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I've found in my book that: $$\liminf_{n\to\infty} \ x_{n} = \sup\{\inf\{x_{k}:k\geq n \}:n \in \mathbb{N}\}$$ $$\limsup_{n\to\infty} \ x_{n} = \inf\{\sup\{x_{k}:k\geq n \}:n \in \mathbb{N}\}$$

If $X_n=\{x_{k}:k\geq n \}$ we define $s_n=\inf X_n$. I understand that $s_n$ is an increasing sequence but how are we sure that we will be able to find the supremum and that this will be the same as the limit of $\liminf_{n\to\infty} \ x_{n}$? Thanks

Jelly Belly
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  • No, we cant. If the sequence is chaotic we cant predict it evolution. Many recursive functions are chaotic. – Masacroso Dec 06 '15 at 14:25
  • For this question, shouldn't you assume you already know all of the elements in the sequence? If you are given a sequence in which you know the elements, even if they were produced randomly, wouldn't that sequence have a limit sup? – Frank Dec 10 '18 at 17:38

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Firstly you need to be working in the affinely-extended real line, so that supremum and infimum of any sequence always exist. $\def\nn{\mathbb{N}}$

Let $y_n = \sup(\{ x_k : k \in \nn_{\ge n} \})$.

By definition $\limsup_{n\to\infty} x_n = \lim_{n\to\infty} y_n$.

But $(y_n)_{n\in\nn}$ is a decreasing sequence and hence by monotone convergence theorem (suitably extended to the extended real line) we get:

$\lim_{n\to\infty} y_n = \inf(\{ y_n : n \in \nn \})$.

Similarly for limit inferior.

user21820
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  • Could you please explain how did you use the monotone convergence theorem in this case? – Jelly Belly Dec 06 '15 at 14:48
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    @JellyBelly: Any monotonic sequence converges, so the inf must be the limit. It is not hard to prove this fact. In fact, usually the proof of the monotone convergence theorem is by showing that the inf is the limit! – user21820 Dec 06 '15 at 14:56
  • How do you know that any monotonic sequence converges? Which theorem did you use to say this? – Jelly Belly Dec 06 '15 at 15:00
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    @JellyBelly: That is the monotone convergence theorem! It says (for the affinely-extended reals) that every monotonic sequence converges. And as I already said, the proof of this theorem is via showing exactly what I claimed. If you're not familiar with the monotone convergence theorem, then try to prove it yourself. It's not that hard. – user21820 Dec 06 '15 at 15:04
  • Thanks! Maybe I know the theorem with other name. I'm not learning math it in english. I was using the definition of Wikipedia. – Jelly Belly Dec 06 '15 at 15:07
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    @JellyBelly: Ah yes language is an unfortunate barrier sometimes. The wikipedia article doesn't mention the version for affinely-extended reals, but it's easy to get it from the standard one because an unbounded increasing sequence must converge to $\infty$ by definition, and its supremum is also $\infty$ by definition. – user21820 Dec 07 '15 at 06:01
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If $(x_{n})$ is bounded, then of course the decreasing sequence $(\sup_{k \geq n}x_{k})_{n \geq 1}$ is bounded below and the increasing sequence $(\inf_{k \geq n}x_{k})_{n \geq 1}$ is bounded above; hence the corresponding $\inf$ and $\sup$ exist by axiom.

By the way, the usual treatment is to define $\limsup_{n}x_{n} := + \infty$ if $(x_{n})$ is unbounded from above and $\liminf_{n}x_{n} := -\infty$ if $(x_{n})$ is unbounded from below.

Yes
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