Sufficient conditions for a = lim inf xn

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SUMMARY

The discussion focuses on proving that for a bounded below sequence \( x_n \) in \( \mathbb{R} \), if conditions (i) and (ii) hold, then \( a = \liminf x_n \). Condition (i) states that for every \( \epsilon > 0 \), there exists an \( N \) such that \( a - \epsilon < x_n \) for all \( n \geq N \). Condition (ii) asserts that for every \( \epsilon > 0 \) and \( M \), there exists an \( n > M \) such that \( x_n < a + \epsilon \). The conclusion drawn is that if \( b < a \) is a cluster point, then it leads to a contradiction with condition (i), confirming that \( a \) is indeed the limit inferior of the sequence.

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samkolb
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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?
 
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Well, suppose b < a is a cluster point. Doesn't that give you a problem with (i).?
 
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.
 
Try e = (a - b)/2. Can't you show all but finitely many of the xn are bounded away from b?
 
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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