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Let $\{a_n\}$ be a sequence. I understand $\limsup_{n \rightarrow \infty}{a_n}$ is the greatest value of convergent subsequences and $\liminf_{n \rightarrow \infty}{a_n}$ is the smallest value of convergent subsequences. For example $$a_n = \{1, 2, 3, 1, 2, 3, 1, 2, 3, ...\}$$ $a_n$ has subsequences that converge to $\{1, 2, 3\}$ so that $$\limsup_{n \rightarrow \infty}{a_n} = 3 \quad \liminf_{n \rightarrow \infty}{a_n} = 1$$

My trouble is understanding how to make that interpretation connect to the definitions $$\limsup_{n \rightarrow \infty}{a_n} = \inf_n{ \sup_{m\geq n}{a_m}} \\ \liminf_{n \rightarrow \infty}{a_n} = \sup_n{ \inf_{m\geq n}{a_m}}$$

Can someone try and connect the two ideas together for me to have a greater understanding of the definition? Thank you.

The text I'm using is Russell Gordon's Real Analysis, A First Course and Richard F. Bass' Real Analysis for Graduate Students.

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$\{\sup_{m\geq n} x_m\}$ is decreasing in $n$, so you can expect that its limit will be a lower bound on $\sup_{m\geq n} x_m$. Furthermore, it must be the greatest lower bound, because otherwise we can write an $x\in \mathbb{R}$ that separates $\lim_{n\to \infty} \sup_{m\geq n} x_m$ (the definition of $\limsup_{n\to \infty} x_n$) from all $\sup_{m\geq n} x_m$. This an application of the monotone convergence theorem. Therefore, $$\limsup_{n\to \infty} x_n = \lim_{n\to \infty}\sup_{m\geq n} x_m = \inf_n\sup_{m\geq n} x_m$$ The same logic can be applied to find $\liminf_{n\to \infty} x_n = \sup_n\inf_{m\geq n} x_m$.

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Let $L=\lim_{k\to \infty}a_{n_k}$ denote the greatest subsequential limit of $(a_n)$ in $[-\infty, \infty]$. Here's a proof that $L =\inf_n{\sup_{m\geq n}{a_m}}$.

Let's prove first that $L\leq \inf_n{\sup_{m\geq n}{a_m}}$. Consider some $n\geq 1$. Since $n_k$ increases to $\infty$, there is some $K$ such that $k\geq K \implies {n_k}\geq n$. For any $k\geq K$, we have ${n_k}\geq n$, hence $\sup_{m\geq n}a_m \geq a_{n_k}$. Letting $k\to \infty$ in the last inequality yields $\sup_{m\geq n}a_m \geq L$. Since this is valid for all $n$, by definition of $\inf$, we have $L\leq \inf_n{\sup_{m\geq n}{a_m}}$.

Let's prove next that $L\geq \inf_n{\sup_{m\geq n}{a_m}}$. It suffices to prove that $\inf_n{\sup_{m\geq n}{a_m}}$ is a subsequential limit. I suppose that $\inf_n{\sup_{m\geq n}{a_m}}$ is finite and I leave the infinite case to you. It suffices to prove that $\forall \epsilon, \forall N, \exists n\geq N, |a_n-\inf_n{\sup_{m\geq n}{a_m}}|\leq \epsilon$. Let $\epsilon >0$ and $N\geq 1$. By definition of $\inf$, there is some $n_0\geq N$ such that $|\sup_{m\geq n_0}a_m - \inf_n{\sup_{m\geq n}{a_m}}|\leq \epsilon/2$. By definition of $\sup$, there is some $n_1\geq n_0$ such that |$a_{n_1}-\sup_{m\geq n_0}a_m|\leq \epsilon/2$. Then $|a_{n_1}-\inf_n{\sup_{m\geq n}{a_m}}|\leq \epsilon$ and we're done.

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