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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 ofRn.5.1 Compact Spaces and Subspaces
Denition 5.1LetAbe a subset of the topological spaceX. Anopen coverforA is a collectionOof open sets whose union containsA. Asubcoverderived from the open coverOis a subcollectionO0ofOwhose union containsA. Example 5.1.1LetA= [0;5] and consider the open coverO=f(n1;n+ 1)jn=1;::: ;1g:
Consider the subcoverP=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 ofOthat coversA. Denition 5.2A topological spaceXiscompactprovided that every open cover ofXhas a nite subcover.
This says that however we writeXas a union of open sets, there is always a nite subcollectionfOigni=1of these sets whose union isX. A subspaceAofXiscompact ifAis a compact space in its subspace topology. Since relatively open sets in the subspace topology are the intersections of open sets inXwith the subspaceA, the denition of compactness for subspaces can be restated as follows. Alternate Denition:A subspaceAofXiscompactif and only if every open cover ofAby open sets inXhas a nite subcover. 4344 CHAPTER 5. COMPACTNESSExample 5.1.21. Any space consisting of a nite number of points is compact.
2. The real lineRwith the nite complement topology is compact.
3. An innite setXwith 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 integern, sinceO=fB(0;n)jn=
1;::: ;1gis an open cover with no nite subcover.
A sequence of setsfSng1n1isnestedifSn+1Snfor each positive integern. Theorem 5.1 (Cantor's Nested Intervals Theorem)Iff[an;bn]g1n=1is a nested sequence of closed and bounded intervals, then\1n=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 anddthe greatest lower bound of the right endpoints. The existence ofcanddare guaranteed by the Least Upper Bound Property. Now, by property (iii),cbnfor alln, socd. Since a ncdbn, then [c;d][an;bn] for alln. Thus,\1n=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 thatc=dandcis the one point of the intersection.Theorem 5.2The interval[0;1]is compact. Proof:LetObe an open cover. Assume that [0;1] is not compact. Then either [0;12] or [12;1] is not covered by a nite number of members ofO. Let [a1;b1] be the half that is not covered by a nite number of members ofO. Apply the same reasoning to the interval [a1;b1]. One of the halves, which we will call [a2;b2], is not nitely coverable byOand has length14. We can continue this reasoning inductively to create a nested sequence of closed intervalsf[an;bn]g1n=1, none of which is nitely coverable byO. Also, by construction, we have that b nan=12n;c1999, David Royster Introduction to TopologyFor Classroom Use Only
5.1. COMPACT SPACES AND SUBSPACES 45so 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 intervalO2Owithp2O. 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 byOsince we just covered it with one set fromO. This contradiction shows that [0;1] is nitely coverable byOand is compact.Compactness is dened in terms of open sets. The duality between open and closed sets and ifC=XnO, Xn \ 2IC 2IO leads us to believe that there is a characterization of compactness with closed sets. Denition 5.3A familyAof subsets of a spaceXhas thenite intersection propertyprovided that every nite subcollection ofAhas non-empty intersection. Theorem 5.3A spaceXis compact if and only if every family of closed sets inX with the nite intersection property has non-empty intersection. This says that ifFis a family of closed sets with the nite intersection property, then we must have that\ FC 6=;. Proof:Assume thatXis compact and letF=fCj2Igbe a family of closed sets with the nite intersection property. We want to show that the intersection of all members ofFis non-empty. Assume that the intersection is empty. Let O=fO=XnCj2Ig.Ois a collection of open sets inX. Then, 2IO2IXnC=Xn\
2IC =Xn ;=X: Thus,Ois an open cover forX. SinceXis compact, it must have a nite subcover; i.e., X=n[ i=1O i=n[ i=1(XnCi) =Xnn\ i=1C i: This means that\ni=1Cimust be empty, contradicting the fact thatFhas the nite intersection property. Thus, ifFhas the nite intersection property, then the intersection of all members ofFmust be non-empty.The opposite implication is left as an exercise.c
1999, David Royster Introduction to TopologyFor Classroom Use Only
46 CHAPTER 5. COMPACTNESSIs compactness hereditary? No, because (0;1) is not a compact subset of [0;1]. It
isclosed hereditary. Theorem 5.4Each closed subset of a compact space is compact. Proof:LetAbe a closed subset of the compact spaceXand letObe an open cover ofAby open sets inX. SinceAis closed, thenXnAis open and O =O[ fXnAg is an open cover ofX. SinceXis compact, it has a nite subcover, containing only nitely many membersO1;::: ;OnofOand may containXnA. SinceX= (XnA)[n[
i=1O i; it follows that An[ i=1O iandAhas a nite subcover.Is the opposite implication true? Is every compact subset of a space closed? Not
necessarily. The following though is true. Theorem 5.5Each compact subset of a Hausdor space is closed. Proof:LetAbe a compact subset of the Hausdor spaceX. To show thatAis closed, we will show that its complement is open. Letx2XnA. Then for eachy2A there are disjoint setsUyandVywithx2Vyandy2Uy. The collection of open sets fUyjy2Agforms an open cover ofA. SinceAis compact, this open cover has a nite subcover,fUyiji= 1;:::;ng. Let U=n[ i=1U yiV=n\ i=1V yi: Since eachUyiandVyiare disjoint, we haveUandVare disjoint. Also,AUand x2V. Thus, for each pointx2XnAwe have found an open set,V, containingxwhich is disjoint fromA. Thus,XnAis open, andAis closed.Corollary 6LetXbe a compact Hausdor space. A subsetAofXis compact if
and only if it is closed. The following results are left to the reader to prove. Theorem 5.6IfAandBare disjoint compact subsets of a Hausdor spaceX, then there exist disjoint open setsUandVinXsuch thatAUandBV. Corollary 7IfAandBare disjoint closed subsets of a compact Hausdor spaceX, then there exist disjoint open setsUandVinXsuch thatAUandBV.c1999, David Royster Introduction to TopologyFor Classroom Use Only
5.2. COMPACTNESS AND CONTINUITY 475.2 Compactness and Continuity
Theorem 5.7LetXbe a compact space andf:X!Ya continuous function fromXontoY. ThenYis compact.
Proof:We will outline this proof. Start with an open cover forY. Use the continuity offto pull it back to an open cover ofX. Use compactness to extract a nite subcoverforX, and then use the fact thatfis onto to reconstruct a nite subcover forY.Corollary 8LetXbe a compact space andf:X!Ya continuous function. The
imagef(X)ofXinYis a compact subspace ofY. Corollary 9Compactness is a topological invariant. Theorem 5.8LetXbe a compact space,Ya Hausdor space, andf:X!Ya continuous one-to-one function. Thenfis a homeomorphism.