Let $M$ be a transitive proper class model of $ZFC$. I want to prove a few absoluteness results:
- $(\omega^\omega)^M$ = $\omega ^\omega \cap M$
By the way when this happens do we say $\omega^\omega$ is absolute? Or is that only when $A^M = A$?
Anyway, in $M$ we have that $\omega$, and being a function is absolute, so $(\omega ^ \omega)^M = \{x \in M: (x : \omega \rightarrow w)^M\} = \{x \in M: x: \omega \rightarrow \omega\} = \omega^\omega \cap M$
Is that fine?
- Let $T$ be a tree on $\omega \times \omega$ and $T \in M$. Then $(T$ is a tree on $\omega \times \omega)^M$
Again being a function and $\omega$ are absolute so elements of the tree (functions from naturals to $\omega)$ are absolute. Moreover ordered pairs are absolute as well.
$(T$ is a tree on $\omega \times \omega)$ $\iff$ $(\forall (s, t) \in T , \forall n \in \omega, n < dom(s) \rightarrow (s|n, t|n) \in T)$
But since $T \in M$, domains, restrictions, $\omega$, ordered pairs are absolute, and we are only quantifying over bounded sets ($T$ and $\omega$), the "being a tree" formula is $\Delta^{ZFC}_1$ and so absolute. This is in fact stronger than what was asked to prove since it was asked to simply prove downward reflection.
I'm not sure about this one. Feels handwavey. Thoughts?
- Suppose $A \subseteq \omega$ is arithmetic (i.e, is a definable set in the structure $\mathbb{N} = (\omega, +, ., <, e, 0, 1))$, then $A \in M$.
Given any sentence $\varphi$ in the language of arithmetic above, "$\mathbb{N} \vDash \varphi$" is absolute since it is bounded over $\omega$ and satisfaction is recursively defined, etc. Since every element of $\omega$ is definable we don't have to worry about formulas, just sentences.
Now let $\varphi$ define $A$ above. We may then use Comprehension in $M$ to define $\{a \in \omega^M : (\mathbb{N} \vDash \varphi(a))^M\} = \{a \in \omega: \mathbb{N} \vDash \varphi(a)\} = A$, and so $A \in M$.
I would appreciate some feedback on these. Thank you for your help.
${^\omega}\omega\cap M=({^\omega}\omega)^M$ happens when the formula $x\in {^\omega}\omega$ is absolute between $V$ and $M$. This is the same for subset relations (so $\mathcal{P}^M(x)=\mathcal{P}(x)\cap M$, although powerset function itself is not absolute.) I think your argument is fine. I also think your proof on the absoluteness of trees also looks fine.
For satisfaction relation over $\mathbb{N}$, I want to point out that this is absolute between $M$ and $V$ not because it is recursively defined, but its defining formula is bounded. It is right that $\models$ over $\mathbb{N}$ is recursively defined. But a more important point is that its recursion ends up with $\omega$-th stage of recursion, and every element and formulas involving the definition is $\Delta_0$.
If the recursion involves with more complex formula, then the resulting object need not be absolute (like $V_\alpha$.)