Is there a maximal ordered field? What about $\mathbb R$?

Question

In my real analysis class today, we revisited the fact that there is only one complete ordered field (up to isomorphism): $\mathbb R$. But this induced several thoughts I began wondering about.

I remembered that rationals are the "minimal ordered fields" (in the sense that they are the initial objects in some appropriate category, and hence are also isomorphic). And this led me to ask if there are any "maximal ordered fields".

I have seen Eric Wofsey's answer that the answer is $\mathbb R$ if we impose Archimedean-ness too, but can we say anything about the general case?

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I have studied category theory at a very rudimentary level only.

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Edit: My reasoning that $\mathbb R$ should be the maximal ordered field

Consider the following innocent argument of mine: Start with any ordered field $F$, and then complete it by Dedekind cuts. In this way, we'd have constructed a complete ordered field (and hence, $\mathbb R$), in which our $F$ can be embedded canonically.

I am yet to attempt a proof of this argument, and probably I will stump at some point because of Joe's answer and the existence of surreals (of which I have no idea about).

Adding this to my weekend's to-think list!

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To give this question a closure:

I tried defining Dedekind cuts over a general ordered field using the usual definition. I was almost able to prove that the cuts form an additive group with the usual definition of cut addition–only showing that $A + (-A)\supseteq 0^*$ (where $0^*$ is the zero cut, and $-A$ is as defined usually) gave me hell, and I had to succumb to assuming Archimedean property to show this fact.

But this raised another question: If $0^*\subseteq A + (-A)$ for all cuts $A$, then can we conclude Archimedean property on the parent field?

Answer 1

Let me give you two seemingly contradictory answers:

There is no Maximal Ordered Field

We can prove that there is no maximal ordered field. Based on your post, I think some of this will be new to you, but I want to keep things brief and high level. There’s a theorem in mathematical logic called the “Lowenheim-Skölem Theorem”. This theorem has a trivial corollary that says the following. Suppose we have a mathematical structure that can be defined by “first order sentences”. For example, Groups, Rings, Fields, Ordered Fields, Graphs, etc. Suppose we also can prove that there is an infinite example of our structure (like an infinite group, or $\mathbb{Q}$ as an ordered field). Then there are examples of our structure of all infinite cardinalities.

From the above corollary to Lowenheim-Skölem, there are ordered fields of all infinite cardinalities. Since there is no largest cardinality, there can be no largest ordered field.

Why doesn’t this argument apply to complete ordered fields? “completeness” isn’t a first order property. Hence the Lowenheim-Skölem Theorem does not apply.

There is a Maximal Ordered Field

The above argument assumes we are defining an ordered field as a set with additional structure. In set theory, there are collections of sets that are “too big” to be sets. We call such collections “proper classes”. For example, there is no set of all sets, but there is a class of all sets. Suppose we allow the underlying collection of objects for our ordered field to be a proper class.

Then we can define the field of Surreal Numbers! This ordered field has the property that every ordered field defined on a set can be embedded inside it.

This can be formalized in set theories other than ZFC, like NBG set theory.

Conclusion

Whether or not there is a “maximal” ordered field is going to depend on what exactly we mean by “maximal ordered field” and our underlying set theory.

Regarding the Comments

Here I’ll try to clear up some of the concerns raised in the comments:

“Joe, so the surreal numbers are not really a set, but a class?” — Yes. The underlying collection of objects on which the surreal field is defined is a proper class.

“Why is completeness a second order property when we have the axiom of power set?” — This is a subtle point, and to really understand what’s going on you’re going to need to learn some mathematical logic. If you’re interested in learning this, Chris Leary’s book is an exceptionally gentle treatment. Enderton’s book is also consider canonical, but it’s harder.

When we say “completeness is a second order property”, we mean as a property of ordered fields. We can only quantify over elements of the field. You are absolutely correct that in set theory, we can “hack this” by declaring there is a power set. But to quantify over the power set of elements of an ordered field is different than quantifying over elements of the field itself. Specifically, when quantifying over elements of the ordered field, there is no sentence we can build out of quantifications, and, or, not, $+,\cdot, <, 0,1$ that corresponds to completeness.

From the perspective of mathematical logic, our ordered field axioms are purely “syntactic”. We then find sets in a meta-theoretic set theory and interpretations of $0,1,+,\cdot,<$ that satisfy those axioms. The axioms we write down exist in their own domain. They can’t appeal to the power set of a model of the axioms.

The point you raise, is, in some sense though, the way we prove that the “real numbers” that we can construct in set theory are complete. It is also how we are able to talk about “complete ordered fields” in set theory.

Followup Question

You are correct that Dedekind completing a non-Archimedean field will not give you a field. You then made the following proposition:

Proposition: Let $\mathbb{F}$ be an ordered field. Let $0^* := \{ x \in \mathbb{F} : x < 0 \}$ denote the zero-cut. If for all Dedekind cuts $A \subseteq \mathbb{F}$ there is an additive inverse $-A$ such that $A + (-A) =0^*$, then $\mathbb{F}$ is Archimedean.

Proof: This proposition is correct. Indeed, it is easy to demonstrate that the contrapositive is true. Suppose that $\mathbb{F}$ is not Archimedean. Consider the following set: $$X := \{x \in \mathbb{F} : \text{ there is a } q \in \mathbb{Q} \text{ such that } x < q \}$$ Since $\mathbb{F}$ is non-Archimedean, $X$ is non-empty and forms a left Dedekind cut. Further, it is easy to see that $X$ has no additive inverse. Suppose that there’s another cut $Y$ such that $X + Y = 0^*$. I leave the details to be filled in by you, but we may following the following outline of a proof:

1. There are $x \in X$ and $y \in Y$ such that $x + y > -1$.
2. By the definition of $X$, there is a $q \in \mathbb{Q}$ such that $q + y > -1$. It follows that $q + 1 + y > 0$.
3. $q + 1 \in \mathbb{Q} \subseteq X$, so $q + 1 + y \in X + Y = 0^*$.

This is a contradiction. Thus $X$ has no additive inverse.

Answer 2

You're looking for the surreal numbers: https://en.wikipedia.org/wiki/Surreal_number

Answer 3

HINT:

Let $K$ be an ordered field. There exists an extension of the order of $K$ to $K(t)$ :
$$P(t)/Q(t)>0$$ if the quotient of the leading coefficients of $P$, $Q$ is positive.