Part A. Metric and Topological Spaces

Part A Metric and Topological Spaces 1

Lecture A1 Introduction to the Course This course is an introduction to the basic ideas of topology and metric space theory for first-year graduate students. Topology studies the idea of continuity in the most general possible context. As a separate subject, it takes its origin from the pioneering papers of Poincaré at the end of the 19th century, and its development is one of the central stories of 20th century pure mathematics. (The history by Dieudonné, A History of Algebraic and Differential Topology, 1900-1960, is available free to Penn State members online. but that history begins roughly where this course leaves off.) But the idea of continuity is so central that topological methods crop up all over mathematics in analysis of course (including its computational twin, numerical analysis); in geometry; in pure algebra; even in number theory. So, if you re a first-year grad student, you most likely need to take this course (and maybe even Math 528 as well). In this lecture we ll just describe the course protocol and prerequisites we ll get started properly next time. There is a recommended course textbook Introduction to Topology by Gamelin and Greene, which is available in a cheap Dover reprint for 15 bucks. However, more important than that are the online course materials, which are available in two ways: Through the course website, which is http://www.math.psu.edu/roe/527-fa10/index.html Through the course page on ANGEL, Penn State s course management system, at http://cms.psu.edu Registered students (i.e., those taking this course for a grade) will want to use the ANGEL site which will contain homework assignments, quizzes and supporting material, which all need to be taken on time in order to receive credit. The AN- GEL page also contains the detailed course syllabus, which contains Universityrequired information about exam schedules, academic integrity requirements, office hours, and suchlike. Please be sure to read this information carefully. Detailed lecture notes for each lecture will be posted 24 hours in advance on these sites. To succeed in the course you need to read and study the notes before the lecture to which they refer. This is very important! Most lectures will contain one or more in-class exercises. You are expected to attempt these exercises before the next class session. Students may be called 2

on to present their solutions in class. You grade your own exercises using the ANGEL system, and this self-assessment will form a (small) component of your final in-class grade. Exercise A1.1. Here is your first in-class exercise. Find the corresponding assessment item on ANGEL (it s called inclassaug23 ), and enter your response ( I fully understand this ). You must complete this assignment before tomorrow evening, when it will expire. Now for some information about the prerequisites for the course, that is, things that I will assume that you know. There are four groups of these. 1. Number systems and their properties: specifically the natural numbers N, integers Z, rational numbers Q, real numbers R, and complex numbers C. Courses in analysis or foundations of mathematics often contain an account of procedures for constructing all or some of these, but we ll just take them as given. The most important of these number systems for us is the system R of real numbers. The basic properties of the reals are summarized by Proposition A1.2. R is an archimedean complete ordered field. What does this mean? (a) A field basic properties of additional, multiplication, subtraction, division commutative, associative, distributive laws. (b) ordered. There is an order relations < with the expected properties. In particular x 2 0 for all x. (c) Archimedean no infinite or infinitesimal real numbers. There is a unique (injective) homomorphism of rings Z R taking 1 to 1. (Proof?) The archimedean property is that every element of R is smaller than (the image of) some element of Z. (d) complete no holes, contrast Q. There are various ways of expressing completeness. The standard way is the Least Upper Bound Axiom. Let S be any nonempty set of real numbers. A number a R is an upper bound for S if, for all x S, x a. Let U S be the set of all upper bounds for S. The least upper bound axiom says that, if S and U S are both nonempty, then U S has a least member. This is called the least upper bound for S and written sup S. 3

Example A1.3. Let S = {x R : x 2 < 2}. Then U S = {a R : a > 0, a 2 2}. U S has a least member, namely sup S = 2. The corresponding statement with R replaced everywhere by Q would be false. Exercise A1.4. (Cantor s nested interval theorem) Let [a n, b n ], n = 1, 2,... be a sequence of closed intervals in R that are nested in the sense that [a n+1, b n+1 ] [a n, b n ] and whose lengths b n a n tend to zero. Show that the intersection n [a n, b n ] contains one and only one real number c. Exercise A1.5. Use the previous exercise to show that the real numbers are uncountable, i.e. cannot be listed as c 1, c 2,.... Hint: Suppose such a listing is possible. Pick an interval [a 1, b 1 ] of length at most 1 that doesn t contain c 1, then a subinterval [a 2, b 2 ] of length at most 1 2 that doesn t contain c 2, and so on. Apply the nested interval theorem. What can you say about the limit point c? 4

Lecture A2 Metric Spaces We continue with our list of prerequisites for the course (we discussed number 1 last time.) 2. Set-theoretic language such as unions (A B), intersections (A B), complements (A\B), subsets (A B) and so on. In topology it is often necessary to deal with infinite unions and intersections. Let F be a family of sets (that is, a set whose members are subsets of some other set). The intersection of the family, written F F F or just F, is the set of objects that belong to every member of the family F : in symbols x F F F, x F. Similarly the union of the family, written F F F or just F, is the set of objects that belong to some member of the family F : in symbols x F F F, x F. De Morgan s Laws extend to this context: if F is a family of subsets of some set X, then X \ F = X \ F F F F F and the same with intersection and union reversed. 3. Quantified statements and their proofs. We already met some quantifiers in the previous section: things like and. Throughout pure mathematics, but especially analysis and topology, one meets concepts that are built up of layers of quantifiers nested together in complicated ways. For instance, a function f : R R is continuous at a given point x 0 if ɛ > 0 δ > 0 x R x x 0 < δ f(x) f(x 0 ) < ɛ. It is necessary that you can grasp what such a complicated statement says, and that you can understand how the form of the statement dictates the shape of its proof. For instance the statement above begins ɛ > 0 so its proof has to begin Let ɛ > 0 be arbitrary, followed by an argument that establishes the rest of the statement for a given arbitrary value of ɛ. You could call a simple pattern like this a proof skeleton corresponding to the for all quantifier. 5

Exercise A2.1. Give some examples of other proof skeletons corresponding to other parts of a statement, e.g.,, and so on. Exercise A2.2. Write down in symbols (as above) what it is for the function f to be not continuous at x 0. 4. Axiomatic method This is the standard operating procedure of modern mathematics. We understand specific examples by fitting them into a general theory. The general theory consists of a collection of axioms, such as those for a group or a vector space. There are many examples which realize the axioms, and we develop a theory that applies to all of them. We will study topology from this point of view. We ll begin with the axioms for metric spaces. Definition A2.3. A metric space is a set X equipped with a function d: X X R (called a metric or distance function) such that: (i) d(x, x ) 0 for all x, x ; moreover, d(x, x ) = 0 if and only if x = x. (ii) d(x, x ) = d(x, x) for all x, x (symmetry). (iii) d(x, x ) d(x, x ) + d(x, x ) for all x, x, x (triangle inequality). The motivating example is the metric d(z, w) = z w on C or R. Definition A2.4. Let (X, d X ) and (Y, d Y ) be metric spaces. An isometry from X to Y is a bijection f : X Y such that for all x 1, x 2 X. d Y (f(x 1 ), f(x 2 )) = d X (x 1, x 2 ) For the usual metric on the plane, the isometries are just the congruences of Euclidean geometry. Two metric spaces that are related by an isometry are equivalent from the point of view of metric space theory. Here are some more examples of metric spaces. Example A2.5. The vector space R n can be made into a metric space by defining the distance between points x = (x 1,..., x n ) and y = (y 1,..., y n ) to be d(x, y) = ( x 1 y 1 2 + + x n y n 2) 1/2. (The proof that this does indeed satisfy the triangle inequality is a standard exercise.) The same formula also makes C n into a metric space. These are called Euclidean spaces. 6

Example A2.6. One can also define metrics on R n by d (x, y) = ( x 1 y 1 + + x n y n ) and d (x, y) = sup { x 1 y 1,..., x n y n }. These are different metrics from the standard one (though, as we ll see later, they are in a certain sense equivalent.) Example A2.7. Let X be any set. The discrete metric on X is defined by { 0 (x = y) d(x, y) = 1 (x y) Example A2.8. Let A be a finite set (the alphabet) and consider the set A n of n-tuples of elements of A, which we think of as n-letter words in the alphabet A. Define a distance on A n by d(x, y) = #{i : 1 i n, x i y i }; in other words, the number of positions in which the two words differ. This Hamming distance was introduced in 1950 to give a technical foundation to the theory of error-correcting codes (Hamming, Error-detecting and error-correcting codes, Bell System Technical Journal 29(1950), 147 160.) Example A2.9. Let X be any metric space and let Y be a subset of X. Then the distance function on X, restricted to Y, makes Y into a metric space; this metric structure is called the subspace metric on Y. Example A2.10. Here is an infinite-dimensional example. Consider the collection C[0, 1] of all continuous functions [0, 1] C. We can define a metric by d(f, g) = sup{ f(t) g(t) : t [0, 1]}. Convergence of a sequence of functions in this metric is called uniform convergence. Definition A2.11. A subset U of a metric space X is open if for every x U there is ɛ > 0 such that the entire ball is contained in U. B(x; ɛ) := {x X : d(x, x ) < ɛ} 7

The triangle inequality shows that every ball is open, so there are plenty of open sets. A set F whose complement X \ F is open is called closed. Note carefully that closed does not mean the same as not open. Many sets are neither open nor closed, and some may be both. 8

Lecture A3 Open and Closed Sets We finished last time with an important definition: A subset U of a metric space X is open if for every x U there is ɛ > 0 such that the entire ball is contained in U. B(x; ɛ) := {x X : d(x, x ) < ɛ} Example A3.1. In a discrete metric space (Example A2.7), the open ball B(x, 1 2 ) is just the set {x}. Consequently, in a discrete space, every subset is open. Example A3.2. Consider the three metrics on R n defined in examples A2.5 and A2.6. Each open ball in any one of these metrics contains an open ball (with the same center but possibly different radius) in any one of the other metrics. Consequently, these three metrics all have exactly the same open sets. That is what is meant by saying that they are equivalent. Lemma A3.3. The union of any collection of open sets is open. The intersection of a finite collection of open sets is open. The empty set, and the entire metric space X, are open. Proof. Let F be a collection of open subsets of a metric space X and let U = F be the union of the family. If x U, then there is some V F such that x V. Since V is open, there is ɛ > 0 such that B(x; ɛ) V. Since V U, we also have B(x; ɛ) U. Thus for any x U there exists ɛ > 0 such that B(x; ɛ) U; which is to say that U is open. Now let F = {U 1,..., U n } be a finite collection of open sets and let U = F = U1 U n. If x U, then for each i = 1,..., n there is ɛ i > 0 such that B(x; ɛ i ) U i. Let ɛ = min{ɛ 1,..., ɛ n } > 0. Then B(x; ɛ) U i for all i, and thus B(x; ɛ) U. Thus for any x U there exists ɛ > 0 such that B(x; ɛ) U; which is to say that U is open. Equivalently, using de Morgan s laws, we have Lemma A3.4. The intersection of any collection of closed sets is closed. The union of a finite collection of closed sets is closed. The empty set and the entire metric space X are closed. 9

Let X be any metric space, and let S be a subset of X. Definition A3.5. The interior of S (in X), denoted S, is the union of all the open subsets of X that are included in S. In symbols, we have S = { U : U S and U open in X }. The interior of S is an open set (because it is the union of a family of open sets) and (from the definition) any open subset of X that is included in S is also included in S. Thus, the interior of S is just the biggest open subset of X that is included in X. In particular, S is open if and only if it is equal to its own interior. Dually, we have Definition A3.6. The closure of S (in X), denoted S, is the intersection of all the closed subsets of X that include S. The closure of S is a closed set, and it is the smallest closed set that includes S; in particular, S itself is closed iff it is equal to its own closure. Finally, Definition A3.7. The boundary of S (in X), denoted X, is the set-theoretic difference S = S \ S. The boundary of S is the intersection of two closed sets (S and X \ S ), and it is therefore closed. Example A3.8. Let X = R and let S = ( (0, 1) Q ) (2, 3]. Then S = (2, 3), S = [0, 1] [2, 3], and S = [0, 1] {2, 3}. Exercise A3.9. Show that if A, B are subsets of X, and A B, then A B and A B. Now show that for any subset S of X, ( (S ) ) = ( S ). One more remark about open sets. Let X be a metric space and let Y be a subset of X. As we said earlier, Y can be considered as a metric space in its own right (with the metric that it inherits from X). What is then the relation between the open subsets of the new space Y and the open subsets of the original space X? 10

Proposition A3.10. Let X be a metric space, Y a metric subspace (as above). Then a subset V Y is open in Y iff it can be written V = U Y for some U X open in X. Proof. Notice the following relationship between balls in Y and in X:if y Y then B Y (y; r) = B X (y; r) Y. Let V be open in Y. Then for each y V there is r y > 0 such that B Y (y; r y ) V. Let U = y Y B X(y; r y ). This is a union of open subsets of X, so open in X, and ( ) U Y = B X (y; r y ) Y = B Y (y; r y ) = V. y Y y Y This proves one direction of the if and only if statement in the proposition; the other direction (which is easier) is an exercise. The definition of continuity is translated in the natural way to the metric space context. Definition A3.11. A function f : X Y between metric spaces is continuous at x X if for every ɛ > 0 there is δ > 0 such that d(f(x), f(x )) < ɛ whenever d(x, x ) < δ. It is continuous if it is continuous at every x X. But there is a very important alternative characterization in terms of open sets. Theorem A3.12. Let X and Y be metric spaces. Then f : X Y is continuous iff, for every open U Y, the inverse image is open in X. f 1 (U) := {x X : f(x) U} Proof. Suppose that f is continuous and let U Y be open. Let x f 1 (U); then f(x) U so by definition of open there is ɛ > 0 such that B(f(x); ɛ) U. By definition of continuous there is δ > 0 such that if x B(x; δ) then f(x ) B(f(x); ɛ) U. But this means that B(x; δ) f 1 (U). Thus the set f 1 (U) is open. Conversely, let x X, ɛ > 0 and suppose that f satisfies the condition in the theorem. In particular, we may consider U = B(f(x); ɛ), an open set such that x f 1 (U). Our hypothesis tells us that f 1 (U) is open, which means that there is a δ > 0 such that B(x; δ) f 1 (U). We have shown that whenever x B(x; δ), f(x ) B(f(x); ɛ). This gives us continuity. 11

Lecture A4 Continuity and sequences In the previous lecture we saw that a map f : X Y between metric spaces is continuous iff f 1 (U) is open (in X) whenever U is open (in Y ). Remark A4.1. It is equivalent to say that f 1 (F ) is closed in X whenever F is closed in Y. But in general nothing can be said about the behavior of the direct image f(s) of an open or a closed set S under a continuous map. Definition A4.2. A map f : X Y between metric spaces is called a homeomorphism if it is a bijection and both f and f 1 are continuous. (Equivalently, f is a continuous map with a continuous inverse.) If there is a homeomorphism between X and Y, then we say that these spaces are homeomorphic. A homeomorphism is a topological equivalence (it is easy to check that the relation of homeomorphism is an equivalence relation in the sense of algebra). In topology, we consider homeomorphic spaces to be essentially the same. A fundamental question in topology is to devise processes for answering the question, Are two concretely given spaces homeomorphic or not? Example A4.3. The map x x(1 + x 2 ) 1 2 is a homeomorphism from R to the open interval 9 1, 1). Its inverse is y y(1 y 2 ) 1 2. Example A4.4. The map t (cos 2πt, sin 2πt) is a continuous bijection from the half-open interval (0, 1] onto the unit circle in R 2, but it is not a homeomorphism. Example A4.5. Let (X, d) be a metric space and let (X, d ) be the same set equipped with the discrete metric. Then the identity map (X, d ) (X, d) is a continuous bijection, but it is usually not a homeomorphism. Example A4.6. Are R n and R m homeomorphic if m n? This and related questions were problematical in the early days of topology. (Note that there are continuous maps of R m onto R n even if m < n space-filling curves.) But the answer, as expected, is indeed no, as was shown by Brouwer and others at the beginning of the 20th century. We ll now study the properties of sequences. A sequence in a metric space X is a mapping from the natural numbers N to X: we ll follow the usual abbreviation of referring to the sequence (x n ) rather than the sequence which maps the natural number n to x n X. 12

Definition A4.7. Let (x n ) be a sequence in the metric space X. We say (x n ) converges to x X if for every ɛ > 0 there is an integer N such that d(x, x n ) < ɛ whenever n > N. We write then x = lim n x n. Exercise A4.8. Let T be the metric space {1, 1, 1,..., 0} (with the metric it inherits as a subspace of R). Show that a sequence (x n ) in the metric space X is 2 3 convergent if and only if there exists a continuous function f : T X such that f(1/n) = x n ; and in that case f(0) = lim n x n. Proposition A4.9. A function f : X Y between metric spaces is continuous at x if and only if, whenever (x n ) is a sequence converging to x, the sequence (f(x n )) converges to f(x). Proof. Suppose that f is continuous at x and let ɛ > 0 be given. By definition of continuity there is δ > 0 such that d(x, x ) < δ implies d(f(x), f(x )) < ɛ. By definition of convergence there is N such that for all n > N, d(x, x n ) < δ. Thus d(f(x), f(x n )) < ɛ and f(x n ) converges to f(x). Suppose that f is not continuous at x. Then there is ɛ > 0 such that for all δ > 0 there is x with d(x, x ) < δ and d(f(x), f(x ) ɛ. Let x n be a value of x corresponding to δ = 1/n. Then x n x, but d(f(x), f(x n )) ɛ for all n, hence f(x n ) does not converge to f(x). Remark A4.10. This result illustrates how sequences can be used to probe the topological properties of metric spaces. At the same time, it shows how the usefulness of these probes depends on the countable local structure of a metric space specifically, on the fact that any ball around x X contains one of the countably many balls B(x; 1/n). When we come to consider more general topological spaces, this countable local structure may not be available, and then sequences will be of less use. Definition A4.11. Let X be a metric space, A X. A point x X is a limit point of A if it is the limit of a sequence of distinct points of A. Lemma A4.12. x is a limit point of A if and only if every open set containing x also contains infinitely many points of A. Proof. Suppose that x is a limit point of A. Then there is a sequence (x n ) of distinct points of A, converging to x. Let U be an open set containing x. Then there is some ɛ > 0 such that B(x; ɛ) U. By definition of convergence, there 13

is N > 0 such that x n B(x; ɛ) U for all n > N. In particular, U contains infinitely many of the {x n }, all of which are members of A. Conversely suppose that every open set containing x also contains infinitely many points of A. Define a sequence (x n ) inductively as follows: x 1 is any point in B(x; 1) A, and, assuming that x 1,..., x n 1 have been defined, let x n be any point in B(x; 1/n) A that is distinct from the (finitely many!) points x 1,..., x n 1. Then (x n ) is a sequence of distinct points of A and it converges to x. Proposition A4.13. The closure of a subset A is the union of A and the set of all its limit points. In particular, A is closed if and only if it contains all its limit points. Proof. A point x belongs to the closure of A iff it is not in the interior of X \A. By definition of interior, this is the same as to say that each ball B(x; ɛ) must meet A. By the previous lemma, then, any limit point of A must belong to A. Conversely, suppose that x A \ A. Define a sequence (x n ) inductively as follows: x 1 is any point in B(x; 1) A, and, assuming that x 1,..., x n 1 have been defined, let x n be any point in B(x; δ n ) A where δ n = min{d(x 1, x),..., d(x n 1, x)}. Then (x n ) is a sequence of distinct points of A and it converges to x, so x is a limit point of A. Definition A4.14. An isolated point of A is a point of A that is not a limit point of A. A closed subset of a metric space is perfect if it has no isolated points. Exercise A4.15. Show that a A is isolated if and only if there is ɛ > 0 such that B(a; ɛ) A = {a}. Exercise A4.16. A closed ball in a metric space is a set B(x; r) := {x X : d(x, x ) r}. Show that a closed ball is closed. Must every closed ball be the closure of the open ball of the same center and radius? 14

Lecture A5 Compactness I ll take for granted the notion of subsequence of a sequence of points in a matric space. (Formally speaking, if a sequence in X is a map N X, a subsequence of the given sequence is the result of composing it with a strictly monotonic map N N.) Lemma A5.1. Every sequence of real numbers has a monotonic subsequence. Proof. Let (x n ) be a sequence of real numbers. Suppose that it has no monotonic increasing subsequence. Then there must be some n such that x m < x n for all m > n. (Otherwise, we could build a monotonic increasing subsequence by induction.) But now take this n and call it n 1. Apply the previous argument to the tail of the original sequence beginning at n 1, i.e. the subsequence x n1 +1, x n1 +2,.... This also has no monotonic increasing subsequence, so by the same argument there is n 2 > n 1 such that x m < x n2 for all m > n 2. Continuing in this way by induction we get a monotonic decreasing subsequence x n1, x n2,.... It is a standard consequence of completeness that every bounded, monotonic sequence of real numbers is convergent. Thus we get the Bolzano-Weierstrass theorem: every bounded sequence of real numbers has a convergent subsequence. Definition A5.2. A Cauchy sequence in a metric space is a sequence (x n ) with the following property: for every ɛ > 0 there exists N > 0 such that, for all n, m > N, d(x n, x m ) < ɛ. A metric space is called complete if every Cauchy sequence in it converges. A Cauchy sequence is necessarily bounded. Moreover, if a Cauchy sequence has a convergent subsequence, then the whole sequence is in fact convergent. These facts combine with the Bolzano-Weierstrass theorem to show that every Cauchy sequence of real numbers converges: i.e., R is (Cauchy) complete. Definition A5.3. A metric space X is (sequentially) compact iff every sequence of points of X has a subsequence that converges in X. Proposition A5.4. Every closed, bounded subset of R n or C n is sequentially compact. (A subset is bounded if it is contained in some ball.) 15

Proof. For R this is just the Bolzano-Weierstrass theorem. For R n, this follows from the fact (easily proved) that a sequence of points in R n is convergent if and only if each of its coordinate sequences is convergent. Proposition A5.5. If A is a subset of any metric space X, and A is compact (in its own right), then A is bounded and closed in X. Proof. Let x X be a limit point of A. Then there is a sequence (a n ) in A converging to x. But, by compactness, (a n ) has a subsequence converging in A. Thus x A, and A is closed. Suppose that A is not bounded. Then one can construct by induction a sequence (a n ) in A such that d(a n, a 0 ) > 1 + d(a n 1, a 0 ) for n 1. The triangle inequality shows that d(a n, a m ) > 1 for n m. Clearly (a n ) has no convergent subsequence. Exercise A5.6. Let f : X Y be continuous and surjective. Show that if X is compact then Y is compact also. What has this got to do with the familiar calculus principle that a continuous function on a closed bounded interval is itself bounded and attains its bounds? Despite the above evidence, it s not in general true that closed and bounded equals compact. (Counterexample?) We shall analyze this in detail. Definition A5.7. Let X be a metric space. An open cover U for X is a collection (finite or infinite) of open sets whose union is all of X. A Lebesgue number for U is a number δ > 0 such that every open ball of radius δ is a subset of some member of U. Theorem A5.8. (Lebesgue) Every open cover of a (sequentially) compact metric space has a Lebesgue number. Proof. Suppose that U does not have a Lebesgue number. Then for every n there is x n X such that B(x n ; 1/n) is contained in no member of U. If X is compact, the sequence (x n ) has a subsequence that converges, say to x. Now x belongs to some member U of U, since U is a cover. Thus there is ɛ > 0 such that B(x; ɛ) U. There is n > 2/ɛ such that d(x n, x) < ɛ/2. But then which is a contradiction. B(x n ; 1/n) B(x n ; ɛ/2) B(x; ɛ) U 16

Proposition A5.9. The following conditions on a metric space X are equivalent. (a) Every sequence in X has a Cauchy subsequence. (b) For every δ > 0 there is a finite cover of X by balls of radius δ. In this case we say X is totally bounded (some writers say precompact). Proof. (a) implies (b): Choose x 1 X, and then inductively choose x n X such that d(x n, x m ) δ when m < n, so long as this is possible. The process must terminate because if it didn t it would produce a sequence with no Cauchy subsequence. When it does terminate, with x N say, it does so because the balls B(x n ; δ), n = 1,..., N, cover X. (b) implies (a): Notice the following implication of (b): given any δ > 0, any sequence in X has a subsequence all of whose members are separated by at most δ (call this a δ-close subsequence ). Let (x n ) be any sequence in X. Let (x 1 n) be a 2 1 -close subsequence of (x n ), let (x 2 n) be a 2 2 -close subsequence of (x 1 n), and so on by induction. Then (x n n) is a Cauchy subsequence of the original sequence. A metric space X is said to be (covering) compact if every open cover of X has a finite subcover. Proposition A5.10. The following conditions on a metric space X are equivalent: (a) X is sequentially compact; (b) X is complete and totally bounded; (c) X is covering compact. Proof. It is easy to see that (a) and (b) are equivalent. Suppose (a). Let U be an open cover of X. Let δ > 0 be a Lebesgue number for U (which exists because of Theorem A5.8). Since X is totally bounded it has a finite cover by δ-balls. But each such ball is a subset of a member of U ; so U has a finite subcover. In the other direction, suppose (c). Let (x n ) be a sequence without convergent subsequence. Then for each x X there is some ɛ x such that x n / B(x; ɛ x ) for all but finitely many n. The B(x; ɛ x ) form a cover of X. Picking a finite subcover we obtain the contradiction that x n / X for all but finitely many n. 17

Remark A5.11. For general topological spaces, the notions covering compact and sequentially compact are not equivalent, and covering compact is usually the most appropriate one. Exercise A5.12. A map f : X Y between metric spaces is uniformly continuous if for each ɛ > 0 there is δ > 0 such that d(f(x), f(x )) < ɛ whenever d(x, x ) < δ. (The extra information beyond ordinary continuity is that δ does not depend on x.) Show that if X is compact, every continuous f is uniformly continuous. Hint: use Theorem A5.8. 18

Lecture A6 Compactness and Completeness (The lecture will begin with a review of the basic results about compactness, from the previous section). Definition A6.1. Let X be a compact metric space. Then C(X) denotes the space of all continuous functions X C. It is a metric space equipped with the metric (Compare Example A2.10.) d(f, g) = sup{ f(x) g(x) : x X}. Compactness of X ensures that the supremum exists (why)? We can also consider the space C R (X) of continuous real-valued functions. Proposition A6.2. For a compact X, the metric spaces C(X) and C R (X) are complete. Proof. Most completeness proofs proceed in the same three-stage way: Given a Cauchy sequence, identify a candidate for its limit; show that the candidate is in the space in question; and show that the sequence approaches the candidate limit in the metric of the space in question. Let (f n ) be a Cauchy sequence in C(X). Then, for each x X, f n (x) is a Cauchy sequence in C. Since C is complete this sequence converges. Denote its limit by f(x). We show that f is a continuous function on X. Fix x X and let ɛ > 0 be given. Because (f n ) is Cauchy there is N such that for n, n N we have f n (x) f n (x) < ɛ/3 for all x. Take n = N and let n to find that f N (x) f(x) ɛ/3 for all x. Now, f N is continuous at x so there is δ > 0 such that f N (x) f N (x ) < ɛ/3 whenever x x < δ. It follows that, whenever x x < δ, f(x) f(x ) f(x) f N (x) + f N (x) f N (x ) + f N (x ) f(x ) < ɛ. This shows that f is continuous. Finally to show that f n f in C(X) we must prove that for every ɛ > 0 there is N such that f n (x) f(x) < ɛ for all x whenever n > N. But in fact we already proved this in the previous paragraph. 19

It is an interesting and important question what are the compact subsets of the complete space C(X). The answer is given by the Ascoli-Arzela theorem. Closed and bounded is not enough. 20

Lecture A7 Applications of completeness Recall that a metric space X is said to be complete if every Cauchy sequence in X converges. The space (0, 1) is not complete, whereas the homeomorphic space R is complete. This proves that completeness is not a topological property, i.e., is not preserved under homeomorphism. Exercise A7.1. A uniform homeomorphism between metric spaces is a homeomorphism f : X Y such that both f and f 1 are uniformly continuous. Show that if X is complete and there is a uniform homeomorphism X Y, then Y is complete. Proposition A7.2. Let X be a complete metric space and let U X be an open set. Then U is homeomorphic to a complete metric space (it is topologically complete). Proof. Let f : U R be the function defined by f(x) = 1 inf{d(x, y) : y X \ U}. This is a well-defined, continuous function with the property that f(x n ) whenever x n is a sequence in U that converges to some x X \ U. Now let V be the metric space with the same points as U but with the metric d V (x, x ) = d(x, x ) + f(x) f(x ). It is easy to see that this is indeed a metric. Moreover, it is complete: if (x n ) is a Cauchy sequence for the metric d V, then it is also a Cauchy sequence for the metric d (so it converges in X, say to x X) and, in addition, the values f(x n ) are bounded (so the limit x in fact belongs to V ). Finally I claim that the identity map U V is a homeomorphism. Suppose x U and let ɛ > 0 be given. Since f is continuous, there is δ 1 > 0 such that d(x, x ) < δ 1 implies f(x) f(x ) < ɛ/2. Put δ = min{δ 1, ɛ/2)}. Then if d(x, x ) < δ, we have d V (x, x ) < ɛ 2 + f(x) f(x ) < ɛ. So the identity map U V is continuous, and the continuity of the inverse is easy. 21

Remark A7.3. A subset of a metric space is called a G δ -set (German: Gebeit- Durschnitt) if it is the intersection of countably many open sets. For example, the irrational numbers form a G δ subset of R. Generalizing the above construction, it can be shown that any G δ subset of a complete metric space is topologically complete. In particular, the topology on the irrational numbers can be given by a complete metric. But the same is not true for the rational numbers, as will follow from the Baire category theorem. Remark A7.4. A countable union of closed sets is called a F σ -set (French: fermé, somme). This is the only known example of dual mathematical concepts being described using dual languages. Definition A7.5. Let X be a metric space. A mapping f : X X is a (strict) contraction if there is a constant a < 1 such that d(f(x), f(x )) ad(x, x ) for all x, x X. Note that a contraction must be continuous. Theorem A7.6. (Banach) A contraction on a complete metric space has a unique fixed point (a point x such that f(x) = x). Proof. A fixed point is unique because if x, x are two such then d(x, x ) = d(f(x), f(x )) ad(x, x ), which implies d(x, x ) = 0. To prove existence, start with any x 0 X and define x 1 = f(x 0 ), x 2 = f(x 1 ) and so on. If d(x 0, x 1 ) = r then d(x n, x n+1 ) a n r and so d(x n, x n+k ) (a n + + a n+k 1 )r an r 1 a which tends to 0 as n. Thus (x n ) is a Cauchy sequence, which converges to a point x. We have so x is a fixed point. f(x) = lim f(x n ) = lim x n+1 = x Exercise A7.7. Banach s fixed point theorem can be extended as follows: if f : X X is a map and some power f N = f f is a strict contraction, then f has a unique fixed point. Prove this. Banach s fixed point theorem is one of the most important sources of existence theorems in analysis. Another important source of existence theorems, which also uses completeness, is the Baire category theorem. 22

Definition A7.8. A subset of a metric space is dense if its closure is the whole space Example A7.9. The rational numbers Q are dense in R. (It is often important that R has a countable dense subset. A space with this property is called separable.) Theorem A7.10. (Baire) In a complete metric space, the intersection of countably many dense open sets is dense. Proof. Let x X and let ɛ > 0. Let {U n }, n = 1, 2,..., be dense and open. Choose x 1 U 1 B(x; ɛ/2). There is r 1 such that B(x 1 ; r 1 ) U 1 ; without loss of generality take r 1 < ɛ/4. Since U 2 is dense there is x 2 U 2 B(x 1 ; r 1 ). Proceed inductively in this way choosing x n and r n such that x n U n B(x n 1 ; r n 1 ), B(x n ; r n ) U n, r n < r n 1 /2. Then (x n ) is a Cauchy sequence, say converging to a point x. Since x n B(x j ; r j ) for all j n, the point x belongs to B(x j ; r j ) for every j, and hence to Un. Moreover, d(x, x ) < ɛ. Since x and ɛ were arbitrary, U n is dense. Remark A7.11. The name category theorem comes from the following (traditional) terminology. A set A is nowhere dense if the complement of its closure is dense. A set is of first category if it is the countable union of nowhere dense sets. Baire s theorem then says that in a complete metric space the complement of a set of first category (sometimes called a residual set) is dense. Remark A7.12. Notice that this gives us another proof that R is uncountable (compare Exercise A1.5). For the complement of a point in R is a dense open set. More generally, any complete, perfect (Definition A4.14) metric space is uncountable. For in a perfect metric space, the complement of any point is dense. Exercise A7.13. Consider the complete metric space C R [0, 1] of continuous realvalued functions on [0, 1]. For a, b [0, 1] show that the set of functions f which are nondecreasing on the interval [a, b] is nowhere dense in C R [0, 1]. Using Baire s theorem, deduce that there exist functions f C R [0, 1] that are nowhere monotonic, that is, monotonic on no subinterval of [0, 1]. 23

Lecture A8 Normed Spaces Definition A8.1. Let V be a vector space (real or complex). A norm on V is a function V R, denoted v v, such that (a) v 0 for all v, and v = 0 iff v = 0. (b) λv = λ v, where λ is a scalar (real or complex). (c) v + v v + v. Clearly, d(v, v ) = v v is then a metric. Because of (b), this metric has an affine structure not present in a general metric. Example A8.2. Let x = (x 1,..., x n ) be a vector in R n or C n. The expressions all define norms. x 1 = x 1 + + x n x 2 = ( x 1 2 + + x n 2) 1/2 x = max{ x 1,..., x n } Exercise A8.3. Let a and b be positive real numbers and let p > 1. Show that inf{t 1 p a p + (1 t) 1 p b p : t (0, 1)} = (a + b) p. Hence (or otherwise) show that the expression x p = ( x 1 p + + x n p ) 1/p is also a norm on R n or C n. Example A8.4. If V = C(X), we may define f = sup{ f(x) : x X}. This is a norm, and it gives rise to the usual metric on C(X). Since it is complete (theorem A6.2) it is a Banach space as defined below. 24

Definition A8.5. If a normed vector space is complete in its metric, it is called a Banach space. Proposition A8.6. A linear mapping T : V W between normed vector spaces is continuous if and only if there is a constant k such that (one then says that T is bounded). T v k v Proof. If T is bounded then T u T v k u v so T is continuous. If T is continuous then there is some δ > 0 such that u < δ implies T u < 1. Take k = 1/δ. Definition A8.7. The best constant k in the above proposition, that is the quantity sup{ T v : v 1} is called the norm of T and denoted T. Exercise A8.8. Show that, with the above norm, the collection L(V, W ) of bounded linear maps from V to W is a normed vector space, and that it is a Banach space if W is a Banach space. Exercise A8.9. Show that the norm of linear maps is submultiplicative under composition, i.e. S T S T. 25

Lecture A9 Differentiation in Normed Spaces The basic idea of differentiation is that of best linear approximation. Normed spaces provide a systematic way to express this. Notation A9.1. Let f be a function defined on some ball B(0; ɛ) in a normed space V and having values in another normed space W. We shall write f(h) = o( h ) to mean that the limit lim h 0 ( h 1 f(h) ) exists and equals zero. Similarly for expressions like f(h) = g(h) + o(h). Exercise A9.2. Suppose that T : V W is a linear map. Show that T (h) = o( h ) if and only if T = 0. Now let f : V W be a continuous (need not be linear) map between normed vector spaces. In fact, we need only suppose f is defined on some open subset U of V. Definition A9.3. With above notation, let x V. We say that f is differentiable at x if there is a bounded linear map T : V W such that f(x + h) = f(x) + T h + o( h ). By the exercise, T is unique if it exists. It is called the derivative of f at x and written Df(x). Example A9.4. If V = W = R then every linear map V W is multiplication by a scalar, i.e. we identify L(V, W ) = R. Under this identification our definition of the derivative corresponds to the usual one from Calculus I. Example A9.5. If V = R m and W = R n then L(V ; W ) is the space of n m matrices. The matrix entries of the derivative of f are the partial derivatives of the components of f as defined in Calculus III. Remark: The existence of the partial derivatives does not by itself imply differentiability in the sense of our definition above. This just shows that our definition is right and partials are wrong! Example A9.6. If V = R and W is arbitrary, the space of (bounded) linear maps L(V ; W ) can be identified with W itself. Under this identification the derivative of f is given by the usual formula f(x + h) f(x) Df(x) = lim. h 0 h 26

Exercise A9.7. Show that the derivative of a linear map is always equal to the map itself. Proposition A9.8. (Mean value theorem) Suppose that the function f is differentiable throughout a ball B(x 0 ; r) V, and that Df(x) k for all x B(x 0 ; r). Then f(x) f(x 0 ) k x x 0 for all x B(x 0 ; r). Proof. Let ɛ > 0. By definition of the derivative each y B(x 0 ; r) has a neighborhood U y B(x 0 ; r) such that f(y ) f(y) (k + ɛ) y y y U y. Now we will creep along the line segment [x 0, x] B(x 0 ; r). Let x t = (1 t)x 0 + tx, t [0, 1], and let T be the set of all those t [0, 1] such that f(x s ) f(x 0 ) (k+ɛ) x s x 0 for all s t. Clearly, T is an interval and the displayed equation above shows that it contains [0, θ) for some θ > 0. Let τ = sup T (0, 1] and let y = x τ. If τ < 1 then there exists δ > 0 such that f(x t ) f(y) (k + ɛ) x t y t (τ 2δ, τ + 2δ). For any c [0, 1] we have τ cδ T and so f(x τ+cδ ) f(x 0 ) f(x τ+cδ ) f(x τ ) + f(x τ ) f(x τ cδ ) + f(x τ cδ ) f(x 0 ) (k + ɛ) ( x τ+cδ x τ + x τ x τ cδ + x τ cδ x 0 ) = = (k + ɛ) ( x τ+cδ x 0 ). The final equality is because the three vectors appearing here are positive multiples of the same vector. If follows that τ +δ T, contradicting the definition of τ as the supremum. Therefore it is impossible that τ < 1 and the theorem is proved. Remark A9.9. From the proof one sees that the theorem is true if we replace the ball by any region that is star-shaped about x 0. It is easy to see that the derivative of f ± g is Df ± Dg. Another familiar calculus result that generalizes easily is the chain rule. 27

Proposition A9.10. (Chain rule) Let V, W, X be normed vector spaces, and let x V. Let f : V W be differentiable at x and let g : W X be differentiable at y = f(x). Then g f is differentiable at x and D(g f)(x) = Dg(y) Df(x). Note that f need only be defined near x, and g need only be defined near y. Proof. Let k(h) = f(x + h) f(x) = Df(x) h + o( h ). Note that there is a constant A such that k(h) A h for small h. Now write giving the result. (g f)(x + h) (g f)(x) = g(y + k(h)) g(y) = Dg(y) k(h) + o( k(h) ) = Dg(y) Df(x) h + o( h ) Remark A9.11. Note that the derivative of f, Df, is itself a function having values in a normed space (namely the space L(V, W )). Thus we can differentiate it and define second and higher derivatives if we require. Exercise A9.12. (Open-ended) Formulate precisely the advanced calculus result that the mixed derivatives are symmetric, that is 2 f/ x y = 2 f/ y x, and prove it (under suitable hypotheses) for maps between normed vector spaces. Let F be a function R V V, where V is a normed vector space. Let [a, b] R. A function f : [a, b] V is a solution to the differential equation defined by F if it is differentiable and f (t) = F (t, f(t)) for all t [a, b]. The value f(a) V is called the initial condition for the solution. Theorem A9.13. Suppose that V is a Banach space and suppose also that the function F is continuous and satisfies a Lipschitz condition, that is, there exists a constant C > 0 such that F (t, v) F (t, v ) C v v. Then for any a R and any v 0 V there exists a nontrivial interval [a, b] on which there is a solution to the differential equation f (t) = F (t, f(t)) with initial condition f(a) = v 0 ; moreover, this solution is unique. 28

Proof. We need to use some facts about integration for V -valued functions. It isn t the job of this course to teach you about integration theory for that see Math 501 so I will just state the various facts that are needed. Moreover, in a first course on integration these facts are usually proved only for real or complex valued functions; you will need to take it on faith that the same facts are true for functions having values in a normed vector space V. Finally, we shall use the fact that if V is a Banach space (that is, complete) then the space of continuous functions [a, b] V, equipped with the sup norm, is also complete. This is a fact that we have already proved for real or complex valued functions, and the proof in the general case is exactly the same. We will use the fundamental theorem of calculus: a function f C[a, b] is a solution of the differential equation if and only if it is a solution of the equivalent integral equation f(s) = v 0 + s a F (t, f(t)) dt, t [a, b]. We will show that a solution to this equation exists (for b sufficiently close to a) by applying Banach s contraction mapping theorem (A7.6) to the space C[a, b] of continuous functions on [a, b] (with values in V, if you like). Let I : C[a, b] C[a, b] be defined by the right hand side of the display above, that is, (I f)(s) = v 0 + s a F (t, f(t)) dt, t [a, b]. If f, g C[a, b] then the integrands in I f and I g differ by C f g at most, where C is the Lipschitz constant, and therefore I f I g C(b a) f g by standard estimates on integrals. Therefore, if (b a) < C 1, the map I is a strict contraction and Banach s fixed point theorem provides a unique solution to I f = f, which is a solution to the differential equation with the specified initial condition. 29

Lecture A10 The inverse function theorem Let V, V be normed vector spaces. The space L(V, V ) is then a normed vector space also. Say T L(V, V ) is invertible if it has an inverse which is a bounded linear map from V to V. Proposition A10.1. Let V, V be Banach spaces. Then the invertibles form an open subset of L(V, V ), and the inverse operation T T 1 is continuous (on this subset) from L(V, V ) to L(V, V ). Proof. Without loss of generality V = V. Look first at a nbhd of the invertible operator I. Suppose S < 1. For y V consider the map φ y : V V, v y Sv. Since S < 1 this map is a strict contraction, so it has a unique fixed point x (Theorem A7.6). This fixed point is an x such that (I + S)x = y. By the triangle inequality, y (1 S ) x, so x (1 S ) 1 y and the map y x is bounded. We have shown that if S < 1, I+S is invertible. Moreover y (I + S) 1 y = S(1 + S) 1 y S (1 S ) 1 y which shows that the map S (I + S) 1 is continuous at S = 0. We have shown that the identity is an interior point of the set of invertibles and that the inverse is continuous there. However, left multiplication by a fixed invertible is a homeomorphism from the set of invertibles to itself; so the same results apply to any point of the set of invertibles. Exercise A10.2. Let V be a normed space and let W be the normed space L(V, V ). Show that the mapping i: T T 1 is differentiable (where defined) on W, and that its derivative is Di(T ) H = T 1 HT 1. 30

Definition A10.3. A map f from an open subset of a normed space V to a normed space W is continuously differentiable or C 1 if it is differentiable (everywhere) and the map x Df(x) is continuous. The inverse function theorem says that under suitable conditions, if Df(x) is invertible then f is locally invertible near x. Theorem A10.4. Let f be as above, defined near x V, and suppose that f is continuously differentiable and that Df(x) L(V, W ) is invertible. Suppose moreover that V, W are Banach spaces. Then there is an open set U containing x such that f is a bijection of U onto f(u), which is an open set containing f(x), and such that its inverse g : f(u) U is also continuously differentiable. Proof. It is an application of Banach s fixed point theorem A7.6. We will construct the inverse function by looking at the fixed points of a suitable map. Let A = Df(x) 1. For y W define a map φ y : V V by φ y (z) = z + A (y f(z)). A fixed point of φ y is a solution to f(z) = y. We have Dφ y (z) = I A Df(z) = A ( Df(x) Df(z) ). Since Df is continuous there is r > 0 such that if z U = B(x; r) then Dφ y (z) < 1. From the mean value theorem we conclude that if F is a closed 2 subset of U, and if we know for some reason that φ y maps F to F, then φ y is actually a contraction of this complete metric space and so has a unique fixed point. Fix z 0 U and let y 0 = f(z 0 ). Thus y 0 f(u) and we want to prove that there is some ɛ > 0 such that B(y 0 ; ɛ) f(u); this will show that f(u) is open. Choose δ > 0 such that the closed ball B(z 0 ; δ) is contained in U. Choose ɛ = 1 2 A 1 δ. If z B(z 0 ; δ) and y B(y 0 ; ɛ) then we may compute φ y (z) z 0 = φ y (z) φ y0 (z 0 ) φ y (z) φ y (z 0 ) + φ y (z 0 ) φ y0 (z 0 ) 1 2 z z 0 + A(y y 0 ) 1 2 δ + 1 2 δ = δ using the mean value theorem A9.8. We conclude that, for these y, the map φ y is a contraction of the complete metric space B(z 0 ; δ) and thus has a (unique) fixed point there. We have shown that each y 0 f(u) has an ɛ-neighborhood contained 31

in f(u); thus, f(u) is open. The contraction property of the φ y ensures that each y f(u) has only one inverse image in U; thus f is a bijection of U onto f(u). Now to show that the inverse function g = f 1 : f(u) U is differentiable at y = f(x) write g(y + h) g(y) = u(h), say. The calculations of the previous paragraph show that u(h) 2 A h for h small. Then h = f(g(y + h)) y = f(g(y) + u(h)) y = Df(x) u(h) + o( h ). Apply A = Df(x) 1 to get u(h) = A h + o( h ). This gives differentiability of g with Dg(f(x)) = Df(x) 1. Continuous differentiability now follows from the continuity of the inverse operation on the space of bounded linear maps (Proposition A10.1). 32

Lecture A11 Topological Spaces We have been emphasizing that up to homeomorphism properties of metric spaces depend only on their open sets. This motivates a further abstraction which removes the numerical notion of metric entirely. Definition A11.1. A topology on a set X is a family T of subsets of X with the following three properties: (i) If U α is any family of members of T, then the union α U α is also a member of T ; (ii) If U 1,..., U n is a finite family of members of T then the intersection n i=1 U i is also a member of T ; (iii) The sets and X are members of T. A set equipped with a topology is called a topological space. The members of T are called open subsets of the topological space X. Notice that the open subsets of a metric space automatically satisfy (i) (iii); thus, every metric space carries a topology (called its metric topology). Different metrics can however give rise to the same topology (as we have already seen); moreover, there are topologies that do not arise from any metric. Example A11.2. The discrete topology on X has T = P(X): every subset is open. This topology arises from the discrete metric. Example A11.3. The indiscrete topology on X has just two open sets, and X. This topology does not arise from a metric as soon as X has more than one point. (Why not?) Example A11.4. The cofinite topology on X consists of together with all the cofinite subsets of X (a subset is cofinite if its complement is finite). It is natural to say that a closed set is one whose complement is open, so that the closed sets in the cofinite topology are just the finite sets together with the whole space. Notice that this topology has many fewer open (or closed) sets than the topologies with which we are familiar. For example, in the cofinite topology on an infinite set, any two nonempty open sets have a nonempty intersection. A similar but more sophisticated example is the next one. 33