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Math Help - Dedekinds cut Question

  1. #1
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    Dedekinds cut Question

    Question Let S and T be nonempty sets of real numbers such that every real number is in S or T and if s \exists S and t \exists T, then s<t. Prove that there is a unique real number \beta such that every real number less than \beta is in S and every real number greater than \beta is in T. (A decomposition of the reals into two sets with these properties is a Dedekind cut. This is known as Dedekind's theorem.)

    I'm not really sure how to approach the problem.First of all, from what i understand if we have an open interval of real numbers, in this case (s,t), there are many rational and irrational numbers between s and t ; because the rational and irrationals are dense in the reals.
    So what's all this talk about \beta been unique ? What's going on ?

    Secondly, the question does not specify if \beta is in any of the sets. I used that assumption in my won answer.


    What I did:
    ( I doubt this is correct)

    The inequality s < t \foralls&t would mean that the smallest member in t is an upper bound of S.

    Taking the smallest real number in T, say t_{0}, the following is true if T is bounded.

    \beta\leq t
    and
    t_{0}<\beta + epsilon for some \epsilon>0

    This implies that inf T = \beta

    Because s<t \beta is an upper bound of S and possible inf S.

    Supposed

    s\leq\beta \forall s\exists S
    then

    s_{0}>\beta-\epsilon_{0}for some \epsilon_{0}>0

    Which implies supS = \beta

    So this means that s \leq \beta \leq t




    When i do it this way it seems more correct because sup S and inf T is unique because they are bounded and according to the theorem i have here every non emepty set of real numbers that is bounded from above or below, has a unique inf or sup.
    Last edited by justchillin; August 10th 2009 at 01:22 PM.
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  2. #2
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    Quote Originally Posted by justchillin View Post
    Question [i]Let S and T be nonempty sets of real numbers such that every real number is in S or T and if s \in S and t \in T, then s
    I don't know if this is Dedekind at work, but the question appears to have been cut.
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    that was good.
    Fixed.
    Can you help me now?
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  4. #4
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    What you did was more or less the right approach, but it has one serious error, right at the start, where you talk about "the smallest real number in T". There is no reason to suppose that T contains a smallest member. (For example, T might consist of all the strictly positive real numbers, with S consisting of all the negative real numbers together with 0.)

    What you should do is to start by saying that T is bounded below (because S is nonempty, and every member of S is a lower bound for T). So T has an inf, call it \beta. [That's the key point of the proof: T may not have a smallest member, but it does have an inf.]

    If x is any real number greater than \beta then there exists a real number t\in T with \beta<t<x (otherwise x would be a lower bound for T, contradicting the fact that \beta is the greatest lower bound for T). Therefore x\notin S and hence x\in T.

    On the other hand, if y is any real number less than \beta then y\notin T and hence y\in S.
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    Quote Originally Posted by Opalg View Post
    What you did was more or less the right approach, but it has one serious error, right at the start, where you talk about "the smallest real number in T". There is no reason to suppose that T contains a smallest member. (For example, T might consist of all the strictly positive real numbers, with S consisting of all the negative real numbers together with 0.)

    What you should do is to start by saying that T is bounded below (because S is nonempty, and every member of S is a lower bound for T). So T has an inf, call it \beta. [That's the key point of the proof: T may not have a smallest member, but it does have an inf.]

    If x is any real number greater than \beta then there exists a real number t\in T with \beta<t<x (otherwise x would be a lower bound for T, contradicting the fact that \beta is the greatest lower bound for T). Therefore x\notin S and hence x\in T.

    On the other hand, if y is any real number less than \beta then y\notin T and hence y\in S.
    Thanks a lot. I'm actually studying this stuff on my own so it's a bit hard for me.

    Funny enough my book actually warned me about talking about the smallest positive real number.

    If i understand you right as long as i know that T is bounded from below i know that inf T exist . And from there i can go on with the proof ?!


    Btw i still confused about this :
    I'm not really sure how to approach the problem.First of all, from what i understand if we have an open interval of real numbers, in this case (s,t), there are many rational and irrational numbers between s and t ; because the rational and irrationals are dense in the reals.
    So what's all this talk about been unique ? What's going on ?
    What about the density of the rationals and irrationals ?
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  6. #6
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    Quote Originally Posted by justchillin View Post
    What about the density of the rationals and irrationals ?
    Please Start a new thread for a new question.
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    Quote Originally Posted by Plato View Post
    Please Start a new thread for a new question.
    It's still in reference to the OP so its not really a new question.

    justchillin, I'm not sure what you are getting at with density. As far as the \beta being unique, the idea behind these cuts is to construct the real numbers as partitions of the rational numbers. In other words for each real \beta the corresponding "cut" can be defined to be the rational numbers less than \beta. We can then define addition and multiplication and get something isomorphic to the real numbers as we know them. Then the uniqueness property now makes sense. We want each cut to uniquely represent each real number. I hope that clarifies things somewhat.
    Last edited by siclar; August 10th 2009 at 04:11 PM. Reason: Typo
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    Quote Originally Posted by siclar View Post
    It's still in reference to the OP so its not really a new question..
    I disagree with that.
    You said yourself, "justchillin, I'm not sure what you are getting at with density."
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    Quote Originally Posted by siclar View Post
    It's still in reference to the OP so its not really a new question.

    justchillin, I'm not sure what you are getting at with density. As far as the \beta being unique, the idea behind these cuts is to construct the real numbers partitions of the rational numbers. In other words for each real \beta the corresponding &quot;cut&quot; can be defined to be the rational numbers less than \beta. We can then define addition and multiplication and get something isomorphic to the real numbers as we know them. Then the uniqueness property now makes sense. We want each cut to uniquely represent each real number. I hope that clarifies things somewhat.
    I don't understand what you mean when you say &quot; these cuts is to construct the real numbers partitions of the rational numbers &quot;. What do you mean by partitions of rational numbers ? From what i understand we cut an interval of real numbers into two and then show that they are both bounded ( at the cut location) either from above or below . Is that what you mean ? Btw i got another question about the Dedekind cut, i don't really understand what i need to do. It says : Using properties (A)-(H) (In my book this is the basic properties of real numbers like commutative, associative,distributive laws.... the inverse of a number and the tricotomy property and transitive property ) of the real numbers and taking Dedekind theorem ( As stated in the previous question i posted) as given, show that every nonempty set U of real numbers that is bounded above has a supremum. The hint is : Let T be the set of upper bounds of U and S be the set of real numbers that are not upper bounds of U. What do i do? I mean i guess i can't use the completeness axiom. So how to i arrive at this result using only thr basic properties of the real numbers and the Dedekind cut ?
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    Corrected my typo: yeah its constructed as partitions of rational numbers. But now you obviously are venturing into a new question which merits a new thread which I'm sure will be echoed soon by Plato.
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  11. #11
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    I created a new thread so you can post in it. Sorry plato.
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