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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 $\aleph_1$?

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    $\begingroup$ Is the weight of an ordered field just the weight of the topology on the underlying set of the field induced by the ordering, or is it something else? $\endgroup$ – Noah Schweber Nov 9 '14 at 16:47
  • $\begingroup$ Pending clarification of the definition of weight, would a real-closed field whose order structure is an $\eta_{\alpha+1}$-set of cardinality $\omega_{\alpha+1}$ work? $\endgroup$ – Avshalom Nov 9 '14 at 18:04
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    $\begingroup$ Steve Simpson claims in this FOM thread (personal.psu.edu/t20/fom/postings/9905/msg00071.html) that the field $\text{NO}(\kappa)$ of all surreal numbers of hereditary size at most $\kappa$ is a real-closed ordered field of size $2^\kappa$ and density $\kappa$. It seems to me that this isn't quite right, since $\text{No}(\kappa)$ would include all ordinals up to $\kappa^+$, and so couldn't have a dense set of size $\kappa$, but I believe that it will be fruitful to look at birth-initial segments of the surreals for an example. $\endgroup$ – Joel David Hamkins Nov 9 '14 at 18:51
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    $\begingroup$ The weight of an ordered field is just the weight of the interval topology on this field. It is equal to the smallest cardinality of an order dense subset. $\endgroup$ – Taras Banakh Nov 9 '14 at 19:59
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    $\begingroup$ Taras, you may write an answer and accept your own answer to make the question "closed". $\endgroup$ – Anton Klyachko Nov 9 '14 at 20:31
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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:

  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 $\kappa$.

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

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Browsing through MathOverflow I have found some answers to related questions, which shed light on my question too. It turned that my question has been considered in the literature. The most appropriate sources are: H.J. Keisler, Six classes of theories, J.Austral Math. Soc. 21 (1976), 257-266, A. Chernikov, I. Kaplan, S. Shelah: http://arxiv.org/abs/1205.3101, and A. Chernikov, S. Shelah: http://arxiv.org/abs/1308.3099. According to Keisler, the supremum of cardinalities of ordered fields containing a dense subset of size $\kappa$ is equal to $ded(\kappa)$ the supremum of cardinalities of linearly ordered spaces containing a dense subset of size $\le\kappa$. The cardinal $ded(\kappa)$ is contained in the interval $(\kappa,2^\kappa]$ and it is consistent that $ded(\kappa)<2^\kappa$ for any cardinal $\kappa$ of uncountable cofinality. On the other hand, by a recent result of Chernikov and Shelah (cited above), $2^\kappa\le ded(ded(ded(ded(\kappa))))$ (which is more than sufficient for my purposes).

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  • $\begingroup$ I got to this question because I realized that the claim I’m making in mathoverflow.net/a/188447 about $\kappa^+=2^\kappa$ is not obvious. Your claim that the supremum of cardinalities of ordered fields with a dense subset of size $\kappa$ is $\mathrm{ded}(\kappa)$ would solve the issue, but I do not understand how you came to it. There is no such statement in Keisler’s paper. His statement about the stability function of OF boils down to the rather trivial fact that $\mathrm{ded}(\kappa)$ is the sup of the numbers of Dedekind cuts on ordered fields of size $\kappa$. $\endgroup$ – Emil Jeřábek supports Monica Dec 1 '14 at 18:13
  • $\begingroup$ ... However, most Dedekind cuts on an OF cannot be realized in an ordered field extension where the original field is a dense subset. This only works for cuts satisfying the condition in mathoverflow.net/a/140962 . $\endgroup$ – Emil Jeřábek supports Monica Dec 1 '14 at 18:16
  • $\begingroup$ (I now changed the claim in the linked post to something easily verifiable: if $\kappa^{<\kappa}=\kappa$, there exists an OF of size $2^\kappa$ with a dense subfield of size $\kappa$.) $\endgroup$ – Emil Jeřábek supports Monica Dec 1 '14 at 19:36

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