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I have to show that if $C, D \subset (0,\infty)$ which is bounded above then $VCD=\{x \in (0,\infty)| \exists c \in C, \exists d \in D s.t. x=cd\}$ is bounded above.

I have been starring at Upper limit for a while and I was thinking about negating everything would help, but I wasn't able to really to so much about this. Any hints would be appreciated

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    $\begingroup$ Hint : show that $\sup(VCD) \leq \sup(C) \times \sup(D)$. $\endgroup$ – Delta-u Apr 18 '18 at 20:21
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Completeness Axiom of $\mathbb{R}$ states that for every upper bounded subset in $\mathbb{R}$, supremum exists and is finite.

Thus, $M_1= \sup C$ and $M_2 = \sup D$ exist and are finite. Now by definition,

$VCD= \{cd : c \in C , d \in D \}$ and by definition $c \le M_1$ and $d \le M_2$ we have, $cd \le M_1M_2$ and hence from the definition of supremum $\sup VCD \le M_1 M_2$

PS: I don"t have reputations to comment. This answer was meant to be a comment.

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