Problem Set 2: Solutions Math 201A Fall 2016 Problem 1 ...

1 Problem Set 2: SolutionsMath 201A:Fall 2016 Problem 1.(a) Prove that a closed subset of a complete metric spaceis complete. (b) Prove that a closed subset of a compact metric space iscompact. (c) Prove that a compact subset of a metric space is closed (a) IfF Xis closed and (xn) is a Cauchy sequence inF, then (xn)is Cauchy inXandxn xfor somex XsinceXis complete. Thenx FsinceFis closed, soFis complete. (b) Suppose thatF XwhereFis closed andXis compact. If (xn)is a sequence inF, then there is a subsequence (xnk) that convergestox XsinceXis compact.

2 Thenx FsinceFis closed, soFiscompact. Alternatively, If{G X: I}is an open cover ofF,then{G : I} Fcis an open cover ofX. SinceXis compact,there is a finite subcover ofXwhich also coversF, soFis compact. (c) LetK Xbe compact. If (xn) is a convergent sequence inKwithlimitx X, then every subsequence of (xn) converges tox. SinceKiscompact, some subsequence of (xn) converges to a limit inK, sox KandKis closed. Suppose thatKis not bounded, and letx1 K. Then for everyr >0there existsx Ksuch thatd(x1,x) r. Choose a sequence (xn) inKas follows. Pickx2 Ksuch thatd(x1,x2) 1.

3 Given{x1,x2,..,xn},pickxn+1 Ksuch thatd(x1,xn+1) 1 + max1 k nd(x1,xk).By the triangle inequality,d(xk,xn+1) d(x1,xn+1) d(x1,xk) 1for every 1 k follows thatd(xm,xn) 1 for everym6=n, so (xn) has no Cauchysubsequences, and therefore no convergent subsequences, soKis a subset of a metric spaceXwith closure A. Definethe interiorA and boundary AofAbyA = {G A:Gis open}, A= A\A .(a) Why isA open and Aclosed?(b) Prove thatX\ A= (X\A) .(c) Prove thatAis closed if and only if A A, andAis open if and onlyif A Ac.(d) IfAis open, does it follow that ( A) =A?Solution (a) A union of open sets is open soA is open, and an intersection ofclosed sets is closed so Aand A= A (A )care closed.

4 (b) Note thatx A if and onlyB (x) Afor some >0. Ifx Ac,thenB (x) Acfor some >0 since Acis open. Since A A, wehave Ac Ac, soB (x) Ac, meaning thatx (Ac) . It follows that Ac (Ac) . For the reverse inclusion, note that ifx (Ac) , then thereexists >0 such thatB (x) Ac, soxis not the limit of any sequenceinA, meaning thatx Ac. It follows that (Ac) Ac, so (Ac) = Ac. (c) IfAis closed, then A=Aso A=A (A )c A. For the converse,note that sinceA A A, we have A= ( A A ) ( A (A )c) =A A A, then it follows that A A, soA= A, meaning thatAisclosed. (d) This is not true in general.

5 For example, defineA R2byA={(x,y) :x2+y2<1}\{(x,0) : 0 x <1}.ThenAis open, but ( A) ={(x,y) :x2+y2<1}6= a metric space with a dense subsetA Xsuch thatevery Cauchy sequence inAconverges inX. Prove thatXis Let (xn) be a Cauchy sequence inX. SinceAis dense inX, we canchoose a sequence (an) inAsuch thatd(xn,an) 0 asn . (Forexample, choosean Asuch thatd(xn,an)<1/n.) Given any >0, there existsM Nsuch thatd(xn,an)< /3 andd(xm,xn)< /3 for allm,n > M, sod(am,an) d(am,xm) +d(xm,xn) +d(xn,an)< ,which shows that (an) is a Cauchy sequence inA. It follows that (an)converges to somex X.

6 Given any >0, there existsN Nsuch thatd(xn,an)< /2 andd(an,x)< /2 for alln > N, sod(xn,x) d(xn,an) +d(an,x)< ,which shows that (xn) converges toxand proves thatXis :X X Rbe the discrete metric on a setX,d(x,y) ={1 ifx6=y,0 ifx= are the compact subsets of the metric space (X,d)?Solution A subset ofXis compact if and only if it is finite. Every finite set is compact. IfF={x1,x2,..,xn} Xand{G X: I}is an open cover ofF, thenxk G kfor some k I, so{G 1,G 2,..,G n}is a finite subcover ofF. Alternatively, every sequence inFhas aconstant subsequence, which converges to a point inF.}

7 Conversely, ifF Xis infinite andGx={x}, then{Gx:x F}is an open cover ofFwith no finite subcover. Alternatively, ifFisinfinite, then there is a sequence (xn) inFwithxm6=xnfor allm6=n,sod(xm,xn) = 1 form6=n, and (xn) has no Cauchy or rough heuristic is that compact sets have many properties incommon with finite sets. For example, finite sets have the finite the Banach space of real sequences (xn) such thatxn 0 asn with the sup-norm (xn) = supn N|xn|. Is the closedunit ballB={(xn) c0: (xn) 1}compact?Solution The closed unit ball inc0is not compact.

8 For example, letek= ( nk) n=1 nk={1 ifn=k0 ifn6=kdenote the sequence whosekth term is one and whose other terms arezero. Thenek c0since limn nk= 0, andekbelongs to the closedunit ball inc0since ek = 1. However, ej ek = 1 for everyj6=k,so (ek) k=1has no Cauchy or convergent subsequences similar argument using the Riesz lemma shows that the closedunit ball in any infinite-dimensional normed space is not compact in the metric (or topological) spaceXis disconnected if there arenon-empty open setsU,V Xsuch thatX=U VandU V= . A spaceis connected if it is not disconnected.}

9 A spaceXis totally disconnected ifits only non-empty connected subsets are the singleton sets{x}withx X.(a) Show that the interval [0,1] is connected (in its standard metric topology).(b) Show that the setQof rational numbers is totally (a) Suppose for contradiction that [0,1] =U VwhereU,Varenonempty, disjoint open sets in [0,1]. We assume that 0 Uwithoutloss of generality. Leta= sup{x [0,1] : [0,x) U}. Since 0 UandUis open, wehave [0, ) Ufor some >0, so 0< a 1. If 0< b < a, then [0,b) Usince, by the definition of the supremum, there existsb < c < asuchthat [0,c) U.]]]]

10 (It also follows that [0,a) = 0<b<a[0,b) U, so thesupremum is attained and, in fact,{x [0,1] : [0,x) U}= [0,a].) Ifa U, then (a ,a+ ) Ufor some >0, but then [0,a+ ) U,contradicting the definition ofa. On the other hand, ifa V, then(a ,a] Vfor some 0< < a, but then [0,b)6 Ufora < b a,also contradicting the definition ofa. (b) LetA Qbe any subset of the rational numbers with at least twoelements. Choosex,y Awithx6=y. The irrational numbers aredense inR, so there existsz R\Qsuch thatx < z < y. LetU= ( ,z) A, V= (z, ) ,Vare open sets in the relative topology onA.]]]]