Proof that a well-ordering is $\omega$-saturated only if it is finite

I wanna prove that an infinite well-ordering is not $$\omega$$-saturated. How can I do this? Does the proof involve Lowenheim-Skolem theorem?

• The title and body of your question say different things. Nov 22, 2020 at 14:32
• I think I corrected the title.
– Jack
Nov 22, 2020 at 14:34
• Yep, that's better. Nov 22, 2020 at 14:35
• I mean, the type of arbitrarily long decreasing chain is an obvious giveaway. Nov 22, 2020 at 14:37
• Do you know the theorem that $\omega$-saturated models realize types in $\omega$-many free variables? Nov 22, 2020 at 14:38

3 Answers

As indicated in Noah's and Atticus's answers, and my comment, it's probably best to understand this fact as an application of the theorem that $$\omega$$-saturated models realize all consistent types in $$\omega$$-many free variables, so an $$\omega$$-saturated infinite linear order realizes the type of a countable descending sequence. But in this answer, I'll try to explain how you could answer the question without using the more general theorem.

Let $$(M;<)$$ be an infinite $$\omega$$-saturated linear order. We would like to show that $$M$$ is not a well-order. To do that, we would like to build an infinite descending chain in $$M$$. Let's try to do this by induction. We start by picking an element $$a_0$$. Then, having chosen $$a_n$$, we pick some $$a_{n+1}.

There's an obvious problem with this strategy: What ensures that there is some $$a_{n+1}? That is, if we accidentally pick $$a_n$$ to be the least element of the order, we get stuck. Saturation is no help: if $$a_n$$ is the least element of $$M$$, then $$M\models \forall x\, \lnot (x < a_n)$$, so the formula $$x is just inconsistent.

Ok, so instead let's pick our sequence by induction, making sure to never pick the least element. But now what ensures that there is some $$a_{n+1} which is not the least element? If $$a_n$$ has only one element below it in the order, we again get stuck. Clearly, we're in trouble if we ever pick an element $$a_n$$ with only finitely many elements below it in the order.

New idea: Make sure that each element of the sequence has infinitely many elements below it. So we start by picking an element $$a_0$$ with infinitely many elements below it. Then, having chosen $$a_n$$ with infinitely many elements below it, we pick some $$a_{n+1} such that $$a_{n+1}$$ has infinitely many elements below it. But what ensures that there is some $$a_{n+1} with infinitely many elements below it? Now saturation is comes into play: This condition amounts to realizing a certain type. Details after the break.

For each natural number $$n>0$$, let $$\varphi_n(x)$$ be the formula $$\exists y_1\dots\exists y_n\,(y_1 Let $$\Phi(x)$$ be the partial type $$\{\varphi_n(x)\mid n\in \mathbb{N}\}.$$ This partial type says "$$x$$ has infinitely many elements below it".

We construct a descending sequence $$(a_n)_{n\in \mathbb{N}}$$ from $$M$$ by induction, ensuring that each $$a_n$$ realizes $$\Phi(x)$$.

In the base case, we show that $$\Phi(x)$$ is consistent by compactness: Given any finite subset $$\{\varphi_n\mid 1\leq n\leq N\}$$, we can pick any $$N+1$$ elements of $$M$$, listed in increasing order $$b_1<\dots (here we use the fact that $$M$$ is infinite). Then taking $$x = b_{N+1}$$, the elements $$b_i$$ serve as witnesses for the variables $$y_i$$ in the formulas $$\varphi_n$$ with $$n\leq N$$. Since $$\Phi(x)$$ is consistent and $$M$$ is $$\omega$$-saturated, we can find $$a_0\in M$$ realizing $$\Phi(x)$$.

Now suppose by induction that we have picked $$a_n$$ realizing $$\Phi(x)$$. We show that the partial type $$\Phi(x)\cup \{x is consistent by compactness: Given any finite subset $$\{\varphi_n\mid 1\leq n\leq N\}$$, recall that $$a_n$$ satisfies $$\varphi_{N+1}$$. So we can find $$b_1<\dots in $$M$$. Then taking $$x = b_{N+1}$$, we have $$x < a_n$$, and the elements $$b_i$$ serve as witnesses for the variables $$y_i$$ in the formulas $$\varphi_n$$ with $$n\leq N$$. Since $$\Phi(x)\cup \{x is consistent and $$M$$ is $$\omega$$-saturated, we can find $$a_{n+1}\in M$$ realizing $$\Phi(x)\cup \{x.

A useful fact is that an $$\aleph_0$$-saturated structure realizes any (partial) type in $$\omega$$-many variables with parameters from a finite set. Ie, if $$\mathfrak{M}$$ is an $$\aleph_0$$-saturated $$\mathcal{L}$$-structure, $$A\subseteq M$$ is a finite subset, and $$\Sigma(v_i)_{i\in\omega}$$ is a finitely satisfiable set of $$\mathcal{L}_A$$ formulas in $$\omega$$-many free variables, then $$\Sigma(v_i)_{i\in\omega}$$ is realized in $$\mathfrak{M}$$. See eg here for a proof. (The analogous result holds for all infinite cardinals.)

This fact allows us to show your result. Let $$(M,\leqslant)$$ be an $$\aleph_0$$-saturated well-order, and consider the collection of formulas $$\Sigma(v_i)_{i\in\omega}=\{v_i>v_j\}_{i; in other words, $$\Sigma(v_i)_{i\in\omega}$$ expresses that the $$v_i$$ form a strictly descending chain. Since well orders are total orders, if $$M$$ is infinite then $$\Sigma(v_i)_{i\in\omega}$$ is finitely satisfiable, and hence by $$\aleph_0$$-saturation is satisfied by elements $$m_0>m_1>\dots$$ of $$M$$. But now the set $$\{m_i\}_{i\in\omega}$$ is a subset of $$M$$ without a least element, contradicting that $$<$$ is a well-order.

The class of well-orderings has a very nice property: it's exactly the class of structures which $$(i)$$ satisfy a particular first-order theory (the theory of linear orders) and $$(ii)$$ do not have a certain countable "bad configuration." Re: $$(ii)$$, the relevant "bad configuration" is:

An infinite descending sequence.

The "saturation slogan" is, "all possible configrations exist." Of course it's more complicated than that - how much saturation you have determines the size of the conigurations the slogan applies to - but the basic idea is sound. This suggests the following line of attack: show that in an infinite $$\omega$$-saturated linear order, the countable "bad configuration" above must occur. This will be a corollary of a more general fact:

Suppose $$\mathcal{M}$$ is $$\omega$$-saturated and $$p$$ is a finitely satisfiable $$\omega$$-type over $$\mathcal{M}$$. Then $$p$$ is realized in $$M$$.