In 1854, Karl Weierstrauss gave an example of a continuous function which was nowhere di erentiable: cos(3 n x) 2 n. sin(3 n x), 2

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1 Why non-pictured analysis? CHAPTER 1 Preliminaries f is continuous at x if lim f(x + h) = f(x) h!0 and f(x + h) f(x) f is di erentiable at x if lim h!0 h Then but Di erentiability =) continuity, continuity 6=) di erentiability. exists. In 1854, Karl Weierstrauss gave an example of a continuous function which was nowhere di erentiable: F (x) = 1X n=0 cos(3 n x) 2 n. Then term by term di erentiation gives 1X 3 n sin(3 n x), 2 n=0 which diverges when x is not a multiple of. 1

2 2 1. PRELIMINARIES 1.1. Sets and Functions (Knowledge of this material is assumed.) Notation. N = {1, 2, 3,... }, the natural numbers. Z = {0, 1, 1, 2, 2,... }, the integers. Q = { m : m, n 2 Z, n 6= 0}, the rational numbers. n R, the real numbers. A\B the complement of B relative to A (other books may use A B or B 0 if A is understood) D(f) domain of f R(f) range of f Let f : A! B be a function with domain A and R(f) B: 1) If E A, f(e) = {f(x) : x 2 E} is the direct image of E under f. 2) If H B, f 1 (H) = {x 2 A : f(x) 2 H} is the preimage of H under f. For two sets A and B, A = B () A B and B A. Notation. p =) q means statement p implies statement q (if p, then q). p () q means p =) q and q =) p (p and q are equivalent statements).

3 1.2. MTHEMATICAL INDUCTION Mthematical Induction Well-Ordering Property (WO) of N an axiom Every non-empty subset of N has a least element. () If S N and S 6= ;, {z} 9 m 2 S {z} 3 there exists such that Principle of Mathematical Induction (MI) - an axiom Let S N 3 (1) 1 2 S; (2) if k 2 S, then k S. Then S = N. Applied Version of MI Let n 0 2 N and P (n) be a statement 8n (1) P (n 0 ) is true; m apple k {z} 8 k 2 S. for every n 0, n 2 N. Suppose (2) 8k n 0, P (k) true (the induction hypothesis) =) P (k + 1) true. Then P (n) is true 8n n 0.

4 4 1. PRELIMINARIES Problem (Page 15 #8). Prove that 5 n 4n 1 is {z divisible by 16 } P (n) 8n 2 N. Proof. Let S N 3 5 n 4n 1 is divisible by S since = 0 is divisible by 16. Suppose k 2 S, i.e. 5 k 4k 1 is divisible by 16 (induction Hypothesis). Then 5 k+1 4(k + 1) 1 = 5 k+1 4k 5 = (5 k+1 20k 5) + 16k = is divisible by 16, so k S. Thus, by math induction, S = N. 5 (5 k 4k 1) +16k {z } divisible by 16

5 1.2. MTHEMATICAL INDUCTION 5 Problem (Page 15 #10). Find and prove a formula for P (n) = (2n 1)(2n + 1). Solution. P (1) = 1 3, P (2) = = 6 15 = 2 5, P (3) = = Conjecture: P (n) = n 2n + 1. Proof. Let S N 3 P (n) = = = 3 7 n 2n S since P (1) = = 1 3 = Suppose k 2 S, i.e., Then P (k) = (2k 1)(2k + 1) = k 2k + 1. P (k + 1) = = (2(k + 1) 1 (2(k + 1) (2k 1)(2k + 1) + 1 (2k + 1)(2k + 3) = so k S. k 2k (2k + 1)(2k + 3) = 2k2 + 3k + 1 (2k + 1)(2k + 3) = (2k + 1)(k + 1) (2k + 1)(2k + 3) Thus, by math induction, S = N. = (k + 1) 2(k + 1) + 1,

6 6 1. PRELIMINARIES Problem (Page 15 #4). Prove 2 n < n! 8n 4, n 2 N. Proof. Let P (n) be (2 n < n!). P (4) is true since 2 4 = 16 < 24 = 4!. Suppose P (k) is true, i.e., 2 k < k!, k 4. Then 2 k+1 = 2 k 2 < k! 2 < k!(k + 1) = (k + 1)!, so P (k + 1) is true. Thus P (n) is true 8n 4, n 2 N. Homework Pages 15-16#3, 6, 13. Principal of Strong Induction (SI) - an axiom. Let S N 3 (1 ) 1 2 S, (2 ) if {1, 2,..., k} S, then k S. Then S = N.

7 1.2. MTHEMATICAL INDUCTION 7 A Bit of Logic (Appendix A) The implication or conditional: (P =) Q) (if P, then Q) (P implies Q) The implication is logically equivalent to its contrapositive: The converse of P =) Q is Q =) P. (not Q) =) (not P ) The double implication or biconditional: (P () Q) (P if and only if Q) (P i Q) P () Q is true when P and Q are either both true or both false. Proof by Contradiction: If C is a contradiction (a statement that is always false, e.g., 1=0), then are logically equivalent. P =) Q and P and (not Q) =) C

8 8 1. PRELIMINARIES Theorem (1). W O () MI () SI. Proof. (W O =) MI). Suppose S N S and (k 2 S =) k S) [so we need to prove S = N], but S 6= N. Then N\S 6= ;, so by W O N\S contains a least element, say m. Since 1 2 S, m 6= 1, so m > 1 and m 1 2 N. Where is m 1? Since m m 1 2 S. 1 < m and m is the least element of N\S, Then (m 1)+1 = m 2 S, contradicting that m 2 N\S. Since our assumption that S 6= N led to this contradiction, S = N, proving MI.

9 (MI =) W O) Suppose ; 6= S N 1.2. MTHEMATICAL INDUCTION 9 [we need to prove S has a least element] and S does not have a least element. Let T be the set of natural numbers that every element of S is greater than or equal to. [We use induction to show T = N] 1 2 T, since if x 2 S, x 2 N, and so x 1. Suppose k 2 T (induction hypothesis). Then 8x 2 S, x k [so where is k + 1?], so k 62 S (since it would be a least element if it were in S). Thus, 8x 2 S, x > k =) x k + 1 =) k T. Then, by MI, T = N Since S 6= ;, 9n 2 S =) n 2 N =) n N = T. But then, since n 2 S and n T, n > n + 1, a contradiction. Since assuming S does not have a least lement led to this contradiction, S does have a least element and W O is proved.

10 10 1. PRELIMINARIES (MI =) SI) We are given that MI is true. Let S N S and ({1, 2,..., k} S =) k S). [To show S = N by using MI.] Suppose k 2 S. Now 1 2 S =) {1} S =) 2 2 S =) {1, 2} S =) 3 2 S =) {1, 2, 3} S. Continuing, within a finite number of steps, we get {1, 2,..., k} S =) k 2 S =) k S (by MI). Thus, by MI, S = N as SI is proved. (SI =) MI) We are given that SI is true. Let S N S and (k 2 S =) k S). [Tos show S = N by using SI.] Suppose {1, 2,..., k} 2 S. Then k 2 S =) k S. Thus S = N by SI, so MI is proved. As a result of this equivalence, once one of the three is taken as an axiom, the two others follow as Corollaries.