MAT 570 REAL ANALYSIS LECTURE NOTES PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 Contents. Sets 2 2. Functions 5 3. Countability 7 4. Axiom of choice 8 5. Equivalence relations 9 6. Real numbers 9 7. Extended real numbers 8. Metric spaces 2 9. Separability 6 0. Sequences of real numbers 7. Continuity 2 2. Completeness 24 3. Compactness 27 4. The Cantor set 3 5. Connectedness 34 6. Function spaces 36 7. Measures 4 8. Outer measure 46 9. Lebesgue-Stieltjes measures 5 Date: August 9, 204.
2 PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 20. Measurable functions 57 2. Integrating nonnegative functions 6 22. Integrating extended real-valued functions 65 23. Convergence in measure 74 24. Product measures 75 25. Iterated integrals 76 26. Complete products 80 27. Signed measures 82 28. Differentiation 89 29. Extension to complex values 98 We start with a few sections (up through Section 7, roughly) of material that should be review, whose purpose is mainly to establish notation.. Sets A set is a collection of objects, called the elements or members of the set. If A is a set and x is an object, we write x A to mean x is an element of A, and x / A to mean x is not an element of A. The empty set, denoted, is the set with no elements, so that the statement x is false for every object x. A is a subset of B, written A B, if every element of A is an element of B, and A B means A is not a subset of B. A equals B, written A = B, if A and B have the same elements, equivalently A B and B A. A is a proper subset of B, written A B, if A B and A B. {x, x 2,..., x n } denotes the set whose elements are x, x 2,..., x n, and {x} is a singleton. {x : P (x)} denotes the set whose elements are the objects x for which the statement P (x) is true. Slight variation: {x A : P (x)} denotes the set of elements x of the set A satisfying the statement P (x). Officially, set should be an undefined term, but this level of informality is appropriate for our purposes.
MAT 570 REAL ANALYSIS LECTURE NOTES 3 natural numbers: N = {, 2, 3,... } integers: Z = {0, ±, ±2,... } { n } rational numbers: Q = k : n, k Z, k 0 real numbers: R complex numbers: C = {x + iy : x, y R}. bounded intervals. (a, b) = {x R : a < x < b} [a, b) = {x R : a x < b} (a, b] = {x R : a < x b} [a, b] = {x R : a x b} (open) (left half-closed, or right half-open) (left half-open, or right half-closed) (closed). unbounded intervals. (a, ) = {x R : x > a} [a, ) = {x R : x a} (, b) = {x R : x < b} (, b] = {x R : x b} (, ) = R (open) (closed) (open) (closed) (open). The power set of X is P(X) = {A : A X}. If X is understood and A X, the complement of A is A c = {x X : x / A}. The intersection of A and B is A B = {x : x A and x B}, the union of A and B is A B = {x : x A or x B}, the difference of A and B, or the relative complement of B in A, is A \ B = {x : x A and x / B}, and the symmetric difference of A and B is A B = (A \ B) (B \ A). Associative Laws. Distributive Laws. De Morgan s Laws. : A (B C) = (A B) C A (B C) = (A B) C. A (B C) = (A B) (A C) A (B C) = (A B) (A C). (A B) c = A c B c (A B) c = A c B c.
4 PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 Families of sets. If C P(X) then the intersection of C is C = A = {x : x A for all A C}, A C and the union of C is C = = {x : x A for some A C}. A C Distributive Laws. A B = B) B C B C(A A B = B). B C B C(A De Morgan s Laws. ( ) c A = A C A C A c ( ) c A = A c. A C A C Indexed families of sets. If I is a set and A i is a set for each i I then the intersection of the indexed family of sets {A i } i I is A i = {x : x A i for all i I}, i I and the union of {A i } i I is A i = {x : x A i for some i I}, i I If I = N we sometimes write intersections as A i or A A 2, and if I = {,..., n} we sometimes write n A i or A A n, and similarly for unions. i= i=
MAT 570 REAL ANALYSIS LECTURE NOTES 5 Distributive Laws. A B i = B i ) i I i I(A A B i = B i ) i I i I(A De Morgan s Laws. ( A c i A i)c = i I i I ( A i)c = A c i. i I i I A and B are disjoint if A B =, otherwise A intersects B. A family C of sets is pairwise disjoint if A B = for all A, B C with A B. An indexed family {A i } i I of sets is pairwise disjoint if A i A j = for all i, j I with i j. A partition of A is a pairwise disjoint family of nonempty subsets of A whose union is A. 2. Functions A function from A to B is a rule f that associated to every element x of A a unique element f(x) of B, called the value of f at x or the image of x under f. 2 We write f : A B to mean that f is a function from A to B. The domain of f is A, and the range of f is {f(x) : x A}. f is one-to-one, or -, or injective, if for all x, y A, x y implies f(x) f(y). f is onto, or surjective, if the range of f is all of B. f is bijective, or - onto, or a - correspondence, if it is both injective and surjective, in which case the inverse of f is the function f : B A defined by f (y) = x if and only if f(x) = y. If f : A B and g : B C, the composition of f and g is the function g f : A C defined by (g f)(x) = g(f(x)). Proposition 2.. Let f : A B and g : B C. () If f and g are -, then g f is -. (2) If f and g are onto, then g f is onto. (3) If g f is -, then f is -. (4) If g f is onto, then g is onto. 2 Again, officially functions should be defined using set theory, but this informality is ok for us here.
6 PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 If f : A B and C A, the restriction of f to C is the function f C : C B defined by f C (x) = f(x) for x C. We sometimes write x f(x) for a function f. A sequence is a function whose domain is N. If x is a sequence, we usually write x n for the value x(n), and call it the nth term of x. Moreover, we usually write the sequence as {x n } = {x n } n N = {x n } n=. A sequence in X is a function x : N X. We sometimes refer to a function x with domain {,..., n} as a finite sequence, and write it as {x i } n i=. If {n < n 2 < } in N, then {x nk } k= is a subsequence of {x n}, and coincides with the composition k n k x nk. Let f : A B. If C A, the image of C under f is If D B, the inverse image of D under f is Proposition 2.2. Let f : A B. f(c) = {f(x) : x C}. f (D) = {x A : f(x) D}. () If {C i } i I is a family of subsets of A, then ( ) f C i = f(c i ) i I i I ( ) f C i f(c i ). i I (2) If {D j } j J is a family of subsets of B, then ( ) f D j = f (D j ) j J j J ( ) f D j = f (D j ). j J j J i I (3) If C A then (4) If D B then and moreover f (f(c)) C, with equality if f is -. f(f (D)) D, with equality if f is onto, f (D c ) = f (D) c.
MAT 570 REAL ANALYSIS LECTURE NOTES 7 3. Countability If f : A B, we say f(a) is an image of A, and a - image of A if f is -. Sets A and B have the same cardinality, or are equivalent, written A B or A = B, if B is a - image of A. Example 3.. The power set P(X) of any set X is equivalent to the set {0, } X of functions from X to {0, }, via the - onto function A χ A : P(X) {0, } A, where χ A : X {0, } is the characteristic function of A, defined by { if x A χ A (x) = 0 if x / A. Theorem 3.2. Equivalence if sets is: () reflexive: A A; (2) symmetric: A B implies B A; (3) transitive: A B and B C imply A C. A is finite if A = or A {,..., n} for some n N, otherwise A is infinite. A is countably infinite if A N. A is countable if it is finite or countably infinite, otherwise it is uncountable. Theorem 3.3. Every subset of a finite set is finite. Theorem 3.4. Every image of a finite set is finite. Theorem 3.5. Every finite union of finite sets is finite. Theorem 3.6. Every finite product of finite sets is finite. Theorem 3.7. No finite set is equivalent to a proper subset of itself. Theorem 3.8. Every subset of a countable set is countable. Theorem 3.9. Every image of a countable set is countable. Theorem 3.0. N 2, Z, and Q are countable. Theorem 3.. Every countable union of countable sets is countable. Theorem 3.2. Every finite product of countable sets is countable. Theorem 3.3. Every infinite set contains a countably infinite subset. Theorem 3.4. Every infinite set is equivalent to a proper subset of itself. Theorem 3.5. {0, } N is uncountable. Theorem 3.6. R, and more generally any interval of nonzero length, is uncountable.
8 PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 4. Axiom of choice Axiom of Choice. If {A i } i I is a family of nonempty sets, then there exists a function f : I i I A i such that f(i) A i for all i I. The Cartesian product of a family {A i } i I of sets is A i x(i) Ai for all i I}. i I A i = {x : I i I We call x i I A I an I-tuple, we write x i = x(i) and call this the ith coordinate of x, and we frequently denote x by (x i ) i I or just (x i ). If I = {,..., n} we write n i= A i = A A n and (x,..., x n ), and call this latter an n-tuple. Elements of A B are called ordered pairs 3 If all the A i s are equal to some fixed set A, we write A I and A n for the products. Example 4.. The set of all sequences in a set A is A N. Thus, the Axiom of Choice says that if I and every A i are nonempty then i I A i is nonempty too. A relation on a set A is a subset of A 2. A relation R on A is: reflexive if (x, x) R for all x A; antisymmetric if (x, y) R and (x, y) R imply x = y, for all x, y A; transitive if (x, y) R and (y, z) R imply (x, z) R, for all x, y, z A; a partial order if is reflexive, antisymmetric, and transitive. A is a partially ordered set if it is equipped with a partial order. We usually use a symbol like for the partial order, and we say x is less than or equal to y, written x y, to mean (x, y) is an element of the partial order. Let A be a partially ordered set. A is totally ordered if for all x, y A we have either x y or y x. A chain in A is a subset that is totally ordered in the induced relation. An upper bound for B A is an element x A such that y x for all y B. An element x A is maximal if for all y A we have x y implies x = y. Zorn s Lemma. If A is a nonempty partially ordered set in which every chain has an upper bound, then A has a maximal element. It is an important theorem of set theory that the Axiom of Choice and Zorn s Lemma are equivalent. 3 fine print: this is not to be taken as an official definition of ordered pair ordered pairs are actually more basic, and are used in the official definition of function (which we did not record here)!
MAT 570 REAL ANALYSIS LECTURE NOTES 9 5. Equivalence relations A relation R on a set A is symmetric if for all x, y A, (x, y) R implies (y, x) R. A relation on a set A is an equivalence relation if it is reflexive, symmetric, and transitive. We usually use a symbol like for the equivalence relation, and say x is equivalent to y, written x y, to mean that (x, y) is an element of the equivalence relation. The equivalence class of x is [x] = {y A : x y}. Theorem 5.. The set A/ defined by {[x] : x A }, i.e., the set of equivalence classes, is a partition of A. Conversely, if P is a partition of A then the relation on A defined by x y if and only if there exists C P such that x, y C is an equivalence relation on A such that A/ = P. In this way, there is a - correspondence between the set of equivalence relations on A and the set of partitions of A. 6. Real numbers Field Axioms. For all x, y, z R, () x + (y + z) = (x + y) + z and x(yz) = (xy)z (associative laws). (2) x + y = y + x and xy = yx (commutative laws). (3) 0 + x = x and x = x, and 0 (additive and multiplicative identities). (4) x + ( x) = 0 and, if x 0, xx = (additive and multiplicative inverses). (5) x(y + z) = xy + xz (distributive law). x y is defined as x + ( y), and, if y 0, x y is defined as xy. Order Axioms. For all x, y, z R, () Exactly one of x < y, y < x, or x = y is true. (2) x < y and y < z imply x < z. (3) x < y implies x + z < y + z. (4) x < y and 0 < z imply xz < yz. x y is defined as x < y or x = y, x > y is defined as y < x, and x y is defined as y x.
0 PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 Completeness Axiom, or the Least Upper Bound Property. Every nonempty subset of R that is bounded above has a least upper bound. Supporting definitions (more than are needed for the statement above): Definition 6.. If A R and x R, then () x is an upper bound of A if y x for all y A; (2) x is a lower bound of A if x y for all y A; (3) A is bounded above if it has an upper bound, otherwise it is unbounded above; (4) A is bounded below if it has a lower bound, otherwise it is unbounded below; (5) A is bounded if it is bounded above and bounded below; (6) x is a maximum, or greatest element, of A if x A and y x for all y A; (7) x is a minimum, or least element, of A if x A and x y for all y A; (8) a least upper bound, or supremum, of A is a least element of the set of all upper bounds of A; (9) a greatest lower bound, or infimum, of A is a greatest element of the set of all lower bounds of A. It s convenient to extend some of the above terminology to functions: Definition 6.2. A function f : A R is bounded above, bounded below, or bounded if its range f(a) has the same property. Similarly, the terms max f, min f, sup f, and inf f mean the corresponding quantity associated with f(a). Lemma 6.3. For all A R, () If A has a maximum, it is unique, and is written max A. (2) If A has a minimum, it is unique, and is written min A. (3) If A has a supremum, it is unique, and is written sup A or lub A. (4) If A has an infimum, it is unique, and is written inf A or glb A. Proposition 6.4. Every nonempty subset of R that is bounded below has a greatest lower bound. Consequences of the Completeness Axiom. Proposition 6.5 (Archimedean Principle). For all x R there exists n N such that n > x. Proposition 6.6 (Density of the Rationals). For all a, b R, if a < b then there exists x Q such that a < x < b. Proposition 6.7 (Density of the Irrationals). For all a, b R, if a < b then there exists x R \ Q such that a < x < b. Proposition 6.8. For all x R there exists a unique n Z such that n x < n +. Remark 6.9. The Well-Ordering Principle, which states that every nonempty subset of N has a smallest element, might seem like it belongs with the above results, but it is actually a much more basic fact, equivalent to the Principle of Mathematical Induction, which is part of the Peano Axioms for the natural numbers.
MAT 570 REAL ANALYSIS LECTURE NOTES Distance. The absolute value of x R is { x if x 0 x = x if x < 0. The distance between x, y R is d(x, y) = x y. Proposition 6.0. For all x, y R, () x 0, and x = 0 if and only if x = 0. (2) xy = x y. (3) x + y x + y ( triangle inequality). (4) x y x y. (5) d(x, y) 0, and d(x, y) = 0 if and only if x = y. (6) d(x, y) = d(y, x). (7) d(x, y) d(x, z) + d(z, y) ( triangle inequality). Definition 6.. Let x R. () The positive part of x is (2) The negative part of x is x + = x = { x if x 0 0 if x < 0. { x if x 0 0 if x > 0. Proposition 6.2. For all x, y R, () x +, x are the unique nonnegative numbers a, b such that x = a b and ab = 0. (2) x = x + + x. 7. Extended real numbers The set of extended real numbers is R = R {, }, where and are two fixed objects not in R. We (partially) extend arithmetic to R as follows: for x R, () x + = + x =. (2) x = + x =. (3) x( ) = ( )x = and x( ) = ( )x = if x > 0. (4) x( ) = ( )x = and x( ) = ( )x = if x < 0. (5) + = and =. (6) ( ) = ( )( ) = and ( ) = ( )( ) =. (7) 0 = 0 = 0 (this last one requires come care, but is convenient in measure and integration theory).
2 PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 And we extend inequalities to R as follows: for all x R, < x <. Proposition 7.. Every subset of R has a sup and an inf. Moreover, () sup = and inf =. (2) A R is unbounded above (in R) if and only if sup A =. (3) A R is unbounded below (in R) if and only if inf A =. 8. Metric spaces Definition 8.. A metric on a set X is a function d : X 2 R such that for all x, y, z X, () d(x, y) 0, and d(x, y) = 0 if and only if x = y; (2) d(x, y) = d(y, x); (3) d(x, y) d(x, z) + d(z, y) (triangle inequality). A metric space is a set equipped with a metric. Example 8.2. R with d(x, y) = x y. Example 8.3. Any set X becomes a metric space with the discrete metric { if x y d(x, y) = 0 if x = y. Definition 8.4. A norm on a vector space X (with scalar field R) is a function x x : X R such that for all x, y X, () x 0, and x = 0 if and only if x = 0; (2) cx = c x for all c R; (3) x + y x + y (triangle inequality). Example 8.5. R n with the Euclidean norm x = n x 2 i. Proposition 8.6. Every normed space becomes a metric space with d(x, y) = x y. Definition 8.7. Every subset A of a metric space X becomes a metric space, called a subspace of X, with metric d A 2. Definition 8.8. Let X be a metric space, let x X, and let r > 0. () The open ball of radius r centered at x is B r (x) = {y X : d(x, y) < r}. (2) The closed ball of radius r centered at x is B r (x) = {y X : d(x, y) r}. Definition 8.9. Let X be a metric space and A X. A is open if for all x A there exists r > 0 such that B r (x) A, and A is closed if A c is open. Proposition 8.0. Let X be a metric space. i=
MAT 570 REAL ANALYSIS LECTURE NOTES 3 () and X are both open and closed. (2) Open balls are open, and closed balls are closed. In particular, singletons are closed. (3) Every union of open sets is open, and every intersection of closed sets is closed. (4) Every finite intersection of open sets is open, and every finite union of closed sets is closed. Proof. () Obvious. (2) Let x X and r > 0. To see that B r (x) is open, let y B r (x). Put s = r d(x, y). Since y B r (x), we have d(x, y) < r, so s > 0. For all z B s (y), d(x, z) d(x, y) + d(y, z) < d(x, y) + r d(x, y) = r, so z B r (x). Thus B s (y) B r (x), and we have shown that B r (x) is open. Now we show that the closed ball B r (x) is closed. By definition, we have to show that the complement B r (x) c is open. Let y B r (x) c, and put s = d(x, y) r. Since y / B r (x), we have d(x, y) > r, so s > 0. For all z B s (y), d(x, z) d(x, y) d(y, z) > d(x, y) (d(x, y) r) = r, so z B r (x) c. Thus B s (y) B r (x) c, and we have shown that B r (x) c is open, and hence B r (x) is closed. (3) Let {U i } i I be a family of open subsets of X, and let x i I U i. Then we can choose j I such that x U j. Since U j is open, we can choose r > 0 such that B r (x) U j. Since U j i I U i, we have B r (x) i I U i, and we have shown that i I U i is open. The statement about intersections of closed sets now follows from De Morgan s Laws. (4) Let U,..., U n be open, and let x n i= U i. For each i =,..., n, since x U i and U i is open, we can choose r i > 0 such that B ri (x) U i. Put r = min{r,..., r n }, which exists in (0, ) since {r,..., r n } (0, ) is finite. Then for every i =,..., n we have so and we have shown that n i= U i is open. B r (x) B ri (x) U i, B r (x) n U i, As above, the statement about finite unions of closed sets now follows from De Morgan s Laws. Definition 8.. The closure of A X is i= A = {B : B is closed and B A}. Proposition 8.2. A is the smallest closed set containing A, and in particular A is closed if and only if A = A.
4 PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 Proof. A is closed and contains A, being an intersection of closed sets containing A. Moreover, if B A is closed, then B A by construction. Thus, A is the smallest closed set containing A, and now the other statement follows immediately. Proposition 8.3. x A if and only if for all ε > 0 there exists y A such that d(x, y) < ε. Proof. First assume that x A. Arguing by contradiction, suppose that there exists ε > 0 such that B ε (x) A =. Then B ε (x) c is a closed set containing A, and so contains A, and therefore x / A, which is a contradiction. Conversely, assume the statement regarding ε. Arguing by contradiction, suppose x / A. Since A is closed, there exists ε > 0 such that B r (x) A =, which is a contradiction. Definition 8.4. x is a limit point of A if for all ε > 0 there exists y A such that 0 < d(x, y) < ε. Corollary 8.5. A = A {all limit points of A}, and in particular A is closed if and only if it contains all its limit points. Definition 8.6. A is dense in X if A = X. Example 8.7. Both Q and R \ Q are dense in R. Lemma 8.8. A is dense in X if and only if every nonempty open set intersects X. Convergence. Definition 8.9. A sequence {x n } in a metric space X converges to x, written x n x if for all ε > 0 there exists k N such that d(x n, x) < ε for all n k. Lemma 8.20. If x n x and x n y, then x = y. Proof. Arguing by contradiction, suppose {x n } converges to distinct points x, y. Then d(x, y) > 0, so because x n x there exists k N such that d(x n, x) < d(x, y)/2 for all n k, and similarly there exists j N such that d(x n, y) < d(x, y)/2 for all n j. Let m = max{k, j}. Then which is a contradiction. d(x, y) d(x, x m ) + d(x m, y) < d(x, y) 2 + d(x, y) 2 = d(x, y), Definition 8.2. If x n x then x is the limit of {x n }, written x = lim x n = lim n x n. Definition 8.22. A sequence converges, or is convergent, if it has a limit, otherwise it diverges, or is divergent. Proposition 8.23. {x n } has a subsequence converging to x if and only if for all ε > 0 and k N there exists n k such that d(x n, x) < ε.
MAT 570 REAL ANALYSIS LECTURE NOTES 5 Proof. First assume that {x ni } is a subsequence of {x n } converging to x, and let ε > 0 and k N. Choose i k such that d(x ni, x) < ε, and note that n i i k. Conversely, assume that statement regarding ε. Choose n N such that d(x n, x) <. Then choose n 2 n + such that d(x n2, x) < /2. Continuing inductively, for every k = 2, 3,... choose n k n k + such that d(x k, x) < /k. Then so {x nk } is a subsequence converging to x. n < n 2 < and d(x nk, x) k 0, Proposition 8.24. If x n x then every subsequence converges to x. Proof. Assume that x n x and {x nk } is a subsequence of {x n }. Let ε > 0. Choose j N such that d(x n, x) < ε for all n j. Then for all k j we have n k k j, so d(x nk, x) < ε, and we have shown that lim k x nk = x. Proposition 8.25. If x n x then there exists ε > 0 and a subsequence {y k } of {x n } such that d(y k, x) ε for all k N. Proof. Assume that x n x, and choose ε > 0 that violates the condition in the definition of x n x. Then we can choose n such that d(x n, x) ε. Then we can choose n 2 > n such that d(x n2, x) ε. Continuing inductively, for every k > we can choose n k > n k such that d(x nk, x) ε. Then n < n 2 <, so {x nk } is a subsequence with the desired property. Proposition 8.26. Let X be a metric space, A X, and x X. Then x A if and only if there is a sequence {x n } in A such that x n x. Thus, A is closed if and only if whenever {x n } is a sequence in A that converges in X we have lim x n A. Proof. First assume that x A. For every n N we can choose x n A such that d(x n, x) < /n. Since d(x n, x) 0, we have x n x. Conversely, assume that {x n } is a sequence in A such that x n x, and let ε > 0. Then we can choose k N such that d(x k, x) < ε. Since x k A, we have shown that x A. Proposition 8.27. x n x in X if and only if d(x n, x) 0 in R. Product spaces. Lemma 8.28. If (X i, d i ) are metric spaces for i =,..., n, then there is a metric d on n i= X i given by d(x, y) = max {d i(x i, y i )}. i=,...,n Moreover, a sequence {x k } in n i= X i converges to x if and only if for each i =,..., n the coordinate sequence {x k (i)} k= converges to x(i). In the above lemma, we use the alternative notation x(i) for the ith coordinate of an n-tuple x, to avoid confusion with the indexing of the sequences. Definition 8.29. With the above notation, d is the product metric on n i= X i, and ( n i= X i, d) is the product space.
6 PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 9. Separability Throughout, X is a metric space. Definition 9.. X is separable if it has a countable dense set. Example 9.2. R is separable. Definition 9.3. A base for X is a family B of open subsets such that every open subset of X is a union of members of B. Example 9.4. The family of all open balls in X is a base for X. Definition 9.5. X is second countable if it has a countable base. Example 9.6. The family of all open intervals with rational endpoints is a countable base for R. Theorem 9.7. X is separable if and only if it is second countable. Proof. First assume that X is separable, and let D be a countable dense subset. Let V X be open and x V. Since V is open we can choose r > 0 such that B r (x) V. Then choose n N such that /n < r/2. Then since D is dense we can choose t D such that d(t, x) < /n. Then for all y B /n (t) we have so d(y, x) d(y, t) + d(t, x) < n + n < r, x B /n (t) B r (x) V. Thus {B /n (t) : t D, n N} is a countable base for X. Conversely, assume that X is second countable, and let B be a countable base for X. Without loss of generality each set U B is nonempty, so we can choose x U U. Then D := {x U : U B} is countable, and we will show that it is dense in X. It suffices to show that if V X is nonempty and open then D V. There exists U B such that U V, and then we have x U D V. Proposition 9.8. Every subset of a separable metric space is separable. Proof. Let X be a separable metric space and Y X. Since X is separable, it has a countable base B. It is an exercise to show that the countable family {U Y : U B} is a base for Y. Thus Y is second countable, and therefore separable. Definition 9.9. If A X and C P(X), then C is a cover of A, or covers A, if A C. If C is a cover of A, a subcover of C is a subfamily of C that also covers A. A cover of A is open if its members are all open sets. Theorem 9.0 (Lindelöf s Theorem). If X is separable, then every open cover of X has a countable subcover.
MAT 570 REAL ANALYSIS LECTURE NOTES 7 Proof. Let X be separable, and let C be an open cover of X. Since X is separable, we can choose a countable base {V n } n N for X. For each x X we can choose U x C and n x N such that x V nx U x. Then S := {n x : x X} is countable. For each n S we can choose U n C such that V n U n. Then X = V nx = V n U n, x X n S n S so {U n } n S is a countable subcover of C. 0. Sequences of real numbers Definition 0.. A sequence {x n } in R diverges to, written x n or lim x n =, if for all M R there exists k N such that x n > M for all n k. Similarly for x n. If {x n } converges in R, or diverges to or, we sometimes say it converges in R. Proposition 0.2. If x n x and y n y, then: () x n ± y n x ± y. (2) x n y n xy. (3) x n /y n x/y if y 0. With careful interpretation, the above proposition holds even if x or y is infinite, as long as the forbidden forms and 0 are avoided. Remark 0.3. If {x n } is a sequence in R and x R, then x n x in R if and only if both of the following hold: For all a > x there exists k N such that x n < a for all n k. For all a < x there exists k N such that x n > a for all n k. Similarly, there is a subsequence {y k } of {x n } with y k x in R if and only if both: For all a > x and k N there exists n k such that x n < a. For all a < x and k N there exists n k such that x n > a. Definition 0.4. A sequence {x n } in R is () increasing, or nondecreasing, if x x 2. (2) decreasing, or nonincreasing, if x x 2. (3) strictly increasing if x < x 2 <. (4) strictly decreasing if x > x 2 >. (5) monotone if it is increasing or decreasing. (6) strictly monotone if it is strictly increasing or strictly decreasing.
8 PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 Theorem 0.5. Every monotone sequence in R converges in R. More precisely, if {x n } is increasing then lim x n = sup x n = sup x n = sup{x n : n N}, n and if {x n } is decreasing then lim x n = inf x n = inf n x n = inf{x n : n N}. Proof. Suppose {x n } is increasing, and let x = sup x n. To show that x n x in R, it suffices to verify the following two statements: () for all a < x there exists k N such that a < x n for all n k; (2) for all b > x there exists k N such that x n < b for all n k. For (), since x is the least upper bound of {x n : n N}, we can choose k N such that a < x k, and then, since {x n } is increasing, for all n k we have a < x k x n. For (2), just note that for all n N we have x n x < b. On the other hand, if {x n } is decreasing, just multiply everything by and use what we have already proved. Definition 0.6. Let {x n } be a sequence in R. () The lim sup of {x n } is (2) The lim inf of {x n } is lim sup x n = lim sup x n = lim x n := inf sup x n. n k n k lim inf x n = lim inf n x n = lim x n := sup k inf x n. n k Lemma 0.7. If {x n } is a sequence in R, then the sequence {sup n k x n } k= so ( ) lim sup x n = lim sup x n. k n k Similarly, the sequence {inf n k x n } k= is increasing, so ( ) lim inf x n = lim inf x n. k n k is decreasing, Proposition 0.8. Let {x n } be a sequence in R and let x R. Then lim sup x n = x if and only if: () for all a > x there exists k N such that x n < a for all n k, and (2) for all a < x and all k N there exists n k such that x n > a.
Similarly for lim inf x n. MAT 570 REAL ANALYSIS LECTURE NOTES 9 Proof. First assume that lim sup x n = x. To show (), let a > x. For each k N let s k = sup x n. n k Since x is the greatest lower bound of {s k : k N}, we can choose k N such that s k < a, and then for all n k we have x n s k < a. On the other hand, to show (2), let a < x and k N. Since x = inf k s k, we have a < s k. Since s k is the least upper bound of {x n : n k}, we can choose n k such that a < x n. Conversely, assume () (2). With the above notation, we must show that x is a lower bound of {s k : k N}, and there is no greater lower bound. For the first, we argue by contradiction: suppose that k N and s k < x. By (2) we can choose n k such that x n > s k. But this contradicts s k = sup n k x n. Finally, we must show that if a > x then a is not a lower bound of {s k : k N}. Choose b such that a > b > x. By () we can choose k N such that x n < b for all n k, and hence s k = sup x n b < a. n k This proves the statement concerning lim sup. For lim inf, just multiply everything by and use what we have already proved. Proposition 0.9. For all sequences {x n }, {y n } in R, if x n y n for all n N then lim sup x n lim sup y n and lim inf x n lim inf y n. Proof. Since x n y n for all n, for every k we have and so lim sup x n = inf k sup n k x n sup y n, n k sup x n inf n k k sup y n = lim sup y n. n k The statement regarding lim inf follows upon multiplying by. Proposition 0.0. For any sequence {x n } in R, () lim sup x n = if and only if {x n } is unbounded above. (2) lim inf x n = if and only if {x n } is unbounded below. (3) lim sup x n = if and only if x n. (4) lim inf x n = if and only if x n.
20 PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 Proof. () Since a > is impossible, by Proposition 0.8 we have lim sup x n = if and only if for all a R and all k N there exists n k such that x n > a, i.e., if and only if every tail {x n : n k} is unbounded above. But this is equivalent to the sequence {x n } itself being unbounded above. (2) Just take negatives and use (). (3) Since x < is impossible, we have lim sup x n = if and only if for all a R there exists k N such that x n < a for all n k, i.e., x n. (4) Just take negatives and use (3). Definition 0.. Let {x n } be a sequence in R, and let x R. We say x is a subsequential limit of {x n } if some subsequence of {x n } converges (in R) to x. Proposition 0.2. Let {x n } be a sequence in R. () lim sup x n is the largest subsequential limit of {x n }. (2) lim inf x n is the smallest subsequential limit of {x n }. (3) {x n } converges in R if and only if lim inf x n = lim sup x n. Proof. () Let x = lim sup x n. First we prove that x is a subsequential limit. It suffices to show that for all a < x and k N there exists n k such that x n > a, and for all a < x and k N there exists n k such that x n < a, but these follow immediately from Proposition 0.8. Now we show that there is no larger subsequential limit. Let b > x, and chose a such that b > a > x. It follows from Proposition 0.8 that b is not a subsequential limit. (2) Just take negatives and use (). (3) If x n x in R, then x is the only subsequential limit of {x n }, and so lim inf x n = lim sup x n. Conversely, if lim inf x n = lim sup x n, then, letting x be this common value, then, by Proposition 0.8, for all a > x there exists k N such that x n < a for all n k, and for all a < x there exists k N such that x n > a for all n k; these two properties imply that x n x. Corollary 0.3. Every closed nonempty subset of R has a max if it is bounded above, and a min if it is bounded below. Series. Given a sequence {a n } in R, the series with nth term a n, written n= a n or a n, is the sequence {s n }, where n s n = is the nth partial sum of the series a n. If the series converges (that is, if the sequence {s n } of partial sums converges), its limit is the sum of the series: a n = lim s n, n= i= a i
MAT 570 REAL ANALYSIS LECTURE NOTES 2 and we also write this equality if the series diverges to or. If a n 0 for all n, then {s n } is increasing, so a n either converges or diverges to, and in the first case we write n= a n <. A series n= a n converges absolutely if n= a n <, in which case it converges, and a n a n, n= n= and moreover every rearrangement has the same sum, i.e., a n = n= for any bijection k n k : N N. an converges conditionally if it converges but not absolutely, in which case it has rearrangements that diverge, and moreover every real number is the sum of some rearrangement. If I is a nonempty set and a i [0, ] for all i I, we define { } a i = sup a i : F I is finite. i I i F Let P = {i I : a i > 0}. If P is uncountable, then i a i =. If P is countably infinite, and n i n : N P is any bijection, then a i = a in. i I Throughout, X, Y, Z are metric spaces. k= n= a nk. Continuity Definition.. If f : X Y and x X, then f is continuous at x if for all ε > 0 there exists δ > 0 such that for all y X, d(x, y) < δ implies d(f(x), f(y)) < ε. Proposition.2. f : X Y is continuous at x X if and only if x n x in X implies f(x n ) f(x). Proof. First assume that f is continuous at x, and let x n x in X. Let ε > 0. Choose δ > 0 such that d(x, y) < δ implies d(f(x), f(y)) < ε. Then choose k N such that d(x n, x) < δ for all n k. Then d(f(x n ), f(x)) < ε for all n k, and we have shown that f(x n ) f(x). Conversely, suppose that f is discontinuous at x, and chose ε > 0 that contradicts the definition of continuity. For each n N we can choose x n X such that d(x n, x) < /n and d(f(x n ), f(x)) ε. Then x n x and f(x n ) f(x). Proposition.3. If f, g : X R are continuous at x, then so are: () f + g;
22 PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 (2) cf for every c R; (3) fg; (4) f g if g is never 0. Proof. Let x n x in X. Then and similarly for cf, fg, and f/g. (f + g)(x n ) = f(x n ) + g(x n ) f(x) + g(x) = (f + g)(x), Proposition.4. If f : X Y, g : Y Z, x X, f is continuous at x, and g is continuous at f(x), then g f is continuous at x. Proof. If x n x in X, then f(x n ) f(x) in Y, and so (g f)(x n ) = g(f(x n )) g(f(x)) = (g f)(x). Definition.5. If f : X Y is continuous if it is continuous at every point of x. Example.6. f = (f,..., f n ) : X (Y,..., Y n ) is continuous if and only if the coordinate functions f i : X Y i are continuous for i =,..., n. Theorem.7. f : X Y is continuous if and only if f (U) is open for every open U Y. Proof. First assume that f is continuous, and let U Y be open. To show that f (U) is open, let x f (U). Then f(x) U, and since U is open we can choose ε > 0 such that B ε (f(x)) U. By continuity, we can choose δ > 0 such that d(x, y) < δ implies d(f(x), f(y)) < ε, i.e., y B δ (x) implies f(y) B ε (f(x)). Then B δ (x) f (U). Conversely, assume the condition regarding open sets. Let x X and ε > 0. Then B ε (f(x)) is open in Y, so f (B ε (f(x))) is open in X. We have x f (B ε (f(x))), so we can choose δ > 0 such that B δ (x) f (B ε (f(x))). Thus d(x, y) < δ implies f(y) B ε (f(x)), i.e., d(f(x), f(y)) < ε. Corollary.8. f : X Y is continuous if and only if f (A) is closed for every closed A Y. Proof. Take complements and apply the properties of the inverse-image set function B f (B) : P(Y ) P(X). Definition.9. Let A R, f : A Y, and x A. Then f is right continuous at x if for all ε > 0 there exists δ > 0 such that for all y A, x < y < x + δ implies d(f(x), f(y)) < ε, and similarly for left continuous. Remark.0. If A R and x A, then f : A Y is continuous at x if and only if it is both right and left continuous at x. Definition.. If A R, then f : A R is () increasing, or nondecreasing, if x < y implies f(x) f(y).
MAT 570 REAL ANALYSIS LECTURE NOTES 23 (2) decreasing, or nonincreasing, if x < y implies f(x) f(y). (3) strictly increasing if x < y implies f(x) < f(y). (4) strictly decreasing if x < y implies f(x) > f(y). (5) monotone if it is increasing or decreasing. (6) strictly monotone if it is strictly increasing or strictly decreasing. Proposition.2. If I R is an interval, f : I R is monotone, and f(i) is an interval, then f is continuous. Proof. Let x I. By symmetry it suffices to show that f is right continuous at x, and without loss of generality x < sup I. Furthermore, multiplying by if necessary, without loss of generality f is increasing. Case. f(a) f(x) for all a > x. Then, since f is increasing, f is constant on I [x, ), and so is right continuous at x. Case 2. f(a) > f(x) for some a > x. Let ε > 0, and without loss of generality assume that ε/2 < f(a) f(x). Since f is increasing and f(i) is an interval, we have f ( [x, a] ) [ = f(x), f(x) + ε ]. 2 Thus we can choose b [x, a] such that f(b) = f(x)+ε/2. Put δ = b x. Then x < y < x+δ implies f(y) f(x) = f(y) f(x) f(b) f(x) = ε 2 < ε. Proposition.3. If I R is an interval and f : I R is monotone, then f has at most countably many discontinuities, which are all jump discontinuities, i.e., places where the left-hand limit 4 f(x ) differs from the right-hand limit f(x+). Proof. We only handle the case where f is increasing and I = (a, b) is bounded and open; the other cases are left as an exercise. Let D be the set of discontinuities of f. Since f is monotone, it has one-sided limits at each element x (a, b), and f is discontinuous at x if and only if f(x ) < f(x+). Since f is increasing, the intervals {(f(x ), f(x+))} x D are pairwise disjoint, and ( ) f(x+) f(x ) f(b) f(a) <. x D Since f(x+) f(x ) > 0 for all x D, D must be countable. Definition.4. f : X Y is uniformly continuous if for all ε > 0 there exists δ > 0 such that for all x, y X, d(x, y) < δ implies d(f(x), f(y)) < ε. Definition.5. f : X Y is Lipschitz if there exists k > 0 such that d(f(x), f(y)) kd(x, y) for all x, y X. Proposition.6. Every Lipschitz function is uniformly continuous, and every uniformly continuous function is continuous. 4 you might want to review limits (both two-sided and one-sided) from calculus
24 PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 Example.7. If f : (a, b) R has bounded derivative 5 then it is Lipschitz, by the Mean Value Theorem. Example.8. If X is N with the discrete metric, then f : X N defined by f(n) = n (where in the codomain N has the usual metric from R) is uniformly continuous but not Lipschitz. Example.9. f : (0, ) R defined by f(x) = /x is continuous but not uniformly continuous. Here s a very special type of Lipschitz function: Definition.20. A function f : X Y is isometric if d(f(x), f(y)) = d(x, y) for all x, y X. If there is an isometric bijection between X and Y then they should be considered indistinguishable as metric spaces. Here is a slightly weaker, but still fairly strong, relation between metric spaces: Definition.2. A continuous bijection whose inverse is also continuous is a homeomorphism, and two metric spaces are homeomorphic if there is a homeomorphism between them. Remark.22. If f : X Y is a homeomorphism, then f(a) is open or closed if and only if A is, and f(x n ) f(x) if and only if x n x. Throughout, X, Y are metric spaces. 2. Completeness Definition 2.. A X is bounded if there exist x X and r > 0 such that A B r (x). A function f : S X (where S is any set) is bounded if f(s) is bounded. In particular, a sequence {x n } in X is bounded if {x n : n N} is a bounded subset of X. Remark 2.2. A X is bounded if and only if its diameter sup x,y A d(x, y) is finite. Example 2.3. A subset of R is bounded in this new sense if and only if it is bounded in the old sense, i.e., is bounded above and below. Example 2.4. A subset A R n is bounded if and only if sup x i < for i =,..., n, x A if and only if there exist a i b i in R for i =,..., n such that n A [a i, b i ]. i= The following is an example of a common theme in mathematics: generalization often involves turning a theorem into a definition. 5 you might want to review derivatives from calculus
MAT 570 REAL ANALYSIS LECTURE NOTES 25 Definition 2.5. A sequence {x n } in a metric space X is Cauchy if for all ε > 0 there exists k N such that d(x n, x j ) < ε for all n, j k. Proposition 2.6. Every convergent sequence is Cauchy, and every Cauchy sequence is bounded. Proof. First let x n x, and let ε > 0. Choose k N such that d(x n, x) < ε/2 for all n k. Then for all n, j k we have d(x n, x j ) d(x n, x) + d(x, x j ) < ε 2 + ε 2 = ε. Now let {x n } be Cauchy. Choose k N such that d(x n, x j ) < for all n, j k. Then for all i N we have d(x i, x k ) max{d(x, x k ),..., d(x k, x k ), }. Definition 2.7. A metric space is complete if every Cauchy sequence converges. Definition 2.8. A Banach space is a normed space for which the associated metric space is complete. Example 2.9. A product space n i= X i is complete if and only if X i is complete for every i =,..., n. Theorem 2.0. R is complete. Proof. Let {x n } be a Cauchy sequence in R. Then {x n } is bounded, and so lim sup x n R, and {x n } has a subsequence converging to this real number. But any Cauchy sequence with a convergent subsequence must itself converge (a routine triangle inequality argument). Example 2.. The subspace Q of R is not complete. Proposition 2.2. Every closed subset of a complete metric space is complete, and every complete subset of a metric space is closed. Proof. First assume that X is complete and A X is closed. Let {x n } be a Cauchy sequence in A. Then {x n } is also Cauchy in X, so because X is complete there exists x X such that x n x. Since A is closed, we have x A. Thus {x n } converges in A, and we have shown that A is complete. For the other part, assume that A X is complete. Let {x n } be a sequence in A, and assume that x n x in X. Since A is complete, {x n } converges in A. Since the limit of a convergent sequence is unique, we have x A. Thus A is closed. Proposition 2.3. If f : X Y is uniformly continuous and {x n } is a Cauchy sequence in X, then {f(x n )} is Cauchy. Consequently, if f is - onto and both f and f are uniformly continuous, then X is complete if and only if Y is complete. Proof. Exercise.
26 PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 Example 2.4. The function f : R (, ) defined by f(x) = x + x is bijective and uniformly continuous, and f : (, ) R is continuous but not uniformly continuous. R is complete, but (, ) is not. Definition 2.5. Two metrics d and d 2 on a set X are equivalent if there exist constants a, b > 0 such that ad (x, y) d 2 (x, y) bd (x, y) for all x, y X. Two equivalent metrics d, d 2 on X give two metric spaces (X, d ) and (X, d 2 ) that have the same open sets (although not the same open balls), closed sets, convergent sequences, and continuous functions, and one is complete if and only if the other one is. Example 2.6. Let and 2 be norms on a vector space. Then the associated metrics are equivalent if and only if the norms are equivalent in the following sense: there exist a, b > 0 such that a x x 2 b x for all x X. Example 2.7. The Euclidean norm on R n is equivalent to the product norm x = max i=,...,n x i. It turns out that for every metric space X there is an isometric function from X onto a dense subset of a complete metric space X. We can identify X with its image under f and regard it as a dense subset of X. If X is another such space, there is an isometric bijection of X to X that extends the identity map on X. All this is a routine, but tedious, exercise. Thus X is essentially unique, and we call it the completion of X. Here s one construction of a completion: put an equivalence relation on the set of all Cauchy sequences on X by {x n } {y n } if d(x n, y n ) 0, and let X be the set of all equivalence classes. Define d ( [{x n }], [{y n }] ) = lim d(x n, y n ), and embed X into X using constant sequences. The particular construction is not important; the universal properties are all we need. Example 2.8. [a, b] is the completion of (a, b), and R is the completion of Q. Proposition 2.9. Let X be the completion of X, let Y be complete, and let f : X Y be uniformly continuous. Then f extends uniquely to a continuous function f : X Y, and f is in fact uniformly continuous. Proof. Exercise.
MAT 570 REAL ANALYSIS LECTURE NOTES 27 3. Compactness Throughout, X, Y are metric spaces. Definition 3.. A X is compact if every open cover of A has a finite subcover. Proposition 3.2. Every closed subset of a compact metric space is compact, and every compact subset of a metric space is closed. Proof. First lset X be compact and let A X be closed. Let U be an open cover of A. Then U {A c } is an open cover of X, so by compactness we can find a finite subcover F. Then F \ {A c } is a finite subfamily of U that covers A. Now let A X be compact. Claim: for each x A c there exist disjoint open sets U x, V x such that A U x and x V x. To see this, fix x A c, and for every y A choose disjoint open sets R y, S y such that y R y and x S y (for example, use open balls centered at y, x with radius d(x, y)/2). The family {R y } y A is an open cover of A, so there is a finite subset B A such that A y B R y. Put U x = y B R y and V x = y B S y. Then U x V x = because each R y is disjoint from every S y and hence is disjoint from V x. Also, V x is open because B is finite, and x V x because x S y for every y. Now we see that, in particular, each x A c is contained in an open set V x disjoint from A. Thus A c is open, so A is closed. Remark 3.3. We ll see later that compactness in metric spaces can be characterized using sequences, which allows for shorter proofs of certain results like the preceding proposition. Definition 3.4. A family of sets has the finite intersection property if every finite subfamily has nonempty intersection. Proposition 3.5. X is compact if and only if every family of closed subsets with the finite intersection property has nonempty intersection. Proof. First assume that X is compact, and let {A i } i I be a family of closed subsets with the finite intersection property. Put U i = A c i for each i I. Then {U i } i I is a family of open subsets, and no finite subfamily covers X, by the finite intersection property and De Morgan s Laws. Thus {U i } i I cannot cover X, otherwise there would be a finite subcover by compactness. Therefore, again by De Morgan s Laws, i I A i. The converse can be proved by reversing the above argument. More precisely, assume the condition regarding the finite intersection property, and let {U i } i I be an open cover of X. Put A i = Ui c for each i I. Then {A i } i I is a family of closed subsets, and i I A i = by De Morgan s Laws. Thus {A i } i I cannot have the finite intersection property, so we can choose a finite set F I such that i F A i =. Then {Ui c } i I is a finite subcover, again by De Morgan s Laws.
28 PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 Definition 3.6. X is sequentially compact if every sequence in X has a convergent subsequence. Definition 3.7. A X is totally bounded if for all ε > 0 there exist x,..., x n A such that A n i= B ε(x i ). Remark 3.8. Note that the finite subset {x i } n i= of A in the above definition has the property that for all y A there exists i such that d(x i, y) < ε. Such a set is called a finite ε-net in A. Proposition 3.9. If A X is totally bounded, then it is bounded and separable, every sequence in A has a Cauchy subsequence, and every subset of A is totally bounded. Proof. Let A X be totally bounded. We can choose a finite -net {x i } n i= in A. Each ball B (x i ) is bounded, and hence so is the union n i= B (x i ), and therefore so is the smaller set A. To see that A is separable, for each n N choose a finite (/n)-net S n in A. Put D = n= S n. Then D is a countable subset of A, and we will show that it is dense. Let x A and ε > 0. We can choose n N such that /n < ε, and then there exists y S n such that d(x, y) < n < ε. Thus x D. Now let {x n } be a sequence in A. There exists a finite -net S in A, and then there exists t S such that d(x n, t) < /2 for infinitely many n, and hence there is a subsequence {x n } n= of {x n } such that d(x n, x j ) < for all n, j N. Similarly, there is a subsequence {x 2n } n= of {x n } n= such that d(x 2n, x 2j ) < /2 for all n, j N. Continuing inductively, we get a sequence of successive subsequences {x kn } n= such that, for every k N, {x k+,n } n= is a subsequence of {x kn } n= and d(x kn, x kj ) < /k for all n, j N. Then {x kk } k= is a subsequence of {x n}. Let ε > 0, and choose l N such that /l < ε. Then for all k, m l we have d(x kk, x mm ) min{k, m} l < ε. Thus {x kk } is Cauchy. Finally, let C A and ε > 0. Choose a finite ε/2-net F in A. Put D = {x F : B ε/2 (x) C }, and for each x D choose y x C such that d(x, y x ) < ε/2. Then C x D B ε/2 (x) x D B ε (y x ), and we have shown that C is totally bounded. Theorem 3.0. For A X, the following are equivalent: () A is compact.
MAT 570 REAL ANALYSIS LECTURE NOTES 29 (2) A is sequentially compact. (3) A is complete and totally bounded. Proof. () (2) Assume that A is compact, and let {x n } be a sequence in A. Arguing by contradiction, suppose that {x n } has no subsequential limit in A. Then for all x A we can choose r x > 0 such that S x := {n N : x n B rx (x)} is finite. By compactness there is a finite subset F A such that A x F B rx (x). But x n A for all n N, so N = x F S n, which is finite, and this is a contradiction. (2) (3) Assume that A is sequentially compact. To see that A is complete, let {x n } be a Cauchy sequence in A. By sequential compactness, {x n } has a convergent subsequence. But then {x n } converges itself since it is Cauchy. To see that A is totally bounded, we argue by contradiction. Suppose ε > 0 and A has no finite ε-net. In particular, A. So, we can choose x A. Then we can choose x 2 A such that d(x 2, x ) ε. Then choose x 3 A such that d(x 3, x i ) ε for i =, 2. Continuing inductively, we get a sequence {x n } in A such that d(x n, d k ) ε for all n k. But then {x n } has no Cauchy subsequence, and hence no convergent subsequence, which is a contradiction. (3) () Assume that A is complete and totally bounded. Let U be an open cover of A. Since A is totally bounded, it is separable, so by Lindelöf s Theorem, without loss of generality U = {U n } n=. Arguing by contradiction, suppose U has no finite subcover. Then for every n we can choose x n A\ j<n U j. Again by total boundedness, {x n } has a Cauchy subsequence, which must converge by completeness. Thus {x n } has a subsequential limit x. Since U is a cover we can choose j N such that x U j. Since x is a subsequential limit, there exists n > j such that x n U j, which is a contradiction. Theorem 3.. Every continuous image of a compact set is compact. Proof. Let {y n } be a sequence in Y. For each n N we can choose x n X such that f(x n ) = y n. By compactness, {x n } has a convergent subsequence {z k }, and then by continuity {f(z k )} is a convergent subsequence of {y n }. Remark 3.2. Here s an alternative proof, using open covers: Let X be compact and f : X Y be continuous and onto. Let U be an open cover of Y. Then by continuity {f (U) : U U} is an open cover of X, so by compactness there is a finite subfamily F U such that {f (U) : U F} covers X, and hence F covers Y. Proposition 3.3. A product space n i= X i is compact if and only if every X i is compact.