Is this $\epsilon$-conditon for $\limsup$ false?

Question

The following are equivalent:

(i) $a=\limsup s_n$,

(ii) $\forall \varepsilon>0$ there are only a finite number of natural numbers $n$ such that $a+\varepsilon<s_n$ but there are an infinite number such that $a-\varepsilon< s_n$

I don't believe this result is true. If (ii) is true, $\exists N$ such that if $n\ge N$, then $a-\varepsilon<s_n\le a+\varepsilon.$ This means that $a=\lim s_n$.

On the other hand, assuming (i) is true, $a=\limsup s_n= \lim_{N\to \infty}\sup_{n\ge N}s_n\iff(\forall\varepsilon\exists M:N>M\Rightarrow|\sup_{n\ge N}s_n-a<\varepsilon.|)$ So shouldn't the $\varepsilon$-condition in (ii) involve $\sup_{n\ge N}s_n$ rather than $s_n?$

Answer 1

If (ii) is true, $\exists N$ such that if $n\ge N$, then $a-\varepsilon<s_n\le a+\varepsilon.$

This is incorrect. There may be infinitely many $n$ such that $a-\epsilon\geq s_n$. For instance, maybe $a-\epsilon<s_n$ whenever $n$ is odd but $a-\epsilon\geq s_n$ whenever $n$ is even. You don't know that $a-\epsilon<s_n$ for all but finitely many $n$, only for infinitely many $n$.

Answer 2

If 2) is true, there exists $N$ such that $s_{n}\leq a+\epsilon$ for all $n\geq N$. But it is not necessarily that $s_{n}>a-\epsilon$ for all $n\geq N$, rather, it is that $s_{n}>a-\epsilon$ for infinitely many $n\geq N$.

Anyway, we still have $\sup_{n\geq N}s_{n}\leq a+\epsilon$ and $\sup_{n\geq N}s_{n}>a-\epsilon$. On the other hand, $\sup_{n\geq M}s_{n}\leq\sup_{n\geq N}s_{n}\leq a+\epsilon$ for all $M\geq N$. Also for such $M$, $\sup_{n\geq M}s_{n}>a-\epsilon$ since $s_{n}>a-\epsilon$ for infinitely many $n$.

So we conclude that $|\sup_{n\geq M}s_{n}-a|\leq\epsilon$ for all $M\geq N$, so $\limsup_{n\rightarrow\infty}s_{n}=\lim_{M\rightarrow\infty}\sup_{n\geq M}s_{n}=a$.