notes2_metric spaces

Chapter 2 Metric Spaces 2.1 Denition and Some Examples Denition 2.1 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 2.1.1 The real line, Ris a metric space using the standard distance func- tion, the absolute value: d(a;b) =jabj. The above properties are standard proofs about the absolute value function. Example 2.1.2 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 2.1.3 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 2.1 (Cauchy-Schwarz Inequality) For any points a;b2Rn ja
bjkakkbk: Theorem 2.2 (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 2.1.4 [The Taxicab Metric] Dene a function d0:R2R2!Ras follows. Ifx= (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 2.1.5 [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 2.1.6 [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 2.1.7 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 2.1.8 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 2.2 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 2.3 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 2.4 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 2.3 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 2.5 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 2.2.1 Letf: (X;d)!(Y;d0) be given by f(x) =bfor allx2Xwhere b2Yis 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 2.2.2 Let 1X: (X;d)!(X;d) denote the identity map from Xto itself given by 1 X(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 2.2.3 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 2.6 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 2.7 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 2.8 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 2.4 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 2.5 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 2.9 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 2.1 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 thatxn2Vwhenevern>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 2.6 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 2.7 The open subsets of a metric space (X;d)have the following proper- ties: 1.Xand;are open sets. 2. The union of any 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 3. The intersection of a nite family of open sets is open. 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 2.8 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 2.10 Let(X;d)be a metric space and Aa subset ofX. 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 2.2 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 2.3 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 2.9 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 2.11 LetAbe a subset of a metric space (X;d). A pointx2Ais 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 2.6.1 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 2.12 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 2.6.2 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 2.10 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 2.11 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 2.13 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 2.6.3 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