# If $\{a_n\}$ is bounded sequence , prove that $\limsup\limits_{n\to \infty} a_n=-\liminf\limits_{n\to \infty}(-a_n)$

If $\{a_n\}$ is bounded sequence, prove that $$\limsup_{n\to \infty} a_n=- \liminf _{n\to \infty} (-a_n)$$

My Attempt:

\begin{align} \liminf_{n \to \infty} (-a_n)&=\lim_{n \to \infty}\inf(-a_n,-a_{n+1},....)\\ &=\lim _{n \to \infty} (-\sup (a_n,a_{n+1},....))\\ &=-\lim _{n \to \infty}\sup (a_n,a_{n+1},....) \end{align}

hence

$\limsup\limits_{n\to \infty} a_n=-\liminf\limits_{n\to \infty}(-a_n)$

Is this attempt right?

• If you have already proven that $\sup A=-\inf(-A)$ for an upward bounded set $A$ then yes. Also note that \liminf, \limsup, \inf and \sup are existing commands. – Arthur Aug 30 '17 at 5:25
• @Arthur...what i understand with your comment is first we need to prove $\sup A=-\inf(-A)$ – Inverse Problem Aug 30 '17 at 5:30
• Question body and question title do not match. – Kenny Lau Aug 30 '17 at 5:32
• @KennyLau...sorry i edited now thanks for point me – Inverse Problem Aug 30 '17 at 5:34
• If the main purpose of your question is to ask for verification and criticism of your own proof (as opposed to asking for any proof of this claim), then you should use (proof-verification) tag. See also the tag-info. – Martin Sleziak Aug 30 '17 at 7:55