I'm trying understand the definitions of limit superior and limit inferior of a sequence (xn) in extended real numbers. I know that lim 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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