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 $s_n&lt;l+\epsilon$ for $n&gt;N_1$ and $t_n&lt;m+\epsilon$ for $n&gt;N_2$. Also  $\{s_n\}$ and $\{t_n\}$ are bounded sequences, sequence $\{s_n.t_n\}$ is also bounded.  But $$s_nt_n&lt;lm+l\epsilon +m\epsilon +\epsilon^2$$  How to prove that. I have no idea how to prove. Please help.
The question is based on the following exercise in real analysis:     Assume that $A,B\subset{\Bbb R}$ are both bounded and $x&gt;0$ for all $x\in A\cup B$. Show that $$ \sup(AB)=\sup A\sup B $$   where    $$ AB:=\{ab\in{\Bbb R}:a\in A, b\in B\}. $$   Since $0&lt;a\leq\sup A$ and $0&lt;b\leq\sup B$ for all $a\in A$ and $b\in B$, we have $$ ab\leq\sup A\sup B $$ for all $ab\in AB$ which implies that $\sup AB\leq\sup A\sup B$. I have trouble with another direction: $$ \sup AB\geq\sup A\sup B $$ I was trying to show that for every $\epsilon &gt;0$, $\sup AB-\epsilon \geq \sup A\sup B$. If one uses the definition of supremum, one has the estimates that for every $\epsilon&gt;0$, $$ \sup A-\epsilon\leq a, \quad \sup B-\epsilon\leq b $$ for some $a\in A,\ b\in B$. It follows that $$ \sup A\sup B\leq (a+\epsilon)(b+\epsilon)=ab+\epsilon(a+b)+\epsilon^2\leq \sup AB+\epsilon (a+b)+\epsilon^2 $$ which seems quite close to what I want. How can I go on?
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