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The book I'm reading showed a the promise without proving it, after the bolanzo Weirestrass theorem. The theorem is : Let $\left\{{a_n}\right\}^\infty _{n=1}$ , $\left\{{b_n}\right\}^\infty _{n=1}$ be real number sequences Prove: $$\scriptsize\ \lim_{\overline{n\to\infty}}a_n+\lim_{\overline{n\to\infty}}b_n\le\lim_{\overline{n\to\infty}}(a_n+b_n)\le\lim_{\overline{n\to\infty}}a_n+\overline{\lim_{n\to\infty}}b_n\le\overline{\lim_{n\to\infty}}(a_n+b_n)\le\overline{\lim_{n\to\infty}}a_n+\overline{\lim_{n\to\infty}}b_n$$

The first third and fifth are obvious by signing $$\overline{\lim_{\overline{n\to\infty}}}a_n=\overline{\underline{A}} , \overline{\lim_{\overline{n\to\infty}}}b_n=\overline{\underline{B}} $$ and get and solve it easily, what I can't understand how to prove is whats between those, my attempt at proving that $$\overline{\lim_\overline{n\to\infty}}(a_n+b_n)=\overline{\lim_\overline{n\to\infty}}a_n+\overline{\lim_\overline{n\to\infty}}b_n$$ by bolanzo Weiresterass we get that there exist sub sequences $a_{n_k},b_{n_k},a_{n_i},b_{n_i}$ such that $$\lim_{n\to\infty}a_{n_k}=\overline{A},\lim_{n\to\infty}a_{n_i}=\underline{A},\lim_{n\to\infty}b_{n_k}=\overline{B},\lim_{n\to\infty}b_{n_i}=\underline{B}$$ thus, $$\overline{\lim_{\overline{n\to\infty}}}(a_n+b_n)= \lim_{n\to\infty}(a_{n_{k,i}}+b_{n_{k,i}})=\lim_{n\to\infty}a_{n_{k,i}}+\lim_{n\to\infty}b_{n_{k,i}}=\underline{\overline{A}}+\overline{\underline{B}}$$ but then why would hthey write ≤ and not =?

EXtra: how do I prove $$\overline{\lim_\underline{n\to\infty}}(-a_n)=-\overline{\lim_\overline{n\to\infty}}a_n$$

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    This is a duplicate of https://math.stackexchange.com/questions/70478/properties-of-liminf-and-limsup-of-sum-of-sequences-limsup-s-n-limin, which should fully answer your question. I'd only add that that while you correctly write e.g. $\varliminf a_n = \lim a_{n_i}$ and $\varliminf b_n = \lim b_{n_i}$, it doesn't follow that $\varliminf\left(a_n+b_n\right) = \lim\left(a_{n_i}+b_{n_i}\right)$. As to your extra question, it follows straightforwardly from the fact that $\sup(S) = \inf(-S)$ for any set $S$, which you should be able to prove easily. – Prasiortle Mar 01 '24 at 20:38
  • In that question they are dealing with only the right part of the inequality, I'm seeking to also solve the left part – Someguyalive Mar 02 '24 at 00:10
  • For the middle part, see https://math.stackexchange.com/questions/205346/how-to-prove-these-inequalities-liminfa-n-b-n-leq-liminfa-n-limsup, and for the leftmost part, see https://math.stackexchange.com/questions/3467700/prove-liminfa-nb-n-geq-liminfa-n-liminfb-n-for-bounded-sequences-a. – Prasiortle Mar 02 '24 at 09:01

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