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Math 55b Lecture Notes Evan Chen Spring 2015 This is Harvard College's famous Math 55b, instructed by Dennis Gaitsgory. The formal name for this class is \Honors Real and Complex Analysis" but it generally goes by simply \Math 55b". It is an accelerated one-semester class covering the basics of analysis, primarily real but also some complex analysis. The permanent URL ishttp://web.evanchen.cc/ evanchen/coursework. html, along with all my other course notes. Special thanks to W Mackey for providing notes on the several day that I slept through class. Contents 1 January 29, 2015 5 1.1 De nition and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.2 Continuous Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.3 Forgetting the Metric Topological Spaces . . . . . . . . . . . . . . . . . 7 1.4 Intermediate Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 8 2 February 3, 2015 10 2.1 Sequential Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.2 Cauchy Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.3 R is Complete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.4 1-countable and Hausdor . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 3 February 5, 2015 12 3.1 Completion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 3.2 Why no compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.3 Sequential Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.4 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 4 February 12, 2015 16 4.1 Warm-Up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 4.2 \Applied" Math . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 4.3 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 4.4 Convergent Things . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 4.5 Convergence Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 a 4.6 The Series n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 4.7 Exponential Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 5 February 17, 2015 22 1 5.1 L norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 1Evan Chen (Spring 2015) Contents 5.2 De ning the integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 5.3 Uniform Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 5.4 De ning the Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 5.5 Properties of Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 5.6 Riemann Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 6 February 19, 2015 27 6.1 Bonus Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 6.2 PSet 2, Problem 7b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 6.3 PSet 3, Problem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 6.4 PSet 3, Problem 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 6.5 PSet 3, Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 6.6 Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 6.7 PSet 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 7 February 24, 2014 31 7.1 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 7.2 Di erentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 7.3 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 7.4 More \applied math" . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 7.5 Higher Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 8 February 26, 2015 35 8.1 Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . 35 8.2 Pointwise converge sucks . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 8.3 Uniform Convergence of Functions . . . . . . . . . . . . . . . . . . . . . . 36 x 8.4 Applied Math Di erentiating e . . . . . . . . . . . . . . . . . . . . . . 37 n 8.5 Di erentiable Paths inR . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 9 March 3, 2015 39 9.1 Di erentiable forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 9.2 Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 9.3 Higher-Order Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 10 March 5, 2015 42 10.1 Inverse function theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 10.2 Proof of the Inverse Function Theorem . . . . . . . . . . . . . . . . . . . . 43 10.3 Step 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 10.4 First Step . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 10.5 Third Step . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 10.6 Second Step . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 11 March 10, 2015 47 11.1 Completing the Proof of Inverse Function Theorem . . . . . . . . . . . . . 47 11.2 Implicit Function Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 48 11.3 Di erential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 11.4 Lawns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 12 March 12, 2015 52 13 March 24, 2015 55 13.1 Midterm Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 2Evan Chen (Spring 2015) Contents 13.2 PSet Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 13.3 Review of Exterior Products . . . . . . . . . . . . . . . . . . . . . . . . . 59 13.4 Di erential Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 14 March 26, 2015 61 14.1 More on Di erential Forms . . . . . . . . . . . . . . . . . . . . . . . . . . 61 14.2 Topological Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 14.3 Smoothness of Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 14.4 Cotangent space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 15 March 31, 2015 65 15.1 Boxes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 15.2 Supports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 15.3 Partitions of Unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 15.4 Existence of Partitions of Unity . . . . . . . . . . . . . . . . . . . . . . . . 66 15.5 Case 1: A is Compact . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 15.6 Case 2: A is a certain union . . . . . . . . . . . . . . . . . . . . . . . . . . 68 15.7 Case 3: A open . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 15.8 Case 4: General A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 16 April 2, 2015 71 16.1 Integration on Bounded Open Sets . . . . . . . . . . . . . . . . . . . . . . 71 16.2 This Integral is Continuous . . . . . . . . . . . . . . . . . . . . . . . . . . 72 16.3 Open Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 17 April 7, 2015 76 17.1 Integration Over Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . 76 17.2 Change of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 17.3 Integration Over Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . 79 18 April 9, 2015 81 18.1 Digression: Orientations on Complex Vector Spaces . . . . . . . . . . . . 81 18.2 Setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 18.3 Integration on Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 18.4 Stoke's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 19 April 14, 2015 86 20 April 16, 2015 88 20.1 Complex Di erentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 20.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 20.3 Inverse Function Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 20.4 Complex Di erentiable Forms . . . . . . . . . . . . . . . . . . . . . . . . . 90 20.5 Cauchy Integral Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 20.6 Computation when f = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 20.7 General Proof of Cauchy's Formula . . . . . . . . . . . . . . . . . . . . . . 92 21 April 21, 2015 93 22 April 28, 2015 96 22.1 Riemann Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 96 22.2 Proofs of First Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 3Evan Chen (Spring 2015) Contents 22.3 Proof of Hurwitz's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 98 22.4 A Theorem We Tried to Use But Couldn't . . . . . . . . . . . . . . . . . 99 4Evan Chen (Spring 2015) 1 January 29, 2015 §1 January 29, 2015 Grading system for 55B with' 55A. We'll be more or less following Baby Rudin. §1.1 De nition and Examples This subsection is copied from my Napkin project. De nition 1.1. A metric space is a pair (M;d) consisting of a set of points M and a metric d :MMR . The distance function must obey the following axioms. 0 • For any x;y2M, we have d(x;y) =d(y;x); i.e. d is symmetric. • The function d must be positive de nite which means that d(x;y) 0 with equality if and only if x =y. • The function d should satisfy the triangle inequality: for all x;y;z2M, d(x;z) +d(z;y)d(x;y): Abuse of Notation 1.2. Just like with groups, we will abbreviate (M;d) as just M. Example 1.3 (Metric Spaces ofR) (a) The real lineR is a metric space under the metric d(x;y) =jxyj. (b) The interval 0; 1 is also a metric space with the same distance function. (c) In fact, any subset S ofR can be made into a metric space in this way. 2 Example 1.4 (Metric Spaces ofR ) 2 (a) We can make R into a metric space by imposing the Euclidean distance function p 2 2 d ((x ;y ); (x ;y )) = (x x ) + (y y ) : 1 1 2 2 1 2 1 2 2 (b) Just like with the rst example, any subset ofR also becomes a metric space 2 after we inherit it. The unit disk, unit circle, and the unit square 0; 1 are special cases. 2 Example 1.5 (Taxicab onR ) 2 It is also possible to place the following taxicab distance onR : d ((x ;y ); (x ;y )) =jx xj +jy yj: 1 1 2 2 1 2 1 2 For now, we will use the more natural Euclidean metric. (One can also use max instead of a sum.) 5Evan Chen (Spring 2015) 1 January 29, 2015 n Example 1.6 (Metric Spaces ofR ) We can generalize the above examples easily. Let n be a positive integer. We de ne the following metric spaces. n (a) We let R be the metric space whose points are points in n-dimensional Euclidean space, and whose metric is the Euclidean metric p 2 2 d ((a ;:::;a ); (b ;:::;b )) = (a b ) + + (a b ) : 1 n 1 n 1 1 n n This is the n-dimensional Euclidean space. n n (b) The open unit ballB is the subset ofR consisting of those points (x ;:::;x ) 1 n 2 2 such that x + +x 1. 1 n n1 n (c) The unit sphereS is the subset ofR consisting of those points (x ;:::;x ) 1 n 2 2 such that x + +x = 1, with the inherited metric. (The superscript n 1 1 n n1 indicates that S is an n 1 dimensional space, even though it lives in 1 2 n-dimensional space.) For example, S R is the unit circle, whose distance between two points is the length of the chord joining them. You can also think n of it as the \boundary" of the unit ball B . Example 1.7 (Discrete Space) Let S be any set of points (either nite or in nite). We can make S into a discrete space by declaring the following distance function. ( 1 if x6=y d(x;y) = 0 if x =y: IfjSj = 4 you might think of this space as the vertices of an regular tetrahedron, 3 living inR . But for larger S it's not so easy to visualize. . . Example 1.8 (Graphs are Metric Spaces) Any connected simple graph G can be made into a metric space by de ning the distance between two vertices to be the graph-theoretic distance between them. (The discrete metric is the special case when G is the complete graph on S.) Example 1.9 (Function Space) We can let M be the space of integrable functions 0; 1R and de ne the metric R 1 by d(f;g) = jfgj dx. 0 §1.2 Continuous Maps This is again largely excerpted from my Napkin project. Abuse of Notation 1.10. For a function f and its argument x, we will begin abbrevi- ating f(x) to just fx when there is no risk of confusion. 6Evan Chen (Spring 2015) 1 January 29, 2015 In calculus you were also told (or have at least heard) of what it means for a function to be continuous. Probably something like A function f :RR is continuous at a point p2R if for every " 0 there exists a  0 such thatjxpj =) jfxfpj". Question 1.11. Can you guess what the corresponding de nition for metric spaces is? All we have do is replace the absolute values with the more general distance functions: this gives us a de nition of continuity for any function MN. De nition 1.12. Let M = (M;d ) and N = (N;d ) be metric spaces. A function M N f :MN is continuous at a point p2M if for every " 0 there exists a  0 such that d (x;p) =) d (fx;fp)": M N Moreover, the entire function f is continuous if it is continuous at every point p2M. Notice that, just like in our de nition of an isomorphism of a group, we use both the metric of M for one condition and the metric of N on the other condition. Example 1.13 Let M be any metric space and D a discrete space. When is a map f : D M continuous? Any map DM is continuous. 1 Proof. Take an open ball of radius . 2 Example 1.14 2 3 The mapRR by x7x is continuous. So is x7x . Proof. Homework. Example 1.15 n n Let X =R with one product metric and let Y =R with another product metric. Then id :XY is continuous. Thus we will generally not be pedantic about the choice of metric. §1.3 Forgetting the Metric Topological Spaces Again excerpted from the Napkin project. De nition 1.16. A topological space is a pair (X;T ), whereX is a set of points, and T is the topology, which consists of several subsets of X, called the open sets of X. The topology must obey the following four axioms. • ? and X are both inT . • Finite intersections of open sets are also inT . • Arbitrary unions (possibly in nite) of open sets are also inT . Abuse of Notation 1.17. We refer to the space (X;T ) by just X. (Do you see a pattern here?) 7Evan Chen (Spring 2015) 1 January 29, 2015 Example 1.18 We can declare all sets are open: a discrete space is a topological space in which every set is open. Example 1.19 Declare U open if and only if8x2U, the ball of radius r centered at x is contained in U. Proof. Check all the axioms. Blah. Proposition 1.20 Let M be a metric space. Then for all x2M and r 0, the ball B(x;r) =fyjd (x;y)rg M is open. Proof. Pick y in the ball. Let t =d(x;y). Then tr, so pick " with t +"r. You can check using the triangle inequality that B(y;")B(x;r). Example 1.21 0; 1 is not open since no ball at 0 is contained inside it. De nition 1.22. A subset Y is closed i XY is open. De nition 1.23. A function f :XY of topological spaces is continuous at p2X if the pre-image of any neighborhood of fp is also a neighborhood of p. With some e ort, we can show this is the same de nition of continuity as with metric spaces. §1.4 Intermediate Value Theorem Theorem 1.24 (IVT) Let f : 0; 1R be continuous such that f(0) 0 and f(1) 0. Then9a2 0; 1 such that f(a) = 0. This theorem is not cheap, and requires the following theorem. Theorem 1.25 Let AR be bounded above. Then there exists a least upper bound y2R. Proof of IVT. Long and boring. Just draw a picture. The main point is to take y to be the least upper bound of the a for which f(a) 0. 8Evan Chen (Spring 2015) 1 January 29, 2015 \This is just a game of quanti ers. To do this, all you have to do is be sober, which is not a problem since you are all under 21. . . . I cannot do this past 6PM, not because I'm not sober, but because I am old." Dennis Gaitsgory 9Evan Chen (Spring 2015) 2 February 3, 2015 §2 February 3, 2015 Didn't attend class. §2.1 Sequential Continuity Continuous functions send convergent sequences to convergent sequences, limits sent to limits. The converse is true in metric spaces. §2.2 Cauchy Completeness So far we can only talk about sequences converging if they have a limit. But consider the p sequence x = 1, x = 1:4, x = 1:41, x = 1:414, . . . . It converges to 2 inR, of course. 1 2 3 4 But it fails to converge inQ. And so somehow, if we didn't know about the existence of R, we would have no idea that the sequence (x ) is \approaching" something. n That seems to be a shame. Let's set up a new de nition to describe these sequences whose terms eventually get close to each other, but don't necessarily converge to a point. De nition 2.1. Letx ;x ;::: be a sequence which lives in a metric spaceM = (M;d ). 1 2 M We say it is Cauchy if for any " 0, we have d (x ;x )" M m n for all suciently large m and n. Note that, unlike the rest of this chapter, this is a notion which applies only to metric mak make e sure sure spaces. In a general topological space there is not a good enough notion of \distance" to make this de nition work. Question 2.2. Show that a sequence which converges is automatically Cauchy. (Draw a picture.) Now we can de ne the following. De nition 2.3. A metric space M is complete if every Cauchy sequence converges. Most metric spaces aren't complete, likeQ. But it turns out that every metric space can be completed by \ lling in the gaps" somehow, resulting in a space called the completion of the metric space. The construction is left as an (in my opinion) fun problem. It's a theorem thatR is complete. To prove this I'd have to de neR rigorously, which I won't do here (yet). In fact, there are some competing de nitions ofR. It is sometimes de ned as the completion of the spaceQ. Other times it is de ned using something called Dedekind cuts. For now, let's just accept thatR behaves as we expect and is complete. Example 2.4 (Examples of Complete Sets) (a) R is complete. (b) The discrete space is complete, as the only Cauchy sequences are eventually constant. (c) The closed interval 0; 1 is complete. n (d) R is in fact complete as well. (You are welcome to prove this by induction on n.) 10Evan Chen (Spring 2015) 2 February 3, 2015 Example 2.5 (Non-Examples of Complete Sets) (a) The rationalsQ are not complete. 1 (b) The open interval (0; 1) is not complete, as the sequence x = is Cauchy but n n does not converge. §2.3 R is Complete Theorem 2.6 (Bolzano-Weirestra) Any sequence in 0; 1 has a convergent subsequence. §2.4 1-countable and Hausdor Most of the \nice" properties of metric spaces carry over to 1-countable Hausdor general topological subspaces. 11Evan Chen (Spring 2015) 3 February 5, 2015 §3 February 5, 2015 Last time we de ned Cauchy sequences. Cool. Today we will complete a metric space Theorem 3.1 Let X be a metric space. (a) There exists an X and an isometric embedding  : X X so that X is complete and im is dense in X. (b) Any isometric embedding f : X Y with Y complete factors uniquely through X via an isometric embedding. (In other words, X is universal with this property.) (c) In (b), if f\(X) is dense than f is an isometry. 0 0 Here a dense set X X means that every neighborhood of X contains a point of X . Also, f\(X) just meansff(x)jx2Xg. Corollary 3.2 The completion ofQ isR. Proof. We have a dense isometryQ,R. Okay, it's not actually that straightforward, you need some properties of R. . . We will assume in what follows thatR is complete. §3.1 Completion Consider the set of all Cauchy sequncees of X. We de ne a metric on them by  (fx g;fy g) = lim ((x ;y )): n n n n n Actually, we rst need to show that this limit exists. Lemma 3.3 Iffx g,fy g are Cauchy, then the sequence (x ;y )2R asn = 1; 2;::: is Cauchy. n n n n Proof. By \Quadrilateral Inequality", j(x ;y )(x ;y )j(x ;x ) +(y ;y ): n n N N n N n N Using the "=2 trick completes the proof. Assuming thatR is complete (COUGH COUGH), this now implies that lim ((x ;y )) n n n exists. You can check it also satis es the Triangle Inequality. Unfortunately, it's easy to nd examples with (fx g;fy g) = 0, with (x ) =6 (y ). So n n n n we \mod out" by this. Hence, we de ne C(X) as the set of all Cauchy sequences modded out by the relation x y if (fx g;fy g) = 0. The rest is left as homework. n n n n 12Evan Chen (Spring 2015) 3 February 5, 2015 §3.2 Why no compactness Lemma 3.4 A function f : 0; 1R is bounded above. Proof. Otherwise we get a sequence of points x , x , . . . such that f(x ) m for all 1 2 m m. Then we can nd a convergent subsequence using Bolzano-Weirestra. This breaks sequential continuity. Theorem 3.5 Any function f : 0; 1R has a global maximum. Proof. By the lemma,f\0; 1 is bounded above and we can take the least upper boundy. We claim this is actually in the limit. If not, then we can construct a sequence x ;x ;::: 1 2 1 such that f(x )y . Take a convergent subsequence by Bolzano-Weirestra. Then m m (x ) converges to some x2 0; 1, but then we must have f(x) =y. m This sucks. Compactness is better. Rawr. Second Proof, from Jane. As before, there is a least upper bound y. Assume for contra- diction that the bound is never obtained. Then the function 0; 1R by 1 x7 yf(x) is an unbounded continuous function on 0; 1, impossible. §3.3 Sequential Compactness De nition 3.6. A topological space is sequentially compact if every sequence has a convergent subsequence. Proposition 3.7 Let f :XY . If X is sequentially compact then f(X) is sequentially compact. Proof. Trivial. If (fx ) is a sequence, take a converge subsequence of (x )2X. n n Theorem 3.8 (Heine-Borel) AR is sequentially compact if and only if it closed and bounded. We can now kill the original maximum theorem. Given f : 0; 1R, its image inR is bounded and closed and we can easily use this to show that f has a maximum. To prove Heine-Borel, we rst prove one direction in greater generality. 13Evan Chen (Spring 2015) 3 February 5, 2015 Lemma 3.9 0 0 Let Y Y , where Y is sequentially compact and Y is 1-countable and Hausdor . 0 Then Y is closed in Y . Proof. BecauseY is 1-countable, we can use the sequence de nition of closed in the tricky 0 0 direction: it suces to prove that if y ;y ;y with y 2Y then y2Y . But y has 1 2 i i 0 a convergent subsequence to some y . It had also better converge to y. By Hausdor , 0 y =y . Lemma 3.10 0 0 Suppose Y  Y for some topological spaces, with Y a Hausdor space. If Y is 0 closed and Y is sequentially compact then Y is sequentially compact. 0 Proof. Triviality: given y in Y use sequential compactness of Y to force some subse- n 0 0 quence to converge to y2Y . Since Y is closed, y2Y . Proof of Heine-Borel. First, suppose A is bounded and closed; by bounded-ness it lives in some A a;b. By Bolzano-Weirestra, the set a;b is sequentially compact. By our lemma, A a;b is sequentially compact. For the converse, suppose A is sequentially compact. We did a lemma that show A is closed. Hence A is bounded. §3.4 Compactness Apparently Gaitsgory does not like sequential compactness QQ. De nition 3.11. A topological space is compact if every open cover has a nite subcover. Proposition 3.12 If f :XY is continuous and Y is compact then f(X) is compact. Proof. Tautological using the de nition of continuity. We can mirror this for sequential compactness now. Lemma 3.13 0 0 Let Y be a Hausdor space and Y a compact subspace. Then Y is closed. 0 0 Proof. We will show that for any y2 Y nY , there is a neighborhood U  Y nY 0 0 containingy. For everyy 2Y , we can use the Hausdor condition to nd neighborhoods 0 0 U 03y andV 03y. By de nition,Y is covered byU 0, and we can nd a nite subcover y y y n 0 Y  U y i i=1 Now take n \ V 3y: y i i=1 14Evan Chen (Spring 2015) 3 February 5, 2015 0 This is a neighborhood of y2Y disjoint from Y . Proposition 3.14 If X is rst-countable and X is compact, then it is sequentially compact. Proof. Suppose not. Letfx g be a sequence with no convergent subsequence. Using n 1-countable, we nd that for every x2X, there exists a U contains only nitely many x S elements of the sequence. But then we can take a nite subcover U . x Proposition 3.15 If X is second-countable, then sequentially compact implies compact. Here second countable means there is a countable base. 15Evan Chen (Spring 2015) 4 February 12, 2015 §4 February 12, 2015 \Today we will be doing applied math" Gaitsgory \This can't be happening" \Applied by my standards". §4.1 Warm-Up Lemma 4.1 Let a , b be sequences converging to a and b. Then a +b a +b, ab ab. i i i i i i 2 Proof. The maps +; :R R are continuous. §4.2 \Applied" Math Also known as: cooking up stupid bounds. Also known as: suppose a x b for all n, where a and b are the locations of n n n n n two cops and x is the location of a drunkard. n Now a surprisingly nontrivial statement. Example 4.2 n Letjaj 1. Then the sequence (a ) converges to 0. n n Proof. WLOG a 0 (the other case is easy). We bound the sequence a between two guys. Put the estimate     n 1 1 n 1 : a a From this we deduce 1 1 n 0a   : 1 n 1 a Example 4.3 p n For any real number a 0, the sequence ( a) converges to 1. n p n The existence of a follows from the Intermediate Value Theorem. Proof. It suces to consider a 1, since otherwise we can consider 1=a. Let x be such n n that x =a for each n. Then n   n a 1a 1 + n a so 1x 1 + . n Example 4.4 p n The sequence ( n) converges to 1. n 1 n Proof. 1n (1 + ) by Bernoulli's inequality. n 16Evan Chen (Spring 2015) 4 February 12, 2015 §4.3 Series We'll put curly brackets around series for emphasis. De nition 4.5. A seriesfa g converges if the sequence b =a + +a converges. n n 1 n Here is a better notion. De nition 4.6. The seriesfa g converges absolutely ifb =jaj++ja j converges. n n 1 n Example 4.7 n Forjaj 1 the seriesfa g converges absolutely. n n+1 1a 1 n Proof. 1 + +a = . 1a 1a Lemma 4.8 If the seriesfa g converges then the sequence (a ) converges to 0. n n Remark 4.9. The converse is false; for example, take the harmonic series. Proof. The partial sums S converge as sequences to some b. Then (S S ) is a n n+1 n n convergent sequence to bb = 0. Lemma 4.10 The seriesfa g converges if and only if8" 0 there is a natural N such that n n 2 X a " i i=n 1 for all n ;n N. 1 2 Proof. The condition just says that the partial sums are a Cauchy sequence. Since R is complete, that's equivalent to partial sums converging. Lemma 4.11 A series converging absolutely also converges. Proof. Use the triangle inequality in the previous lemma in the form X X " jaj  a : i i 17Evan Chen (Spring 2015) 4 February 12, 2015 §4.4 Convergent Things Theorem 4.12 Consider a convergent seriesfa g. n (a) Iffa g converges absolutely, then for any permutation  :NN, the series n fa g converges absolutely to the same value. n (n) (b) Iffa g converges but not absoluetly, then for any L2R we can permute the n sequence  :NN so thatfa g converges to L. (n) This is why absolutely convergent series are better than non-convergent series. Lemma 4.13 A monotonically increasing sequence b converges if and only if it's bounded above. n Proof. One direction's clear. Now assume it's increasing and bounded by L. By com- pactness on b ;L we get a convergent subsequence and we're happy. 1 §4.5 Convergence Tests Proposition 4.14 Letfa g have positive terms. Thenfa g converges if and only if there existsA such n n that n X a A n i=1 for all n. Proof. Let S be the partial sums and apply the previous lemma. n Corollary 4.15 Letfa g andfb g have positive terms. If all partial sums of a are at most the n n n partial sums of b , then convergence of b implies convergence of a . n n n Proposition 4.16 (Cauchy Root Test) Letfa g be a series. n p n (a) Assume lim sup ja j 1. Thenfa g converges absolutely. n n p n (b) Assume lim sup ja j 1. Thenfa g diverges. n n p n Proof. For part (a), let a = lim sup ja j. Let" such that a +" 1. For all but nitely n many n's, p n a a +": n n So for suciently large n we have a  (a +") , done. n 18Evan Chen (Spring 2015) 4 February 12, 2015 For part (b), note that in nitely many terms are actually greater than 1 which is impossible. No conclusion when the lim sup equals 1. Proposition 4.17 (Ratio Test) Letfa g be a series and assume that n a n+1 lim sup 1: a n Thenfa g converges absolutely. n Proof. Let r be the lim sup and let " 0 such that r +" 1. For suciently large n we haveja j (r +")ja j and bound the sequence above. n+1 n Example 4.18 n x The seriesfa g given by a = converges absolutely. n n n Proof. Direct application of ratio test. a §4.6 The Series n c In homework, we will de ne what x means for real numbers x and c. In fact I can tell you: def c q x = limx : qc q2Q c Now let's consider the sequence a =n , where c 0 is a real number. The Ratio and n Root tests both fail. Here is the answer. Theorem 4.19 (Zeta Series) c The sequencefa g given by a =n converges for c 1 and diverges for c 1. In n n particular the harmonic series 1 1 1 + + +::: 1 2 3 is divergent. The idea is the \powers of two" estimate, which we prove in full generality here. Lemma 4.20 Letfa g be a series of positive real numbers which is monotically decreasing, and let n n n b = 2 a : n 2 Thenfb g converges if and only iffa g converges. n n 19Evan Chen (Spring 2015) 4 February 12, 2015 The lemma immediately implies the theorem on zeta series. §4.7 Exponential Function We de ne 1 X x def n exp(x) = : n n=0 We showed earlier that this always converges. Lemma 4.21 (Physicist's Lemma) Letfa g andfb g be absolutely convergent series. De ne n n n X c = ab : n i ni i=0 P P Thenfc g converges to the product of a  b . n n n Proof. We want to show that there for an" 0 there exists anN such that whennN 2n X c AB " i i=0 where A and B are sums offa g andfb g. n n For suciently large N, n n X X 1 a b AB  ": n n 3 i=0 i=0 So we want to estimate the value of 2n n n X X X c a b : i n n i=0 i=0 i=0 Expanding, we nd that the quantity is actually X X X X X X ab + ab = b a + b a i j i j j i i j ni2n nj2n 1jn ni2n 1in nj2n jn in So it suces to prove that for large enough n we have X X 1 b a ": j i 3 1jn ni2n Using absolute convergency, X X jbj jaj j i 1jn ni2n "=1000 The left term is at most B, while the right term can be made to be at most , as B desired. You can actually weaken the condition to just one series being absolutely convergent. 20