If liminf exists, is the sequence is bounded below?

Question

Let $\{a_n\}$ ($n \in {\mathbb Z}_+$ and $a_n \in {\mathbb R}$) be a sequence and \begin{align} \liminf_{n\to \infty} a_n > -\infty. \end{align}

Does it mean $\{a_n\}$ is bounded below with a finite number? or \begin{align} \inf_{n \in {\mathbb Z}_+} a_n > -\infty. \end{align}

My intuition tell me this is true, but how to prove it?

Answer 1

HINT: Recall that

$$\liminf_{n\to\infty}a_n=\lim_{m\to\infty}\inf_{n\ge m}a_n\;.$$

Say that this limit is $L$. Then there is an $m\in\Bbb N$ such that $a_n\ge L-1$ for $n\ge m$; why? Now use the fact that any finite set of real numbers is bounded to show that the sequence is bounded below.

Answer 2

If you have \begin{align} \liminf_{n\to \infty} a_n:=\lim_{n\to \infty}(\inf_{m\geq n}a_m)=a > -\infty. \end{align} then your sequence $a_n$ is indeed bounded. Only a finite number of elements actually exceed this limit and the smallest of those (the minimum exists) is your lower bound of the sequence. So you really have $$ \liminf_{n\to \infty} a_n\neq\inf_{n\geq1}a_n $$ Why only a finite number? Let's just assume infinitely many $(a_{n_k})$ would exceed the limit inferior $a$, then they would have to be infinitely often smaller than the limit inferior itself and therefore $$ \lim_{n\to \infty}\inf_{m\geq n}a_m>\lim_{n_k\to \infty}\inf_{s\geq n_k}a_{s} $$ which would lead to a contradiction since $(a_{n_k})$is especially a subsequence of $(a_n)$. So for $n\leq n_k$ $\inf_{m\geq n}a_m\leq\inf_{s\geq n_k}a_{s} $.