Sequence: Show that $\limsup(s_n.t_n) ≤ \limsup(s_n) · \limsup(t_n)$ Suppose $\{s_n\}$ and $\{t_n\}$ are bounded sequences of nonnegative numbers. Show that as $n\rightarrow \infty$ $$\limsup(s_n.t_n) ≤ \limsup(s_n) · \limsup(t_n)$$
My attempt:
Let $\limsup(s_n)=l$ and $\limsup(t_n)=m$ then $s_n<l+\epsilon$ for $n>N_1$ and $t_n<m+\epsilon$ for $n>N_2$. Also  $\{s_n\}$ and $\{t_n\}$ are bounded sequences, sequence $\{s_n.t_n\}$ is also bounded. 
But $$s_nt_n<lm+l\epsilon +m\epsilon +\epsilon^2$$
How to prove that. I have no idea how to prove. Please help.
 A: $$
\begin{align}
\limsup_{n\to\infty}s_nt_n
&=\lim_{n\to\infty}\sup_{k\ge n}s_kt_k\tag{1}\\
&=\lim_{n\to\infty}\sup_{\substack{j\ge n\\k\ge n\\j=k}}s_jt_k\tag{2}\\
&\le\lim_{n\to\infty}\sup_{\substack{j\ge n\\k\ge n}}s_jt_k\tag{3}\\
&=\lim_{n\to\infty}\sup_{j\ge n}s_j\sup_{k\ge n}t_k\tag{4}\\
&=\lim_{n\to\infty}\sup_{j\ge n}s_j\lim_{n\to\infty}\sup_{k\ge n}t_k\tag{5}\\
&=\limsup_{n\to\infty}s_n\limsup_{n\to\infty}t_n\tag{6}
\end{align}
$$
Explanation:
$(1)$: definition
$(2)$: rewrite in terms of two variables
$(3)$: the sup over a smaller set is less than the sup over a larger set
$(4)$: separate the sups
$(5)$: limit of a product is the product of the limits
$(6)$: definition

Clarification
$$
\begin{align}
\sup_{\substack{j\ge n\\k\ge n}}s_jt_k
&=\lim_{m_1,m_2\to\infty}\max_{\substack{m_1\ge j\ge n\\m_2\ge k\ge n}}s_jt_k\tag{7}\\
&=\lim_{m\to\infty}\max_{\substack{m\ge j\ge n\\m\ge k\ge n}}s_jt_k\tag{8}\\
&=\lim_{m\to\infty}\max_{m\ge j\ge n}s_j\max_{m\ge k\ge n}t_k\tag{9}\\
&=\lim_{m\to\infty}\max_{m\ge j\ge n}s_j\lim_{m\to\infty}\max_{m\ge k\ge n}t_k\tag{10}\\
&=\sup_{j\ge n}s_j\sup_{k\ge n}t_k\tag{11}
\end{align}
$$
Explanation:
$\phantom{1}(7)$: write sup as a limit of max
$\phantom{1}(8)$: limit of any subsequence is the limit of the sequence
$\phantom{1}(9)$: the greatest product is the product of the greatest elements
$(10)$: limit of a product is the product of the limits
$(11)$: definition
A: You've shown that for any $\epsilon > 0$: there exists a sufficiently large $N$ such that $n>N$ implies that
$$
s_nt_n < lm + (l+m)\epsilon + \epsilon^2
$$
Thus, for any such $n$, we have $\sup_{k \geq n} s_k t_k < lm + (l+m)\epsilon + \epsilon^2$.  Thus,
$$
\limsup_{n \to \infty} s_nt_n = \lim_{n \to \infty} \sup_{k \geq n} s_k t_k \leq lm + (l+m)\epsilon + \epsilon^2
$$
Howevever, since $\epsilon>0$ is arbitrary, we may conclude that 
$$
\limsup_{n \to \infty} s_nt_n \leq lm + (l+m)(0) + (0)^2 = lm
$$
as desired.
A: Let $s_{k_n}t_{k_n}$ be the subsequence of $s_nt_n$ which converges to $\limsup s_nt_n$.
As $s_{k_n}$ is bounded, it possesses a converging subsequence $s_{k'_n}\to s$, and as $t_{k'_n}$ is bounded, it possesses a converging subsequence $t_{k''_n}\to s$.
Clearly,
$$
st=\lim_{n\to\infty}s_{k''_n}t_{k''_n}
= \lim_{n\to\infty}s_nt_n=  \limsup_{n\to\infty}s_nt_n .
$$
At the same time
$$
s=\lim_{n\to\infty}s_{k''_n}\le\limsup_{n\to\infty}s_n  \quad\text{and}\quad
t=\lim_{n\to\infty}t_{k''_n}\le\limsup_{n\to\infty}t_n.
$$
Hence
$$
\lim_{n\to\infty}s_nt_n=st\le 
\big(\limsup_{n\to\infty}s_n\big)\big( \limsup_{n\to\infty}t_n\big).  
$$
