Compactness in metric spaces - UCL

1 MATHEMATICS 3103 (Functional Analysis)YEAR 2012 2013, TERM 2 HANDOUT #2: Compactness OF metric SPACESC ompactness in metric spacesThe closed intervals [a, b] of the real line, and more generally the closed bounded subsetsofRn, have some remarkable properties, which I believe you have studied in your course inreal analysis. For instance:Bolzano Weierstrass bounded sequence of real numbers hasa convergent can be rephrased as:Bolzano Weierstrass theorem (rephrased).LetXbe any closed boundedsubset of the real line. Then any sequence (xn) of points inXhas a subsequenceconverging to a point ofX.(Whyis this rephrasing valid? Note that this property doesnothold ifXfails to be closedor fails to be bounded why?) And here is another example:Heine Borel covering of a closed interval [a, b] or moregenerally of a closed bounded setX R by a collection of open sets has afinite theorems are not only interesting they are also extremely useful in applications, aswe shall see.

2 So our goal now is to investigate the generalizations of these concepts to begin with some definitions: Let (X, d) be a metric space. AcoveringofXis acollection of sets whose union isX. Anopen coveringofXis a collection ofopensetswhose union isX. The metric spaceXis said to becompactif every open covering hasa finite abstracts the Heine Borel property; indeed, the Heine Boreltheorem states that closed bounded subsets of the real line are can rephrase Compactness in terms of closed sets by makingthe following observation:IfUis an open covering ofX, then the collectionFofcomplementsof sets inUis a collectionof closed sets whose intersection is empty (why?); and conversely, ifFis a collection of closedsets whose intersection is empty, then the collectionUof complements of sets inFis an opencovering. Thus, a spaceXis compact if and only if every collection of closed sets withanempty intersection has a finite subcollection whose intersection is also empty.

3 Or, passing tothe contrapositive, we can put it another way by making the following definition: a collectionFof sets is said to have thefinite intersection propertyif every finite subcollection ofFhas a nonempty intersection. We have then shown:1Or more formally: If (U ) I(whereIis some index set) is a collection of open sets ofXsatisfying IU =X, then there exists a finite subsetJ Isuch that JU = metric spaceXis compact if and only if every collectionFof closedsets inXwith the finite intersection property has a nonempty far so good; but thus far we have merely made a trivial reformulation of the definitionof Compactness . Let us go farther by making another definition:A metric spaceXis said to besequentially compactif every sequence (xn) n=1ofpoints inXhas a convergent subsequence. This abstracts the Bolzano Weierstrass property;indeed, the Bolzano Weierstrass theorem states that closed bounded subsets of the real lineare sequentially finally, let us make another definition: A metric space (X, d) is said to betotallybounded(orprecompact) if, for every >0, the spaceXcan be covered by a finite familyof open balls of radius.

4 (You could alternatively use closed balls and get the same concept why?) Another way of saying this is: A metric space (X, d) is totally bounded if, forevery >0, there exists a finite subsetA Xsuch thatd(x, A)< for allx X. (Why isthis equivalent?) Any such finite subset is called an then have the following fundamental theorem characterizing compact metric spaces :Theorem ( Compactness of metric spaces )For a metric spaceX, the following areequivalent:(a)Xis compact, every open covering ofXhas a finite subcovering.(b) Every collection of closed sets inXwith the finite intersection property has a nonemptyintersection.(c) IfF1 F2 F3 ..is a decreasing sequence of nonempty closed sets inX, then n=1 Fnis nonempty.(d)Xis sequentially compact, every sequence inXhas a convergent subsequence.(e)Xis totally bounded and have already proved the equivalence of (a) and (b). Let us now prove (b) = (c)= (d) = (e) = (a).Proof of (b)= (c).This is trivial, since adecreasingsequence ofnonemptyclosedsets obviously has the finite intersection property.

5 (Why? Ifn1, .. , nkare given indices,what isFn1 Fn2 .. Fnk?) Proof of (c)= (d).Let (xn) be a sequence of points inX, and letFnbe the closure ofthe set{xn, xn+1, xn+2, ..}. The family of sets{Fn}is decreasing ( F2 F3 ..),and all the setsFnare nonempty and closed. Therefore, by (c), the set n=1 Fncontains2 Some authors ( Kolmogorov Fomin) define an -net to be any finite subsetA Xsuch thatd(x, A) for allx X, withnon-strictinequality. This changes slightly the statements of proofsbut makes noessential difference can you see why?2at least one pointa. Then it is easy to see that (xn) contains a subsequence convergingtoa: for instance, setn1= 1 and then letnkbe the smallest integer> nk 1such thatd(xnk, a)<1/k; such an integer exists becauseabelongs to all the setsFn. (You shouldmake sure you understand this last step.) Proof of (d)= (e).To prove thatXis complete, let (xn) be any Cauchy sequenceinX. By (d), (xn) contains a subsequence converging to some pointa X.

6 But then, byProposition (see last week s notes), the whole sequence (xn) converges toa. This showsthatXis suppose thatXis not totally bounded, there exists a number >0 such thatXhas no finite covering by open balls of radius . Then we can define a sequence (xn) n=1ofpoints inXhavingd(xi, xj) for alli6=j, by the following inductive construction: Firstletx1be any point inX. Then, supposing thatx1, .. , xn 1have been chosen, we know thatthe union of the open balls of centerxi(1 i n 1) and radius is not the whole space,hence we can choose a pointxnsatisfyingd(xi, xn) for alli < n. When we are done, wehaved(xi, xj) for alli6=j(why?). On the other hand, the sequence (xn) cannot haveany convergent subsequence; for if it had a subsequence (xnk) converging toa, then therewould exist an integerk0such thatd(xnk, a)< /2 for allk k0, and hence by the triangleinequalityd(xnk, xnk )< for allk, k k0, contrary to the definition of the sequence (xn).

7 3 Proof of (e)= (a).Suppose thatXis not compact, we have an open covering(U ) IofXsuch that no finite subfamily is a covering ofX. We will define a sequence(xn) n=1of points inX, as follows: First choose an -net with = 1/2 (this is possiblebecauseXis totally bounded), and letx1be any element of that -net with the propertythat no finite subfamily of (U ) Iis a covering ofB(x1,1/2). [Such an element has to exist,because if every ball of radius 1/2 centered at a point of the -net had a finite subcoverfrom (U ) I, then the whole spaceXwould have a finite subcover from (U ) I(why?).]Next choose an -net with = 1/4, and letx2be any element of that -net satisfyingB(x1,1/2) B(x2,1/4)6= and having the property that no finite subfamily of (U ) Iis a covering ofB(x2,1/4). [Such an element has to exist, because if every ball of radius1/4 centered at a point of the -net and having nonempty intersection withB(x1,1/2) hada finite subcover from (U ) I, thenB(x1,1/2) would have a finite subcover from (U ) I(why?)]

8 ] Continue analogously: at thenth stage, choose an -net with = 1/2n, and letxnbe any element of that -net satisfyingB(xn 1,1/2n 1) B(xn,1/2n)6= and having theproperty that no finite subfamily of (U ) Iis a covering ofB(xn,1/2n).It follows from this construction thatd(xn 1, xn) 12n 1+12n 12n 2( )(why?) and hence that, form < n,d(xm, xn) d(xm, xm+1) +d(xm+1, xm+2) +..+d(xn 1, xn)( )3 Another way of stating this argument is: The sequence (xn) clearly cannot have any Cauchy subsequence;therefore, it cannot have any convergent 12m 1+12m+..+12n 2( ) 12m 2,( )which shows that (xn) is a Cauchy sequence inX. SinceXis complete, the sequence (xn)converges to some pointa let 0 Ibe an index such thata U 0(why must such an index exist?). Thereexists >0 such thatB(a, ) U 0. By the definition ofa, there exists an integernsuchthatd(xn, a)< /2 and also 1/2n< /2 (why?). The triangle inequality then shows thatB(xn,1/2n) B(a, ) U 0( )(why?). But this contradicts the fact that no finite subfamily of (U ) Iis a covering ofB(xn,1/2n).

9 Whew! Warning:For general (nonmetrizable) topological spaces , Compactness isnotequivalentto sequential also have the following easy fact:Proposition totally bounded metric space (and in particular everycompact met-ric space) is totally bounded, then there exists for eachna finite subsetAn Xsuchthat, for everyx X,d(x, An)<1/n. Now letA= n=1An. The setAis either finite orcountably infinite (why?); and for eachx Xwe haved(x, A) d(x, An)<1/n, henced(x, A) = 0, hencex A(why?). This proves thatAis dense inX. Intuitively, a separable space is one that is well approximated by a countable subset ,while a compact space is one that is well approximated by a finite subset . (Albeit in aslightly different sense of well approximated in the two cases.)A subsetAof a metric spaceXis said to becompactifA, considered as a subspace ofXand hence a metric space in its own right, is compact. We have the following easy facts,whose proof I leave to you:Proposition (a) A closed subset of a compact space is compact.

10 (b) A compact subset of any metric space is closed.(c) Afiniteunion of compact sets is course, aninfiniteunion of compact sets need not even be closed (give an example!); andeven when it is closed, it need not be compact (give another example!).The Heine Borel (or Bolzano Weierstrass) theorem of elementary real analysis can berestated as follows:Proposition ( Compactness of subsets inR)A subsetA Ris compact if andonly if it is closed and corresponding result forRnis an easy consequence:Proposition ( Compactness of subsets inRn)A subsetA Rnis compact if andonly if it is closed and compact subset ofRnis obviously closed and bounded (why?), so we needonly prove the converse. Moreover, every bounded subset ofRnis contained in a cube[ M, M]nfor someM < , so by Proposition (a) we need only prove that [ M, M]nis compact. But this follows from the Heine Borel (or Bolzano Weierstrass) theorem forRcombined with the fact that any finite Cartesian product of compact spaces is compact, astraightforward result that you will prove in Problem 2 of Problem Set #2.