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- Aug 10th 2009, 11:22 AM #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 and t , then s<t. Prove that there is a unique real number such that every real number less than is in S and every real number greater than 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 been unique ? What's going on ?

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

( I doubt this is correct)

What I did:

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

Taking the smallest real number in T, say , the following is true if T is bounded.

and

for some

This implies that

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

Supposed

then

for some

Which implies supS =

So this means that

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.

- Aug 10th 2009, 12:40 PM #2

- Aug 10th 2009, 01:23 PM #3

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- Aug 10th 2009, 01:47 PM #4
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 .*[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 then there exists a real number with (otherwise x would be a lower bound for T, contradicting the fact that is the greatest lower bound for T). Therefore and hence .

On the other hand, if y is any real number less than then and hence .

- Aug 10th 2009, 01:56 PM #5

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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 ?

- Aug 10th 2009, 03:13 PM #6

- Aug 10th 2009, 03:52 PM #7

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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 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 the corresponding "cut" can be defined to be the rational numbers less than . 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.

- Aug 10th 2009, 03:58 PM #8

- Aug 10th 2009, 04:09 PM #9

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I don't understand what you mean when you say " these cuts is to construct the real numbers partitions of the rational numbers ". 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 ?

- Aug 10th 2009, 04:13 PM #10

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- Aug 10th 2009, 04:15 PM #11

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