- #1

Mr Davis 97

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## Homework Statement

Prove the following: Suppose S and T are non-empty, bounded subsets of (0,+∞) and let ST be the set, ST = {st: s∈S and t∈T}. Then sup(ST) = sup(S) sup(T)

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Hint: Notice that if 0 < y < sup(S)sup(T), then there exists r ∈ (0,1) such that y = (rsup(S))(rsup(T)). After finding such r, consider the inequalities rsup(S) < sup(S) and rsup(T)< sup(T).

## Homework Equations

## The Attempt at a Solution

First, we know that ##\sup (S) \sup (T)## is an upper bound of ##ST##, because we have individually ##\forall s \in S ~ s \le \sup S## and ##\forall t \in T ~ t \le \sup T##, which taken together impleies that ##\forall st \in ST ~ st \le \sup (S) \sup (T)##.

Second, we need to show that if ##b < \sup (S) \sup (T)##, then ##b## is not an upper bound. Here is where I'm stuck. Is there is where I use the hint? I don't really see how the hint can be used...

I think I might have the second part. By definition of the supremum, we know that ##\forall \epsilon \exists s \in S ## where ##\sup (S) - \epsilon < s##. Also, we have ##\forall \epsilon \exists t \in T ## where ##\sup (T) - \epsilon < t##. Combining these two statements, we have that ##(\sup T - \epsilon)(\sup S - \epsilon) = \sup T \sup S - \epsilon ' < st##. Since we already showed that \sup T \sup S, this shows that it is the supremum. Is this correct?

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