Proving that a set is nonempty and bounded above.by silvermane Tags: analysis, bound, proof, set theory, supremum 

#1
Oct310, 03:08 PM

PF Gold
P: 117

1. The problem statement, all variables and given/known data
Let A = {x: x^2 + 3x + 2 <0}. Prove that this set is nonempty 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 proofwriting tips. Thank you so much for your help and tips in advance! :) 



#2
Oct310, 03:25 PM

P: 851

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 that1e is not even an upper bound of the set. Do the same for the infima with a slight modification. 



#3
Oct310, 03:37 PM

P: 851

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. 



#4
Oct310, 03:37 PM

P: 617

Proving that a set is nonempty 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. 



#5
Oct310, 03:44 PM

P: 851

@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. 



#6
Oct310, 04:41 PM

PF Gold
P: 117

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 :) 


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