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Boot Camp: Real Analysis Lecture Notes

Lectures by Itay NeemanNotes by Alexander Wertheim

August 23, 2016

Introduction

Lecture notes from the real analysis class of Summer 2015 Boot Camp, delivered byProfessor Itay Neeman. Any errors are my fault, not Professor Neemans. Corrections arewelcome; please send them to [firstinitial][lastname]@math.ucla.edu.

Contents

1 Week 1 31.1 Lecture 1 - Construction of the Real Line . . . . . . . . . . . . . . . . . . . . 31.2 Lecture 2 - Uniqueness of R and Basic General Topology . . . . . . . . . . . 71.3 Lecture 3 - More on Compactness and the Baire Category Theorem . . . . . 111.4 Lecture 4 - Completeness and Sequential Compactness . . . . . . . . . . . . 16

2 Week 2 202.1 Lecture 5 - Convergence of Sums and Some Exam Problems . . . . . . . . . 202.2 Lecture 6 - Some More Exam Problems and Continuity . . . . . . . . . . . . 252.3 Lecture 7 - Path-Connectedness, Lipschitz Functions and Contractions, and

Fixed Point Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.4 Lecture 8 - Uniformity, Normed Spaces and Sequences of Functions . . . . . 34

3 Week 3 393.1 Lecture 9 - Arzela-Ascoli, Differentiation and Associated Rules . . . . . . . . 393.2 Lecture 10 - Applications of Differentiation: Mean Value Theorem, Rolles

Theorem, LHopitals Rule and Lagrange Interpolation . . . . . . . . . . . . 453.3 Lecture 11 - The Riemann Integral (I) . . . . . . . . . . . . . . . . . . . . . 513.4 Lecture 12 - The Riemann Integral (II) . . . . . . . . . . . . . . . . . . . . . 58

4 Week 4 654.1 Lecture 13 - Limits of Integrals, Mean Value Theorem for Integrals, and In-

tegral Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 654.2 Lecture 14 - Power Series (I), Taylor Series, and Abels Lemma/Theorem . . 724.3 Lecture 15 - Stone-Weierstrass and Taylor Series Error Approximation . . . . 804.4 Lecture 16 - Power Series (II), Fubinis Theorem, and exp(x) . . . . . . . . . 87

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5 Week 5 955.1 Lecture 17 - Some Special Functions and Differentiation in Several Variables 955.2 Lecture 18 - Inverse Function Theorem, Implicit Function Theorem and La-

grange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 985.3 Lecture 19 - Multivariable Integration and Vector Calculus . . . . . . . . . . 99

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1 Week 1

As per the syllabus, Week 1 topics include: cardinality, the real line, completeness, topology,connectedness, compactness, metric spaces, sequences, and convergence.

1.1 Lecture 1 - Construction of the Real Line

Todays main goal will be the construction of the real numbers. We will take the construc-tion of N,Z, and Q for granted.

Lets start with a fact. The rationals form a dense linear order with no endpoints.Unpacked, this means:

(i) Dense: For all x and y, there exists z such that x < z < y

(ii) Linear: For all x and y, either x < y or x = y or y < x

(iii) No endpoints: For all x, there exists y such that y < x; for all x, there exists y suchthat y > x

It turns out that every countable dense linear order with no endpoints is isomorphic to(Q;

Proof. By the uniqueness of the cardinality of finite sets, we have {0, 1, . . . , n 1} 6={0, 1, . . . ,m1}, so by CSB, we must have {0, 1, . . . , n1} {0, 1, . . . ,m1} or {0, 1, . . . ,m1} {0, 1, . . . , n 1}. It must be the latter, since inclusion is clearly an injection from{0, 1, . . . , n 1} to {0, 1, . . . ,m 1}

Note that one doesnt really need the strength of CSB to prove the Pigeonhole Principle;a direct argument can be made. We have the following (nearly) immediate corollary.

Corollary 1.1.6. For finite A,B, if A ( B, then A 6= B.

Example 1.1.7. Note that the above corollary fails for infinite sets. Indeed, N and N \ {0}are indeed equinumerous via the map n 7 n+ 1.

Now we wll talk a bit about countable sets.

Definition 1.1.8. A set A is countable if A is equinumerous with N, i.e. A = N.

Example 1.1.9. Here are some familiar faces which are countable:

(i) N is countable; just take the identity map.

(ii) Z is also countable. One can biject N with Z as follows:

0, 1, 2, 3, 4, . . .

0, 1,1, 2,2, . . .

(iii) More generally, if A and B are countable sets, then A B is also countable.

Claim 1.1.10. N N is countable.

Proof. Proof 1: There is an obvious injection from N to NN given by n 7 (0, n). On theother hand, (n,m) 7 2n3m is an injection from N N to N by the fundamental theorem ofarithmetic, so by CSB, N = N N.Proof 2: Picture the elements of N N as a square lattice, e.g. by identifying N Nwith the corresponding set of points in the Cartesian plane. Starting at the first element ofthis square (the origin, as it were), and for the kth element on the bottom row, count up kelements and over k 1 elements to the left. That is, we define a bijection f : N N Nso that f(k2 + 1), . . . , f((k + 1)2) lists the elements of the (k + 1) (k + 1) square minuselements of the k k contained inside it.

Corollary 1.1.11. If A and B are countable, then so is AB.

Proof. Fix bijections fA : N A and fB : N B. Fix n 7 (h1(n), h2(n)) a biijection fromN to N N. Then n 7 (fA(h1(n)), fB(h2(n))) is a bijection from N to AB.

Corollary 1.1.12. Q is countable.

Proof. There is an injection from N to Q given by inclusion. Also, there is an injection fromQ to Z Z by mapping each element p/q Q (in lowest terms) to (p, q) Z Z. By theprevious corollary, since Z is countable, Z Z is countable, so there is an injection fromZ Z to N, whence composing injections, we obtain an injection from Q to N. ApplyingCSB, were done.

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Now we may return to our claim stated earlier.

Theorem 1.1.13. Every countable dense linear order with no endpoints is isomorphic to(Q; 0 Q such that for all q Q, our claim is false.Taking q = 0, we find 2 6 2, and by the above proposition, since Q, the inequality isstrict. Further, if (n)2 < 2 for any n > 1 N, then taking q = n, we find (q + )2 6 2,i.e. ((n + 1))2 6 2. Again, since n + 1 is rational, we must have strict inequality, i.e.((n + 1))2 < 2. Hence, by induction, (n )2 < 2 for all n N. This is impossible, forexample, taking the smallest n greater than the rational number 2/.

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Now we will construct the real numbers, using equivalence classes of strictly increasingsequences of bounded rational numbers, which we will naturally identify with their supre-mums.

Definition 1.1.16. A sequence (an)n=0 is strictly increasing if for all n,m N, n an and n N,m Nsuch that am > bn. In colloquial terms, the sequences (an) and (bn) are interleaved.

Proposition 1.1.17. is an equivalence relation on E.

Proof. It is straightforward to verify the three necessary conditions.

(1) (an) (an), since (an) is strictly increasing

(2) is symmetric by the requirement for equivalence

(3) is also transitive by a (layered) application of the requirement for equivalence

For (an) E, let [an be the equivalence class of (an), which formally is the set {(bn)|(bn) (an)}. By moving to equivalence classes, we have [an] = [bn] if and only if (an) (bn), i.e.we translate equivalence to equality. Let E be the set of equivalence classes. Define < onE by setting [an] < [bn] if k N such that n N, an < bk. Informally, [an] < [bn] if theterms of (bn) eventually bound the terms of (an).There are two things to check here, namely that < is well-defined, and < is a linear orderon E.Well-defined: If (an) (an), (bn) (bn), suppose there exists k N such that n N, an < bk. Ten take l such that bk < bl. Then for all n N, we have m N such thatan < am. So a

n < am < b

l, so < is well-defined.

Linear order on E: This is precisely what we have rigged in our definition of the equiva-lence relation on E. That is, if [an] 6

Corollary 1.1.19. R is a dense linear ordering.

Proof. This is clear, since Q R!

Proposition 1.1.20. R has no endpoints.

1.2 Lecture 2 - Uniqueness of R and Basic General TopologyToday, we will talk about the properties which characterize R, as well as some generaltopology.

Definition 1.2.1. An order (L;

isomorphic to Q. Let f : D Q witness this. We will extend f to an isomorphism from Lto R.For every x L, let x = {u | u D, u 6 x} L. Since x is a bounded set in L, andsince D has no endpoints, there exists d D such that d > x. Since f is order preserving,f(d) > f(u) for every u x, so f(u) must be an upper bound for f(x) in R. Thus, wemay put f(x) = supR(f(x)) by the Dedekind completeness of R.Note that if x < y L, then there exist z1, z2 D such that x < z1 < z2 < y by thedensity of D in L. Since f is order-preserving on the elements of D, f(z2) > f(z1) > f(u)for every u x, so f(z2) > f(z1) > f(x). However, since z2 y, f(z1) < f(z2) 6 f(y), sof(x) < f(y). This estalishes that f is order-preserving and injective on L.Fix z R and let A = {p Q | p 6 z}. Let x be the sup in L of B = f1(A) D; x existssince L is Dedekind complete. Note B x, since B D, and b 6 x for each b B, soz = supR(f(B)) 6 supR(f(x)) = f(x). Suppose z < f(x); then there is an element u xsuch that z < f(u) 6 f(x). Take q Q such that z < q < f(u); then f1(q) < u 6 x, sincef1 is order preserving on Q, and f1(q) > b for each b B. So f1(q) is an upper boundfor B, but f1(q) < x, a contradiction. So f(x) = z, and hence f is surjective.

Topological Spaces

Definition 1.2.6. A topological space is a pair (X,T) where X is a set, T is a collectionof subsets of X, and T satisfies:

(1) T, X T

(2) V1, . . . , Vn T impliesni=1 Vi