Can we always find the infimum and supremum of a sequence?

Question

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

Answer 1

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.

Answer 2

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.