Let $F$ be an ordered field.

What is the least ordinal $\alpha$ such that there is no order-embedding of $\alpha$ into any bounded interval of $F$?

  • 1
    $\begingroup$ This depends entirely on the field. What kind of an answer are you looking for? $\endgroup$ Mar 4 '16 at 16:11
  • 1
    $\begingroup$ I wonder if the answer, in general, can be specified in terms of certain properties of the ordered field, like its cofinality. $\endgroup$
    – Haidar
    Mar 4 '16 at 16:15
  • 3
    $\begingroup$ The issue of "bounded" seems moot, since if an ordinal $\beta$ embeds into a field at all, then it embeds into a bounded interval, by first translating to the positives and then composing with $x\mapsto -\frac 1x$. $\endgroup$ Mar 4 '16 at 18:10
  • 2
    $\begingroup$ As Joel David Hamkins said, this ordinal is not bounded by any function of cofinality. One can also prove that this ordinal is regular so it is a cardinal. I wonder if it can be that $2^{\alpha} < |F|$? $\endgroup$
    – nombre
    Mar 6 '16 at 13:32
  • 1
    $\begingroup$ @nombre It can’t be, because of the Erdős–Rado theorem; this is mentioned in my answer. $\endgroup$ Mar 10 '16 at 18:18

This parameter of a field does not equal any its common cardinal characteristic that I could think of, though it is related in several ways.

Let me first introduce some notation. Assume $F$ is an ordered field. As noted in the comments, if an ordinal embeds in $F$, it embeds in every interval $(a,b)$ of $F$, so we can simply put

$$o(F)=\min\{\alpha\in\mathrm{Ord}:\alpha\text{ does not embed in }F\}.$$

As also noted in the comments, $o(F)$ is a regular uncountable cardinal. We can further consider:

  • cardinality $|F|$

  • density $d(F)$ (= least cardinality of a dense subset), which also equals the weight of $F$ as an ordered topological space, and the cellularity of $F$ (maximal number of disjoint open intervals); these three invariants coincide for any bi-ordered group

  • cofinality $\def\cf\mathit{cf}\cf(F)$

These parameters satisfy

$$\cf(F)\le d(F)\le|F|\le2^{d(F)}.$$

Clearly, $\cf(F)$ embeds in $F$. On the other hand, an embedding of $\alpha$ in $F$ gives a family of $|\alpha|$ disjoint open intervals, thus

$$\tag{1}\cf(F)^+\le o(F)\le d(F)^+.$$

The Erdős–Rado theorem $(2^\kappa)^+\to(\kappa^+)^2_2$ implies that a linear order of size larger than $2^\kappa$ contains a well-ordered or inverse well-ordered subset of size $\kappa^+$, thus


where $o(F)^-=\kappa$ if $o(F)=\kappa^+$ is a successor cardinal, and $o(F)^-=o(F)$ otherwise.

Even better, let $D(F)$ be the Dedekind–MacNeille completion of $F$ (i.e., the set of Dedekind cuts of $F$, ordered by inclusion). The Erdős–Rado argument applies to $D(F)$, even though it is not a field. Since $F$ is dense in $D(F)$, any ordinal that embeds in $D(F)$ also embeds in $F$. Thus,


This appears to be essentially all one can say. Some examples:

  • $o(F)$ can be as large as permitted by (1). As explained in Joel’s answer, for any $\kappa$ and regular $\lambda\le\kappa$, there is a field $F$ of cofinality $\lambda$ such that $\kappa$ embeds in $F$; by taking its subfield generated by a copy of $\kappa$ and a cofinal sequence, we can assume $|F|=\kappa$, thus

$$\cf(F)=\lambda, \qquad |F|=d(F)=\kappa, \qquad o(F)=\kappa^+.$$

  • $o(F)$ can be as small as permitted by (2). Let $\kappa$ be a cardinal such that $\kappa=2^{<\kappa}$ (i.e., $\kappa=\lambda^+=2^\lambda$, or $\kappa$ is strong limit). By Corollary 2 in https://mathoverflow.net/a/188628, there is a field $F$ of size $|F|=2^\kappa$ with a dense subfield of size $\kappa$; by construction, we can also ensure $\kappa$ embeds in $F$. This makes

    $$d(F)=\kappa,\qquad o(F)=\kappa^+,\qquad |F|=2^\kappa.$$

    Now, let $K$ be the rational function field $K=F(x)$, where $x>F$. Then

    $$\cf(K)=\omega,\qquad o(K)=\kappa^+,\qquad |K|=d(K)=2^\kappa,$$

    using the following easily shown property:

Lemma: If $F$ is an ordered field, the rational function field $F(x)$ with $x>F$ satisfies $\cf(F(x))=\omega$, $|F(x)|=d(F(x))=|F|$, and $o(F(x))=o(F)$.

  • $\begingroup$ Thanks for this very instructive answer. I didn't know the Erdös-Rado theorem. $\endgroup$
    – nombre
    Mar 10 '16 at 18:36
  • $\begingroup$ This was quite helpful! Thanks Emil $\endgroup$
    – Haidar
    Mar 11 '16 at 0:12

Let me address merely the suggestion the OP makes in the comment: whether this ordinal can be specified from the cofinality of the field.

The answer is no, because any ordered field $F$ can be elementarily extended to a field with cofinality $\omega$, or indeed, to a field with any given regular cofinality. To have cofinality $\delta$, simply extend $\delta$ many times, making sure to put a new element on top each time, taking unions at limit stages.

$$F\prec F_1\prec F_2\prec\dots\prec F_\alpha\prec\dots\prec F_\delta$$

The resulting field will have the same cofinality as $\delta$, because the sequence of those points newly added on top at each stage will be cofinal in $F_\delta$. Since the initial field $F$ could have had very large embedded ordinals, which will still embed into the resulting field $F_\delta$, this shows that there are fields with very large embedded ordinals, as large as desired, which nevertheless have cofinality $\omega$, or any desired cofinality.

  • 2
    $\begingroup$ For a striking illustration of Joel’s observation, let $\mathbf{No}$ be the ordered field of surreal numbers, $a$ be an indeterminate where $a>\mathbf{No}$ and $\mathbf{No}(a)$ be the ordered simple transcendental extension of $\mathbf{No}$ generated by $a$ and $\mathbf{No}$. Although $\mathbf{No}(a)$ has cofinality $\omega$, it contains the entire class of ordinals. This ordered field, which is constructible in NBG, is discussed in the author’s JSL (2001: Proposition 3, p. 1240) for reasons unrelated to the question at hand. $\endgroup$ Mar 4 '16 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.