Limit supremum properties I have the following theorem that I'm trying to prove:
Let $(s_n)$ be a bounded sequence and let $m = \lim\sup s_n$. Then the following properties hold:


*

*For every $\varepsilon > 0$ there exists a natural number $N$ such that $n\geq N$ implies that $s_n < m + \varepsilon$.

*For every $\varepsilon > 0$ and for every $i\in\mathbb{N}$ there exists an integer $k > i$ such that $s_k > m-\varepsilon$.
What I have so far is the following:

Suppose that for a bounded sequence $(s_n)$, $m = \lim\sup s_n$. Then no number larger than $m$ can be a subsequential limit of $(s_n)$. That is, given any $(s_{n_k})$ and any $\varepsilon > 0$, $\lim_{k\to\infty} s_{n_k} < m+\varepsilon$. Thus there can only be a finite number of terms as large as $m+\varepsilon$, otherwise a subsequence of those terms would have a limit greater than $m$.  

The thing I'm struggling with is defining an $N$ such that $n\geq N$ implies $s_n < m + \varepsilon$. Any pointers on how to define such $N$ would be great!
 A: I will define $(a_m^+)_{m=n}^{\infty}$ to be the sequence $(c_m)_{m}^{\infty}$ where $c_k=\text{sup} (a_N)_{N=k}^{\infty}$ (in other words this sequence consists of the suprerum of the sequence $(a_m)_{m=n}^{\infty}$ with the first element of the sequence shifted each time.) The definition of $\limsup s_n$ is then $\inf(s_{N}^+)_{N=m}^{\infty}$. 
(1) We always have $m+\epsilon > m ~
\forall \epsilon > 0$, so that $m+\epsilon>\inf (s_N^+)_{N=m}^{\infty}$.
Now if $m+\epsilon < s_k^+ ~ \forall k\geq m$ , then $ m+\epsilon<\inf(s_N^+)$, a contradiction
So $\exists t\geq m$ such that $m+\epsilon > s_t^+$, or $m+\epsilon > \sup (s_t)_{t}^{\infty}$, and by the definition of supremum, we conclude $m+\epsilon > s_n ~\forall n \geq t$.
(2) One always has $m-\epsilon < m ~\forall \epsilon >0$. Thus, 
$m-\epsilon<\inf(S_N^+)_{N=m}^{\infty}$ and thus by the definition of infimum, 
$m-\epsilon<s_N^+ \forall N\geq m$, fix $N$
Or, $m-\epsilon <\sup (s_m)_{m=N}^{\infty}$.
Now, if $s_k \leq m-\epsilon ~\forall k\geq N$ it would imply $\sup (s_k)_{k=N}^{\infty}=m-\epsilon < \sup (s_m)_{m=N}^{\infty}$, contradiction!
Thus, $\exists j\geq N$ for which $m-\epsilon < s_j$ holds and we are done
