rational numbers are dense( dedekind cuts)

Question

First of all, is the title correct?

I want to prove the following proposition

Let $A$ be a dedekind cut

$\forall\epsilon>0$ $\exists p \in A,q\in A^c$ such that $q-p < \epsilon$

please give me some hints

Answer 1

Let $p_0\in A$ and $q_0\in A^c$, and define the sequences $p_n, q_n$ recursively as follows: For $n \geq 1$, look at $r = \frac{p_{n-1}+q_{n-1}}{2}$. If $r \in A$, set $p_n = r, q_n = q_{n-1}$. If $r \in A^c$, set $p_n = p_{n-1}, q_n = r$. We see that $p_n, q_n$ are both rational in either case, and also that $p_n \in A, q_n \in A^c$.

This gives $$ q_n - p_n = \frac{q_0 - p_0}{2^n} $$ so for any $\epsilon > 0$, and any starting point $p_0, q_0$, there will be an $N$ such that $q_N-p_N

As for the title, this is true by the above proof. For the Dedekind cut (real number) $A$, and any $\epsilon$, we find a rational $p$ with $|A-p| <\epsilon$