Prove that sup(ST) = sup(S)sup(T)

[Mr Davis 97]

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)
.
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?

 

[andrewkirk]

I also do not find the hint helpful. But I think you're on a promising track here:

we have that $(\sup T - \epsilon)(\sup S - \epsilon) ...$.

If you expand that product you will get an expression that should be able to be made arbitrarily close to $\sup S\sup T$, via suitable choice of $\epsilon$.

 

[Mr Davis 97]

I also do not find the hint helpful. But I think you're on a promising track here:

If you expand that product you will get an expression that should be able to be made arbitrarily close to $\sup S\sup T$, via suitable choice of $\epsilon$.

What I did was let $\epsilon ' = \epsilon (\sup T +\sup S) - \epsilon ^2$. How could I show that this could be any positive real number?

 

[andrewkirk]

What I did was let $\epsilon ' = \epsilon (\sup T +\sup S) - \epsilon ^2$. How could I show that this could be any positive real number?

You don't need to. All you need do is show that $\epsilon'$ can be made arbitrarily small.

 

[Mr Davis 97]

You don't need to. All you need do is show that $\epsilon'$ can be made arbitrarily small.

Well. if $\epsilon (\sup T + \sup S) - \epsilon ^2 > 0$ then $\epsilon < \sup T + \sup S$. Does this show that $\epsilon '$ can be made arbitrarily small? As long as $\epsilon \in (0, \sup T + \sup S)$, then $\epsilon '$ is positive.

 

[andrewkirk]

Well. if $\epsilon (\sup T + \sup S) - \epsilon ^2 > 0$ then $\epsilon < \sup T + \sup S$.

That's heading off track. We have a formula that expresses $\epsilon'$ in terms of $\epsilon$ and $\epsilon$ is universally quantified so we can make it as close to 0 as we wish. I'm sure if you look at the formula for $\epsilon'$ you'll soon see how to make it as close to 0 as you need by choosing $\epsilon$ small enough. If you can make $\epsilon'$ as small as you wish then you can make $st$ as close to $\sup S\sup T$ as you wish, so there cannot be an upper bound below $\sup S\sup T$.

 

[Mr Davis 97]

That's heading off track. We have a formula that expresses $\epsilon'$ in terms of $\epsilon$ and $\epsilon$ is universally quantified so we can make it as close to 0 as we wish. I'm sure if you look at the formula for $\epsilon'$ you'll soon see how to make it as close to 0 as you need by choosing $\epsilon$ small enough. If you can make $\epsilon'$ as small as you wish then you can make $st$ as close to $\sup S\sup T$ as you wish, so there cannot be an upper bound below $\sup S\sup T$.

Okay, I'm pretty much done with the proof now. I just have one question. How could I prove that $\epsilon ' = \epsilon (\sup S + \sup T - \epsilon)$ could in fact get as close to zero as I wish? I mean, it's intuitively clear. If we let $a = (\sup S + \sup T - \epsilon)$, then clearly $\epsilon ' = \epsilon a$ can be made as close to zero as I wish, since $a > 0$ and $\epsilon$ can get as close to zero as we wish, as long as $\epsilon < \sup S + \sup T$. Is this enough to prove it, or is there something more formal that I have to do?

 

[andrewkirk]

Okay, I'm pretty much done with the proof now. I just have one question. How could I prove that $\epsilon ' = \epsilon (\sup S + \sup T - \epsilon)$ could in fact get as close to zero as I wish?

A strategy that usually works in these cases - where the control variable $\epsilon$ occurs more than once in the formula whose value one is trying to make small - is to try to show that the formula is $\leq$ a formula that is $constant \times \epsilon$. Can you see a way to do that with the above?

The proof would take the form of letting $\eta>0$ be the number that we want to make $\epsilon'$ smaller than. Then say 'assume $\epsilon <$ [some formula involving $\sup S,\sup T,\eta$]' and then work through the inequalities to prove that $\epsilon'<\eta$ if $s,t$ are within $\epsilon$ of $\sup S,\sup T$ respectively..

Choosing the constraint on $\epsilon$ carefully can result in a proof that doesn't contain any hand-waving and that is shorter and more intuitive than one that does.