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Show infimum

  1. Jul 23, 2011 #1
    Description attached.
    Solution attached.

    Is my solution correct?
    Thank you for your help.

    Attached Files:

  2. jcsd
  3. Jul 23, 2011 #2
    I don't understand the relevance of the solution from third line and onwards. First two lines are correct. Then you should go like:

    Consider x in R s.t. x > 0. Then x is not a lower bound for S since x is not less than or equal to all the elements of x; specifically x > 0 while 0 is in S. Thus 0 is the infimum of S.
  4. Jul 23, 2011 #3
    Hi. Thank you for your help.

    Is there a specific reason you chose to work with x in R in order to prove the problem?
    Probably I indeed wrote the nonsense, but I followed the book's solution for the supremum. It said that infimum is solved similarly. For the supremum, the book used not an x, but an outsider v to prove that v is not the lowest upper bound. Likewise, I chose an outsider t to prove that t is not the greatest lower bound.
    Thank you for your help.
  5. Jul 23, 2011 #4
    I believe that it is correct. What the 3rd line and onwards shows is:
    while we can obviously see that 0 is a lower bound to the set, if any other lower bound were to exist, it cannot be larger than 0 ( by proof from the link ). Thus, if t is a lower bound to our set S, t <= 0 and so 0 is our inf.
    It can be worded better though maybe
  6. Jul 23, 2011 #5
    Right, this is the meaning I was trying to express in my proof. I just don't know how successful I was in that.
  7. Jul 23, 2011 #6
    How to show that the set does not have upper bounds?

    I said that the set is not bounded above, thus it does not have upper bounds. There is no u such that for any x in R, x is less or equal to u.
  8. Jul 23, 2011 #7
    If this were for an assignment ( especially if you are assumed to be new at proofs ), I would be a bit more "wordy" -- especially since the question asks you to state "in detail".
    So, you must say things like, " suppose t is an arbitrary lower bound.." and show how it relates to your problem, and how what you said even proves anything ( i.e. "therefore, by definition, if t <= c for all lower bounds t.." )
  9. Jul 23, 2011 #8
    Try a proof by contradiction. Sometimes, if something is super obvious but you cannot prove it somehow, a contradiction proof might work.

    ( i.e. answer the question: can something bound the set? what happens if you "try"? )
  10. Jul 23, 2011 #9
    I took another look and your solution was fine beside a small mistake. In the fourth line it should be [itex]x' \in S_{1}[/itex] not [itex]x'\in \mathbb{R}[/itex]. Then solution is fine.
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