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:


*

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


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


*$\forall a\in\alpha,b\in\beta,\quad a<b$

*$\alpha$ 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?
 A: 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:


*

*Begin by finding the "$\beta$" in this context. That is, what is the set that satisfies those conditions for the $\alpha$ we just defined?

*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.


*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.
