# A model satisfying the existence of a model

This problem comes from Jech exercise 12.11:

If $$\kappa$$ is an inaccessible cardinal then $$V_\kappa \models$$ there is a countable model of ZFC.

I see this question partially answered here: Jech, "Set theory" exercises 12.11 - Is my proof right?

But I still have a question. What does it mean for $$V_\kappa \models (\mathfrak{U} \models$$ ZFC) ? If my impression is right only a "$$\sigma$$" should follow the "$$\models$$" where $$\sigma$$ is a sentence or formula in the language of set theory. How do you turn "$$\mathfrak{U} \models$$ ZFC" into a sentence in the language set theory.

In the post linked above the author only showed that $$(\varphi^{\omega,E})^{V_\kappa}$$ held for each axiom of ZFC, but is that the same thing as showing (or all that is required to show) $$V_\kappa \models (\omega, E)$$ is a model of ZFC ?

Fix a transitive model $$\textbf{K}$$ of ZF (possibly with $$\mathbf{K}$$ a proper class).

When I say that a statement $$\Phi(a_1, ..., a_n)$$ in the language of set theory is absolute, I mean that $$\forall a_1,\in \textbf{K} \ldots \forall a_n \in \textbf{K}, \Phi^{\textbf{K}}(a_1, \ldots, a_n) \iff \Phi(a_1, \ldots, a_n)$$. Here, $$\Phi^{\textbf{K}}(a_1, \ldots, a_n)$$ is the statement $$\Phi(a_1, \ldots, a_n)$$ with all quantifiers relativised to $$\mathbf{K}$$.

First of all, given a vocabulary $$V$$ and a statement $$\phi(x_1, ..., x_n)$$ in the language $$V$$, and given a structure $$M$$ and values $$a_1, ..., a_n \in M$$, we know how to define $$M, x_1 \gets a_1, ..., x_n \gets a_n \models \phi(x_1, ..., x_n)$$. In particular, this is done by induction on $$\phi(x_1, ..., x_n)$$. Note that here, $$V$$, $$\phi(\ldots)$$, and $$M$$ are objects within set theory.

The key here is to verify that the statement $$M, x_1 \gets a_1, ..., x_n \gets a_n \models \phi(x_1, ...., x_n)$$ is absolute for all statements $$\phi$$ in the vocabulary in any transitive model of ZF. This is easy to prove, since the statement involves only $$\Delta_0$$ quantifiers.

Given a set $$S$$ of sentences, we say a structure $$M$$ models $$S$$ if and only if $$\forall s \in S, M \models S$$. Once again, this statement involves only a $$\Delta_0$$ quantifier, so it is absolute.

We also know how to define the vocabulary $$K$$ of ZFC. It's just $$V = \{\in\}$$. And we know how to define the set $$S$$ of axioms of ZFC. You should verify that both $$V$$ and $$S$$ are absolute. This is pretty straightforward, but a bit tedious.

Therefore, the statement "$$M$$ models ZFC" is absolute.

Now suppose that there is an inaccessible cardinal $$\kappa$$. Then ZFC is consistent. Then by the downward Lowenheim-Skolem theorem, there is a countable model $$M$$ of ZFC. Since $$M$$ is countable, we can place it into bijection with $$\mathbb{N}$$, and thus we can obtain a model $$(\mathbb{N}, \in_\mathbb{N})$$ of ZFC.

Now $$\mathbb{N} \in V_\kappa$$, and also $$P(\mathbb{N}) \in V_\kappa$$. Since $$V_\kappa$$ is transitive, $$\in_\mathbb{N} \in V_\kappa$$. Therefore, $$(\mathbb{N}, \in_\mathbb{N}) \in V_\kappa$$.

Since $$V_\kappa$$ is a transitive model of ZFC, we see that the statement $$(\mathbb{N}, \in_\mathbb{N}) \models ZFC$$ is absolute for $$V_\kappa$$. Since the statement is true, $$((\mathbb{N}, \in_\mathbb{N}) \models ZFC)^{V_\kappa}$$ holds. And of course, $$V_\kappa \models \mathbb{N}$$ is countable.

Therefore, we see that $$V_\kappa \models$$ there exists a countable model of ZFC.

Note that therefore, it's actually the case that if there is an accessible cardinal, then ZFC + ("ZFC is consistent") is consistent. This means that there is a countable model of ZFC + ("ZFC is consistent"). And therefore, $$V_\kappa \models$$ there is a countable model of ZFC + ("ZFC is consistent"). Therefore, ZFC + "ZFC is consistent" + "(ZFC + ("ZFC is consistent")) is consistent". We can repeat this ad infinitum.

So the existence of an inaccessible cardinal is much, much stronger than the consistency of ZFC.

• That note is very interesting. Thank you.
– JDC
Commented Oct 23, 2021 at 0:32
• I'm trying to understand how we can define the model relation. How is V={∈} an object of set theory, for ∈ isn't an object in set theory? Also how is ϕ(…) an object of set theory? Do you use a trick like Godel numbering? I can see that if those are objects of set theory then you can define ⊨ as relation that is an object in set theory and you could tell if a model satisfied a formula if the parameters paired with the formula was an object in the relation e.g. "((a1,..., an),ϕ[a1,...,an])∈ ⊨"
– JDC
Commented Oct 23, 2021 at 0:41
• @JDC Formally, we can take absolutely any definable object in ZFC and define $\in$ to be that object. A statement $\phi$ is just a sequence of characters with certain properties. So yes, it basically is Godel numbering. Keep in mind that according to the “ZFC philosophy”, all math must be done by encoding things as sets. So you should always know how to take math you’re doing and encode it into set theory if you’re working with ZFC. Commented Oct 23, 2021 at 1:16