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 \liminf and the \limsup of some examples of sequences (x_n), including the following examples: x_n=(-1)^n; x_n=(-1)^n (n+1)/n = (-1)^n(1+\frac1n), and \Jds x_n=\frac{2n^2+4n+1}{3n^2+7}.
I don't want to repeat everything I said there. Instead I want to look at the connection between \liminf, \limsup, 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 (x_n) converges if and only if \liminf_{n \to \infty} x_n = \limsup_{n \to \infty} x_n\, in which case we have \lim_{n\to\infty}x_n = \liminf_{n \to \infty} x_n = \limsup_{n \to \infty} x_n\,.
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 \Rbar.
I'll start by recalling the definitions of \liminf and \limsup for bounded sequences of real numbers. Below I'll often use (x_n) as an abbreviation for a sequence (x_n)_{n=1}^\infty.
Let (x_n) be a bounded sequence of real numbers. So there are real numbers a and b with the property that, for all n \in \N, we have a \leq x_n \leq b. We then have
a \leq \inf\{x_n:n\in \N\} \leq \sup\{x_n:n\in \N\} \leq b\,.
For each n\in \N, set
s_n=\inf \{x_n,x_{n+1},x_{n+2},\dots\} \Jand S_n= \sup \{x_n,x_{n+1},x_{n+2},\dots\}\,.
We note that the sequence (s_n) is noncreasing while (S_n) is nonincreasing and, for all n \in \N, we have
a \leq s_1 \leq s_n \leq S_n \leq S_1 \leq b\,.
Since the sequences (s_n) and (S_n) are monotone and bounded, they both converge in \R. Indeed (s_n) converges to \sup\{s_1,s_2,\dots\}, and (S_n) converges to \inf\{S_1,S_2,\dots\}\,.
We define \liminf_{n \to \infty} x_n = \sup\{s_n:n\in\mathbb{N}\}= \lim_{n\to \infty} s_n\,,and \limsup_{n \to \infty} x_n = \inf\{S_n:n\in\mathbb{N}\}= \lim_{n\to \infty} S_n\,. For convenience, set s=\liminf_{n \to \infty} x_n and S=\limsup_{n \to \infty} x_n.
We'll work with the same assumptions, definitions and notation for the rest of this post.
Set A=\{x \in \R: (x_n) \textrm{ has at least one subsequence which converges to }x\,\}. We then claim the following:
- A \neq \emptyset (essentially the Bolzano-Weierstrass theorem), and in fact we have s\in A and S \in A;
- A has minimum element s and maximum element S.
We start by proving claim 1. We show that s \in A (the proof for S is similar). First recall that the sequence (s_n) is monotone increasing, and
s=\sup\{s_n:n\in\mathbb{N}\}= \lim_{n\to \infty} s_n\,.\tag{$1$} We choose inductively a sequence of natural numbers n_1<n_2<n_3<\cdots with the property that, for each k \in \N, we have |x_{n_k}-s| < 1/k.To choose n_1, note that by (1) (and the definition of \sup) there exists an m\in \N with s-1<s_m \leq s. (You can take the first such m if you like).
Then, by the definition of s_m as an infimum, there exists n\geq m such that s_m\leq x_n < s_m+1. (You can take the first such n if you like.)
We have s-1<s_m\leq x_n <s_m+1 \leq s+1\,. Thus |x_n-s|<1, and we can take n_1=n for this n.
Now let k \in \N with k\geq 2, and suppose that we have chosen n_1<n_2<\cdots<n_{k-1} already. To choose n_k, we again start by using (1): we choose m\in\N such that s-\frac{1}{k} < s_m \leq s\,. (You can take the first such m if you like.)
Set m'=\max\{m,n_{k-1}+1\} (this is one way to make sure that we are past n_{k-1}), and note that we still have
s-\frac1k < s_{m'} \leq s\,.
By the definition of s_{m'} as an infimum, there exists n\geq m' such that s_{m'}\leq x_n < s_{m'}+\frac{1}{k}\,. (You can take the first such n if you like.)
Then we have n>n_{k-1} (we were careful above) and s-\frac1k<s_{m'}\leq x_n <s_{m'}+\frac1k \leq s+\frac1k\,. Thus |x_n-s|<1/k, and we can take n_k=n for this n.
The inductive choice may proceed, and gives us a subsequence (x_{n_k})_{k=1}^\infty with the property that x_{n_k} \to s as k\to \infty. Thus s \in 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 \in \R with x<s. Then, by (1), there exists m \in \N with s_m>x. But then, for all k \geq m, we have x_k \geq s_m. It follows that no subsequence of (x_n) can converge to x, and so x \notin A. Thus s is the minimum element in A, as claimed.
The proof for S is similar.\quad\square
If you like my notion of 'absorption', you might like to think of that last bit another way. For each m \in \N, the interval [s_m,\infty) absorbs the sequence (x_n) (by stage m) and hence also absorbs every subsequence of (x_n). If such a subsequence converges, the limit must also be in [s_m,\infty).
Thus for all x \in A, we must have, for all m \in \N, that x \in [s_m,\infty). It follows that x \geq \sup_{m} s_m = s.
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