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 $(\omega,\kappa)$.
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Is there an ordered field all of whose unfilled Dedekind cuts have mismatched cofinality types?
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Does it change things to consider only models $\mathbb{R}^\ast$ of nonstandard analysis, where the transfer principle holds?
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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.
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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!
Best Answer
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.