Analog of the real numbers made from ordinals

Question

The proper class $On$ is in many ways an extension of the natural numbers. And just as the naturals can be completed to form the integers by the Grothendieck completion, the ordinals(with the natural sum and product) can be made into an abelian group. The integers are then made into the rationals using the field of fractions construction and the ring made from the ordinals can also undergo the same process.

So far, I see no problem with the construction, but I know that the only ordered complete field is that if the real numbers. Seeing as this is so, how would the Cauchy completion of the field of fractions violate the field axioms?

Also, would $Frac(Gro(On))$ be useful for anything?

Answer 1

Edit: it has been pointed out to me that I failed to properly read the question. My answer applies to the ordinary sum and product, but does not apply to the “natural” sum and product, which is what the question was about. Please see the answer of @lonza leggiera. However, since the answer is still interesting, I will leave it up.

There are a few major flaws in this.

First, the addition and multiplication operations are not commutative. This means that the Grothendieck completion does not apply, since Grothendieck completion requires a commutative monoid.

We can try a more general version of the construction. The Grothendieck completion is the left adjoint of the forgetful functor $AbGroup \to AbMonoid$. We can try instead the left adjoint $F$ of the forgetful functor $U : Group \to Monoid$. Unfortunately, this will produce the zero group.

To see why, consider that we would have some monoid homomorphism $\eta_{On} : On \to U(F(On))$. Fix an ordinal $\alpha$. We can find some ordinal $\beta$ such that $\alpha + \beta = \beta$; specifically, take the smallest infinite $\beta$ such that $|\alpha| < |\beta|$. Then $\eta_{On}(\alpha) + \eta_{On}(\beta) = \eta_{On}(\beta)$, so $\eta_{On}(\alpha) = 0$. So all the elements of $On$ are sent to $0$. So $\eta_{On}$ factors through the subgroup inclusion $0 \to F(On)$, which means $F(On)$ is the zero group.

Of course, the fact that multiplication is also not commutative (nor is it distributive) means that even if you could get off the ground with constructing a group which contains the ordinals as a submonoid under addition, you would have to rework the ring part of the construction.

Answer 2

Since the ordinals are (order isomorphic to) a subclass of the surreal numbers (which already form a Field), and surreal addition and multiplication coincide with the natural addition and multiplication of ordinals, your field of fractions will turn out to be a subField of the surreal numbers.

It seems to me that trying to apply Cauchy completion to the surreal numbers would be fraught with difficulty. Here's a relevant discussion (which I've only glanced through). There's also a brief relevant discussion on pp.$26$-$27$ of the second edition of Conway's On Numbers and Games.

Addendum

An observation more pertinent to your question is that the notion referred to by the word "complete" in the the term "complete ordered field" is different from that which it refers to in the term "complete metric space". Cauchy completion of an ordered field will produce an object which is complete in the second sense (with respect to a notion of "metric" that has been suitably generalised if necessary), but not necessarily the first.

A complete ordered field must be Archimedean, and an ordered field that is not Archimedean cannot be made so by Cauchy completion. The field of fractions of the ordinals under natural addition and multiplication is not Archimedean, so any Cauchy completion of it therefore can't be a complete ordered field.