In the past few hours I'm trying to find an example where $$\sup\{a_nb_n|n\in N\}> \sup\{a_n|n\in N\}\sup\{b_n|n\in N\}$$
But I just can't find it, can someone see what I miss here?
EDIT: $a_n,b_n$ are both bounded from above and below
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The One
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2your example does not work. You get $1>1$. – EQJ Nov 29 '14 at 13:27
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How about
$$a_n = \begin{cases} -2, & \text{if $n$ is even} \\ -3, & \text{if $n$ is odd} \end{cases}$$
$$b_n = \begin{cases} -3, & \text{if $n$ is even} \\ -2, & \text{if $n$ is odd} \end{cases}$$
Then you get $6>4$.
Rory Daulton
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Let $A_n = \{ \frac{1-n}{n} \} $ and $B_n = \{ \sin(n) \} $.
Notice $\sup A_n = 0 $ and $\sup B_n = 1 $ and $\sup (A_n B_n) = 2 $
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Is it obvious that $\sup\left({\sin(n)\colon n\in \mathbb N}\right)=1$? – Git Gud Nov 29 '14 at 14:14
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are you asking me this question or you are asking yourself or you are asking to the asker ? – Nov 29 '14 at 14:18
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To be honest, the fact that $| \sin x | \leq 1 $ gives the intution. However, since $n$ is integer, $\sin n $ never hits $1$. So, I would say the proof of this fact is not as easy. I dont know how to prove it right now. – Nov 29 '14 at 14:27
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