I faced this claim in a paper and cannot get it: "Assuming CH, every maximal Hardy field is isomorphic to $(\bf{No}(\omega_1), \partial_{\omega_1})$"

I encountered this claim in a paper and cannot get it: "Assuming CH, every maximal Hardy field is isomorphic to $(\bf{No}(\omega_1), \partial_{\omega_1})$"

I cannot get how it could be true. This paper is cited for this result.

As I understand it, any germ(growth rate) of a real function at infinity is mapped to countable surreals ad vice versa.

There are no countable surreals beyond those germs.

But what if we allow a generalization of the Hardy fields to include functions with countably-infinite derivatives or polynomials with coefficients in countable surreals? Would not they be greater than the germs of all real functions, yet will be countable surreals themselves?

I was also thinking about the theory of numerosity. It is a measure of subsets of reals satisfying the inclusion-exclusion principle, also known as Euclid's principle and surreal-valued.

Any surreal-valued element of a Hardy field can be mapped to a sequence of reals by dividing the derivative of the corresponding germ into blocks of area 1 and putting $a_k$ into the center of mass of that segment, thus the germ becomes a smoothed counting function for that sequence.

But then what about a numerosity of a dense set, such as rationals or dyadic rationals? According to the Euclid's principle they are greater than any Hardy field, but at the same time smaller than the numerosity of reals. This contradicts the claim that there are no countabvle surreals that are greater than all elements of a maximal Hardy field.

The surreals are said to be a maximal ordered field, so the numerosity of rationals should belong there. But if it is greater than any Hardy field element, so not countable, what it should be?... Uncountable? Numerosity of a countable set is uncountable? This makes no sense.|

I discussed it with Gemini-3.1-pro and it concluded that it is indeed uncountable. Unfortunately, the discussion was lost.

P.S. A more clear re-formulation of the central argument by AI:

1. The Hardy Field Equivalence: By the theorem, the maximal Hardy field $\mathcal{H}$ is isomorphic to $\mathbf{No}(\omega_1)$. This means every growth rate (germ) in $\mathcal{H}$ corresponds to a "countable surreal" (a surreal born before stage $\omega_1$).
2. Your Construction: For any given germ $f \in \mathcal{H}$ (which corresponds to some countable surreal $s$), we can find a discrete subset $S \subset \mathbb{Q}$ whose counting function outgrows $f$. Because $\mathbb{Q}$ is dense, we can pack as many points of $S$ as we want into any interval to make its counting function arbitrarily large, while $S$ remains discrete.
3. Euclid's Principle: According to the theory of numerosity, if $S \subset \mathbb{Q}$, then $\mathfrak{n}(S) < \mathfrak{n}(\mathbb{Q})$.
4. The Conclusion: Since we can construct a discrete $S \subset \mathbb{Q}$ to beat any given germ in the Hardy field, $\mathfrak{n}(\mathbb{Q})$ must be strictly greater than the numerosity of every such sequence. Therefore, $\mathfrak{n}(\mathbb{Q})$ is strictly greater than every element in $\mathbf{No}(\omega_1)$.

PP.S. The crux of this issue has been clarified in the discussion with the AI here (the first few replies are the most important).