Real Analysis and Multivariable Calculus: Graduate Level Problems and Solutions Igor Yanovsky 1 Real Analysis and Multivariable Calculus Igor Yanovsky, 2005 2 Disclaimer: This handbook is intended to assist graduate students with qualifying examination preparation. Please be aware, however, that the handbook might contain, and almost certainly contains, typos as well as incorrect or inaccurate solutions. I can not be made responsible for any inaccuracies contained in this handbook. Real Analysis and Multivariable Calculus Igor Yanovsky, 2005 3 Contents 1 Countability 5 2 Unions, Intersections, and Topology of Sets 7 3 Sequences and Series 9 4 Notes 13 4.1 Least Upper Bound Property . 13 5 Completeness 14 6 Compactness 16 7 Continuity 17 7.1 Continuity and Compactness . 18 8 Sequences and Series of Functions 19 8.1 Pointwise and Uniform Convergence . 19 8.2 Normed Vector Spaces . 19 8.3 Equicontinuity . 21 8.3.1 Arzela-Ascoli Theorem . 21 9 Connectedness 21 9.1 Relative Topology . 21 9.2 Connectedness . 21 9.3 Path Connectedness . 23 10 Baire Category Theorem 24 11 Integration 26 11.1 Riemann Integral . 26 11.2 Existence of Riemann Integral . 27 11.3 Fundamental Theorem of Calculus . 27 12 Di®erentiation 30 12.1 R ! R ..................................... 30 12.1.1 The Derivative of a Real Function . 30 12.1.2 Rolle's Theorem . 30 12.1.3 Mean Value Theorem . 30 12.2 R ! Rm .................................... 31 12.3 Rn ! Rm ................................... 31 12.3.1 Chain Rule . 34 12.3.2 Mean Value Theorem . 35 @ @f @ @f 12.3.3 @x ( @y ) = @y ( @x )........................... 36 12.4 Taylor's Theorem . 37 12.5 Lagrange Multipliers . 40 Real Analysis and Multivariable Calculus Igor Yanovsky, 2005 4 13 Successive Approximations and Implicit Functions 41 13.1 Contraction Mappings . 41 13.2 Inverse Function Theorem . 41 13.3 Implicit Function Theorem . 44 13.4 Di®erentiation Under Integral Sign . 46 Real Analysis and Multivariable Calculus Igor Yanovsky, 2005 5 1 Countability The number of elements in S is the cardinality of S. S and T have the same cardinality (S ' T ) if there exists a bijection f : S ! T . card S · card T if 9 injective1 f : S ! T . card S ¸ card T if 9 surjective2 f : S ! T . S is countable if S is ¯nite, or S ' N. Theorem. S; T 6= Á. 9 injection f : S,! T , 9 surjection g : T ³ S. Theorem. Q is countable. Proof. Need to show that there is a bijection f : N ! Q. Since N ⊆ Q, card N · card Q, and therefore, 9f that is injective. To show card N ¸ card Q, construct the following map. The set of all rational numbers can be displayed in a grid with rows i = 1; 2; 3;::: and columns j = 0; ¡1; 1; ¡2; 2; ¡3; 3;:::. j Each aij, the ij'th entry in a table, would be represented as i . Starting from a11, and assigning it n = 1, move from each subsequent row diagonally left-down, updating n. This would give a map g : N ! Q, which will count all fractions, some of them more than once. Therefore, card N ¸ card Q, and so g is surjective. Thus, card N = card Q, and Q is countable. Theorem. R is not countable. Proof. It is enough to prove that [0; 1) ½ R is not countable. Suppose that the set of all real numbers between 0 and 1 is countable. Then we can list the decimal representations of these numbers (use the in¯nite expansions) as follows: a1 = 0:a11a12a13 : : : a1n ::: a2 = 0:a21a22a23 : : : a2n ::: a3 = 0:a31a32a33 : : : a3n ::: and so on. We derive a contradiction by showing there is a number x between 0 and 1 that is not on the list. For each positive integer j, we will choose jth digit after the decimal to be di®erent than ajj: x = 0:x1x2x3 : : : xn :::, where xj = 1 if ajj 6= 1, and xj = 2 if ajj = 1. For each integer j, x di®ers in the jth position from the jth number on the list, and therefore cannot be that number. Therefore, x cannot be on the list. This means the list as we chose is not a bijection, and so the set of all real numbers is uncountable. (Need to worry about not allowing 9 tails in decimal expansion: 0:399 ::: = 0:400 :::). 1 injective = 1-1: f(s1) = f(s2) ) s1 = s2. 2surjective = onto: 8t 2 T; 9s 2 S, s.t. f(s) = t. Real Analysis and Multivariable Calculus Igor Yanovsky, 2005 6 Problem (F'01, #4). The set of all sequences whose elements are the digits 0 and 1 is not countable. Let S be the set of all binary sequences. We want to show that there does not exist a one-to-one mapping from the set N onto the set S. Proof. 1) Let A be a countable subset of S, and let A consist of the sequences s1; s2;:::. We construct the sequence s as follows. If the nth digit in sn is 1, let the nth digit of s be 0, and vice versa. Then the sequence s di®ers from every member of A in at least one place; thus s2 = A. However, s 2 S, so that A is a proper3 subset of S. Thus, every countable subset of S is a proper subset of S, and therefore, S is not countable. Proof. 2) Suppose there exists a f : N ! S that is injective. We can always exhibit an injective map f : N ! S by always picking a di®erent sequence from the set of sequences that are already listed. (One way to do that is to choose a binary representation for each n 2 N). Suppose f : N ! S is surjective. Then, all sequences in S could be listed as s1; s2;:::. We construct the sequence s as follows. If the nth digit in sn is 1, let the nth digit of s be 0, and vice versa. Then the sequence s di®ers from every member of the list. Therefore, s is not on the list, and our assumption about f being surjective is false. Thus, there does not exist f : N ! S surjective. Theorem. card (A) < card (P (A))4. Proof. card (A) · card (P (A)), since A can be injectively mapped to the set of one- element sets of A, which is a subset of P (A). We need to show there is no onto map between A and P (A). So we would like to ¯nd a thing in P (A) which is not reached by f. In other words, we want to describe a subset of A which cannot be of the form f(a) for any a 2 A. Suppose jAj = jP (A)j. Then there is a 1-1 correspondence f : A ! P (A). We obtain a contradiction to the fact that f is onto by exhibiting a subset X of A such that X 6= f(a) for any a 2 A. For every a 2 A, either a 2 f(a), or a2 = f(a). Let X = fa 2 A : a2 = f(a)g. Consider a 2 A. If a 2 f(a), then a2 = X, so f(a) 6= X. If a2 = f(a), than a 2 X, so f(a) 6= X. Therefore, X 6= f(a), 8a 2 A, a contradiction. Therefore, card (A) < card (P (A)). Theorem. Suppose that f : [0; 1] ! R is an increasing function. Show that f can have at most a countable number of discontinuities. Proof. Let E = fx 2 [0; 1] : f is discontinuous at xg. Given any x 2 E, we know that limt!x¡ f(t) < limt!x+ f(t) and, using this fact, we choose r(x) 2 Q such that limt!x¡ f(t) < r(x) < limt!x+ f(t). ) We have de¯ned a 1 ¡ 1 function r : E ! Q. 3A is a proper subset of B if every element of A is an element of B, and there is an element of B which is not in A. 4Power set of a set S is the set whose elements are all possible subsets of S, i.e. S = f1; 2g, P (S) = 2S = fÁ; f1g; f2g; f1; 2gg. jP (A)j = 2n, if jAj = n. Real Analysis and Multivariable Calculus Igor Yanovsky, 2005 7 2 Unions, Intersections, and Topology of Sets Theorem. Let E® be a collection of sets. Then [ \ c c ( E®) = (E®): ® ® S c T c S Proof. Let A = ( E®) and B = ( E®). If x 2 A, then x2 = E®, hence x2 = E® for c T c any ®, hence x 2 E® for every ®, so that x 2 E®. Thus A ½ B. c S Conversely, if x 2 B, then x 2 E® for every ®, hence x2 = E® for any ®, hence x2 = E®, S c so that x 2 ( E®) . Thus B ½ A. Theorem. S a) For any collection G of open sets, G is open. ® T® ® b) For any collection F® of closed sets, ® F® is closed. Tn c) For any ¯nite collection G1;:::;Gn of open sets, i=1 Gi is open. Sn d) For any ¯nite collection F1;:::;Fn of closed sets, i=1 Fi is closed. S Proof. a) Put G = ® G®. If x 2 G, then x 2 G® for some ®. Since x is an interior point of G®, x is also an interior point of G, and G is open. b) By theorem above, [ \ \ [ c c c c ( F®) = (F®) ) ( F®) = (F®); (2.1) ® ® ® ® c T and F® is open.