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 $F$?



Which ordinals can be embedded into an ordered field?
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:



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> **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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