# Linear orders with only short cuts

Here's a question from Shelah's book Classification Theory. Given an order $I$, we consider cuts of the form $(A,B)$ where for all $a \in A, b \in B$ we have $a < b$ and $A \cup B = I$. The cofinality of a cut $(A,B)$ is the pair $(\lambda, \kappa)$ where $\lambda$ is the cofinality of $A$ and $\kappa$ is the cofinality of $B$ in the reverse order.

The reader is asked to show in Exercise VII.1.7 that given any order $J$ with $|J| \leq \kappa$ there is an order $I$ with $J \subseteq I$ and $|I| = \kappa$ so that if $(A,B)$ is a cut of $I$ then the cofinality of $(A,B)$ is $(\aleph_{0}, \aleph_{0})$. In other words, given any order of size at most $\kappa$, it can be expanded to an order of size $\kappa$ so that every cut has countable cofinality.

Does anyone know how to show this? In the absence of any hypothesis on $\kappa$ it is difficult to see how one could kill off all uncountable cuts in an order of size $\kappa$ without increasing the cardinality.

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I don´t think the statement is true as it is. If $J=\omega_1+1$ then for any $I \supseteq J$ (of whatever cardinality) you can define a cut as: $x \in A$ iff $x \in I$ and $\exists \alpha \in \omega_1 (x<\alpha)$ and $B=I \setminus A$. This cut can't have cofinality $(\omega, \omega)$.
But this is not a "proper" cut if the $\omega_1$ sequence has a supremum in $I$. – François G. Dorais Sep 29 '11 at 23:32
Yes, this is a convincing counterexample. On a second reading, I realized that I misinterpreted the question. The point is not that the only cuts have countable cofinality, but that all cuts with cofinality $(\lambda, \kappa)$ with $\lambda$ and $\kappa$ uncountable actually have cofinality &(\lambda, \lambda)& - that is, all cuts with uncountable cofinality have the same cofinality in the lower cut and in the upper cut with the reverse order. – nick Sep 30 '11 at 23:10