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Completeness Axiom

  1. Jun 8, 2009 #1
    I'm having a little touble understanding application of the completeness axiom to certain subsets of real numbers. In a problem in a book (Fundamentals of Real Analysis by Haggarty), it asks you to show that the set S={a + b*sqrt(2) : a,b are rational} is not complete. As a hint, it tells you to show that sqrt(3) is not in the set. That's fine. What I don't understand is how it follows, from this last fact, that the set is not complete.
    I know the Completeness axiom applies to bounded sets, so how can it be applied here. I tried following a similar approach to proving that Q is not complete, but I'm not sure if it's correct. It went something like this:
    Take a subset of S such that a + b*sqrt(2) < sqrt(3), a,b rational. Then obviously the supremum (which is sqrt(3)) is not part of S and so it's not complete. Is this valid? It seems really trivial to me...
    If anyone could enlighten me, that would be great.
  2. jcsd
  3. Jun 8, 2009 #2
    He means "complete" in the sense that every Cauchy sequence has a limit. So all you need to do is find a Cauchy sequence that converges to something outside S, say [tex]\sqrt{3}[/tex].
    Last edited: Jun 8, 2009
  4. Jun 9, 2009 #3
    Well, completeness hasn't been defined using sequences yet in this part of the book, so I'd rather not use methods that aren't required. The only things given are the axioms for a complete ordered field and some properties of sets (Archimedean Postulate and density of Q).
  5. Jun 9, 2009 #4
    is it not enough to show that a subset of the set {a + b*sqrt(2) < sqrt(3) : a,b rational} does not have a least upper bound within the set?
    So for example if you let a=0, the subset is {b*sqrt(2) < sqrt(3) : b rational} which is equivalent to {b < sqrt(3/2) : rational}, which does not have a rational least upper bound. Or is that not valid?
  6. Jun 9, 2009 #5
    That's enough, and valid.
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