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Repeating the question from the title:

Question. Let $(a_i)_{i=1}^{\infty}$ be a sequence. How to show that $\liminf_{n \rightarrow \infty} ≤ \limsup_{n \rightarrow \infty}$?

So far I've only managed to show that the inequality holds in one specific case, that is, the case where $\limsup_{n \rightarrow \infty}$, $\liminf_{n \rightarrow \infty}$ are the elements of sequences $(a_{i}^{+})_{i=1}^{\infty}$ and $(a_{i}^{-})_{i=1}^{\infty}$, respectively (in this case the proposition holds, since for arbitrary $N,M ≥ 1$, we have $\sup(a_i)_{i=N}^{\infty} ≥ \inf(a_i)_{i=M}^{\infty}$, even if $N ≠ M$).

I've tried to generalize by using contradiction, but it led me nowhere. More concretely, if for the sake of contradiction we assume that $\liminf_{n \rightarrow \infty} > \limsup_{n \rightarrow \infty}$, then it must be the case that $\liminf_{n \rightarrow \infty} > a_n$ for all $n≥N$ and some $N$ (this is the result from the previous exercise in the textbook). This doesn't lead to any contradiction though, since there are sequences where limit inferior is larger than all elements of the sequence past a specific point (e.g., consider the sequence $0.9,0.99,0.999,\cdots$; limit inferior in this case is $1$, which is larger than any element of the sequence)

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  • $\begingroup$ Let $i\in\mathbb{N}$ be arbitrary and note that $\inf_{k\geq i}a_{k}\leq \sup_{k\geq i} a_{k}$. Then take $\lim_{i\to\infty}$ of both sides. You should also be assuming something about $(a_{i})_{i}$ being bounded so that the suprema and infima exist at each $i$. Note that this is true in the extended reals where these values exist as extended real numbers. $\endgroup$
    – JWP_HTX
    May 9, 2021 at 23:14
  • $\begingroup$ Alternatively, in your proof by contradiction, if $\liminf_{n\to\infty} a_n>\limsup_{n\to\infty} a_n$, then there exists a real number $R$ such that $\liminf_{n\to\infty} a_n>R>\limsup_{n\to\infty} a_n$. Now you should be able to use your idea to reach a contradiction. $\endgroup$ May 9, 2021 at 23:45

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