Representing the set of rationals $\mathbb{Q}$ as a germ or surreal number

Question

Let us define natural equivalence between elements of Hardy fields and integrals of Dirac comb-like functions.

Let us assume a natural embedding of Hardy field into surreal numbers ($[x]=\omega$). Define an infinitesimal-free germ to be a sum of a purely infinite surreal number and a real number ($\mathbb{J}+\mathbb{R}$). In other words, its Conway normal form has no terms of negative order. Let $[f(x)]$ be an infinitesimal-free germ, for instance, $[\ln x]$ (its Conway normal form is $\omega^{1/\omega}$, so it belongs to $\mathbb{J}$).

We naturally postulate that such germ is equivalent to the integral $[f(x)]=f(a)+\int_a^\infty f'(t) \, dt$, with arbitrary real $a$, in our case, $$\ln \omega=[\ln x]=\ln 1+\int_1^\infty (\ln t)' \, dt=\int_1^\infty \frac1t \, dt.$$

Now, we take the last integral and separate it into pieces of area equal to $1$, in our case: $$\int_1^\infty \frac1t \, dt=\int_1^e \frac1t \, dt+\int_e^{e^2} \frac1t \, dt + \cdots = \sum_{k=0}^\infty \int_{e^k}^{e^{k+1}}\frac1t \, dt.$$

Now, we replace each interval with a Diract Delta, located at the center of mass of each interval, in our case, at $s_k=e^{k+1}-e^k$, and swap the integral and the sum: $$\int_0^\infty \sum_{k=0}^\infty \delta(t-(e^{k+1}-e^k)) \, dt = \int_0^\infty \sum_{k=0}^\infty \delta(t-s_k) \, dt.$$ The last integral is an integral of a Dirac Comb-like function, and we call it numerosity of the set $S=\{s_k\}$. Indeed, it goes over all the positive real line and adds $1$ each time it encounters an element of $S$.

So, each element of the Hardy field can be represented as such (generally, divergent) integral of a Dirac-Comb-like function, or as a numerosity of a non-dense set.

An inverse process is also possible, given a non-dense (and without accumulation points) set $\{s_k\}$ of positive reals one can obtain a germ corresponding to it, but the procedure is more complicated and involves interpolation via Newton series.

Now, the theory of surreal numbers says that all countable surreal numbers are contained in $\bf{No}(\omega_1)$, which is isomorphic to a Hardy field of germs.

But we know that the set of all (positive) rationals is countable. So, there should be a surreal number and a germ corresponding to it. And a non-dense set with equal numerosity.

That non-dense set necessarily should include non-rational numbers, because according to Euclid's principle, the numerosity of a part is smaller than the numerosity of a whole set, so it cannot be a subset of rationals.

So, can we find such germ at infinity, surreal number and non-dense set that are equal to the set of positive rationals?

Or should we conclude, there is a surreal number that is countable but does not belong to $\bf{No}(\omega_1)$? Can $\omega_1$ be countable?

Answer 1

Here is some hypothesis about to what surreal number the numerosity of rationals may correspond. It is not entirely rigorous, and I point out, where.

Let's start with the definition of Dirichlet function, which is the indicator function of $\mathbb{Q}$:

$\mathbf {1} _{\mathbb {Q} }(x)=\lim _{k\to \infty }\left(\lim _{j\to \infty }\left(\cos(k!\pi x)\right)^{2j}\right)$

The numerosity of rationals is the numerosity of the values at which this function is equal to $1$. And it is equal to $1$ where $\cos^2(k!\pi x)$ is equal to $1$.

We know that the numerosity of maximums of $\cos^2(\pi x)$ is equal to the numerosity of periodic lattice with interval $1$, that is, the numerosity of integers (its maximums are exactly at integers), that is, $2\omega$. The numerosity of the maximums of $\cos^2(a\pi x)$ is $a$ times greater as the density of the lattice is greater as well, so it is $2a\omega$.

Now we can rewrite $\lim_{k\to\infty}k!$ as $\prod_{k=1}^\infty k$ so to get the numerosity of the maximums as $2\omega\prod_{k=1}^\infty k$.

The next step needs verification from the point of view of the rules of surreal numbers, but it looks plauseble that $\omega\prod_{k=1}^\infty k=\omega!$. If so, we get the numerosity of rationals as $\bf{N}(\mathbb{Q})=2\omega!$.

The numerosity of the rationals on the unit-length interval $[0,1)$ would be $\bf{N}(\mathbb{Q}_{[0,1)})=\bf{N}(\mathbb{Q})/(2\omega)=\Gamma(\omega)=\prod_{k=1}^\infty k$.