When substructures become elementary substructure

Question

It is known that given $M\subseteq N$ structure and a substructure such that $M\equiv N$, we do not necessarily have $M\prec N$.

Similarly we can have $M_0\subseteq M_1\subseteq M_2$ such that $M_0\equiv M_1\equiv M_2$ that are not elementary substructure to one another.

Is there a chain where this stops to be true? For example, if I have $\kappa$ that is measurable such that $(M_\mu\mid\mu\in(\kappa\cap Card))$ chain of substructures, all with the same theory $T$ (for convenient sake, let's fix some countable language), and (to have somewhat tame behavior) all are bounded subsets of $\kappa$, does there exists cardinals $\alpha<\beta<\kappa$ such that $M_\alpha\prec M_\beta$ (I would imagine that if 1 such pair exists, there will reflect downwards and we will have at least stationary chain of elementary substructures)?

I played a bit with the problem and I think that the answer is yes. Let $\kappa$ be measurable, and $\mathcal U$ be $\kappa$-complete normal non-principal ultrafilter on $\kappa$, and $j$ be the induced elementary embedding on it's ultrapower.

First let's assume that $|M_\lambda|\ge\lambda$ for $\cal U$-large set $X$, as if $|M_\lambda|<\lambda$ for $\cal U$-large set then it is trivial by normality and the fact that $\kappa$ is a strong limit.

Let $Y=\{\mu\mid \text{there exists $\alpha\in X$ such that }\mu=|M_\mu|\}$ and for $i\in Y$ let $Y_i$ be first model of size $i$, $Y$ satisfy "for all $\mu$ we have $\mu\in Y\iff (\mu\in X\land \mu\text{ is larger than every $X_i$ for $i<\mu$})$". Because $\kappa\in j(X)$, the ultrapower computers the $X_i$ for $i<\kappa$ correctly and the ultrapower thinks $X_{\kappa}\ge\kappa$, then $\kappa\in j(Y)$.

In summery we have $Y\subseteq \kappa$ a $\cal U$-large set such that $(Y_i\mid i\in Y)$ is a chain of elementary equivalent substructures such that $|Y_i|=i$.

My next step was to try to reflect downwards some properties on $Y$ to somehow get a fixed point on "approximate"-Skolem hull operation. Basically fix some Mahlo, or limit of Mahlos, $\nu\in Y$ and look at $(Hull_{Y_{\nu}}(Y_\mu)\mid \mu\in Y\cap\nu)$ and hope it will somehow lend on some $Y_\mu\subseteq Y_\nu$, but I couldn't really find a way to do this.

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Does what I wrote can get me somewhere? If it is indeed true, is there some tighter conditions (as my idea was basically to reflect down to $Y$ all of the LCA I want, so it is obviously an overkill if this method indeed works)?

Answer 1

Nothing like this ever holds, even for very simple theories.

Take for example the language consisting of a single binary relation symbol $E$, and the $\{E\}$-theory $T$ that says that $E$ is an equivalence relation with infinitely many classes, all but one of which have exactly two elements and one of which has exactly one element. Given an ordinal $\alpha$, let $\mathcal{M}_\alpha$ be the $\{E\}$-structure with underlying set $$(\alpha\times 2)\cup\{(\alpha,0)\}$$ where $E$ is interpreted as agreement on the left coordinate. Then for all infinite ordinals $\alpha<\beta$ we have $\mathcal{M}_\alpha\subseteq\mathcal{M}_\beta$ and $\mathcal{M}_\alpha\equiv\mathcal{M}_\beta$ (each is a model of $T$, which is complete), but $\mathcal{M}_\alpha\not\preccurlyeq\mathcal{M}_\beta$ (they disagree about which is the "special" $E$-class).

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That said, things get much more interesting if we replace "is an elementary substructure of" with "can be elementarily embedded in." Vopenka's principle says that every proper class of structures in the same language contains one that elementarily embeds into another. While these days generally believed to be consistent, VP is a very strong large cardinal axiom; its full strength is not immediately obvious but it is easy to see that it fails assuming the axiom of constructibility.

Answer 2

As Noah point out, without controlling the domain of the models, there is a little chance to ever have a situation like what you describe.

If we can have some restrictions on the domains we can construct situations with that property, for example:

Let $(A_i\mid i\le\lambda)$ be a continuous chain of substructures whose domain is a set of Ordinals over a fixed language (assume $|L|\le|A_0|\ge\aleph_0$) such that $|A_\lambda|$ is uncountable regular. (Note that it forces $|A_\lambda|=\lambda$)

Taking your idea of taking Hulls: let $(B_i\mid i <\lambda)$ defined as $B_i=Hull_{A_\lambda}(A_i)$, then we have for all $i<\lambda$ some $i<j<\lambda$ such that $A_i\subseteq B_i\subsetneq A_j$.

The reason $B_i\neq A_j$ is from cardinality arguments, and for $B_i\subseteq A_j$, $|B_i|=|A_i|$ so $B_i$ is bounded in $A_\lambda$ (as set of ordinals), by $(A_i)$ being continuous there is some $A_j$ that covers $\sup B_i$.

Let $f(0)$ be the minimal such $j$, for $A_0$, $f(1)$ be the minimal such $j$ for $A_{f(0)}$ and so on.

Let $\kappa$ be the limit of $f(i)$, because $(A_i)$ is continuous we have $\bigcup_{i<\kappa} A_i=\bigcup_{i<\omega}A_{f(i)}=A_\kappa$ but $A_{f(i)}\subseteq B_{f(i)}\subseteq A_{f(i+1)}$ so $A_\kappa=\bigcup_{i\in\omega}B_{f(i)}$, so $A_\kappa\prec A_\lambda$.

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The requirement of being continuous is the important requirement, it forbid to have new definable elements at every stage, only (potentially) the successors. Regularity is just a tool to guarantee that we don't accidentally define our $A_\lambda$.

In the situation of measurable cardinal, even if we require $M_\mu\subseteq \mu$ like you suggested in the comments, we don't have reason to believe our collection will have some continuous subcollection.

In fact, if $\cal U$ is a normal $\kappa$ complete ultrafilter on $\kappa$, $Y\in\cal U$ and $Y$ is a set of cardinals such that $Y_i\subseteq i$, then we can reflect being regular and get some ${\cal U}\ni Y'\subseteq Y$ set of regular cardinals, in particular it is very discontinuous.

But we do have a different tool, normality!

Let's add the assumption that $Y\subseteq \bigcup Y_i$. Note that we can assume that for all $i<j<\kappa$ we have $Y_j\cap\sup Y_i=Y_i$ (this is from the fact $\kappa$ is ineffable and the added assumption)

Let $\phi(x;y)$ and $z\in Y_{\min Y}^n$, define $g_{\phi,z}(\beta)$ be the minimal $u$ such that $Y_\beta\models \phi(u;z)$ (otherwise let it be $\min Y_{\min Y}$), this is a regressive function, so there exists $\cal U$-big set $U_{\phi,z}$ such $g_{\phi,z}$ is constant on. Let $U_0=\bigcap_{\phi\in WFF, z\in Y_{\min Y}^{<\omega}}U_{\phi,z}$, this is a $\cal U$-big set.

If $\min Y\in U_0$ we are done.

Otherwise build $U_1$ using $\min U_0$ instead of $\min Y$ and so on for finite stages and let $U_\omega$ be the same but starting from $\min\bigcap U_i$

It is simple to verify that $\min U_\omega=\min\bigcap U_i$ so we are done!

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As a last note, we can actually state the above as a particular version of Vopenka's principle Noah mentioned:

Let $\kappa$ be measurable and $(A_i\mid i\in\kappa\cap Card)$ be a collection of structures of some fixed small language such that $|A_i|\le i$, and let $h_i:A_i\to i$ be any list of encodings, then there exists $i<j$ such that $h_i''A_i\prec h_j''A_j$.

This is true because there exists only $2$ to the power of the language many theories, so fix some normal $\kappa$ complete ultrafilter $\cal V$ on $\kappa$ and you will find a $\cal V$-big set of indices with the same theory, and from there everything will go smoothly.