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Sufficient conditions for a = lim inf xn 
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#1
Dec109, 06:51 PM

P: 37

This is part of a theorem which is left unproved in "Elementary Classical Analysis" by Marsden and Hoffman.
Let xn be a sequence in R which is bounded below. Let a be in R. Suppose: (i) For all e > 0 there is an N such that a  e < xn for all n >= N. (ii) For all e > 0 and all M, there is an n > M with xn < a + e. Show that a = lim inf xn. (Definition: When xn is bounded below, lim inf xn is the infimum of the set of all cluster points of xn. If xn has no cluster points, then lim inf xn = + infinity. If xn is not bounded below, then lim inf xn =  infinity.) I was able to use (i) and (ii) to show that a is the limit of a subsequence of xn, hence a is a cluster point. So to show that a = lim inf xn, it is sufficient to show that a is a lower bound for the set of cluster points. This is what I can't do. Any suggestions? 


#2
Dec209, 11:44 AM

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Thanks
PF Gold
P: 7,650

Well, suppose b < a is a cluster point. Doesn't that give you a problem with (i).?



#3
Dec209, 12:18 PM

P: 37

So I set e = a  b in (i) and get b < xn for all n>=N. I don't see the problem with this.
I think this implies all cluster points x satisfy b <= x, which means b <= a, but now I'm back where I started. 


#4
Dec209, 12:41 PM

HW Helper
Thanks
PF Gold
P: 7,650

Sufficient conditions for a = lim inf xn
Try e = (a  b)/2. Can't you show all but finitely many of the x_{n} are bounded away from b?



#5
Dec209, 01:45 PM

P: 37

I think I get it now.
Setting e = (a b)/2 in (i) gives (a + b)/2 < xn for all n >= N ==> (a  b)/2 < xn  b for all n >= N ==> no subsequence of xn can converge to b, since (a  b)/2 > 0 ==> b is not a cluster point. thanks! 


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