# Upper Bounds

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#1
If I tell you that: sup x ∈ (0,∞) i.e. is the supremum of the set of values for each value of x > 0.

Then, if this set of values is strictly bounded above by 1/n, why does this imply that the supremum of the set is strictly less than 1/n?

I thought, OK, so let's label the set S = x ∈ (0,∞). Then, we have that all the values in S < 1/n. So, we have that any upper bound of S must be >= 1/n. Hence, any supremum must be smaller than any upper bound, so it must be <= 1/n. So, I had sup(S) = 1/n, and so I wrote sup(S) <= 1/n, and then I used the Sandwich theorem. This was my intuition.

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6 months ago
#2
(Original post by Takeover Season)
If I tell you that: sup x ∈ (0,∞) i.e. is the supremum of the set of values for each value of x > 0.

Then, if this set of values is strictly bounded above by 1/n, why does this imply that the supremum of the set is strictly less than 1/n?
Of itself, it doesn't.

You could actually work out the maximum value of over the interval and it comes to , and hence

I thought, OK, so let's label the set S = x ∈ (0,∞). Then, we have that all the values in S < 1/n.
Not sure why you called it by another name, S; seemed fine.

So, we have that any upper bound of S must be >= 1/n.
No, this isn't true. You can only claim this if you know that 1/n is the supremum (least upper bound). And in fact it isn't. It's just an upper bound.
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#3
(Original post by ghostwalker)
Of itself, it doesn't.

You could actually work out the maximum value of over the interval and it comes to , and hence

Not sure why you called it by another name, S; seemed fine.

No, this isn't true. You can only claim this if you know that 1/n is the supremum (least upper bound). And in fact it isn't. It's just an upper bound.
Hi, thank you for your reply. Sorry, I meant S is the set of values of over the interval and is the supremum of this set of values. Now, we have shown that each value in the set S .

Oh, so now I understand. I can only say that any upper bound of S either the supremum of S or the maximum of S, and as = max/sup(S), I cannot make that conclusion.
However, it would've been right instead to say that: any upper bound of S right?

Oh yes, so that means as each value in the set S is bounded above by , then it must be that is an upper bound of S. As the supremum of a set S is less than or equal to every upper bound of a set, then .

Then, we have and the conclusion follows.
Last edited by Takeover Season; 5 months ago
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5 months ago
#4
(Original post by Takeover Season)
Hi, thank you for your reply. Sorry, I meant S is the set of values of over the interval and is the supremum of this set of values. Now, we have shown that each value in the set S .
Sorry, I missed that distinction. My bad.

Oh, so now I understand. I can only say that any upper bound of S either the supremum of S or the maximum of S, and as = max/sup(S), I cannot make that conclusion.
However, it would've been right instead to say that: any upper bound of S right?
Yes.

Oh yes, so that means as each value in the set S is bounded above by , then it must be that is an upper bound of S. As the supremum of a set S is less than or equal to every upper bound of a set, then .

Then, we have and the conclusion follows.
Well we can go a bit further. Knowing that we can say with the strict inequality - in this case. And that justifies its usage in the given text.

But without that, or some similiar working, we can only say
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#5
(Original post by ghostwalker)
Sorry, I missed that distinction. My bad.

Yes.

Well we can go a bit further. Knowing that we can say with the strict inequality - in this case. And that justifies its usage in the given text.

But without that, or some similiar working, we can only say
No problem.

Oh right, thank you - that makes perfect sense! Yes, the given working wasn't in the text, so I was confused as to how the strict inequality appeared, but I assume they expect me to do that working out.

Thanks a lot!
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