Prove that $\limsup s_n \le\limsup t_n$ The question is: 
Let $s_n$ and $t_n$ with $ n \ge 1 $ be two bounded sequences. 
Assume that there exists a natural number $N$ such that for every $ n \ge N $, we have $ s_n \le t_n $. 
Prove that $$ \lim_{n\to\infty} \sup s_n \le \lim_{n\to\infty} \sup t_n. $$ 
I think its obvious, but the part of proving it is kinda tricky.
Can someone please help?
 A: One definition of $\limsup$ of a sequence $\{ r_{k} \}$ is as follows. Define
$$
              R_{n} = \sup \{ r_{n}, r_{n+1}, r_{n+2},\ldots \}.
$$
Whether finite or infinite,
$$
             R_{1} \ge R_{2} \ge R_{3} \ge ....
$$
converges downward to some finite or infinite limit defined as $\limsup_{n}r_{n}$. For and $n \ge N$, the least upper bound $T_{n}$ of $\{ t_{n},t_{n+1},t_{n+2},\ldots\}$ is an upper bound for $\{ s_{n},s_{n+1},s_{n+2}\}$ because $s_{k} \le t_{k}$ for $k \ge n \ge N$. Therefore the least upper bound $S_{n}$ of $\{ s_{n},s_{n+1},s_{n+2},\ldots\}$ must satisfy $S_{n} \le T_{n}$. So the argument is now reduced to the case of monotone sequences:
$$
   \limsup_{n} s_{n}  = \lim_{n} S_{n} \le \lim_{n} T_{n} = \limsup_{n} t_{n}.
$$
The infinite cases can be dealt with directly. If $\limsup_{n} t_{n}=\infty$, there is nothing to show. If $\limsup_{n} s_{n}=-\infty$, there is nothing to prove. In all other cases, both $\limsup$ terms are finite, and $\{ S_{n} \}$, $\{ T_{n} \}$ are non-decreasing, convergent sequences with $S_{n} \le T_{n}$ for $n \ge N$, a case which is assumed to be known.
A: I'll use the following property:

For a sequence $(a_n)_{n\in\mathbb N}$ the $\limsup a_n$ is  the largest limit point of $(a_n)_{n\in\mathbb N}$, where $a$ is a limit point if there is a subsequence $(a_{k_n})_{n\in\mathbb N}$ such that $\lim_{n\to\infty} a_{k_n}=a$.
Assume that $\limsup s_n>\limsup t_n$.
Then there is a subsequence $(s_{k_n})_{n\in\mathbb N}$ of $(s_{n})_{n\in\mathbb N}$ such that $\lim_{n\to \infty} s_{k_n}=\limsup s_n>\limsup t_n$.
Therefore $s_{k_n}>\limsup t_n$ for all large $n$ so $t_{k_n}>\limsup t_n$ for all large $n$.
Now find a subsequence of $t_{k_n}$ that converges ($t_{k_n}$ is bounded) to derive a condradiction.
A: On the contrary let $S = \limsup s_{n} > \limsup t_{n} = T$. Then by definition of $\limsup$ we have the following:
1a) For any $\epsilon > 0$ there are infinitely many values of $n$ for which $s_{n} > S - \epsilon$.
1b) For any $\epsilon > 0$ we have $s_{n} < S + \epsilon$ for all sufficiently large values of $n$.
2a) For any $\epsilon > 0$ there are infinitely many values of $n$ for which $t_{n} > T - \epsilon$.
2b) For any $\epsilon > 0$ we have $t_{n} < T + \epsilon$ for all sufficiently large values of $n$.
Put $2\epsilon = S - T > 0$ so that $T + \epsilon = S - \epsilon$. Now for all sufficiently large values of $n$ we have $t_{n} < T + \epsilon = S - \epsilon$ and hence $s_{n} \leq t_{n} < S - \epsilon$. This contradicts the fact that $s_{n} > S - \epsilon$ for infinitely many values of $n$. Hence we must have $S \leq T$.
