Revisiting liminf and limsup

This post is also available in PDF form from my WordPress Blog at https://explainingmaths.wordpress.com/2023/09/01/revisiting-liminf-and-limsup/

My recent series of posts about Fatou's Lemma for sums was aimed primarily at mathematics students at 3rd-year undergraduate level or above. So I think I should write a post more suitable for first-year undergraduate students. So I'm going to have another look at the topic of lim inf and lim sup for bounded sequences of real numbers (though most of what I say can be generalised to sequences, or even nets, of extended real numbers).

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:

Many undergraduates have difficulty understanding the notions of the lim inf and lim sup of a sequence.

and a bit later

One approach that can help a little is to explain that lim infn(xn) is the minimum of all the possible limits of subsequences of the sequence (xn), and similarly for lim sup, with max in place of min.

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 infnxn=lim supnxn in which case we have limnxn=lim infnxn=lim supnxn.

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 nN, we have axnb. We then have
ainf{xn:nN}sup{xn:nN}b.
For each nN, 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 nN, we have
as1snSnS1b. 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 infnxn=sup{sn:nN}=limnsn,and lim supnxn=inf{Sn:nN}=limnSn. For convenience, set s=lim infnxn and S=lim supnxn.

We'll work with the same assumptions, definitions and notation for the rest of this post.

Set A={xR:(xn) has at least one subsequence which converges to x}. We then claim the following:

  1. A (essentially the Bolzano-Weierstrass theorem), and in fact we have sA and SA;
  2. A has minimum element s and maximum element S.

We start by proving claim 1. We show that sA (the proof for S is similar). First recall that the sequence (sn) is monotone increasing, and

s=sup{sn:nN}=limnsn. We choose inductively a sequence of natural numbers n1<n2<n3< with the property that, for each kN, we have |xnks|<1/k.

To choose n1, note that by (1) (and the definition of sup) there exists an mN with s1<sms. (You can take the first such m if you like).

Then, by the definition of sm as an infimum, there exists nm such that smxn<sm+1. (You can take the first such n if you like.)

We have s1<smxn<sm+1s+1.  Thus |xns|<1, and we can take n1=n for this n.

Now let kN with k2, and suppose that we have chosen n1<n2<<nk1 already. To choose nk, we again start by using (1): we  choose mN such that s1k<sms. (You can take the first such m if you like.)

Set m=max{m,nk1+1} (this is one way to make sure that we are past nk1), and note that we still have 

s1k<sms.

By the definition of sm as an infimum, there exists nm such that smxn<sm+1k. (You can take the first such n if you like.)

Then we have n>nk1 (we were careful above) and s1k<smxn<sm+1ks+1k. Thus |xns|<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 xnks as k. Thus sA 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 xR with x<s. Then, by (1), there exists mN with sm>x. But then, for all km, we have xksm. It follows that no subsequence of (xn) can converge to x, and so xA. 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 mN, 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 xA, we must have, for all mN, that x[sm,). It follows that xsupmsm=s.

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