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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 $F$?

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    $\begingroup$ This depends entirely on the field. What kind of an answer are you looking for? $\endgroup$ Mar 4 '16 at 16:11
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @nombre It can’t be, because of the Erdős–Rado theorem; this is mentioned in my answer. $\endgroup$ Mar 10 '16 at 18:18
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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

$$\tag{2}|F|\le2^{o(F)^-},$$

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,

$$|D(F)|\le2^{o(F)^-}.$$

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 $o(F(x))=o(F)$.

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  • $\begingroup$ 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
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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.

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    $\begingroup$ 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

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