1
$\begingroup$

Suppose $T$ is a tree on $\omega \times \omega \times \delta$ for some ordinal $\delta$ is a homogeneous tree (with some coherent set of measures witnessing homoegeneity). ($T$ can have additional properties if it is helpful for getting the tree $S$ with the property I want below.)

Under the appropriate large cardinal assumptions, the set

$(\forall y \in {}^{\omega}\omega)p[T] = \{x \in {}^\omega\omega : (\forall y)(\exists f)((x,y,f) \in [T])\}$

is also homogeneous Suslin. (I believe this result is essentially part of Martin-Steel proof of projective determinacy.) So there is some homogeneous tree $S$ so that $p[S] = (\forall y)p[T]$.

My question is that if $\mathbb{P}$ is a very small forcing (say size $2^{\aleph_0}$), can one choose $S$ so that $1_{\mathbb{P}} \Vdash p[S] = (\forall y)p[T]$?


I think this result is true if $T$ is a homogeneous tree representing a projective set (See my earlier question Universally Baire Tree Representation of Projective Sets). I have not seen a proof of generic absoluteness for projective sets, but I think this may be one way of getting projective absoluteness for forcings of certain sizes. My question is essentially whether some type of generic absoluteness holds for formulas of the form $(\forall y)A(x)$ where $A$ is a homogeneous Suslin set.

Thanks for any information or references about this question.

$\endgroup$
1
1
$\begingroup$

Assuming $\lambda$ is a limit of Woodin cardinals, $\delta_0,...,\delta_n,...$, then the pointclass $Hom_{<\lambda}$ is closed under $\forall^{\mathbb{R}}$. This indeed follows from Martin-Steel.

I think the kind of absoluteness you're looking for is the "tree production lemma" of Woodin. The tree production lemma gives absolutely complementing trees to $\{x:\varphi(x)\}$ and $\{x:\neg \varphi(x)\}$,i.e universally Baire representations, given some generic absoluteness holds for $\varphi$. The required generic absoluteness follows mainly from the stationary tower forcing. Stationary tower forcing also implies generic absoluteness for the $\varphi$ defining say $\mathbb{R}^{\#}$. So we have generic absoluteness for theory fo $L(\mathbb{R})$. In particular, $Hom_{\infty}$ is closed under sharps, assuming there is a proper class of Woodin cardinals. This material is in section 4 of Steel's paper the derived model theorem. It is also in Paul Larson's book.

In any case, assuming there are arbitrarily large Woodin cardinals, $(\Sigma^2_1)^{Hom_{\infty}}$ sets are absolute for set forcing and $L(\mathbb{R})$-truth can be reduced to $(\Sigma^2_1)^{Hom_{\infty}}$-truth, so this is another way to obtain the sort of absoluteness you want. (this is in section 5 of Steel's paper).

But there are limits to the above. I think we don't have absoluteness for statements of the form $\forall^{\mathbb{R}}(\Sigma^2_1)^{Hom_{\infty}}$ for the above assumptions...

$\endgroup$
2
  • $\begingroup$ Thanks very much for your reference. Just one question: I could not find explicitly the definition of $(\Sigma_2^1)^{\text{Hom}_{<\lambda}}$ in Steel's paper, but I get the impression that they are formulas of the kind that appear in Theorem 5.1 of Steel's paper. For my question, if $A$ is homogeneously Suslin, both $A$ and $(\forall y)A$ are $(\Sigma_2^1)^{\text{Hom}_{<\lambda}}$ formulas and so Theorem 5.1 would give the desired absoluteness. Is this the idea? Thanks $\endgroup$
    – William
    Feb 9 '16 at 18:42
  • $\begingroup$ @William Exactly, that's the idea. The more sets you can show are $Hom_{\infty}$, the more formulae you can express in a $(\Sigma^2_1)^{Hom_{\infty}}$ way and this buys you more absoluteness. $\psi$ is $(\Sigma^2_1)^{Hom_{\infty}}$ if it is of the form $\exists A\in Hom_{\infty} (HC,\in,A)\models \varphi$, some $\varphi$. $(HC,\in,A)\models \varphi$ is projective. This is theorem 5.1 in Steel's paper. 5.6 explains the reduction of $L(\mathbb{R})$-truth to $(\Sigma^2_1)^{Hom_{\infty}}$-truth. The part about the tree production is not really relevant, it just offers uB representation. $\endgroup$ Feb 9 '16 at 22:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.