Dedekind Cuts and Rationals

Question

I have a question regarding how you might construct the reals from the rationals by taking Dedekind cuts.

My basic understanding is that a Dedekind cut is a bipartition of the rationals such that the two partitions $X, Y$ satisfy certain properties:

$\forall x \in X,\; \exists y \in X \text{ such that } x < y$

$\forall x \in Y, \; \exists y \in Y \text{ such that } y < x$

$\forall x,y, \; x<y \text{ and } y \in X \Rightarrow x \in X$

$\forall x,y, \; x<y \text{ and } x \in Y \Rightarrow y \in Y$

$\forall x,y, x<y \Rightarrow \text{ either } x \in X \text{ or } y \in Y$

$X \text{ and } Y$ are disjoint

$X \text{ and } Y$ are both non-empty.

A Dedekind cut is then used to represent a real number $z$ that can be see as the "mid-point" of this bipartition of the rationals in the sense that $\forall x \in X, \forall y \in Y, \; x < z < y$

In most descriptions of defining the reals as the set of all Dedekind cuts of the rationals, these inequalities are strict. However, doesn't that mean that the rationals cannot be defined in this way? Am I meant to take this to mean that the irrationals are defined using these Dedekind cuts and that the reals are to be treated as the union of this set of irrationals and the set of rationals?

If that is the case, doesn't that mean the irrationals and rationals in this construction of the reals are different kinds of objects in that they would have different ranks? I would imagine that is something inconvenient and would ideally want to be avoided.

Answer 1

Note that it is not part of your axioms that $X\cup Y = \Bbb Q$. For instance, the rational number $0$ is given by $$ X = \{q\in \Bbb Q\mid q<0\}\\ Y = \{q\in \Bbb Q\mid q>0\} $$ and the number $0$ isn't contained in either of them. On the other hand, the axiom $$\forall x,y, x<y \Rightarrow \text{ either } x \in X \text{ or } y \in Y$$ implies that $\Bbb Q\setminus(X\cup Y)$ has at most one element.

The usual definition of Dedekind cuts that I've come across does have $X\cup Y = \Bbb Q$, but it also allows $Y$ to have a least element, which your axioms do not allow. Specifically, $$\forall x \in Y, \; \exists y \in Y \text{ such that } y < x$$ says that $Y$ has no least element.

Answer 2

The problem is that these inequalites are not necessarily strict. For instance, it is stated here that the second set (your $Y$) may have a smallest (rational) number $q$, in which case the Dedekind cut $(X,Y)$ correspondes to the rational number $q$.