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If $L = \lim \sup a_n$ and $l = \lim \inf a_n$ then given any $\varepsilon > 0$ there is a natural number $N$ such that $l - \varepsilon < a_n < L + \varepsilon$ whenever $n > N$.

This seems like it would be really easy to prove but I'm having trouble with it. My understanding is that $\limsup a_n = \sup C$ and $\liminf a_n = \inf C$ where $C$ is the set of cluster points of $\{a_n\}$. Since $L$ and $l$ are real numbers $C \neq \emptyset$ and so we can deduce that $\{a_n\}$ is bounded above and below. So it has a supremum and infimum. But I'm just not seeing where to go from here. Any hints or proofs would be greatly appreciated!

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  • $\begingroup$ Is it possible that $\limsup a_n < \sup C{\;}$ or $\limsup a_n > \sup C{\;}$ ? $\endgroup$
    – M. Strochyk
    Jan 16, 2013 at 19:09
  • $\begingroup$ I don't think so. By assigning $\liminf$ and $\limsup$ to variables I think it exempts them from being $\infty$ and $-\infty$ which is the only other things they can be. Right? $\endgroup$
    – Robert Cardona
    Jan 16, 2013 at 19:12

1 Answer 1

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Assume that it isn't true. Then there exists $\varepsilon>0$ such that for every $N\in\mathbb N$ we are able to find a number $n_N$ such that $n_N>N$ and $l-\varepsilon \ge a_{n_N}$ or $a_{n_N} \ge L+\varepsilon$. Now can you find a subsequence of $a_n$ which converges to a number outside of $\left[l,L\right]$ or diverges to $\pm\infty$?

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  • $\begingroup$ Well we have $L + \varepsilon \leq a_{n_N} \leq l - \varepsilon$ and so $\{a_{n_N}\}$ is bounded so by the Bolzano-Weierstrass Theorem it has a convergent subsequence, which must be within $[L + \varepsilon, l - \varepsilon]$ correct? $\endgroup$
    – Robert Cardona
    Jan 16, 2013 at 19:22
  • $\begingroup$ I think it should be written $(-\infty, l - \varepsilon] \cup [L + \varepsilon, \infty)$? $\endgroup$
    – Robert Cardona
    Jan 16, 2013 at 19:30
  • $\begingroup$ @Robert: keep in mind that $l\le L$ so it is that $L+\varepsilon \le a_{n_N}$ or $l-\varepsilon \ge a_{n_N}$. But infinitely many $a_n$ will satisfy one of them, let's say $a_{n_N}\le l - \varepsilon$. And this sequence will have limit point less than $l$ (if it's bounded then by Bolzano-Weierstrass like you said, or $-\infty$ if it's unbounded) $\endgroup$
    – Adam
    Jan 16, 2013 at 19:35

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