I'm trying understand the definitions of limit superior and limit inferior of a sequence $(x_n)$ in extended real numbers. I know that $\lim \sup$ and $\lim \inf$ are the limits of the sequences defined by $a_m := \sup \{ x_k \ ; \ k \geq m \}$ and $b_m := \inf \{ x_k \ ; \ k \geq m \}$ respectively, the sequences $(a_m)$ and $(b_m)$ are decreasing and increasing respectively and the $\lim \sup$ and $\lim \inf$ always exist since the sequence is defined on extended real numbers, finally, I know that if $(x_n)$ is a bounded sequence, then
$$\lim_{n \rightarrow \infty} \sup \{ x_k \ ; \ k \geq m \} = \inf_m \left\{ \sup \{ x_k \ ; \ k \geq m \} \right\} = \inf_m a_m$$
because $(a_m)$ is a decreasing sequence and this sequence is bounded (otherwise, $(x_n)$ would be unbounded).
Analogously,
$$\lim_{n \rightarrow \infty} \inf \{ x_k \ ; \ k \geq m \} = \sup_m \left\{ \inf \{ x_k \ ; \ k \geq m \} \right\} = \sup_m b_m$$
because $(b_m)$ is a increasing sequence and this sequence is bounded (otherwise, $(x_n)$ would be unbounded).
My doubt is why makes sense $\lim_{n \rightarrow \infty} \sup \{ x_k \ ; \ k \geq m \} = \inf_m \left\{ \sup \{ x_k \ ; \ k \geq m \} \right\}$ and $\lim_{n \rightarrow \infty} \inf \{ x_k \ ; \ k \geq m \} = \sup_m \left\{ \inf \{ x_k \ ; \ k \geq m \} \right\}$ when $(x_n)$ is unbounded (i.e., $(x_n)$ doesn't has upper bound, lower bound or both)? I would like to know how to argue to prove this equalities.
Thanks in advance!
Answer
If the sequence has no upper bound then $\sup(\{x_k\mid k\geq m\})=+\infty$ for every $m$ so that:
- $\lim_{n\to\infty}\sup(\{x_k\mid k\geq m\})=+\infty$
- $\inf(\{\sup(\{x_k\mid k\geq m\})\mid m=1,2,\dots\})=\inf(\{+\infty,+\infty,\dots\})=+\infty$
From this we conclude that they are equal again (and both equalize $+\infty$).
Same story for $\liminf$ where $-\infty$ takes the place of $+\infty$.
Actually this takes place in the extension $\overline{\mathbb R}=\{-\infty\}\cup\mathbb R\cup\{+\infty\}$.
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