Revisiting liminf and limsup
I said quite a lot about this in a previous post on my WordPress blog in the post
https://explainingmaths.wordpress.com/2009/01/10/an-application-of-absorption-to-teaching-lim-inf-and-lim-sup/and there was a lot of discussion in the comments below the post. In particular, in those comments, I discussed the lim inf and the lim sup of some examples of sequences (xn), including the following examples: xn=(−1)n; xn=(−1)n(n+1)/n=(−1)n(1+1n), and xn=2n2+4n+13n2+7.
I don't want to repeat everything I said there. Instead I want to look at the connection between lim inf, lim sup, convergent subsequences and the Bolzano-Weierstrass Theorem.
Quoting from the WordPress post mentioned above:
and a bit later
That isn't the approach I took in that post, so I thought that I'd revisit that approach now.
If we can establish those claims, the Bolzano-Weierstrass Theorem will follow (every bounded sequence of real numbers has at least one convergent subsequence).
These claims also fit with the standard result that a bounded sequence of real numbers (xn) converges if and only if lim infn→∞xn=lim supn→∞xn in which case we have limn→∞xn=lim infn→∞xn=lim supn→∞xn.
If you don't want to restrict yourself to bounded sequences, the same results hold for all sequences of extended real numbers, using a suitable notion of convergence in ¯R.
I'll start by recalling the definitions of lim inf and lim sup for bounded sequences of real numbers. Below I'll often use (xn) as an abbreviation for a sequence (xn)∞n=1.
Let (xn) be a bounded sequence of real numbers. So there are real numbers a and b with the property that, for all n∈N, we have a≤xn≤b. We then have
a≤inf{xn:n∈N}≤sup{xn:n∈N}≤b.
For each n∈N, set
sn=inf{xn,xn+1,xn+2,…}andSn=sup{xn,xn+1,xn+2,…}.
We note that the sequence (sn) is noncreasing while (Sn) is nonincreasing and, for all n∈N, we have
a≤s1≤sn≤Sn≤S1≤b.
Since the sequences (sn) and (Sn) are monotone and bounded, they both converge in R. Indeed (sn) converges to sup{s1,s2,…}, and (Sn) converges to inf{S1,S2,…}.
We define lim infn→∞xn=sup{sn:n∈N}=limn→∞sn,and lim supn→∞xn=inf{Sn:n∈N}=limn→∞Sn. For convenience, set s=lim infn→∞xn and S=lim supn→∞xn.
We'll work with the same assumptions, definitions and notation for the rest of this post.
Set A={x∈R:(xn) has at least one subsequence which converges to x}. We then claim the following:
- A≠∅ (essentially the Bolzano-Weierstrass theorem), and in fact we have s∈A and S∈A;
- A has minimum element s and maximum element S.
We start by proving claim 1. We show that s∈A (the proof for S is similar). First recall that the sequence (sn) is monotone increasing, and
s=sup{sn:n∈N}=limn→∞sn. We choose inductively a sequence of natural numbers n1<n2<n3<⋯ with the property that, for each k∈N, we have |xnk−s|<1/k.To choose n1, note that by (1) (and the definition of sup) there exists an m∈N with s−1<sm≤s. (You can take the first such m if you like).
Then, by the definition of sm as an infimum, there exists n≥m such that sm≤xn<sm+1. (You can take the first such n if you like.)
We have s−1<sm≤xn<sm+1≤s+1. Thus |xn−s|<1, and we can take n1=n for this n.
Now let k∈N with k≥2, and suppose that we have chosen n1<n2<⋯<nk−1 already. To choose nk, we again start by using (1): we choose m∈N such that s−1k<sm≤s. (You can take the first such m if you like.)
Set m′=max{m,nk−1+1} (this is one way to make sure that we are past nk−1), and note that we still have
s−1k<sm′≤s.
By the definition of sm′ as an infimum, there exists n≥m′ such that sm′≤xn<sm′+1k. (You can take the first such n if you like.)
Then we have n>nk−1 (we were careful above) and s−1k<sm′≤xn<sm′+1k≤s+1k. Thus |xn−s|<1/k, and we can take nk=n for this n.
The inductive choice may proceed, and gives us a subsequence (xnk)∞k=1 with the property that xnk→s as k→∞. Thus s∈A as claimed.
The proof for S is entirely similar (interchanging inf and sup), so we have proved claim 1. (In particular this proves the Bolzano-Weierstrass Theorem at the same time.)
For claim 2, we show that s is the minimum element in A. (The proof that S is the maximum element is again similar.)
Let x∈R with x<s. Then, by (1), there exists m∈N with sm>x. But then, for all k≥m, we have xk≥sm. It follows that no subsequence of (xn) can converge to x, and so x∉A. Thus s is the minimum element in A, as claimed.
The proof for S is similar.◻
If you like my notion of 'absorption', you might like to think of that last bit another way. For each m∈N, the interval [sm,∞) absorbs the sequence (xn) (by stage m) and hence also absorbs every subsequence of (xn). If such a subsequence converges, the limit must also be in [sm,∞).
Thus for all x∈A, we must have, for all m∈N, that x∈[sm,∞). It follows that x≥supmsm=s.
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