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Homework Help: Prove that the least upper bound of a set of a set of integers is

  1. Sep 21, 2010 #1
    Problem Statement:
    Prove that the least upper bound of a set of integers is an integer.

    Using well ordered principle this is very trivial. However, is there another way?

    ANY comments or ideas relating to the topic would be highly appreciated.

    It is assumed that the set mentioned is bounded above.

    The mobile version of PF has bugs so the code that was supposed to be pre-generated for this thread did not show up :-(
  2. jcsd
  3. Sep 21, 2010 #2


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    I don't think this is trivial with the well ordering principal... You can look at all integers larger than your set and find a smallest one, but that will definitely not be the supremum. If you have a different approach I'd be interested in seeing it.

    I think proof by contradiction is good here: Suppose the supremum is not an integer. This means that its fractional part is non-zero. Can you find a contradiction with that?
  4. Sep 21, 2010 #3
    I tried that approach but I hit a wall. It depends on how you define integers.

    In my approach I showed that there is a contradiction if sup(S) was not an integer by using the archimedean property, that is, I can always find a better upper bound.
  5. Sep 21, 2010 #4
    hmm may i ask, completeness axiom say " every nonempty set of real number that bounded from above have a least upperbound" does this correct conversely??

    because sometimes i see when a "definition" they define "if then" but i find it always true conversely
  6. Sep 21, 2010 #5
    I know what you mean; I found in a lot of theorems if can be replaced with iff. However, that doesn't apply here. Regarding the converse.....
    For the completeness axiom it is the case that every set that has a least. upper bound must be bounded above.
  7. Sep 22, 2010 #6


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    The fractional part of 1.9999... is not .9999....., it's 0. The fractional part is literally defined to be x-[x] where [x] is the largest integer smaller than or equal to x. So in this case, [1.999...]=2, x-[x]=0 as expected and the fractional part is not dependent on how you represented the number
  8. Sep 22, 2010 #7
    I guess I was overcomplicating things then; it is possible to do it that way. In fact, that was my original solution but I did not trust it.

    The way I did it also works, right ?

    Anyway the whole point of posting this question was to find another way of proving this theorem without well ordering.

    So I must ask can this be done without well ordering ?
  9. Sep 22, 2010 #8
    Does anyone have any other input?
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