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Definition of the supremum

  1. Oct 28, 2012 #1
    i was trying to formalize the definition of the supremum in the real Nos (supremum is the least upper bound that a non empty set of the real Nos bounded from above has ) but the least upper part got me stuck.

    Can anybody help?
  2. jcsd
  3. Oct 28, 2012 #2


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    If epsilon>0 what can you say about the set T of all t such that
  4. Oct 28, 2012 #3
    There is a t belonging to T such that : ##a-\epsilon<t\leq t##, where a= supremum.

    But i did not ask for the formalization of that theorem ,which we can prove by using the definition of the supremum
  5. Nov 1, 2012 #4


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    What type of formalization do you want, epsilon-delta or some other? The most obvious (and silly) would be
    let S be a set of real numbers
    let U(S)={x in R|x>=s for all s in S} be the set of all upper bounds of S

    thus sup(S) is the unique real number such for any real number x either x>=sup(S) or there exist s in S such that s>=x

    This is one of those occasions where we have n equivalent statements so we make one the definition and arbitrarily the other n-1 become trivial theorems.
  6. Nov 1, 2012 #5
    The epsilon delta type
  7. Nov 2, 2012 #6


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    There you go

    sup(S) is the unique real number such that
    for all s in S sup(s)>=s
    for all epsilon>0
    there exist t in S such that
  8. Nov 2, 2012 #7
    We want that only in symbols no words.

    Again this is a theorem of the definition i asked in my original post,but anyway lets see how this can be trasfered into logical symbols
  9. Nov 2, 2012 #8


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    There is a possible problem here. If you want to express it in first order predicate logic, this is not possible, since we need to express two types of objects, numbers and sets of numbers, while first order predicate logic only deals with one type of objects.
  10. Nov 3, 2012 #9
    Yes,you are right we need 1st and 2nd order predicates.

    But ifyou could express it in 1st order predicates ,i would be very interested to see.
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