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I am trying to prove the second part of the following question stated as

  1. If $\{a_n\}$ has a subsequence which is bounded below by $\alpha$ then $\limsup a_n\geq \alpha$.

  2. If $\{a_n\}$ does not have a subsequence which is bounded below, then $\limsup a_n=-\infty$.

Here is my solution for the first question;

Let $\{a_{{n}_{k}}\}\subset\{a_n\}$ be bounded below by $\alpha$. Then, $\exists\;\alpha\in \Bbb{R}:\alpha\leq \{a_{{n}_{k}}\},\;k\geq1$ $$\implies \alpha\leq a_{{n}_{k}}\leq \sup_{j \ge n_k}\{a_{j}\},\forall\;k\geq1$$ $$\implies \alpha\leq\lim_{k\to \infty} \sup_{j \ge n_k}\{a_{j}\}\leq \lim_{n\to \infty}\sup\{a_{{n}}\}$$ $$\implies \alpha\leq \lim_{n\to \infty}\sup\{a_{{n}}\}$$

Please, can anyone help me with the second question?

Omojola Micheal
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1 Answers1

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If $a_n$ has no subsequence which is bounded below, then $$\forall\alpha\in\mathbf R\space \forall N \space \exists n > N(a_n<\alpha) $$ for if not, the sequence $a_N,a_{N+1},\dots$ would be bounded below by $\alpha$. So for every $\alpha$ we can construct a subsequence whose sup is less than $\alpha$, and from there you should have plain sailing.

saulspatz
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