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If S is supremum of set A

  1. Feb 6, 2009 #1
    Hello,
    I found the definition. If S is supremum of set A, then
    a) [tex]\forall x\in A:x\leq S[/tex]
    b) [tex]\forall\varepsilon>0\;\exists x_0\in A:S-\varepsilon<x_0[/tex]

    Now let define set [tex]A=\{1,2,3,4,5\}[/tex]. Is number 5 supremum of set A? Condition a) is satisfied, but b) is problem. If [tex]\varepsilon=0.1[/tex], there isn't x_0 in the set A such that
    [tex]5-0.1<x_0[/tex]

    Could you explain that?
     
  2. jcsd
  3. Feb 6, 2009 #2

    quasar987

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    Re: supremum

    What's wrong with x_0=5 ?
     
  4. Feb 6, 2009 #3
    Re: supremum

    As quasar said above, x0 is chosen to be 5.

    The supremum of a finite set is always the same as the maximum of that set. The supremum, in a sense, can be thought of as a generalization of the idea of maximum.
     
  5. Feb 6, 2009 #4
    Re: supremum

    The supremum of a set is either an element of the set or a limit that the elements tend to. Of course in this finite set the supremum is going to be an element of the set.
     
  6. Feb 6, 2009 #5

    HallsofIvy

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    Re: supremum

    The explanation is that the definition you give is wrong. (a) is the definition of "upper bound". In order that an upper bound be a "supremum" (also called "least upper bound") it must be the smallest of all upper bounds. In this case, every number less than or equal to 5 is an upper bound so 5 is the "least upper bound" or "supremum".
     
  7. Feb 6, 2009 #6

    Office_Shredder

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    Re: supremum

    Halls, the definition consists of both parts (a) and (b). (a) says that it's an upper bound. (b) says that it's the least such upper bound (notice that (b) doesn't make S an upper bound by itself)
     
  8. Feb 6, 2009 #7

    HallsofIvy

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    Re: supremum

    But the whole point is that, while if (a) and (b) are both true, then S is a supremum, it can happen, as in the example given here, that a number is a supremum without (b) being true.
     
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