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R is complete?

  1. Aug 3, 2008 #1
    1. The problem statement, all variables and given/known data

    I'm reading a book on analysis independently. There is a Theorem that R is complete, i.e. any Cauchy sequence of real numbers converges to a real number.

    He lets a1, a2, a3, ... be a Cauchy sequence, then considers the the set:

    S = { x an element of R : x ≤ an for an infinite number of positive integers n }

    and the proof shows that lim an = supS.

    I'm baffled at what the set S is supposed to be. The proof won't work if it is the intersection of sets { x : x ≤ an } for all n, nor union of such sets. It can't be the limit of an because this is a proof of it's existence.
  2. jcsd
  3. Aug 3, 2008 #2
    The set S seems explicitly described.

    x is in S if and only if x is less than or equal to infinitely many of the a_i.
  4. Aug 4, 2008 #3
    If R is not complete, then there exists a convergent sequence that converges to number that is not real.

    So to show R is complete, you must show ALL convergent sequences converge to a real number. We already know all convergent sequences are Cauchy, so if you show all Cauchy sequences in R converge to a number in R, then you have shown all convergent sequences converge to a number in R which by def means R is complete.

    If you already knew the above sorry =b

    By axiom (I believe, I am rusty), R has the least upperbound (lub) property. That means if I have a sequence with an upper bound, it will have a least upper bound, and what's important for this proof is that by "have a least upper bound" we mean that not only does the lub exist, but it is a real number.

    What's the goal? To show a_n is is a real number.

    Well, your proof comes up with a bounded (by above) set, which gives it a lub that is real. Then it turns out that a_n is the lub, so a_n is real. Which is the goal.

    If you also already knew the above then bourbaki probably explained it.. I can only help with an example:

    a_n := 1/n ie 1, 1/2 , 1/3, 1/4 (just a random convergent sequence).. then

    S = R^- U { 0 } (all the negative real numbers and 0).

    You can see easily that the lub is 0 and that a_n converges to 0.

    what if we let it go to 0 from the left?
    a_n = 0 - 1/n ie -1, -1/2, -1/3 etc

    S = (-infty,0) // why not 0?

    Anyways,.. again the lub is 0 and this is another example of an S.

    So the quick answer is S is a trick used to solve the problem, and someone (I hope!) spent a long time figuring it out.. I only hope because it would have taken me personally a long time, if ever, to prove it like this.
  5. Aug 4, 2008 #4


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    How, exactly are you defining "convergent" in this case? I would have said, rather, that there exist Cauchy sequences that do not converge.

    again, since you are working in the real number system, there are no numbers except real numbers. How are you defining "converge" here?

  6. Aug 4, 2008 #5
    I like your wording better.

    If R is not complete, then there exists a Cauchy sequence that does not converge.

    Thanks :)
  7. Aug 4, 2008 #6
    Thanks, perfect. Sometimes just hearing something said with different wording helps people get unstuck.
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