1. ## 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 $\forall$s&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.

2. Originally Posted by justchillin
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.

3. that was good.
Fixed.
Can you help me now?

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 $\beta$. [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 (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 $y\in S$.

5. Originally Posted by Opalg
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 (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 ?!

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 ?

6. Originally Posted by justchillin
What about the density of the rationals and irrationals ?

7. Originally Posted by Plato
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.

8. Originally Posted by siclar
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."

9. Originally Posted by siclar
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 ?

10. 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.

11. I created a new thread so you can post in it. Sorry plato.