If there is a sequence of structures which are in $K=\text{Mod}(\Phi),\Phi\subseteq FO(\tau)$ then $\mathfrak{B}\in \mathcal{K}$

Question

Let $(\mathfrak{A_m})_{m\in\mathbb{N}}$ be a sequence of structures in $\mathcal{K}$ for which we also know $\mathfrak{B}\equiv_m \mathfrak{A_m}$

I am not sure if I can say it like that but if I take a sentence $\psi\in\Phi$ then $\psi$ has finitely many quantifiers, say $m$. And because $\mathfrak{A_m}\models\Phi$, $\mathfrak{A_m}\models \psi$ (?). And therefore since $\mathfrak{B}\equiv_m\mathfrak{A_m}$ we have $\mathfrak{B}\models \psi$. And because we can do that for every $\psi\in\Phi$ we get $\mathfrak{B}\models\Phi (?)\Rightarrow \mathfrak{B}\in \mathcal{K}$. Is my proof correct? I am unsure at certain parts I marked them wih (?) - would be helpful if someone could explain me why I am allowed to say this.

If not is this proof correct?

Assume $\mathfrak{B}\notin \mathcal{K}$ then there must be at least one $\psi\in\mathcal{K}$ such that $\mathfrak{B}\not\models\psi$. If $\psi$ has $m$ quantifiers then $\mathfrak{A_m}\models \Phi$ by assumption and thus also $\mathfrak{A_m}\models \psi$ and then also $\mathfrak{B}\models \psi$ which is a contradiction.

What do you think of my proofs? Can I really say that given an axiom system $\Psi$ and a model $\mathfrak{A}$ for that axiom system then $\mathfrak{A}$ is also a model for every sentence in the axiom system?

Answer 1

Okay, so regarding your first proof...

• $\mathfrak{A_m}\models\Phi \implies \mathfrak{A_m}\models \psi$ follows from definition of $\mathfrak{A_m}\models\Phi$; recall that the latter just means that $\mathfrak{A_m}\models \psi$ for all $\psi \in \Phi$.
• Again, by definition, $\mathfrak{B}\models\Phi $ just means that $\mathfrak{B}\models \psi$ for all $\psi \in \Phi$, and since you can give the same kind of argument for each $\psi$, $\mathfrak{B}\models\Phi $ indeed follows.

So your first proof is fine.

With respect to your second proof, I'm assuming that instead of "$\psi\in\mathcal{K}$" you meant "$\psi \in \Phi$". After stating that $\mathfrak{B}\not\models\psi$ I would be slightly more picky and I would continue the argument as follows:

"Since $\psi \in \Phi$ and $\mathfrak A_m \in \mathcal K$ for all $m \in \mathbb N$, $\mathfrak A_m \models \psi$ for all $m \in \mathbb N$. In particular, if $\psi$ has $k$ quantifiers, $\mathfrak A_k \models \psi$, and since $\mathfrak A_k \equiv_k \mathfrak B$, we have that $\mathfrak B \models \psi$, contradicting $\mathfrak{B}\not\models\psi$."

See the difference between the above and your proof? I have used the same information, but I've been painfully pedantic and I stated precisely where I have used each bit of the given data. I also have changed the letter I used to denoted the number of quantifiers of $\psi$ to not mix it with the indexing letter ($m$ in this case).

To finish this and answering your last question: yes, a model of an axiom system is a model of each sentence of this axiom system, by definition of model of an axiom system.