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Homework Help: Can the well ordering on N be proven from completeness and archimedean property ?

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data
    I am just curious ?

    I have a feeling that completeness or the archimedean property relies on well ordering but I am not entirely sure.

    However, completeness funishes a supremum or infimum for any subset of R that is bounded above or below, respectively.

    3. The attempt at a solution

    [tex]N \subset R[/tex]

    So if S is any non empty subset of N then, S is a subset of R.

    If S bounded below, it has a infima in R.

    By the archimedean principle we can find can an integer that is greater or equal to the infima which would be in S.

    Is there something that prevents me from doing this ? Like completeness relying on well ordering of N.
  2. jcsd
  3. Sep 18, 2010 #2


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    I am not sure what you are asking. "Well ordering" of the natural numbersw does NOT follow from the completeness of the real numbers nor the Archimedean property because the natural numbers can be defined independently of the real numbers- in fact, normally, one defines the natural numbers first, then the integers from them, then the rational numbers from them, and the real numbers from the rational numbers. The "well ordering" property- every set of natural numbers has a smallest member follows from the "induction" axiom.

    I notice, on re-reading, that, while you titled this "Can the well ordering on N be proven from completeness and archimedean property?", in your post you ask the opposite question- do "completeness" and the "archimedian property" depend on well ordering. The answer is no because we can construct other sets of numbers, for example the set of rational numbers, that have all the properties of the real numbers except completeness.
    Last edited by a moderator: Sep 18, 2010
  4. Sep 18, 2010 #3

    My question should have been "does the construction of R depend on well ordering ".

    Basically to .....

    1)Prove the well ordering principle by using completeness and the archimedean principle ( That is what the sketch of a proof in my original post was)
    2) If the archimedean property depened on the truth of well ordering then I obviously can not use it in the proof.

    From what I understand the well ordering is equivalent to complete and mathematical induction. So it would seem that I can "prove" the well ordering principle with completeness and the archimedean principle.

    My question about completeness depending on well ordering was retarded; I wasn't thinking when I aksed that question.

    Is my thinking correct? I sketched the idea of my "proof " in my original post. Is it valid ?
  5. Sep 18, 2010 #4
    Wait, I don't even need the archimedean property.

    Since, [tex] S \subset N \Rightarrow S \subset R[/tex]

    By completeness, inf(S) exist since S is bounded below eg 1 is a lower bound of S.

    The fact that inf(S) is an integer is clear.

    In fact, inf(S) is in S.

    Isn't that the well ordering principle ?
  6. Sep 18, 2010 #5


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    You got me confused! Why is it "clear" that inf(S) is an integer? Assuming that it is, why does it follow, without using the "well ordering principal" that inf(S) is in S?
  7. Sep 18, 2010 #6
    I am working on a proof to show that the least upper bound of any set of integers is an interger.

    I showed that if inf(S) was not an integer then I looked at it's decimal expansion and found that I could select epsilon small enough that I could find a better upper bound which means inf(S) is an integer.

    But then I discovered that it depends on how the integers are defined.
    0.99999999999999.... =1
    If not then there is some [tex] s_{0}[/tex]
    [tex]s_{0} < inf(S) +1 \Rightarrow s_{0}-1<inf(S)\leq s_{0}[/tex]
    Inf(S) must equal [tex]s_{0}[/tex] since there is no integer between [tex]s_{0}[/tex] and

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