NOTES FOR MAT 570, REAL ANALYSIS I, FALL Contents

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1 NOTES FOR MAT 570, REAL ANALYSIS I, FALL 2016 JACK SPIELBERG Contents Part 1. Metric spaces and continuity 1 1. Metric spaces 1 2. The topology of metric spaces 3 3. The Cantor set 6 4. Sequences 7 5. Continuous functions 9 6. Uniform continuity Cauchy sequences and complete metric spaces Compactness Continuity and compactness Connectedness Continuity and connectedness Convergence of functions Banach spaces Compactness in function space Uniform approximation and the Stone-Weierstrass theorem Uniform convergence and the interchange of limits 29 Part 2. Some problems Problems 32 Part 3. Appendices The Riemann integral The Darboux approach Measure zero and the Riemann integral The fundamental theorem of calculus Infinite series Series of functions Power series Conditional convergence Bounded linear maps 54 1

2 NOTES FOR MAT 570, REAL ANALYSIS I, FALL Part 1. Metric spaces and continuity 1. Metric spaces Much of what we do in analysis ultimately comes down to measuring the distance between two real numbers. We use the absolute value for this: x y is the distance between the numbers x and y. There are many other situations where we use the distance between points in an essential way. For example, the Pythagorean theorem is used to define the usual distance between points in R 2, and even in R n. One of the wonderful abstractions of XXth century mathematics is a generalization of this notion of distance. In fact, it isn t too hard to notice that everything we use distance for in advanced calculus (e.g. limits, continuity, etc.) relies only on a few very coarse aspects of the distance function. The following definition sets these out precisely, and gives the basic setting for the first part of this course. Definition 1.1. Let X be a set. A metric on X is a function d : X X R such that (1) d(x, y) 0 for all x, y X (positivity). (2) d(x, y) = 0 if and only if x = y (definiteness). (3) d(x, y) = d(y, x) (symmetry). (4) d(x, y) d(x, z) + d(z, y) (triangle inequality). Example 1.2. The usual metric on R (or on C) is defined by d(x, y) = x y. Remark 1.3. Two common variations of the triangle inequality are easily proved as exercises: (1) d(x, y) d(x, z) d(y, z). (2) d(x, y) d(x, z 1 ) + d(z 1, z 2 ) + + d(z n 1, z n ) + d(z n, y). Many important examples of metric spaces arise from norms on vector spaces. We will consider vector spaces over R and over C. Sometimes we may write F for the scalar field, if the choice of real or complex scalars doesn t matter. Definition 1.4. Let V be a vector space (over R or C). A norm on V is a function : V R such that (1) x 0 (positivity). (2) x = 0 only if x = 0 (definiteness). (3) cx = c x for all c F (and x V ) (homogeneity). (4) x + y x + y (triangle inequality). Remark 1.5. If is a norm on V, there is an associated metric on V given by d(x, y) = x y. (Exercise. Prove the triangle inequality.) Definition 1.6. Let (X, d) be a metric space. A subset Y X is called bounded if there are a point x 0 X and a positive real number M such that d(y, x 0 ) M for every point y Y. Example 1.7. (Function space) Let S be a nonempty set. The bounded functions from S to R k (or C k ) are defined by B(S, R k ) = {f : S R k : the range of f is bounded}. It is easy to see that B(S, R k ) is a vector space over R (with point-wise operations). The uniform norm is defined on B(S, R k ) by f u = sup x S f(x). Exercise 1.8. Prove that the uniform norm really is a norm. (Mainly, prove the triangle inequality.) Example 1.9. R n = R R (n factors) can be thought of as B ( {1,..., n}, R ). Similarly, we can think of C n as B( ( {1,..., n}, C ). The uniform norm here is usually denoted. Thus (x 1,..., x n ) = max 1 i n x i. Another important way of producing a norm on a vector space is by means of an inner product. Definition Let V be a vector space over R or C. An inner product on V is a function, : V V F such that (1) x, x 0. (2) x, x = 0 only if x = 0. (3) x, y = y, x. (4) ax + by, z = a x, z + b y, z for all a, b F (and x, y, z V ).

3 2 JACK SPIELBERG Remark Notice that the complex conjugate in property 3 of Definition 1.10 is irrelevant in the case of a real vector space. Property 4 is called linearity in the first variable. By properties 3 and 4 it follows that an inner product on a real vector space is also linear in the second variable. In the case of a complex vector space, these two properties imply conjugate linearity in the second variable: z, ax + by = a z, x + b z, y for all a, b C (and x, y, z V ). It also follows from these properties that 0, y = y, 0 = 0 for all y V. Example (1) The standard inner product on R n is x, y = x 1 y x n y n. (2) The standard inner product on C n is x, y = x 1 y x n y n. Note that the use of the complex conjugate is necessary in order to satisfy property 1 of Definition Remark This is a good place to mention that one can consider C n just as a metric space. In that setting, it is the same thing as R 2n, in the obvious way. However, if we consider C n as a vector space over the complex field, then we are allowed to multiply a vector by any complex number, e.g. by i. This operation is not permissible on R 2n. The standard inner product also shows how different the spaces C n and R 2n are. Exercise Verify that in any complex inner product space, x + y, x + y = x, x + y, y + 2Re x, y. Theorem (Cauchy-Schwartz inequality) Let ( V,, ) be an inner product space. x, x 1/2 y, y 1/2. Then x, y Proof. By Remarks 1.11, the inequality holds if either x or y equals zero. So suppose that both are non-zero. For any a, b C, we have 0 ax + by, ax + by = a 2 x, x + b 2 y, y + 2Re ( ab x, y ). Using the polar form of the complex number x, y, we may write x, y = x, y e it, for some (real number) t. Then also e it x, y = x, y. Now let a = y, y 1/2 and b = x, x 1/2 e it. We get 0 y, y x, x + x, x y, y 2Re ( y, y 1/2 x, x 1/2 e it x, y ) = 2 x, x y, y 2 x, x 1/2 y, y 1/2 x, y. Divide by 2 x, x 1/2 y, y 1/2, and rearrange the inequality to finish the proof. Corollary Let V be an inner product space. For x V let x = x, x 1/2. Then is a norm on V. Proof. We will prove the triangle inequality, leaving the verification of the other properties of a norm as an exercise. Let x, y V. Then by the Cauchy-Schwartz inequality, x + y 2 = x + y, x + y = x, x + y, y + 2Re x, y = x 2 + 2Re x, y + y 2 x x y + y 2 = ( x + y ) 2. Example The usual norm on R n or C n arises from the usual inner product. The corresponding metric space is usually referred to as n-dimensional (real or complex) Euclidean space. We note the following important inequalities for the Euclidean norm (proof by squaring). Remark Let x F n. Then for any i, x i x x x n. Definition Let (X, d) be a metric space, and let Y X. If we restrict the metric d to points of Y then Y becomes a metric space, called a subspace of X. Example The circle (or torus) is a subspace of Euclidean space: T = { (x, y) R 2 : x 2 + y 2 = 1 }. (Thus, for example, d ( (1, 0), (0, 1) ) = 2.) We can also think of T as the set of complex numbers of modulus one: {z C : z = 1}.

4 NOTES FOR MAT 570, REAL ANALYSIS I, FALL It is very important to remember that, while pictures can give a lot of valuable intuition, they are not a substitute for a proof. In this course, you may never use a picture as part of a proof (though they can be included to help explain what you are doing). The next two examples illustrate some of the more bizarre kinds of metric spaces. The first is much more frequently encountered than you might imagine the first time you see it. It provides a counterexample to many obvious facts about metric spaces that are not actually true. The point is this: any theorem that we prove about metric spaces must be true for all metric spaces. In particular, it will be true for the metric spaces in the next examples. Example Let X = 1 {0, 1} = { (x 1, x 2,...) : x i {0, 1} for all i }. (Thus X is the set of all sequences of 0 s and 1 s. We mention in passing that the set X is uncountable. In fact, X is equivalent, in the sense of cardinality, to P(N), the set of all subsets of N.) We define a metric on X as follows. For x, y X with x y, the set {i : x i y i } is non-empty. By the well-ordering principle, it has a least element. We set k(x, y) = min{i : x i y i }. Then we define d(x, y) = { 1 k(x,y), if x y 0, if x = y. We claim that d is a metric on X. The proofs of positive definiteness and symmetry are immediate. We will verify the triangle inequality. In fact, we will prove something stronger, called the ultrametric inequality. Lemma For any x, y, z X, d(x, y) max { d(x, z), d(y, z) }. Proof. We will write k(x, x) = as a kind of shorthand. (But notice that then we have that d(x, y) < d(u, v) if and only if k(x, y) > k(u, v), for any points x, y, u, v X.) Now let x, y, z X. If d(x, y) d(x, z) then the inequality holds. So suppose that d(x, y) > d(x, z). Then k(x, y) < k(x, z). Since x i = z i for i < k(x, z), we have that z k(x,y) = x k(x,y) y k(x,y). Therefore k(y, z) k(x, y), and hence d(x, y) d(y, z). Therefore max { d(x, z), d(y, z) } = d(y, z) d(x, y). The following is another example of a metric space that varies from what our intuition suggests. This one often seems like a stupid metric space... well, it is stupid, but it is also a metric space. Every theorem about metric spaces must be true for it, and hence any statement that is not true for this example, cannot be proven using only the axioms of a metric space. Example Let S be any set. The discrete metric on S is defined by { 1, if x y, d(x, y) = 0, if x = y. Remark It is easy to see that the discrete metric on a set with n points can be realized as a subspace of Euclidean n-space. It is a little harder to find a natural setting for the discrete metric on N. The discrete metric on R is a useful counterexample to keep in mind. 2. The topology of metric spaces Definition 2.1. Let (X, d) be a metric space, and let a X, r > 0. The open ball with center a and radius r is the set B r (a) = { x X : d(x, a) < r }. The closed ball with center a and radius r is the set B r (a) = { x X : d(x, a) r }. Example 2.2. In R, B r (a) = (a r, a + r) and B r (a) = [a r, a + r]. You should sketch the pictures of the open and closed balls in R 2. These pictures are extremely useful as intuition when proving things. But it is NEVER permissible to use a picture as a substitute for a proof. The next definitions are REALLY, REALLY important. Definition 2.3. Let (X, d) be a metric space, and let E X. (1) E is an open set if for each x E there is r > 0 such that B r (x) E. (2) E is a closed set if E c is an open set.

5 4 JACK SPIELBERG Proposition 2.4. In a metric space, open balls are open sets and closed balls are closed sets. Proof. Let a X and r > 0, and let x B r (a). We need to find an open ball centered at x (with some positive radius) that is completely contained in B r (a). We know that d(x, a) < r, since x B r (a). Let s = r d(x, a). Then s > 0. We claim that B s (x) B r (a). To prove this, let y B s (x). Then d(y, x) < s. Then d(y, a) d(y, x) + d(x, a) < s + d(x, a) = r d(x, a) + d(x, a) = r, and hence y B r (a). Therefore B r (a) is an open set. The proof that closed balls are closed sets is left as an exercise. Proposition 2.5. The following hold in a metric space. (1) The union of any collection of open sets is open. (2) The intersection of a finite collection of open sets is open. (3) The intersection of any collection of closed sets is closed. (4) The union of a finite collection of closed sets is closed. Proof. These are important exercises. (The last two parts follow from the first two by DeMorgan s laws (and the notation for families of sets).) Exercise 2.6. Along with proving the above proposition, explain why your proof of (2) does not extend to infinitely many open sets. Example 2.7. (1) In any metric space X, X and are both open and closed. (It is a fairly deep fact (Corollary 11.8) that if X = R n then these are the only sets that are simultaneously open and closed.) (2) A singleton set in a metric space is a closed set. Proof. Let x X and y {x} c. Then y x, so r = d(x, y) > 0. Then B r (y) {x} c. (3) A finite subset of a metric space is a closed set. (4) In R n, sets of the form {x : x i > c}, {x : x i < c}, are called open half-spaces, and are open sets. Sets of the form {x : x i c}, {x : x i c} are called closed half-spaces, and are closed sets. Proof. Since a closed half-space is the complement of an open half-space, it is enough to prove openness of open half-spaces. If H = {x : x i > c} and y H, let r = y i c > 0. If z B r (y) then y i z i z i y i d(z, y) < r. Then z i = y i (y i z i ) > y i r = c, and hence z H. Thus B r (y) H, and we have shown that H is open. The proof for the other kind of open half-space is left as an exercise. Definition 2.8. An open box in R n is a set of the form (a 1, b 1 ) (a n, b n ), where a i b i for each i. Closed boxes in R n are defined similarly, by including all finite endpoints of the interval factors of the Cartesian product. We note that an open box is a finite intersection of (at most 2n) open half-spaces, and hence is an open set. Similarly, closed boxes are closed sets. It is important to remember that, while the complement of an open set is a closed set, the opposite of open is not closed many (most, even) sets are neither open nor closed. Exercise 2.9. Give an example of a subset of R that is neither open nor closed, and prove it. We next introduce the operations of interior and closure. These provide important open and closed sets associated with arbitrary subsets of a metric space. Definition Let X be a metric space and let E X. The interior of E is the set The closure of E is the set int (E) = { U : U E and U is open }. E = { K : K E and K is closed }. Remark (1) E is closed if and only if E = E. (2) E is open if and only if E = int(e).

6 NOTES FOR MAT 570, REAL ANALYSIS I, FALL Remark We observe that (1) int(e) is an open set, and is the largest open set contained in E. (2) E is a closed set, and is the smallest closed set containing E. (3) E = ( int(e c ) ) c, and int(e) = (Ec ) c. Proof. The first two items follow immediately from the definitions. For the third item, we have E = {K : E K, K closed} (E) c = {K c : E K, K closed} = {K c : K c E c, K c open} = {U : U E c, U open} = int (E c ). Taking complements of both sides yields the first formula. If we apply the first formula to E c, and take complements of both sides, we obtain the second formula. The above definitions are abstract, in that they don t give an explicit criterion to use to decide if a point does or does not belong to the interior or closure of a set. We now give such criteria. (Note the different quantifiers used for the two characterizations.) Proposition (1) x int(e) if and only if there is r > 0 such that B r (x) E. (2) x E if and only if for every r > 0, B r (x) E. Proof. (1) This is almost instantly obtained from the definition, and we leave the details as an exercise. (2) We note that x (E) c if and only if x Int(E c ), by Remark 2.12(3). But this is true if and only if there is r > 0 such that B r (x) E c, by part (1). But this is true if and only if there is r > 0 such that B r (x) E =. By negating the first and last items in this chain of equivalent statements, we find that x E if and only if for all r > 0, B r (x) E. Example It is worth thinking about the above definitions and results in the context of some examples. In R, (1) int ( (a, b] ) = (a, b). (2) int (Z) =. (3) int (Q) =. (4) (0, 1] = [0, 1]. (5) Z = Z. (6) Q = R. The following terminology is common, and expresses a useful notion. Definition Let x be a point of a metric space. A neighborhood of x is a set N such that x int (N). Thus a neighborhood of x is a set that contains an open set which contains x, or a set that contains some open ball with center x. For example, [0, 1] is a neighborhood of 1/2, but not of 0. Definition Let X be a metric space, and let f : X R k. The support of f is the closure of the set of points where f is nonzero, and is denoted supp(f). Thus supp(f) = {x X : f(x) 0}. Definition (1) Let X be a metric space, and let E X. E is dense in X if E = X. (2) The metric space X is separable if it contains a countable dense subset. Example (1) The set Q of rational numbers is dense in R. Since Q is countable, R is a separable metric space. (2) The set R \ Q of irrational numbers is also dense in R. Since the irrationals form an uncountable set, they are not a countable dense subset of R.

7 6 JACK SPIELBERG It might seem tempting to try to describe the new points sucked in by the closure operation, i.e. the points of E that are not already in E. However it turns out to be much more useful to describe the property that brings them into the closure. This property may apply also to some points already in E. Definition Let X be a metric space, E X, and a X. The point a is an accumulation point of E (also called by some people limit point or cluster point) if for every r > 0, the intersection E B r (a) is infinite. We write E for the set of accumulation points of E. Example Let X = R. (1) {1, 1 2, 1 3,...} = {0}. (2) {0, 1, 1 2, 1 3,...} = {0}. (3) Z =. (4) Q = R. Exercise Prove the equalities in the previous example. Note that E E this follows from Proposition 2.13(2). Therefore E E E. In fact, the two sides are equal, which fact is the content of the next result. Proposition E = E E. Before proving the proposition, we give a lemma that may seem surprising at first. (Namely, the existence of infinitely many points is equivalent to the existence of one point as long as it isn t that point.) Lemma a E if and only if for each r > 0, ( E \ {a} ) B r (a). Proof. ( ): We prove the contrapositive. Suppose that for some r > 0, ( E \ {a} ) B r (a) =. Then E B r (a) {a}, a finite set. Hence a E. ( ): ( We again prove the contrapositive. Let a E. Then there is r > 0 such that E B r (a) is finite. Let E Br (a) ) \ {a} = {x 1, x 2,..., x m }. Let s = min i d(a, x i ) > 0. Then ( E \ {a} ) B s (a) =. Proof. (of Proposition 2.22) We already proved in the comments before the proposition. For, let a E. If a E then clearly a E E. So suppose that a E. Let r > 0. By Proposition 2.13(2) we have E B r (a). But since a E this implies that ( E \ {a} ) B r (a). By Lemma 2.23, a E. Corollary E is closed if and only if E E. Remark Let X be a metric space. Recalling Definition 1.6, we see that a subset E X is bounded if there is a ball in X that contains E. It doesn t matter whether the ball is open or closed. X itself is called bounded if it is a bounded subset of itself. (In this case we say that the metric is bounded.) 3. The Cantor set In this section we introduce the first interesting set that most people come across. Let F 0 = [0, 1], F 1 = [0, 1 3 ] [ 2 3, 1], F 2 = [0, 1 9 ] [ 2 9, 1 3 ] [ 2 3, 7 9 ] [ 8 9, 1], and so on. Recursively, F n is obtained by removing the open middle third from each subinterval of F n 1. Thus F n is the disjoint union of 2 n closed intervals, each of length 3 n. F n is closed, nonempty, and F n F n+1. Definition 3.1. The Cantor set, C, is the set n=0 F n. It is a good idea to draw a picture. It isn t hard to see that C is nonempty: all the endpoints of the closed subintervals making up the F n s belong to C. Still, this set of endpoints is a countable set. In fact, C is much bigger, as we will now see. Recall the space X of Example 1.21: X = 1 {0, 1} is the set of all sequences of 0 s and 1 s. We will prove that C X. Definition 3.2. We define f : X C as follows. Let x = (x 1, x 2,...) X. For each n define a closed interval I n (x) recursively by I 0 (x) = [0, 1] { left piece of I n (x) F n+1, if x n+1 = 0, I n+1 = right piece of I n (x) F n+1, if x n+1 = 1.

8 NOTES FOR MAT 570, REAL ANALYSIS I, FALL Then I 0 (x) I 1 (x). Let us write I n (x) = [a n, b n ]. The nesting of these intervals implies that a 1 a 2 b 2 b 1. Let α = sup{a 1, a 2,...} and β = inf{b 1, b 2,...}. We claim that n=0 I n(x) = [α, β]. To see this, we first note that since a n α β b n for all n, [α, β] n=0 I n(x). On the other hand, if x n=0 I n(x), then a n x b n for all n. Hence x is an upper bound for the set of a n s, and a lower bound for the set of b n s. Thus α x β. This proves the claim. Finally, since b n a n = 3 n, we have β α 3 n for all n. Therefore α = β. It follows that n=0 I n(x) is the singleton set {α}. We define f by setting f(x) = α. More precisely, the above argument allows us to describe f as follows: {f(x)} = I n (x). Proposition 3.3. f is bijective. n=0 Proof. We first show that f is injective. Let x, y X with x y. Let k = k(x, y) (recall Example 1.21). For i < k, x i = y i, so that I i (x) = I i (y). Since x k y k, I k (x) and I k (y) are two disjoint subintervals of I k 1 (x) = I k 1 (y). Since f(x) I k (x) and f(y) I k (y), we must have f(x) f(y). We now show that f is surjective. Let t C. Then t F n for all n. For each n, let I n be the subinterval of F n containing t. Since I n and I n+1 are subintervals of F n and F n+1, respectively, then either I n I n+1 or I n I n+1 =. Since both contain t, we must have I n I n+1. Thus we must have n=0 I n = {t}. Now let { 0, if I n is the left piece of I n 1 F n, x n = 1, if I n is the right piece of I n 1 F n. Letting x = (x 1, x 2,...) X, we see that I n = I n (x) for all n, so that t = f(x). Corollary 3.4. R, C, X, and P(N) are equivalent sets. In particular, R is uncountable. Proof. In Example 1.21 we mentioned that X P(N). (Briefly, the map x X {n N : x n = 1} is a bijection.) In Proposition 3.3 we saw that X C. We finish the proof by showing that R C. Since C R we have C R. By the Cantor-Bernstein theorem, it suffices to show that R C. Since C P(N), it suffices to show that R P(N). But since N Q, we know that P(N) P(Q). Thus we will be finished if we can show that R P(Q). We do that as follows. We define a function g : R P(Q) by g(t) = {q Q : q < t}. If s t are distinct points of R, say s < t, then by the density of Q in R there exists q Q with s < q < t. Then q g(t) and q g(s), and we have g(s) g(t). Hence g is one-to-one. 4. Sequences Definition 4.1. Let X be a set. A sequence in X is a function x : N X. Remark 4.2. We usually write x n instead of x(n), but the latter notation is often useful too. We sometimes write (x n ) n=1, or (x n ), for x. It is important to remember that in this notation, n is a dummy variable it is the argument of the function x. (So, in particular, there is nothing special about the letter n used as the argument it will often be convenient to use a different letter.) Some texts use curly braces instead of parentheses, but we will avoid this notation, for the following reason. The range of the sequence x is the subset {x n : n N} of X. This is often referred to as the set of terms of (x n ). It is important to distinguish between the sequence itself (which is a function from N to X), and its set of terms (which is a subset of X). While we are on the subject of the subtlety of the notation for sequences, let me point out a common mistake to guard against. What should we make of the following statement (taken from more than one actual homework paper!): Let (x n ) be a sequence, and let (x i ) be another sequence.? Of course, this deserves a quantity of red ink, but you should think carefully about the precise error. (And PLEASE don t make this mistake too.) Definition 4.3. Let (x n ) be a sequence in a metric space X, and let a X. (x n ) converges to a if for every ε > 0, there exists n 0 N such that for all n n 0 we have d(x n, a) < ε. We write x n a (as n ) to indicate that (x n ) converges to a.

9 8 JACK SPIELBERG Lemma 4.4. A sequence in a metric space converges to at most one point. Proof. Suppose that x n a and x n b. Let ε > 0. There exist n 1, n 2 N such that d(x n, a) < ε/2 for all n n 1, and d(x n, b) < ε/2 for all n n 2. Let n = max{n 1, n 2 }. Then d(a, b) d(a, x n ) + d(x n, b) < ε/2 + ε/2 = ε. Since d(a, b) < ε for all ε > 0, it follows that a = b. Definition 4.5. If x n a, a is called the limit of (x n ), and we write lim n x n = a. We say that (x n ) converges if it has a limit; otherwise it diverges. Proposition 4.6. Let X be a metric space, let E X, and let a X. (1) a E if and only if there is a sequence in E converging to a. (2) a E if and only if there is a sequence in E \ {a} converging to a. (3) E is closed if and only if every sequence in E that converges in X has its limit in E. Proof. We prove part of the proposition, and leave the rest as an exercise. (1) ( ): Let a E. By Proposition 2.13(2), for each n N we have E B 1/n (a). Choose x n E with d(x n, a) < 1/n. Then x n a. Remark 4.7. Sequences are an important tool for studying metric spaces. One can think of a sequence as a kind of probe a function from N to the space picks out a certain countable subset in a manner indexed by the natural numbers. It is also useful to use sequences as tools to study a sequence itself. This leads to the next definition. Definition 4.8. Let x be a sequence in a set X, and let n be a strictly increasing sequence in N. (Thus n : N N satisfies n 1 < n 2 < n 3 <.) Then x n is another sequence in X. It is called a subsequence of x. Remark 4.9. The terms of the subsequence x n may be denoted (x n) i = x ( n(i) ) = x n(i) = x ni. Thus we may write x n = ( x ni ) i=1. The idea of a subsequence is pretty simple, but the notation can lead to lots of silly mistakes, against which you should be on guard. For example, let (x n ) be a sequence. The expression x 50 makes sense it is the 50th term of the sequence. Now let ( ) x ni be a subsequence. The expression xn50 makes sense it is the 50th term of the subsequence, and equivalently, it is the n 50 th term of the original sequence. However, the expression x 50i does not make sense. If we try to interpret it, we first realize that it is the value of the function x at the argument 50 i. So 50 i must be an element of the domain of x, namely a natural number. Now 50 i must be the value of the function 50 at the argument i. But this is nonsense 50 is not a function, so it can t be evaluated at the argument i. Here is another example to keep in mind. Suppose that we have a bunch of sequences in X. Say that x 1, x 2,... are all sequences (i.e. we have a sequence of sequences). How should we write the terms of the nth sequence? We have that x n : N X, so we can write x n = ( x n (i) ), using function notation for x i=1 n. If you want to use a subscripted i, one way to keep things clear is to use more parentheses: x n = ( ) (x n ) i. i=1 This makes it clear that i is the argument of the function x n, and not the argument of n (which is not a function). The more usual way is to use two subscripts: x n = (x ni ) i=1, and this is what we will do when we are faced with this situation. Writing it out longhand for clarity gives x n = (x n1, x n2, x n3,...). Note that it is necessary to write so clearly that the reader does not mistake the second subscript for a sub-subscript, and hence for a subsequence. Here is a simple result about subsequences. Proposition Let (x n ) be a convergent sequence in a metric space. Then every subsequence of (x n ) is also convergent, and has the same limit. Remark Before proving the proposition, we observe that if n : N N is strictly increasing, then n i i for all i. This is easily proved by induction on i, and we omit the proof. We do point out that equality is possible. In fact, letting n i = i for all i shows that every sequence is a subsequence of itself. Proof. (of Proposition 4.10) Let x n a, and let (x ni ) be a subsequence. We will show that x ni a. Let ε > 0. Since x n a, there is m such that d(x n, a) < ε whenever n m. Now if i m, then n i m, by the remark, so that d(x ni, a) < ε. Thus x ni a (as i ).

10 NOTES FOR MAT 570, REAL ANALYSIS I, FALL Remark It is clear from the definition that convergence or divergence of a sequence is unaffected if finitely many terms are changed. Convergence, divergence, the limit if convergent, are examples of properties of a sequence that depend only on the ultimate behavior of the sequence. In fact, such properties are the only ones that are important for sequences. One way to describe this is by means of tails of a sequence. If (x n ) is a sequence, the nth tail is the subsequence (x i ) i=n. Thus, if the sequence converges to L, then every tail of the sequence also converges to L. We sometimes say that a property holds eventually for a sequence if it holds for some tail. A sequence in a metric space is called bounded if it is a bounded function (compare with Example 1.7), that is, if its set of terms is a bounded set. A sequence is unbounded if it is not bounded. The proof of the next result is a good exercise, but it will also follow from some later results. Lemma Let (x n ) be a convergent sequence (in some metric space). Then (x n ) is bounded. 5. Continuous functions Definition 5.1. Let (X, d) and (Y, ρ) be metric spaces, f : X Y a function, and x 0 X. f is continuous at x 0 if for every ε > 0 there exists δ > 0 such that for every x X, if d(x, x 0 ) < δ then ρ ( f(x), f(x 0 ) ) < ε. f is continuous if it is continuous at each point of X. In the previous definition we have used different letters to denote the metrics on the domain and codomain of a function. Generally, we will not make such a distinction. The reader should be able to decide when the same letter is being used for two different purposes. Remark 5.2. Here are some equivalent formulations of continuity at a point x 0. (1) For every ε > 0 there exists δ > 0 such that f ( B δ (x 0 ) ) B ε ( f(x0 ) ). (2) For every open ball C with center f(x 0 ), there exists an open ball B with center x 0 such that f(b) C. (3) For every ε > 0 there exists δ > 0 such that B δ (x 0 ) f 1( B ε ( f(x0 ) )). Exercise 5.3. If f is continuous at x 0, then there is an open ball B centered at x 0 such that f is bounded on B. Example 5.4. (If the proof is not given, it is an exercise.) (1) Let f : R R be given by f(x) = x 2. Then f is continuous. Proof. Let x 0 R, and let ε > 0. Then for any x R, f(x) f(x0 ) = x 2 x 2 0 if x x 0 < 1, then if in addition x x 0 < ε/(1 + 2 x 0 ), then = x x 0 x + x 0 x x 0 ( x x x 0 ) ; x x 0 ( x 0 ) ; < ε. Now choose δ > 0 such that δ < min { 1, ε/(1 + 2 x 0 ) }. Then x x 0 < δ implies that x 2 x 2 0 < ε. (2) Define the identity function id : X X by id(x) = x. id is continuous. (3) Fix y 0 Y. Define f : X Y by f(x) = y 0 for all x X. Then f is continuous. (f is called a constant function.) (4) Define χ Q : R R by { 1, if x Q χ Q (x) = 0, if x Q. χ Q is discontinuous at each point of R.

11 10 JACK SPIELBERG (5) (The Hermite function) Define h : R R by { 1 h(x) = n, if x = m n in lowest terms, where m, n Z with n > 0 0, if x R \ Q. Then h is continuous at each irrational number, and discontinuous at each rational number. The proof is a nice exercise. (It is interesting to consider the opposite continuity behavior.) (6) We define the coordinate projections on R n (or C n ), π i : R n R, by π i (x) = x i. The π i are continuous (by Remark 1.18). Earlier we said that sequences are an important tool for studying objects in analysis. As evidence, we now show how to use sequences to characterize continuity of a function between metric spaces. Theorem 5.5. Let X and Y be metric spaces, and let f : X Y be a function. f is continuous if and only if for every convergent sequence x n a in X, we have that f(x n ) f(a) in Y. (Thus f is continuous if and only if it preserves convergent sequences, and maps the limit of a convergent sequence to the limit of the image sequence.) Proof. The forward direction is straightforward, and we leave it as an exercise. For the reverse direction we prove the contrapositive. Suppose that ( f is not ) continuous at a. Then there is ε > 0 such that for every δ > 0 there is x B δ (a) with f(x) B ε f(a). We apply this to δ = 1/n: thus there is a sequence (xn ) in X such that d(x n, a) < 1/n and ρ ( f(x n ), f(a) ) ε. But then clearly x n a while f(x n ) f(a). Remark 5.6. In the previous theorem, it is equivalent to state the result as: for every convergent sequence (x n ) in X, the sequence (f(x n )) in Y converges. The fact that the image sequence must converge to the image of the limit follows. We didn t mention this before, but the word topology has a technical meaning: the topology of a metric space is the collection of all the open subsets of the space. A property of the space is topological if it can be defined just by using the open sets. It is very important to know that continuity of functions is a topological property. Theorem 5.7. f : X Y is continuous if and only if for every open set V Y, the inverse image f 1 (V ) is open in X. Proof. (= ): ( Let V Y be open. Let x 0 f 1 (V ). Then f(x 0 ) V. Since V is open there is ε > 0 such that B ε f(x0 ) ) V. By Remark 5.2(3) there is ( δ > 0 such that B δ (x 0 ) f 1 (V ). Hence f 1 (V ) is open. (= ): Let x 0 X and let ε > 0. Since B ε f(x0 ) ) is open, then f 1( ( B ε f(x0 ) )) is open. Since x 0 f 1( ( B ε f(x0 ) )) there is δ > 0 such that B δ (x 0 ) f 1( ( B ε f(x0 ) )). Therefore f is continuous at x 0 (by Remark 5.2(3)). The proof of the following corollary is an exercise. Corollary 5.8. f : X Y is continuous if and only if for every closed set V Y, the inverse image f 1 (V ) is closed in X. This is a good place to introduce the notion of sameness for metric spaces. First, the definition: Definition 5.9. Let X and Y be metric spaces. A homeomorphism from X to Y is a function f : X Y which is bijective, continuous, and such that its inverse function f 1 is continuous. Two metric spaces are called homeomorphic if there exists a homeomorphism from one to the other. Homeomorphic metric spaces have the same topological structure and properties. It is colloquial to describe this by saying that one space can be deformed into the other by bending and stretching without tearing. Here are some simple examples. Example (1) Any two open disks in R 2 are homeomorphic. (2) Any two closed disks in R 2 having positive radii are homeomorphic. (3) No open disk in R 2 is homeomorphic to any closed disk in R 2. (This is not obvious.) (4) Every open ball in R n is homeomorphic to every open box in R n. (5) The unit circle T = {x R 2 : x = 1} is not homeomorphic to the unit interval [0, 1] R. (Again, it isn t so obvious how to prove this.)

12 NOTES FOR MAT 570, REAL ANALYSIS I, FALL Example Recall the function f : X C from Definition 3.2, where X = 1 {0, 1} is as in Example 1.21, and C is the Cantor set (Definition 3.1). We will show that f and f 1 are continuous functions. First some notation. If (a 1, a 2,..., a n ) n 1 {0, 1}, let Z(a 1,..., a n ) = {x X : x i = a i for 1 i n}. Such sets are called cylinder sets. Note that cylinder sets are simultaneously closed and open (i.e. clopen see Definition 10.3): Z(a 1,..., a n ) = B 1/n (x) = B 1/(n+1) (x) for any x Z(a 1,..., a n ). Note also that f ( Z(a 1,..., a n ) ) = C I n (x) (again for any x Z(a 1,..., a n )), which is a clopen subset of C (recall the definition of I n (x) from Definition 3.2). Thus these two families of clopen subsets are paired by the function f. Since every open subset of X is a union of open balls, i.e. of cylinder sets, and every open subset of C is a union of subsets of the form C I n (x) (an exercise!), it follows from Theorem 5.7 that f and f 1 are continuous. The proofs of the next two results are easy, and so are left as exercises. Corollary (of Theorem 5.7) Let X be a metric space. f : X R is continuous if and only if f 1( (a, b) ) is open for all a < b in R. Equivalently, f : X R is continuous if and only if {f < a} and {f > a} are open for all a R. Theorem Let f : X Y and g : Y Z be functions between metric spaces, and let x 0 X. If f is continuous at x 0, and g is continuous at f(x 0 ), then g f is continuous at x Uniform continuity Continuity is a locally defined property. Suppose that f : X Y is continuous. If ε > 0 is given, and if a point x 0 X is given, then continuity of f at x 0 provides a positive number δ with a certain property (Definition 5.1). The local-ness is expressed in the order of the quantifiers in that definition (and as we have rephrased it above): the number δ need only do its job for the one point x 0 already chosen. In fact, this means that δ (perhaps slightly modified) works throughout some ball centered at x 0. A(n open) ball centered at x 0 is a neighborhood of x 0. A property is local if each point has a neighborhood in which the property holds. A globally defined property, on the other hand, is one that holds everywhere. Continuity would be globally defined if the same δ worked for all points of X. Not all continuous functions have such a strong form of continuity; those that do have a special name. Definition 6.1. Let X and Y be metric spaces, and let f : X Y be a function. f is uniformly continuous if for every ε > 0 there exists δ > 0 such that for all x 1, x 2 X, if d X (x 1, x 2 ) < δ then d Y ( f(x1 ), f(x 2 ) ) < ε. Note that the only difference between this definition and the definition of continuity on X is in the order in which the point and the δ are specified. Some examples will help to clarify this. Example 6.2. (1) Let f : [ 10, 10] R be given by f(t) = t 2. Then f is uniformly continuous. Proof. Let ε > 0 be given. Let δ = ε/20. If t 1, t 2 [ 10, 10] are such that t 1 t 2 < δ, then f(t 1 ) f(t 2 ) = t 2 1 t 2 2 = t 1 + t 2 t 1 t 2 < ( t 1 + t 2 ) δ 20δ = ε. (2) Let g : R R be given by g(t) = t 2. Then g is not uniformly continuous. Proof. We choose ε = 1. Let δ > 0 be given. Choose t > 1/δ, and let s = t + δ/2. Then s t = δ/2 < δ, while s 2 t 2 = s t s + t = (δ/2)(2t + δ/2) > δt > 1 = ε. Therefore g is not uniformly continuous. (3) Let h : (0, 1) R be given by h(t) = sin(1/t). Then h is not uniformly continuous. Proof. We choose ε = 2. Let δ > 0 be given. Choose n > 1/ δ. Let s = 2/[(2n + 1)π] and let t = 2/[(2n + 3)π]. Then s t = 2 ( 1 π 2n ) = 2 ( ) 2 1 2n + 3 π (2n + 1)(2n + 3) n 2 < δ. But h(s) h(t) = 1 ( 1) = 2 ε. Therefore h is not uniformly continuous.

13 12 JACK SPIELBERG Definition 6.3. Let X and Y be metric spaces, and f : X Y. We say that f is Lipschitz continuous (or just Lipschitz) on X if there is a constant C > 0 such that for all x, x X, we have d(f(x), f(x )) Cd(x, x ). The number C is called a Lipschitz constant for f. Proposition 6.4. Let f : X Y be Lipschitz. Then f is uniformly continuous. The proof is left as an exercise. Remark 6.5. It is easy to see that a Lipschitz function is indeed a continuous function. continuous functions are Lipschitz. (See problem 50.) But not all Remark 6.6. If we think about the case of a function f : R R, we see that a Lipschitz constant for f is the same thing as a bound on the size of the (absolute value of the) difference quotient of f between any two (distinct) points of R. Thus f is Lipschitz if and only if there is a bound on the slopes of all secant lines connecting points of the graph of f. If in addition, we assume that f is differentiable, then any limit of slopes of secant lines is bounded by the Lipschitz constant; thus the derivative is bounded. 7. Cauchy sequences and complete metric spaces It may not have seemed important at the time, but the definition of convergence for a sequence has an unfortunate limitation. Namely, in order to check the definition, it is necessary to have the limit in hand. In order to use sequences as a tool to study spaces, it would be very helpful to be able to give an internal characterization of convergence, one that doesn t refer to the limit itself. This motivation is not possible to carry out in general, but the idea that came from it is very important. Definition 7.1. Let (X, d) be a metric space. A sequence (x n ) in X is Cauchy if for every positive real number ε, there exists n 0 N such that for all m, n n 0 we have d(x m, x n ) < ε. Informally, we say that the sequence is Cauchy if its terms can be made close to each other merely by requiring them to be far enough out in the sequence. It is an exercise in the logic of quantifiers to convince yourself that the definition captures precisely the idea behind this informal statement. The following lemma provides many examples of Cauchy sequences. Lemma 7.2. A convergent sequence is Cauchy. Proof. Let (x n ) be convergent, with limit x. Let ε > 0 be given. By the definition of convergence there is n 0 such that for all n n 0, d(x n, x) < ε/2. Then if m, n n 0 we have d(x m, x n ) d(x m, x) + d(x, x n ) < ε/2 + ε/2 = ε. Therefore (x n ) is Cauchy. Example 7.3. Here is an example of a non-cauchy sequence in R: let x n = n. (Exercise: prove that it s 1 not Cauchy.) But successive terms do get close to each other: x n x n+1 = n+ n+1 < 2 n. Example 7.4. Here is an example of a Cauchy sequence that does not converge. Let X = (0, 1) with the usual metric gotten from R. The sequence (1/n) in X is Cauchy but not convergent. (Remember the definition of convergence (Definition 4.5): the limit has to belong to the metric space.) Example 7.5. Here is a more interesting example of a non-convergent Cauchy sequence. Let V be the vector space of all finite real sequences: V = { (x 1, x 2,...) : x i R, and there exists i 0 such that for all i > i 0, x i = 0 }. We define a norm on V by x = ( ) 1/2 i=1 x2 i (note that the sum is actually finite). It s easy to see that this is a norm: the properties defining a norm only involve finitely many vectors at a time, and then the required property actually occurs in some Euclidean space, where we already know that the properties hold. Now, let v n = ( 1 2, 1 4, 1 8,..., 1 2 n, 0, 0, 0,...) V. If m < n, we have v n v m 2 1 = (0, 0,..., 0, 2 m+1,..., 1, 0, 0,...) 2 2n = n i=m+1 ( 1 ) 2 1 = 2 i 4 m+1 n m 1 i=0 1 4 i < 1 4 m.

14 NOTES FOR MAT 570, REAL ANALYSIS I, FALL Thus (v n ) is Cauchy in V. But we claim that (v n ) does not converge. To prove this, let y = (y i ) be an arbitrary vector in V. There is k such that y i = 0 for i > k. For n > k, y v n 2 = (y i v ni ) 2 = i=1 n ( yi 1 ) 2 ( yk+1 2 i 1 ) 2 1 = 2 k+1 4 k+1. i=1 Thus d(v n, y) 2 (k+1) for all n > k. Therefore v n y. Definition 7.6. A metric space is called complete if every Cauchy sequence converges. Theorem 7.7. R n is complete. We will give the proof after a couple of lemmas about Cauchy sequences in general metric spaces. Lemma 7.8. A Cauchy sequence is bounded. Proof. Let (a n ) be a Cauchy sequence. Then there is L such that d(a m, a n ) < 1 for all m, n L. Let R = max { d(a 1, a L ),..., d(a L 1, a L ) } + 2. Then d(a n, a L ) < R for all n, and hence (a n ) is bounded Lemma 7.9. A Cauchy sequence having a convergent subsequence is convergent. Proof. Let (a n ) be a Cauchy sequence, and let (a ni ) be a convergent subsequence, with limit c. We claim a n c. Let ε > 0. Since (a n ) is Cauchy there is L such that d(a m, a n ) < ε/2 for all m, n L. By the definition of convergence, there is i 0 such that d ( a ni, c ) < ε/2 for all i i 0. Let i 1 i 0 be such that n i1 L. Then for any n n i1 we have d(a n, c) d(a n, a ni1 ) + d(a ni1, c) < ε 2 + ε 2 = ε. Hence a n c. Proof. (of Theorem 7.7) We first show that R is complete. Let (a n ) be a Cauchy sequence in R. By Lemma 7.8 we know that (a n ) is bounded. It follows from the Bolzano-Weierstrass Theorem that (a n ) has a convergent subsequence. Then by Lemma 7.9 we know that (a n ) converges. Thus R is complete. Now it follows easily from Remark 1.18 that R n is complete (the details are left as an exercise). We now discuss the completion of a metric space. We will not prove the main theorem, but it is a doable exercise for someone with enough interest. Definition Let X be a metric space. A completion of X is a metric space Y and an isometric map f : X Y such that f(x) is dense in Y. For example, a completion of Q can be given by the inclusion of Q into R. Theorem Every metric space has a completion. Any two completions of a metric space are equivalent in the following sense: if f i : X Y i are completions of X, for i = 1, 2, then there is a unique isometric map g : Y 1 Y 2 such that g f 1 = f 2. Here is the idea of the proof of existence. Let S be the set of all Cauchy sequences in X. Define a relation on S by (x n ) (z n ) if lim n d(x n, z n ) = 0. It can be verified that is an equivalence relation. Let Y = X/sim, the set of equivalence classes. Define ρ : Y Y R by ρ ( [(x n )], [(z n )] ) = lim n d(x n, z n ). (Thus to compute the value of ρ on a pair of equivalence classes one chooses representatives of the classes, and evaluates the above limit.) It can be verified that this limit always exists, and does not depend on the choice of representatives of the equivalence classes. Further, it can be verified that ρ is a metric on Y. Define f : X Y by f(x) = [(x)], the equivalence class of the constant sequence (x) n=1. Finally, it can be verified that (Y, f) is a completion of X. Here is the idea of the proof of uniqueness. Let (Y i, f i ), i = 1, 2, be two completions. Let E = f 1 (x) Y. Define h : E Y 2 by h(f 1 (x)) = f 2 (x). Since f 1 and f 2 are isometric maps, h is uniformly continuous. By an argument very similar to that in Problem 21b there is a continuous map g : Y 1 Y 2 such that g E = h. But then for x X we have g f 1 (x) = g(f 1 (x)) = h(f 1 (x)) = f 2 (x), so g f 1 = f 2.

15 14 JACK SPIELBERG 8. Compactness Compactness is probably the most important concept in analysis. It can be described in various ways. The right way is not necessarily the easiest to understand. Before we give the definition, here is some motivation for why it is reasonable. The basic problem that compactness addresses is the transition from local information to global information. That may sound cryptic, and it is meant to be a catchy phrase that will become more intelligible as you get more used to these ideas. But it isn t hard to see what it is about. Local (near a point) means in some open ball centered at that point. Here is a simple example of using this terminology. If a function is continuous at a point, then it is bounded in some open ball centered at that point. Thus if a function is continuous on a set, it is bounded locally on that set: each point in the set has a neighborhood on which the function is bounded. On the other hand, global (on a set) means on the whole set. A function is globally bounded if it is bounded on its domain, i.e. if it is a bounded function. Is every continuous function bounded? Of course not! For example, a non-constant polynomial on R is continuous, but not bounded. Local boundedness does not generally imply global boundedness. However if the domain of the polynomial is taken to be a closed bounded interval, then the extreme value theorem from calculus implies that the polynomial is bounded on the interval. The great insight was that it is a property of the domain that lets us pass from local boundedness to global boundedness, and this property is called compactness. Now, recall what the word local means: in a neighborhood of a point. A property holds locally on a set if for each point, there is an open ball centered at the point such that the property holds in that ball. If the set is infinite, this will give an infinite collection of open balls, one for each point. We could obtain the property globally if we had a finite collection of balls instead of an infinite collection. Compactness of the set means that we can always reduce to a finite collection. You might notice that a lot of mathematics seems to proceed in this way: what would we like to have? Let s give a name to the situation where we have what we want. Now let s analyze the situation to see what exactly we were asking for. In fact, compactness can be described in a variety of ways that seem very different. That means that we can prove that a space is compact using an easy description. Then we can use compactness via a complicated description. OK, with that as motivation, here is the precise definition. Definition 8.1. Let X be a set. A cover of X is a collection of sets whose union contains X. If U is a cover of X, a subcover of U is a subcollection of U that is also a cover of X. Example 8.2. (1) The set of all open intervals is a cover of R. (2) { (a, b) : a < b, a, b Z } is a subcover of example (1). Definition 8.3. Let X be a metric space, and let E X. An open cover of E is a cover of E whose elements are open subsets of X. Definition 8.4. Let X be a metric space, and let E X. E is compact if every open cover of E has a finite subcover. Example 8.5. (1) Example 8.2(1) is an open cover of R having a finite subcover. (2) Example 8.2(2) is an open cover of R not having a finite subcover. In particular, it follows that R is not compact. Example 8.6. (1) Finite sets are compact. (The proof is an exercise.) (2) {0, 1, 1/2, 1/3,...} is a compact subset of R. (The proof is an exercise.) (3) [0, 1] is a compact subset of R (this is a special case of Corollary 8.30). Proof. Let U be an open cover of [0,1]. Let E = { x [0, 1] : [0, x] is finitely covered by U }. Note that 0 E, so E. Let c = sup E. Then c [0, 1]. We first claim that c E. To see this, choose U 0 U with c U 0. Then there exists r > 0 such that (c r, c + r) U 0. By the definition of supremum, there is y E with y > c r. By definition of E there is a finite subcollection V U with [0, y] V. But then V {U 0 } is a finite subcollection of U covering [0, c], proving that c E. Now we note that, in fact, V {U 0 } covers [0, a] for any number a between c and c + r. Thus if c < 1 we could find a larger element of E than c, contradicting its status as supremum. So we have shown that c = 1. Thus [0, 1] is finitely covered by U.