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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