3
$\begingroup$

I'm new here, so I don't know how to do the fancy symbols. Sorry...

This is for my intro. to adv. math class, and I've been struggling this entire semester. I kinda understand the concept being asked, but I have no idea how to go about proving it.

It was explained that a rational cut is: Let $x$ be an element of $\mathbb{Q}$. The set $\{z \in \mathbb{Q} | z < x\}$ is a rational cut. It is denoted by $x^*$.

My professor defines a Dedekind cut as:

  1. $\{\alpha,\beta\}$ is a partition of $\mathbb{Q}$, that is,

$$\alpha\cup\beta=\mathbb{Q},\quad \alpha\ne \varnothing\ne\beta,\quad \alpha\cap\beta=\varnothing$$

  1. $\forall a\in\alpha,b\in\beta,\quad a<b$

  2. $\alpha$ has no maximum element in $\mathbb{Q}$. In other words,

$\neg\exists x\in\mathbb{Q}$ such that $\alpha=\{y\in\mathbb{Q}|y\le x\}$

So how can I prove that a rational cut is a Dedekind cut?

$\endgroup$
  • $\begingroup$ Are you sure you aren't being asked to show that every rational cut uniquely determines a Dedekind cut? Also, your `In other words' statement of 3 also doesn't make sense. I think what you mean to say is something along the lines of $\neg \exists x (x\in \mathbb{Q} \wedge \forall y (y\in \alpha \rightarrow y\leq x))$. $\endgroup$ – Hayden Apr 15 '15 at 4:36
  • $\begingroup$ Technically the way you wrote it is close, but not exactly what they meant. Your formulation says, informally, that "there is no rational upper bound of $\alpha$", which of course isn't true in a Dedekind cut. I suggested a correction to fix the $\LaTeX$ and made that statement more clear while retaining the notation their professor actually used. $\endgroup$ – Nico Apr 15 '15 at 5:24
3
$\begingroup$

Just to clarify the definition of the Dedekind cut you gave, $\alpha$ is the actual Dedekind cut and $\beta$ is its complement in $\mathbb{Q}$. That wasn't clearly stated in your formulation and it is important to know that to proceed.

Now let $x\in\mathbb{Q}$ and define $\alpha=\{q\in\mathbb{Q}|q<x\}$. Now your goal is to show that $\alpha$ satisfies the three axioms you spelled out in your question. Here are a few hints:

  1. Begin by finding the "$\beta$" in this context. That is, what is the set that satisfies those conditions for the $\alpha$ we just defined?
  2. Prove the second axiom holds.
    • Why is it that something in $\alpha$ must be smaller than something in its complement? This relies on having created a good, solid representation of what $\beta$ is.
  3. Try assuming that there is some maximum rational number in $\alpha$. Can you find an example in $\alpha$ that is bigger? If so, you will have proved (by contradiction) that there is no maximum.

I was intentionally vague to promote thought. Let me know in a comment if you need more of a hint.

$\endgroup$
  • $\begingroup$ Thank you. Let me attempt this. I'm not really sure how to go about finding a beta. Would it be proper to say that beta = {z element Q | x < z}, where x is an element in Q? $\endgroup$ – David Apr 15 '15 at 18:01
  • $\begingroup$ @david close! if you take the union of $\alpha$ and $\beta$, do you get all of $\mathbb{Q}$? (Hint: no) how can you tweak your definition of $\beta$ to account for this? $\endgroup$ – Nico Apr 16 '15 at 18:35
  • $\begingroup$ Oh, I think I got it. I say that x is < or equal to z. I tried it out, and I think it works! The only thing I'm having trouble with is the last part, 3. I'm assuming an m $\in\mathbb{Q}$. How can I go and use this to prove that there is no max element. Someone told me that I have to make 3 cases: m>x, m<x, and m=x. But I feel like there should be an easier way... $\endgroup$ – David Apr 17 '15 at 20:12
  • $\begingroup$ @david There certainly is! Actually, the fact that $m\in\alpha$ tells you automatically that $m<x$ (by definition). So I'll give you the tools and shove you in the right direction: as you said, let $m\in\mathbb{Q}$ and consider your rational number $x$. I claim that there is rational number between these two (in fact, between any two rational numbers). Can you prove it? $\endgroup$ – Nico Apr 18 '15 at 21:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.