notes5 compactness

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 5.1 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 5.1.1 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 5.2 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. 43 44 CHAPTER 5. COMPACTNESS Example 5.1.2 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 5.1 (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 5.2 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 5.1. COMPACT SPACES AND SUBSPACES 45 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 5.3 A family Aof subsets of a space Xhas the nite intersection property provided that every nite subcollection of Ahas non-empty intersection. Theorem 5.3 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 46 CHAPTER 5. COMPACTNESS Is compactness hereditary? No, because (0 ;1) is not a compact subset of [0 ;1]. It isclosed hereditary . Theorem 5.4 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 5.5 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 5.6 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 5.2. COMPACTNESS AND CONTINUITY 47 5.2 Compactness and Continuity Theorem 5.7 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 5.8 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 5.4 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 5.5 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 48 CHAPTER 5. COMPACTNESS Theorem 5.9 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 5.3.1 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 5.3. LOCALLY COMPACT AND ONE-POINT COMPACTIFICATIONS 49 Thisfis a one-to-one continuous function from (0 ;1)1onto the unit circle. By Theorem 5.8, this is a homeomorphism. c 1999, David Royster Introduction to Topology For Classroom Use Only