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 $\kappa$.

**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:

$|F|=\kappa$,

$(0,1)_F$ contains $\kappa$ disjoint (nondegenerate) intervals,

$F$ has an increasing cofinal subsequence $\{u_\alpha:\alpha<\lambda\}$,

$F$ 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 $\kappa$.

**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 $\kappa^\lambda\le(2^\nu)^\lambda=2^\nu$.

acceptyour own answer to make the question "closed". $\endgroup$ – Anton Klyachko Nov 9 '14 at 20:31