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

  1. Oct 3, 2010 #1

    silvermane

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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! :)
     
  2. jcsd
  3. Oct 3, 2010 #2
    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.
     
  4. Oct 3, 2010 #3
    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.
     
    Last edited: Oct 3, 2010
  5. Oct 3, 2010 #4
    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.
     
    Last edited: Oct 3, 2010
  6. Oct 3, 2010 #5
    @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.
     
  7. Oct 3, 2010 #6

    silvermane

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    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

    :)
     
  8. Oct 3, 2010 #7

    silvermane

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    Yeah, it's much easier than I thought too. I think I need to read the question more fully
    :P
     
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