notes4_compact topo spaces

Topology MATH 4181 Fall Semester 1999 David C. Royster UNC Charlotte November 11, 1999 Chapter 1 Introduction to Topology 1.1 History1 Topology is thought of as a discipline that has emerged in the twentieth century. There are precursors of topology dating back into the 1600's. Gottfried Wilhelm Leibniz (1646{1716) was the rst to foresee a geometry in which position, rather than magnitude was the most important factor. In 1676 Leibniz use the term geometria situs(geometry of position) in predicting the development of a type of vector calculus somewhat similar to topology as we see it today. The rst practical application of topology was made in the year 1736 by the Swiss mathematician Leonhard Euler (1707{1783) in the K onigsberg Bridge Problem. Carl F. Gauss (1777{1855) predicted in 1833 that geometry of location would become a mathematical discipline of great importance. His studey of closed surfaces such as the sphere and the torus and surfaces much like those encountered in multi- dimensional calculus may be considered as a harbinger of general topology. Gauss was also interested in knots, which are of current interest today in topology. The word topology was rst used by the German mathematics Joseph B. Listing (1808{1882) in the title of his book Vorstudien zur Topologie (Introductory Studies in Topology ), a textbook published in 1847. Listing book dealt with knots and sur- faces but failed to generate much interest in either the name or the subject matter. Throughout much of the nineteenth and early twentieth centuries, much of what now falls under the auspices of topology was studied under the name of analysis situs (analysis of position). Bernard Riemann (1826{1866) was the rst mathematician to foresee topology in the generality it has achieved today. He initiated the study of connectivity of a surface, or the arrangement of holes in a surface. He used concepts in which the number of dimensions exceeded three, which at that time was generally conceded to be the maximum number of dimensions involved with any geometric object. Present-day topology can be traced to two primary sources: the development of non-euclidean geometry and the process of putting calculus on a rm mathematical foundation. 1The information here is taken from Principles of Topology by Fred H. Croom. 2 1.2. SETS AND SET OPERATIONS 3 1.2 Sets and Set Operations We need some basic information about sets in order to study the logic and the ax- iomatic method. This is not a formal study of sets, but consists only of basic deni- tions and notation. Bracesfandgare used to name or enumerate sets. The roster method for naming sets is simply to list all of the elements of a set between a pair of braces. For example the set of integers 1, 2, 3, and 4 could be named f1;2;3;4g: This does not work well for sets containing a large number of elements, though it can be used. The more common method for this is known as the set builder notation . A property is specied which is held by all objects in a set. P(x), read P of x , will denote a sentence referring to the variable x. For example, x= 23 xis an odd integer. 1x4. The set of all objects xsuch thatxsatisesP(x) is denoted by fxjP(x)g: The setf1;2;3;4gcan be named fxj1x4; x2Zg=fx2Zj1x4g: From hence forth, the words object, element , and member mean the same thing when referring to sets. Sets will be denoted mainly by capital Roman letters and elements of the sets by small letters. The following have the same meaning: a2A ais in setA ais a member of set A ais an element of set A Likewise,a62Ameans that aisnotan element of set A. Ais a subset ofBif every element of Ais also an element of B. The following have the same meaning: AB Every element of Ais an element of B c 1999, David Royster Introduction to Topology For Classroom Use Only 4 CHAPTER 1. INTRODUCTION TO TOPOLOGY Ifa2A, thena2B Ais included in B BcontainsA Ais a subset of B Note that a set is always a subset of itself. IfAandBare sets, then we say that A=BifAandBrepresent the same set: A=B AandBare the same set AandBhave the same members ABandBA The set which contains no elements is known as the empty set , and is denoted by ;. Note that for each set A,;A. The intersection of two sets AandBis the set of all elements common to both sets. The intersection is symbolized by A\Borfxjx2Aandx2Bg. The union of two sets AandBis the set of elements which are in AorBor both. The union is symbolized by A[Borfxjx2Aorx2Bg. 1.2.1 Universal Sets and Compliments When we are working in an area or on a certain problem, we always have a frame of reference in which we are working called a universal set . In our geometry course, it will be the set of points that lie on a plane. In calculus we consider the set of real numbers, the set of real functions, the set of dierentiable functions, and the set of continuous functions as universal sets. The complement of a setAis dened to be the set of all elements of the universal set which are not in A, and is symbolized by CA=A0=Ac. Note that A[Acis always the universal set, while A\Ac=;. The set dierence of the setsAandBis dened to be all of those elements in A which are not in B. It is denoted by AnB=fx2Ajx3Bg: Note thatAnBandBnAwill usually be dierent, and that even though AnB=; it need not follow that A=B. c 1999, David Royster Introduction to Topology For Classroom Use Only 1.3. PRODUCTS SETS 5 1.3 Products Sets LetXandYbe sets. The set of all ordered pairs f(x;y)jx2X and y2Ygis theproduct set XY, or Cartesian product ordirect product ofXandY. A slice of this product set is fxgYorXfygfor a given x2Xory2Y. Examples of common product spaces are the plane R2=RR, 3-spaceR3=RR2, a right circular cylinder, S1[0;1], or the torus, S1S1. Theorem 1 LetXandYbe sets and let A;CXandB;DY. a)A(B\D) = (A\B)(A\D). b)A(B[D) = (A[B)(A[D). c)A(YnD) = (AY)n(AD). d)(AB)\(CD) = (A\C)(B\D). e)(AB)[(CD)(A[C)(B[D). f)(XY)n(AB) = (X(YnB))[((XnA)Y) The concept of the product of two sets can be extended to more than two factors. IffXign i=1is a nite collection of sets, then their product is X1X2


Xn=nY i=1Xi=f(x1;x2;:::;xn)jxi2Xifor eachi= 1;2;:::;ng: For an innite collection of sets, the product is dened by 1Y i=1Xi=f(x1;x2;x3;:::)jxi2Xifor eachi= 1;2;:::g: 1.4 Functions Afunctionf:X!Yis a rule which assigns to each x2Xa uniquey2Yand we sayy=f(x). Ify=f(x) thenyis called the image ofxandxis called the preimage ofy. The setXis the domain offandYis the range orcodomain off. LetAX. The setf(A) =fy2Yjy=f(x) for somex2Agis called the image ofA. The setf(X) is called the image off. ForBY, the set f1(B) =fx2Xjf(x)2Bg is the inverse image ofBunderf. The set of points =f(x;f(x))2XYjx2Xg is called the graph of the function f. c 1999, David Royster Introduction to Topology For Classroom Use Only 6 CHAPTER 1. INTRODUCTION TO TOPOLOGY A functionf:X!Yisinjective if for distinct elements x1;x22X,f(x1)6=f(x2) inY. Another way to think of this is to say that fis injective if f(x1) =f(x2) implies thatx1=x2. Iff(X) =Y, the function fis said to be surjective . A function that is surjective and injective is called a bijection . In this case we have thatf:X!Yis a bijection provided that each member of Yis the image underfof exactly one member of X. In this case the inverse function f1:Y!X exists assigning to each element y2Yits unique preimage x=f1(y) inX. The identity function iX:X!Xfrom a set Xto itself is the function dened byiX(x) =xfor allx2X. This function is often denoted by 1 X. Iff:X!Yandg:Y!Zare functions, then the composite function gf:X! Zis dened by gf(x) =g(f(x)), forx2X. Denition 1 LetXbe a set. A sequence inXis a function f:Z+!Xwhose domain is the set of all positive integers, Z+or the set of positive integers less than or equal to some given positive integer N. The sequence is called nite if its domain isf1;2;:::;Ngand innite if its domain is all positive integers. 1.5 Equivalence Relations LetXbe a set. A relationRonXis a subset of XX. If (x;y)2Rwe will say thatxis related to ybyRand to write xRy. A relationRon a setXis called re exive ,symmetric , ortransitive if it satises the corresponding property below. (a)The Re exive Property :xRx for allx2X. (b)The Symmetric Property : IfxRy, thenyRx. (c)The Transitive Property : IfxRy andyRz, thenxRz. Denition 2 Anequivalence relation on a setXis a relation on Xwhich is re exive, symmetric, and transitive. Denition 3 Letdenote an equivalence relation on X. Forx2Xthe set [x] = fy2Xjyxgis calle dth e equivalence class of x. Proposition 1 LetXbe a set and letdenote an equivalence relation on X. a)x2[x]for eachx2X. b)xyif and only if [x] = [y]. c)x6yif and only if [x]\[y] =;. d) Forx;y2X,[x]and[y]are either identical or disjoint. c 1999, David Royster Introduction to Topology For Classroom Use Only 1.6. CARDINALITY 7 1.6 Cardinality We are often interested in how big sets are in relation to one another. Clearly, we can tell the dierence in sizes of two nite sets, but how do we dierentiate between two innite sets? Are there dierent sizes of innite sets? How do we compare sets to tell if one has a greater number of members? Denition 4 a setAisnite ifAis empty or if there is a bijection between Aand the set of integers from 1toNfor some positive integer N. In the latter case, Ais said to have Nmembers. If a set is not nite, it is called innite . Denition 5 A set is denumerable orcountably innite if there is a bijection between the set and the positive integers. A set which is either nite or denumerable is called countable . A set which is not countable is called uncountable . Lemma 1 a) Each subset of a nite set is nite. b) Each subset of a countable set is countable. c) Each set which contains an innite set is innite. d) Each set which contains an uncountable set is uncountable. Example 1 1. The setZ+[f0gof all non-negative integers is countable. The bijection is f:Z+[f0g!Z+given by f(n) =n+ 1;n2Z+[f0g: 2. The set of all integers, Zis countable. The bijection g:Z!Z+is given by g(n) =( 2n1 ifnis positive 2n ifnis negative 3. The product set Z+Z+is countable. One method is to use the Cantor Diagonalization Method to count the ordered pairs ( m;n). A second method is to dene the function g:Z+Z+!Z+by g(m;n) = 2m3n;(m;n)2Z+Z+: Now,gis not surjective, but the Fundamental Theorem of Arithmetic on the unique factorization into primes guarantees that the function gis injective. Thus, there is a bijection from Z+Z+to a subset of Z+. Since every subset of a countable set is countable, we have that Z+Z+is countable. c 1999, David Royster Introduction to Topology For Classroom Use Only 8 CHAPTER 1. INTRODUCTION TO TOPOLOGY Theorem 2 a) IffAigN i=1is a nite collection of nite sets, then bothSN i=1AiandQN i=1Aiare nite. (Finite unions and nite products of nite sets are nite.) b) IffAig1 i=1is a countable collection of countable sets, thenS1 i=1Aiis countable. (Countable unions of countable sets are countable.) c) IffAigN i=1is a nite collection of countable sets, then bothQN i=1Aiis countable. (Finite products of countable sets are countable.) Why didn't we claim that a countable product of countable sets is countable? Mainly because it is not true, as is seen in the following example. Example 2 LetAi=f0;1gfori= 1;2;:::. Let U=1Y i=1Ai=f(a1;a2;a3;:::)jai= 0 or 1g: Assume that Uis countable. Then there is a bijection f:Z+!U. For an element a= (a1;a2;a3;:::)2Uwe shall refer to a1as the rst coordinate, a2as the second coordinate, and so forth. We can list all of the elements in Uusing the bijection f. They areff(1);f(2);F(3);:::g. Consider the following element in U. Denex= (x1;x2;x3;:::) as follows: xi=( 0 if theithcoordinate of f(i) is 1 1 if theithcoordinate of f(i) is 0 Then, we have that for each positive integer i,x6=f(i) for they dier in the ith coordinate. This means that xis not in the exhaustive list of elements we have in our bijection. That is this bijection is not surjective. This contradiction show us that U cannot be countable. Theorem 3 The set of rational numbers is countable. Proof: There are several ways of proving this. One method is to use the Cantor Diagonalization Method to count the rationals. (Method 1 ) List the positive rational numbers in rows where the rst row consists of all positive rational numbers with 1 as a denominator, the second row lists all positive rational numbers with 2 as a denominator, and so on: 1 12 13 14 1::: 1 22 23 24 2::: 1 32 33 34 3::: 1 42 43 44 4::: ............... c 1999, David Royster Introduction to Topology For Classroom Use Only 1.6. CARDINALITY 9 Clearly, we have each positive rational number in here numerous times, but the diagonalization method will still show that there are a countable number of elements in this array. The positive rationals form a subset of this array, thus there must be a countable number of positive rationals. This will yield that there are a countable number of rational numbers. (Method 2 ) Every rational number can be expressed uniquely in lowest terms as m=n wheremandnare integers with no common positive divisor other than 1, and nis positive. Consider the function m=n7!(m;n) from the set of rational numbers intoZZ. This function is injective since the ordered pair ( m;n) determines only one rational number m=n. Thus, the set of rational numbers is equivalent to a subset of the countable set ZZ, and is hence countable. Theorem 4 The set of real numbers is uncountable. Proof: We will make use of the example of the countable product above. Each element inQ1 i=10;1 is a sequence consisting of 0's and 1's. Each of these sequences represents a unique real number between 0 and 1, by the correspondence (a1;a2;a3;:::)7!0:a1a2a3:::: This is a one-to=one correspondence. Thus, the set of real numbers between 0 and 1 representable as a decimal using only 0's and 1's is an uncountable set. Thus, R contains an uncountable set and hence is uncountable. Theorem 5 The set of irrational numbers is uncountable. Proof: Since the set of real numbers is the union of the set of rational numbers and the set of irrational numbers, if the set of irrationals were countable, then we would have that the real numbers are countable. That failing to be true, implies that the irrationals must be uncountable. c 1999, David Royster Introduction to Topology For Classroom Use Only Chapter 2 Metric Spaces 2.1 Denition and Some Examples Denition 6 LetXbe a set and d:XX!R+a function satisfying the following properties. For all x;y;z2X, a)d(x;y) = 0 if and only if x=y. b)d(x;y) =d(y;x). c)d(x;z)d(x;y) +d(y;z). Thendis called a metric ordistance function onXandd(x;y)is called the distance fromxtoy. The setXwith a metric dis called a metric space and is denoted by (X;d). Note that these properties are modeled on the distance functions that we have on RandR2. Doing so we usually call property (c) the Triangle Inequality . Example 3 The real line, Ris a metric space using the standard distance function, the absolute value: d(a;b) =jabj. The above properties are standard proofs about the absolute value function. Example 4 The plane,R2, with the usual Euclidean distance formula is a metric space. IfP= (x1;y1) andQ= (x2;y2), then d(P;Q) =p (x2x1)2+ (y2y1)2: Example 5 These are special cases of the general Euclidean n-space, Rn=f(a1;a2;::: ;an)jai2Rg: 10 2.1. DEFINITION AND SOME EXAMPLES 11 The distance formula here is the usual distance formula for Euclidean n-space: d((x1;x2;:::;xn);(y1;y2;::: ;yn)) = nX i=1(xiyi)2!1=2 : dis called the usual metric onRn. To show that dis a metric, we need two standard results about vectors in Rn. First, leta2Rn. The normkakis the distance from ato the origin O= (0;0;:::; 0): kak=d(a;O) = nX i=1a2 i!1=2 : Theorem 6 (Cauchy-Schwarz Inequality) For any points a;b2Rn ja
bjkakkbk: Theorem 7 (The Minkowski Inequality) For any points a;b2Rn ka+bkkak+kbk: The distance between two points is given by d(a;b) =kabk. The rst two conditions making da metric are easily seen to be satised. We only need check the Triangle Inequality. Let x;y;z2Rn d(x;z) =kxzk=kxy+yzk kxyk+kyzk =d(x;y) +d(y;z) Example 6 [The Taxicab Metric] Dene a function d0:R2R2!Ras follows. If x= (x1;x2) andy= (y1;y2), then d0(x;y) =jx1y1j+jx2y2j: This is called the taxicab metric because the distance is measured along line segments parallel to the coordinate axes. Clearly,d0(x;x) = 0 and if d0(x;y) = 0, thenjx1y1j+jx2y2j= 0 which meansjx1y1j= 0 andjx2y2j= 0. This implies that x1=y1andx2=y2, andx=y. Because of the basic properties of the absolute value, it is obvious that d0(x;y) =d0(y;x). The Triangle Inequality follows because of the validity of the Triangle Inequality with the absolute value on the real line. What is the following set? U=fx= (x1;x2)2R2jd0(x;O) = 1g: We can dene an analogous metric, called the taxicab metric, on Rn. d0(x;y) =nX i=1jxiyij: c 1999, David Royster Introduction to Topology For Classroom Use Only 12 CHAPTER 2. METRIC SPACES Example 7 [The Max Metric on Rn] Another metric for Rnis given by taking the largest of the dierences of the coordinates of xandy. d00(x;y) = maxfjxiyign i=1: Example 8 [The Discrete Metric] For any set X, dene d(x;y) =( 0 ifx=y 1 ifx6=y This denes a metric on X, called the discrete metric . It is usually of little use, except for counterexamples. It does show, though, that every set can be assigned a metric. Example 9 LetC[a;b] denote the set of all continuous real-valued functions dened on the interval [ a;b]. Forf;g2C[a;b] dene (f;g) =Zb ajf(x)g(x)jdx: The fact that is a metric follows from the usual properties of the Riemann integral. This metric measures the distance between two functions to be the area between the two graphs from x=atox=b. Example 10 For the set C[a;b] dene0by 0(f;g) = lubfjf(x)g(x)jjx2[a;b]g: The metric is called the supremum metric or the uniform metric forC[a;b]. It measures the distance between fandgto be the supremum of the vertical distances from points ( x;f(x)) to (x;g(x)) on the graphs of fandgon the closed interval [ a;b]. Denition 7 A numberuis an upper bound for a setAof real numbers provided thataufor alla2A. If there is a smallest upper bound u0forA, that is an upper bound that is less than or equal to all other upper bounds for A, thenu0is called theleast upper bound or supremum ofA. The least upper bound for a set Ais denoted by lubAorsupA. Denition 8 A number`is an lower bound for a setAof real numbers provided that`afor alla2A. If there is a largest lower bound `0forA, that is a lower bound that is less than or equal to all other lower bounds for A, then`0is called the greatest lower bound or inmum ofA. The greatest lower bound for a set Ais denoted by glbAorinfA. c 1999, David Royster Introduction to Topology For Classroom Use Only 2.2. CONTINUOUS FUNCTIONS 13 A very basic property of the real numbers is included in the following two state- ments: The Least Upper Bound Property : Every non-empty set of real numbers which has an upper bound has a least upper bound. The Greatest Lower Bound Property : Every non-empty set of real numbers which has a lower bound has a greatest lower bound. We will accept the rst property as an axiom of the real number system. The second property follows from the rst. Denition 9 Let(X;d)be a metric space and let Abe a non-empty subset of X. Iffd(x;y)kx;y2Aghas an upper bound, then Ais said to be bounded , and lubfd(x;y)kx;y2Agis called the diameter ofA. For completeness, we dene the diameter of the empty set to be 0. If the set Xis bounded, then we call (X;d)a bounded metric space. Ifx2X, then the distance fromxtoAis dened by d(x;A) = glbfd(x;y)jy2Ag: Theorem 8 Letf(Xi;di)gn i=1be a nite collection of metric spaces and let X=nY i=1Xi: For each pair of points x= (x1;x2;:::;xn),y= (y1;y2;:::;yn)inX, letd:XX! Rbe dened by d(x;y) = nX i=1(di(xi;yi))2!1=2 : Then (X;d)is a metric space. The metric ddened above is called the product metric onX. 2.2 Continuous Functions In topology we are concerned with how spaces are changed when stretched, bent, twisted and modied | but not torn. We do so by studying the maps that do so. Our friend here is the continuous map. In your study of calculus, you saw that continuous functions did many things. At the time you were more interested in special continuous functions | the dierential functions. We here are more interested in the more general function. In calculus, we saw that a continuous function was one that did not do too much damage to the domain in the range. By this, we mean that if two points were close in the domain, then their images were not too far apart in the image. We saw this intuitively through looking at graphs and looking at limits. To insure specicity, we need the denition of continuity due to Cauchy and Weierstrauss. It is one with which you are familiar. c 1999, David Royster Introduction to Topology For Classroom Use Only 14 CHAPTER 2. METRIC SPACES Denition 10 Letf: (X;d)!(Y;d0)be a function between two metric spaces. Let a2X. We say that fis continuous at aif given any  >0there is a > 0so thatd0(f(x);f(a))<  wheneverd(x;a)< . We say that fis continuous if it is continuous at a2Xfor alla2X. This clearly depends on the metric in each of the two spaces. A change of met- ricmight change the continuity of the given function. Will it? Is continuity that dependent on the metric in the domain or the range? Let's check two well-known functions that we think should be continuous and make certain that they are continuous under this denition. Example 11 Letf: (X;d)!(Y;d0) be given by f(x) =bfor allx2Xwhereb2Y is a constant. This is just the constant function . To show that fis continuous, we need to show that if we are given any >0, then we can nd a  >0 so that whenever d(x1;x2)<thend0(f(x1);f(x2))<. In this case, this is easy. This is because d0(f(x1);f)(x2)) =d0(b;b) = 0< for any choice ofx1;x22X. Thus, it does not matter what we may choose for . You could take =or= 1. Regardless, whenever d(x1;x2)<thend0(f(x1);f(x2)) = 0<, and we are done. Example 12 Let 1X: (X;d)!(X;d) denote the identity map from Xto itself given by 1X(x) =x. We claim that this function is continuous. Again, to show this we are given an >0. We then need to nd a  >0 so that wheneverd(x1;x2)<thend(1X(x1);1X(x2))<. However, since 1 X(x1) =x1and 1X(x2) =x2, it is easy to see that if we take , then ifd(x1;x2)< it follows thatd(1X(x1);1X(x2)) =d(x1;x2)<. Thus, 1 Xis a continuous function. Example 13 This time we will be working with the same underlying set, but we will place a dierent metric on it. Will this make a dierence? LetX=Rnwith the usual metric. Let Y=Rnwith the maximum metric, d00((x1;x2;:::;xn);(y1;y2;:::;yn)) = max 1infjxiyijg: Deneh: (X;d)!(Y;d00) byh(x1;x2;:::;xn) = (x1;x2;:::;xn). It is the identity map on the underlying set, but it does not carry the same metric information. Is h continuous? Is h1continuous? It turns out that both are continuous! To prove this, let's rst look at h1: (Y;d00)! (X;d). We are given an >0. We need to nd a >0 so that if d00((x1;x2;:::;xn);(y1;y2;:::;yn))< thend(h1(x1;x2;:::;xn);h1(y1;y2;:::;yn))<. c 1999, David Royster Introduction to Topology For Classroom Use Only 2.3. OPEN SETS AND CLOSED SETS 15 To say that d00((x1;x2;:::;xn);(y1;y2;:::;yn))<means thatjxiyij<for all i= 1;:::;n . Thus, d(h1(x1;x2;:::;xn);h1(y1;y2;:::;yn)) =d((x1;x2;:::;xn);(y1;y2;:::;yn)) (2.1) = nX i=1jxiyij2!1=2 (2.2) < nX i=12!1=2 =pn (2.3) Thus, we need pn< , or take<pn. Now, to show that his continuous we are given an  >0. We need to nd so that whenever d((x1;x2;:::;xn);(y1;y2;:::;yn))<, we have that d00((x1;x2;:::;xn);(y1;y2;:::;yn))<: To say that d((x1;x2;:::;xn);(y1;y2;:::;yn))<means that (Pn i=1jxiyij2)1=2< . Thus, each of the dierences jxiyijmust be less than , and the largest of these dif- ferences is still less than . Thus, in order for d00((x1;x2;:::;xn);(y1;y2;:::;yn))<, we need only choose < . 2.3 Open Sets and Closed Sets Denition 11 Let(X;d)be a metric space, a2X, andr>0a positive real number. Theopen ball Bd(a;r)withcenteraandradiusris the set Bd(a;r) =fx2Xjd(a;x)<rg: When there is only one metric under consideration, we will simplify the notation to B(a;r). Denition 12 Let(X;d)be a metric space and let a2X. A subset NXis a neighborhood of aif there is a >0so thatB(a;)N. The collection Naof all neighborhoods of a point a2Xis called a complete system of neighborhoods of the pointa. Denition 13 A subsetUof a metric space (X;d)is an open set with respect to the metric dprovided that Uis a union of open balls. The family of all open sets dened in this way is called the topology for Xgenerated by d. A subsetCX is said to be closed (with respect to d) if its complement XnCis an open set (with respect tod). c 1999, David Royster Introduction to Topology For Classroom Use Only 16 CHAPTER 2. METRIC SPACES Thus, a neighborhood of aand an open set containing aneed not be the same thing. However, if Uis an open set containing a, thenUis a neighborhood of a. Theorem 9 The following statements are equivalent (TFAE) for a subset Uof a metric space (X;d). a)Uis an open set; b) for each x2Uthere is an x>0so thatB(x;x)U. c) for each x2U,d(x;XnU)>0, ifU6=X. Proof: What this means is that Statements (a) and (b) are equivalent, (b) and (c) are equivalent, and (a) and (c) are equivalent. We can show this by proving that (a) is equivalent to (b) and then that (b) is equivalent to (c). In condition (c) we will assume that U6=Xsince the distance from the empty set is not dened. Assume that Uis an open set and let x2U. SinceUis the union of open balls, thenx2B(a;r)U. Thend(x;a)< r. We want to center an open ball at xand have it contained in U. Choosexrd(x;a). ThenB(x;x)B(a;r) for the following reason: If y2B(x;x), d(y;a)d(y;x) +d(x;a)<x+d(x;a)rd(x;a) +d(x;a) =r: Thus,B(x;x) is an open ball of positive radius centered at xand contained in U. Thus (a) =)(b). To show that ( b) =)(a), since each x2Ulies in an open ball contained in U, Uis the union of these open balls. To see that ( b) =)(c), letB(x;x)U. Then any point within distance xof xis inU, so the distance from xtoXnUmust be at least x. Thus,d(x;XnU)>0 for eachx2U. Assuming that (c) holds, d(x;XnU) =x>0 depending on x. This means that the distance from xto a point outside Umust be at least x, so any point within distancexofxmust be in U. This means B(x;x)U. Note that we have just shown that for each a2Xand for each  >0, the open ballB(a;) is a neighborhood of each of its points. 2.3.1 Neighborhoods and Continuous Functions How do we plan to use this information? While our denition of continuity is precise, it requires some specicity and does not look generalizible. What I mean by this is that the denition seems to rely specically on the denition of the metric, and it will be hard to realign our denition when we have to move away from metric spaces. Theorem 10 Lef: (X;d)!(Y;d0).fis continuous at a2Xif and only if for each neighborhood Moff(a)there is a corresponding neighborhood Nofa, such that f(N)M; c 1999, David Royster Introduction to Topology For Classroom Use Only 2.4. LIMITS 17 or equivalently Nf1(M): Proof: First, let's suppose that fis continuous at a2Xand letMbe a neigh- borhood of f(a). This means that for some  > 0Bd0(f(a);)M. Sincefis continuous at a2Xwe know that we can nd  > 0 so that if d(x;a)<  then d0(f(a);f(x))<. This says that Bd(a;)f1(M) and we already have seen that Bd(a;) is a neighborhood of a2X. Thus, if fis continuous, we have found a corresponding neighborhood to M. Now, suppose that for any neighborhood, M, off(a) we can nd a neighborhood Nofaso thatf(N)M. Let >0 be given to you. You must nd a  >0 so that whenever d(a;x)< we haved0(f(a);f(x));. Now, let M=Bd0(f(a);).M is a neighborhood of f(a), so we know that there is a neighborhood NXofaso thatf(N)M. SinceNis a neighborhood of a, it must contain a -ball centered at a,Bd(a;)N, by the denition of a neighborhood. Thus, if we have x2Bd(a;), thenf(x)2M=Bd0(f(a);). The other way of writing this is: if d(a;x)< then d0(f(a);f(x))<. Therefore, fis continuous at a2X. 2.4 Limits Recall that a sequence is just a function a:Z+!(X;d). We want to discuss what happens to the sequence as we let ngo to innity; in other words what happens to the sequence as we look further and further into the range of a. Let us rst recall the denitions in the real numbers and then try to set them up so that we can easily generalize them to arbitrary metric spaces. Letfaigbe a sequence of real numbers. A real number Lis said to be the limit of the sequencefangif, given any >0, there is a positive integer Nsuch that whenever n>N ,janLj<. In this case we say that the sequence converges to Land write lim n!1an=L: How can we generalize this to an arbitrary metric space? It should not be hard, because all we used in the denition was the distance function in the real numbers. We will use the distance function in our metric space similarly. Denition 14 Letfxngbe a sequence in the metric space (X;d). We say that this sequencefxngconverges to x2Xif given any  >0there is a positive N2Z+so that whenever n>N ,d(x;xn)<. In this case we will write limxn=x. Lemma 2 Let(X;d)be a metric space and fxngbe a sequence in X. Then limxn= x2Xif and only if for each neighborhood Vofxthere is an integer N > 0so that xn2Vwhenevern>N . c 1999, David Royster Introduction to Topology For Classroom Use Only 18 CHAPTER 2. METRIC SPACES This is nothing but applying the denitions of convergence and neighborhood, and its proof will be omitted. IfSis a set of innite points and there is at most a nite number of elements ofSfor which a certain statement is false, thent he statement is said to be true for almost all ofS. Thus, we may phrase the above lemma by saying that the sequence fxngconverges to xif each neighborhood of xcontains almost all of the poins of the sequence. One reason for looking a sequences is the concept of continuity. In calculus we dene a function to be continuous at a2Rif the following conditions were met: 1. limx!af(x) exists; 2.f(a) exists; 3. limx!af(x) =f(a). It suces to check this for all sequences appproaching a(a fact to be proven later). Thus, we nd that we can show that fis continous at aif for each sequence fxng!a, we have thatff(xn)g!f(a). We are able to extend this result to arbitrary metric spaces. Theorem 11 Letf: (X;d)!(Y;d0).fis continous at a point a2Xif and only if whenever limxn=xwe have limf(xn) =f(x). The proof is straightforward. Proof: Assume that fis continuous and let fxng!xinX. Let >0 and let M=Bd0(f(x);)2V. There is a neighborhood UofxinX, such that f(U)M. SinceUis a neighborhood there is a >0 so thatBd(x0)U. Now,fxng!xso for thisthere is a positive integer Nso that whenever n>N we havexn2Bd(x0)U. Thus,f(xn)2f(U)M. Therefore, for any neighborhood Moff(x) there is a positive integer Nso that whenever n>N we haved0(f(x);f(xn)<, which implies that the sequence ff(xn)gconverges to f(x). To prove the other direction, 2.5 Open Sets and Closed Sets Revisited Remember that we dened an open set as a set that is the union of open balls. A set is closed if its complement is open. Theorem 12 The open subsets of a metric space (X;d)have the following properties: 1.Xand;are open sets. 2. The union of any family of open sets is open. 3. The intersection of a nite family of open sets is open. c 1999, David Royster Introduction to Topology For Classroom Use Only 2.5. OPEN SETS AND CLOSED SETS REVISITED 19 Proof: These are straightforward. 1. The whole space Xis open since it is the union of all open balls with all possible centers and radii. The empty set is open since it is the union of the empty collection of open balls. 2. IffUj2Agis a family of open sets in X, then each Uis a union of open balls. ThenS 2AUis the union of all of the open balls that comprise each U and is hence open. 3. LetfUiji= 1;:::;ngbe a nite collection of open sets and let x2Tn i=1Ui. Then, by our previous theorem there exist i,i= 1;:::;n so thatBd(x;i)Ui and n\ i=1Bd(x;i)n\ i=1Ui: Let= minfiji= 1;:::;ng. Then,Tn i=1Bd(x;i) =Bd(x;). Thus,Bd(x;) is an open ball centered at xand contained inTn i=1Ui. Thus,Tn i=1Uiis open. Theorem 13 The closed subsets of a metric space (X;d)have the following proper- ties: 1.Xand;are closed sets. 2. The intersection of any family of closed sets is closed. 3. The union of a nite family of closed sets is closed. This follows from our previous theorem and complements. Denition 15 Let(X;d)be a metric space and Aa subset of X. A pointx2Xis alimit point oraccumulation point ofAprovided that every open set containing xcontains a point of Adistinct from x. The set of limit points of Ais called its derived set, denoted by A0. Lemma 3 Let(X;d)be a metric space and Aa subset of X. A pointx2Xis a limit point of Aif and only if d(x;Anfxg) = 0 . Lemma 4 A subsetAof a metric space (X;d)is closed if and only if Acontains all its limit points. c 1999, David Royster Introduction to Topology For Classroom Use Only 20 CHAPTER 2. METRIC SPACES Proof: LetAbe closed and let xbe a limit point of A. Ifx62AthenXnAis an open set containing xbut containing no other point of A. Thus,xcould not be a limit point of A. This means that if xis a limit point of A, then it must be a member ofA. Now suppose that Acontains all of its limit points. To show that Ais closed, we must show that XnAis open. Ifx2XnA, thenxis not a limit point of A. Thus, there is some open set Uxcontaining xbut no other point of A. ThenXnAis the union of all of these sets. Hence, XnAis open and Ais closed. What is the connection between limit points and the limit of a sequence? Theorem 14 Let(X;d)be a metric space and Aa subset of X. 1. A point x2Xis a limit point of Aif and only if there is a sequence of distinct points ofAwhich converges to x. 2. The set Ais closed if and only if each convergent sequence of points of Acon- verges to a point of A. Corollary 1 Letxbe a limit point of a subset Aof a metric space X. Then every open set containing xcontains innitely many members of A. 2.6 Interior, Closure, and Boundary Denition 16 LetAbe a subset of a metric space (X;d). A point x2Ais an interior point ofAif there is an open set Uwhich contains xand is contained in A;x2UA. The interior ofA, denoted int A, is the set of all interior points of A. Note that for the open set Uin the denition, every point of Uis an interior point ofA. Thus, the interior of Acontains every open set contained in Aand is the union of this family of open sets. This means two things: 1. the interior of a set Ais an open set, and 2. the interior of a set Ais the largest open set contained in A. Item (2) above means that if Uis open and UA, thenUintA. Example 14 LetX=Rwith the usual metric. 1. Fora;b2Rwitha<b int(a;b) = int[a;b) = int(a;b] = int[a;b] = (a;b): c 1999, David Royster Introduction to Topology For Classroom Use Only 2.6. INTERIOR, CLOSURE, AND BOUNDARY 21 2. The interior of a nite set is empty, since such a set cannot contain any open interval. 3. The interior of the set of irrational numbers is empty, since each open interval must contain some rational number. Likewise, the interior of the set of rationals is empty. If the rationals contained an open interval, then the set of rationals would have to be uncountable, since an open interval is uncountable. 4. int;=;; intR=R. Denition 17 TheclosureAof a subset of a metric space (X;d)is the union of the setAand the set of its limit points: A=A[A0 whereA0is the derived set of A. Example 15 LetX=Rwith the usual metric. 1. Fora;b2Rwitha<b (a;b) =[a;b) =(a;b] =[a;b] = [a;b]: 2. The closure of a nite set is itself, since the set of limit points of a nite set is empty. 3. The closure of the set of rational numbers is R. Likewise, the closure of the set of irrationals is R. Since every open interval contains both rational and irrational numbers. 4.;=;;R=R. While the interior of a set is the largest open set contained in the set, the closure has a similar property described in the next theorem. Theorem 15 IfAX, thenAis a closed set and is a subset of every closed set containingA. This says that the closure of a set is the smallest closed set containing the set. Proof: To show that Ais closed, we need to show that it contains all of its limit points. Suppose that x62A. Then there is an open set Ucontaining xso that U\A=;. Now, this means that Ucannot contain a limit point of Aeither, since if an open set contains a limit point of Ait must contain some other point of Aalso. Thus,Ucontains no point of A, soxis not a limit point of A. This means that all of the limit points of Amust be contained in A. Thus,Ais closed. Suppose now that Fis a closed subset of XandAF. Then we can show that AFand, sinceFcontains all of its limit points, then F=F[F0=F. Thus, AFfore every closed set FcontainingA. c 1999, David Royster Introduction to Topology For Classroom Use Only 22 CHAPTER 2. METRIC SPACES Since this shows that Ais the smallest closed set containing A, we can easily show thatAis the intersection of all closed sets containing A. Theorem 16 LetAbe a subset of the metric space (X;d). 1.Ais open if and only if A=intA. 2.Ais closed if and only if A=A. Denition 18 LetAbe a subset of the metric space (X;d). A pointx2Xis a boundary point ofAprovided that x2A\XnA. The set of boundary points of A is called the boundary ofAand is denoted by @A. The industrious reader will readily work to show that the following statements are equivalent for a subset AofXand a points xin the metric space ( X;d). 1.x2@A, 2.x2(AnintA), 3. Every open set containing xcontains a point of Aand a point of XnA. 4. Every neighborhood of xcontains a point of Aand a point of XnA. 5.d(x;A) =d(x;XnA) = 0. 6.x2A\XnA. Example 16 1. LetX=Rwith the usual metric. For a;b2Rwitha<b @(a;b) =@[a;b) =@(a;b] =@[a;b] =fa;bg: 2. InRn @B(a;) =fx2Rnjd(a;x) =g: 3. The boundary of the set of all points in Rnhaving only rational coordinates is Rn. 4. For any metric space ( X;d), @;=@X=;: c 1999, David Royster Introduction to Topology For Classroom Use Only Chapter 3 Topological Spaces 3.1 Denition and Some Examples We want to generalize the concepts that we developed in studying the metric spaces. We want to remove our reliance on a distance function. We were able to dene most of what we wanted to do in metric spaces by dening our concepts in terms of the open sets. This was especially true of our study of continuous functions. We will use the results that we proved about open sets as our basis for the gener- alization. We will dene open sets as sets that satisfy certain conditions. Denition 19 LetXbe a set and Ta family of subsets of Xsatisfying the following properties. a) The set Xand;belong to T, b) The union of any family of members of Tis a member of T. c) The intersection of any nite family of members of Tis a member of T. ThenTis called a topology forXand the members of Tare called open sets . The ordered pair (X;T)is called a topological space , or simply a space. If we use the terminology open sets instead of member of T, then the denition of a topological space may be restated as follows: A family of subsets of Xis a topology forXmeans that: a) BothXand;are open sets. b) The union of any family of open sets is an open set. c) The intersection of any nite family of open sets is open. 23 24 CHAPTER 3. TOPOLOGICAL SPACES Example 17 The usual topology for the real line Ris the topology generated by its usual metric. We shall refer to the real line with the usual topology as simply the real line orR. Example 18 The usual topology onRnis the topology generated by the usual metric onRn. It is also the topology generated by the taxicab metric and the max metric. Thus, the usual topology does not distinquish the metric determining it from the other two. We shall refer to Rnwith the usual topology as Euclidean n-space , or simplyRn. Example 19 For any set Xwe take T= 2Xto be the set of all subsets of X. This clearly satises all of the properties of a topology, since we have included every possible subset in the topology. This is called the discrete topology . Note that it is the topology generated by the discrete metric. Also, note that this is the largest possible collection of open subsets of X. Example 20 At the opposite extreme, we may take T=f;;Xg. This is called the trivial topology , orindiscrete topology , onX. This is the smallest collection of open sets onX. Example 21 LetXbe a set. We shall take Tto consist of;,X, and all sets Uso thatXnUis a nite set. Then Tis a topology on Xcalled the conite topology , or nite complement topology . This is really of interest only when Xis an innite set. WhenXis a nite set, this is the same as the discrete topology. Denition 20 A subsetFof a topological space Xisclosed ifXnFis an open set. Theorem 17 The closed sets of a topological space Xhave the following properties: a)Xand;are closed. b) The intersection of any family of closed sets is closed. c) The union of any nite family of closed sets is closed. Denition 21 Let(X;T)be a topological space and let AX. A pointxinXis alimit point ofAif every open set containing xcontains a point of Adistinct from x. The set of limit points of Ais called the derived set ofA, denotedA0. c 1999, David Royster Introduction to Topology For Classroom Use Only 3.2. INTERIOR, CLOSURE AND BOUNDARY 25 Example 22 LetX=fa;b;c;dg. LetT0be the indiscrete topology; T1, the discrete topology; T2=f;;X;fag;fbg;fa;bgg; and T3=f;;X;fag;fbg;fcg;fa;bg;fa;cg;fb;cg;fa;b;cgg. The reader should verify that T2andT3are topologies on X. LetA=fa;bg,B=fcg, andC=fdg. We want to nd the limit points of these sets in the dierent topologies. T0T1T2T3 AX;fc;dgfdg Bfa;b;dg;fdgfdg Cfa;b;cg;fcg; Theorem 18 A subsetAof a topological space Xis closed if and only if Acontains all of its limit points. This is no surprise, and is proven exactly the way in which we proved it earlier in a metric space. We were careful there not to use the distance function, but to use the open sets. Denition 22 LetXbe a topological space and let fxngbe a sequence of points in X. We say thatfxngconverges to the point x2X, orxis the limit of the sequence, if for each open set Ucontaining xthere is a positive integer Nso thatxn2Ufor allnN. Sequences are not as fundamental in general topological spaces as they are in metric spaces. The following example may show why. Example 23 ConsiderRwith the conite topology. Let fxngbe any sequence of real numbers. Let a2Rbeany real number. Then fxngconverges to a, because if Uis any open set containing a, thenRnUis a nite set. Since fxngis an innite set, we must have that innitely many members of fxnglie inU. Thus, there is a positive integer Nsuch that if nN xn2U. Thus,fxngconverges to a. However, awas any arbitrary real number. This means that fxngconverges to every real number . What is more (and maybe worse) is that fxngwas an arbitrary sequence. This means that every sequence converges to every real number. There are no non- convergent sequences and sequences do not have unique limits. 3.2 Interior, Closure and Boundary Just as before we will dene the interior, closure, and the boundary. Denition 23 LetAbe a subset of the topological space X. A pointx2Ais an interior point ofAif there is an open set Uso thatx2UA.Ais called a neighborhood ofx. The interior ofA, denoted by A, is the set of all interior points ofA. c 1999, David Royster Introduction to Topology For Classroom Use Only 26 CHAPTER 3. TOPOLOGICAL SPACES Theclosure ,A, of A is the union of Aand it set of limit points: A=A[A0: A pointx2Xis aboundary point ofAifx2A\XnA. The set of boundary points ofAis called the boundary ofAand is denoted by @A. Theorem 19 For any subsets A;B of a topological space X a) The interior of Ais the union of all open sets contained in Aand is the largest open set contained in A. b)Ais open if and only if A=A. c) IfAB, thenAB. d)(A\B)=A\B. Proof: We will oer only a proof for (d). The others follow from closely from what we did in the case of a metric space. SinceA\Bis a subset of both AandB, then by (c) ( A\B)A\B. Now, A\Bis an open set and is a subset of A\B. Thus by (a), A\B(A\B). This completes the proof. Theorem 20 For any subsets A;B of a topological space X a) The closure of Ais the intersection of all closed sets containing in A and is the smallest closed set containing in A. b)Ais closed if and only if A=A. c) IfAB, thenAB. d)(A[B) =A[B. We leave this to the reader to prove. Theorem 21 LetAbe a subset of a topological space X. a)@A=A\XnA=@(XnA). b)@A,A, and (XnA)are pairwise disjoint sets whose union is X. c)@Ais a closed set. d)A=A[@A. e)Ais open if and only if @A(XnA). c 1999, David Royster Introduction to Topology For Classroom Use Only 3.3. BASIS FOR A TOPOLOGY 27 f)Ais closed if and only if @AA. g)Ais open and closed if and only if @A=;. Proof: Parts (a){(d) follow immediately from the denitions. (e) IfAis open then A=A. Now by (b) Aand@Aare disjoint. Thus, Aand @Aare disjoint. This implies that @AXnA. Now, if@AXnAthen no point ofAis a boundary point of A. Thus, every point of Ais an interior point of Aand A=A. Thus,Ais open. (f) This follows from our duality of open and closed sets. (g) IfAis both open and closed, then @AA\(XnA) =;. If@A=;then clearly@AA| meaning Ais closed | and @A(XnA) | meaning Ais open. Denition 24 A setAin a topological space Xisdense ifA=X. IfXhas a countable dense set, then Xis aseparable space. Example 24 1. The reals with the usual topology is separable, since the rationals are dense. 2. Euclidean n-space is separable, since the set of points having only rational coordinates is dense and countable. 3. The reals with the conite topology is separable, since every countable innite set is dense. Denition 25 A subsetBof a spaceXisnowhere dense if(B)=;. Note that a nite subset of a metric space is nowhere dense. In other topological spaces, we will see more interesting examples. 3.3 Basis for a Topology It appears that a topology can be relatively large. In fact, for an innite set the discrete topology consists of all subsets of the space, so it would be prohibitive to have to check all subsets. We have seen though that we can get by with just checking some of the sets. For the discrete topology we have usually only checked the singleton sets. For a metric space we were able to do everything we wanted by working with the open balls. In fact we dened all open sets in terms of the open balls. Can we do this in general? Can we nd a certain collection of subsets that will generate all of the elements of the topology, just like the open balls generate the metric topology? The answer is yes, because we can take Tas this generating set. This begs the answer, because we are looking for a smaller collection than the whole topology. c 1999, David Royster Introduction to Topology For Classroom Use Only 28 CHAPTER 3. TOPOLOGICAL SPACES Denition 26 Let(X;T)be a topological space. A basisBforTis a subcollection ofTwith the property that each member of Tis a union of members of B. The members of Bare called basic open sets andTis the topology generated by B. Example 25 Most of what we have seen is based on metric spaces. 1. The collection of all open intervals is a basis for the usual topology on the reals. 2. The collection of all open balls is a basis for the metric topology on the metric space (X;d). 3. For any set Xthe collection of all singleton sets fxgis a basis for the discrete topology. Denition 27 Let(X;T)be a topological space. A local basis ata2Xis a subcollection BaofTsuch that a)abelongs to each member of Ba, and b) each open set containing acontains a member of Ba. Denition 28 A spaceXisrst countable if there is a countable local basis at each point of X. The space Xissecond countable if the topology for Xhas a countable basis. Note that every second countable space is rst countable because if there is a countable basis B, then the number of these sets containing any given point a2X is at most countable. Theorem 22 Every second countable space is separable. Proof: LetXbe a second countable space with a countable basis B. LetAbe a set formed by choosing one element from each non-empty element of B. Each point ofXis a limit point of some point in Aby the denition of a basis. Thus, Ais dense inX. Theorem 23 a) Every metric space is rst countable. b) Every separable metric space is second countable. The proof is left to the reader. We have been starting with a topology and asking if there is a basis for it. We could be starting with a collection of open sets and asking if it forms a basis for a topology. Not every collection of open sets will work. When is a collection of open sets a basis for a topology on X? c 1999, David Royster Introduction to Topology For Classroom Use Only 3.4. CONTINUOUS FUNCTIONS 29 Theorem 24 A family Bof subsets of a set Xis a basis for a topology on Xif and only if both of the following hold: a) The union of members of BisX. b) For each B1;B22Bandx2B1\B2, there is a member BxofBsuch thatx2BxB1\B2. Example 26 [Sorgenfrey Line] Let Bbe the collection of all half-open intervals of Rof the form [ a;b),a < b . Clearly, the union of all of these intervals is R. If we take two of these sets and intersect them, we can nd another of these sets in the intersection. Thus, these sets form a basis for a topology on the real line, called the half-open interval topology T00forR.Rwith this topology is called the Sorgenfrey line. The Sorgenfrey line has the property that it is rst countable and separable, but not second countable. 3.4 Continuous Functions We were able to move away from the epsilon-delta denition of continuity of a function in between two metric spaces by using open balls. We will use this as our starting point for general topological spaces. Denition 29 A function f: (X;T)!(Y;T0)iscontinuous at a pointa2Xif for each open set VinYcontaining f(a)there is an open set UXcontaining a so thatf(U)V, or equivalently Uf1(V). Now, we are more interested in the situation where the function is continous at every point of X. This means that for every open set VinYand every point a2X withf(a)2V, there is an open set UaXwitha2Uaandf(Ua)V. Equivalently, we havea2Uaf1(V). This means that for each point in f1(V) we can nd an open set containing that point and contained in f1(V). Thus,f1(V) must be open inXfor each open set VY. This leads us to a more general denition. Denition 30 A function f: (X;T)!(Y;T0)iscontinuous if for each open set VinY f1(V)is an open set in X. Theorem 25 Letf:X!Ybe a function on the topological spaces XandYand leta2X. The following are equivalent. a)fis continuous at a. b) For each open set V2Ycontaining f(a), there is an open set UinX such thata2Uf1(V). c) For each neighborhood Voff(a),f1(V)is a neighborhood of a. c 1999, David Royster Introduction to Topology For Classroom Use Only 30 CHAPTER 3. TOPOLOGICAL SPACES The proof is left to the reader. Theorem 26 Letf:X!Ybe a function of topological spaces. The following are equivalent. (i)fis continuous. (ii) For each closed subset CY,f1(C)is closed in X. (iii) For each subset AX,f(A)f(A). (iv) There is a basis Bfor the topology of Yso thatf1(B)is open inXfor each basic open set B2B. Proof: To show that ( i) implies ( ii) will require the duality between open and closed sets. IfCYis closed, the YnCis open inY. Sincefis continuous, f1(YnC) is open inX. HenceXn(f1(YnC)) is open. If x2Xn(f1(YnC)) thenf(x)62YnC, orf(x)2C. Thus,Xn(f1(YnC))f1(C). The opposite inclusion is clear. Thus, f1(C) =Xn(f1(YnC)) is closed. A similar analysis shows that ( ii) implies ( i). To show that ( ii) implies ( iii), letAX. Thenf(A) is a closed subset of Y. Hence,f1(f(A)) is a closed subset of X. Now,Af1(f(A)) soAf1(f(A)). Thus,f(A)f(A). To show that ( iii) implies ( ii), letCbe a closed subset of Y. Then, f(f1(C))ff1(C)CC sof1(C)f1(C) makingf1(C) a closed set. For the last equivalence, ( i) clearly implies ( iv). We need to prove the opposite implication. Let Obe an open set in Y. Then by the denition of a basis, O=[2IB for some subcollection fBg2Iof the basis B. Then f1(O) =f1 [ 2IB! =[ 2If1(B): Since each f1(B) is open in Xand the union of any family of open sets is open, thenf1(O) is open in Xandfis continuous. Theorem 27 Iff:X!Yandg:Y!Zare continuous functions, then gf:X! Zis continuous. Denition 31 A function f:X!Yis ahomeomorphism if a)fis one-to-one, ( injective ) b)fis onto, ( surjective ) c 1999, David Royster Introduction to Topology For Classroom Use Only 3.5. SUBSPACES 31 c)fis continuous, d)f1is continuous. Topological spaces are topologically equivalent orhomeomorphic if there home- omorphism from ffromXontoY. The continuity of the function and its inverse are extremely important! A prop- ertyPof topological spaces is a topological property ortopological invariant provided that if space Xhas property P, then so does every space which is homeo- morphic to X. Theorem 28 Separability is a topological property. Theorem 29 First countability and second countability are topological properties. Denition 32 A topological space is metrizable provided that the topology on Xis generated by a metric. Theorem 30 Metrizability is a topological property. SinceRand (0;1) are homeomorphic, the property of being a bounded metric space is not a topological property. Likewise, distance is not a topological invariant. 3.5 Subspaces Let (X;T) be a topological space and let Abe a subset of X. The relative topology orsubspace topology T0onAdetermined by Tconsists of all sets of the form O\Afor whichOis an open set of T. T0=fO\AjO2Tg: The members of T0are called relatively open sets inA, and (A;T0) is called a subspace of (X;T). Note that this is actually a topology for A. ;=;\A A =X\A; so both;andAare open in A. IffUgare open in A, thenU=O\Aand [ U=[ (O\A) = [ alphaO! \A is relatively open since the union of any family of open sets is open in X. For any nite family of open sets Ui=Oi\A, we have n\ i=1Ui=n\ i=1(Oi\A) = n\ i=1Oi! \A is relatively open since the intersection of any nite family of open sets is open in X. c 1999, David Royster Introduction to Topology For Classroom Use Only 32 CHAPTER 3. TOPOLOGICAL SPACES Lemma 5 Let(a;T0)be a subspace of the topological space (X;T). A subset Fof Ais closed in the subspace topology on Aif and only if F=C\Afor some closed subsetCofX. Example 27 1. The closed interval [ a;b] witha<b is a subspace of Rwith the usual topology. The open sets containing aare sets of the form [ a;c) with a<c<b . 2. The subset of Rn+1consisting of all ( n+ 1)-tuples ( x1;x2;::: ;xn;xn+1) with xn+1= 0 is homeomorphic to Rn 3. Leta<b<c<d . LetA= [a;b][(c;d) be considered as a subspace of the real line. Then the subset [ a;b] ofAis both relatively open and relatively closed. It is clearly closed because [ a;b] = [a;b]\Aand [a;b] is closed in the real line. It is open because for 0 <  < cb, [a;b] = (a;b+)\Y. Thus, we see that since ( c;d) is the complement of this set that is both relatively open and relatively closed, then we see that ( c;d) is both relatively open and relatively closed. Denition 33 A property Pto topological spaces is hereditary provided that if X has property P, then every subspace of Xhas this property. Example 28 1. First countability and second countability are hereditary proper- ties. IfXhas a countable basis, then intersecting these basis elements with A will give a countable basis for the subspace topology. First countable is similar. 2. Separability is not hereditary. Let AR2consist of the x-axis and the point a= (0;1). Dene a topology TonAby taking the empty set and all subsets ofAthat contain the singleton set fag. Then (X;T) is separable because the singleton setfagis dense. Every point except ais a limit point offag. However, the subspace topology on R(thex-axis) is the discrete topology, so ( R;T0) is not separable. 3.6 Hausdor Spaces A topological space Xis aHausdor space if for each pair of distinct points a;b2Xthere exist disjoint open sets UandVsuch thata2Uandb2V. Example 29 1. Every metric space is Hausdor. We proved this as a homework problem, but it is simple. Let r=d(a;b) and then take two open balls of radius r=2 centered at aandbrespectively. 2. The real line with the co-nite topology is not Hausdor. Likewise, the real line with the co-countable topology is not Hausdor. c 1999, David Royster Introduction to Topology For Classroom Use Only 3.6. HAUSDORFF SPACES 33 3. Take any set with more than one point and give it the indiscrete (trivial) topol- ogy. It is not Hausdor. 4. The space in Example 2 is not Hausdor, because you can never separate a from any other point. 5. [The Zariski Topology ] Letnbe a positive integer and consider the family P of all polynomials in nreal variables x1;x2;:::;xn. Forp2PletZ(p) denote its solution set in Rn: Z(p) =f(x1;x2;:::;xn)2Rnjp(x1;x2;:::;xn) = 0g: LetBbe the collection of sets that are complements of some Z(p) for some p2P. This forms a basis for a topology on Rncalled the Zariski topology . For the real line, n= 1, this is just the co-nite topology. This is because each nite set of real numbers is the solution set for some polynomial in one real variable. If A=fa1;a2;:::;ang, then p(x) = (xa1)(xa2):::(xan) is a polynomial with Aas its solution set. Likewise, the set of solutions of a polynomial in one real variable of dimension nis at mostn. Forn>1 this is not the co-nite topology. For example, the line y=ainR2 is the solution set to the polynomial in two variables p(x;y) =ya: Note that this is not a nite set. However, each nite set can serve as the solution set of a polynomial. Now,Rnwith the Zariski topology is not Hausdor. Assume that P= (a1;a2;:::;an) andQ= (b1;b2;:::;bn) are two distinct points in Rn. Theorem 31 1. The property of being a Hausdor space is topological and hered- itary. 2. In a Hausdor space a sequence fxng1 n=1cannot converge to more than one point. Proof: We will prove 1 only and leave the other to the reader. Suppose that Xis Hausdor and f:X!Yis a homeomorphism. For a6=b2Y, we have that f1(a) andf1(b) are distinct points in X. Thus, there are disjoint open setsU;VXso thatf1(a)2Uandf1(b)2V. Hence,f(U) andf(V) are disjoint open sets of Ycontainingaandbrespectively. To show that Hausdor is hereditary, assume that Xis Hausdor and that AX. Leta6=b2A. Then,a6=b2Xand there are disjoint open sets U;VXso that a2Uandb2V. Then,U\AandV\Aare disjoint relatively open subsets of A containingaandbrespectively. c 1999, David Royster Introduction to Topology For Classroom Use Only Chapter 4 Connectedness 4.1 Connected and Disconnected Spaces It is easier to dene disconnected than connected , and we will do so. Denition 34 A topological space Xisdisconnected orseparated if it is the union of two disjoint, non-empty open sets. Such a pair A;B of subsets of Xis called aseparation ofX. A space is connected if it is not disconnected. A subspace YofXisconnected provided that it is a connected space when assigned the subspace topology. Example 30 1. A discrete space with more than one point is disconnected. 2. Any set with the indiscrete topology is connected, since there do not exist two non-empty open sets. 3. LetAbe the set of non-zero real numbers with the subspace topology. Then A is disconnected since ( 1;0) and (0;1) form a separation. 4. LetB=R2nRbe the plane minus the real axis. Bis disconnected. 5.X= [0;1][[2;3] is disconnected. 6. LetXbe the set of real numbers with the addition of a point (0 ;1), and let the topology TforXconsist of;,X, and all subsets of Xwhich contain a. Now, Xis connected because every open set contains the point a. On the other hand, as a subspace of ( X;T) the real line is assigned the discrete topology and is disconnected. These last few examples show that the property of being connected is not heredi- tary. 34 4.2. HOW TO TELL IF A SPACE IS CONNECTED? 35 Example 31 The real lineRwith the usual topology is connected. Suppose other- wise that R=A[B whereA\B=;are both open and non-empty. Since A=RnBandB=RnA we see that AandBare also both closed. Consider two points a;b2Rwitha2A andb2B. We may assume that a<b . PutA+=A\[a;b]. ThenA+is a closed and bounded subset of R. Thus it must contain its least upper bound c. Now,c6=bsinceAandBhave no points in common. Thus, c<b . Thus,Acontains no points of ( c;b], placing (c;b]B. Thus, c2B. Now,Bis closed, so c2B. This implies that c2A\B, contradicting the assumption that AandBare disjoint. Thus, Ris connected. 4.2 How to Tell If a Space is Connected? Denition 35 Non-empty subsets AandBareseparated sets ifA\B=A\B=;. Theorem 32 The following are equivalent for a topological space X. 1.Xis disconnected. 2.Xis the union of two disjoint, non-empty closed sets. 3.Xis the union of two separated sets. 4. There is a continuous function from Xonto the discrete two-point space f0;1g. 5.Xhas a proper subset Awhich is both open and closed. 6.Xhas a proper subset Asuch that A\(XnA) =;: Proof: First, we will show that (1) ,(2). Assume that Xis disconnected. Then there are two nonempty, disjoint open sets AandB, so thatX=A[B. This means thatXnA=BandXnB=A. Since both AandBare open, we now have that both AandBare closed, and the result follows. The proof of the opposite implication is analogous. (1),(3): Assume that Xis disconnected. We have then that there are two nonempty, disjoint open sets AandB, so thatX=A[B. We just proved that AandBare closed. Thus, A=AandB=B. Hence, A\B=A\B=;=A\B=A\B: c 1999, David Royster Introduction to Topology For Classroom Use Only 36 CHAPTER 4. CONNECTEDNESS Again, the opposite implication is analogous. (1),(4): Assume that Xis disconnected. Dene a function f:X!f0;1gby f(x) =( 0 ifx2A 1 ifx2B: fis clearly onto. Since we have placed the discrete topology on f0;1g, we only need to checkf1(f0g) andf1(f1g) to check that fis continuous. Since f1(f0g) =A is open and f1(f1g) =Bis open, we see that fis continous. Assume that there is a continous, onto function f:X!f0;1g. Sincef0;1ghas the discrete topology, the sets f0gandf1gare open and disjoint. Let A=f1(f0g) andB=f1(f1g). These then form a separation for X. (1),(5): Assume that Xis disconnected and that AandBform a separation of X. ThenAis a proper subset which is both open and closed. If UXis a proper subset which is both open and closed, then V=XnUis open and UandVform a separation of X. (1),(6): Assume that Xis disconnected and AandBform a separation of X. ThenXnA=B,A=A, and (XnA) =B=B=XnA. Hence A\(XnA) =;: If such a set Aexists, then AandXnAform a separation of X. Corollary 2 The following statements are equivalent. 1.Xis connnected. 2.Xis not the union of two disjoint, non-empty closed sets. 3.Xis not the union of two separated sets. 4. There is not continuous function from Xonto a discrete two point space f0;1g. 5. The only subsets of Xthat are both open and closed are Xand;. 6.Xhas no proper subset Aso that A\(XnA) =;: Theorem 33 LetXbe a connected topological space and let f:X!Ybe a contin- uous surjective function. Then Yis a connected space. Proof: We will prove that if Yis not connected, then Xis not connected, completing the proof by proving the contrapositive. If Yis disconnected, let AandBform the separation of Y. ThenY=A[Band the sets f1(A) andf1(B) c 1999, David Royster Introduction to Topology For Classroom Use Only 4.2. HOW TO TELL IF A SPACE IS CONNECTED? 37 (a) are open sets since fis continous; (b) are disjoint since fis a function; (c) are non-empty since fis surjective; and (d) have union Xbecause X=f1(Y) =f1(A[B) =f1(A)[f1(B): Thus, ifYis disconnected, then Xis disconnected. Corollary 3 Iff:X!Yis a continuous function, then f(X), the image of f, is connected. Lemma 6 A subspace Yof a spaceXis disconnected if and only if there are open setsUandVinXsuch that U\Y6=;; V\Y6=;; U\V\Y=;; YU[V: Theorem 34 IfYis a connected subspace of X, thenYis connected. Proof: Suppose that Yis connected. Consider a continuous function f:Y!f0;1g. We must show that fis not surjective. We know that the restriction fjYis not surjective, so that fmapsYeither tof0gorf1g. Assume that f(Y) =f0g. Sincef is continuous, we know that f(Y)(f(Y)) =f0g=f0g; sofis not surjective. Thus, Yis connected. Corollary 4 LetYbe a connected subspace of XandZa subspace so that YZ Y. ThenZis connected. Example 32 Each interval of the real line is connected. We know that each open interval of the real line is connected, being homeomorphic to the real line. A non- degenerate closed interval is the closure of an open interval, and is hence connected. A degenerate closed interval is a single point, which is connected. Any other interval is trapped between an open interval and its closure, so it is connected. The empty set, which is an interval, is connected because there are no non-empty subsets. c 1999, David Royster Introduction to Topology For Classroom Use Only 38 CHAPTER 4. CONNECTEDNESS Example 33 [The Topologist's Sine Curve ] LetA=f(0;y)j1y1gand B=f(x;y)j0<x1; y= sin(=x)g. PutT=A[B.Tis called the topologist's sine curve . Note that Bis connected, since it is the continuous image of (0 ;1]. Note also thatT=B, so thatTis connected. –101 We should never believe that the union of two connected sets is connected, since one of our rst examples of a disconnected space was two disjoint intervals. What would it take for a union of connected sets to be connected? Theorem 35 LetXbe a space and let fAj2Igbe a family of connected subsets ofXfor which\2IAis not empty. Then [2IAis connected. Proof: We shall use Lemma 6 to prove that Y=[2IAis connected. Suppose thatUandVare open sets of Xso that: U\Y6=;; U\V\Y=;; YU[V: We need to then show that Y\V=;. This will show that Yis connected. Since U\Y6=;,Umust contain some point of some A, for some 2I. SinceAis connected, then AU. Ifb2\2IA, thenbmust be in A, sob2U. Thus,U contains a point bin eachA,2I. SinceAis connected, the AUfor each 2I. Thus, Y=[ 2IAU soV\Y=;. Various variants of this theorem are as follows: c 1999, David Royster Introduction to Topology For Classroom Use Only 4.2. HOW TO TELL IF A SPACE IS CONNECTED? 39 Lemma 7 LetXbe a space,fAj2Iga family of connected subsets of X, and Ba connected subset of Xsuch that for each 2I,A\B6=;. ThenB[([2IA) is connected. Lemma 8 LetfAngbe a sequence of connected subsets of a space Xsuch that for each integer n1,Anhas at least one point in common with one of the preceding setsA1;::: ;An1. Then[1 n=1Anis connected. Denition 36 Aconnected component , orcomponent , of a topological space is a connected subset CofXwhich is not a proper subset of any connected subset of X. Theorem 36 For a topological space X 1. Each point x2Xbelongs to exactly one component. The component Cxcon- tainingxis the union of all the connected subsets of Xwhich contain x, and is thus the largest connected subset of Xcontaining the point x. 2. For points x;y2Xthe components CxandCyare either disjoint or identical. 3. Every connected subset of Xis contained in a component. 4. Each component is a closed set. 5.Xis connected if and only if it has one component. 6. IfCis a component of XandAandBform a separation for X, thenCis a subset ofAor a subset of B. Example 34 1. For the subspace X= (0;1)[(2;3) ofR, there are two compo- nents, (0;1) and (2;3). Both components are closed sets with respect to the subspace topology. 2. In a discrete space, each component consists of only one point. 3. For the set Qof rational numbers with the subspace topology, each component consists of only one point. Note, however, that the subspace topology is not the discrete topology. Example 35 LetXbe the subspace of R2consisting of the following sequence: X=f(x;y)j0x1; y =1 n;n= 1;:::;1g[ 0;1 2 [1 2;1 Now, [0;1 2) is the component of Xcontaining 0 and (1 2;1] is the component of X containing 1. Thus, 0 and 1 belong to dierent components. However, for any separation of Xinto disjoint non-empty open sets AandBwhose union is X, both 0 and 1 belong to Aor toB. c 1999, David Royster Introduction to Topology For Classroom Use Only 40 CHAPTER 4. CONNECTEDNESS Denition 37 A spaceXistotally disconnected if each component of Xconsists of a single point. Thus, a discrete space is totally disconnected, as is the subspace of rational num- bers in the real line. 4.3 Applications of Connectedness You have been using the properties of connectedness in Calculus without knowing it. The Intermediate Value Theorem as well as several xed point theorems all depend on connectedness. Theorem 37 (Intermediate Value Theorem) Letf: [a;b]!Rbe a continuous real-valued function on a closed interval, [a;b]. Lety0be a real number between f(a) andf(b). Then there is number c2[a;b]for whichf(c) =y0. Proof: The interval [ a;b] is connected, so f([a;b]) is connected, since fis continuous. Thus,f([a;b]) is an interval in R. Therefore, any number, y0, betweenf(a) andf(b) must be in the image f([a;b]). This means that y0=f(c) for somec2[a;b]. You have used the following corollary numerous times. Corollary 5 Letf: [a;b]!Rbe continuous for which one of f(a)andf(b)is positive and the other is negative. Then the equation f(x) = 0 has at least one root betweenaandb. Theorem 38 (Fixed Point Theorem) Letf: [a;b]![a;b]be a continuous func- tion on the closed interval [a;b]. Then there is a c2[a;b]so thatf(c) =c. Proof: Iff(a) =aor iff(b) =bthen we are done, so assume that f(a)6=aand f(b)6=b. Thus,a < f (a) andf(b)< bsince the range of fis [a;b]. Dene a new functiong: [a;b]!Rby g(x) =f(x)x; x2[a;b]: Theng(a) =f(a)a >0 andf(b) =f(b)b <0. andgis continuous. Thus, by the corollary, there is a c2[a;b] so thatg(c) = 0 orf(c)c= 0. Thus, there is a c2[a;b] so thatf(c0 =c. Iff:X!Xis a function, then we say that xis axed point offiff(x) =x. A topological space Xhas the xed point property if every continuous function f:X! Xhas at least one xed point. We can restate the above theorem as follows: Theorem 39 Every closed and bounded interval has the xed-point property. c 1999, David Royster Introduction to Topology For Classroom Use Only 4.4. PATH-CONNECTED SPACES 41 Theorem 40 The xed-point property is a topological invariant. Example 36 The real line does not have the xed-point property. An example is f(x) =x= 1. Thus, no open interval has the xed-point property. The intervals of the form [a;b), (a;b], (1;b], and [a;1) also do not have the xed point property. Then-sphere,Sn,n1, does not have the xed-point property since the function g(x) =xhas no xed point. 4.4 Path-Connected Spaces Our initial idea of connectedness is really best summed up in the denition of path connected. We usually think of a space as being connected if we can get from any one point to another without leaving the space. This is a stronger property than connectedness, though. We useIto denote the unit interval, [0 ;1]. Denition 38 Apath in a spaceXis a continuous function f:I!X. The points f(0)andf(1)are the endpoints of the path. The path fis called a path from f(0) tof(1). Iffis a path in X, the reverse path fis the path dened by f(t) =f(1t); t2I: Denition 39 A spaceXispath connected if for any pair of points x;y2Xthere is a path in Xwith initial point xand terminal point y. A subspace AofXispath connected provided that Ais path connected with its subspace topology. Example 37 Every interval on the real line is path connected. For a;b2Kdene the path by p(t) =a(1t) +bt; t2I: Example 38 The generalizations of the interval to subsets of Rnare called convex sets. A setCofRnis convex if for any two points, a;b2C, the line segment joining aandblies entirely in C. Since a line segment denes a path, each convex set is path connected. Theorem 41 Every path connected set is connected. Proof: Suppose that Xis path connected and let a2X. For each x2XletCx denote the image of the path from atox. Since each Cxis connected and a2Cxfor allx2X, then by our previous theorem X=[ x2XCx is connected. c 1999, David Royster Introduction to Topology For Classroom Use Only 42 CHAPTER 4. CONNECTEDNESS Example 39 The Topologist's Sine Curve, T, is connected but it is not path connnected. Thus, while in Rconnected and path connected are identical, in Rnthere are sets that are connected, but not path connected. All is not lost though. Theorem 42 Every open, connected subset of Rnis path connected. There are analogous results for path connectedness as for connectedness. Likewise, one denes the path component of a spaceXas a path connected subset of Xwhich is not a proper subset of any path connected subset of X. 4.5 Locally Connected and Locally Path Connected Spaces The terms connected and path connected are global properties, i.e., properties that apply to the whole space. Local topological properties are characteristics of a space "near" a particular point. Denition 40 A topological space Xislocally connected at a pointp2Xif every open set containing palso contains a connected set which contains p. The space Xis locally connected if it is locally connected at each point. Theorem 43 LetXbe a topological space. a)Xis locally connected at a point p2Xif and only if there is a local basis atpconsisting of connected open sets. b)Xis locally connected if and only if it has a basis of connected open sets. Example 40 1. Any interval in Ris connected and locally connected. 2.Rnis connected and locally connected for each integer n0. 3. The subspace [0 ;1][[2;3] is locally connected, but not connected. 4. LetX=f(x;0)j0x1g,Y=f(0;y)j0y1g, and Z=1 n;y
n2Z+;0y1 . LetC=X[Y[Z.Cis called the topolo- gist's comb . The topologist's comb is obviously connected, being path connected. A path can run up and down the vertical tines and across the base. However, Cis not locally connected at any point (0 ;t), 0<t1, since small open sets containing such points consist of collections of open vertical intervals. 5. The setQof rational numbers is neither connected nor locally connected. c 1999, David Royster Introduction to Topology For Classroom Use Only 4.5. LOCALLY CONNECTED AND LOCALLY PATH CONNECTED SPACES43 Thus, from these examples, we see that one property does not imply the other. This is commonly true with global and local properties, but even this statement is not infallible. Recall that second countability does imply rst countability. Theorem 44 A spaceXis locally connected if and only if for each open subset OX, each component of Ois an open set. Denition 41 A spaceXislocally path connected at a pointp2Xif every open set containing pcontains a path connected set containing p. The space Xislocally path connected if it is locally path connected at each point. Theorem 45 LetXbe a topological space. a)Xis locally path connected at a point p2Xif and only if there is a local basis at pconsisting of path connected open sets. b)Xis locally path connected if and only if it has a basis of path connected open sets. Theorem 46 A spaceXis locally path connected if and only if for each open subset OX, each path component of Ois an open set. Theorem 47 IfXis a connected, locally path connected space, then Xis path con- nected. Proof: For each point x2X, letPxdenote the path component of Xto whichx belongs. Since Xis an open set, Theorem 46 shows that each Pxis open. Recall that path components are either disjoint or identical. For a particular point a2X, suppose that Pa6=X. ThenPaand the union of allPxfor whichx62Paare disjoint, non-empty open subsets of Xwhose union is X. This would imply that Xis disconnected. Since Xis connected, we have that Pa=XandXmust be path connected as well. c 1999, David Royster Introduction to Topology For Classroom Use Only Chapter 5 Compactness Compactness is the generalization to topological spaces of the property of closed and bounded subsets of the real line: the Heine-Borel Property. While compact may infer "small" size, this is not true in general. We will show that [0 ;1] is compact while (0;1) is not compact. Compactness was introduced into topology with the intention of generalizing the properties of the closed and bounded subsets of Rn. 5.1 Compact Spaces and Subspaces Denition 42 LetAbe a subset of the topological space X. An open cover forA is a collection Oof open sets whose union contains A. Asubcover derived from the open cover Ois a subcollection O0ofOwhose union contains A. Example 41 LetA= [0;5] and consider the open cover O=f(n1;n+ 1)jn=1;::: ;1g: Consider the subcover P=f(1;1);(0;2);(1;3);(2;4);(3;5);(4;6)gis a subcover of A, and happens to be the smallest subcover of Othat covers A. Denition 43 A topological space Xiscompact provided that every open cover of Xhas a nite subcover. This says that however we write Xas a union of open sets, there is always a nite subcollectionfOign i=1of these sets whose union is X. A subspace AofXiscompact ifAis a compact space in its subspace topology. Since relatively open sets in the subspace topology are the intersections of open sets in Xwith the subspace A, the denition of compactness for subspaces can be restated as follows. Alternate Denition: A subspace AofXiscompact if and only if every open cover ofAby open sets in Xhas a nite subcover. 44 5.1. COMPACT SPACES AND SUBSPACES 45 Example 42 1. Any space consisting of a nite number of points is compact. 2. The real line Rwith the nite complement topology is compact. 3. An innite set Xwith the discrete topology is not compact. 4. The open interval (0 ;1) is not compact. O=f(1=n;1)jn= 2;::: ;1gis an open cover of (0 ;1). However, no nite subcollection of these sets will cover (0;1). 5.Rnis not compact for any positive integer n, since O=fB(0;n)jn= 1;::: ;1gis an open cover with no nite subcover. A sequence of sets fSng1 n1isnested ifSn+1Snfor each positive integer n. Theorem 48 (Cantor's Nested Intervals Theorem) Iff[an;bn]g1 n=1is a nested sequence of closed and bounded intervals, then \1 n=1[an;bn]6=;. If, in addition, the diameters of the intervals converge to zero, then the intersection consists of precisely one point. Proof: Since [an+1;bn+1][an;bn] for eachn2Z+, the sequencesfangandfbngof left and right endpoints have the following properties: (i)a1a2


an:::andfangis an increasing sequence; (ii)b1b2


bn:::andfbngis a decreasing sequence; (iii) each left endpoint is less than or equal to each right endpoint. Letcdenote the least upper bound of the left endpoints and dthe greatest lower bound of the right endpoints. The existence of canddare guaranteed by the Least Upper Bound Property. Now, by property ( iii),cbnfor alln, socd. Since ancdbn, then [c;d][an;bn] for alln. Thus,\1 n=1[an;bn] contains the closed interval [c;d] and is thus non-empty. If the diameters of [ an;bn] go to zero, then we must have that c=dandcis the one point of the intersection. Theorem 49 The interval [0;1]is compact. Proof: LetObe an open cover. Assume that [0 ;1] is not compact. Then either [0;1 2] or [1 2;1] is not covered by a nite number of members of O. Let [a1;b1] be the half that is not covered by a nite number of members of O. Apply the same reasoning to the interval [ a1;b1]. One of the halves, which we will call [a2;b2], is not nitely coverable by Oand has length1 4. We can continue this reasoning inductively to create a nested sequence of closed intervals f[an;bn]g1 n=1, none of which is nitely coverable by O. Also, by construction, we have that bnan=1 2n; c 1999, David Royster Introduction to Topology For Classroom Use Only 46 CHAPTER 5. COMPACTNESS so the diameters of these intervals goes to zero. By the Cantor Nested Intervals Theorem, we know that there is precisely one point in the intersection of all of these intervals; p2[an;bn], for alln. Sincep2[0;1] there is an open interval O2Owithp2O. Thus, there is a positive number, >0 so that (p;p+)O. LetNbe a positive integer so that 1 =2N<. Then since p2[aN;bn] it follows that [an;bn](p;p+)O: This contradicts the fact that [ aN;bN] is not nitely coverable by Osince we just covered it with one set from O. This contradiction shows that [0 ;1] is nitely coverable byOand is compact. Compactness is dened in terms of open sets. The duality between open and closed sets and if C=XnO, Xn \ 2IC! =[ 2IO leads us to believe that there is a characterization of compactness with closed sets. Denition 44 A family Aof subsets of a space Xhas the nite intersection property provided that every nite subcollection of Ahas non-empty intersection. Theorem 50 A spaceXis compact if and only if every family of closed sets in X with the nite intersection property has non-empty intersection. This says that if Fis a family of closed sets with the nite intersection property, then we must have that\ FC6=;. Proof: Assume that Xis compact and let F=fCj2Igbe a family of closed sets with the nite intersection property. We want to show that the intersection of all members of Fis non-empty. Assume that the intersection is empty. Let O=fO=XnCj2Ig.Ois a collection of open sets in X. Then, [ 2IO=[ 2IXnC=Xn\ 2IC=Xn;=X: Thus,Ois an open cover for X. SinceXis compact, it must have a nite subcover; i.e., X=n[ i=1Oi=n[ i=1(XnCi) =Xnn\ i=1Ci: This means that \n i=1Cimust be empty, contradicting the fact that Fhas the nite intersection property. Thus, if Fhas the nite intersection property, then the intersection of all members of Fmust be non-empty. The opposite implication is left as an exercise. c 1999, David Royster Introduction to Topology For Classroom Use Only 5.1. COMPACT SPACES AND SUBSPACES 47 Is compactness hereditary? No, because (0 ;1) is not a compact subset of [0 ;1]. It isclosed hereditary . Theorem 51 Each closed subset of a compact space is compact. Proof: LetAbe a closed subset of the compact space Xand let Obe an open cover ofAby open sets in X. SinceAis closed, then XnAis open and O=O[fXnAg is an open cover of X. SinceXis compact, it has a nite subcover, containing only nitely many members O1;::: ;OnofOand may contain XnA. Since X= (XnA)[n[ i=1Oi; it follows that An[ i=1Oi andAhas a nite subcover. Is the opposite implication true? Is every compact subset of a space closed? Not necessarily. The following though is true. Theorem 52 Each compact subset of a Hausdor space is closed. Proof: LetAbe a compact subset of the Hausdor space X. To show that Ais closed, we will show that its complement is open. Let x2XnA. Then for each y2A there are disjoint sets UyandVywithx2Vyandy2Uy. The collection of open sets fUyjy2Agforms an open cover of A. SinceAis compact, this open cover has a nite subcover,fUyiji= 1;:::;ng. Let U=n[ i=1UyiV=n\ i=1Vyi: Since each UyiandVyiare disjoint, we have UandVare disjoint. Also, AUand x2V. Thus, for each point x2XnAwe have found an open set, V, containing x which is disjoint from A. Thus,XnAis open, and Ais closed. Corollary 6 LetXbe a compact Hausdor space. A subset AofXis compact if and only if it is closed. The following results are left to the reader to prove. Theorem 53 IfAandBare disjoint compact subsets of a Hausdor space X, then there exist disjoint open sets UandVinXsuch thatAUandBV. Corollary 7 IfAandBare disjoint closed subsets of a compact Hausdor space X, then there exist disjoint open sets UandVinXsuch thatAUandBV. c 1999, David Royster Introduction to Topology For Classroom Use Only 48 CHAPTER 5. COMPACTNESS 5.2 Compactness and Continuity Theorem 54 LetXbe a compact space and f:X!Ya continuous function from XontoY. ThenYis compact. Proof: We will outline this proof. Start with an open cover for Y. Use the continuity offto pull it back to an open cover of X. Use compactness to extract a nite subcover forX, and then use the fact that fis onto to reconstruct a nite subcover for Y. Corollary 8 LetXbe a compact space and f:X!Ya continuous function. The imagef(X)ofXinYis a compact subspace of Y. Corollary 9 Compactness is a topological invariant. Theorem 55 LetXbe a compact space, Ya Hausdor space, and f:X!Ya continuous one-to-one function. Then fis a homeomorphism. 5.3 Locally Compact and One-Point Compacti- cations Is it always possible to consider a topological space as a subspace of a compact topolog- ical space? We can consider the real line as an open interval (they are homeomorphic). Can we always do something of this sort? Denition 45 A spaceXislocally compact at a point x2Xprovided that there is an open set Ucontaining xfor whichUis compact. A space is locally compact if it is locally compact at each point. Note that every compact space is locally compact, since the whole space Xsatises the necessary condition. Also, note that locally compact is a topological property. However, locally compact does not imply compact, because the real line is locally compact, but not compact. Denition 46 LetXbe a topological space and let 1denote an ideal point, called thepoint at innity , not included in X. LetX1=X[1 and dene a topology T1onX1by specifying the following open sets: (a) the open sets of X, considered as subsets of X1; (b) the subsets of X1whose complements are closed, compact subsets of X; and (c) the set X1. The space (X1;T1is called the one point compactication ofX. c 1999, David Royster Introduction to Topology For Classroom Use Only 5.3. LOCALLY COMPACT AND ONE-POINT COMPACTIFICATIONS 49 Theorem 56 LetXbe a topological space and X1its one-point compactication. Then a)X1is compact. b)(X;T)is a subspace of (X1;T1). c)X1is Hausdor if and only if Xis Hausdor and locally compact. d)Xis a dense subset of X1if and only if Xis not compact. Proof: a) Any open cover OofX1must have a member Ucontaining1. Since the complement X1nUis compact, it has a nite subcover fOign i=1derived fromO. Thus,U;O 1;:::;Onis a nite subcover of X1. b) The fact that ( X;T) is the subspace topology in ( X1;T1) basically follows from the denition of the extended topology. It also requires that we look at what open sets containing the point at innity look like. One such set isU=X1itself andU\X=Xis open inX. The second type is a subset of X1so thatXnUis closed and compact in X. In this case U\Xis open since its complement is closed. c) Suppose that X1is Hausdor. Then Xis Hausdor since the property is hereditary. Now, let p2X. SinceX1is Hausdor, there are open, disjoint sets UandVinX1so that12Uandp2V. Thus,VX1nU and this latter set is closed and compact in X. HenceVX1nU, soV is compact, since it is a closed subset of a compact set. Thus, Xis locally compact at p. Now, suppose that Xis Hausdor and locally compact. To show that X1 is Hausdor, we only need to be able to separate 1from any point in p2X. SinceXis locally compact, there is an open set Oso thatp2O andOis compact. Then OandX1nOare two disjoint open sets in X1 containingpand1respectively. d) IfXis compact, thenf1g is an open set in X1, sincef1g =X1nX. Thus,1is not a limit point of X, andX6=X1. Hence,Xis not dense. IfXis not dense in X1, thenX=X, since162X. Hence,f1g is open inX1. Thus,Xis compact. Example 43 What is the one-point compactication of the open interval (0 ;1)? You can dene a function f: (0;1)1!S1by f(t) =( (cos(2t);sin(2t)) if 0<t< 1 (1;0) if t=1 c 1999, David Royster Introduction to Topology For Classroom Use Only 50 CHAPTER 5. COMPACTNESS Thisfis a one-to-one continuous function from (0 ;1)1onto the unit circle. By Theorem 55, this is a homeomorphism. c 1999, David Royster Introduction to Topology For Classroom Use Only