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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?

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  • $\begingroup$ What have you tried? Have you tried to write out the definition of $\liminf$ in terms of quantifiers ($\exists,\forall$, etc)? $\endgroup$ – Dan Robertson Oct 21 '15 at 0:06
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    $\begingroup$ If $\liminf_{n\to\infty} a_n = \ell$ then $$\forall\epsilon>0\quad\exists N\quad\forall n\ge N\quad |\ell - \inf_{m>n} a_n | < \epsilon$$ Or in even more basic terms, $$\forall\epsilon>0\quad\exists N\quad\forall N'\ge N\quad\exists \ell' \quad(\forall \delta>0 \quad\exists n\ge N' \quad |a_n-\ell'|<\delta)\;\text{and}\;(\forall n\ge N' a_n>\ell) \;\text{and}\; |\ell-\ell'|<\epsilon.$$ Can you get anywhere with that? $\endgroup$ – Dan Robertson Oct 21 '15 at 0:20
  • $\begingroup$ $\liminf$ means the $\inf$ of $\{a_n\}$ when $n \to \infty$. The sequence is not necessarily convergent. $\endgroup$ – Ryan Oct 21 '15 at 0:20
  • $\begingroup$ @DanRobertson You can define the problem with $\varepsilon-\delta$ language, but it is the same meaning as I wrote. But thanks anyway to point out that :-) $\endgroup$ – Ryan Oct 21 '15 at 0:51
  • $\begingroup$ In many straightforward analysis problems, writing the statement in terms of quantifiers is often useful; normally the proof of the statement (or its contrapositive) can then be done by the translation of $\forall$ to "Let" and then constructing suitable values whenever you see $\exists.$ You just do the obvious (only) thing at each step. $\endgroup$ – Dan Robertson Oct 21 '15 at 0:55
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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.

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  • $\begingroup$ Thanks a lot. This techique can be also employed to prove "convergent sequence is bounded". $\endgroup$ – Ryan Oct 21 '15 at 0:27
  • $\begingroup$ @Ryan: It can indeed. You’re very welcome. $\endgroup$ – Brian M. Scott Oct 21 '15 at 0:28
  • $\begingroup$ cc; @BrianM.Scott If liminf is finite, then "eventually" the "lim"s settle down, they exist from some point $N$ on, i.e. lower bounds exist. There just isn't "room enough" between $0$ and $N$ for the sequence to be unbounded, before it finally behaves. $\endgroup$ – BrianO Oct 21 '15 at 0:44
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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} $.

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  • $\begingroup$ Thank you very much for your answer. Sorry, this system can only accept one answer. But I vote your answer. Thank you again! $\endgroup$ – Ryan Oct 21 '15 at 0:57
  • $\begingroup$ @Ryan You're welcome, glad I could help. $\endgroup$ – user190080 Oct 21 '15 at 0:59

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