### What’s So Special About the Rationals?

#### Posted by David Corfield

I posed a question over at Math Overflow which netted me one useful answer from Andrew Stacey (along with my first nice question badge). Let’s see if the Café can help me with what I really want to know.

So, the rationals form the unique dense linearly ordered set without endpoints of countably infinite cardinality. On top of this we can build up compatible structure all the way to an ordered topological field. The rationals form the field of fractions of the integers, and the prime field for characteristic $0$.

For some additional steps on the way, Wikipedia tells us

By virtue of their order, the rationals carry an order topology. The rational numbers also carry a subspace topology. The rational numbers form a metric space by using the metric $d(x, y) = |x - y|$, and this yields a third topology on $\mathbb{Q}$. All three topologies coincide and turn the rationals into a topological field. The rational numbers are an important example of a space which is not locally compact. The rationals are characterized topologically as the unique countable metrizable space without isolated points. The space is also totally disconnected. The rational numbers do not form a complete metric space; the real numbers are the completion of $\mathbb{Q}$. In addition to the absolute value metric mentioned above, there are other metrics which turn $\mathbb{Q}$ into a topological field.

So the question now is whether we can see how the bits of pieces of the structure of the rationals fit together? What are the dependences?

For a simple case of what I’m looking for, take the integers with addition $\langle \mathbb{Z}, + \rangle$. As the free abelian group on one generator, a ring structure comes for the ride, since we can show that it is a monoid object in the category of abelian groups, using

$Hom_{Ab}(F\{*\}), F\{*\}) \cong Hom_{Set}(\{*\}), U F\{*\}).$

So what about the rationals? From the fact that $\langle \mathbb{Q}, \lt \rangle$ is the Fraïssé limit of the category of finite linearly ordered sets and order preserving injections we get density and homogeneity. Could we see how to get a $\langle \mathbb{Q}, + \rangle$-torsor structure to emerge here?

One thought as to what next: I guess we might consider the Fraïssé limit of finite *pointed* linearly ordered sets, which should be, I think, $\langle \mathbb{Q}, 0, \lt \rangle$.

## Re: What’s So Special About the Rationals?

The fourth or fifth thought that pops into my head is that “since” there’s exactly one two-ways of injectively monotonically mapping $[n]$ to $[n+1]$ such that the two ways are nowhere equal, there should also be (essentially) one way of mapping $\langle \mathbb{Q}, < \rangle$ to itself with no stable bounded intervals.

Hmm… I can get my head around

this, though: by trichotomy etc., for any rational $x$, $\langle ]-,x[, > \rangle$ and $\langle ]x,-[,<\rangle$ are both countable infinite d.l.o.w/o.e.p.; so they’re isomorphic; choose such an isomorphism $\phi$. Then $\phi \cup \phi^{-1}\cup \{(x,x)\}$ is an order isomorphism of $\langle \mathbb{Q},<\rangle$ and $\langle \mathbb{Q},>\rangle$, fixing $x$. Similarly, there’s (far more than) one such — say $\psi$ — fixing $y$ for any $y>x$. Then $\psi\phi$ is an order isomorphism mapping $x<y$ to some $z>y$ (and now you know how I write compositions); similarly $\phi^{-1}\psi^{-1}$ maps $y$ to some $w< x$, and hence $\phi^{-1}\psi^{-1}(x)<x$. The union $U$ of the intervals $](\psi\phi)^{-n}(x),(\psi\phi)^n(x)[$ is again countable infinite dense linear without endpoints, and stable under $\psi\phi$, which acts as an order automorphism and with no stable (relatively) bounded intervals.The argument currently playing in my head says I should look at the Euclidean algorithm to generate — from $1$, ${\cdot}+1$, and $(\cdot)^{-1}$ — all positive rationals, and then take their negatives … and then argue from some nonsense that $\mathrm{ad}_1$ as described, and any order antimorphism $\mathrm{inv}$ of $U^+= U\cap ]0,-[$ fixing $\mathrm{ad}_1(0)$ together generate a useful monoid such that the orbit of $1=\mathrm{ad}_1(0)$ is a dense order and has no endpoints. It’s then this orbit that becomes the (positive) rationals; but again it’ll be isomorphic to (half of) whatever d.l.o.w/o.e.p. you started with, by structural uniqueness.