Property of Rational Numbers

Question

Is there a name for the property of the rational ordered field that given any enumeration of the set of rational numbers used to define this ordered field, meaning a bijection between the rational numbers and the natural numbers $~f:\mathbb{N}\to\mathbb{Q}$, the following is true:

$\forall n\in\mathbb{N},\exists q\in\mathbb{Q},\forall m \in\{1,2,\dots,n\},(~f(m)\lt q\lt 0~)\lor(~0\lt q\lt f(m)~)$.

The reason I ask, is that I want to know what to call this kind of density of an ordered field (where you can enumerate, but not in order of magnitude).

Answer 1

It sounds to me like you're interested in the order type of $\mathbb Q$. Two totally ordered sets are of the same order type iff there is a strictly increasing bijection between them. $\mathbb N$ and $\mathbb Q$ are of different order types, which is reflected in the fact that for example, between any two rationals there is another rational, but this is not the case for $\mathbb N$. $\mathbb N$ also has different order type to $\mathbb Z$, since $\mathbb Z$ is unbounded in both directions, whereas $\mathbb N$ is only unbounded in one direction.

Every ordered field contains a copy of $\mathbb Q$, because every ordered field is of characteristic zero, and every field of characteristic zero contains $\mathbb Q$. Thus no ordered field is order-isomorphic to $\mathbb N$.

You may also be interested in the notion of a dense order. Every ordered field is dense, because if $a < b$, $\frac {a+b} 2$ is between $a$ and $b$. Proof:

$$\frac {a+b} 2 - a = \frac {b - a} 2 > 0$$

$$b - \frac {a + b} 2 = \frac {b - a} 2 > 0$$

...since $b-a>0$; and $\frac 1 2>0$, since $1$ is positive in any ordered field, therefore so is $2$, so if $\frac 1 2$ were negative, then $1$ would be as well.