to find some N such that for all n N, x n < ɛ. Then x n < ɛ and we can take N = N. (b) If (x n ) x, then (x n x) 0 and by part (a), x n x 0. In symbols, 0 N N n N x n x < ɛ. () 7

8 Using some algebra we can prove that for x n, x 0, 0 x n x x n x. (3) The proof is similar to the proof of the triangular inequality. Combining properties () and (3), we conclude that there exists some N such that for all n N, x n x x n x < ɛ. This proves that ( x n ) x. Problem.3.3 Prove the Squeeze Theorem. Proof. We know that and x n y n z n for all n N lim n x n = lim n z n = l. From this last equation we can also derive that lim n (z n x n ) = 0. In symbols, 0 N N n N, x n l < ɛ () 0 N N n N, z n l < ɛ () 0 N 3 N n N 3, z n x n < ɛ. (3) We need to prove that 0 N 4 N n N 4, y n l < ɛ. By the triangular inequality, y n l y n x n + x n l z n x n + x n l Using properties () and (3) for ɛ, we can find N, N 3 such that for all n N, N 3, z n x n < ɛ and x n l < ɛ. Combining together we get that for all n N 4 = max{n, N 3 }, y n l z n x n + x n l < ɛ + ɛ = ɛ. This proves that (y n ) l and finishes the proof. Problem.3.5 Prove that (x n ), (y n ) are both convergent at the same limit iff the shuffled sequence z n = (x, y, x, y,...) is convergent. Proof. Since (x n ), (y n ) are subsequences of (z n ) the right-to-left direction is immediate. We need to prove the left-to-right direction. Assume that lim n x n = lim n y n = l. In symbols, 0 N N n N, x n l < ɛ () 0 N N n N, y n l < ɛ () 8

9 We need to conclude that 0 N 3 N n N 3, z n l < ɛ. Using () and () we can find some N, N such that for all n max N, N, x n l < ɛ and y n l < ɛ. Observe that z = x, z 3 = x, z 5 = x 3,... and in general z n = x n+ odd. Similarly, z = y, z 4 = y, z 6 = y 3,... and in general z n = y n even., for n:, for n: Let N 3 = max{n, N }. Then for all n N 3, n+ n max{n, N }. n+ Since z n is either equal to x n+ or y n and both indexes, n are greater than max{n, N }, it follows that for all n N 3, z n l < ɛ, which proves that (z n ) converges to l. Problem.3.6 (a) Prove that if b n b, then b n b. (b) Can we conclude that if b n b, then b n b? Proof. (a) Since b n b, b n b. By Exercise.3., part (b), b n b. But b n = b n and b = b and the result follows. (b) The conclusion is false. Consider b n = (,,,,,,...) and b =. Then b n = (,,,,,...) which converges to b, but b n is a divergent sequence. Problem.3.8 Proof. (a) Let x n = (,,,,,,...) and y n = (,,,,,,...). Then both (x n ) and (y n ) are divergent, but (x n + y n ) = (0, 0, 0, 0, 0, 0,...) and (x n + y n ) converges to 0. (b) This is impossible, by theorem.3.3, part (ii). (c) Let b n = n. Then (b n) 0, but b n = n diverges. (d) By theorem.3. (b n ) is bounded. Since (a n ) is unbounded, (a n b n ) can not be bounded. (e) Let (a n ) = (0, 0, 0, 0, 0, 0,...) and (b n ) = (,,,,,,...). Then (a n b n ) = (0, 0, 0, 0, 0, 0,...) and both (a n ) and (a n b n ) are convergent, while (b n ) is divergent. Problem.3.9 Answer. Theorem.3.4 does not remain true is we replace all inequalities with strict inequalities. E.g. for a n = n > 0, we can not conclude that lim n a n > 0. 9