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Proving that a set is non-empty and bounded above.

by silvermane
Tags: analysis, bound, proof, set theory, supremum
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silvermane
#1
Oct3-10, 03:08 PM
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1. The problem statement, all variables and given/known data
Let A = {x: x^2 + 3x + 2 <0}. Prove that this set is non-empty and bounded above. What is the least upper bound? Is it bounded below?

3. The attempt at a solution
Well, solving for the zeros and understanding that between the zeros, we satisfy our values of x's, I have that the set is (-2,-1) which has both an upper bound and lower bound. The least upper bound would be -1 as well.

I understand what's it's asking, but I'm having trouble writing a proof and would like it if I could perhaps have proof-writing tips.

Thank you so much for your help and tips in advance! :)
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╔(σ_σ)╝
#2
Oct3-10, 03:25 PM
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I suppose for you could do following.

Show that -1+ e (epsilon) cannot be the suprema because there is no element in the set greater than -1 + e - e/2. Which is a property of the suprema of a set . That is, if supA is the supremum of a set there exist x in A such that x>supA- e.

What this would succeed in showing is that any number greater than -1 is not a supremum. You should also argue that-1-e is not even an upper bound of the set.



Do the same for the infima with a slight modification.
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#3
Oct3-10, 03:37 PM
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Well so much for reading the question lol. You don't even need to prove anything regarding infima and suprema :-) You could proof it for fun if you like.

All you have to prove is that is non empty which is easy. Consider x= - 1/2.

Then you have to show it bounded above. This can be done by showing that
any element greater than 1 or 2 (pick any) is not in the set. This implies they are upper bounds.

JonF
#4
Oct3-10, 03:37 PM
P: 617
Proving that a set is non-empty and bounded above.

A isn’t empty because: -1.5 is in the set.

Consider 1^2 + 3*1 + 2 = 6, so clearly 1 isn’t in A. Polynomials are continuous functions so 1 then either has to be an upper or lower bound. Since -1.5 is in the set 1 must be an upper bound.

Let f(x) = x^2 + 3x + 2, Let b = -1 + ε. Consider b^2 + 3b + 2. = (-1 + ε)^2 +3(-1 + ε) +2 = 1 -2ε + ε^2 - 3 + 3 ε +2 = ε^2 + ε.
Thus if c > -1 then it isn’t the least upper bound, since -1 is an upper bound. If c < -1 then clearly an ε for b in A can be chosen where c < b. Thus the suprema is -1.
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#5
Oct3-10, 03:44 PM
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@JonF

Please don't give full solutions to problems. It does not help people learn.

Btw I think you made an error in your post. You said -1.5 is a upper bound; I don't know what you meant since it is neither an upper or lower bound:-).

And both of us did not read the question fully. We are not supposed to prove the suprema.
silvermane
#6
Oct3-10, 04:41 PM
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@JonF

Please don't give full solutions to problems. It does not help people learn.

Btw I think you made an error in your post. You said -1.5 is a upper bound; I don't know what you meant since it is neither an upper or lower bound:-).

And both of us did not read the question fully. We are not supposed to prove the suprema.
Haha wow! Thank you for the helpful hints, and concern about learning for one's self. :)))

Thank you as well Jon, though I didn't need a straight answer. I like to work myself and have that great feeling of "look what I can do" lol

:)
silvermane
#7
Oct3-10, 04:43 PM
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Quote Quote by ╔(σ_σ)╝ View Post

And both of us did not read the question fully. We are not supposed to prove the suprema.
Yeah, it's much easier than I thought too. I think I need to read the question more fully
:P


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