# Metric Spaces Joseph Muscat 2003 (Last revised May 2009)

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1 1 Distance J Muscat 1 Metric Spaces Joseph Muscat 2003 (Last revised May 2009) (A revised and expanded version of these notes are now published by Springer.) 1 Distance A metric space can be thought of as a very basic space having a geometry, with only a few axioms. Metric spaces are generalizations of the real line, in which some of the theorems that hold for R remain valid. Some of the main results in real analysis are (i) Cauchy sequences converge, (ii) for continuous functions f(lim n x n ) = lim n f(x n ), (iii) continuous real functions are bounded on intervals of type [a, b] and satisfy the intermediate value theorem, etc. At first sight, it is difficult to generalize these theorems, say to sequences and continuous functions of several variables such as f(x, y). That is the aim of this abstract course: to show how these theorems apply in a much more general setting than R. The fundamental ingredient that is needed is that of a distance or metric. This is not enough, however, to get the best results and we need to specify later that the distance be of a nice type, called a complete metric. In what follows the metric space X will denote an abstract set, not necessarily R or R n, although these are of the most immediate interest. When we refer to points we do not necessarily refer to geometrical points, although this is how most of us visualize them. They may in fact be sequences, functions, images, sounds, signals, etc. Definition with the properties for all x, y, z X. A distance or metric on a metric space X is a function d : X 2 R + (x, y) d(x, y) (i) d(x, y) = 0 x = y (ii) d(y, x) = d(x, y) (iii) d(x, y) d(x, z) + d(y, z)

2 1 Distance J Muscat Example On N, Q, R, C, and R N, one can take the standard Euclidean distance d(x, y) := x y. Check that the three axioms for a distance are satisfied (make use of the fact that a+b a + b ). Note that for R n, the Euclidean i=n i=1 x i y i 2. distance is d(x, y) := One can define distances on other more general spaces, e.g. the space of continuous functions f(x) with x [0, 1] has a distance defined by d(f, g) := max x [0,1] f(x) g(x). The space of shapes (roughly speaking, subsets of R 2 having an area) have a metric d(a, B) := area of (A B A B). In all these cases we get an idea of which elements are close together by looking at their distance Exercises 1. Show that if x 1,..., x n are n points, then d(x 1, x n ) d(x 1, x 2 ) d(x n 1, x n ). 2. Verify that the metric defined on R 2 does indeed satisfy the metric axioms. 3. Show that if d is a metric, then so are the maps D 1 (x, y) := 2d(x, y) and D 2 (x, y) := d(x,y), but that d(x, 1+d(x,y) y)2 need not be a metric. Hence a metric space can have several metrics. 4. Show that if X, Y are metric spaces with distances d X and d Y, then X Y is also a metric space with distance ( ) ( ) x1 x2 D(, ) := d X (x 1, x 2 ) + d Y (y 1, y 2 ). y 1 y 2 (Note that for R 2, this metric is not the Euclidean one.) 5. Show further that ) D(( x1, y 1 ( x2 y 2 is also a metric for X Y. ) ) := max(d X (x 1, x 2 ), d Y (y 1, y 2 )) 6. Prove that the function d(m, n) := 1/m 1/n, on the natural numbers, is a metric.

3 1.1 Balls J Muscat 3 7. Show that in fact the axioms (i) and (iii) imply both d(x, y) 0 and axiom (ii). 8. Show that d(x, y) d(x, z) d(y, z). 9. Consider the set X of bytes, i.e., sequences of 0s and 1s of length 8. The Hamming distance between two bytes is defined as the number of places where their bits differ. Show that this is a distance on X. 1.1 Balls Definition An (open) ball is the set B r (a) := { x X : d(x, a) < r } where a X is its center and r > 0 its radius. Note: The ball is the surroundings of the point a. One can consider and X to be balls with r = 0 and r = respectively, but this is unusual. Although we call it a ball, one must remove any preconceptions of it being round etc. In R, every ball B r (a) = { x R : x a < r } = (a r, a + r) is just an open interval. In particular, B 1/2 (a) = (a 1/2, a + 1/2). Conversely, any open interval of the type (a, b) is a ball in R, with center (a +b)/2 and radius (b a)/2. In R 2, the ball B r (a) is the disk with center a and radius r without the circular perimeter. In N, the ball B 1/2 (m) = { m }, and B 2 (m) = { m 1, m, m + 1 }. Definition A point x of a set A is called an interior point of A when ɛ > 0 B ɛ (x) A. A point x (not in A) is an exterior point of A when ɛ > 0 B ɛ (x) X A. All other points of X are called boundary points. A point x of A is called an isolated point when there is a ball B ɛ (x) which contains no points of A other than x itself.

4 1.1 Balls J Muscat 4 Definition A set A is open in X when all its points are interior points. Note: An interior point of A can be surrounded completely by a ball inside A ; open sets do not contain their boundary. We also say that A is a neighborhood of a when a is an interior point of A. The empty set is open by default, because it does not contain any points. The whole space X is also open because any ball about any point of X is a subset of X Example Show that (a, b) is open in R, but that [a, b] and { a } are not. Theorem A Balls are open sets in X. Proof Let x B r (a) be any point in the given ball. This means that d(x, a) < r. Let ɛ := r d(x, a) which is positive. Then B ɛ (x) B r (a) since for any y B ɛ (x), d(y, a) d(y, x) + d(x, a) < ɛ + d(x, a) = r. Therefore y B r (a). Open sets, however, need not be balls. Note that we need to specify that A is open in X. For example we have seen that { m } is open in N (since it is a ball in N) but not open in R (see previous example). Theorem B A set is open it is the union of balls.

5 1.1 Balls J Muscat 5 Proof Let A be an open set. Then x A ɛ > 0 x B ɛ (x) A which implies that A = {x} B ɛ (x) A. x A x A Hence A = x A B ɛ(x), a union of balls. Conversely, let A = i B r i (a i ) be a union of balls, and let x be any point in A. Then x is in at least one of these balls, say, B r (a). But balls are open and hence x B ɛ (x) B r (a) A. A is therefore open. One can study open sets without reference to balls or metrics in the subject of topology. The basic properties of open sets are: Theorem C Any union of open sets is open. Any finite intersection of open sets is open. Proof Consider i A i where A i are all open. Given any x i A i, it must lie in at least one of these sets A i which is open. Therefore x B r (x) A i i A i showing that the union is open. For the second part it is enough to consider the intersection of two open sets A B (the rest can be done by induction). Let x A B. Then x A and x B, both sets being open. Therefore there are open balls B r1 (x) A and B r2 (x) B. Pick the smaller of these two balls, say the one with radius r 1. Then, x B r1 (x) B r1 (x) B r2 (x) A B.

6 1.1 Balls J Muscat Example In R the intersection of two open intervals is either empty or else another open interval, both of which are open. Any collection of open intervals is open. Note that an infinite intersection of open sets need not be open. For example, in R, consider the open intervals ( 1/2 n, 1/2 n ) which are nested one inside another. Their intersection is just the set { 0 } (prove this!) which is not open in R. Theorem D Distinct points in X can be separated by disjoint open balls. x y r > 0 B r (x) B r (y) = Proof If x y then d(x, y) > 0. Let r := d(x, y)/2. Then B r (x) B r (y) = since otherwise, z B r (x) B r (y) d(x, z) < r and d(y, z) < r which is a contradiction. d(x, y) d(x, z) + d(y, z) < 2r = d(x, y) Exercises 1. If r 1 < r 2, show that B r1 (x) B r2 (x). 2. Show that in R, the sets (a, ) and (, a) are open sets. 3. Show that in R 2, (i) the half-plane S = { (x, y) R 2 : y > 0 } and (ii) the rectangles R = { (x, y) R 2 : a < x < b, c < y < d } are open sets. 4. Show that the interval (a, b) on the x-axis, is open in R but not open in R 2.

7 1.1 Balls J Muscat 7 5. Find an example in R 2 in which the infinite intersection of open sets is not open. 6. Show that for any metric space X, the set X { x } is open in X. 7. Prove properly by induction, that the finite intersection of open sets is open. 8. Show that if { x } are open sets in X for all points x X, then all subsets of X are also open in X. In particular, all subsets of N are open in N. 9. Given any set X, let d(x, y) = { 0 x = y 1 x y. Show that d is a metric (it is called the discrete metric), for which the open balls are just the sets { x } and X. Deduce that all the subsets of X with this metric are open in X. 10. * Show that the set of interior points of A is the largest open set inside A, i.e., if A denotes the set of interior points of A, and V A is an open set, then V U.

8 2 Closed Sets and Convergence J Muscat 8 2 Closed Sets and Convergence 2.1 Closed Sets Definition A set F is closed in a space X when X F is open in X Example On R the set [a, b] is closed, since R [a, b] = (, a) (b, ) is the union of two open sets, hence itself open. Similarly [a, ) and (, a] are closed in R. N is closed in R, but Q is not. On any metric space X, the sets X and are closed since X = X and X X = are open in X. Proposition 2.1 The sets { x } are closed in X. Proof. Exercise. Theorem A The finite union of closed sets is closed. Any intersection of closed sets is closed. Proof X (F G) = (X F ) (X G) which is the intersection of two open sets, hence open. X i F i = i (X F i ) which is the union of open sets, hence open.

9 2.1 Closed Sets J Muscat Exercises 1. Let U be an open set. Show that X U is closed in X. 2. Show that the closed ball { x X : d(x, a) r } is closed. 3. Let U be an open set and F a closed set in X. Show that U F is open and F U is closed. 4. Show that a finite collection of points { a 1,..., a N } form a closed set. 5. Find a set in R which is (i) neither open nor closed; (ii) both open and closed. 6. Find an example in which the infinite union of closed sets (i) is not closed; (ii) is closed. 7. Determine which of the following sets in R are closed, open, neither or both: (i) (0, ); (ii) Z; (iii) n [0, 2 1/n); (iv) R; (v) [0, 1] n=2 { 1/n }; (vi) [0, 1] { 5 }. 8. * Let the Cantor set be defined as follows. Start with the closed interval [0, 1]; remove the middle interval (1/3, 2/3) to end up with two closed intervals [0, 1/3] [2/3, 1]. For each of these intervals remove the middle interval of each to end up with four closed intervals [0, 1/9] [2/9, 1/3] [2/3, 7/9] [8/9, 1]. Show that if we continue this process indefinitely we end up with a closed set. Definition A) such that A limit point of a set A is a point b (not necessarily in ɛ > 0 a b a A B ɛ (b) In other words, a limit point is one that cannot be separated from its set it can be an interior point or part of the boundary. Every point of A is either a limit point or an isolated point. Theorem B A set is closed it contains all its limit points.

10 2.1 Closed Sets J Muscat 10 Proof Let F be a closed set, and let x be a limit point of it. x / F x X F x B ɛ (x) X F which shows that B ɛ (x) F =, a contradiction since x is a limit point of F. Conversely, suppose F contains all its limit points, and let x / F. Then x is not a limit point of F. Therefore, ɛ > 0 y x y / F B ɛ (x) i.e., F B ɛ (x) { x } X F x B ɛ (x) X F which means that x can be surrounded by a ball inside X F. Therefore X F is open, which by definition means that F is closed. Definition the closure of A Any set A can be closed by adding its limit points to form A = A {limit points of A} Note: the set of limit points of a set is sometimes called its derived set Exercises 1. Find the limit points and closure of Z and Q in R. 2. Show that (a, b) = [a, b]; { 1/n } = { 1/n } {0}. 3. Can a set not have limit points? Can an infinite set not have limit points? 4. A set A is said to be dense in X when A = X. Show that any ɛ-ball of X must contain elements of A. Prove that Q is dense in R. 5. Let A be a bounded non-empty set in R. Show that sup A is a limit point of both A and R A. 6. Show that x A inf a A d(x, a) = 0.

11 2.2 Convergence J Muscat Show that any point of A must be either a limit point of A or an isolated point of A. 8. * Show that A B = A B; A B A B; A is closed A = A. 9. * Show that A is the smallest closed set containing A i.e., if F A is a closed set, then A F. 10. ** Show that the limit points of A belong to A so that A = A. 11. ** Let the decimal expansion of x be 0.n 1 n 2 n Show that is closed in R. { x : n n k k 5 k } 2.2 Convergence Definition when A sequence (x n ) in the metric space X converges to x, written x n x as n, or lim n x n = x, ɛ > 0 N n N x n B ɛ (x). In other words, any neighborhood of x contains all the sequence from N onwards. This definition generalizes the definition of convergence for real sequences the expression x n x < ɛ generalizes to d(x n, x) < ɛ which is the same as x n B ɛ (x). Proposition 2.2 A sequence can only converge to one point. Proof Suppose x n x and x n y as n. Suppose that x y then they can be separated by two disjoint balls B r (x) and B r (y). But N 1 n > N 1 x n B r (x)

12 2.2 Convergence J Muscat 12 and N 2 n > N 2 x n B r (y). Therefore for n > N 1, N 2 we have that x n B r (x) B r (y) = which is a contradiction. Theorem C If x n F converges to x, then x F. If (x n ) is a convergent sequence in a closed set F, then lim n x n F. Proof If x is in F then we re done. So suppose that the limit x F. Take any ball B ɛ (x) about x. Since x n converges to x, we can find an integer N such that n N x n B ɛ (x). Hence x is a limit point of F and so is in F. The corollary follows immediately from the theorem Exercises 1. Show that if x is a limit point of the set A then there is a sequence (a n ) in A which converges to x. (Hint: take balls around x, with radii decreasing to 0) 2. Show that if x n x then d(x n, x m ) 0 as n, m. ( ) ( ) xn x 3. Show that a sequence in X Y converges to if, and only y n y if, x n x and y n y. (Recall the distance defined on X Y.) 4. Show that if d 1 (x, y) λd 2 (x, y) (λ > 0) and the sequence (x n ) converges with respect to d 2, then it converges with respect to d Try to generalize the definition of lim x x0 f(x) from the case of real functions to the case of functions f : X Y on metric spaces.

13 2.3 Completeness J Muscat Completeness When does a sequence converge/diverge? Let us take some examples of divergent sequences: (n) diverges in R because the set { n } seems to be too large; (1/n) diverges in the set (0, 1) because the point it s trying to converge to, 0, is not in the set; (( 1) n ) diverges because it keeps alternating between two points; (x n ) where x n+1 = xn x n in Q diverges because were it to converge then the limit would satisfy x = x/2 + 1/x which implies that x 2 = 2 which we know cannot be satisfied by any rational number. We seem to be able to make an intuitive distinction between sequences that do not converge at all (truly divergent), and those that seem to converge outside the set if we restrict ourselves to X however, it fails to converge because the metric space does not contain the limit. The same phenomenon can occur for other more general spaces e.g. a sequence of continuous functions can converge to a discontinuous function. It is possible to distinguish between the two ideas rigorously as follows: Definition A Cauchy sequence is one such that ɛ > 0 N n, m > N d(x n, x m ) < ɛ. In other words, lim n,m d(x n, x m ) = 0. Intuitively speaking, a Cauchy sequence is one which is trying to converge. Convergent sequences are always Cauchy (see exercises above) but there are examples of Cauchy sequences which do not converge (e.g. the case of the rational sequence above, which tries to converge to 2). Definition converges. A metric space is complete when every Cauchy sequence It follows that the set of rational numbers Q is not complete; but it will be shown that the set of real numbers R is complete. This is the main reason why we use the real numbers rather than the rationals in most applications of calculus. Note that every metric space X can be completed i.e., there is a complete metric space X which contains X; for example Q = R and (0, 1) = [0, 1]. Proposition 2.3

14 2.3 Completeness J Muscat 14 For a complete metric space X, any subset F is complete F is closed in X. Proof Let F X be complete i.e., any Cauchy sequence in F converges to a limit in F. Let x be a limit point of F. Then as shown in an exercise above, we can find a sequence (x n ) in F which converges to x, x n x. But convergent sequences are Cauchy and F is complete. Hence (x n ) converges to a point in F i.e., x F. Thus F contains all its limit points and is therefore closed. Conversely, let F be a closed set in X and let (x n ) be a Cauchy sequence in F. Then (x n ) is a Cauchy sequence in X, which is complete. Therefore x n x for some x X. But x must be a limit point of F and so x F. Thus any Cauchy sequence of F converges in F. Corollary If X is a complete metric space, then a Cauchy sequence x n F converges to a limit in F. Theorem D The set of real numbers R is complete. Proof (From Real Analysis course) Let (x n ) be a Cauchy sequence, that is x n x m 0 as n, m. Set ɛ = 1, then there must be an N 0 beyond which x n x m < 1; let y 0 := x N0. Now set ɛ = 1/2, then there must be an N 1, with N 1 > N 0, beyond which x n x m < 1/2; let y 1 := x N1. Continuing this way we get a sequence y k for which y k+1 y k < 1/2 k. Now consider the sum y 0 y K = y 0 y 1 + y 1 y y K : K (y k y k+1 ) k=0 K y k y k+1 k=0 K 2 k < 2. k=0

15 2.3 Completeness J Muscat 15 The middle sum is increasing but bounded above, and so must converge. By comparison, the first sum must converge to, say, y; set x := y 0 y. Then y K y 0 y = x as K. That is, we have shown that the subsequence y K converges to x; it is left as an exercise to show that the full sequence x n converges to x. It follows that any closed subset of R is complete. For example, any Cauchy sequence in [0, 1] must converge to a limit in [0, 1]; and any Cauchy sequence in the Cantor set must converge there as well. However a Cauchy sequence of rational numbers need not converge to a rational number because Q is not closed in R. Theorem E If X, Y are complete metric spaces then so is X Y. d( Proof ), ( xn y n ( xm y m ( xn ) Let be a Cauchy sequence in X Y. This means that ) y n ) = d X (x n, x m ) + d Y (y n, y m ) 0 as n, m. In particular d(x n, x m ) 0, so that the sequence (x n ) is a Cauchy sequence in X, which is ( complete. ) ( ) We therefore get x n x X, and similarly, y n y Y. Now xn x d(, ) = d y n y X (x n, x) + d Y (y n, y) 0 as n, so that the Cauchy ( ) xn sequence converges in X Y. y n Corollary Hence R n = R... R and C are complete.

16 2.3 Completeness J Muscat Exercises 1. Show that the interval (a, b] is incomplete, but [a, b] is complete. 2. Show that if a Cauchy sequence (x n ) has a convergent subsequence x ni x as i, then (x n ) must converge to x. 3. A set A is called bounded when all the distances between points in A are bounded i.e., d(x, y) R for x, y A. Show that Cauchy sequences are bounded. Hence we get the following set inclusions: {set of convergent sequences} {set of Cauchy sequences} {set of bounded sequences}. 4. In the proof of the completeness of X Y we used one particular distance d X + d Y. Show that the other distance max(d X, d Y ) is also complete (if d X and d Y are complete).

17 3 Continuity J Muscat 17 3 Continuity Definition A function f : X Y is continuous when x X ɛ > 0 δ > 0 d X (x, y) < δ d Y (f(x), f(y)) < ɛ Since d(a, b) < r b B r (a), this can be rewritten as x X ɛ > 0 δ > 0 y B δ (x) f(y) B ɛ (f(x)) or even as x X ɛ > 0 δ > 0 fb δ (x) B ɛ (f(x)). Note also that a function is discontinuous at x if the opposite holds i.e., ɛ > 0 δ > 0 y B δ (x) f(y) / B ɛ (f(x)). Proposition 3.1 A function f : X Y is continuous for every open set V in Y, f 1 V is open in X. Proof Let f be a continuous function and let V be an open set in Y. We want to show that U := f 1 V is open in X. Let x be any point of U. Then f(x) V, which is open. Hence and so In other words, f(x) B ɛ (f(x)) V, δ > 0 fb δ (x) B ɛ (f(x)) V. δ > 0 B δ (x) f 1 V = U. Conversely, assume the second statement holds for all open sets V. In particular apply it to the open set B ɛ (f(x)). Then f 1 B ɛ (f(x)) is open in X, that is, δ > 0 x B δ (x) f 1 B ɛ (f(x))

18 3 Continuity J Muscat 18 which implies as required. δ > 0 fb δ (x) B ɛ (f(x)), Note that if U is open in X, then fu need not be open in Y. Proposition 3.2 If f : X Y and g : Y Z are continuous, then so is g f : X Z. Proof Let W be any open set in Z. Then g 1 W is an open set in Y, and so f 1 g 1 W is an open set in X. But this set is precisely (g f) 1 W. Note that f continuous need not imply that f 1 is continuous. Theorem A f is continuous f( lim n x n ) = lim n f(x n ). Proof Let f be a continuous function and let (x n ) be a sequence converging to x in the domain. We want to show that f(x n ) f(x) in the codomain as n. Consider the neighborhood B ɛ (f(x)) of f(x). Then, since f is continuous, δ > 0 fb δ (x) Bɛ(f(x)). But x n x means N > 0 n > N x n B δ (x)

19 3 Continuity J Muscat 19 which implies that f(x n ) fb δ (x) B ɛ (f(x)). Conversely, suppose f is not continuous. Then there is a point x such that ɛ > 0 δ > 0 fb δ (x) B ɛ (f(x)). In particular ɛ > 0 n fb 1/n (x) B ɛ (f(x)). Therefore we can find points x n B 1/n (x) for which f(x n ) / B ɛ (f(x)) i.e., f(x n ) f(x) while x n x. We thus see that continuous functions preserve convergence, a central concept in metric spaces; that is, if a sequence converges in X, and it is mapped continuously to Y, then it still converges. In a sense, they correspond to homomorphisms of groups and rings, which preserve the group and ring operations. The following concept is then analogous to an isomorphism between metric spaces. Definition A homeomorphism between two metric spaces X and Y is a map φ : X Y such that φ is bijective (1-1 and onto) φ is continuous φ 1 is continuous In other words, two spaces are homeomorphic when one can be obtained continuously and continuously invertibly from the other; in effect when convergence in one space is equivalent to convergence in the other. The most vivid example is of one space being continuously deformed into another homeomorphic to it; the classic tongue-in-cheek example is that a teacup is homeomorphic to a doughnut! Exercises 1. Show that the constant functions f(x) := y 0 Y are continuous. 2. Find a continuous function f : R R and an open set U R such that fu is not open.

20 3 Continuity J Muscat Let When is χ A continuous? χ A (x) = { 1 x A 0 x / A 4. Show that if F is a closed set in Y and f : X Y is a continuous function, then f 1 F is closed in X. 5. Show that if f, g : X R are continuous functions, then so are the functions f + g and λf. 6. Show that { x X : f(x) > 0 } is open in X when f : X R is a continuous function. 7. Reprove the proposition that g f is continuous using distances rather than open sets. 8. Show that if x n x and y n y then d(x n, y) d(x, y) and d(x n, y n ) d(x, y). 9. Show that continuous functions need not map Cauchy sequences to Cauchy sequences. 10. Find an example where f is continuous and bijective but f 1 is not. 11. Show that X is homeomorphic to Y is an equivalence relation on metric spaces, so that we can partition metric spaces into equivalence classes. 12. Show that (0, 1) is homeomorphic to (a, b). 13. Let f : Q Q be a continuous function. Show that there is only one way of extending it to the reals as a continuous function f : R R. Show further that the only function f : Q Q satisfying f(x + y) = f(x) + f(y) is f(x) = λx. Deduce that there is only one continuous function f : R R with this property. 14. * Let f : A Y and g : B Y be continuous functions with A, B X. Suppose that the two functions agree on A B, i.e., x A B f(x) = g(x). Show that the function { f(x) x A h(x) = g(x) x B is well-defined and continuous on A B.

21 3.1 Applications J Muscat * A function f : X Y is called isometric when f preserves distances i.e., d Y (f(x), f(y)) = d X (x, y) x, y X. Show that onto isometric maps are continuous, invertible and have a continuous inverse (i.e., they are homeomorphisms). 3.1 Applications Definition A function is called a contraction when there is a constant 0 k < 1 such that x, y X, d(f(x), f(y)) k d(x, y). If follows that f is continuous, because d(x, y) < δ := ɛ/k d(f(x), f(y)) < ɛ. Theorem B The Banach contraction mapping theorem Let X be a complete metric space, and suppose that f : X X is a contraction map. Then f has a unique fixed point x = f(x). Proof Consider the iteration x n+1 = f(x n ) with x 0 = a any point in X. Note that d(x n+1, x n ) = d(f(x n ), f(x n 1 )) kd(x n, x n 1 ); hence, by induction, d(x n+1, x n ) k n d(x 1, x 0 ). (x n ) is a Cauchy sequence since d(x n, x m ) d(x n, x n 1 ) d(x m+1, x m ) (k n k m )d(x 1, x 0 ) km 1 k d(x 1, x 0 ) which converges to 0 as m, n. Hence x n x, and by continuity of f, x = lim n x n+1 = lim n f(x n ) = f( lim n x n ) = f(x).

22 3.1 Applications J Muscat 22 Moreover, the rate of convergence is given by d(x n, x) kn 1 k d(x, x 0). Suppose there are two fixed points x = f(x) and y = f(y); then d(x, y) = d(f(x), f(y)) kd(x, y) so that d(x, y) = 0 since k < 1.

23 4 Connectedness J Muscat 23 4 Connectedness Definition A set A is disconnected when it can be divided into (at least) two disjoint non-empty subsets A = B C and each subset can be covered exclusively by an open set (U, V respectively) i.e., B U, C U =, C V, B V =. A set is called connected otherwise Example Two distinct points are disconnected by Theorem 1D. Single points are always connected because they cannot be separated into two non-empty sets. Similarly the empty set is connected. Any subset of the natural numbers is disconnected except the single points {n} and the empty set. The set of rational numbers Q is disconnected e.g. Q (, 2) ( 2, ). Theorem A The connected subsets of R are precisely its intervals. Proof Let A be a connected subset of R. Let a, b A and let a < x < b. Then if x / A, it follows that (, x) (x, ) is a non-trivial open partition of A, which is a contradiction for a connected set. Hence x A, which means that A is an interval. For the converse, first consider an interval [a, b]. Suppose it to be disconnected. Then it can be split up into disjoint subsets B and C, each covered exclusively by U and V respectively. Suppose b belongs to B U say. Consider the open set V [a, b] (open in [a, b]). Then this set is bounded above by b and is non-empty (because it contains the non-empty set C). Hence by the least upperbound property of the real numbers, V [a, b] has a least upperbound α b. Now α ought to belong to either U or V. If it belongs to U then we should find an ɛ-ball around it contained completely in U; but this would make α ɛ a smaller upperbound. If instead it belongs to V, then

24 4 Connectedness J Muscat 24 we can find an ɛ-ball around it contained completely in V ; but then there would be points of V larger than α. As both of these are contradictory, we conclude that the set [a, b] is connected. Now consider any other interval. The intervals = (a, a) and { a } = [a, a] are always connected. So suppose the interval contains at least two points and is disconnected. Then we can find two points contained in separating open sets, a U and b V. But this means that the set [a, b] is disconnected using the same open sets U and V, which is false. Hence every interval is connected. Theorem B If f : X Y is continuous and A X is a connected set, then fa is a connected subset of Y. Proof Suppose f A is disconnected into the non-empty disjoint sets B and C covered exclusively by the open sets U and V. That is, Then, fa = B C U V ; U C = V B =. A = f 1 B f 1 C f 1 U f 1 V ; f 1 U f 1 C = f 1 V f 1 B =. Moreover f 1 B and f 1 C are disjoint and f 1 U and f 1 V are open sets. Hence f A disconnected A disconnected. This means that the property of connectedness is preserved by continuous maps. In particular a path which is a continuous function γ : [0, 1] X has a connected image. For example, straight line segments, circles and parabolas in R 2 are connected. Corollary The Intermediate Value Theorem. If f : [a, b] R is continuous and f(a) < c < f(b) then there exists an x such that a x b and f(x) = c.

25 4 Connectedness J Muscat 25 Proof We have proved that [a, b] is a connected set and hence f[a, b] is also connected in R. But connected subsets of R are just the intervals. Now f(a), f(b) f[a, b] and so c f[a, b] i.e., c = f(x) for some x [a, b]. Similarly a continuous function f : [0, 1] 2 R has the same property: if f(a) < c < f(b) then there is an x [0, 1] 2 such that c = f(x). By the above theorem, this remains true if we substitute [0, 1] 2 by any connected subset of R 2. Proposition 4.1 If A and B are connected sets and A B then A B is connected. Proof Suppose A B splits up into two parts covered exclusively by open sets U and V. Then A itself would split up into the two parts A U and A V were these non-empty. But A is known to be connected hence one of these must be empty, say A U =. Similarly for B, B V =. In other words, A V and B U. Hence A B A U = Exercises 1. Show that the only connected subsets of Q are the single points or the empty set. 2. Let y be a positive real number. Use the intermediate value theorem to show that n y exists. 3. Generalize the last proposition to the case when there are an infinite number of connected sets A i with i A i. 4. Show that R 2 is connected by considering the radial lines and applying the previous exercise. More generally show that any vector space over the real numbers is connected. 5. Similarly show that (a, ) R is connected. Show further that R 2 { x } is connected but that R { x } is not. Deduce that R and R 2 are not homeomorphic. 6. Use the same idea to show that [a, b] is not homeomorphic to [a, b).

26 4 Connectedness J Muscat Show that if a set contains a non-trivial proper subset which is both open and closed then it is disconnected. Equivalently, a connected set cannot contain a non-trivial open and closed proper subset. 8. Do balls have to be connected? Consider the space X = (, 1) (1, ). Find a ball in this space which is not connected. 9. Let A be a connected metric space and let f : A N be a continuous function. Show that f must be a constant function. 10. Suppose that there is an x 0 X and an r > 0 such that d(x 0, y) r y. Moreover there is another point x 1 such that d(x 0, x 1 ) > r. Show that X must be disconnected. 11. * Show that A is connected if A is.

27 5 Compactness J Muscat 27 5 Compactness Definition A set B is bounded when R > 0 x, y B d(x, y) < R. That is, a set is bounded when the distances between any two points in the set have an upperbound. The least such upperbound is called the diameter of the set: Example. diamb = sup{d(x, y) : x, y B}. The set [a, b) is bounded in R because d(x, y) b a for any two elements x, y [a, b); in fact diam[a, b) = b a. The set R is unbounded because d(0, x) = x can be made as large as needed. Note: Boundedness is a property of how large a set is. However it is not a good metric property: if B is bounded and f is a continuous function, then fb need not be bounded (see Ex. 4). Thus a set may be bounded in one metric space X, but not remain so when X is deformed continuously to another. Proposition 5.1 A set B is bounded when it can be covered by a ball, R > 0, a X, B B R (a). Proof. (Left as an exercise) Hence, the set [0, 1) (2, 3) is bounded because it can be covered by the ball B 4 (0). Definition A set B is totally bounded when ɛ > 0 a 1... a N B N B ɛ (a i ). That is, a set is totally bounded when it can be covered by a finite number of ɛ-balls, however small their radii ɛ. i=1

28 5 Compactness J Muscat Example. The set [0, 1] is totally bounded because it can be covered by the balls B ɛ (nɛ) for n = 0,..., N where N > 1/ɛ. Proposition 5.2 A totally bounded set is bounded. Proof Let B be a totally bounded set. We can therefore cover B with a finite number of ɛ-balls, in particular with 1-balls. B N B 1 (a i ). i=1 Since there are only a finite number of these balls, we can find the maximum distance between their centers. D := max{d(a i, a j )}. Now, given any two points x, y B, they must be covered by two of these balls B 1 (a I ) and B 1 (a J ), say. Therefore, using the triangle inequality twice, d(x, y) d(x, a I ) + d(a I, a J ) + d(a J, y) 1 + D + 1. That is, D + 2 is an upperbound for the distances between points in B. In one of the exercises, you are asked to show that, in R, bounded sets are totally bounded (i.e., boundedness and totally boundedness coincide for the real line). So a totally bounded set also need not be preserved by a continuous function. Definition A set K is said to be compactif K i N B ɛi (a i ) i = i 1,..., i N : K B ɛik (a ik ). k=1 In plain English, a set is compact when given any cover of balls of varying sizes, we can find a finite subcollection of them that still cover the set. This is a stronger property than totally-boundedness.

29 5 Compactness J Muscat Example. The set [0, 1) is not compact. For example, the cover of balls B 1 1/n (0) for n = 2,... has no finite subcover. Similarly the sets R and N are not compact. On the other hand we will soon see that the sets [a, b] are compact in R. By an open cover we mean a cover of open sets, and a subcover is a subcollection of a cover. Theorem A A set is compact any open cover of it has a finite subcover. Proof Let K be the compact set and let the open sets A i cover it i.e., K i A i. Now, by theorem 1B, the open sets A i are composed of balls. Therefore it follows that K is inside a union of (a large number of) balls, i.e., it is covered by these balls. Therefore there is a finite number of these balls B ɛ1 (a 1 ),..., B ɛn (a N ) which still cover the set K. Each of these balls is inside one of the open sets A i. Therefore there is a finite number of these open sets (those which contain these balls) that still cover the compact set K. Conversely, suppose K is such that any open cover of it has a finite subcover. In particular if we can cover it with (open) balls, then we can find a finite subcollection of them which still cover K. We show next that compactness is a metric property: it is preserved by continuous functions. Theorem B If K is compact and f is a continuous function, then fk is compact. Proof Let A i be an open cover for fk. We would like to show that a finite subcollection of them still covers fk.

30 5 Compactness J Muscat 30 From fk i A i we can deduce K f 1 i A i = i f 1 A i. But f 1 A i are open sets since f is continuous (Prop ). Therefore the right-hand side is an open cover of K, which is compact. Therefore, by Theorem 5A, a finite number of these open sets will do to cover K. It follows that K N f 1 A i. i=1 fk N A i. That is, a finite number of the original open sets A i will cover the set fk, which is therefore compact. i=1 Theorem C Compact sets are closed and bounded. Proof Compact sets are bounded. Let K be a compact set. For any fixed point a X, consider the balls B n (a) for n = 1, 2,.... These balls cover the set K (in fact they cover the whole space X). Therefore a finite number of these balls will cover K. But these balls are nested in each other, so that it follows that the largest of these finite number of balls, say B N (a), contains all the others. Hence B N (a) contains K, which is therefore bounded. Compact sets are closed in X. Let x X K. We would like to surround it by a ball outside K. From Theorem 1D, x can be separated from y K by open balls B ry (x) and B ry (y). Since y B ry (y), these latter balls cover K. But K is compact, so that a finite subcollection of these balls still cover K, K B r1 (y 1 )... B rn (y N ).

31 5 Compactness J Muscat 31 Now let U := B r1 (x)... B rn (x). Since this involves a finite intersection of open sets, then U is also an open set. Moreover U K = since z U z B ri (x) for i = 1,..., N Therefore, z / B r1 (y 1 )... B rn (y N ) K. x U X K. Theorem D A closed subset of a compact set is itself compact. Proof Let F be a closed subset of a compact set K. Let the open sets A i cover F. Then K F (X F ) i A i (X F ). The right-hand side is the union of open sets since X F is open when F is closed. But K is compact and therefore a finite number of these open sets will do to cover it i.e., N K A i (X F ). Hence i=1 F N A i, i=1 since X F is outside F and doesn t cover it Exercises 1. Show that B is bounded if and only if R > 0, a X, B B R (a). 2. Show that any subset of a bounded set is bounded; the union of two (or finite number of) bounded sets is bounded; repeat the exercise with totally bounded sets.

32 5 Compactness J Muscat Show that a Cauchy sequence is bounded. Deduce that an unbounded sequence cannot converge. 4. Find an example of a bounded set B R and a continuous function f : B R such that fb is not bounded. 5. Show that if diama R and A B R (a) then A B 2R (a). 6. Show that, in R, a bounded set is totally bounded. (Hint: first show that a ball is totally bounded). 7. Theorem 5C shows that a compact set is bounded. Show further that a compact set is totally bounded. (Hint: consider ɛ-balls centred about all the points of the compact set). 8. Show that R is not compact by finding an open cover that does not have a finite subcover. 9. Show that [0, 1] is not compact in Q by finding an infinite open cover that does not have a finite subcover. (Hint: you need to consider an irrational number in between 0 and 1). 10. Let X be a space with the discrete metric of Ex Show that a subset K is compact K consists of a finite number of points. 11. Show that compact sets behave like closed sets i.e., any intersection and any finite union of compact sets are compact. (Hint: use theorem 1D) 12. (Cantor) Let K n be a decreasing sequence of nonempty compact sets. Show that if n K n = then X K n form an open cover of K 1. Deduce that n K n is compact and nonempty. Show moreover that if diamk n 0 then n K n consists of a single point. 13. * Show that for a metric space with the distance function D 2 of Ex , any set is bounded. Show that not all sets are totally bounded and deduce that a bounded set need not be totally bounded. 14. * Let K precompact mean that K is compact. Show that any subset of a precompact set is precompact, and that continuous images of precompact sets are precompact.

33 5.1 Compactness in R J Muscat Compactness in R Theorem E (Heine-Borel) The closed interval [a, b] is compact in R. Proof Let [a, b] i A i be an open cover of the set. Suppose there is no finite subcover, and we seek to obtain a contradiction. Divide the set [a, b] in two [a, (a + b)/2] and [(a + b)/2, b]. One of the two (and possibly both) does not admit a finite subcover: call it [a 1, b 1 ]. Repeat this process of dividing to get nested intervals [a n, b n ] of length (b a)/2 n which do not admit finite subcovers. Now (a n ) is an increasing real sequence bounded above by b, while (b n ) is decreasing bounded below by a. They therefore both converge, say, a n x and b n y with x y since a n b n. In fact x = y since a n x y b n and b n a n 0. This limit x is in the set [a, b] and is therefore covered by some open set A i0. This implies that we can surround x with an ɛ-ball x B ɛ (x) A i0. But a n x implies that there is an N such that a N B ɛ (x); similarly for b n, so that [a N, b N ] B ɛ (x) A i0. But [a N, b N ] was not supposed to be covered by a finite number of A i s. This implies that there must be a finite subcover to start with. Exercise: Generalize this to closed rectangles [a, b] [c, d] in R 2, by repeatedly dividing it into four sub-rectangles and adapting the same argument. Can you extend this further to closed cuboids in R 3? Corollary In R, a set is compact it is closed and bounded.

34 5.2 Compactness in a general Metric Space J Muscat 34 Proof That a compact set is closed and bounded is shown in Theorem 5C. Conversely, let F R be a closed and bounded set. By its boundedness F B R (a) [a, b]. Hence F is a closed subset of a compact set and by 5C this implies that it is itself compact. Exercise: Generalize this to show that any closed and bounded subset of R 2, R 3 etc. is compact. Corollary Let f : [a, b] R be a continuous function. Then its image f[a, b] := {f(x) : a x b} is bounded, and the function attains its bounds, i.e., c [a, b] x [a, b] f(x) f(c). Proof [a, b] is a compact set. Its continuous image f[a, b] is therefore also compact, and by 5C is bounded i.e., f[a, b] B R (0) i.e., f(x) R. Moreover compact sets are closed and so contain their limit points. In particular it contains sup f[a, b] i.e., sup f[a, b] = f(c) for some c [a, b]. More generally, a continuous function f : K R on a compact set will be bounded and attain its bounds. 5.2 Compactness in a general Metric Space We have seen that in the space of real numbers, a subset is compact if and only if it is closed and bounded. But this is not true for every metric space. Instead, we can prove the following: Theorem F A set is compact it is complete and totally bounded.

35 5.2 Compactness in a general Metric Space J Muscat 35 Proof The proof will be divided into several parts: Every infinite subset of a compact set K has a limit point in K: Let A be a subset of a compact set K, and suppose it has no limit points in K. This means that any x K is not a limit point, and so we can find balls B ɛ (x) which do not contain elements of A unless they are at the center. In particular, this is an open cover of K. But K is compact and so must have a finite subcover of these, K B ɛ1 (x 1 )... B ɛn (x N ). N N A K A B ɛi (x i ) = A B ɛi (x i ) {x 1,..., x N }. i=1 i=1 So A is a finite set. Equivalently, infinite subsets of K must have at least one limit point in K. The property that every infinite subset of a set has a limit point is termed the Bolzano-Weierstraß property (BW-compact for short). So we have shown here that compact sets are BW-compact. K BW-compact K complete: Let (a n ) be a Cauchy sequence in a given BW-compact set K. If the set { a n } is infinite, then it has a limit point x in K. Now, from exercises and , we can deduce that there is a subsequence of (a n ) which converges to x, and, since (a n ) is Cauchy, that the whole sequence (a n ) converges to x. Otherwise, the set { a n } is finite, and the sequence (a n ) must repeat itself. The fact that the sequence is Cauchy implies that the sequence must eventually be constant a n = a N for n N, so that it converges to a N in K. K compact K totally bounded: This is trivially true, since for any ɛ > 0, each x K belongs to B ɛ (x). Hence, K x K B ɛ(x), which reduces to a finite subcover since K is compact i.e., K N i=1 B ɛ(x i ), making K totally bounded. K complete and totally bounded K compact: This part of the proof is really a generalization of the Heine-Borel theorem. Let K be a complete and totally bounded set. Suppose K is covered by open sets V i, and that no finite number of these open sets is enough to cover K. Since K is totally bounded, we can find a finite number of balls of radius 1 which cover K, N K B 1 (y i ). i=1

36 5.2 Compactness in a general Metric Space J Muscat 36 If each of these balls had a finite subcover from the open cover { V i }, then so would K. So at least one of these balls cannot be covered by a finite number of V i s; let us call this ball B 1 (x 1 ). Now consider B 1 (x 1 ) K, also totally bounded. We can again cover it with a finite number of balls of radius 1/2, one of which does not have a finite subcover, say B 1/2 (x 2 ). Repeat this process to get a nested sequence of balls B 1/2 n(x n ), none of which have a finite subcover. The sequence (x n ) is Cauchy since d(x n, x m ) < 1/2 n (for m > n), and K is complete, hence x n x in K. But x is covered by some open set V i. Hence x B ɛ (x) V i. Moreover since 1/2 n 0 and x n x, we can find an N such that 1/2 N < ɛ/2 and d(x N, x) < ɛ/2, so that B 1/2 N (x N ) B ɛ (x) V i, which contradicts the way that the balls B 1/2 n(x n ) were chosen. Note that complete subsets are always closed, and totally bounded sets are always bounded. Hence the statement K compact K closed and bounded is really just a special case of this theorem. In general, however, closed and bounded sets are not complete and totally bounded, although we have seen that this is true in R N. Theorem G K is compact every infinite subset of K has a limit point in K. Proof One implication has already been proved. Moreover, we have also shown that BW-compactness implies completeness, so we only need to prove that BW-compact sets are totally bounded to show the converse. Let K be a (non-empty) BW-compact set. Let ɛ > 0 and let a 1 be any point in K. Either B ɛ (a 1 ) covers K (and we re done i.e., K would be bounded) or if not, there is some point a 2 with d(a 2, a 1 ) ɛ. Repeating the argument

37 5.2 Compactness in a general Metric Space J Muscat 37 for a 2, either B ɛ (a 1 ) B ɛ (a 2 ) cover K or else they omit some point, say a 3. Continuing like this we get a sequence of points a n each covered by B ɛ (a n ) satisfying d(a n, a m ) ɛ. Suppose this process does not stop and we end up with an infinite set { a n }. This has a limit point x since K is BW-compact, which implies that some of these a n are close to x within ɛ/2. But by the choice of the a n, d(a n, a m ) ɛ. This leads to a contradiction unless the process actually stops in a finite number of steps and the B ɛ (a n ) cover the whole of K Exercises 1. Show that a finite set of points is BW-compact. 2. Show directly that a BW-compact set is closed. 3. Show that [a, b] is not compact in Q by finding an infinite set of rational numbers in [a, b] that do not have a limit point. 4. Let γ : [0, 1] R 2 be a continuous curve. Show that there is a maximum and a minimum distance of points on the curve to the origin. Give an example to show that this is false if [0, 1] is replaced by (0, 1). 5. Show that the Bolzano-Weierstraß property is equivalent to the property that every sequence has a convergent subsequence. 6. Show directly that a continuous function maps a BW-compact set to a BW-compact set. 7. Show that if X, Y are compact metric spaces then so is X Y. (Hint: show that X Y is BW-compact.) 8. Let X be a complete metric space. Show that a subset K is compact it is closed and totally bounded. Deduce that for a complete metric space, a set is precompact (defined in ) if and only if it is totally bounded. 9. * Let X be a totally bounded space. Consider finite covers of 1 n -balls and let A be the set of all the centers of all these balls. Show that A is countable and A = X.

38 A Answers J Muscat 38 A Answers Ex (7) 0 = d(x, x) d(x, z) + d(x, z) = 2d(x, z) d(x, y) d(x, x) + d(y, x) = d(y, x) d(x, y) Ex (4) There are points in R 2, such as (x, ɛ) (a < x < b), which are arbitrarily close to points in the interval (x, 0). (5) For example, n ( 1/n, 1/n) = { 0 }. Ex (5) For example, (i) [0, 1[ and (ii) R. (6) For example, (i) n=1 [0, 1 1/n] = [0, 1), (ii) n=0 { n } = N. (7) (i) open, (ii) closed, (iii) neither, (iv) both, (v) neither, (vi) closed. Ex (1) Z = Z, Q = R. (3) Yes to both, e.g. N in R. Ex (2) For example, f(x) := x 2 with U = ( 1, 1), then fu = [0, 1). (3) For R (or R N ), never. But on disconnected spaces, such as Z, χ A may be continuous (e.g. A = { n }). { x x 0 (10) For example, take X := R (0, 1] and f(x) := x 1 x > 1 ; then f : X R is bijective and continuous but f 1 is discontinuous at 0. Ex (8) For example, B 4 (2). Ex (4) For example, f(x) := 1/x maps (0, 1) (1, ). Ex (4) For example, any spiral curve, such as γ(x) := e 1/x (cos(1/x), sin(1/x)), approaches but never reaches the origin, so has no minimum distance.

39 B Summary J Muscat 39 B Summary B.1 Definitions 1. A distance (or metric) on a metric space X is a function d : X 2 R + with the properties (x, y) d(x, y) i d(x, y) = 0 x = y ii d(y, x) = d(x, y) iii d(x, y) d(x, z) + d(y, z) 2. A ball is a set B r (a) = {x X : d(x, a) < r}. 3. An open set, A, in X is a set such that x A ɛ > 0 B ɛ (x) A. Every point of A can be surrounded by a ball inside A. 4. A closed set, F, in X is a set such that X F is open in X. Every point not in F can be surrounded by a ball outside F. 5. A limit point of a set A is a point b such that ɛ > 0 a b a A B ɛ (b) Every ball around b contains points of A apart from b. 6. The closure of a set A is the set A = A {limit points of A}. 7. A sequence (x n ) converges to x, x n x means ɛ > 0 N n > N x n B ɛ (x). The sequence of points eventually comes arbitrarily close to the limit. 8. A Cauchy sequence is one such that ɛ > 0 N n, m > N d(x n, x m ) < ɛ. Eventually the points get arbitrarily close to each other.

40 B.1 Definitions J Muscat A complete metric space is one in which every Cauchy sequence converges. 10. A continuous function f : X Y is one for which x X ɛ > 0 δ > 0 y B δ (x) f(y) B ɛ (f(x)) Points sufficiently close to x are mapped to points arbitrarily close to f(x). 11. Two metric spaces X and Y are homeomorphic when there is a map φ : X Y which is bijective (1-1 and onto), continuous, and its inverse is continuous. 12. A connected set is one such that A U V with U, V open sets A U V or A U = or A V =. It cannot be separated by two non-trivial open sets. 13. A bounded set B is one for which R > 0 x, y B d(x, y) < R. It can be covered by a single ball. 14. A totally bounded set B is one for which ɛ > 0 a 1... a N B It can be covered by a finite number of ɛ-balls. 15. A compact set K is one for which N B ɛ (a i ). i=1 K i N A i i = i 1,..., i N : K A ik. k=1 Any cover of open sets (balls) has a finite subcover. 16. A BW-compact set K is one for which every infinite subset of K has a limit point in K.

41 B.2 Main Theorems J Muscat 41 B.2 Main Theorems 1. Balls are open sets. 2. Any union and any finite intersection of open sets is open. 3. A set is closed when it contains its limit points. 4. A Cauchy sequence in a set A converges to a limit in A, for a complete metric space; closed sets of complete spaces are complete. 5. f : X Y continuous and V open in Y f 1 V open in X. 6. A function f is continuous f(lim n x n ) = lim n f(x n ). 7. The connected sets of R are the intervals. 8. A continuous function maps a connected set to a connected set. 9. The union of non-disjoint connected sets is connected. 10. Compact sets are closed and bounded. 11. A continuous function maps a compact set to a compact set. 12. (Heine-Borel theorem) The compact sets of R n are the closed and bounded sets. (In particular [a, b] is compact.) 13. A set is compact it is BW-compact it is complete and totally bounded.

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