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# Class Note for MATH 623 at UMass

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This 2 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Massachusetts taught by a professor in Fall. Since its upload, it has received 17 views.

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Date Created: 02/06/15

Some fact about sup inf lim sup and lim inf 1 Supremum and In mum For a set X of real numbers7 the number sup X7 the supremum ofX or least upper bound of X is de ned by 1 ForalleX7z 2 For any 6 gt 0 there exists x such that z gt g 7 6 and the in mum of X ian is de ned similarly 2 Sequences accumulation points lim sup and lim inf Let 3311 be a sequence of real numbers A point z is called an accumulation point of if there exists a subsequence which converges to x A well known theorem is Theorem 21 Boltzano Weierstrass Theorem Any sequence which is bounded above ie there eaists M such that an S M for all n has at least one accumulation point Note by the way that this theorem is what is needed to prove that compact sets in R or Rd are exactly the sets which are closed and bounded It is instructive to have a look at the proof Proof Consider the set X xg in nitely many an are gt This set is bounded and we de ne sup X The number g is nite and by de nition for any 6 gt 0 only nitely many an satisfy an 2 E and in nitely many an such that an 2 g 7 6 Therefore there are in nitely many an in the interval g 7 6 6 For any integer k take 6 i one construct the subsequence in the following inductive way Choose n1 such am E g 7 17 g 17 then inductively choose nk gt nk1 such that znk E g 7 1h7 The sequence ank il converges to g I This proof exhibits not any accumulation point7 but the largest accumulation point and it is called the limit superior of the sequence We denote it by g lim sup an sup xg in nitely many an are gt x Hoe Using sequences which are bounded below and the inf instead of the sup one de nes lim inf an inf xg in nitely many an are lt x Hoe which is the smallest accumulation point of the sequence 3 Properties of hm sup and hm inf Trivially we have hm inf xn 3 lim sup xn naoo H00 and lim xn x if and only if hm inf xn hm sup z Hoe 4 00 naoo We also have hm supxn y S hm sup xn lim sup yn hm infzn y 2 hm inf xn hm inf yn You should prove this and note that the inequalities can be strict Find such examples The lim sup and lim inf can also be written as follows lim sup xn lim sup xk inf sup xk 4 00 400 an 21 an hm inf xn hm inf xk sup inf xk naoo 71700 kzn gt1 an We prove this for lim sup Note rst that yn E sukan M is a decreasing sequence7 ie7 yn S y and thus it is convergent and we have hmH00 yn infH00 y Let g denote this limit7 then for any 6 gt 0 there exists N such that for all n 2 N we have Synsupxk 6 an Using the right inequality for n N shows that at most nitely many Q are bigger than g 6 On the other using the de nition for the sup and left inequality for n N we can nd 711 2 N such that xm gt g 7 6 Using the left inequality for 711 1 we can nd 712 such that zm gt g 7 E and thus there exists in nitely many bigger than g 7 6 This shows that g limsupyH00 x

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