In my analysis class, our professor showed us a "proof" that shows that every ordered field $ \mathbb{F}$ thats also complete holds the Archimedean property. The proof was trivial, and consisted of showing that for any $x\in \mathbb{F}$ there exists an $ n \in \mathbb{N}$ such that $x<n$. While I was reviewing my journal I came upon this proof and felt that this was wrong since we're assuming that we can compare the elements in $\mathbb{F}$ to the natural numbers. I haven't seen fields and rings yet in school, only groups and vector spaces. However I'm aware that there are fields that consists of elements other than numbers, so I'm thinking that there should be a field where this doesn't hold….Unfortunately the class is very boring and non rigorous, so the only example I have thought of so far is the the field constructed by Dedekind cuts since it represents $ \mathbb{R}$ but I'm not completely sure since thats something I self learned through Rudins book and might not understand Dedekind cuts as I deeply as I probably should.
Does every complete ordered field $ \mathbb{F}$ hold the Archimedean property
analysisfield-theoryreal-analysis
Related Solutions
Any ordered field $F$ has characteristic $0$, so it contains a copy of $\mathbb{Z}$; by the universal property of the quotient field, the ring monomorphism $\mathbb{Z}\to F$ lifts to a monomorphism $\mathbb{Q}\to F$. We can identify $\mathbb{Q}$ with its image, so it's not restrictive to assume that $\mathbb{Q}\subseteq F$.
It's not really difficult: if $m/n\in\mathbb{Q}$, then we send it to $$ \frac{f(m)}{f(n)}\in F $$ where $f\colon \mathbb{Z}\to F$ is the (unique) monomorphism. Is this a field homomorphism? Just a check.
Now we come to the order. First of all, positive integers are positive in $(F,\prec)$: if $n>0$, then $$ n=\underbrace{1+1+\dots+1}_{\text{$n$ times}} $$ and therefore $0\prec n$. Conversely, if $n<0$, then $$ n=-(\,\underbrace{\,1+1\dots+1}_{\text{$-n$ times}}\,) $$ and so $n\prec0$.
Any element of $\mathbb{Q}$ can be represented as $m/n$ with $n>0$, because $a/b=(-a)/(-b)$, where $a,b\in F$, $b\ne0$. So, let $0\prec m/n$ in the ordering of $F$, with $n>0$. Then, by the properties of ordered fields, $$ 0\prec n\cdot\frac{m}{n}=m $$ and therefore $m>0$. So a rational which is positive in $(F,\prec)$ is also positive in the usual order. A rational which is negative in $(F,\prec)$ is the opposite of a positive rational (in both orders).
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:
- There are $x \in X$ and $y \in Y$ such that $x + y > -1$.
- By the definition of $X$, there is a $q \in \mathbb{Q}$ such that $q + y > -1$. It follows that $q + 1 + y > 0$.
- $q + 1 \in \mathbb{Q} \subseteq X$, so $q + 1 + y \in X + Y = 0^*$.
This is a contradiction. Thus $X$ has no additive inverse.
Best Answer
Lemma: Every ordered field contains a copy of the natural numbers, all of which are positive.
Proof: We know that $1$ (the multiplicative identity) is not equal to $0$ - that's a standard field axiom. Since this is an ordered field, $1$ is either positive or negative. But then, $1\cdot 1=1$ is either the product of two positive numbers or the product of two negative numbers. Either way, it's positive.
From $1>0$, we get $2=1+1>1>0$, $3=2+1>2>0$, and so on. By induction, the natural numbers (defined as $n$-fold sums of $1$) are all positive. Since they're positive, they're not equal to zero. Also, as a strictly increasing sequence, none of them is equal to any other. That's it.
With this lemma, it makes sense to talk about the Archimedean property in any ordered field, and to prove the property if we also have completeness.