I want to prove that the two definitions of $\limsup_{n\to\infty}$ are equivalent.
Article in question : https://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior
Essentially i want to prove that for a sequence $(a_n)_n$ in $\mathbb{R}$ whose set of all limit points is $E$, that this equation works $$\lim_{n\to\infty}\sup_{m\ge n}a_m = \sup E$$
Let $B=\lim_{n\to \infty}\sup_{m\ge n}a_m.$ Let $b_n=\sup_{m\ge n}a_m.$
(1).If $B\in \Bbb R:$ The sequence $(b_n)_{n\in \Bbb N}$ is decreasing (i.e. $n'>n\implies b_{n'}\le b_n)$ and converges to $B.$
Take any $\epsilon\in \Bbb R^+.$ Take $ n_1$ such that $n\ge n_1\implies B+\epsilon> b_n.$ For any $n\ge n_1$ there exists $n'\ge n$ such that $a_{n'}>b_n-\epsilon$ by def' of $b_n.$ And if $n'\ge n$ then $a_{n'}\le b_n,$ also by def'n of $b_n.$ And $B+\epsilon > b_{n'}$ if $n'\ge n\ge n_1.$ So we have $$B+\epsilon \ge b_{n'}\ge a_{n'}>b_n-\epsilon\ge B-\epsilon.$$ And $n'\ge n,$ with no upper bound on $n.$
So for any $\epsilon >0$ the set $\{n'\in \Bbb N: |a_{n'}-B|<\epsilon\}$ is not empty and has no upper bound in $\Bbb N$. So there exists a subsequence of $(a_n)_{n\in \Bbb N}$ converging to $B.$
Suppose $(a_n)_{n\in \Bbb N}$ had a subsequence converging to $C>B.$ Take $r\in \Bbb R^+$ such that $B<B+r<C.$ Then there are infinitely many $m$ such that $a_m>B+r.$ But then $ b_n\ge B+r$ for every $n$ (by the def'n of $b_n)$ implying $B= \lim_{n\to \infty}b_n\ge B+r>B,$ which is absurd.
Note: We allow the possibility $C=\infty,$ with the usual convention that "$(x_n)_{n\in \Bbb N}$ converges to $\infty$" means that for each $r\in \Bbb R,$ the set $\{n\in \Bbb N: x_n\le r\}$ is finite (i.e. bounded above in $\Bbb N)$.
(2). If $B=\infty:$ Then $\sup_n a_n=b_1=\infty.$ Because $b_n\le b_1$ for all $n\in \Bbb N,$ so if $b_1 <\infty$ then $B=\lim_{ n\to \infty}b_n\le b_1<\infty.$
So $\{a_n: n\in \Bbb N\}$ is unbounded above in $\Bbb R$, so $(a_n)_{n\in \Bbb N}$ has a subsequence converging to $\infty.$ It can't have a subsequence converging to $more$ than $\infty.$