If m S then there exists anx S such thatm x m which implies thatm is an

If m s then there exists anx s such thatm x m which

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(Heine-Borel Theorem)A subsetSofRis compact if and only if it is closedand bounded.Proof:SupposeSis compact. LetGbe the collection of open intervalsIn= (-n, n), n= 1,2, . . ..ThenGis an open cover ofS. SinceSis compact,Gcontains a finite subcoverIn1, In2, . . ., Ink.Letm= maxni. ThenSIm, and for allxS,|x| ≤m. ThereforeSis bounded.To show thatSis closed, we must show thatScontains all its accumulation points. Supposethatpis an accumulation point ofSand supposep /S.For each positive integern,letGn= [p-1/n, p+ 1/n]c. Since the complement of a closed interval is an open set,Gnis an open 11
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(Bolzano-Weierstrass Theorem)IfSRis a bounded infinite set, thenShas at least one accumulation point.Proof:LetSbe a bounded infinite set and suppose thatShas no accumulation points. ThenSis closed (vacuously), andSis compact. For eachxS, letNxbe a neighborhood ofxsuch thatSNx={x}. The set of neighborhoodsNx, xSis an open cover ofS. SinceSis compact,this open cover has a finite subcoverNx1, Nx2, . . ., Nxk.ButS[Nx1Nx2. . .Nxk] ={x1, x2, . . ., xk}which implies thatSis finite, a contradiction. Exercises 1.5 1. True – False.Justify your answer by citing a theorem, giving a proof, or giving a counter-example.(a) Every finite set is compact.(b) No infinite set is compact.(c) If a set is compact, then it has a maximum and a minimum.(d) If a set has a maximum and a minimum, then it is compact.(e) IfSRis compact, then there is at least one pointxRsuch thatxis anaccumulation point ofS.(f) IfSRis compact andxis an accumulation point ofS,thenxS.2. Show that each of the following subsetsSofRis not compact by giving an open cover ofSthat has no finite subcover.(a)S= [0,1)(b)S=N(c)S={1/n:nN}3. Prove that the intersection of any collection of compact sets is compact.4. Prove that ifSRis compact andTis a closed subset ofS,thenTis compact. 12
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