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Apostol's Mathematical Analysis 1.6

  1. Apr 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that every nonempty set of positive integers contains a smallest member. This is called the well-ordering principle.


    3. The attempt at a solution
    I'm just starting out with analysis, so I'm not too sure about the format of proofs. Here goes:

    Proof. First suppose the set S of positive integers is finite. Then assume S contains no smallest member. It follows that for every x in S, there exists an infinite number of members of S less than x. This contradicts our assumption that S is finite. Thus every finite set of integers contains a smallest member.

    Any help/critique is appreciated.
     
  2. jcsd
  3. Apr 2, 2009 #2

    HallsofIvy

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    What properties of positive integers do you have to work with?
     
  4. Apr 4, 2009 #3
    Sorry for the late reply. I have the definition of an inductive set and the definition of a positive integer: a real number which belongs to every inductive set.
     
  5. Apr 4, 2009 #4
    Oh, also some of the field axioms for real numbers:
    1. Exactly one of the relations x = y, x < y, x > y holds.
    2. If x > y and y > z, then x > z.
     
  6. Apr 4, 2009 #5
    So, for the non-empty set S of positive integers, one of the relations x < y or x > y holds between any two members. This means that each member in S is greater than or less than each of the other members. Also, if some member x of S is less than some member y, and y is less than some member z, then x is less than z. It follows that there is a member less than every other member in S.

    Is this clear enough? Intuitively, I see why the theorem is true, but I don't think I present my thoughts very rigorously.
     
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