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Prove that $\liminf (-a_n) = -\limsup (a_n)$.

by $\liminf$ and $\limsup$ I mean the limit supremum and limit infimum of a sequence $a_n$.

do you guys have any hints?

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    $\begingroup$ Have you tried using the definition of lim inf and lim sup? $\endgroup$ Nov 26, 2015 at 21:01
  • $\begingroup$ your'e right, I deleted the example. $\endgroup$
    – F1sargyan
    Nov 26, 2015 at 21:05
  • $\begingroup$ $-\liminf (-a_n)=-\lim_{n\to\infty} \inf_{m \geq n} (-a_m) = \lim_{n\to\infty} - \inf_{m\geq n} (-a_m) = \lim_{n \to\infty} \sup_{m \geq n} -(-a_m) = \lim_{n\to\infty} \sup_{m \geq n} (a_m) = \limsup (a_n)$. The only step that really needs justification is switching from the $\inf$ to the $\sup$. Can you justify that step? $\endgroup$
    – kccu
    Nov 26, 2015 at 21:07
  • $\begingroup$ @Justpassingby what do you mean by using the definition? the definition I know is that lim sup is the supremum of the partial limits set of a sequence $a_n$ and for lim inf a similar definition, did u mean something else? $\endgroup$
    – F1sargyan
    Nov 26, 2015 at 21:08
  • $\begingroup$ @F1sargyan Working from the definitions is essentially what I did in my comment. $\endgroup$
    – kccu
    Nov 26, 2015 at 21:09

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I assume that the sequence $(a_n)$ is bounded. You want to show $$ \sup_m\inf_{n\geq m}(-a_n)=-\inf_m\sup_{n\geq m}(a_n)\tag{*} $$

Show first that for any bounded (real) sequence $(b_n)$: $$ \sup_n (-b_n)=-\inf_n b_n\tag{**} $$ Now use (**) to show (*).

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