# Products of Suprema

Icebreaker
A and B are bounded sets. $$C = \{ab | a \in A, b \in B\}$$
Show that (Sup A)(Sup B) = Sup C.

I tried to do it as follows,

$$\alpha = Sup A \Rightarrow \forall \epsilon > 0, \exists a \in A s.t. \alpha - \epsilon < a$$

$$\beta= Sup B \Rightarrow \forall \epsilon > 0, \exists b \in B s.t. \beta - \epsilon < b$$

$$\alpha\beta < ab$$

$$\alpha\beta - \alpha\epsilon - \beta\epsilon + \epsilon^2 < ab$$

No matter what I set epsilon to, I can't isolate the final epsilon. Any help?

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Is it possible to prove that $$\alpha\epsilon + \beta\epsilon - \epsilon^2 >0$$?