# Proving that a rational cut is a Dedekind cut

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?

• 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 $\neg \exists x (x\in \mathbb{Q} \wedge \forall y (y\in \alpha \rightarrow y\leq x))$. – Hayden Apr 15 '15 at 4:36
• 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. – Nico Apr 15 '15 at 5:24

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.

• 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? – David Apr 15 '15 at 18:01
• @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? – Nico Apr 16 '15 at 18:35
• 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... – David Apr 17 '15 at 20:12
• @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? – Nico Apr 18 '15 at 21:00