Real Analysis - Notes and After Notes Fall 2008

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1 Real Analysis - Notes and After Notes Fall 2008 October 29, Introduction into proof August 20, 2008 First we will go through some simple proofs to learn how one writes a rigorous proof. Let start with the following proposition. Theorem 1. Let m be an odd positive integer. Then m 2 is an odd integer. Since m is an odd number it can be written as m = 2k + 1 for some k integer. Hence m 2 = (2k + 1)(2k + 1) = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1 Since k is an integer 2k 2 + 2k must be an integer. Let p = 2k 2 + 2k. Then and hence it means m 2 is an odd integer. m 2 = 2p + 1 Homework: Prove the similar statement for odd numbers. Theorem 2. Let m be an even integer. Then m 2 is an even integer. Now let us proved the reverse statement. Theorem 3. Let m 2 be an even integer. Then m is an even integer. Proof by contradiction. If m is an odd integer then by the previous theorem m 2 is also an odd integer and that is a contradiction. Homework: Prove the similar statement for odd numbers. Theorem 4. Let m 2 be an odd integer. Then m is an odd integer. Now let us prove the following statement. 1

2 Theorem 5. 2 is an irrational number. Proof by contradiction. Let us assume that 2 is a rational number. It implies that there exists two relatively prime integers m and n such that 2 = m. It follows that n 2 = m2 n 2 m 2 = 2n 2 m 2 is an even number By previous Theorem it follows that m is an even number. Therefore there exists an integer k such that m = 2k. Plugging it back into original identity we have m 2 = (2k) 2 = 4k 2 = 2n 2 Dividing both sides by 2. 2k 2 = n 2 It follows n 2 is an even integer and therefore n is an even integer. Therefore there exists an integer l such that n = 2l. This would mean that m n = 2k 2l = k l which would mean that m and n are not relatively prime and that is in contradiction with our assumption. Hence such m and n do not exist which further means 2 is an irrational number. How do we proof set identities? The general principle is: To prove that for two sets A, B A = B. we have to show two things: x A x B (A B) and x B x A (B A) Here is an example: Prove the following statement Theorem 6. Let A, B be two sets. Then ( A B ) c = A c B c x ( A B ) x A or x B or both 2

3 Assume now that x A. It implies x / A c x / A c B c. Therefore x ( A c B c) c. Analogously in the case of x B. This proofs that A B ( A c B c) c. Now the other way around Let x ( A c B c) c. Then But If on the other hand Hence we proved that x / (A c B c) x / A c or x / B c x / A c x A x A B. x / B c x B x A B. A B = ( A c B c) c Now use the fact that for D = C D c = C c and obtain ( A B ) c = A c B c and we are done. Except, how do we know that D = C D c = C c is right? Well this is for Homework. Homework: Give the other identities. August 25, 2008 The next example is a bit more complex. First we need to introduce the notion of the Image of a set under some mapping. Let f : X Y be a mapping from a set X into a set Y. Further, let A X. Then we define f[a], the image of A under f to be f[a] := {f(x) x A} Theorem 7. Let X and Y be two sets and A X and B X. Let f : X Y Then f [ A B ] = f[a] f[b] Let y f [ A B ]. This implies x A B such thatf(x) = y But Assume that x A. then x A B x A or x B x A y f[a] y f[a] f[b] If x B. then x B y f[b] y f[a] f[b] 3

4 In any case f [ A B ] f[a] f[b]. Now in reverse Let y f[a] f[b]. This implies y f[a] or y f[b] If y f[a] we have y f[a] x A such thatf(x) = y But x A x A B y f [ A B ]. Analogously in the case y f[b] we have y f [ (A B ]. Hence we proved that f [ A B ] f[a] f[b] which yields the identity. 2 Definition of sets Can we defined a set? Set is a fundamental object of mathematics so it surely must have a definition. The truth is not so simple. We discussed the Cantor definition of sets and its fallacies. We then discuss the Zermelo-Fraenkel axiomatic approach. 3 Functions and its properties Definition: A function f : X Y is an one-to-one function if f(x) = f(y) x = y. Definition: A function f : X Y is an onto function if y Y x X f(x) = y. Example: Let f : N N be defined as f(n) = 2 n. Prove that f is an one-to-one function but not onto. Let f(n) = f(m) 2 n = 2 m log 2 (2 n ) = log 2 (2 m ) n = m The other way is trivial. Definition: Two sets A and B are equipotent A B if a function f : A B which is one-to-one and onto. Definition: A segment of natural numbers [1..n] := {1, 2, 3,..., n} where n N 4

5 Definition: A set A has a cardinality n if A [1..n]. Such set is called a finite set. A set is called countable if it is finite or if it is equipotent to N. Otherwise it is called uncountable. The cardinality of a set A is denoted by A. The sets which are infinite but countable have cardinality ℵ 0 (aleph zero). Examples: (a) Show that N {all positive even integers}. (b) Show that Z Q. This proves that a set of all rational numbers is countable. (c) Let PA := {polynomials with integer coefficients}. Show PA is countable. The following theorem about cardinality is given without the proof. Theorem 8. Let A and B be two sets and A B Then A B If a set is countable then it can be written as a list of its elements, i.e., This follows from {x 1, x 2, x 3,..., x n,...} A = N A N f : N Awhichisone to oneandonto. Hence the set A can be written as Setting f(k) = x k for k N we get A = {f(1), f(2), f(3),...} A = {x 1, x 2, x 3,..., x n,...} It follows that to prove a set is countable it is enough to show we can write it as a sequence of its elements. We actually use this to show that Q + is countable. Q + = { 1 1, 1 2, 2 1, 1 3, 2 2, 3 1, 1 4, 2 3, 3 2, 4 1, 1 5,...} For homework prove this really is sequence of all positive rational numbers and that there are no repeats. Hint: Notice the pattern in the sums of denominators and numerators. August 27, 2008 We proved that R < N. The proof is based on Cantorś diagonalization argument. First notice N R N R So we know that R is at least countable. Assume that it is countable. Then we could write it as a sequence of its elements. This could be done in variety of ways. So we need to show that any such attempt would be flawed. How? We choose an arbitrary sequence which represents R. Let us say R = {x 1, x 2, x 3,..., x n,...} 5

6 Every x k can be written as n + 0.d k,1 d k,2 d k,3... where n is a numberś integral part and d k,j is its j-th decimal after the decimal point. Now we construct a number x in such way it would guarantee that x is not on the list and hence the list is not complete. Since we have chosen an arbitrary list, it implies that no list would work and hence R can not be written as a sequence of its elements which can mean only that R is uncountable,.i.e., R > N. So how do we construct x? First let the integral part of x be 0. Then define its j-th decimal a j to be any decimal but d j,j. Now we just have to show that x can not be on the list. But that you can do for homework. Remark: Cantor s argument is elegant and beautiful but not completely rigorous. For more info check the references. 4 Real Numbers Axioms of real numbers: Let R be a set. Let 0 and 1 are two elements of R.Furthermore let + : R R R and : R R R be two binary operations called addition and multiplication respectively and let an order relation to be smaller than < be defined on R R If the following axioms are satisfied than R is a set of real numbers. The Associative Laws If a, b, and c are any numbers, then and a + (b + c) = (a + b) + c The Commutative Laws If a, b, and c are any numbers, then a(bc) = (ab)c. The Roles of 1 and 0 Given any number x we have a + b = b + a and ab = ba. 1x = x and 0 + x = x. Additive and Multiplicative Inverses We state two properties here, one for addition and one for multiplication. The properties stated here make subtraction and division possible in the real number system. a.given any number a, there is at least one number b such that a + b = 0. b.given any number a 0, there is at least one number b such that ab = 1. 6

7 The Distributive Law This property of arithmetic provides a relationship between the operations + and. It says that if a, b, and c are any numbers, then a(b + c) = ab + ac. Properties of the Order < a.given any real numbers a and b, either a < b or b < a or a = b, and not more than one of these three conditions can hold. b.given any real numbers a, b and c, if a < b and b < c, then a < c. c.if a, b, and x are any real numbers and a < b, then we have a + x < b + x. d.if a, b, and x are any real numbers and a < b and x > 0, then we have ax < bx. As usual, if a and b are real numbers, then the assertion a > b means that b < a, the assertion a = b means that either a < b or a = b, and the assertion a = b means that either a > b or a = b. The Axiom of Completeness The axiom of completeness is a precise way of saying that if the real numbers are laid down on a number line in the traditional way, then every point on that number line is occupied by a number. We will discuss it in more details shortly. Cancelation Law for Addition a = b a + x = b + x x, a, b R One side of implication is straightforward (Which one?). The other one has to be proven. First let x be an additive inverse of x, i.e., let x + ( x) = 0. then a + x = b + x a + x + ( x) = b + x + ( x) a + 0 = b + 0 a = b. This completes the proof. Which axioms did we use in the proof? For homework prove the Cancelation Law for Multiplication. Now notice that ( x) = x Why is that? Uniqueness of the inverse a + b = 0 a + c = 0 b = c a + b = 0 = a + c a + b + b = a + c + b 0 + b = a + b + c b = 0 + c b = c Indicate the axioms or laws we used to prove the statement. 7

8 Notice that 0a = 0 was not a part of axiomatic setting of real numbers. Hence we have to prove it. 5 Boundness a0 = a0 a(0 + 0) = a0 a0 + a0 = a0 a0 + a0 + ( (a0)) = a0 + ( (a0)) a0 = 0 Definition 1. Let A R. A number s is an upper bound of set A if s a a A. A number s is an lower bound of set A if s a a A. The greatest lower bound is called infimum of set A and it is denoted by inf A. The lowest upper bound is called supremum of set A and it is denoted by sup A. Examples: Show that A B sup A sup B. Let x A x B sup B x sup B is upper bound for A. This implies sup A sup B. How about the infima. Is it true that A B inf A inf B? A counterexample is very easy to find. Chose A = [1, 2] and B = [0, 5]. It turns out inf A = 1 > inf B = 0? Is it then true that A B inf A inf B Try to prove it: Let x A x B inf B x inf B is lower bound for A. This implies inf A inf B. Done or Q.E.D. 6 Axiom of Completeness - AC Axiom Every non empty set A of real numbers which is bounded above has a supremum. As a direct consequence of AC we can show that Every non empty set A of real numbers which is bounded below has an infimum. AC implies the following characterization of the intervals. Theorem 9. Let S R Then S is an interval u, v S and x R such that u < x < v x S 8

9 The prof consists of four cases: S is only bounded above, S is only bounded bellow, S is unbounded, and S is bounded above and below. We are to prove the last case and the others are for homework. S is bounded implies it has supremum and infimum. Let a = inf S and b = sup S.Let x be a real number such that a < x < b. Then u S such that a < u < x. Why? Since otherwise x is a lower bound of S which is. Further v S such that x < v < b. Why? Since otherwise x is an upper bound of S which is. So we have u < x < v x S (a, b) S Now just check for the end points and show nothing else can be in S and we are done. 7 Archimedean Property of system R Every non-empty set of integers that is bounded above must have the largest number Same is true for bounded below. Every non-empty set of integers that is bounded below must have the smallest number Let w = sup S. If w S we are done. Hence let us assume w / S.This implies w 1 is not an upper bound of S. Further m S such that w 1 < m < w and m is not an upper bound(why?) which implies n S such that w 1 < m < n < w 0 < n m < 1 The last implication is a contradiction with the fact that n m is an integer. A corollary of this theorem is Theorem 10. Set Z is unbounded. Proof for homework A fact of life for real numbers reciprocals Theorem 11. Let x > 0. Then n N such that 0 < 1 n < x. 9

10 Let x > 0. Since Z is unbounded there is n N such that 1 x < n(why?). This implies x > 1 n. A set of all rational numbers is dense Theorem 12. Q is dense, i.e. for every two real numbers a and b there q Q such that a < q < b. a < b b a > 0 1 n N such that n < b a a + 1 n < b Now let S = {x Z x > a} = {x Z x > na}. n S is non-empty and it is bounded below. Why? Hence it has and it contains its infimum. Let m = inf S. It follows m n > a but m 1 a n This further implies m n = m n n a + 1 n < b a < m n < b. Hence m n is the rational number we were looking for. Q.E.D. September 10, A few exercises We start with the following fact Theorem 13. Let A be a bounded set or real numbers. Then δ > 0 a A such that sup(a) a < δ Since A is bounded it has a supremum. Now fix δ > 0. If there is no a A such that sup A a < δ it would imply sup A δ a a A 10

11 This would mean sup A δ is an upper bound and also smaller than sup A which is a contradiction. Q.E.D. Powers of real numbers We have seen in some earlier course how to define b q for b > 0 and q Q. We assume as axioms that Now we define (1) b 0 = 1 (2) b 1 = b (3) b x b y = b x+y x, y Q b r = sup{b q q Q, q < r} for all irrational numbers r. In this exercise we want to show that b r b s = b r+s r, s R Any rational number q < r + s can be written as q = q 1 + q 2 where q 1 and q 2 are rational numbers such that q 1 < r and q 2 < s. This implies b q 1+q 2 = b q 1 b q 2 < b r b s b r b s is an upper bound for {b q q Q, q < r + s} and therefore To prove equality we will assume that Let b r+s b r b s b r+s < b r b s b r b s b r+s = ɛ Now it is true that δ > 0 there is q 1 < r and q 2 < s such that b r b q 1 < δ Then b s b q 2 < δ b r b s b q 1+q 2 = b r b s b q 1 b q 2 = b r b s b q 1 b s + b q 1 b s b q 1 b q 2 b s b r b q 1 + b q 1 b s b q 2 Since b q 1 < b r, b r b q 1 < δ and b s b q 2 < δ we have Furthermore if we choose δ = b r b s b q 1+q 2 = b s δ + b r δ = δ( b s + b r ) ɛ 2( b s + b r ) we have the estimate b r b s b q 1+q 2 < ɛ 2 11

12 Finally, and estimate yield b r b s b r+s = ɛ b q 1+q 2 > b r+s This is a contradiction because b r+s is supremum. Supremum of a real function Let f : X R. We define supremum of f over the set X to be sup f := sup {f(x) x X} X Is it true that sup X f + g = sup X f + sup X g? It is not clear. Let us try an example. Let X = [0, 1] and f(x) = x 2 and g(x) = 1. 1+x Then sup X f = 1 and sup X g = 1. Hence sup X f + sup X g = 2. We will now show that x x 3 2 x [0, 1]. x x 3 2 x3 + x 2 2x 1 < 0 Notice that the left hand side is 0 for x = 1. This means we can factor (x 1). We obtain x x 3 2 (x 1)(x2 + 4x + 1) < 0 Since x 1 0 for x [0, 1] and x 2 + 4x + 1 > 0 we proved the inequality. Hence x x 3 2 < 2 sup X f + g < sup X f + sup g X September 15, Topology of the set of real numbers Neighborhoods Given a real number x and a set A of real numbers., we say that A is a neighborhood of x if there exists a number δ > 0 such that ( x δ, x + δ ) A. Simple consequence is If A is a neighborhood of x and A B, then B is a neighborhood of x as well. A fact of life: An interval (a, b) is a neighborhood of all its elements. 12

13 Theorem 14. Let U and V be two neighborhoods of the same point x then U V is also a neighborhood of x. U and V nbhds of x δ 1, δ 2 > 0 such that (x δ 1, x + δ 1 ) U and (x δ 2, x + δ 2 ) V. Let δ := min{δ 1, δ 2 }. Then also This implies and U V is a nbhd of x. (x δ, x + δ) (x δ 1, x + δ 1 ) U (x δ, x + δ) (x δ 2, x + δ 2 ) V (x δ, x + δ) U V Theorem 15. The intersection of finitely many neighborhoods of a number x is the neighborhood of x. Open and Closed sets A set O of real numbers is an open set if it is a neighborhood for all of its elements. Hence an interval (a, b) is an open set. How about [a, b]? It is not! Why? A set A is a closed set if R \ A is an open set. Theorem 16.. (a) The intersection of finitely many open sets is open. (b) The union of finitely many closed sets is closed. (c) The union of any family (finite or infinite) of open sets is open. (d) The intersection of any family (finite or infinite) of closed sets is closed. (a) Let O j, j = 1, 2,..., N be a finite collection of open sets. Let x N j=1 O j. Then x O j and O j is a nbhd of x for all j = 1, 2,..., N. Hence there exist δ j for j = 1, 2,..., N such that Let δ := min{δ j, j = 1, 2,..., N}. Then we have (x δ j, x + δ j ) O j (x δ, x + δ) (x δ j, x + δ j ) O j (x δ, x + δ) This means N j=1 O j is a nbhd for x. Since x was chosen arbitrary, we have N j=1 O j is an open set. (d) Let F α be a closed set for all α Λ. This means Fα c is a collection of open sets. Now ( ) c ( ) ( ) α Λ F α = α Λ Fα c = a union of infinite open sets N j=1 O j 13

14 From (b) we have that such union is open and we proved (d). Closure, Interior and Boundary of sets A closure of a set A is a set of numbers x such that for every neighborhood U of x We denote the closure of A by A. U A. An interior of a set A is a set of numbers x for which there exists a neighborhood U of x such that U A. We denote the interior of A by Å. The boundary of a set A, denoted by A is defined as The last requirement is equivalent to or to A := A \ Å. A := A A c A := {x every neighborhood N of x contains element of A and elements of A c } September 21, 2008 Theorem 17.. (a) Let x A δ > 0, (x δ, x + δ) A. (b) A B A B. (c) A is closed. (d) Let A be closed set then A = A. (e) A = A. (f) A = A Å (g) Let A be an open set A A =. (h)the closure of a set A is the smallest closed set containing A. Hence the closure of a set A is an intersection of all closed sets which contain A. Proofs: (a) directly from the definition. ( ) Let N x be a nbhd of x. Then δ > 0 such that (x δ, x+δ) N x. Then by assumption (x δ, x + δ) A N x A x A. (b) Let x A and let N x be a nbhd of x. Then N x A N x B A B. 14

15 (c) We need to show that A c is open. Let x A c by negation of (a) that This yields δ > 0 such that (x δ, x + δ) A = (x δ, x + δ) A c Let y (x δ, x + δ) A. Then (x δ, x + δ) is a nbhd of y and hence (x δ, x + δ) A = (x δ, x + δ) A c A c is open A is closed. (d) Assume A A. then x A \ A. Now A closed implies A c is open and hence a nbhd of x. From (a) it follows A c A which is a contradiction. (e) A is a closed set. From (f) A = A Theorem 18. Let S be a closed and bounded above set of real numbers. Then S contains the largest element. Let q = sup S. Assume q / S q S c. Since S is closed S c is open. Hence δ > 0 (q δ, q + δ) S c r (q δ, q + δ) q δ < r < q This implies r is an upper bound for S which is a contradiction. Limit Point of a set A point x is a limit point of a set A if for every neighborhood N of x the set N (A {x}). L(S) is a set of limit points of a set S. Remark:L(S) S. Example: Let S = (0, 1) {2}. Find L(S) and S. Compare. Answer:L(S) = [0, 1], but S = [0, 1] {2}. Remark: S = S L(S). Examples: (1) L(Q) = R. (2) L(Z =. 15

16 Theorem 19. Let S be a set of real numbers and x a real number. Then U a nbhd of x U S is finite x / L(S). Before the proof let us see two direct corollaries of this theorem: (1) Finite sets does not have limit points. (2) x L(S) ( U x U U S is infinite. Let U S \ {x} = {y 1, y 2,..., y n }. Set δ = min{ x y 1, x y 2,..., x y n }. Then we have (x δ, x + δ) S \ {x} = which means we found a nbhd of x which does not contain any members of S. This means x / L(S). 10 Sequences The sequence {x n } where n = 1, 2, 3,... can be understood as a function f : N R. For a given set S we define that a sequence {x n } is eventually in S if n N x k S k > n The {x n } is frequently in S if for infinitely many k N x k S. Examples: x n := ( 1) n Let S = [0, 1]. Is x n is eventually in S? Frequently? x n := 1 n Let S = [ 0.01, 0.01]. Is x n is eventually in S? Frequently? Limit A x is a limit of the sequence {x n } if Partial limit A x is a partial limit of the sequence {x n } if U is a nbhd of x {x n }is eventually in U U is a nbhd of x {x n }is frequentally in U 16

17 Theorem 20. Suppose that (x n ) is a sequence of real numbers and that x is a real number. The following conditions are equivalent: 1. x n x as n 2. ɛ > 0 (x n ) is eventually in the interval (x ɛ, x + ɛ). 3. ɛ > 0 N N x n (x ɛ, x + ɛ) n > N. 4. ɛ > 0 N N x n x < ɛ n > N. (2) (1). Let U be an arbitrary nbhd of x. Then ɛ > 0 such that (x ɛ, x + ɛ) U. By (2) (x n ) is eventually in (x ɛ, x + ɛ) and therefore in U. (4) (1). This directly follows from the fact that x (x ɛ, x + ɛ) x n x < ɛ The rest of the implications proof is left for exercise. Theorem 21. The limit of a sequence is unique. Assume (x n ) has two limits x and y. Then for any given ɛ > 0 we can find N an integer such that x n x < ɛ and y n y < ɛ 2 2 It follows x y = x x n + x n y x x n + x n y < ɛ 2 + ɛ 2 = ɛ Since this is true for every positive ɛ it follows x y = 0 x = y. Examples: (1) x n = 1 + ( 1)n. Show the limit is 1. n n (2). Show the limit is 0. n Show the limit is 2. (3) 2n 2 n 2 +2n+1 (4) n 3 3n 3 5n+1 Show the limit is 1 3 (1) Fix ɛ > 0. Then x n 1 = 1 + ( 1)n n. Let me fill the details in for eg. (1) and (4). 1 = ( 1)n n = 1 n = ( ) Choose N an integer such that N > 1. Such N exists since N is unbounded. This yields ɛ 1 < ɛ and hence for every n > N we have 1 < 1. All together with (*) we have that N n N x n 1 = 1 n < 1 N < ɛ since ɛ has been arbitrarily chosen we proved it for all ɛ > 0. 17

18 (4) Fix ɛ > 0. Then x n 1 n 3 = 3 3n 3 5n = 3n3 (3n 3 5n + 1) 3 3(3n 3 5n + 1) 5n 1) = = 5 n 1 ) 5 n < 2 = ( ) 3(3n 3 5n + 1) 3 3n 3 5n + 1 3n 3 5n + 1 In the last estimate we used the facts that 5 < 2 and n 1 < n for all n integers. We 3 5 continue by dividing the numerator and denominator by n. 1 1 ( ) < 2 3n < 2 = ( ) 3n n In the last step we used the fact that <. Pulling out 3 as a common something+ 1 something n factor and using the difference of squares we obtain For n > 2 we have that 1 n 5 3 ( ) < n < 3. Thus n + 1 ( ) < 2 n = ( ) < 2 1 n So finally we got Choose N such that 2 N < ɛ and we have that x n 1 < 2 3 n x n 1 < 2 3 n < 2 N < ɛ n > N Theorem 22. Every convergent sequence of real numbers is bounded. Let (x n ) be a convergent sequence and x its limit. Then for ɛ = 1 we can find an integer N such that x n x < 1 n > N x 1 < x n < x + 1 Now let y = max (x + 1, x 1, x 2,..., x N ). We then have x n < y n N Similarly we prove (x n ) is bounded below. Theorem 23. Suppose that (x n ) and (y n ) are sequences of real numbers, that the sequence (x n ) is bounded, and that the sequence (y n ) converges to 0. Then x n y n 0 as n. 18

19 First (x n ) is bounded means there is M > 0 such that (x n ) < M Since (y n ) 0 and ɛ M such that It follows is a positive number for any given ɛ > 0 we can find N an integer y n < ɛ M n, n > N x n y n < M y n < M ɛ M = ɛ Theorem 24. Suppose that (x n ) and (y n ) are sequences of real numbers that converge, respectively, to numbers x and y. Then we have: 1. x n + y n x + y 2. x n y n x y 3. x n y n xy 4. x n yn x y providing y 0 (1) Fix ɛ > 0. Then for ɛ 2 find N 1 and N 2 such that x n x < ɛ 2 Choose N = max (N 1, N 2 ). n > N 1 and y n y < ɛ 2 n > N 2 x n + y n (x + y) = x n x + y n y < x n x + y n y < ɛ 2 + ɛ 2 = ɛ Here we used the triangle inequality. The latter inequality is true for all n > N. (2) For HWK. (3) Since (y n ) is convergent there is M > 0 such that y n < M. Fix ɛ > 0. Now x n y n xy = x n y n xy n + xy n xy < x n y n xy n + xy n xy < y n x n x + x y n y Now we can see that we have to find N 1 such that x n x < ɛ for all n > N 2M 1 and N 2 such that y n y < ɛ for all n > N 2 x 1. This is possible since both sequences are convergent. Finally, choose N = max (N 1, N 2 ). Thus we have x n y n xy < y n x n x + x y n y < M ɛ 2M + x (4) For HWK. Q.E.D. ɛ 2 x = ɛ n > N Theorem 25 (The Sandwich Theorem). Suppose that (x n ), (y n ), and (z n ) are sequences of real numbers and that the inequality x n y n z n holds for all sufficiently large integers n. Suppose that x is a real number and that both of the sequences (x n ) and (z n ) converge to x. Then the sequence (y n ) must also converge to x. 19

20 (1) Fix ɛ > 0. Then find N such that The inequality x n y n z n implies x n x < ɛ n > N and z n x < ɛ n > N x n x y n x z n x ɛ < x n x y n x z n x < ɛ y n x < ɛ n > N Theorem 26. Every monotone and bounded sequence is convergent. Let x = sup (x n ). Then for any given ɛ there is x k (x ɛ, x+ɛ). Since the sequence is monotone we have x k x n x n > k Hence limx n = x. 11 Limits and infinity is not a number but if cautious we can work with it almost as it is a number. We extend the set of real numbers by adding and. The infinite arithmetic then has the following rules: + = = + x = k = + ( ) = ( ) = + x = k ( ) = where x is any real number and k a positive real number. Next we would like to describe the topology of extended R. A set U is a nbhd of if it contains (w, ) where w is some real number. A set U is a nbhd of if it contains (, w) where w is some real number. A sequence (x n ) is convergent to if U nbhd of (x n ) is eventually in U 12 Limits and Topology Theorem 27. If (x n ) S then every partial limit is in S. Let x be a partial limit. Let U be an arbitrary nbhd of x. Then (x n ) is frequently in U and U S. Hence x S. Corollary 28. Let S be a closed set and (x n ) S such that lim x n = x. Then x S. 20

21 Trivial. The limit is a partial limit. Theorem 29. If S is any set and x S then there is a sequence (x n ) such that lim x n = x. Since S is closed, a set (x 1, x + 1 ) S is non-empty for every n N. Choose an n n arbitrary element of this set to be x n. Then we have for all n. Hence lim x n = x. x n x < 1 n 13 Cantor Intersection Theorem and Bolzano-Weierstrass Theorem Theorem 30. Every sequence of real numbers has a monotone subsequence. There are two cases: either the sequence is bounded or not. Let us first assume that sequence is bounded by M, i.e., x n < M Then there is m = inf (x n ). Yet again we have two choices either infimum is in the sequence or not. If it is, choose x n1 to be the first element of the subsequence. Continue the process again considering the sequence (x n ) n > n 1 Find the infimum of the new sequence,let us call it m 2. We know that m 2 > m. Find the element of the sequencex n2 such that x n2 = m 2 and continue the process by considering the sequence (x n ) n > n 2. If you are able to continue this process infinitely many times you will produce a monoton increasing subsequence. If you are stopped at any point then you have to start from the beginning using the reduced sequence for which the previous construction failed and follow the second subcase. Second subcase: If infimum is not attained in the sequence, there are infinitely many elements of the sequence in the (m, m + 1). Choose x k n k in such way that This subsequence is obviously monotone. The unbounded case is left for HWK. x nk (m, m + 1 k ) x n k x nk 1. Theorem 31 (Cantor Intersection Theorem). Let (F n ) be a sequence of non-empty bounded closed sets which are nested,i.e., Then n F n is non-empty. F n F n+1 n N. 21

22 Choose x n F n. The sequence (x n ) is contained in F 1 and since F 1 is bounded the sequence is bounded as well. The previous theorem guarantees us we can choose a subsequence (x nk ) which is monotone and bounded. Hence it converges. Let lim x nk = x. The claim is x n F n. To show this we choose an arbitrary F n. From the construction of the sequence (x n ) we know N N x nk F n n k > N. F n closed implies the limit of any sequence is contained in F n so in particularly the sequence (x nk ). Hence x F n This implies x F n n N and therefore x n F n. This means the n F n is not empty. Theorem 32 (Bolzano -Weierstrass Theorem). Every bounded infinite set of real numbers has a limit point. We start by constructing a sequence (x n ) in S, such that x n x k whenever n k. First choose x 1 to be an arbitrary element of S. Then choose x 2 S \{x 1 }, x 3 S \{x 1, x 2 }, and so on. The sequence (x n ) has a monotone subsequence (y n ). Now, (y n ) is monotone and bounded hence it converges to some x. We claim x is a limit point of the set S. Let U be a nbhd of x. Since x is a limit of (y n ), U contains -many elements of (y n ) therefore This means x is a limit point of S. U (S \ {x}) In our discussions in two previous theorems we used this fact: Theorem 33. Every bounded sequence has a convergent subsequence. Examples: (a) Define diamf = sup { x y x, y F }. Let F n be a nested sequence of closed bounded sets, such that diam(f n ) 0. Then n F n = {x}. Assume that x, y n F n and x y > ɛ > 0. This means diamf n > ɛ > 0 for all n positive integers. Hence diam(f n ) 0.A contradiction. (b) Let F 1 = [0, 1]. To define F n we split F n 1 in two halfs and we choose randomly one of the halfs to be F n. Then n F n = {x}. What is x? 14 Cauchy sequence A sequence (x n ) is a Cauchy sequence if ɛ > 0 N N x n x m < ɛ n > N m > N 22

23 Theorem 34. A sequence is a Cauchy sequence iff it is convergent. one direction was for takehome. Let us proof the other direction. Let (x n ) be a Cauchy sequence. Then for ɛ = 1 we can find N an integer such that x n x m < 1 for all n, m > N. This mean that x n (x N 1, x N + 1). Define s = sup { x 1, x 2,..., x N + 1 }. We have that (x n ) is bounded by s. It is left to show that it has only one partial limit and hence it is convergent. Assume there two partial limits x and y. Then there is an ɛ > 0 such that x y > ɛ. Now for this ɛ and for every N we can find x n and x m such that x n x < ɛ 10 x n y < ɛ 10 Hence x n x m > y ɛ 10 x ɛ 2ɛ 2ɛ = x y > x y > 8ɛ 10 Since it is true for all N it implies (x n ) is not a Cauchy sequence and that is a contradiction. 15 Examples: The idea is that we try to do this examples in class avoiding to follow some slick proofs from a book. This should show the students a process of creating mathematics or maybe more precise doing mathematics. The process which is hard has many wrong turns, and mane drafts before we reduce the solution to some slick argument in few lines. To obtain a result one must get his hands dirty, so to speak. Here I would like to indicate only a few remarks as afterthoughts. Example 1. Let x n+1 = 3 + x n. Show that (x n ) is bounded and monotone and therefore convergent. Example 2. Fibonacci sequence Let F 0 = F 1 = 1. Further, let F n+1 = F n + F n 1 The (F n ) is a Fibonacci sequence. Now we define G n = F n+1 F n Show the (G n ) is bounded and monotone and therefore convergent. As afterthought here is the slicker way to prove the convergence of (G n ). First we noticed G n > 0 1 G n > 0 G n+1 = G n 1 n. Then G n 1 1 G n 1 G n+1 = G n = 2 23

24 Hence we proved that 1 G n 2 n. In class we observed that (G n ) is an alternating sequence. We will prove that using the mathematical induction. First the basis of induction: G 0 = 1 < G 2 = 3 2 G 1 = 2 > 5 3 = G 3 For the step of induction we assume that: G 2n < G 2n 2 and G 2n 1 > G 2n 3. We need to show that G 2n+2 < G 2n and G 2n+1 > G 2n 1. First notice that G 2n < G 2n 2 1 > 1 It follows G 2n G 2n 2. G 2n+1 = > = G 2n 1 G 2n G 2n 2 Analogously we show the other claim, G 2n+1 > G 2n 1 1 < 1 G 2n+1 G 2n 1 Hence G 2n+2 = G 2n+1 < G 2n 1 = G 2n This proves that (G n ) consists of two monotone sequences both of which are bounded and therefore convergent. Computing the limit in both cases shows that both of them converge to the same limit, the golden ratio number φ = Example 3. Geometric Sequence and series Let q be a real number. Then a geometric sequence is define to be x n = q n. We will first show that q n is unbounded for q > 1. Assume q n is bounded. Hence it has the supremum M. Then for ( M, M) there is q an integer N such that But then after multiplication by q M q < qn < M M < q N+1 which is a contradiction. Hence (q n ) is unbounded. Next we will show that q n is bounded for q < 1. For q > 0 we have 1 > q > q 2 > q 3 >. Hence the sequence is bounded above by 1 and below by 0. Forq < 0 the sequence is bounded by 0 above and by 1 from below. Next we show that q n is monotone. trivial? Finally, we show q n 0 when q < 1. 24

25 A geometric sum s n = 1 + q + q 2 + q q n. A geometric series s is defined as s := q n := lim s n n=0 Show that s = 1 1 q. s n qs n = (1 + q + + q n ) (q + q q n+1 ) = 1 q n+1 s n = 1 qn+1 1 q lim s n = lim 1 qn+1 1 q = 1 1 q Cantor set A construction of the Cantor set We start with C 0 = [0, 1]. Then split C 0 into thirds and drop the middle one. We obtain C 1 = [0, 1 3 ] [ 2 3, 1]. We denote the first interval by C 10 and the second interval by C 11. We continue by splitting each of the intervals into thirds and dropping the middle one. Then the next level C 2 = [0, 1 9 ] [ 2 9, 3 9 ] [ 6 9, 7 9 ] [ 8 9, 1]. We denote the four intervals by C 20, C 21, C 22, C 23. The procedure continues by again splitting each of the intervals into thirds and dropping the middle one. We created (C n ) a sequence of sets. Show that (C n ) is a nested sequence of non-empty bounded closed sets. Visualize the whole process in the form of a tree. How many sets on each level? What is the diameter of the intervals on each level? Following a path in the tree we create another nested sequence of bounded closed sets which diameter is converging to 0. Hence by the Cantor Intersection Sets Theorem this sequence converges to a point. Describe the points in the Cantor sets. 16 Function limits Let S be a set of real numbers and that f : S R. Given a number a that is a limit point of S and given a real number λ, we say that λ is a limit of the function f at the number a and we write lim x a f(x) = λ if, for every neighborhood V of the number λ, it is possible to find a neighborhood U, U S, of the number a such that the condition f(x) V holds for every number x U \ {a}). Another way of stating this definition is to say that for every neighborhood V of the number λ it is possible to find a neighborhood U of the number a such that f[u \ {a})] V. 25

26 Technical version of this definition is lim f(x) = λ ɛ > 0 δ > 0 0 < x a < δ f(x) λ < ɛ x a For the proof look into Lewin s book. Example: Prove that lim 3x + 1 = 7 x 2 Fix ɛ > 0. We need to find δ > 0 such that 3x < ɛ whenever x 2 < δ. We have Hence if we choose δ = ɛ 3 3x = 3x 6 = 3 x 2 we would have 3x = 3 x 2 < 3δ = 3 ɛ 3 = ɛ Let us try another example: Prove that lim x 5 x2 = 25 Fix ɛ > 0. We need to find δ > 0 such that x 2 25 < ɛ whenever x 5 < δ. We have x 2 25 = x + 5 x 5 The expression x + 5 = x = x < x Hence we choose δ = min{ ɛ, ɛ } Then x 2 25 = x + 5 x 5 < x x 5 < δ δ < ɛ 2 + ɛ 2 = ɛ Remark: The ɛ-δ game is a game of small numbers hence it is enough to prove it for 0 < ɛ < 1. Then we can estimate x + 5 < 11 and the proof is somewhat simpler. Theorem 35. The limit of a function is always unique. Assume that lim f(x) = λ and lim f(x) = µ x a x a We will show that for every ɛ > 0 λ µ < epsilon. This implies λ = µ. Fix ɛ > 0. λ µ = λ f(x) + f(x) µ λ f(x) + f(x) µ = ( ) For epsilon there is δ > 0 such that f(x) λ < ɛ 2 and f(x) µ < ɛ 2 Hence λ µ < ( ) < ɛ 2 + ɛ 2 = ɛ Hence λ = µ and the limit is unique. 26

27 17 Continuity of real functions A function f : Ω R is continuous on Ω if U, open set, Ω f 1 (U)is an open set. This is a general topological definition of continuity. On the number line this definition is equivalent to A function is continuous at the point x Ω if ɛ δ x y < δ f(x) f(y) < ɛ Given x Ω and ɛ > 0, the interval V = (f(x) ɛ, f(x) + ɛ) is an open set. Hence f 1 (V ) is an open set in Ω. Since x f 1 (V ) there is δ > 0 such that U = (x δ, x + δ) f 1 (V ). This implies f(u) V ( x y < δ f(x) f(y) < ɛ) A function is continuous on a set if it is continuous at every point of the set. Examples: Let f(x) = 1. Show that f is continuous at every point of the domain. x Let us start with specific example, x = 3. Fix ɛ > = y 3 δ < y 3y 3(3 δ) = ( ) We need to find δ > 0 such that δ 3(3 δ) < ɛ δ < ɛ(9 3δ) δ(1 + 3ɛ) < 9ɛ δ < 9ɛ 1 + 3ɛ 9ɛ Hence we have to choose δ smaller than. This implies 1 + 3ɛ ( ) < ɛ General case is analogous. Do It for Homework. For HWk also show that Theorem 36. A function f : Ω R is continuous at point a if For Hwk. lim f(x) = f(a) x a 27

28 18 Algebraic Rules for Limits and Continuity Theorem 37. Let Then (a) (b) lim f(x) = λ, lim x a g(x) = µ. x a lim f(x) + g(x) = λ + µ x a lim f(x) g(x) = λµ x a (c) lim c f(x) = cλ c x a R (d) lim x a 1 f(x) = 1 λ providing λ 0 Theorem 38. Let f : Ω R be continuous at point a Ω. Let g : Ω R be continuous at point a Ω. Then (a) f + g(x) is continuous at point a. (b) f g(x) is continuous at point a. (c) λf(x) is continuous at point a for every real number λ. (d) f (x) is continuous at point a providing g(a) 0. g 28

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