A question on the upper bound of the limit supremum My question is:

Let $(x_n)$ be a bounded sequence of real numbers.
  Prove that for every $\epsilon > 0$ and every $N\in\mathbb{N}$ there are $n_1, n_2\geq N$ such that 
  $$\limsup_{n\rightarrow\infty}x_n \leq x_{n_{1}}+\epsilon, \qquad x_{n_{2}}-\epsilon\leq \liminf_{n\rightarrow\infty} x_n. $$

Thank you for any help in advance. 
 A: Suppose that given $\epsilon>0$ and $N\in\mathbb{N}$, 
$x_{n_{1}}+\epsilon<v$ for every $n_{1}\ge N$, where $v=\lim \sup x_{n}$.
Then $x_{n_{1}}< v-\epsilon$ for $n_{1}\ge N$, and therefore $V_{N}=\sup\{x_{n}:n\ge N\}\le v-\epsilon$. 
Since $v=\displaystyle\lim_{k\to\infty}V_{k}$, this implies that $v\le v-\epsilon$, which gives a contradiction.
A similar argument applies for the $\lim\inf x_{n}$.
A: Hint: According to the definition of $\sup$ and $\inf$ there exist $n_1,n_2 \geqslant N$ with 
$$x_{n_1}>\sup_{n\geqslant N} x_n - \epsilon, \qquad x_{n_2}<\inf_{n\geqslant N} x_n + \epsilon.$$
A: Assume the opposite and derive a contradiction: for example, for the $\limsup$, suppose there were some $\epsilon>0$ and $N\in\mathbb{N}$ such that for every $m\geq N$,
$$\limsup_{n\to\infty}x_n>x_{m}+\epsilon.$$
Compare $\limsup\limits_{n\to\infty}x_n$ with $\sup\limits_{m\geq N}(x_m+\epsilon)$; after some easy manipulations, you'll see that $\limsup\limits_{n\to\infty}x_n$ would have to be greater than itself.
