First Concepts of Topology, Manuais, Projetos, Pesquisas de Matemática

Baixe First Concepts of Topology e outras Manuais, Projetos, Pesquisas em PDF para Matemática, somente na Docsity! FIRST CONCEPTS OF TOPOLOGY THE GEOMETRY OF MAPPINGS OF SEGMENTS, CURVES, CIRCLES, AND DISKS W.G.CHINN ano N.E.STEENROD The Mathematical Association of America New Mathematical Library FIRST CONCEPTS OF TOPOLOGY THE GEOMETRY OF MAPPINGS OF SEGMENTS, CURVES, CIRCLES, AND DISKS Illustrations by George H. Buehler ©Copyright 1966 by The Mathematical Association of America (Inc.) All rights reserved under International and Pan-American Copyright Conventions. Published in Washington, D.C. by The Mathematical Association of America Library of Congress Catalog Card Number: 66-20367 Print ISBN 978-0-88385-618-5 Electronic ISBN 978-0-88385-933-9 Manufactured in the United States of America Note to the Reader This book is one of a series written by professional mathematicians in order to make some important mathematical ideas interesting and under- standable to a large audience of high school students and laymen. Most of the volumes in the New Mathematical Libray cover topics not usually included in the high school curriculum; they vary in difficulty, and, even within a single book, some parts require a greater degree of concentration than others. Thus, while the reader needs little technical knowledge to understand most of these books, he will have to make an intellectual effort. If the reader has so far encountered mathematics only in classroom work, he should keep in mind that a book on mathematics cannot be read quickly. Nor must he expect to understand all parts of the book on first reading. He should feel free to skip complicated parts and return to them later; often an argument will be clarified by a subsequent remark. On the other hand, sections containing thoroughly familiar material may be read very quickly. The best way to learn mathematics is to do mathematics, and each book includes problems, some of which may require considerable thought. The reader is urged to acquire the habit of reading with paper and pencil in hand; in this way mathematics will become increasingly meaningful to him. The authors and editorial committee are interested in reactions to the books in this series and hope that readers will write to: Anneli Lax, Editor, New Mathematical Library, NEW YON UNIVERSITY, THE COURANT INSTITUTE OF MATHEMATICAL SCIENCES, 251 Mercer Street, New York, N. Y. 10012. The Editors NEW MATHEMATICAL LIBRARY I . 2. 3. 4. 5. 6. 7. 8. 9. 10. Numbers: Rational and Irrational by Ivan Niven What is Calculus About? by K K Sawyer An Introduction to Inequalities by E. E Beckenbach and R. Bellman Geometric Inequalities by N. D. Kazarinoff The Contest Problem Book I Annual High School Mathematics Examinations 195CL-1960. Compiled and with solutions by Charles 7: Salkind The Lore of Large Numbers by f? J. Davis Uses of Infinity by Leo Zippin Geometric Transformations I by 1. M. Yaglom, translated by A. Shields Continued Fractions by Carl D. Olds Replaced by NML-34 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. Hungarian Problem Books I and 11, Based on the Eotvos Competitions Episodes from the Early History of Mathematics by A. Aaboe I8944905 and I90Gl 928, translated by E. Rapaport Groups and Their Graphs by E. Grossman and K Magnus The Mathematics o f Choice by Ivan Niven From Pythagoras to Einstein by K. 0. Friedrichs The Contest Problem Book I1 Annual High School Mathematics Examinations 1961-1965. Compiled and with solutions by Charles 7: Salkind First Concepts of Topology by W G. Chinn and N . E. Steenrod Geometry Revisited by H. S. M. Coxeter and S. L. Greitzer Invitation to Number Theory by Oystein Ore Geometric Transformations I1 by I. M. Yaglom, translated by A. Shields Elementary Cryptanalysis-A Mathematical Approach by A. Sinkov Ingenuity in Mathematics by Ross Honsberger Geometric Transformations 111 by I. M. Yuglom. translated by A. Shenitzer The Contest Problem Book 111 Annual High School Mathematics Examinations 19661972. Compiled and with solutions by C. 7: Salkind and J . M. Earl Mathematical Methods in Science by George fdlya International Mathematical Olympiadsl95%1977. Compiled and with solutions by S. L. Greitzer The Mathematics of Games and Gambling by Edward K fuckel The Contest Problem Book IV Annual High School Mathematics Examinations 197S1982. Compiled and with solutions by R. A. Artino. A. M. Gaglione. and N. Shell The Role of Mathematics in Science by M. M. Schiffer and L. Bowden International Mathematical Olympiads I 9 7 w 985 and forty supplementary problems. Compiled and with solutions by Murray S. Klamkin Riddles of the Sphinx by Martin Gardner U.S.A. Mathematical Olympiads 1972-1986. Compiled and with solutions by Murray S. Klamkin Graphs and Their Uses by 0,vstein Ore. Revised and updated by Robin J. W s o n Exploring Mathematics with Your Computer by Arthur Engel Game Theory and Strategy by Philip D. Strafln, JI: Episodes in Nineteenth and Twenthieth Century Euclidean Geometry by Ross Honsberger The Contest Problem Book V American High School Mathematics Examinations and American Invitational Mathematics Examinations 198S1988. Compiled and augmented by George Berzsenyi and Stephen B. Maurer Other titles in preparation Introduction Our purpose in writing this book is to show how topology arose, develop a few of its elements, and present some of its simpler applications. Topology came to be recognized as a distinct area of mathematics about fifty years ago, and its major growth has taken place within the last thirty years. It is the most vigorous of the newer branches of mathe- matics and has been producing strong repercussions in most of the older branches. It got its start in response to the needs of analysis (the part of mathematics containing calculus itnd differential equations). However, topology is not a branch of anidysis. Instead, it is a kind of geometry. It is not an advanced form of geometry such as projective or differential geometry, but rather a primitive, rudimentary form-one which underlies all geometries. A striking fact about topology is that its ideas have pene- trated nearly all areas of mathematics. In most of these applications, topology supplies essential tools and concepts for proving certain basic propositions known as existence theorems. Our presentation of the elements of topology will be centered around two existence theorems of analysis. The first, given in Part I, is funda- mental in the calculus and was known long before topology was recog- nized as a subject. In working out its proof, we shall develop basic ideas of topology. This will show how topology got started, and why it is useful. Our second main theorem, given in Part 11, is a generalization of the first from one to two dimensions. I n contrast to the first, a. topological concept is needed for its formulation. Its proof exhibits that peculiar blending of numerical precision und rough qualitative geometry so char- acteristic of topology. Both theorems have numerous applications. We shall present those having the strongest topological flavor. The beginnings of topology can be found in the work of Karl Weier- strass during the 1860’s in which he analyzed the concept of the limit of a function (as used in the calculus). In this endeavor, he reconstructed 1 2 FIRST CONCEPTS OF TOPOLOGY the real number system and revealed certain of its properties now called “topological”. Then came Georg Cantor’s bold development of the theory of point sets (1874-1895); it provided a foundation on which topology eventually built its own house. A second aspect of topology, called com- binaforiul or algebraic topology, was initiated in the 1890’s by the re- markable work of Henri PoincarC dealing with the theory of integral calculus in higher dimensions. The first aspect, called set-theoretic topol- ogy, was placed on a firm foundation by F. Hausdorff and others during the period 1900-1910. A union of the combinatorial and set-theoretic aspects of topology was achieved first by L. E. J. Brouwer in his investiga- tion (1908-1912) of the concept of dimension. The unified theory was given a solid development in the period 1915-1930 by J. W. Alexander, P. L. Alexandrov, S. Lefschetz and others. Until 1930 topology was called anulysis situs. It was Lefschetz who first used and popularized the name fofiology by publishing a book with this title in 1930. Since 1930 topology has been growing at an accelerated pace. To em- phasize this point we shall mention a few of topology’s achievements. It invaded the calculus of variations through the theory of critical points developed by M. Morse (Institute for Advanced Study, Princeton). It reinvigorated differential geometry through the work on fibre bundles by H. Whitney (Institute for Advanced Study, Princeton), the work on differential forms by G. de Rham (Lausanne), and the work on Lie groups by H. Hopf (Zurich). It enforced a minor revolution in modern algebra through the development of new foundations for algebra and a new branch called homological algebra. Much of this work is due to S. Eilenberg (Columbia University) and S. MacLane (University of Chicago). Topology gave a new lease on life to algebraic geometry via the theory of sheaves and cohomology, and i t has found important a p plications to partial differential equations through the works of J. Leray (Paris) and M. Atiyah (Oxford). Applications of topology have been made to sciences other than mathe- matics, but nearly all of these occur through some intervening mathemati- cal subject. For example, the changes topology has made in differential geometry have initiated topological thinking in relativity theory. Topol- ogy has become a basic subject of mathematics, in fact, a necessity in many areas and a unifying force for nearly all of mathematics. When a non-mathematician asks a topologist, “What is topology?”, “What is it good for?”, the latter is a t a disadvantage because the ques- tioner expects the kind of answer that can be given to analogous questions about trigonometry, namely, trigonometry deals with the determination of angles and is used to solve problems in surveying, navigation, and astronomy. The topologist cannot give such a direct answer; he can say, correctly, that topology is a kind of geometric thinking useful in many areas of advanced mathematics, but this does not satisfy the questioner who is after some of the flavor of the subject. The topologist can then bring out paper, scissors and paste, construct a Mobius band, and cut INTRODUCTION 3 along its center line, or he can take some string and show how three separate loops can be enlaced without being linked. If he feels energetic, he can demonstrate how to take off his vest without removing his coat. These are parlor tricks, each based on a serious mathematical idea which would require a t least several hours to explain. To present these tricks without adequate explanation is to present a caricature of topology. To appreciate topology i t is necessary to take the viewpoint of the mathematician and explore some of its successful applications. Most of these applications have in common that they occur in the proof of an existence theorem. An existence theorem is one which asserts that each of a certain broad class of problems has a solution of a particular kind. Such theorems are frequently the basic structure theorems of a subject. One of our principal aims is to demonstrate the power and flexibility of topology in proving existence theorems. The existence theorem we shall prove in Part I answers the question: When can an equation of the form f (x) = y be solved for x in terms of y ? Here f (x) stands for a function or formula (such as 2- 4 1 + x* ) defined for real numbers x in some interval [a , b ] (such as [ 2 , 41 ), and y denotes a real number (such as v). The problem is: Does there exist a number x in the interval [a, b] such that f (x) = y ? Formu- lated for the example it becomes: Is there an x between 2 and 4 such that 9- 4- = u? 33 We emphasize that we are not asking for methods of finding the value or values of x in special cases. Instead we are seeking a broad criterion, applicable to each of many different problems, to determine whether or not a solution exists. Once the criterion assures us that a particular prob- lem has a solution, we can start hunting for it with the knowledge that the search is not in vain. The criterion given by our main theorem (stated in Section 1) requires the notion of the continuity of a function (defined in Section 3). The proof of the theorem (given in Sections 2-8) is based on two topological prop- erties of the interval [a , b] called compactness and connectedness. We give these concepts a thorough treatment because they are basic in modern mathematics. The main theorem of Part I1 is an existence theorem which answers the question: When can a pair of simultaneous equations f(x, y) = a and g(x, y ) = b be solved for x and y in terms of a and b ? A familiar example of such a problem is the pair of simultaneous linear equations x - 2y = 3 and 3 x + y = 5 ; these can be solved readily by elimination. Here is a more difficult problem of the same type: Find a pair of numbers x , y satisfying the two equations 6 EXISTENCE THEOREMS IN DIMENSION 1 [I curve we take whose x-coordinate is less than 2, a higher point can be found by taking one whose x-coordinate is nearer to 2. To avoid such a situation, we shall include the end values -1 and 2. Then 9 + 1 is less than or equal to 5 for all x such that -1 I x I 2, and the function has the maximum value 5 when x = 2. Consider now the problem of starting with a y-value and trying to solve the equation x2 + 1 = y for a corresponding x-value in the interval -1 to 2. If the y-value exceeds the maximum 5, there is surely no solution. This is likewise the case if y is less than the minimum 1. However, if y is between 1 and 5, there is a solution. We can see this from the graph by drawing a horizontal line at a height above the x-axis equal to the y-value. If the line is too high or too low, it does not meet the curve. At a height between 2 and 5, it cuts the curve once, and between 1 and 2, it cuts twice. (A formula for x in terms of y i s x = d y - 1 . ) Figure 1.1 Figure 1.2 As a second example, let f ( x ) be defined by the formula 2x xp +-1 for all x-values such that -3 I x 5 3 . Its graph is shown in Fig. 1.2. It is readily seen by inspecting the equation 2x y = - xS+1 that a positive x-value gives a positive y-value, and a negative x-value gives a negative y-value. Moreover, changing the sign of the x-value changes only the sign of the y-value; hence the curve is symmetric with respect to the origin. The highest point on the curve occurs when x is 1 and then y is 1. To see that this is so, we do a bit of algebraic juggling of the difference 1 - f ( x ) : 911 FIRST EXISTENCE THEOREM 7 x l + 1 - 2x ( x - 1)’ P 2x $ + I 2 + l 2 + 1 * l - f ( x ) = 1-- = Since this last expression can never be negative, it follows that 1 - f ( x ) 2 0, whence f ( x ) 5 1. By symmetry, the lowest point on the curve occurs a t x = -1 and y = -1. It is evident therefore that the equation f ( x ) = y has no solution if y > 1 or if y < -1, but for each y-value such that -1 5 y 5 1, the equation can be solved. (Multiplying both sides of the equation by a? + 1, and solving the resulting quadratic gives x = (1 - d F y ) / y . ) Figure 1.3 Figure 1.4 One might try to generalize from the two examples treated thus far and conjecture that, if f ( x ) is any function defined for z-values such that a 5 x 5 b, then f ( x ) has amaximumvalue M, aminmum value m, and for each y-valuesuch that m 5 y 5 M the equation f ( x ) = y has a solution. Let us test this conjecture by picturing the graphs of several more functions, recalling that functions can be defined by specifying their graphs in any manner we please. :I:”.--; m - . . . : . . . - . - - - : - L q 1-__ /;q Figure 1.5 Figure 1.6 If the graph for f ( x ) is a smooth curve as shown in Fig. 1.3, the con- jecture appears to be true..A horizontal line at a height y between the heights m and M must intersect the curve. Even if the curve had some corners, as in Fig. 1.4, the conjecture still appears to be correct. However, if there is a break in the curve as in Fig. 1.5, the conjecture is seen to be false, because some horizontal lines pass through the break without meeting the graph. Functions whose graphs have such breaks do arise in mathematics in natural ways. They are called discorriinwus functions. 8 EXISTENCE THEOREMS IN DIMENSION 1 [I The graph of such a function may not even have a highest or lowest point as in the example of Fig. 1.6 where, at the break in the graph at x = c, the point (c, f ( c ) ) on the graph is at r. We are prepared now to state the main theorem. THEOREM. If a function j (x) is defined for all red numbers x in some closed interval [a, b], i f f (x) has real numbers as values, and ij it is con- tinuous, then there is a minimum value m and a maximum value M of the function, and, for each y-value in the closed interval [m, MI, the equutwn j (x) = y has at least one solution x in the intend [a, b]. The statement of the theorem is sometimes abbreviated thus: If the real-valued function f(x) is defined and continuous for a 5 x I b, then it has a minimum value, a maximum value, and takes on all values between. The expression “closed interval” means that the endpoints a, b are included as points of the interval, that is, the limitation on x is a I x I b. The expression “open interval” means that the endpoints are excluded. We shall denote the closed interval by [a, b] and the open interval by (a, b). A “half-open” interval includes one endpoint but not the other, thus (a, b ] means a < x 2 b, and [a, b) means a I x < b . How does the theorem help us to decide whether or not we can solve f ( x ) = y in the case of a particular function f(x), known to be con- tinuous, and a particular y-value? If we can determine the minimum m andmaximum M of f(x), we haveonly toaskif m 5 y 5 11.1. In many cases it may be dificult to find m and M. However, i t is usually easy to compute a number of values of the function. If, for some x-value c, we observe that f(c) < y, and for another d, that y < f ( d ) , then the theorem asserts that there is an x in [c, d] (or [d, c] if d < c )such that f(x) = y. For example, if f(x) is d - d m , wehave f(0) = -1, and f(2) = 5. Therefore d - 4- = 2 has a solution in the interval [ O , 21. We must emphasize that the importance of the theorem lies in its generality. It tells us what we can always count on finding in a great variety of circumstances. In numerous special cases like x2 + 1, i t is of no use to us because the facts we are after are readily available. The theorem shows its power as soon as it is applied to complicated functions. But more important, it is the first theorem of a general theory of con- tinuous functions. It must also be emphasized that our formulation of the theorem is in- complete. We have not defined precisely what is meant by the continuity of f ; we have given an intuitive description based on geometric pictures -the graph of f is a curve with no breaks-but this is only a substitu- tion of one undefined term for another. The next two sections lead up to a precise definition of continuity. 821 SETS AND FUNCTIONS 11 it may be larger. Let us consider first some familiar examples of geo- metrically defined functions. A translation of the plane is a function j R2 + R2. It is the result of a rigid and uniform motion in which each point traverses a line segment or vector; the vectors are the paths of the various points, are all parallel, and have the same length and direction. A translation is specified by the vector for just one point because the others can be constructed from it. Thus, if we know that f carries the point p into the point q, then it will carry a point p' into the point q' such that p , q, p', q' form a parallelogram. For example, if f carries the origin (0,O) into the point (2, -3), then it carries (XI, x2) into the point (XI + 2, XP - 3). There- fore f is given by the formulas yl = XI + 2 and y2 = xs - 3. A rotation of the plane is a function f: R2 -+ R2, again resulting from a rigid motion, this time about a fixed point z called the center of the rotation. Each circle with center z is carried by f onto itself; and each ray (half-line including initial point) issuing from a is carried onto another ray. The angle formed by these two rays is called the angle of rotation, and its measure in degrees does not depend on the initial ray. The rotation is specified by its center and angle of rotation. A rejection in a line L of R2 is a function f: R2 + R2; it is a rigid motion that leaves fixed each point of L and interchanges the two sides of L. It is most easily visualized as the result of the rotation in space of the plane about the line L through 180". It can be shown that the result of any rigid motion of the plane onto itself is a translation, a rotation, a reflection, or a reflection followed by a translation. The shapes and sizes of the configurations in the plane are not altered; only their positions and orientations may be changed. These functions are the congruences of elementary geometry. A similarity of the plane is a function f: RS 4 R2 which alters all lengths by the same factor r. As an example, choose a point z of R', set fa = z, and, for each other point x # z, define fx to be the end- point of the segment (or vector) issuing from z which has the same direction and twice the length of the segment from z to x. This f alters all lengths by the factor r = 2. Such an f, with r > 1, is called an expansion centered at a. When r < 1, f is called a contraction. A similarity with r = 1 is one of the rigid motions described above. A similarity with r # 1 always has a fixed point z, and it is the result of a contraction or expansion with center z followed by a rotation about z or a reflection in some line through z. A similarity always carries straight lines into straight lines, and it does not change the measures of angles between lines. It can alter the size, location, and orientation of a configuration, but it does not alter its shape. 12 EXISTENCE THEOREMS IN DIMENSION 1 P Let L be a line in the plane R2. The perpenduuhr projection f: R2 j- L assigns to each point x of R2 the foot fx of the perpendicular from x to L. Let S be the surface of a sphere in R8 with center a. The radial Projection j R3 - a + S assigns to each point x of R3 different from B the point fx where the ray from a through x intersects S. Figure 2.1 Figure 2.2 Figure 2.3 The foregoing examples indicate in part the kind of functions that will interest us. In order to picture such functions and to make significant statements about them, one uses the notions of images and inverse images. If f: X + Y and A C X, then the image of A under f, denoted by fA, is the subset of Y of values fx for all x E A. Precisely, to say that a point y E Y is in fA means that there is a t least one x E A such that fx = y. One can think of fA as the result of applying f to all of A. For example, under a rigid motion or a similarity f: R2 --+ R2, any straight line L of R2 has as its image a straight line fL of R2. Under perpendicular projection f: R2 -+ ‘L, each line segment A of R2 has as its image fA a line segment of L (see Fig. 2.1), or a single point of L in case A is perpendicular to L (Fig. 2.2). Under radial projec- tion f: R3 - z --+ S, each straight line L of R8 not passing through z has as its image fL a great semi-circle of S with endpoints excluded (see Fig. 2.3). The origin of our use of the word “image” in this sense is evident if we consider any reflection of R’ in a plane. If f: X -+ Y and B C Y , then the inverse image of B under f, denoted by f-lB, is the subset of X consisting of points x such that fx E B. Under perpendicular projection R2 + L, the inverse image of a single point y of L is the line perpendicular to L a t y, and the inverse image of a line segment is the strip between the two lines per- pendicular to L a t the ends of the segment. Under radial projection Ra - z + S, the inverse image of a point y of S is the ray from z through y with z deleted. The inverse image of a circular region on S is a solid cone with the vertex deleted. 621 SETS AND FUNCTIONS 13 Let f: X + Y and let A be a subset of X. Then the image of A is also in Y. If B is a subset of Y such that B contains f A , then the function g: A + B defined by gx = fx for all x E A is called the restriction of f to A and B. Most frequently we shall need to restrict only the domain of j , and, in this case, the restriction of f to A and Y is denoted by f I A (read as: f restricted to A ) . For example, if f: R2 + R2 is a translation, and L is a line in RP, then f I L displaces L to a parallel line. If we have two functions f: X + Y and g: Y + 2, then we can compose the two functions to form a new function denoted by g f : X + 2; it attaches to each x in X the element g ( fx) in 2. For example, let f and g be translations R2 + R2 where f moves each point 2 units toward the east, and g moves each point 2 units to the north. Then gj is the translation moving each point 242 units to the northeast. As a second example, let f and g be numerical functions R + R given by the formulas jx = x2+1, gx = 2 - x . Then the compositions gf and jg can be formed, and they are given by the formulas gjx = g ( f x ) = 2 - ( X ' + l ) = 1 - 9 f g x = f ( g x ) = (2 - x)2 + 1 = 5 - 4% + 39. Some simple functions are so inconspicuous that one must be reminded of their presence. We call attention first to the constant functions: a function f: X + Y is a constant function if the image jX is a single point of Y. There is one constant function for each point of Y. Next, we mention the identity functions: for each set X , the function j: X+ X such that f x = x for every x E X is called the identity function of X . Finally, if A C X , the function j: A +. X such that fx = x for every x E A is called the inclusion function. Clearly, the inclusion is the re- striction of the identity function to the subset. Any restriction of a con- stant function is constant. A function f: X + Y is called one-to-one (abbreviated: 1-1) if each point of Y is the image under j of one and only one point of X.t In this case, the function which assigns to each point y in Y the unique point x in X such that f x = y is called the inverse function of f and Some authors say that j : X-+ Y is one-to-one if each point of Y is the image of at most one point of X , allowing thereby that j X may not be all of Y; and in case j X = Y , they say that j is one-to-one and onto. 16 EXISTENCE THEOREMS IN DIMENSION 1 a (d) The image of the equator is a circle on the-plane. How is the radius (e) What is the inverse image of a ray through the south pole? 6. If f is the translation RI --+ R2 given by the formulas n = X I + 3, y2 = x2- 4, and g is thereflectiongiven by = -21 and y2 = ~ 2 , find the formulas for the compositions gf, fg, for the inverses f-l, g-1, (gf)", and the compositions flgl and g'f-'. Is (gf)-l = g'f-? of the image circle related to the radius of the sphere? 7. For the following, let f: X+ Y. (a) If A , B are subsets of Y, prove that f l (A U B) = f-lA U f-lB and f-l(A tl B ) = f-lA n f-lB. (b) Find f-lY andf-lg. (c) If A C B, how are the inversesf-lrl andf1B related? 8. If f: X 4 Y and g: Y+ 2, show that (gf)-'C = f-'(glC) for each C C 2. If f and g are 1-1, show that gl is 1- 1, and (d)" = f-lgl. 3. Neighborhoods and continuity If x and y are two points of Rn, thedistance from x to y means the ordinary straight line distance. It is denoted by d(x, y), and in terms of the coordinates of x and y, it can be computed from the following formula based on a generalization of the theorem of Pythagoras: d(x, y ) d ( x 1 - ~ 1 ) ' + ($2 - y ~ ) * + *'* + ( x n - Y n F * In case $2 = 1, the formula reduces to d ( x , y) = I x - y (the absolute value of x - y ). Actually we shall not use this formula directly. Instead, we shall use only certain properties of the distance function which can be proved using the formula. These are well-known properties, and we list them now without proof. First, if x and y are different points then d ( x , y) > 0. Secondly, d (x, x) = 0 for all x. Next, for all pairs of points x , y, the distance is symmetric: d (x, y) = d (y, x ) . Finally, for any three points x, y, 2, we have d ( x , 8) I d(x, Y) + d b , z ) - This last is called the "triangle inequality" because it asserts that the sum of the lengths of two sides of a triangle exceeds the length of the third side. 831 NEIGHBORHOODS, CONTINUITY 17 Recall that in our preliminary examination of various graphs in Section 1, whether or not the graph had any breaks was vital to our conclusion; it is for this reason that the main theorem specifies that the function must be continuous. The intuitive description of continuity was based on geometric pictures; the precise definition will be given in terms of another concept that we shall now define: the concept of a neighborhood. DEFINITION. Let X beasubsetof R", let x beapoint af X, and let I be a positive real number. Then we define the neighborhood of x in X of radius r to be the set of all points of X whose distance from x is less than I. The neighborhood is denoted by N ( x , I, X ) , and this is abbreviated to N (x , I) whenever the X is understood. Figure 3.1 For example, if X = R" and n = 2, then N ( x , I, R2) is just the interior of a circle with center x and radius r. Similarly the neighbor- hood of x in R8 of radius I, N ( x , I, Rs), is the interior of a sphere with center 2 and radius I, In case n = 1, then N ( x , Y , R ) is just the open interval (x - I, x + r ) with midpoint x and length 2r. Whenever X is not all of R", then N ( x , I, X) is just that part of X which lies inside N(X, I, Rn) (see Fig. 3.1). It is the intersection of X with the neighborhood in Rn: N ( x , r , X) = X n N ( x , r , R") . We come now to the important concept of the continuity of a function. DEFINITION. Let f: X 4 Y be a function such that X C Rm and I' C R" and let x E X . We shall say that f is continuow at x if, for each neighborhood of f x in Y , there is some neighborhood of x in X whose image under f lies in the neighborhood of j x under consideration. To express this condition briefly, we follow the customary notation of the calculus and denote the radii of these neighborhoods by c and 6. Then the requirement can be restated: for each positive number E , there is a positive number 6 such that fly (x , 6, X ) c N (f., e , Y ) We shall say that f is colztintwus if it is continuous at every point of X. 18 EXISTENCE THEOREMS IN DIMENSION 1 [I If we interpret 6 and c as measures of nearness, then the definition may be paraphrased. one can make fx’ be near to fx by requiring x‘ to be sufficiently near to x. An even rougher paraphrase is: a small change in x produces a small change in fx. In topology we are primarily interested in continuous functions, and we abbreviate the expression “continuous function” by the word “mapping” which in turn is often shortened into “map”. Thus stereo- graphic projection is a mapping of the surface of a sphere with the pole deleted onto a plane. However, in order to illustrate what the definition of continuity means, let us examine several examples of discontinuous functions. L Figure 3.2 As a first example, let f: R2 + R2 leave fixed each point of the plane except for a single point, say p, and let fp = q be some other point. To be specific, we can take p to be the origin of coordinates (0,O) and q to be (1, 0). Then f is continuous at all points except p. To see that it is not continuous at p , take B to be half of d(p , q), so that N(q, c) = N(q , 4) is the interior of a circle of radius 3 about (1,O) (see Fig. 3.2). Then no neighborhood of p is mapped into N(q , #), for each neighborhood of p contains points not in N ( q , #), and since these are left fixed by f, their images are not in N(q , 3 ) . (Fig. 3.2 shows the neighborhood A’ (p, f) ; among its points only p has its image j p in N(q , 3) ). Since for e = 3 there is no corresponding 8 > 0 such that j N @ , 8) C N ( q , #), f is not continuous at p. The intuitive geometric picture is that f rips the point p out of the plane, and then pastes it down on q. L L f L Figure 3.3 PI NEIGHBORHOODS, CONTINUITY 21 can nevertheless be obtained with a bit of algebra by considering various similar triangles (see Exercise 9 below). Geometrically, we can obtain a 6 thus: Let t be a point of intersection of S and the sphere with center f x and radius E. The cross-section on the plane through the three points z, f x , and t is shown in Fig. 3.4. L e t 6 be the perpendicular distance from x to the line zt. Then each point of the interior of the sphere N ( x , 6) projects into N ( f x , E, S). Exercises 1. For which triples of points x, y, z will equality hold in the triangle in- equality; that is, d(x, 2) = 4 x , Y) + d ( y , 211 2. If XI is a point of the neighborhood N ( x , r, X ) , show that there is an / > 0 such that N(x’ , r’, x) c N ( x , r , X ) . What is the largest value of r’ which assures this? 3. If, in testing the continuity of a function f: X -+ Y at a point x, a number 6 > 0 has been found which does for E = 4, why will the same 6 do for all E 2 $? 4. If, as in Exercise 3, a 6 > 0 has been found which does for E = 4, why will any smaller value of 6 do for E = $? 5. Divide the plane R2 into two parts A and B where A consists of all points inside and on a circle C with center z and radius 1, and B is the complement of A in R2. Define f : R2 -+ R2 by f l A rotates A about its center through an angle of 90°, and f leaves fixed each point of B. Where is f continuous and where is it discontinuous? At a point of discontinuity, for what values of E are there no corresponding a’s? 6. If L is a line or a plane in R3, state why the perpendicular projection j: R3+ L is continuous. 7. Let S be the spherical surface of radius 1 with center a t the origin of RJ, let p = (0, 0, 1) be the north pole of S , and let f: S - p - + R2 be the stereographic projection from p of S - p onto the equatorial plane. Construct a diagram which shows that f is continuous. Over what part of S - p is f a contracting function? Show that f is 1- 1, and that f-l: R?-+ S - p is also continuous. What is the image under f of the deleted neighborhood N ( p , r , S ) - p for r-values less than l ? Why is it impossible to define fp so that the extended function is continuous? 22 EXISTENCE THEOREMS IN DIMENSION 1 [I 8. If f and g are continuous functions from an interval [a, b] to R, show that f + g and f - g are continuous. Hint: To demonstrate the continuity a t x of hx = f x + gx and of kx = f x - gx, estimate I hx - hx’ 1 and I kx - kx’ 1 with the help of the triangle inequality I(f..tgx)- (fx’fgx’)I 5 Ifx-fx’l+Igx-gx’I for all x, x’ E [u, b]. 9. Let f: R3 - a+ S be the radial projection onto the surface of a sphere from its center s (see Fig. 3.4). Let the radius of S be 1. Show that 6 in the figure is given as a function of e by 6 = d € d m , where d is the distance from x to z. Hint: Drop a perpendicular from s to the chord from fx to t, let B denote half the angle a t z determined by f x and t, and use the identity sin 20 = 2 sin B cos 0. 4. Open sets and closed sets Our objective is to deiine and study a special class of subsets of a set X in R” called open sets of X. They will play a fundamental role in our subsequent work, because the various topological properties of X we shall discuss are readily expressible in terms of the open sets. Also, the condition for a function to be continuous takes on a very simple form when open sets are used. It is not easy to see in advance why the notion of open set should be an important concept. It is a historical fact that it gained recognition slowly. During the early development of topology (1900-1930), a variety of different approaches to the subject were devised and worked out. Attached to these are concepts with names such as: neighborhood spaces, metric spaces, limit points, sequential limits, and closures. At the time it was not clear that these approaches were equivalent; nor could one predict the direction of development and ultimate form of topology. Not until the end of this period did i t gradually become clear that the con- cept of open set is a simple and flexible tool for the investigation of all topological properties. Since then this concept has provided the preferred approach. DEFINITION. Let X be a subset of Rn. A subset U of X is called an open set of X if, for each point x of U, there is some neighborhood of x in X which lies in U. The condition may be restated: for each x E U , there is a number I > 0 such that N ( x , I, X) C U. 641 OPEN SETS, CLOSED SETS 23 As we shall see shortly, open sets are easily found and occur in great variety. Our first class of examples consists of neighborhoods: All neigM0rhood.s are open sets. Figure 4.1 Let xo E X and ro > 0 be given. To prove our assertion we must show that N (xo, 10, X) is an open set of X (see Fig. 4.1) ; this we shall do by showing that for each point x E N ( a , ro, X), there is a positive number I such that the neighborhood N ( x , I, X) is contained in N(xo, YO, X). Set I = ro - d ( x , xo). Now x E N(xo, ro, X) means that d ( x , xo) < ro; hence I must be positive. Let y E N ( z , I, X); then d(x , y) < r. The triangle inequality gives d(xo, Y ) I d(%, x ) + d(x, r) , and since d ( x , y) < I, we have d(xo, r) < d(x0, x ) + 1. But by our definition of r, d(x0, x ) + I = YO, so d ( % , Y ) < Yo. This shows that any point of .V (x , I, X) is a point of LV (a, ro, X), and therefore N (XO, 10, X) is open in X. The following theorems show how we may manufacture additional examples of open sets from those at hand. MOREM 4.1. If U and V areopensetsof X , then theirintersectiolz U t l V is an open set of X . The intersection of any Jinite number of open sets of X is an open set of X . To prove the first statement, let x E U n V. Since x € U and U is open, there is an I > 0 such that N ( x , I, X) C U. Since x € V and V is open, there is an s > 0 such that N ( x , s, X) C V. Let t be the smaller of I and s; it is clear that N ( x , t, X) lies in both U and V , and therefore in U n V . This proves that U n V is open. To prove 26 EXISTENCE THEOREMS IN DIMENSION 1 [I of Rn, then X t l U is an open set of X. Let x E X n U. Since x E U and U is open, there is an r > 0 such that N ( x , I ) C U. Hence x n N ( x , r ) c Xn U. N ( x , ~ , x ) c X n U . But N ( x , r, X ) = X n N(x , I ) , so This says that for each x E X n U there is a neighborhood of x in X contained in X n U; therefore X n U is an open set of X . This is half of what we must prove. Next we must show that an open set V of X can be enlarged to an open set U of Rn such that V = X n U. If V were a neighborhood N ( x , I , X ) , it is clear that the desired enlargement would be N ( x , I , Rn). Now, it is easily verified that any open set V of X is the union of all neighborhoods contained in V, so we construct the desired enlargement of V by enlarging each neighborhood contained in V. We define U to be the union of the collection of neighborhoods N ( x , I , Rn) for all 2 E V and r > 0 such that N ( x , I , X ) C V , and we shall prove that this U fulfills the requirement, that is, we prove that U is open in Rn and that X n U = V. Since each N ( x , I , Rn) is open in Rn, Theorem 4.3 asserts that U is open in Rn. The proof that X fl U = V will be accomplished in two stages: lint we shall show that each element of V is an element of X n U, and then that each element of X n U is an element of V . Since V is open in X , each point x E V has an N ( x , I , X ) C V , and therefore x E U. This shows that V C U. But V is also a subset of X , so V C X t l U. Finally, to show that X n U C V, we note that any point y E X n U is in both X and U. As a point of U, it lies in some N ( x , I, Rn) such that N ( x , I , X ) C Y . Since it also lies in X , it is in X n N ( x , r, R") = N ( x , I , X), and since N ( x , I , X ) C V , it follows that y E V. This completes the proof that X n U = V and also the proof of the theorem. Let us summarize what has been done so far in Section 4. The main property of an open set U of X is its defining property: each x E U has a neighborhood N ( x , r, X) C U. There is little more that can be said about a single open set. Our theorems state properties of the family of all open sets of X , namely, it contains as elements the empty set, X itself, and every neighborhood N ( x , I , X ) ; moreover, it contains the intersection of any finite collection of its elements, and the union of any collection, finite or infinite. We turn now to the concept of closed set. DEFINITION. Let X be a subset of R". A subset A of X is called a closed set of X if its complement in X is an open set of X . Briefly, A isclosedif X - A isopen. 641 OPEN SETS, CLOSED SETS 27 If we refer to the definition of an open set, we obtain the following test for a set A to be closed in X: each point of X - A has a neighbor- hood not meeting A . For example, a set consisting of a single point is always closed in any larger set X ; for, if x is the point and y is any other point, then hT(y, r) does not contain x if r is less than or equal to the distance from x to y. Similarly, a straight line L in a plane or in space is a closed set in any larger set; for, if y is not on L, and r is the distance from y to the nearest point of L, we have that N ( y , I) does not meet L. Each example we have given of an open set yields, on passing to its complement, an example of a closed set. In the example of the rectangle A of Fig. 4.2, the complement of the exterior V is the union of A with its interior U. Since V is an open set, it follows that A U U is a closed set of R2. Similarly, A U V is closed in Ra. Since the union U U V is open and A is the complement of U U V, it follows that A is closed in R2. The relation between a set A in X and its complement X - A is reciprocal: the complement of X - A is A . This correspondence between subsets of X is called d d i t y in X . Open sets and closed sets are dual concepts because the dual of an open set is a closed set, and conversely. This duality between open sets and closed sets enables us to deduce from each theorem we have proved about open sets a true “dual” theorem about closed sets. In working out the form of the dual propositions, we make use of the fact that union and intersection are “dual” operations in the following sense. The complement of the union of two sets is the inter- section of their complements: X - ( A U B ) = ( X - A ) n ( X - B ) . Similarly, the complement of the intersection of two sets is the union of their complements: x- ( A n B ) = ( x - A ) u ( x - B ) . Thus the following four theorems for closed sets correspond to those we have proved for open sets. The theorem about the intersection of two open sets gives us the dual T~OFUZM 4.1’. If A and B are closed sets of X , then their union A U B is a closed set of X . The union of any jinite number of closed sets of X is a closed set of X . To obtain the dual of the proposition that 0 and X are open, we need only observe that @ and X are complementary sets in X: X - @ = X , and X - X = @. 28 EXISTENCE THEOREMS IN DIMENSION 1 P THEOREM 4.2'. Both the empty set 0 and X itself are simuliamusly open sets of X and closed sets of X . An open set of X is usually not a closed set of X, and vice versa. In Section 7 we shall make a careful study of subsets of X which are both open and closed in X. The theorem on the union of any collection of open sets gives as its dual THEOREM 4.3'. The intersection of any collection of closed sets of X (finite or inJinite in number) is a closed set of X . The proposition about open sets of X being the intersections of X with open sets of Rn has as its dual THEOREM 4.4'. If X C Rn, then the wllectwn of closed sets of X coincides with the collection of intersections of X with all closed sets of Rn. Suppose A C X C Rn and A is closed in Rn; then the theorem asserts that A n X is closed in X. Since A n X = A , we obtain COROLLARY. If A is a closed set in Rn, then, for every set X wn- iaining A , A i s a closed set in X . It should not be thought that every set in R n is either open or closed in R"; many sets are neither. The half-open interval (a, b] is neither open nor closed in R. For any point x in (a , b ] other than b, there is some neighborhood of x which lies in the interval. On the other hand, each neighborhood of b contains points which do not lie in (a, b]. In the example of Fig. 4.2, the union of the interior U and a single point of A is neither an open set nor a closed set of R2. We remarked before that i t was not open. It is not closed because, for any point x of A , every S ( x , r ) contains points of U. But by our test for a set to be closed, each point of its complement must have rl neighborhood not meeting the set. We turn now to the formulation of the continuity of a function in terms of open sets. The ease of the formulation suggests how useful open sets will be in treating questions of continuity. THEOREM 4.5. A function f : X * Y is continuous if and only if the inverse image of each open set of I.' is an open set of X . Equirlalently, f is continuous if and only if the inverse image of each closed set of I' i s a closed set of X . UI COMPLETENESS OF REAL NUMBERS 31 10. Prove that for any function f : X -+ Y the complement in X of the inverse image of a set of Y is the same as the inverse image of its com- plement taken in Y; that is, X - f l A = f-l( Y - A ) . 11. Show that each open set of X is the union of some collection of neigh- borhoods in X. 5. The completeness of the real number system The main point of this section is that there are enough real numbers. In greater detail, if by a real number we mean something representable by a decimal expansion (finite or infinite), then there are enough real numbers to fill up the number line completely. The history of mathematics has been marked by a succession of ex- pansions of the number system. First there was prehistoric man with his counting: one, two, three, many. Then came the concept of the unending sequence of positive integers together with a nomenclature and an ab- breviated notation. Next came the fractions or rational numbers, then came the “roots” of algebraic equations or algebraic numbers, then zero and the negative numbers, and finally the transcendental numbers.t At each of these stages, some of those who used numbers became gradually aware of an inadequacy in the concept of numbers as they understood it. After several attempts they finally succeeded in creating new numbers which, when adjoined to the older numbers, removed the inadequacy. Most of us understand thoroughly the need for the integers and rational numbers including their negatives and zero. It is less well understood why these are not enough. It was the school of Pythagoras which discovered that v2 is not a rational number; precisely, there is no fraction whose square is 2. Here is the proof given by Euclid. Suppose, to the contrary, that m/n is a fraction whose square is 2. We can suppose in fact that m / n is in re- duced form, that is, m and n have no common integral factor other than 1. In particular then, they are not both even integers (all common 2’s having been “cancelled”). We write the equation (m/n)* = 2 in the form m2 = 2n2, which tells us that ma is an even integer. Now the square of an odd integer is itself odd (27 + 1 ) Z = 47-2 + 4r + 1 = 2(272 + 2r) + 1 . Since mz is even, it follows that m is even, so that m = 2k for some integer k. If we substitute this value for m in our equation it becomes t For a detailed treatment of the development of the number concept see Volume 1 of the New Mathematical Library, Numbers: Rat iml and Iwa- tional by Ivan Niven. 32 EXISTENCE THEOREMS IN DIMENSION 1 [I 4ka = 2na, hence n2 = 2ka. This means that n2 is even, and there- fore n is even. Thus both m and n are even integers, contradicting the fact that m/n was in reduced form. The contradiction shows that there can be no fraction whose square is 2. The Pythagoreans needed the number d because they were geometers. Starting with a line segment of length d they could construct, using straightedge and compasses, a square of side d. By the theorem of Pythagoras, the length of the diagonal had to be d d. Think of the successive expansions of the number system this way. Starting with a line L and two points on L, called 0 and 1, one can use compasses to mark off successively the remaining integer points 2,3, and - 1, - 2, 0 . By means of another construction involving an auxiliary line, one can divide each interval [n, n + 11 into as many equal parts as desired. Thus all points on the line L with rational co- ordinates can be constructed from the two points 0 and 1. Figure 5.1 Figure 5.2 Now the rational points are densely distributed along L (between any two such points there are infinitely many more distributed uni- formly). It is easy to see how one might assume, without being aware of it, that these rational points were all the points of L. However, the Pythagoreans discovered that the diagonal of a square with side 1, when mirked off on L (see Fig. 5.1), gave a point v2 which was not one of these rational points. What a jolt this must have been to the discoverers! This forced them to create new numbers to correspond to the new points of L arising from such geometric constructions. Unfortunately, inventing the square roots of rationals still did not give enough numbers. Trisecting an angle or duplicating a cube requires the taking of cube roots of rationals, and these are not usually square roots of rationals. So mathematicians were forced into the creation of n-th roots of rationals and of the even larger set of all algebraic numbers. These are the roots of polynomial equations having integer coefficients. About one hundred years ago it was found that the algebraic numbers are not enough; that is, there are points on the number line that do not correspond to algebraic numbers. In particular, the number T (the ratio of the circumference of a circle to its diameter) was proved to be not an algebraic number. One is led to ask When, if ever, will the process end? 051 COMPLETENESS OF REAL NUMBERS 33 The development of the decimal system and of decimal expansions of numbers gave a new viewpoint on these questions. All of the numbers created so far can be represented by their decimal expansions. It should be emphasized that most rationals and all the other numbers have un- ending decimal expansions. It is natural a t this stage to turn things around and to say that any decimal expansion represents a real number. That is to say, we can define the set R of real numbers to be the set of decimal expansions (with the customary convention that an expansion ending in nines represents the same number as another one ending in zeros, for example, 3.26999.0- = 3.27000.-.). This is in fact what we shall do. To justify the procedure we must show that i t brings an end to the game of creating new numbers; the numbers in the set R fill up the line completely. We must explain what is meant by “filling up the line”. Recall the standard method of extracting a square root such as d. Geometrically we are dealing with the graph of the equation y = x2, and we are trying to determine the x-coordinate of the point where the graph crosses the horizontal line y = 2 (see Fig. 5.2). We test first the squares of the first few integers and we find that la = 1 is too small and 22 = 4 is too large. Now if x is made to increase from one positive value to another, its square also increases. This fact tells us that d lies some- where in the interval lo = [l, 21, and the integer part of its decimal expansion is 1. Next we divide the interval ZO into tenths, and square each of thenumbers 1.0, 1.1, 1.2, 1.9, 2.0. We find that (1.4)2 is less than 2 and (1.5)2 is greater than 2. Thus d lies in the interval I1 = C1.4, 1.51, and the decimal expansion of d begins with 1.4. Next we divide I I into ten equal parts, test the squares of the points of the division, and find that d lies in the interval Zz = C1.41, 1.421. Continuing thus we determine an infinite sequence of intervals l o 3 I1 3 IZ 3 * * * 3 Ik 3 squeezing down on d (the symbol 3 is the reverse of C, and “ A 3 B” is read: “ A contains B”). Each interval is a tenth part of the preceding, and the decimal expansion of d can be read off from the decimal expansions of their left-hand endpoints which are 1, then 1.4, next 1.41, etc. Picture now a similar but more general problem; instead of y = x2 consider y = fx, where f is any continuous function which increases in value as x increases, and instead of finding a number x such that .r2 = 2, we wish to solve the equation fx = b where b is some given number. If we can find an initial interval I0 = [n, n + 11 such that fn < b and f ( n + 1) > b, then we can again carry out the procedure of repeatedly dividing intervals into tenths. This gives an infinite se- quence of intervals lo 3 I1 3 * a . 3 Ik 3 - . a . Amalgamating the decimal expansions of the left-hand endpoints of these intervals just as in the case of the number 42, we can construct the decimal expansion of a number a which lies in each of the intervals, and should therefore be a solution of our problem: fa = b. This suggests strongly the con- 36 EXISTENCE THEOREMS IN DIMENSION 1 [I an 5 t for all n. To prove that c 5 bn for all n, suppose to the contrary that bN < c for some integer N . Then t - bN > 0, and we can choose an integer n bigger than N such that 1 10" > -. C - bN Then we have bn I bN and 10-" < c - bN. These combine to give 6, < t - 1 P . This inequality and the one above involving t n imply that bn < cn. Since all the a's precede all the b's, this means that all the a's precede cn. As this is a contradiction, i t follows that c 5 bn for all n. This proves that c E I n for all n, and completes the proof of the theorem. We could proceed now to show precisely how our completeness theorem enables us to solve equations of the type discussed earlier in this section. But all these results are embodied in the main theorem of Part I (see Section 1); so we shall continue with the working out of its proof. Exercises 1. Show that is not a rational number. [Hint: Show that an integer whose square is divisible by 3 is also divisible by 3; or equivalently, show that the square of an integer not divisible by 3 (i.e., of the form 3k + 1 or 3k + 2 ) is not divisible by 3.1 2. Give another proof of the proposition that the equation 2n2 = m2 has no solution in integers m, n, using the theorem that each integer has a unique factorization as a product of prime numbers. (Hint: Compare the number of factors 2 on each side of the equation.) 3. Show that 2/2 is not a rational number by proving (with the aid of the unique prime factorization theorem) that 2na = m3 has no solution in integers. 4. Show thaithe equation x , y other than (0, 0). = 222 has no solution in rational numbers 5. Show that there are no integers k, m and n such that [ ( k + mV2)/nl3 = 2. 6. If S = UI+ a2 + a3 + - - is an infinite series, the finite sums so = 0, s1 = u1, s2 = u1+ u2, a*., S n = UI+ S+ * * * + G, ... $61 COMPACTNESS 37 are called partial sums of the series. Show how the partial sums of the infinite series 1 - -+ 1 1 1 1 - - -+ - - ... + (-1)n-+ 1 0 . ’ 10 102 103 104 10” determine a regularly contracting sequence of intervals. What is the sum of the series? 7. Show how the partial sums of the infinite series determine a contracting sequence of intervals whose intersection is the sum of the series. What is this sum? 8. Show how the process of long division applied to 12.27/3.41 leads to a regularly contracting sequence of intervals. 9. Find to the second decimal place by the method of contracting intervals. 10. Show that each open interval (a, b ) , where a < b, contains a rational number and also an irrational number; show that the set Q of all rational numbers is neither closed nor open in R. 11. Prove the theorem of Dedekind about a “cut” of the real numbers: If A and B are two non-empty subsets of R such that R = A U B and every number of A is less than every number of B, then there is a real number which is either the largest number of A or the smallest number of B. 6. Compactness A subset X of Rm is said to be bounded if i t is contained in some sufficiently large ball; that is, if there is a point xo and a number Y > 0 such that X C :Y(zo, I). Examples of bounded sets are segments, circles, spheres, triangles, etc. Examples of unbounded sets are lines, half-lines, rays, planes, exteriors of circles in Ra, the entire space R”, and the rational numbers. Intuitively, a set is unbounded if one can run off to infinity along the set. A most important and remarkable property possessed by any subset X of Rm that is both closed in Rm and bounded is that, for any con- tinuous mapping f: X +. Rm, the image set j X is also closed and bounded. The main purpose of this section is to prove this fact. The proof is necessarily somewhat indirect; its development passed through 38 EXISTENCE THEOREMS I N DIMENSION 1 [I many stages beginning with the work of Cauchy (1789-1857). In par- ticular, i t embodies the propositions of analysis often referred to as the theorem of Bolzano-Weierstrass and the theorem of Heine-Borel. In the proof to be given, we first show that the property of being closed and bounded is equivalent to another property called "compactness". This is the major part of the argument. Once this is done, i t is easy to show that fX is compact whenever X is compact and f is continuous. We shall lead up to the definition of compactness by displaying a common property of unbounded sets and non-closed sets. Let X be an unbounded subset of R" and xo a point of R". Picture the sequence of neighborhoods X ( x o , r) where the radius r takes on the values r = 1,2,3, - 0 . These form an expanding sequence of open sets whose union is all of R", since, for each x E R", the dis- tance d ( x , x ~ ) is less than I for some sufficiently large integer r. The intersections X n .\'(%, I), r = 1, 2, 3, 0 , form therefore an ex- panding sequence of open sets of X whose union is all of X; but X is not equal to any one of these open sets because X is unbounded. More- over, X is not contained in the union of any finite number of these sets because their union is just the largest one. Now let X be a bounded but non-closed subset of Rm; then there is some point y in the complement Rm - X of X such that each neigh- borhood A'(y, I) contains points of X (see the definition of closed set in Section 4). For each integer k = 1,2,3, * * * , let uk be the exterior of the circle about y of radius l/k. Each uk is an open set of R"; for, if x E u k , then N (x, d (x , y)- 1/&) is a neighborhood of x contained in uk. The uk form an expanding sequence Ul C UZ C * * * , and their union is the complement of y, because for each point x # y there is a k such that (l/k) < d ( x , y). It follows that the intersec- tions X n uk form an expanding sequence of open sets of X whose union is all of X, but X is not equal to any one of the sets because each N ( y , l/k) contains points of X. Moreover, X is not in the union of any finite number of the sets because their union is just the largest one. Thus if X is unbounded or if X is not closed we can find in X an expanding sequence of open sets of X whose union is X, but X is not the union of any finite number of them. This leads us to the definition of compactness. First, however, we need the definition of an "open covering". DEFINITIONS. Let X be a subset of Rm. A collection C of subsets of Rm is called a covering of X if the union of the sets of C contains X; that is, each point of X lies in a t least one of the sets of C. A covering C of X is calledjinib if the number of sets in C is finite. A covering C of X is said to conlain a covering D of X if each set of D is also a set of C. A covering of X is called an open covering of X if each set of the covering is an open set of X. Finally, the space X is called compact if each open covering of X contains a finite covering %I COMPACTNESS 41 xi = ci (i = 1, 2, 0 0 , m) into 2" congruent boxes, with edges half as long as those of B. Any mdimensiod box B is compact. We shall prove this by a process similar to that used for the one- dimensional box, the interval. Assume, to the contrary, that C is an open covering of B which contains no finite covering. We shall construct a contracting sequence of boxes Bo, BI , 0 , Bk, * * * such that BO = B , no one of them is covered by a finite subcollection of C, and, for each k > 0, Bk is one of the 2'" boxes of the subdivision of Bk-1. By hy- pothesis Bo = B is not finitely covered by C. This starts our inductive construction of the sequence. Assuming that Bo, BI, * - * , BLI have been properly chosen, consider the 2" boxes of the subdivision of Bk-1. If each were covered by a finite subcollection of C, then the union of these 2" subcollections would be a finite subcollection of C covering BLI. Since this is impossible, a t least one of these subboxes of Bk-l is not finitely covered by C. Choose Bk to be one such box. This completes the inductive proof of the existence of the sequence Bo, B1, 0 , Bk, 0 . (Fig. 6.1 illustrates the first three stages for m = 2.) We claim that there is a point x E Bk for every R = 1, 2, ' 0 . . To see this, consider the projections of the sequence of boxes on the i-th coordinate axis, for i = 1, 2 , * a * , m. On each axis the projections form a contracting sequence of intervals. Let xi be a number common to all the intervals formed by the projections on the i-th coordinate axis. Then the point x of Rm whose coordinates are ( X I , x2, * * * , xm) is a point of Bk for all k's. Since x E B, there is an open set U of the covering C such that x E U. Hence there is a number r > 0 such that .V(x, r , B ) C U. Let d denote the length of the longest edge of B. Since all edges were bisected at each stage of the construction of the sequence, it follows that d/2k is the length of the longest edge of Bk. By the theorem of Pythagoras, the length of the diagonal of Bk is a t most fi d/2k. Choose an integer k so large that and i t follows that Bk c -y(X, r , B ) c L' ; hence Bk is contained in a single set of C, and this contradicts the fact that Bk is not finitely covered by C. Our supposition that B is not com- pact has led to a contradiction; therefore B is compact. The collection of sets which we can prove to be compact is greatly en- larged by the following useful proposition. THEOREM 6.2. If X is a closed subset of a compact space B, then X is compact. 42 EXISTENCE THEOREMS IN DIMENSION 1 P To prove this, we shall take any open covering C of X and enlarge each member of C so that the enlarged sets form an open covering C‘ of B. We then use the compactness of B to select a finite covering from C’, and observe that the corresponding unenlarged sets form the desired finite covering of X. For each set U of theopen covering C of X, let U‘ = U u (B - X), and let C’ denote the collection of these larger sets U’. First we shall show that U’ is an open set of B. A point x E U’ is either in U or in B - X. If x E B - X, the hypothesis that X is closed in B tells us that there is an I > 0 such that h’(x, I, 9) C B - X C U’. If x E U, the fact that U is open in X means that there is an I > 0 such that N ( x , I, X) C U, and hence N ( x , I, B) C 6’ U (9 - X) = U’. This proves that U’ is open in B. Now let y be any point of B; either y E X or y E B - X. If y E X, then y E U for some U E C so that y E U’ for the corre sponding U’ E C‘. If y E B - X, then y E U’ for all U’ E C’. There- fore C’ is an open covering of B. Since B is compact, a finite number of the sets of C’, say U;, Ug’, - * * , uk’ cover B and hence also X. It follows that the corresponding sets of C, namely, 6’1, UZ, * * = , uk, form a finite covering of X; for every point of X covered by Uj’ is also covered by Uj. This completes the proof of the theorem. We are now able to prove the converse of Theorem 6.1 by enclosing our closed and bounded set in an mdimensional box (which has been shown to be compact) and then applying Theorem 6.2. THEOREM 6.3. Each closed and bounded subset of R” is compact. Let X be closed in Rm and bounded. Since X is bounded, there is a point b E Rm and a number I > 0 such that X C X ( b , I). Let B denote the m-dimensional box with center a t b and with edges all equal to 21; precisely, a point y E Rm is in B if its coordinates (yl, - 0 0 , ym) satisfy Then B contains N ( b , I), and therefore B 3 X; hence X fl B = X. Furthermore X isclosedin B ; for X isclosedin R”, and by Theorem 4.4’, the closed sets of B coincide with the intersections of B with the closed sets of Rm. The desired conclusion, that X is compact, is now a consequence of the preceding theorem. At this point we have established the equivalence of the property of being compact and the property of being closed and bounded in P. We are prepared now to prove the main proposition of this section. bi - I I y, 5 bi + Y for i = 1, - - * , m . THEOREM 6.4. Let X be a compact space and let f: X + Y be con- tinuous; then the image jX is compact. Wl COMPACTNESS 43 F'ROOP. Let C be an open covering of f X . We want to show that C contains a finite covering of fX. For each U E C, consider the inverse image f - W , and let C' be the collection of all such inverse images. Since f is continuous and U is an open set of fX, each f-'U is an open set of X . For each x E X, f x lies in some U E C because C covers j X ; so x lies in the corresponding j-W. Thus C' is an open covering of X . Since X is compact, there is a finite subcollection D' of C' which covers X . The corresponding subcollection D of C is finite and covers f X ; for, if x E f-1U and f-'U is in D', then fx E U where U is in D. Thus C contains a finite covering of fX. This completes the proof that f X is compact. An immediate consequence of Theorem 6.4 is the following. COROLLARY. If X i s a closed and bounded set in R", and i f f : X-- t R" is continuous, then j X is a closed and bounded set in Rn. To relate the preceding work to the main objective of Part I (to prove the theorem of Section l), we must prove an important property of compact sets on a line. THEOREM 6.5. A compact non-empty set X of real numbers has a maximum and a minimum; that is, there are numbers m and M i n X such that m is the smallest number in X and M is the largest number in X . To appreciate the force of the conclusion, note first that the set R of all real numbers has no largest number and no smallest. In fact, any unbounded set Y of numbers must fail to have a maximum or a mini- mum; for, if it had both, any open interval containing both would contain all of Y, and then Y would be bounded. Note next that there are bounded sets which have neither a maximum nor a minimum. For ex- ample, an open interval (a, b ) has neither a largest number nor a smallest number. These examples show that, to achieve the conclusion of the theorem, we must require X to be bounded, and we must impose some additional condition which is not satisfied by an open interval. Since a compact set is bounded and closed, the single condition of compactness guarantees boundedness and rules out the open interval. Let us proceed with the proof of the theorem. Since X is compact, it is bounded; hence there is a closed interval 10 = [ao, bo] which con- tains X. We shall construct a contracting sequence of intervals 10, 11, ---, Zk, 0 . - with the following properties each interval Ik is a half of Ik-1; each Ik contains a t least one point of X ; and finally, the right-hand endpoint bk of Ik is an upper bound of X-that is, for all x E X, we have x 5 bk, k = 0, 1, 2, * * * . Clearly 10 contains a point of X (since X is not empty), and bo is an upper bound of X. Assume that lo, 11, * * * , Ik-1 have been properly selected. Let c be the 46 EXISTENCE THEOREMS IN DIMENSION 1 [I 13. Show that a subset X of R”’ is compact if and only if each covering of X by open sets of P contains a finite covering. 7. Connectedness For the proof of our main theorem of Part I, we need two important topological properties of the closed interval. The first of these, com- pactness, has been treated in Section 6. We shall discuss now the other property called “connectedness”. Some spaces can be divided in a natural way into two or more parts. For example, a space consisting of two non-intersecting lines can be divided into the two lines. As another example, the complement of a circle in the plane consists of two parts, the part inside the circle and the part outside. Again, if p is a point of a line L, then the complement of p in L falls naturally into the two half-lines determined by p (the deletion of p cuts L into two parts). In each of the foregoing examples, the natural division occurs in just one way. The set Q of rational numbers can be divided into parts in many ways. Each irrational number x produces a division of Q into those rationals greater than x and those less than 2. The set of irra- tional numbers can be divided by each rational number in a similar manner. On the other hand, certain sets cannot be divided into parts in any natural way; this is true, for example, of a line, a line segment, a plane, and a circular disk. Of course it is possible to force a division. For example, if I is the in- terval [a, b] and if c is a number such that a < c < b, then c divides I into the two intervals [a, c] and [c, b] . However, since they have c in common, we do not regard this as a proper division. We obtain a proper division by deleting c from one of the sets, say the second. Let A = [a, c ] and B = (c, b ] ; then A U B = I and A tl B = @. We do not regard such a division or “breaking” of I as natural because the set B “sticks” to A a t the point c. If we delete c also from A to overcome this “stickiness”, then A U B # I, but A U B is the complement of c in I; hence this example is similar to the example of the complement of a point p in a line L. The precise notion we need is now stated. DEFINITION. A separation of a space X is a pair A, B of non-empty subsets of X such that A U B = X, A n B = 0, and both A and B are open in X. A space which has no separation is said to be connected. Consider, for example, the complement X of a circle C in the plane. Let A be its interior and B its exterior; that is, A consists of all points $71 CONNECTEDNESS 47 of X whose distance from the center of C is less than the radius of C, and B is the complement of A in X . The conditions for a separation are readily verified. The fact that A and B are open in X is apparent from Fig. 7.1; each point of A has a neighborhood contained in A , and each point of B has a neighborhood in B. 0 Figure 7.1 Let L be a line, p a point of L, and X the complement of p in L. Let A be the set of points of L to the left of p (see Fig. 7.2) and B the set of points to the right of p . Again each point of A has a neighborhood in A , and similarly for B. A B - a P Figure 7.2 Recall now the “forced” division of the interval I = [a, b] into A = [a, c] and B = (c, b ] . If we check the conditions for a separa- tion we find that all except one are satisfied A is not open in I because no neighborhood of the point G E A lies entirely in A . These preliminary considerations indicate that the definitions of a separation and of a connected space express precisely the rough geometric idea we have in mind, and the next theorems will justify these definitions completely. The definition of a separation can be reworded in several equivalent ways. Since A and B are complements of each other in X, each is open if and only if the other is closed. Thus we could equally well require that A and B be closed in X. Also we may drop explicit reference to B, and say that a separation of X is a subset A of X which is both an open and a closed set of X , and which is neither 0 nor X. Then its complement B in X has the same properties. (Recall that 0 and X are both open and closed in X.) Thus, any of the following may serve as definition of a separation A , B of a space X : 1. A and B are non-empty subsets of X such that A U B = X, A n B = 0, and both A and B are open in X ; 2. A and B are non-empty subsets of X such that A U B = X, A n B = 0, and both A and B are closed in X; 48 EXISTENCE THEOREMS IN DIMENSION 1 [I 3. A is a subset of X which is both an open and a closed set of X, and which is neither 0 nor X. It is usually easier to prove that a space is not connected than to prove that i t is connected. In the first case, we need only exhibit a separation and verify that i t is one, while, in the second case, we must prove some- thing about all open sets of X other than @ and X , namely, that each such set is not closed in X . The following theorems not only show that certain simple spaces are connected, but also present a technique for verifying the connectedness of many spaces. THEOREM 7.1. A closed interval of real numbers i s a connected set. a b b a a I i a > b I' b- m a =M M b, A ' - 4 Figure 7.3 Let I be a closed interval of R, and let A be a closed set of I which is neither @ nor I. To prove the theorem we shall show that A is not open in I. Since A # @ and A # I, there is a point a E A and a point b E I - A . Let I' denote the interval [a , b ] (or [b, a ] , if b < a ) . To picture this, imagine the interval I made up of the sub- sets A and B shown in Fig. 7.3 where A is a closed set and B is its complement in I. Since A and I' are closed, so is their intersection A n I'. Since A n I' is also bounded, it is a compact set. By Theorem 6.5, A n I' has a minimum m and a maximum M. If a < b, then m = a since a is the left-hand endpoint of I'. Since the right-hand endpoint b isnotin A , wehave m = a 5 M < b. Itfollows that each neighborhood of M contains numbers between M and b ; these arenotin A , hence A isnotopen. If a > b, then M = a, b < m 5 a, and each neighborhood of m contains numbers between b and m; these are not in A , so again A is not open. This proves that the only subsets of I which are both open and closed are I and @. Therefore I is connected. THEOREM 7.2. If f: X + Y is a continuous map and A , B is a separation of the image fX, then the inverse images -4' = f-lA and B' = f-'B form a separation of X . We must verify that the pair A', B' satisfies each of the conditions for a separation. Since A is not empty, there is a point y E A . Since $71 CONNECTEDNESS 51 THEOREM 7.6. A compact connected set of real numbers is a closed interval. The converse of this theorem, that a closed interval is both compact and connected, has already been proved (see Section 6 and Theorem 7.1). PROOF. Let X be any connected set of real numbers and let a and b be numbers in X with a < b. We prove first that any number c such that a < c < b is also in X . The point c determinesa separa- tion of its complement in R ; let A consist of all numbers less than c, and let B consist of all numbers greater than c. If, contrary to our claim, X did not contain c, i t would be a subset of A U B, and, since X is connected, by Lemma 7.3 i t would lie wholly in A or wholly in B. But X contains a and b, so this is impossible. Thus we have shown that a connected set of real numbers contains all numbers between any two of its nzcmbers. If, in addition, X is compact, Theorem 6.5 asserts that X has a minimum m and a maximum M. It follows that X is precisely the closed interval [m, MI. Exercises 1. State whether each o,f the following sets is connected; if not connected, find a separation. (a) A circle with one point deleted; with two points deleted. (b) An arc of a circle; an arc with its midpoint deleted. (c) A finite set of points; the singleton set consisting of a single point; the empty set. Figure 7.5 R (d) The torus (see Fig. 7.5) (i) with circle P deleted; (ii) with circle Q deleted; (iii) with circles P and Q deleted; (iv) with closed curve R deleted; (v) with two circles of type P deleted; (vi) with two circles of type Q deleted; (vii) with its interior, but with two circles of type P deleted. 52 EXISTENCE THEOREMS IN DIMENSION 1 [I (e) The union of two disjoint circles in the plane; the intersection of these two disjoint circles. (f) Let A , B, C, D be four points on a circle, equally spaced and in order. Letting AB denote the shortest arc from A to B including the end points, answer the question for the following sets: (i) A B U BC; (ii) A B fl BC; (iii) A B U CD; (iv) A B n C D ; (v) A B C U C D A . 2. A subset D of R” is called star-shaped about a point p if, for each point x E D , the line segment p to x lies in D. Show that such a set is connected. 3. Show that each of the following is connected: the surface of a sphere; the interior of a sphere; the exterior of a sphere; the surface of a torus; the interior of a torus; the exterior of a torus. 4. Show by an example that the inverse image of a connected set is not necessarily connected. 5. Show that central projection of a non-diametral chord or of a tangential line segment into a circle is continuous; conclude that circular arcs are connected. 6. Give an example of two connected sets whose intersection is not connected. 7. Explain whether the points in the plane having a t least one rational co- ordinate form a connected set; those having exactly one rational coordi- nate; those having two rational coordinates. If not connected, show a separation. 8. Let X be the set of points.on all circles in the plane with center a t (0,O) and with radius r , where r is a rational number; find a separation of X. 9. Show that a connected set of real numbers is one of eight things: the empty set, R itself, an open or closed half-line, a single point, or an open, closed, or half-closed interva!. 10. Show that the intersection of a contracting sequence of closed intervals is either a single point or a closed interval. 11. Give another proof that any interval I is connected by assuming that there is a separation I = A U B, constructing a contracting sequence of intervals each with one end in A and the other in B, and deducing a contradiction by showing that a point t of their intersection is not in A or in B. TOPOLOGICAL PROPERTIES 53 8. Topological properties and topological equivalences The principal task in these sections is the proof of the main theorem stated in Section 1: If the real-valued function fx is defined and con- tinuous for a < x < b, then i t has a minimum value, a maximum value, and takes on all values between. We have now completed all the work required for the proof; i t is only necessary to assemble its various parts. The following three propositions have been proved: 1. A closed interval of real numbers is a compact and connected set. 2. A continuous image of a compact set is compact, and a continuous 3. A compact and connected set of real numbers is a closed interual. Each of the first two propositions is obtained by uniting an assertion on compactness proved in Section 6 with an assertion on connectedness proved in Section 7. The third proposition is Theorem 7.6. The three propositions together assert that a continuous image in R of a closed interval is itself a closed interval. This is just another way of stating the main theorem. We have done much more than prove our main theorem; we have proved a number of theorems of considerable generality, and we have analyzed the argument so that theorems similar to the main theorem can be obtained without further trouble. For example, the fact that a closed and bounded set is compact (Section 6 ) , together with proposi- tions 2 and 3, enables us to conclude: image of a connected set is connected. If X is a closed, bounded, and connected set in Rn, and if f: X + R is continwus, then the image fX i s a closed interval. One of the many different kinds of closed, bounded, and connected subsets of Rn is the surface of a sphere in R3. Hence a continuous real- valued function defined on a sphere has a maximum value, a minimum value, and takes on all values between. We can see now that the hy- pothesis of the main theorem, that the domain of f is a closed interval, is unnecessarily restrictive; it is enough to require that the domain of f be closed, bounded, and connected. We are now in a position to begin to answer the question: What is topology? DEFINITION. A property of a subset X of Rm is called a topological property if it is equivalent to a property whose definition uses only the notion of open set of X and the standard concepts of set theory (ele- ment, subset, complement, union, intersection, finite, infinite, etc.). 56 EXISTENCE THEOREMS IN DIMENSION 1 [r Another example of a homeomorphism is provided by the stereographic projection of a spherical surface S, with a pole p deleted, onto its equatorial plane P (see Fig. 8.5). The solution of Exercise 7 of Section 3 shows that this defines a topological equivalence between S - p and P. If we restrict the projection to the points of a single great circle C through p , we obtain a topological equivalence of C - p with a line L. Figure 8.5 Having illustrated the concept of topological equivalence, let us return to the discussion of topological properties. The following theorem states the basic relationship involving these concepts. THEOREM 8.1. If a s zhe t X of R” and a szlbset Y of R* are topologically equivalent, then each has ewery topological property possessed by the other. This is obvious because a topological equivalence f: X + Y sets up a 1-1 correspondence between the points of the two sets, and a 1-1 cor- respondence between their subsets ( A C X is associated with fA C Y and B C Y with f-lB C X) in such a way as to make open sets corre- spond to open sets, and to preserve the relations and operations of set theory (for example, A C B in X if and only if fA C f B in Y ) . Any true statement we can make about points of X , subsets of X, open sets of X , and their set-theoretic relations, will yield a true statement if we replace all points and subsets of X by their images in Y . Let us illustrate the argument with the property of being not con- nected. In terms of open sets, this is stated: X has two non-empty open sets A and B such that A U B = X and A n B = a. If we take images under f and use the obvious relations jX = Y, fa = 0, f(A u B ) = f A U j B and f ( A n B ) = f A n fB, then we obtain: Y has two non-empty opensets fA and f B such that f.4 UfB = Y and j A n j B = a. Therefore Y is not connected. We use Theorem 8.1 to show that certain properties of a subset X of Rm are not topological. Since a line segment is topologically equivalent to any other line segment, its length is not a topological property. Since 881 TOp13LOGICAL PROPERTIES 57 a line segment is equivalent to an arc of a circle, its straightness is not a topological property. Since a sphere S with a point p deleted is equiva- lent to a plane P under stereographic projection, the boundedness of S - p is not a topological property. Since P is closed in Ra and S - p is not closed in Ra, the property of P being closed in R9 is not topological. By this time the answer to the question “What is topology?” should be fairly obvious: Topology is the study of the topological properties of point sets. This is a satisfactory answer but i t is not the complete answer. We must include also the topological properties of functions. If f: X 4 Y is a function such that X C Rm and Y C R“, then a property of f is called topological if i t is equivalent to one whose definition uses only the notions: open sets of X and of Y , images and inverse images, and the standard concepts of set theory. For example, continuity is a topological property of a function because Theorem 4.5 states that f is continuous if and only if the inverse image of each open set of Y is an open set of X. It is easy to find topological properties of functions. As another example, the property of being a constant function is a topological property. Again, the property of f: X --+ Y that f-ly is a compact subset of X for every y E Y is a topological property. The complete answer to the original question is that topology is the study of the topological properties of point sets and junctions. In the light of these definitions, let us reexamine our proof of the main theorem as broken down into the three propositions given a t the beginning of this section. The second proposition is pure topology; i t states that a topological property of X (compactness) and a topological property of f (continuity) imply a topological property of fX (compactness). The same holds with connectedness in place of compactness. The first proposi- tion states two topological properties of a familiar object. The third is a converse of the first: the two topological properties, compactness and connectedness, characterize closed intervals among all subsets of R. We can conclude from this that the proof of the main theorem is nearly all topological. Topology has been called rubber geometry. If one attempts to picture those point sets which are topologically equivalent to a particular set X, i t is a good intuitive device to regard X as made of rubber. If X can be deformed into a set Y by stretching here, contracting there, and twisting (but never tearing or gluing different parts together), then X and Y are topologically equivalent. For example, a small spherical sur- face (balloon) can be inflated into a big one, then it can be squeezed to form an ellipsoid, and then i t can be squeezed still more to yield a surface of a dumbbell. Also, an inflated spherical surface can be allowed to con- 58 EXISTENCE THEOREMS IN DIMENSION 1 P tract until its surface fits tautly over the surface of a solid such as a rectangular box or a tetrahedron (Fig. 8.6). Figure 8.6 Since two topologically equivalent point sets have exactly the same topological properties, the topologist regards them as being essentially the same (topologically indistinguishable). This is analogous to the viewpoint in euclidean geometry that two congruent configurations are completely equivalent. A topologist has been defined to be a mathe- matician who can’t tell the difference between a donut and a cup of coffee. Fig. 8.7 shows several intermediate stages of the deformation of a solid donut into a cup. Figure 8.7 In each of the classical geometries there is a concept of equivalent con- figurations. As already noted, in euclidean geometry two configurations are equivalent if they are congruent, in particular, if there is a rigid motion carrying one onto the other. In projective geometry, figures are equivalent if there is a Fojectivity carrying one onto the other. The projectivities include congruences and similarities, and enough additional transformations so that any two triangles are equivalent, and any circle is equivalent to any ellipse (see Fig. 8.8). Figure 8.8 Figure 8.9 $91 A FIXED POINT THEOREM 61 Let coordinates be introduced on the line so that the segment becomes an interval [a, b ] . Then a mapping of the segment into itself is just a continuous function j [a, b ] -+ [a, 61. Define a new function g: [a, b] + R by g x = f x - x for each x E [a, b ] . Thus g measures the directed distance between x and its image fx. It is positive when f x is to the right of x, i.e. fx > x, and negative when f x is to the left of x. We seek a fixed point of f, that is, a point a t which g is zero. If either endpoint is fixed, we have nothing to prove. Suppose neither is fixed. Since f a and f b are in [a, b ] , a < f a and f b < b ; hence g a > 0 and gb < 0. Since g is continuous (it is the difference of two continuous functions, see Exercise 8 of Section 3) , the main theorem asserts that g takes on all values between g a and gb. So g x = 0 for some x E [a, b ] , and this x is the required fixed point of f. Figure 9.1 Theorem 9.1 may be examined from the point of view of the graph of f illustrated in Fig. 9.1. A fixed point off is one whose corresponding point on the graph lies on the diagonal line (i.e. if f x = x, then (x, fx) = (x, x) is on the diagonal). Since a < f a , the point (a, f a ) lies above the diagonal, and similarly (b, f b ) lies below it. Since the diago- nal line disconnects the plane into the points above and those below, and the graph is a connected set, the graph must intersect the diagonal. The function g measures the vertical distance between the graph and the diagonal. Exercises 1. Find the fixed point of the mapping of the interval [0, 11 onto itself de- fined by fx = (1 - 2. Let the mapping of the interval [0, 11 onto itself be defined by fx = 4x - 4x2. (a) Sketch the graph of the function and the diagonal line y = x. (b) Is the mapping 1 - 1 in the interval? (c) Find the fixed points of the mapping. 62 EXISTENCE THEOREMS I N DIMENSION 1 [I 3. Let the mapping of the interval into itself be defined by [0, 11 f x = 2- x + 1. (a) Sketch the graph of the function and the line y = x. (b) Is this mapping 1 - 1 in the interval? ( c ) Find the Gxed points of the mapping. 4. Show that the following property of a set X in R” is a topological 5. Give an example of a mapping of the interval [0, 11 into itself having property: every mapping of X into itself has a fixed point. precisely two fixed points, namely 0 and 1. 6. Give an example of a mapping of the open interval (0 , l ) onto itself having no fixed points. 7 Show that each mapping of a half-open interval onto itself has at least one fixed point. 10. Mappings of a circle into a line A circle has the following striking property: THEOREM 10.1. Every mapping of a circle into a line carries some pair of diametrical poinis into ihe same image point. PROOF. Let fi C - L be a mapping of a circle C into a line L. By introducing coordinates on L, we may consider f as having the real numbers R as its range. Consider a pair of diametrical points p and p’ on C (Fig. 10.1); let their image points on L have coordinates fp = a and fp’ = b, and examine the function g defined by g p = fp- fp’ = a - b . This is a continuous function of p because f is continuous. Moreover, gp’ = fp’ - fp = b - a = - (a - b ) , so the function g is either zero at p and at p’ (in which case p and p’ have the same image under f), or i t has opposite signs at p and at p‘. In the second case, we apply the main theorem to one of the semi- circlesfrom p to p’ toobtain apoint q such that gq = 0 = fq - fq’. It follows that fq = fq’, that is, the diametrical points q and q’ have the same image point. The analog of diametrical points on a circle is antipodal points on an ellipse; these are points located symmetrically with respect to the center $101 MAPPINGS OF CIRCLE INTO LINE 63 of an ellipse. Since a circle is a special case of an ellipse, diametrical is a special case of antipodal. It is therefore appropriate to ask whether a similar theorem holds for antipodal points. b a Figure 10.1 If a circle X and an ellipse Y have the same center z and lie in the same plane, a homeomorphism is most easily constructed by pairing off two points if they lie on the same ray from z. This is essentially the radial projection onto X mentioned in Section 3 and has been proved continuous in that section. If X and Y are not concentric, we arrive a t a homeomorphism by first projecting Y radially onto a circle X’ that has the same center as Y (see Fig. 10.2). Since X and X’ are similar, X is topologically equivalent to X‘ which is in turn topologi- cally equivalent to Y . Furthermore, the homeomorphism Y + X, composed of the radial projection and the similarity, preserves antipodes; that is, if the image of q is p , and if q’ is the antipode of q, then the image of q’ is the antipode p‘ of p . So the theorem on diametrical points holds for the ellipse with the understanding that antipodal points play the same role as diametrical points. C Figure 10.2 Figure 10.3 A similar argument shows that the result holds also for any star- shaped closed curve such as the polygon B in Fig. 10.3. By projecting B radially from the center point z onto the circle C, we obtain a homeomorphism which transforms each pair of points of B on the same line through z into a diametrical pair on C. 66 EXISTENCE THEOREMS IN DIMENSION 1 Ir This is proved by noting that D,* = Dz, so L ( A , x ) = L ( A , x ‘ ) and L(B, x ) = L(B, x ’ ) ; hence X’A = XA and X’B = XB. However, the positive direction for coordinates on D,, is opposite that of D,; hence a x ‘ = - g A x and gBx’ = -gBx, and therefore hx’ = gA2’ - @’ = - g A x + @% = -hx. Now, by Theorem 10.1, there is a point x of C such that hx’ = hx. For this x we have both hx’ = hx and hx’ = -hx; hence hx = 0, and this implies XA = XB, so that L ( A , x ) = L(B, x ) divides both A and B in half by area. PROOF OF (1): Corresponding to a number y, let Ly denote the line perpendicular to D, through the point of D, with coordinate y, and let fy denote the area of the part of A on the positive side of Ly (the side in the direction of increasing y-values). Then f is a real-valued function of a real variable, j R + R. As y varies from -r to -I, the line Ly sweeps once over the interior of C. Picture Ly as a steel needle mounted on a rod D, at right angles. As the mounting traverses the rod from 2’ to x, the needle sweeps across the interior of C. When y = -I, the mounting is at x’, all of A is on the positive side, so f(-r) is the area of A. When y = r, the mounting is at x, all of A is on the negative side, so f r = 0. Figure 11.2 To show that f is continuous, let y, y’ E R with y < y‘. Then fy - fy’ is the area of the part of A between the lines Ly and Ly’. Since this is contained in the rectangular region shown as shaded in Fig. 11.2, it follows that 1 fy - fy’ 1 < 2r I y - y’ I . Corresponding to an c > 0, we take 6 = a/2r. Then, when y’ is in the &neighborhood of y, it follows that fy’ is in the eneighborhood of fy. Therefore f is continuous at y. Since this is true for each y, f is continuous. By the main theorem, as y varies from --I to +r, fy sweeps over all values starting from the area of A down to 0. Hence there is at least one y-value where fy is exactly half the area of A so that Ly = L ( A , x ) cuts A in half. We need to know that there is only one such cut. Suppose $111 PANCAKE PROBLEMS 67 tothecontrary that both Ly and L,,' divide A in half (i.e. fy = fy') and that y # y', say y < y'. The strip Q between Ly and is an open set, and its comp!ement separates into two parts, one containing the positive side of Ly' and the other the negative side of Ly. Since A is connected and contains points in each of the two parts, A must con- tain a point of Q, say p . Since A and Q are open, A n Q is open, so it contains a neighborhood of p. Therefore A n Q has positive area, hence fy > fy'. Since this contradicts fy = fy', we have proved the uniqueness. The existence and uniqueness of L (B, x ) is proved similarly. This completes the proof of (1). Figure 11.3 PROOF OF (2): Since h is the difference g A - g B , it suffices to prove that gA and g B are continuous (see Exercise 8 of Section 3). Let c be a point of C where we wish to show that gA is continuous, and, in ac- cord with the notation above, let CA be the point on the diameter D, where the perpendicular L ( A , c ) cuts A in half (see Fig. 11.3). Let x be a point of C near c. Through the points u and u where L ( A , c ) meets C, draw lines K and K' perpendicular to D,. The line L ( A , c ) divides the interior of C into two parts, U and V. The strip between K and K' separates its complement in the interior of C into two parts, U' and V', such that U' C U and Y' C V. Therefore U' and V' each contain at most half the area of A. It follows that the line L ( A , x ) , perpendicular to D, and dividing A in half, lies in the strip, and so does the point ZA where L ( A , x ) meets D,. Since the circle through CA with center z meets D, inside the strip, it follows that I w - g g A c I < w , where w is the width of the strip. two triangles gives To obtain an estimate of the size of w, notice that the similarity of W d(e , x ) d(% u ) d ( z , x ) ' - = - 68 EXISTENCE THEOREMS IN DIMENSION 1 [I where e is the foot of the perpendicular from x to D,. Since r = d ( z , x ) , thisgives w = - d(u7 d(e , x ) . r Since d(N, v ) I 2r, and d(e, x ) I d(c , x ) , we obtain and therefore w I 2 d ( c , x ) , I gAx - gAc 1 2d(c, x ) . So if t > 0, and x E N ( c , t/2), it follows that I g A X - gAc 1 < € . This shows that g A is continuous. Similarly g B is continuous. This completes the proof of (2) and Theorem 11.1. For our second pancake problem we are asked to cut one pancake into four equal parts with two perpendicular cuts. Figure 11.4 THEOREM 11.2. If A is a bounded region in the plane, then there are two perpendicular lines which divide A into four parts having equal areas. As before we enclose A within a circle C. For each x E C, let L, be the line perpendicular to D, which divides A in half, and let K, be the line parallel to D, which divides A in half. The two lines divide A into four parts which, counting in a counter-clockwise direction (see Fig. 11.4), have areas denoted by P,, Q,, R,, S,. Since L, and K , divide A in half, we have P, i- Q, = R, 4- S, and Q, -b R, = S, i- P , . 0121 ZEROS OF POLYNOMIALS 71 then the graph of y = ax + b is a line that crosses the x-axis a t x = -b/u, so the polynomial has a zero for this value of x. Next, consider the parabola y = x2 + 1 as an example of a polynomial of degree 2 (see Fig. 12.1). The curve lies entirely in the upper half of the coordinate plane. The minimum value of xz + 1 is 1 because for any real number x , x2 2 0; hence the polynomial has no real zero. Similarly, ~ + 1 has no real zero; neither has x4 - 2 s + 5 since x4 - 2x2 + 5 = ( x 2 - 1)'+ 4 never has a value less than 4 (see Fig. 12.2). On the other hand the poly- nomial x2 - 4x + 3 of even degree has the zeros x = 1 and x = 3 (see Fig. 12.3). y = X' ' 2 X 2 + 5 Figure 12.3 --f;ft+f: Figure 12.2 The graph of y = 9 - x + 5 is the curve shown in Fig. 12.4: it crosses the x-axis somewhere between -2 and -1. The polynomial 9 - 2 9 + x + 4 has degree 5; its graph, sketched in Fig. 12.5, crosses the x-axis somewhere between - 1.7 and - 1.6. Figure 12.4 Figure 12.5 In each of our examples, the graph of an odd degree polynomial rises from - m, crosses the x-axis, and eventually goes on to +a. Even degree polynomials have graphs that come down from + 00 and go back to + 00 with a few possible wiggles between, and our examples show that 72 EXISTENCE THEOREMS IN DIMENSION 1 [I some of these never cross the x-axis. The gist of Theorem 12.1 is that this is not the case with polynomials of odd degree; every odd degree poly- nomial with real coefficients has a t least one real zero. To prove Theorem 12.1, it suffices to consider polynomials of the form f(x) = x" + a1xn-' + *.* + a,lx + a,, , for, if the coefficient of the term of highest degree is not 1, we can mul- tiply the polynomial by the reciprocal of this coefficient without changing its zeros. For x # 0, we may write f(x) in the form or f ( x ) = x"q(x) , where 01 an-' an X p - 1 xn q ( x ) = 1 + - + a ' * +-+-. Our method of proof will consist in showing that the polynomial f, of odd degree, is negative for some x, positive for some other x, and con- tinuous. The main theorem will then yield the desired result. Now if x is a number such that the absolute value of each of the terms is less than l/n, then the sum of these n terms is less in absolute value than n/n = 1; this means that h ( x ) is between -1 and +1, and since q ( x ) = 1 + h(x), q ( x ) is positive. To find an x for which this holds, examine each of the numbers and choose a number b greater than all of them. To see that q ( x ) is positive for an x such that I x I 2 b, we observe that the inequalities, n l a l l , ( ~ I u ~ I ) * / ~ , 0 . . 7 (n I an I>'/" , 1x1 > n l a l l , 1x1 > (nlu2t)lP2, . - * 7 I x l > (n I anI)'/" imply For values of x such that I x I 2 b the sign of the polynomial is the sign of x" because f ( x ) = x n q ( x ) and q ( x ) is positive. Since n is odd, xn has the sign of x. Thus the polynomial is positive for x = b and negative for x = -b . 5121 ZEROS OF POLYNOMIALS 73 To apply the main theorem and infer the existence of a zero between -b and b, i t is necessary to show that a polynomial is a continuous function. In Section 3 we showed that any constant function (e.g., a polynomial of degree 0) and any identity function (e.g., the polynomial x of degree 1) are continuous. In Exercise 2 below, it is required to prove that a product of continuous functions is continuous. From this it follows that x2 = x - x is continuous, XS = x 2 * x is continuous and, by an induction, $ is continuous for every k. Since 3 and a constant a are continuous, the same result tells us that any monomial uxk is con- tinuous. Now every polynomial is the sum of its monomial terms, and any sum of continuous functions is continuous (see Exercise 8, Section 3 and answer) ; therefore every polynomial is continuous. Exercises 1. Show that the polynomial 9 is a continuous function. 2. Show that the product of two continuous functions f and g: [a, b ] + R is a continuous function. Hint: I (I.) (gx) - (I%’) (gx’) I = I (fx) (gx - 67%’) + ( f. - fx’) (gx’) I 5 l fx l I @ - @’I+ Ifx-fx’l lgx ’ l . 3. For a polynomial of degree n, what is the determining factor as to whether the polynomial is positive or negative when x is zero? 4. Use the criterion I x 1 > (n I ak to find a number b such that the polynomial f ( x ) = 2 - 2x2 - 3n is positive for x > b and negative for x < - b . Factor the polynomial into linear factors to find the smallest number a such that f(x) > 0 for x > a, and the largest number c such that f ( x ) < 0 for x < c. 5. Use the criterion I x I > (n I ak I)l’k to find a number b such that the polynomial x5 - 3x4 + 12.52 + 2002 - x + 2 is positive for x > b and negative for x < - b . (Notice that the cubic in Exercise 4 was easy to factor into linear factors, but this is not so for the quintic of this problem.) 76 EXISTENCE THEOREMS IN DIMENSION 2 [rr Recall also that we found the concept of the graph of f: [u, b ] + R very useful in explaining the meaning of the main theorem and in making its truth geometrically evident. In the two-dimensional case, we may also speak of the graph of a mapping f: D + P. To see what i t involves, note that a point of the plane P = Ra is represented by two real numbers (XI, s). Its image under j requires two more, (yl, yz). Then the pair consisting of the point and its image is represented by four numbers, and a point of the graph is a point of four-dimensional space. Thus the graph of j is a curved surface in R4. Here then is our first difficulty. To explain our theorems by graphs would require the ability (which none of us has) to visualize a surface in four dimensions. We must therefore adopt a different method of visualizing mappings: the method of picturing images and inverse images as described briefly in Section 2 of Part I. In the remainder of this section, we shall discuss more tomplicated mappings by the same method. Our purpose is to sharpen the geometric intuition, and to indicate the degree of generality of subsequent theorems. I I I I I I Figure 13.1 In Part I, Section 2, we discussed translations, rotations, reflections and similarities as mappings of the plane into itself. A more complicated mapping is one which expands lengths in one direction and contracts them in another. Fig. 13.1 illustrates a mapping j which doubles lengths in the horizontal direction and halves lengths in the vertical direction. Clearly it alters angles and shapes. It maps a circle into an ellipse. Surprisingly it maps any straight line into a straight line. I , _ _ - I _ _ )- _ - - * - - , - - ' I _ - - r - r , I - - ' - - - I' ,' ,' ,' ,' I , I I I I Figure 13.2 Fig. 13.2 illustrates a shearing transformation P + P. Picture a trellis with many horizontal and vertical slats with a nail inserted a t each junction of a horizontal and a vertical slat. Such a structure is not rigid and can slew over, exercising a scissoring action on unwary fingers. A shear also maps circles into ellipses and straight lines into straight lines. 8131 MAPPINGS OF PLANE INTO ITSELF 77 The mappings considered so far are 1 to 1 mappings. We want to con- sider also mappings which are not 1 to 1. Fig. 13.3 illustrates a simple fold of P about a line. This mapping is 1 to 1 along the line of the fold, but every point above the line of the fold is the image of two distinct points of the plane. - L ,- P - f *IL 0 "' Figure 13.4 Fig. 13.4 illustrates a doubling of a plane on itself. A center point z is mapped into itself. Each point of a ray L is also fixed. Each ray issuing from z is mapped rigidly onto a ray issuing from z but forming an angle with L which is twice the original angle. Think of a ray rotating about a a t a constant angular velocity; its image is a ray rotating about z a t twice the velocity. As the first ray completes a half-rotation, its image completes a full rotation. This mapping is 2 to 1 except a t 'z. Each circle with center z is wrapped twice around itself. A similar mapping is obtained for each integer n by multiplying the angular velocity by n. It is n to 1 except a t z. f - Figure 13.5 An even more complicated mapping is one which winds P over P an infinity of times. This is illustrated in Fig. 13.5. The horizontal line 78 EXISTENCE THEOREMS IN DIMENSION 2 R I L is mapped into the single point z, and each vertical line is mapped rigidly on a line through z. As the vertical line moves horizontally at a constant velocity its image spins about z at a constant angular velocity. The figure shows only a half of one rotation. This mapping is QO to 1. The inverse image of z is an entire ray. The inverse image of any other point consists of two rows of isolated points, one above and one below L, successive points of a row being at a fixed distance apart. - flsj!LT Figure 13.6 Mappings that can be described precisely and quickly are usually too simple to illustrate the complexities to be found in general. Fig. 13.6 illustrates a more complicated mapping which we shall not describe in detail. It carries a family of concentric circles into a family of figure eights. In this case the image of P is just a part of P bounded by two rays. Intuitively, this may be thought of as a stringing out of the con- centric circles along a line with each circle given a half-twist. Exercises 1. Construct a mapping f: P+ P by, first, rolling up the plane P onto a cylinder Q so that the lines parallel to the z-axis are parallel to the axis of Q, and then composing this mapping with a perpendicular projection of Q onto P, assuming that the axis of Q is parallel to the z-axis. Describe the image under f of (a) the plane P, (b) a horizontal line y = constant, (c) avertical line x = constant, (d) a sloping line. (e) Describe the inverse image of a point. 2. Let P be a plane through the center of a sphere S. Construct f: P+ P as the composition of, first, the stereographic projection P + S from the pole p , and then the perpendicular projection of S back into P. Describe (a) the image of P, (b) the image of a line L in P, (c) the inverse image of a point of fP. $151 ATTEMPTS AT MAIN THEOREM 81 5. (a) If the last configuration of Fig. 14.2 is cut by removing a thin strip at A , as shown in Fig. 14.3, is the resulting set homeomorphic to a disk? (b) Which combinations of cuts at A , B and C will produce a homeo- morph of a disk? (c) If a configuration has three holes, how many cuts are required to produce a homeomorph of a disk? Figure 14.3 15. Initial attempts to formulate the main theorem Our main existence theorem in two dimensions is analogous to the main theorem in one dimension. It states that, if f: D + P is a mapping of a disk into the plane, then an equation j x = y has a solution x E D for each point y of P which satisfies a certain condition. The formula- tion of this condition will be a bit complicated. We shall approach i t in stages by showing that several simple but plausible conditions are not adequate. In the one-dimensional theorem, where D is a closed interval [a, b ] , the condition on y is that it lie between fa and fb. Now a and b are the extremes of the interval, and separate i t from the rest of the line. In the case of a disk, the extreme points of D are the points of the bounding circle C, and C separates D from the rest of the plane. Thus the con- dition to be formulated might state that y is related in some way to fC. Clearly to say “ y is between fC” is nonsense. If we restate the one-dimensional condition by requiring that y be enclosed by fa and fb, i t conveys the same idea, and the two-dimensional analogy “ y is enclosed by fC” has intuitive meaning. Let us try to formulate this expression precisely. As a first attempt, consider “y is a point of the disk whose boundary is fC”. This is not adequate since, for many mappings f, j C is not a circle. It could easily be an ellipse or a rectangle. As a second attempt, consider “ y is a point of the region whose boundary is fC”. This is better but i t does not allow for an fC which is a figure 8. As a third attempt, we try “y is a point of some bounded region whose boundary is contained in jC” . This seems to be what is wanted until we examine the example of an f: D + P in Fig. 15.1. This mapping is best described in stages pictured from left to right. First, stretch D into a long thin strip E . Next, bend E around into a curved shape F which resembles a thickened three quarters of a circle. Continue this 82 EXISTENCE THEOREMS IN DIMENSION 2 [11 bending until the two tips are made to overlap in the final configuration fD. The point labeled y is not in fD, yet it belongs to a bounded region whose boundary lies in fC. Figure 15.1 This last example exhibits clearly the difficulty we must surmount. What relationship does the point labeled y’ of fD bear to fC that the point y does not bear to fC ? The answer will be given in terms of a new concept we shall develop: the winding number of a curve about a point. We shall see that the winding number of f I C about y’ is not zero, and its winding number about y is zero. This is why fx = y‘ has a soh- tion x € D, but fx = y does not. Exercise 1. Show by an example that the following condition on a point y does not insure that y E fD: if z is thecenterof D, then y and fs liein acon- nected subset of P - f C. 16. Curves and closed curves Heretofore the word “curve” has referred to the graph of a continuous function f: [a, b ] +. R. We need now to use the word in the following broader sense. A curve in the plane is defined to be a mapping cp: [a, b ] + P of some interval of real numbers into the plane. Each number t E [a, b ] can be thought of as an instant of time, and the cor- responding point cpt E P as the location of a moving point a t the time t . Thus a curve may be regarded as the path of a moving point. In par- ticular any curve has an orientation in the sense that the preferred or posi- tive direction along the curve goes from cpa to cpb. This is the direction of motion (of increasing t ) . In pictures of the curves the orientation is indicated by arrowheads as in Fig. 16.1. Notice that we allow a curve to cross itself; that is, the moving point can pass through the same point $161 CURVES, CLOSED CURVES 83 at several different times. Moreover, the moving point may remain at rest for an interval of time. For example, the constant function which maps the entire interval [a, b] into a single point is a curve in our sense. A closed curve is a curve which begins and ends a t the same point: qa = cpb. The line segment L from a point A to a point B in P may be represented as a curve. Recall that any two segments are similar. So if cp: [a, b] + L is the similarity with cpa = A and cpb = B, then Q defines a curve whose image is L. In this example the moving point has a constant velocity. The graph of a continuous function f: [a, b] + R is a curve. We need only set cpt equal to the point whose coordinates are (t, j t ) for each t E [a, b] . A curve of this type does not cross itself, nor is it closed, because tl # t2 implies that cpt~ and cpt2 have different abscissas. Figure 16.1 Figure 16.2 Any circle C is regarded as a closed curve in the following standard way. Let z denote the center of C, and let LO be a fixed ray (half-line including the initial point) issuing from z. Define cpt for t E [0, 13 as follows: & is the point of intersection of Lo with C, and pt is the point of C such that the angle at z between LO and the segment z to cpt is 360t degrees. For example, p ( f ) is the point a t the 90" mark (f the way around). (See Fig. 16.2.) Since there are 360" in a complete circle, we have cpl = &. In this case also, the moving point has a constant speed. The boundary of a rectangle may likewise be regarded as a closed curve. Take an interval [a, e l and divide i t into four subintervals by numbers b, c, d so that a < b < c < d < e. Let the four vertices of the rectangle be A , B , C, D in that order. As in the example above we can deiine cp so that i t maps the intervals [a, b] , [b, c ] , [G, d ] and [d , e l onto the segments AB, BC, CD and DA, respectively. Since cpa = A and cpe = A , thecurveisclosed. Our pictorial illustrations foster the tendency, sometimes misleading, to regard a curve cp as being no more than the image cp[a, b] . So it must be emphasized that the curve is the mapping (c. For example, there is an infinity of distinct standard representations of the circle C as a closed curve, one for each choice of Lo. 86 EXISTENCE THEOREMS IN DIMENSION 2 Exercises 1. In Fig. 17.2, let y be at the distance r d from the center of the circle C of radius r. As we trace the outer arc of C from one point of tangency to the other, through what angle does the ray Lt turn? Through what angle does it turn as we continue around the circle along the inner arc? 2. The complement in P of the closed curve in Fig. 17.8 consists of seven connected regions labeled A , B , C, D, E, F, G. For each region, state the winding number of the closed curve about a point of that region. 3. Do the same as in the preceding problem for the closed curve of Fig. 17.9. Figure 17.8 Figure 17.9 18. Statement of the main theorem By using the concept of winding number, we can formulate now the main theorem of Part 11. THEOREM 18.1. Let f: D * P be a mapping of a disk into the plane, let C be the boundary circle of D, and let y be a point of the plane not on fC. If the winding number of f I C about y is not zero, then y E fD; i.e. there is a point x E D such that fx = y . What follows is a short intuitive proof. Let Y be the radius of C . For each number s such that 0 I s I I , let C, be the circle of radius s concentric with C ; thus C, = C, and CO is the center point z. Let yl be a point of the plane not in fD. Then for every s in [0, r ] , y’ is not on fC, because C, is in D, and so the winding number W ( f I C,, y l ) is defined for every s in [0, r ] . Abbreviate i t by W ( s ) . Consider now the family of closed curves f 1 C. as s decreases from r to 0. Its mem- bers begin with f I C and eventually shriuk down to the constant curve f I Co, i.e. to the point f z . Since f I C, varies gradually as s decreases steadily, it follows that W (s) is a continuous function of s E [O , Y ] . 8181 MAIN THEOREM a7 How does the winding number W (s) vary? The answer is: not at all, because W is a continuous function of s and each value W ( s ) must be an integer; i t cannot jump from one integer value to another without taking on non-integral values between (see the main theorem of Part I ) . Thus W (s) has the same value for all s; in particular, W ( Y ) = W(0) . But W ( 0 ) = 0 because f I Co = fz is the constant closed curve. Therefore f I C, has winding number zero about y' for every point yf not in fD. It follows that W (f 1 C,, y ) # 0 implies that y is in fD; and to say that y is in fD is to say that there is an x E D such that One can see how the argument works in the illustration of Fig. 18.1 by following the successive closed curves f I C, as s decreases; as soon as the two lobes separate (e.g., the third closed curve drawn), y is clearly in the exterior of this closed curve, and therefore the winding number is zero. Notice that this agrees with the result obtained for Fig. 17.5. fx = y. Figure 18.1 Exercises 1. If the closed curve in Exercise 2 of Section 17 is the f l C of a map must lie in fD? f: D+ P, which of the complementary regions A , B , 2. Answer the analogous question for the closed curve of Exercise 3, Section 17. 3. Let f: P+ P be a mapping of the plane into itself that is a simple fold along a diameter of a disk. (a) What is the image of the boundary circle C? of the disk? (b) What is the winding number W ( j l C, y) of points y in the image of the disk? 88 EXISTENCE THEOREMS IN DIMENSION 2 19. When is an argument not a proof? Most people, when they have seen and understood the arguments of the preceding two sections, are convinced that they have seen the truth and that little more need be added to achieve a complete and logical proof. However, a thoughtful reader should spot the gaps in the reason- ing. The main gap occurs in Section 17 ; no precise definition of the winding number was given. It was left to the intuition to decide how many com- plete rotations the ray Lt makes about its endpoint y as t varies from a to b ; it was assumed that our eyes could follow the rotating ray and integrate its motion into a single number of rotations. As is well known, eyesight is not entirely reliable in this respect; for example, we can be misled into thinking we are seeing continuous motion by a sufliciently rapid sequence of still pictures. Fortunately mathematical concepts and deductions are independent of our ability to visualize motion. The situation we must treat is a static one. We have a closed curve cp, a point v not on the curve, and we wish to attach to y and cp an integer called the winding number which agrees with our intuitive notion. This will be done in the next seven sections. A reader, who prefers new ideas and applications to the careful development of an idea already outlined, should skip to Section 27. Before we submerge ourselves in the details of the definition of W(cp, y), let us note that, to complete the statement and proof of the main theorem, we need only (1) define W(p, y ) precisely, (2) show that it is continuous under the type of variation of cp used (3) show that W(p, y) = 0 whenever cp is a constant closed curve. If we were to define W(cp, y) to be 0 for all cp and y, this would satisfy the requirements (l), (2) and (3), and so the proof of the main theorem would be valid for this W , but the conclusion of the theorem would say nothing because there would be no points y such that W ( f I C, y ) # 0. Thus, to make our efforts worthwhile, we require also that (4) W(cp, y ) should be non-zero for certain curves cp and points y; in particular, it should agree with the winding numbers defined intuitively in Section 17. in the intuitive proof presented in Section 18, and 20. The angle swept out by a curve In order to formulate a good definition of winding number, we first consider the more general concept of “angle swept out by a curve