Direct proof of a strongly $\lambda$-homogeneous elementary extension

Question

A structure $M$ is strongly $\lambda$-homogeneous if a partial elementary map $A \to M$ for $A \subseteq M$ of cardinality less than $\lambda$ extends to an automorphism of $M$. For a fixed cardinal $\lambda$, every structure has an elementary extension that is strongly $\lambda$-homogeneous.

The proof of the existence of strongly $\lambda$-homogeneous elementary extensions goes by way of $\lambda$-bigness in the sense of (longer) Hodges. Here, a structure $M$ is $\lambda$-big if $(M, \bar a)$ is resplendent for any tuple $\bar a \in M$ of length less than $\lambda$, and a structure $N$ is resplendent if for any theory $T$ in an expanded language consistent with that of $N$, some expansion of $N$ models $T$. That every structure has an elementary extension that is $\lambda$-big can be proved easily, but the proof requires careful bookkeeping and counting.

Can the existence of strongly $\lambda$-homogeneous elementary extensions be proved more directly and elementarily (from ZFC)? In case it matters, I'm interested in the case where $\lambda = \aleph_0$.

Answer 1

The other standard way to get strongly $\kappa$-homogeneous elementary extensions in ZFC is by way of the theory of special models. A special model $A$ of cardinality $\lambda$ is one that can be written as a union $A = \bigcup_{\kappa<\lambda}A_\kappa$ of an elementary chain indexed by the cardinals less than $\lambda$, such that each $A_\kappa$ is $\kappa^+$-saturated.

This is also explained in Hodges's Model Theory, Section 10.4. The relevant results are Theorem 10.4.2(b):

Let $A$ be an infinite $L$-structure and $\lambda$ a strong limit number $>|A|+|L|$. Then $A$ has a special elementary extension of cardinality $\lambda$.

...and Corollary 10.4.6. Unfortunately, there's a typo here (which is apparent from reading the proof). Corollary 10.4.6 should read:

If $A$ is special of cardinality $\lambda$, then $A$ is strongly $\text{cf}(\lambda)$-homogeneous.

So to get a strongly $\kappa$-homogeneous elementary extension of a model $A$, you find a strong limit cardinal $\lambda>|A|+|L|$ with $\text{cf}(\lambda)\geq \kappa$, for example: $\lambda = \beth_{\kappa^+}(|A|+|L|)$, and build a special model of cardinality $\lambda$.

You can also read about special models in Section 6.1 of A Course in Model Theory by Tent and Ziegler, where the relevant results are Corollary 6.1.3 and Theorem 6.1.6.

The construction of special models is actually fairly elementary - it's not too much harder than the construction of $\kappa$-saturated models. But maybe you don't find it "direct". It might be possible to construct strongly $\kappa$-homogeneous models by a direct transfinite induction argument - in fact I remember working out such an argument many years ago (of course, I'm not at all sure now that it was correct). But any such argument will almost certainly involve much more bookkeeping than the argument via special models. The idea was to add the new automorphisms to the language as function symbols, to make sure their existence is preserved under extensions. But then one has to think about extensions which are elementary in the original language, which also satisfy axioms about the function symbols in the expanded language... it's a bit of a mess.

Answer 2

The following answer is based on the proof idea outlined in Alex Kruckman's answer. It is not a slick or elegant proof, but I think it fits the bill for direct, and I think I managed to do the job without cumbersome amounts of bookkeeping. There is definitely a cumbersome amount of details I have omitted from the transfinite induction, however.

Lemma 1. Let $M$ be a model over the language $L$, and let $\{f_i\}_{i \in I}$ be a family of automorphisms of $M$. Moreover, let $h \colon A \to M$ be a partial elementary morphism from a subset of $M$ to $M$, and $b \in M$. Then, there is an elementary extension $M'$ of $M$ such that i) $h$ extends to an elementary morphism $A \cup \{b\} \to M'$, and ii) each $f_i$ extends to an automorphism of $M'$.

Proof. Consider the language $L'$ given by: The symbols in $L$, plus a unary function symbol for each $f_i$, plus a constant symbol $\bar a$ for each $a \in M$, plus a constant symbol $c$. Next, consider the theory $T'$ given by:

• The $L$-theory of $M$,
• Axioms stating that each $f_i$ is surjective and preserves all function and predicate symbols in $L$ (plus equality),
• Axioms saying that the type of $c$ over $\{\overline{h(a)}\}_{a \in A}$ is the same as the type of $\bar b$ over $\{\bar a\}_{a \in A}$.

By compactness, this theory is consistent, as $M$ itself with an appropriate interpretation of the constant $c$ will be a model of any finite subset of this theory. A model of $T'$ will furnish the desired model $M'$. $\square$

Lemma 2. Under the hypotheses of Lemma 1, there is an elementary extension $M''$ of $M$ such that i) $h$ extends to an elementary morphism $M \to M''$, and ii) each $f_i$ extends to an automorphism of $M''$.

Proof. Enumerate the elements of $M$, and iterate the construction of Lemma 1, at step $\alpha$ extending $h$ to the $\alpha$-th element of $M$. $\square$

Lemma 3. Under the hypotheses of Lemma 1, there is an elementary extension $M'''$ of $M$ such that i) $h$ extends to an automorphism of $M'''$, and ii) each $f_i$ extends to an automorphism of $M'''$.

Proof. Iterate the construction of Lemma 2 $\omega$-many times, alternating between extending $h$ and its inverse, and let $M'''$ be the union of the resulting models.

More precisely, set $M_0 = M$ and $h_0 = h$. For $n$ even, apply Lemma 2 to $M_n$ and $h_n$ to obtain $M_{n+1}$ and $h_{n+1} \colon M_n \to M_{n+1}$. For $n$ odd, apply Lemma 2 to $M_n$ and $h_n^{-1}$, to obtain $M_{n+1}$ and $h_{n+1}^{-1} \colon M_n \to M_{n+1}$. Then, set $M'''$ to be the union of all $M_n$, and let $f \colon M''' \to M'''$ be the function obtained by gluing all $h_n$. The even steps ensure that $f$ is defined on all of $M'''$, and the odd steps ensure that $f$ is surjective. $\square$

Now we have shown that we may add a single automorphism. It remains to add many of them.

Lemma 4. Let $M$ be a model. There is an elementary extension $M^*$ of $M$ such that each partial elementary morphism from a subset of $M$ to $M$ extends to an automorphism of $M^*$.

Proof. Enumerate the partial elementary morphisms $h_\alpha \colon A \to M$, $A \subseteq M$. Then, construct $M^*$ by transfinite recursion. First, set $M_0 = M$. Then, at a successor step $\alpha+1$, apply Lemma 3 to extend $h_\alpha$ to an automorphism $f_\alpha$ of $M_{\alpha+1}$, while ensuring that the $f_\beta$ remain automorphisms for $\beta < \alpha$. At limit stages, take unions. $\square$

Theorem 5. Let $M$ be a model, and $\kappa$ an infinite cardinal. Then, there is an elementary extension $\bar M$ of $M$ such that any partial elementary morphism of $\bar M$ whose domain has less than$\kappa$ elements extends to an automorphism of $M$.

Proof. Iterate the construction of Lemma 4 $\lambda$-many times, where $\mathop{\mathrm{cof}}\lambda \geq \kappa$; for example, $\lambda = \kappa^+$. Let $\{M_\alpha\}_{\alpha < \lambda}$ be the resulting chain of models, and $\bar M = \bigcup_{\alpha < \lambda} M_\alpha$.

Given a partial automorphism $h \colon A \to \bar M$ with $\#A < \kappa$, we will have $\#(A \cup h(A)) < \kappa$, and hence $A \cup h(A)$ will be contained in $M_\alpha$ for some $\alpha < \lambda$. Thus, $h$ extends to an automorphism of $M_{\alpha + 1}$, and will then extend to compatible automorphisms of $M_\beta$ for $\beta \geq \alpha+1$, which in turn induce a natural automorphism of $\bar M$. $\square$

Thanks to Alex Kruckman for pointing out a gap in my proof of Lemma 3, as well as a typo.