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

  • #1
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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)
.
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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Answers and Replies

  • #2
andrewkirk
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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##.
 
  • #3
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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?
 
  • #4
andrewkirk
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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.
 
  • #5
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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.
 
  • #6
andrewkirk
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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##.
 
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  • #7
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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?
 
  • #8
andrewkirk
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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.
 

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