This is a followup question to Sentence equivalent to $\bigwedge_{i=1}^ \infty \sigma_i$ without using infinite conjunctions. We have a language $$\mathcal{L}=\{R\}$$ (with equality) where $$R$$ is a binary relation symbol. Let $$\tau$$ be the sentence saying $$R$$ is reflexive, symmetric, and transitive. For each $$k \geq 1$$, let $$\sigma_k$$ be the sentence saying there is exactly one equivalence class of size $$k$$.

#1: If $$\mathcal{A}, \mathcal{B}$$ are $$\mathcal{L}$$-structures such that $$\{\tau, \sigma_1, \sigma_2, \dots\}$$ is satisfied by both, and if $$\mathcal{A}, \mathcal{B}$$ both have no infinite equivalence classes, then does this mean $$Th(\mathcal{A})=Th(\mathcal{B})$$? [Intuitively, I'm guessing this is true, but I don't understand how to see why formally.]

Assuming the answer to #1 is yes, (say $$Th(\mathcal{A})=Th(\mathcal{B})=T$$) then I have a further dilemma, or "paradox" if you will, that I'm having trouble resolving.

#2: How many complete 1-types of $$T$$ are there?

One wants to say there are $$\aleph_0$$ many, because it seems for each $$n$$ there should exist some complete 1-type $$\Sigma_n(v)$$ such that the formula $$\theta_i(v)$$ saying "$$v$$ is in the equivalence class of size $$i$$" belongs to $$\Sigma_n(v)$$ if and only if $$n=i$$. Moreover, it seems that an arbitrary complete 1-type $$\Sigma(v)$$ must be of the form $$\Sigma_n(v)$$ for some $$n$$, since there are no infinite equivalence classes.

On the other hand, it seems that each $$\Sigma_n(v)$$ is a principal complete 1-type, isolated by $$\theta_i(u)$$. [Correct me if I'm wrong. Once we know which equivalence class $$u$$ is in we know everything there is to know about $$u$$, so in particular we should have $$T \models \forall v(\theta_n(v) \to \phi(v))$$ for any $$\phi \in \Sigma_n$$, right?]

Now there's a theorem which says that a complete theory $$T$$ has infinitely many complete 1-types iff there exists at least some non-prinicpal complete 1-type. So the above two paragraphs seem to contradict each other. If there are $$\aleph_0$$ many complete 1-types, then there should be at least some non-prinicipal one. But if they're all principal, then there can't be $$\aleph_0$$ many. What have I done wrong?

Re: 1, yes - in fact, we have the stronger fact $$\mathcal{A}\cong\mathcal{B}$$ in that case. (More generally, any two $$\{R\}$$-structures satisfying $$\tau$$ with the same number of classes of each cardinality are isomorphic.)

Re: 2, you're missing the fact that "there is no infinite class" is not first-order expressible! The theory $$T$$ in question has the non-principal $$1$$-type corresonding to "$$x$$ is an element of an infinite class," even though this isn't realized in $$\mathcal{A}$$ or $$\mathcal{B}$$, and this reflects the fact that a structure with no infinite classes may still be elementarily equivalent to a structure with some infinite classes. So there's no paradox here.

(In fact, it's a good exercise to show that $$T=Th(\mathcal{A})=Th(\mathcal{B})$$ is simply the deductive closure of $$\{\tau\}\cup\{\sigma_k:k\in\mathbb{N}\}$$.)

Here's a proof that there is only one non-isolated (= non-principal) type and that the $$\Sigma_n$$s are the only isolated types over $$T$$:

Suppose $$p,q$$ are two types over $$T$$ which are not among the $$\Sigma_n$$s. Let $$\mathcal{M}\models T$$ be a model realizing both $$p$$ and $$q$$ via $$a$$ and $$b$$ respectively. WLOG (apply dLS if necessary) $$\mathcal{M}$$ is countable, so a fortiori every class in $$\mathcal{M}$$ is either finite or countably infinite. By assumption on $$p$$ and $$q$$ we know that $$[a]_\mathcal{M}$$ and $$[b]_\mathcal{M}$$ must be infinite. But then there must be an automorphism $$\alpha\in Aut(\mathcal{M})$$ with $$\alpha(a)=b$$. (More generally, whenever $$\mathcal{N}\models T$$ then the automorphism orbit relation is just $$\{(u,v): \vert[u]_\mathcal{N}\vert=\vert[v]_\mathcal{N}\vert\}$$, and this is a good exercise - essentially the same one as in the first paragraph of this answer.)

But this means that $$a$$ and $$b$$ realize the same types in $$\mathcal{M}$$, that is, $$p=q$$. So there is exactly one $$(1-)$$type over $$T$$ besides the $$\Sigma_n$$s. And we already know that this type must then be non-isolated.

• So, if I understand correctly, my mistake was where I said "Moreover, it seems that an arbitrary complete 1-type $\Sigma(v)$ must be of the form $\Sigma_n(v)$ for some $n$, since there are no infinite equivalence classes."? Is everything else correct? Commented Mar 5, 2022 at 22:02
• @Pascal'sWager Yes, that's right. More broadly, "type realized in $\mathcal{X}$" and "type consistent with $Th(\mathcal{X})$" are in general not the same notion, and that's what we're seeing here. Commented Mar 5, 2022 at 22:15
• Ok, thanks. I'm still trying to figure out how many complete types of each type (no pun intended) there are. Are the $\Sigma_n$ ($n \in \mathbb{Z}^+$) the only principal ones, or are there more? And are there multiple non-prinicipal ones or only the single one you mentioned in your answer? Commented Mar 5, 2022 at 22:21
• The $\Sigma_n$s are the only principal ones, and there is only the single non-principal type. I think Lowenheim-Skolem is the easiest way to prove this (show that if $\mathcal{M}\models T$ with $a,b\in\mathcal{M}$ not satisfying any of the $\Sigma_n$s, then there is an elementary substructure $\{a,b\}\subseteq\mathcal{N}\preccurlyeq\mathcal{M}$ with $a$ and $b$ in the same automorphism orbit of $\mathcal{N}$). The $n$-types for $n>1$ are not much more complicated. Commented Mar 5, 2022 at 22:32
• @Pascal'sWager Again, think about automorphisms: show that if $a,b\in\mathcal{M}\models T$ and $\vert[a]_\mathcal{M}\vert=\vert[b]_\mathcal{M}\vert<\aleph_0$ (which happens if they each satisfy the same $\Sigma_n$), then there is an automorphism $\alpha\in Aut(\mathcal{M})$ with $\alpha(a)=b$. Note that this is really just a special case of the second exercise mentioned in my answer, which in turn is basically the same as the first. In general, automorphisms (esp. + elementary submodel/extension arguments) provide a very powerful tool for analyzing types. Commented Mar 6, 2022 at 4:47