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Let $M$ be a transitive infinite countable ZFC model.

If I understand correctly, all elements of $M$ are sets. For example: $0=\emptyset$, $1 = \{ 0 \} = \{ \emptyset \}$, $2 = \{ 0,1 \} = \{ \emptyset,\{ \emptyset \} \}$, $3 = \{ \emptyset, \{ \emptyset \}, \{ \emptyset, \{ \emptyset \} \} \}$ etc.

I also get how we can view relations as sets of ordered pairs, and functions as certains sets of ordered pairs.

But, I how can we view a formula?

For example, suppose we assume that $M$ satisfies the axiom of infinity.

Does it means that this axiom exists in $M$ as an element. Or is the statement $M \models \phi$ (where $\phi$ is the axiom of infinity) exists as an elemtnt?

and if so, what is the set which represents this statement?

Thank you!

EDIT For example, suppose $\mathbb P$ is a poset and that $M \models \exists p \in \mathbb P (p(0)=2 \and p(1)=3)$. The fact that $M$ satisfies this satement, how does $M$ contains this fact. Does it contains it as a set? a function with range $\{ T, F \}$?

user135172
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    If $\mathcal M$ is a model for a formula $\phi$ and if $\phi$ asserts the existence of an objcet (like the axiom of infinity), saying that $\mathcal M \vDash \phi$ means that in the "universe" of $\mathcal M$ there is an objcet (in this case $\omega$) satisfying the axiom $\phi$. – Mauro ALLEGRANZA Mar 06 '15 at 12:15
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    The same for the "universe" $\mathbb N$ of natural numbers. Saying that $\mathbb N \vDash \exists x \forall y (x \le y)$ means that there is a (natural) number (in this case : $0$) that is less-or-equal than any number, i.e. that : $0 \le n$, for any $n \in \mathbb N$. – Mauro ALLEGRANZA Mar 06 '15 at 12:18
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    I assume that you are "speaking of" forcing and $\mathbb P$-names. You can see L.Halbeisen, Combinatorial Set Theory: With a Gentle Introduction to Forcing (2012),page 273-on. $\mathbb P =(P, \le)$ is a partially ordered set contained in the "domain"; that is, a relation. A $\mathbb P$-name is a set of ordered pairs of the form $\langle \overline x,p \rangle$ where $\overline x$ is a $\mathbb P$-name and $p \in P$. Thus, e.g. $\exists p \in \mathbb P(p(0)=2)$ means that the $p$ is an ordered couple such that $p=(0,2) \in \mathbb P$. – Mauro ALLEGRANZA Mar 06 '15 at 12:43
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    In conclusion, saying that $\mathcal M \vDash \exists p \in \mathbb P \phi(p)$, means that in the model $\mathcal M$ (whose domain, as usual in set theory, contains only sets) there is a partially ordered set $\mathbb P=(P,\le)$ such that $P$ contains a set $p$ such that the formula $\phi(p)$ is satisfied in $\mathcal M$. – Mauro ALLEGRANZA Mar 06 '15 at 12:53

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