If $X$ and $Y$ are two sets linearily ordered by $<$, $X$ is called cofinal in $Y$ if $X \subseteq Y$ and and for every $y \in Y$, there is a $x \in X$ with $y < x$.

If $M$ is some model and $\varphi(x)$ is a formula in the corresponding language, possibly with parameters from $M$, then $\varphi(M)$ denotes the subset of $M$ defined by $\varphi(x)$.

Let $L$ be a language containing a binary relation symbol $<$. Let $T$ be a theory in $L$ containing axioms which express that $<$ is a discrete linear order without a greatest element. Let

$\phantom{asdsdsdsd}$ $M_0 \prec M_1 \prec M_2 \prec ...$

be a countable elementary chain of models of $T$ such that each $M_i$ satisfies the following properties:

i) $M_i$ is $\aleph_1$-saturated

ii) card$(M_i)$ $<$ card$(M_{i+1})$

iii) for every $L$-formula $\varphi(x)$, possibly with parameters from $M_i$, $\varphi(M_i)$ is cofinal in $\varphi(M_{i+1})$ (i.e. each definable subset of $M_i$ is cofinal in the "corresponding" definable subset of $M_{i+1}$)

Here is the question: Is the limit $\bigcup_{i\in \mathbb{N}} M_i$ $\aleph_1$-saturated?


The answer is not necessarily.

Let me describe how to make a counterexample.

First, notice that your requirement that $\varphi(M_i)$ is cofinal in $\varphi(M_{i+1})$ is equivalent merely to the assertion that $M_i$ is cofinal in $M_{i+1}$, which is a special case if one uses a trivial formula. The reason is that the theory of discrete endless orders admits an elimination of quantifiers argument by which every assertion is equivalent to a Boolean combination of various assertions about specific finite distances from the parameters. It follows that $\varphi(M_i)$ either has a largest element, in which case this will also be largest in $M_{i+1}$, or else $\varphi(M_i)$ consists of a final segment of $M_i$, in which case it also will in $M_{i+1}$. So your requirement (iii) is asserting merely that $M_i$ is cofinal in $M_{i+1}$.

The quantifier elimination argument also shows that whenever one discrete order is a substructure of another in such a way to preserve the successor operation, it will therefore preserve finite distances and thus be an elementary substructure.

So, let build the counterexample like this. Start with any $\aleph_1$-saturated $M_0$. Fix a point in $M_0$ that we will call $0$, and consider the points of $M_0$ that are finitely far from $0$. This is the $\mathbb{Z}$-block around $0$ in $M_0$. Let $M_0\prec M_1$ be an elementary $\aleph_1$-saturated cofinal extension of $M_0$, of strictly larger size, which has new $\mathbb{Z}$-blocks just above the $0$-block of $M_0$, but below any other $\mathbb{Z}$-blocks of $M_0$. This is easy enough to arrange, since we can easily find saturated elementary extensions with such new elements, and if our extension happend to not be cofinal, we can simply chop it off at the cut determined by $M_0$ and still have an elementary extension, using the elimination of quantifiers result and the fact that the cofinality of $M_0$ must be uncountable since it itself is saturated.

Continuing in this way, we may build $$M_0\prec M_1\prec M_2\prec\cdots$$ as desired, satisfying all your properties, but such that at each stage, we add a new $\mathbb{Z}$ chain of elements in $M_{i+1}$ just above the $0$-chain of $M_0$, and below all other $\mathbb{Z}$-chains above $0$ in $M_i$.

It follows that the union $\bigcup_i M_i$ is not $\aleph_1$-saturated, since the cut determined by the $0$-chain in this union will have cofinality $\omega$ in the space of $\mathbb{Z}$-chains. More specifically, we can pick $x_i\in M_{i+1}$ in one of the newly added $\mathbb{Z}$-chains of $M_i$, so that $\langle x_i\mid i\in\omega\rangle$ is descending, but the only elements in the union model that are below every $x_i$ and above $0$ are finitely close to $0$. Thus, the type corresponding to squeezing a new $\mathbb{Z}$-chain in between $0$ and the $x_i$'s is not realized, and so the union model is not $\aleph_1$-saturated.

Meanwhile, there are other examples where the union model is $\aleph_1$-saturated. To build such an example, start with $M_0$, again $\aleph_1$-saturated. Fix $\omega$ many cuts in $M_0$, in order, in between $\mathbb{Z}$-chains of $M_0$, such that the cut determined at the supremum of these cuts has uncountable descending cofinality. Now make $$M_0\prec M_1\prec M_2\prec \cdots$$ by systematically adding a huge number of new points just in the $i^{th}$ cut at stage $i$. That is, the model $M_{i+1}$ is formed by adding to $M_i$ a huge number of points in the $i^{th}$ cut, and only adding points in that cut. These extensions will be elementary, by the quantifier elimination argument. They will be cofinal extensions, since we're never adding points on top. They will be saturated, provided that what we start with is saturated and what we add is saturated (note that the cut should have uncountable cofinality from at least one direction, and this will help to prove that the filling that cut with the new structure will be saturated). Finally, I claim that the union model is saturated, simply because the new elements in the successive $M_i$ are in totally different cuts of $M_0$ and do not interact in any way that is useful for defining a type, using the fact that the upper bounds of these new elements has uncountable cofinality.

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    $\begingroup$ The assertions in your second paragraph are not true in general. They only hold if the language contains nothing but the ordering (which is however fine for the counterexample). $\endgroup$ – Emil Jeřábek Jun 25 '14 at 22:21
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    $\begingroup$ Yes, I had meant that part to refer only to models with the order relation, which is all my examples require. But still, doesn't this fully answer the question? In some instances, I show, the union is not $\aleph_1$ saturated, and in other instances, it is. $\endgroup$ – Joel David Hamkins Jun 26 '14 at 0:55
  • $\begingroup$ Your counterexample completely answers my question. $\endgroup$ – user49449 Jun 26 '14 at 13:46

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