Topology on rational intervals

Question

Corrected question: I want to find a topology based on the set of all Dedekind Cuts without using the notion of real numbers. Because of the density of $\mathbb{Q}$ I cannot define a bijection from the Cuts to the set of rational numbers. Is there a way to construct such a topology?

Answer 1

Let ${\cal D}$ be the set of all Dedekind cuts of ${\mathbb Q}$, that is to say the set of all non empty proper subsets $A\subsetneq {\mathbb Q}$ downward closed and without upper bound. For $0 < r \in {\mathbb Q}$, define the open ball of radius $r$ centered at $A$ as the set of all $A'\in {\mathcal D}$ such that $A-r \subsetneq A'\subsetneq A+r$. The topology generated by these open balls in ${\mathcal D}$ is the topology of ${\mathbb R}$, defined without using real numbers.

A subset $U\subset {\mathcal D}$ is open in this topology if and only if for every $A\in U$, there exists $r\in {\mathbb Q}$ such that $r > 0$ and $B(A, r)\subset U$.

Answer 2

Let $R=\{A\subseteq\mathbb Q\text{ closed}:b<a\in A\Rightarrow b\in A\}\setminus\{\emptyset,\mathbb Q\}$ be the set of Dedekind cuts.

Say a subset $U\subseteq R$ is open if for each $A\in U$, there exist rational numbers $p<q$ with $(-\infty,p]\subsetneq B\subsetneq(-\infty, q] \Rightarrow B\in U$. Equivalently, a rational $r\in B$ with $p<r<q$ witnesses $B\in U$.