Is it true (in ZFC) that for any regular infinite cardinal $\kappa$ there exists an ordered field of weight $\kappa$ and cardinality $2^\kappa$ (or at least $>\kappa$)?


The field of real numbers shows that for $\kappa=\aleph\_0$ the answer is trivially "yes".
It seems that this question has an affirmative answer in the Constructive Universe.
Is it true in ZFC?


Or the converse is true: there is a model of ZFC in which all ordered fields of weight $\aleph\_1$ have cardinality $\aleph\_1$?



Possible cardinality and weight of an ordered field
On the one hand, if $K$ is an ordered field of weight $\kappa$, then trivially


$$|K|\le\mathrm{ded}(\kappa):=\sup\bigl\{|L|:\text{$L$ is a linear order with dense subset of size $\kappa$}\bigr\}\le2^\kappa.$$


On the other hand, we have the following lower bounds, which in particular show that there is always an OF of weight $\kappa$ and cardinality strictly larger than $\kappa$.



> 
> **Proposition 1:** If $\lambda$ is the least cardinal such that $\kappa^\lambda>\kappa$, there exists an ordered field of cardinality $\kappa^\lambda$ with a dense subfield of cardinality $\kappa$.
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> 
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**Proof:** Note that the condition implies that $\lambda$ is regular, and $\lambda\le\kappa$.


First, we construct an increasing sequence of fields $\{F\_\alpha:\alpha\le\lambda\}$ of cardinality $\kappa$ as follows. Let $F\_0$ be an OF of cardinality $\kappa$ such that $(0,1)$ contains $\kappa$ disjoint nonempty intervals. (Incidentally, it’s easy to show that an OF of weight $\kappa$ always contains $\kappa$ disjoint intervals, but we will not need this.) If $\alpha$ is limit, we put


$$F\_\alpha=\bigcup\_{\beta<\alpha}F\_\beta.$$


For the successor step, since $\kappa^{<\lambda}=\kappa$, there are only $\kappa$ pairs of subsets $A,B\subseteq F\_\alpha$ such that $A<B$, and $|A|,|B|<\lambda$. Thus, using the compactness theorem, there exists an extension $F\_{\alpha+1}\supseteq F\_\alpha$ of size $\kappa$ such that for every such $A,B$, there is an element $c\in F\_{\alpha+1}$ with $A<c<B$, and moreover, there is an element $u\in F\_{\alpha+1}$ such that $u>F\_{\alpha}$.


Using the regularity of $\lambda$, the field $F:=F\_\lambda$ thus constructed satisfies:


1. $|F|=\kappa$,
2. $(0,1)\_F$ contains $\kappa$ disjoint (nondegenerate) intervals,
3. $F$ has an increasing cofinal subsequence $\{u\_\alpha:\alpha<\lambda\}$,
4. $F$ has the $\eta\_\alpha$ property for $\aleph\_\alpha=\lambda$, that is, if $A,B\subseteq F$ are such that $A<B$ and $|A|,|B|<\lambda$, there is $c\in F$ such that $A<c<B$.


Let $\hat F$ be the Scott completion of $F$, which is the largest ordered field extension of $F$ in which $F$ is dense. It suffices to show that $|\hat F|\ge\kappa^\lambda$, i.e., in the terminology of <https://mathoverflow.net/a/140962>, that $F$ has at least $\kappa^\lambda$ good cuts.


We can construct a tree $\{I\_t:t\in\kappa^{<\lambda}\}$ of nondegenerate intervals $I\_t=[a\_t,b\_t]$ so that:


* If $s$ properly extends $t$, then $a\_t<a\_s<b\_s<b\_t$.
* If $t$ and $s$ are incomparable, $I\_t\cap I\_s=\varnothing$.
* If $\mathrm{dom}(t)=\alpha<\lambda$, then $b\_t-a\_t<1/u\_\alpha$.


We build the tree by induction on $\mathrm{dom}(t)$. For $I\_\varnothing$, we can take any interval shorter than $1/u\_0$. For the successor step, if $I\_t$ has already been constructed, where $\mathrm{dom}(t)=\alpha$, we take the sequence of intervals from (2), scale it down into a subinterval of $I\_t$ shorter than $1/u\_{\alpha+1}$, and call it $\{I\_{t\_\smile\beta}:\beta<\kappa\}$. Finally, if $\alpha$ is limit, then
$$A:=\{a\_{t\restriction\beta}:\beta<\alpha\}< B:=\{b\_{t\restriction\beta}:\beta<\alpha\}$$
have size $<\lambda$, hence applying (4) twice, we can find $I\_t=[a\_t,b\_t]$ so that $A<a\_t<b\_t<B$.


Now, the properties of the tree ensure that for any $\tau\in\kappa^\lambda$, the sets
$$A\_\tau=\bigcup\_{\alpha<\lambda}(-\infty,a\_{\tau\restriction\alpha}],\qquad B\_\tau=\bigcup\_{\alpha<\lambda}[b\_{\tau\restriction\alpha},+\infty)$$
form a good cut, and $(A\_\tau,B\_\tau)\ne(A\_{\tau'},B\_{\tau'})$ for $\tau\ne\tau'$.


Note that if we slightly modify the construction of $F\_\alpha$ so that we also take a real closure on each step, then $F$ becomes real closed, in which case $\hat F$ is also real closed.



> 
> **Corollary 2:** If $\nu$ is the least cardinal such that $2^\nu>\kappa$, there exists an ordered (real-closed) field of cardinality $2^\nu$ with a dense subfield of cardinality $\kappa$.
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**Proof:** Put $\mu=2^{<\nu}\le\kappa$ and $\lambda=\mathrm{cf}(\nu)$. An exercise in cardinal arithmetic shows that $\mu^{<\lambda}=\mu$ and $\mu^\lambda=2^\nu$, hence there exists a RCF of size $2^\nu$ with a dense subset of size $\mu$, which we can enlarge to $\kappa$.


Note that for a given $\kappa$, the bound in the Corollary is better than in the Proposition: clearly $\lambda\le\nu$, hence $\kappa^\lambda\le(2^\nu)^\lambda=2^\nu$.


