# What is the spectrum of possible cofinality types for cuts in an ordered field? Or in a model of the hyperreals? Or in a nonstandard model of arithmetic?

I am interested to know the full range of possibilities for the cofinality type of cuts in an ordered field and in other structures, such as nonstandard models of arithmetic.

Definitions. Specifically, in any linear order $\langle L,\leq\rangle$, a cut is a partition $L=A\sqcup B$, such that every element of $A$ is below every element of $B$. The cut is bounded if both $A$ and $B$ are nonempty. The cut is filled if there is a point $a\in L$ that is strictly larger than every point in $A$ and $\leq$ every point in $B$. Otherwise, the cut is unfilled. A Dedekind cut is a bounded cut for which $A$ has no largest member. Thus, the rational line $\langle\mathbb{Q},\leq\rangle$ has filled Dedekind cuts at every rational number and unfilled Dedekind cuts at every irrational number. The cofinality type of an unfilled Dedekind cut $(A,B)$ is $(\kappa,\lambda)$, where $\kappa$ is the length of the shortest increasing sequence unbounded in $A$ and $\lambda$ is the length of the shortest descending sequence unbounded below in $B$. If a cut is filled by a point $a$, then the cofinality type is $(\kappa,\lambda)$, where $\kappa$ is the shortest increasing sequence converging to $a$ and $\lambda$ is the shortest decreasing sequence converging to $a$. Let us say that a cofinality type $(\kappa,\lambda)$ is mismatched if $\kappa\neq\lambda$. The cofinality of an order is the shortest length sequence unbounded in the order. One could also define the downward cofinality to be the shortest length decreasing sequence unbounded below in the order.

In an ordered field $F$, the cofinality of any filled cut is $(\kappa,\kappa)$, where $\kappa$ is the cofinality of the field, since by inverting an unbounded increasing $\kappa$-sequence we produce a descreasing $\kappa$-sequence converging to $0$. The negation of this sequence converges to $0$ from below, and so by translation every point has $\kappa$-sequences from above and below, making all the filled cuts have the same matched cofinality type $(\kappa,\kappa)$.

But what of the unfilled cuts? For example, in a model $\mathbb{R}^\ast$ of nonstandard analysis, the standard cut, determined by the copy of $\mathbb{R}$ inside $\mathbb{R}^\ast$, has lower cofinality $\omega$, since the standard integers are unbounded in it from below, and a simple compactness argument shows that one can arrange that the upper cofinality of this cut is any desired regular cardinal $\kappa$. In particular, if $\kappa$ is uncountable, this would be an unfilled cut whose cofinality type is mismatched $(\omega,\kappa)$.

1. Is there an ordered field all of whose unfilled Dedekind cuts have mismatched cofinality types?

2. Does it change things to consider only models $\mathbb{R}^\ast$ of nonstandard analysis, where the transfer principle holds?

3. What is the full spectrum of possibility for the cofinality types of Dedekind cuts in an ordered field? Please provide examples illustrating the range of what can happen or theorems limiting that range.

4. What is the situation for cuts in nonstandard models of PA?

It is not difficult to construct $\omega_1$-like models of PA, which are uncountable models all of whose initial segments are countable, and all bounded cuts in such a model have type $(\omega,\omega)$. Also, a simple compactness argument can ensure that the standard cut has type $(\omega,\kappa)$ for any regular cardinal $\kappa$, and also one can ensure that the cofinality of the model is any desired such $\kappa$.

I believe that by using order-indiscernible elements, one can arrange that the cofinality of a model of nonstandard analysis is different from the upper cofinality of the standard cut in it. That is, it seems possible that the standard cut can have type $(\omega,\kappa)$ while the cofinality of the model is some other $\delta$. But I am unsure how to use those methods to control all the unfilled cuts of the model.

The Sikorski theorem mentioned in this answer by Ali Enayat seems relevant, and perhaps that paper provides tools to answer some of the questions. That theorem provides ordered fields of size and cofinality $\kappa$, with the property that every $(\kappa,\lambda)$ or $(\lambda,\kappa)$ cut is filled, for any $\lambda$.

My motivation for asking the question is to provide a sharper version of this MO question. A positive answer to question 1 above would produce an ordered field that is not complete, but which nevertheless exhibits the nested interval property (all descending transfinite sequences of closed bounded intervals have nonempty intersection). The point is that if the unfilled cuts all have mismatched cofinalities, then any descending sequence of closed intervals straddling the cut will have its endpoints stabilize on one side or the other, and consequently have nonempty intersection. But perhaps this is just too much to hope for!

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I'm guessing whoever down-voted meant to up-vote and hit down accidentally. Anyways, by "upper cofinality" and "lower cofinality" are you referring to $\kappa$ and $\lambda$, respectively, in a cut of cofinality $(\kappa,\lambda)$? –  Amit Kumar Gupta Aug 10 '11 at 19:47
Never mind, that's clearly what you mean. –  Amit Kumar Gupta Aug 10 '11 at 19:48
As far as I understand you are also interested on "an ordered field that is not complete, but which nevertheless exhibits the nested interval property". I will suggest you take a look at the paper "Quite Complete real closed fields" by Shelah. –  boumol Aug 11 '11 at 1:48
Shelah's "symmetrically closed" fields are precisely those in which all unfilled cuts have mismatched cofinality types, in my terminology. He describes a "symmetric closure" process by which any field can be symmetrically closed, by systematically filling the symmetric cuts, and proving that this process terminates transfinitely. –  Joel David Hamkins Aug 11 '11 at 2:10
Shelah's arxiv article: arxiv.org/PS_cache/math/pdf/0112/0112212v4.pdf –  Joel David Hamkins Aug 11 '11 at 2:10

For any $(\kappa,\lambda)$ it's easy to construct a real closed field with a $(\kappa,\lambda)$-cut. Let $L$ be the linear order with a increasing $\kappa$-chain followed by a decreasing $\lambda$-chain, i.e., $\kappa+\lambda^*$. Let $F$ be the field ${\mathbb Q}(X_l:l\in L)$ where $(X_l:l\in L)$ are algebraically independent. Order $F$ so that each $X_l$ is positive infinite and every power of $X_l^n < X_j$ whenever $l< j$ and let $R$ be the unique real closure of $F$ compatible with the ordering.

The cut of things above the $\kappa$-chain but below the descending $\lambda$-chain is unfilled.

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Thanks very much, Dave. Can one understand the full spectrum of cofinality types of cuts in your model? That is, we seem to get $(\kappa,\lambda)$ and $(\lambda,\kappa)$ by negating, but we must also have $(\omega,\delta)$ for some $\delta$ because of the standard cut, and hence also $(\delta,\omega)$. Can we control it? –  Joel David Hamkins Aug 11 '11 at 21:28
In this model the standard cut is still $(\aleph_0,\aleph_0)$. There is no positive infinite element that's below all of the positive infinite elements in ${\mathbb Q}(X_0)$. Similarly if $\mu<\kappa$ you could get a $(\mu,\aleph_0)$ cut by looking at the cut above all the $X_\alpha$, $\alpha<\mu$ but below $X_{\mu}^{1/n}$ for all $n$. On the other hand there is nothing special about my choice of linear order $L$. You could pick $L$ with lots of different co-finalities of cuts and then you would realize them in the real closure. –  Dave Marker Aug 11 '11 at 22:08
I see; yes, that is quite flexible. Thanks very much! –  Joel David Hamkins Aug 11 '11 at 22:44

Theorem 1. For every pair of infinite cardinals $(\kappa$,$\lambda)$, there is a model $M$ of $PA$ (Peano arithmetic) that has an initial segment $I$ such that $cf(I) = \kappa$ and $dcf(M$ \ $I) =\lambda$, where $dcf$ is "downward cofinality".

The model $M$ can be constructed using the McDowell-Specker-Gaifman machinery of "minimal types"; which an be thought of as Ramsey ultrafilters over the Boolean algebra of the parametrically definable subsets of an ambient model of $PA$. The rough idea is as follows: start with a countable model $M_{0}$ of $PA$ and use the McDowell-Specker theorem to build an elementary end extension $M_{\kappa}$ of $M_0$ such that $cf(M_{\kappa})=\kappa$. Then build a minimal type/Ramsey ultrafilter $\cal{U}$ over $M_{\kappa}$, and let $M$ be $L$-th iterated ultrapower of $M_{\kappa}$ modulo $\cal{U}$, where $L$ is the REVERSE of $\lambda$. Then the desired $I$ is $M_{\kappa}$.

Corollary. For every pair of infinite cardinals $\kappa$ and $\lambda$, there is a real closed field $F$ that has a Dedekind cut $(J,K)$ such that $cf(J) = \kappa$ and $dcf(K)=\lambda$.

Explanation: Given a model $M$ of $PA$, $M$ can define the real closure $F$ of itself (by using the same arithmetical recipe that defines the field of algebraic real numbers within the standard model of arithmetic). Since every positive element of $F$ is less than one away from some member of $M$, no gap of $M$ is filled by an element of $F$. Hence given $I$ as in the theorem, the cut $J$ of $F$ defined as the set of all $x \in F$ such that $x < i$ for some $i \in I$ does the job.

Let me also point out another results which shows that a real closed field constructed as above as the real closure of a model of $PA$ always has a "matched cut". This follows from the following result of Shelah which also appears as Theorem 11.1.1 (p.281) of the Kossak-Schmerl text on models of $PA$.

Theorem 2. For every nonstandard model $M$ of $PA$ there is an infinite cardinal $\kappa$, and there is an initial segment $I$ of $M$ such that $cf(I) = \kappa = dcf(M$ \ $I)$.

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The answer to your question 1 is known to be a positive one. This existence was proved by Shelah in the paper "Quite Complete real closed fields".

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In the 1980's, Mike Canjar proved some results of the sort you're asking about for the special class of models $\mathbb R^*$ that arise as ultrapowers of $\mathbb R$ with respect to nonprincipal ultrafilters on $\omega$ (which I'll call simply "ultrapowers" here).

• If you start with a model of CH and increase the continuum by adding a lot of Cohen reals, then there are ultrapowers with any two prescribed reasonable cardinals as the cofinality of the model and the coinitiality of the cut at the top of the standard part. Here "reasonable" means uncountable, regular, and not greater than the cardinal of the continuum.

• If you again start with a model of CH and increase the continuum by random forcing, then any ultrapower has cofinality $\omega_1$ (because the forcing is $\omega^\omega$-bounding), but you still get all reasonable cardinalities as the coinitiality of the cut at the top of the standard part of ultrapowers.

• It is provable in ZFC that there is an ultrapower in which both the cofinality of the model and the coinitiality of the cut at the top of the standard part are equal to the cofinality of the dominating number.

The relevant papers are "Countable ultraproducts without CH" (Ann. Pure Appl. Logic 37 (1988) 1-79) and "Cofinalities of countable ultraproducts: the existence theorem" (Notre Dame J. Formal Logic 30 (1989) 539-542). If I remember correctly, the first of these results was obtained independently by Judy Roitman at about the same time as Canjar's thesis, which contained the first two results that I quoted.

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This is not an answer, but it's too long for a comment. You can optimize away the distinction between filled and unfilled cuts as follows. Define a cut to be a pair of subsets $A$ and $B$ such that:

• $A$ is lower rounded: $x \in A \iff \exists y \in A \,.\, x < y$
• $B$ is upper rounded: $y \in B \iff \exists x \in B \,.\, x < y$
• $A$ and $B$ are disjoint
• $A$ and $B$ are located: $x < y \implies x \in A \lor y \in B$
• optionally, $A$ and $B$ are bounded: $\exists x \in L \,.\, x \in A$ and $\exists y \in L \,.\, y \in B$.

In essence locatedness makes sure that we "miss out at most one point". We can still define a cut to be filled as one that has a point between $A$ and $B$. But now the cofinalities are all defined the same way, i.e., as cardinalities of shortest unbounded increasing (decreasing) sequences in the lower (upper) cut.

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Sure, this amounts to the same thing. –  Joel David Hamkins Sep 5 '14 at 10:46
My hidden agenda is to get you to use a definition that works constructively, of course. –  Andrej Bauer Sep 5 '14 at 14:44