Related questions have been posted here and here.
Background
I have seen the following four definitions of the upper and lower limits of a sequence from textbooks and MSE posts:
Definition 1$\quad$ [Baby Rudin]$\quad$ The upper limit, $\limsup_{n\to\infty}s_n$, of a given sequence $\{s_n\}$ is the supreme of the set of numbers $x$ in the extended real number system such that $s_{n_k}\to x$ for some subsequence $\{s_{n_k}\}$ of $\{s_n\}$.
Definition 2$\quad$ [Measure Theory by Donald Cohn]$\quad$ The upper limit, $\limsup_{n\to\infty}s_n$, of a given sequence $\{s_n\}$ is defined by \begin{align*} \limsup_{n\to\infty}s_n = \inf_{n}\sup_{m\geq n}s_m = \inf_{n}\left\{\sup\{s_n,s_{n+1},\dots\}\right\}. \end{align*}
Definition 3$\quad$ [MSE]$\quad$ The upper limit, $\limsup_{n\to\infty}s_n$, of a given sequence $\{s_n\}$ is defined by \begin{align*} \limsup_{n\to\infty}s_n = \lim_{n\to\infty}\sup_{m\geq n}s_m = \lim_{n\to\infty}\left(\sup\{s_n,s_{n+1},\dots\}\right). \end{align*}
Definition 4$\quad$ [MSE]$\quad$ A number $t$ in the extended real number system is the upper limit, $\limsup_{n\to\infty}s_n$, of a given sequence $\{s_n\}$ if \begin{align*} \text{for all $s<t$, we have $s<s_n$ for infinitely many $n$'s} \end{align*} and \begin{align*} \text{for all $s>t$, we have $s<s_n$ for finitely many $n$'s.} \end{align*}
I have some questions about the definitions themselves as well as the equivalence among them.
Question 1
I am not sure if I am right about this. I think Definition 3 should have been stated as follows:
Definition 3$\quad$[Revised]$\quad$ The upper limit, $\limsup_{n\to\infty}s_n$, of a given sequence $\{s_n\}$ shall be defined as follows: \begin{align*} \begin{cases} \limsup_{n\to\infty}s_n &= +\infty &\text{if}\ \ \sup\{s_n,s_{n+1},\dots\}\to+\infty,\\ \limsup_{n\to\infty}s_n &= -\infty &\text{if}\ \ \sup\{s_n,s_{n+1},\dots\}\to-\infty,\ \ \text{and}\\ \limsup_{n\to\infty}s_n &= \lim_{n\to\infty}\sup_{m\geq n}s_m &\text{if}\ \ \text{the limit exists}. \end{cases} \end{align*}
My concern about this revised definition is that what should we say about the case when the limit of $\sup_{m\geq n}s_m$ does not exist yet neither $\sup\{s_n,s_{n+1},\dots\}\to\pm\infty$?
Question 2
In one of the link I provided in the beginning, there is a proof of the equivalence between Definition 1 and Definition 3. However, I am not sure if my understanding of that proof is correct, espectially for the proof of $\sup S \leq \limsup_{n\to\infty}s_n$ where the right-hand side of the inequality is in the sense of Definition 3. Moreover, I would like to try to work out all the details. I would really appreciate it if someone could help me check my work!
Proof$\quad$ Let $S$ be the set of numbers $x$ in the extended real number system such that $s_{n_k}\to x$ for some subsequence $\{s_{n_k}\}$ of a sequence $\{s_n\}$. We want to prove that \begin{align*} \sup S = \limsup_{n\to\infty}s_n, \end{align*} where the right-hand-side is in the sense of the revised Definition 3. Let $F_n=\{s_n,s_{n+1},\dots\}$.
We first prove that \begin{align*} \sup S \leq \limsup_{n\to\infty}s_n. \end{align*} Suppose first that $\sup F_n\to+\infty$. Then $\limsup_{n\to\infty}s_n=+\infty$ by definition, and we are done. Suppose next that $\sup F_n\to-\infty$. We want to show that $\sup S = -\infty$, which is true if and only if every subsequence $\{s_{n_k}\}$ of $\{s_n\}$ such that $s_{n_k}\to x\in S$ satisfies $s_{n_k}\to-\infty$. Assume to the contrary that there is a subsequence $\{s_{n_k}\}$ such that $s_{n_k}\to\alpha>-\infty$. For each $n$, let $n_k$ be the smallest integer bigger than or equal to $n$, so that $\{s_{n_k},s_{n_{k+1}},\dots\}\subseteq F_n$. Then $\sup F_n \geq \sup\{s_{n_k},s_{n_{k+1}},\dots\}\geq\alpha$ for each $n$, contradicting the fact that $\sup F_n\to-\infty$. Therefore, $\sup S=-\infty$. Finally, suppose that the limit $\lim_{n\to\infty}(\sup F_n)$ exists. Then for each subsequence $\{s_{n_k}\}$ such that $s_{n_k}\to x\in S$, construct a sequence $\{u_n\}$ as follows: For index $n$ such that $n=n_k$ for some $n_k$ let $u_n=s_n=s_{n_k}$, and for other index $n$ let $u_n=u_{n+1}$. Let $E_n=\{u_n,u_{n+1},\dots\}$. Then for all $n$ we have $E_n\subseteq F_n$. Thus for all $n$ we have $\sup E_n\leq\sup F_n$. Note that $s_{n_k}\not\to+\infty$. If $s_{n_k}\to x=-\infty$, then $u_n\to-\infty\leq\lim_{n\to\infty}(\sup F_n)$. If $\lim_{k\to\infty}s_{n_k}=x\in\mathbb{R}$, then $\lim_{n\to\infty}u_n=\lim_{k\to\infty}s_{n_k}=x$. Then $E_n\supseteq E_{n+1}$ and so $\sup E_n\geq\sup E_{n+1}$ for all $n$. Moreover, $\sup E_n\geq x$ for all $n$. It follows that $\{\sup E_n\}$ is a bounded decreasing sequence, so $\lim_{n\to\infty}(\sup E_n)$ exists. Since $E_n\subseteq F_n$ for all $n$, we have $u_n\leq\sup E_n\leq\sup F_n$ for all $n$, and it follows that \begin{align*} \limsup_{n\to\infty}s_n = \lim_{n\to\infty}(\sup F_n) \geq \lim_{n\to\infty}(\sup E_n) \geq \lim_{n\to\infty}u_n=\lim_{k\to\infty}s_{n_k}=x\in\mathbb{R}. \end{align*} Therefore, for any subsequence $\{s_{n_k}\}$ of $\{s_n\}$ such that $s_{n_k}\to x \in S$ we have $x\leq\lim_{n\to\infty}(\sup F_n)$; that is, $x\leq\lim_{n\to\infty}(\sup F_n)$ for all $x\in S$. Hence, $\sup S\leq\lim_{n\to\infty}(\sup F_n)=\limsup_{n\to\infty}s_n$. This complets the proof of $\sup S \leq \limsup_{n\to\infty}s_n$.
Now we prove the opposite direction \begin{align*} \sup S\geq\limsup_{n\to\infty}s_n. \end{align*} If $\sup S=+\infty$, then there is nothing to prove. If $\sup S=-\infty$, then every subsequence $\{s_{n_k}\}$ of $\{s_n\}$ such that $s_{n_k}\to x\in S$ satisfies $s_{n_k}\to-\infty$. We want to show that $\sup F_n\to-\infty$. If $\sup F_n\to\alpha>-\infty$, then $\sup F_n\geq\sup F_{n+1}$ for all $n$ implies $\sup F_n\geq\alpha$ for all $n$. Hence, there must exists a subsequence $\{s_{n_k}\}$ such that $s_{n_k}\to\alpha$, a contradiction. Therefore, we must have $\sup F_n\to-\infty$, which means $\limsup_{n\to\infty}s_n=-\infty$. Now suppose $\sup S\in\mathbb{R}$. If $\sup F_n\to+\infty$, we have seen that it will imply the existence of a subsequence $\{s_{n_k}\}$ such that $s_{n_k}\to+\infty$, contradicting our assumption of $\sup S\in\mathbb{R}$. If $\sup F_n\to-\infty$, then there is nothing to prove. So suppose that the limit $\lim_{n\to\infty}(\sup F_n)$ exists. For each $n$, let $l_n\in\mathbb{N}$ be such that $l_n\geq n$ and \begin{align*} \sup F_n \geq s_{l_n} \geq \sup F_n - \frac{1}{2^n}. \end{align*} Then, \begin{align*} \lim_{n\to\infty}(\sup F_n) = \lim_{n\to\infty}s_{l_n}. \end{align*} Since $l_n\geq n$ for each $n$, it follows that $\{l_n\}$ forms a sequence of positive integers such that $l_n\to+\infty$. There there is an increasing subsequence $\{l_{n_k}\}$ of $\{l_n\}$. Let $n_k=l_{n_k}$. Since $\{s_{n_k}\}$ is a subsequence of $\{s_{l_n}\}$ which converges to $\lim_{n\to\infty}(\sup F_n)$, it follows that $\lim_{k\to\infty}s_{n_k} = \lim_{n\to\infty}(\sup F_n)$, and so $\lim_{n\to\infty}(\sup F_n)\in S$. Hence, $\lim_{n\to\infty}(\sup F_n)\leq\sup S$. This complets the proof of $\sup S \geq \limsup_{n\to\infty}s_n$.
I am not sure if this is all correct and rigorous. For example, when proving $\sup S\geq\limsup_{n\to\infty}s_n$, I wrote the following without proof:
If $\sup F_n\to\alpha>-\infty$, then $\sup F_n\geq\sup F_{n+1}$ for all $n$ implies $\sup F_n\geq\alpha$ for all $n$. Hence, there must exists a subsequence $\{s_{n_k}\}$ such that $s_{n_k}\to\alpha$, a contradiction."
Honestly, I just think this is intuitively ture, but couldn't figure out how to prove this claim. It would be great if someone can help me out.
Moreover, any other improvement suggestion would be greatly appreciated!
Question 3
This question is about the equivalence of Definition 2 and 3. I was wondering if the equivalence follows from the Monotone Convergence Theorem?
Question 4
How would one provide a proof for the equivalence between Definition 4 and one of the other three definitions?
Thanks a lot in advance!
