Sizes and shapes of Dedekind cuts

Question

My geometric intuition has failed to tell me that there are different sizes and shapes of Dedekind cuts. I realized it in the course of writing this answer only by doing algebra.

If we define a Dedekind cut for present purposes as a set $(-\infty,\alpha)\cap\mathbb Q$ where $\alpha\in\mathbb R$ then $\mathbb Q\cap(-\infty,\alpha)$ and $\mathbb Q\cap(-\infty,\beta)$ are geometrically congruent Dedekind cuts precisely if $\alpha-\beta\in\mathbb Q$. The congruence is $x\mapsto x + \beta-\alpha$.

If $\alpha,\beta\not\in\mathbb Q$ and $\beta-\alpha\not\in\mathbb Q$ and $\beta-\dfrac\alpha2\in\mathbb Q$ then $x\mapsto \dfrac x 2 + \beta-\dfrac \alpha 2$ is a similarity but not a congruence between $(-\infty,\alpha)\cap\mathbb Q$ and $(-\infty,\beta)\cap\mathbb Q$.

Two Dedekind cuts have the same shape if there is a similarity between them and two of the same shape have the same size if there is a congruence between them.

Dedekind cuts of the form $(-\infty, \alpha) \cap \mathbb Q$ when $\alpha \in \mathbb Q$ are self-similar, i.e. similar to a rescaling of themselves.

Is all this just another way of talking about Diophantine approximations, or is there something else of interest to be said about this?

Answer 1

The bigger picture here is that in an ordered field with its native absolute value, any similarity between two subsets is affine. (In this setting, given any two points, a third point is uniquely determined by its distances to those points. This provides a way of extrapolating a similarity from its values at two points)

In the case of Dedekind cuts in the rationals as you define them, since they are bounded above, but not below, this affine map must also be order-preserving. Hence it must have the form $f(x) = p + qx$ with $p \in \mathbb{Q}, q \in \mathbb{Q^+}$.

We have then that a similarity of cuts in $\mathbb{Q}$ extends uniquely to an order-preserving self-similarity of $\mathbb{Q}$, which extends uniquely to an order-preserving self-similarity of $\mathbb{R}$. From there it seems obvious that $\alpha, \beta \in \mathbb{R}$ have similar Dedekind cuts in $\mathbb{Q}$ if and only if there is a rational $p$ and a positive rational $q$ such that $\beta = p + q\alpha$.