# Can the surreal numbers be completed to form an ordered field?

The surreal number line isn’t Cauchy complete as it’s filled with “gaps”. When constructing the real numbers from the rationals, one could take the ring of all Cauchy sequences and take the quotient by identifying the sequences that approach $$0$$ with $$0$$ and getting the real numbers as a result. Could this same procedure be repeated with the Surreal numbers to complete it? If so, would any properties be lost or anything interesting happen as a result?

Great question. let me say a few things that might be interesting.

First, “Cauchy Completeness” isn’t going to be the best way to think about this. We usually think of Cauchy completions for metric spaces. $$\mathbb{Q}$$ is a metric space, so it works, but when we start with the surreal numbers we’re going to have more trouble.

A better way of thinking about the completeness of ordered sets is “Dedekind completeness”. An ordered set $$L$$ is Dedekind complete if and only if every bounded subset has a least upper bound and a greatest lower bound. That is, suppose $$L$$ is Dedekind complete. Then for any bounded $$X \subseteq L$$, there is a “least upper bound” $$\sup X \in L$$ such that for all $$x \in X$$, we have $$x \leq \sup X$$, and if $$a \in L$$ satisfies that $$x \leq x$$ for every $$x \in X$$, then $$\sup X \leq a$$. Similarly we define a greatest lower bound, denoted by “$$\inf$$”.

Instead of looking at Cauchy completions, an analogous way to “fill in the holes” in the rationals is take the Dedekind completion. We define a “left Dedekind cut” of $$\mathbb{Q}$$ to be a pair of sets $$\langle L,R \rangle$$ with the following properties:

1. $$L \leq R$$. that is, for all $$\ell \in L$$ And $$r \in R$$, we have that $$\ell \leq R$$
2. $$L$$ and $$R$$ are disjoint.
3. $$L \cup R = \mathbb{Q}$$
4. If $$\ell \in L$$ and $$x \leq \ell$$, then $$x \in L$$ ($$L$$ is “closed below”)
5. If $$r \in R$$ and $$r \leq x$$, then $$x \in R$$ ($$R$$ is “closed above”)
6. L has no greatest element.

Notice that condition 6 is the only “non-symmetric” condition. R can (and oftentimes will) have a least element. This non-symmetric condition is what is specified by calling it a “left” Dedekind cut.

We may construct the real numbers $$\mathbb{R}$$ as the collection of all Dedekind cuts on $$\mathbb{Q}$$. Without going into too much detail, the spirit of the idea is that a Dedekind cut $$\langle L, R\rangle$$ corresponds to the least upper bound of $$L$$. An example will be illustrative.

Set $$L := \{x \in \mathbb{Q} : x^2 < 2 \}$$, and $$R := \{x \in \mathbb{Q} : x^2 > 2 \}$$. It is easy to see that the pair $$\langle L, R\rangle$$ is a Dedekind cut of $$\mathbb{Q}$$. It is clear that $$L$$ has no least upper bound in $$\mathbb{Q}$$. If it did, that least upper bound would have to square to $$2$$, but we know that $$\sqrt{2}$$ is irrational. This Dedekind cut exactly picks out the “hole” at $$\sqrt{2}$$, and plays that role in the construction of the reals.

To formally put an order on the collection of Dedekind cuts can be accomplished by comparing the left sets. Suppose that $$\langle L, R \rangle$$ and $$\langle L’, R’ \rangle$$ are both Dedekind cuts of $$\mathbb{Q}$$. Then we may say that $$\langle L, R \rangle \leq \langle L’, R’ \rangle$$ if and only if for all $$\ell \in L$$, there is an $$\ell’ \in L’$$ such that $$\ell \leq \ell’$$. It is essential to note that this new order is Dedekind complete. With some work, one may put field operations on this new ordered set, and show that it is isomorphic to the Cauchy-completion of the rationals.

Further notice that we may apply this process to any ordered set. We can take left Dedekind cuts for any ordered set, and build a new ordered set which is Dedekind complete.

We may carry this out with the surreal numbers as well! We’ll get a new ordered class which is Dedekind complete, but it will not get an induced field structure from the field operations on the surreal numbers. Indeed, if it were a field, it would be isomorphic to $$\mathbb{R}$$. However, it is easy to see that we cannot get the reals as the the Dedekind completion of the surreal numbers will not be separable.

Saying much of anything about this new order will be a challenge. You’re going to immediately run head first into set theoretic concerns. Indeed, the surreal numbers themselves can’t even be defined in standard ZFC set theory. Even for more modest “large fields” like the Hyperreals $$^*\mathbb{R}$$, it will be difficult to say much of anything substantial about the Dedekind completion. Indeed, without assuming the Continuum Hypothesis, I don’t think you can pin down the cardinality of the Dedekind Completion of $$^*\mathbb{R}$$.

• I’d like to ask for some clarification: 1) Why is $No$ not a metric space? 2) Why does its Dedikind completion not get a field structure? 3) Because I’m assuming NBG set theory for this question, could any of the last remarks be answered in this new theory? Jun 28 at 21:54
• The surreals are, in a very real sense, simply too big to be a metric space. At least if you want your metric to induce the order topology.
– Joe
Jun 28 at 23:26
• 2) The Dedekind completion can’t get a field structure for the reasons I described above. I’m not sure exactly where it’ll break, but it has to somewhere. For the hyper-reals, addition breaks. see here: math.stackexchange.com/questions/2010740/…
– Joe
Jun 29 at 0:58
• 3) I don’t think that working in NBG will make things easier. I’m pointing out that you aren’t dealing with ZFC sets anymore to show how far beyond our intuitions this ordered class will be.
– Joe
Jun 29 at 0:59
• @Joe: Can the Surreals be made into a Vector Space ( over the Reals or otherwise)?
– MSIS
Aug 9 at 22:44