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In Feng, Magidor, and Woodin "Universally Baire Sets of Reals", they show that if $A$ is a $\mathbf{\Pi}_2^1$ set and $U$ and $V$ are any pair of trees witnessing the universal baireness of $A$, then in all generic extension, one continues to have $A = p[U]$, where $A$ as defined by the $\mathbf{\Pi}_2^1$.

Is this true for $\mathbf{\Pi_3^1}$ (or more generally any higher projective sets)? Rather than any pair of trees as in the $\mathbf{\Pi}_2^1$ case, if one assumes all $\mathbf{\Pi}_3^1$ sets are universally baire, does there exists some pair of trees $T$ and $U$ such that $T$ continues to represent a given $\mathbf{\Pi}_3^1$ set $A$ in all generic extensions? What about, if one fix a particular forcing $\mathbb{P}$, can one find pairs of trees $T$ and $U$ that continue to represent a particular $\mathbb{\Pi}_3^1$ set in $\mathbb{P}$ extensions.


To be a bit more precise. Suppose $\varphi$ is a $\Pi_3^1$ formula possible using some reals as parameters. With appropriate large cardinals, all $\mathbf{\Pi}_3^1$ sets are universally Baire. Hence there are trees $T$ and $U$ such that

$V \models (\forall x)(\varphi(x) \Leftrightarrow x \in p[T])$ and $V \models (\forall x)(\neg \varphi(x) \Leftrightarrow x \in p[U])$

If $\mathbb{Q}$ is a forcing, then $1_\mathbb{Q} \Vdash_{\mathbb{Q}} ...$ the same two statements.

The question is for any $\mathbf{\Pi}_3^1$ formula $\varphi$ can one find trees $T$ and $U$ witnessing universally Baireness with the additional property that:

For all forcings $\mathbb{Q}$, $1_\mathbb{Q} \Vdash (\forall x)(\varphi(x) \Leftrightarrow x \in p[\check T])$.

This is what I mean by a $\mathbf{\Pi}_3^1$ sets being represented by the same tree in all generic extensions.


Thanks for any information that can be provided.

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  • $\begingroup$ As far as I know (for what it's worth) these questions are still open. If I'm wrong, I'd be very interested to hear about it. $\endgroup$ – Trevor Wilson Apr 30 '15 at 6:16
  • $\begingroup$ It is true under large cardinals. So the question is asking if the assumption can be weakened to universal Baireness. Do I understand the question correctly? $\endgroup$ – Yizheng Zhu Apr 30 '15 at 9:51
  • $\begingroup$ If I understand correctly, you are asking the following: UB sets are such that for every forcing there are two trees whose branches in the generic extension form the UB set and its complement. For $\mathbf\Pi^1_2$ we can reverse the quantifiers and find trees fitting for any forcing. Can we do it for higher projective sets, assuming these sets are UB? $\endgroup$ – Asaf Karagila Apr 30 '15 at 10:27
  • $\begingroup$ @Asaf I think that "for some forcing" versus "for every forcing" is not the main point, but rather whether the way of extending the set to a generic extension using trees agrees with the way of extending using the syntactical definition. $\endgroup$ – Trevor Wilson Apr 30 '15 at 16:36
  • $\begingroup$ @Trevor: It might be the underlying issue allowing us to prove this for the case of $\mathbf\Pi^1_2$; but this is really the question as far as I read it. $\endgroup$ – Asaf Karagila Apr 30 '15 at 16:51
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If we assume the existence of large cardinals (as in the latest version of the question) then such trees for $\Pi^1_3$ formulas exist. First let's get a local version, meaning a tree for a given $\Pi^1_3$ formula that works for all posets below a certain cardinality. At the end I will mention how to get trees that work for posets of arbitrary cardinality.


Let $S_1$ be a tree on $\omega \times \omega \times \omega \times \omega$ for a given $\Sigma^1_1$ formula. Let $\kappa$ be a cardinal. Then $S_1$ has a $\mathord{<}\kappa$-absolute complement $T_1$, which is a tree on $\omega \times \omega \times \omega \times \kappa$, namely the tree of attempts to build a real $x \in p[S_1]$ and a rank function on the tree $(S_1)_x$ with values in $\kappa$.

The tree $T_1$ is a tree for the corresponding $\Pi^1_1$ formula, and it works in every generic extension by a poset of size less than $\kappa$. From $T_1$ we obtain in a natural way a tree $S_2$ on $\omega \times \omega \times \kappa$ for the corresponding $\Sigma^1_2$ formula. (This tree $S_2$ is just the Shoenfield tree, but we need to pay attention to how it was constructed.) The tree $S_2$ also works in every generic extension by a poset of cardinality less than $\kappa$.

Now assume that $\kappa$ is measurable. Then by Martin's proof of $\Sigma^1_1$ determinacy, the tree $T_1$ is $\kappa$-homogeneous, which implies that the Shoenfield tree $S_2$ is $\kappa$-weakly homogeneous. Then by Martin and Solovay ("A basis theorem for $\Sigma^1_3$ sets of reals") the tree $S_2$ has a $\mathord{<}\kappa$-absolute complement $T_2$. (This tree $T_2$ is a tree on $\omega \times \omega \times \gamma$ for an ordinal $\gamma$ that is a little bit larger than $\kappa$; its exact value is not very important.)

The Martin–Solovay tree $T_2$ is a tree for the corresponding $\Pi^1_2$ formula, and it works in every generic extension by a poset of cardinality less than $\kappa$. From $T_2$ we obtain in a natural way a tree $S_3$ on $\omega \times \gamma$ for the corresponding $\Sigma^1_3$ formula. The tree $S_3$ also works in every generic extension by a poset of cardinality less than $\kappa$.

Now assume also that there is a Woodin cardinal $\delta < \kappa$. Then by Martin and Steel ("A proof of projective determinacy") the fact that $T_1$ is $\kappa$-homogeneous, and therefore $\delta^+$-homogeneous, implies that the tree $T_2$ is $\mathord{<}\delta$-homogeneous. So the tree $S_3$ is $\mathord{<}\delta$-weakly homogeneous, and applying the Martin–Solovay construction again we get a tree $T_3$ for the corresponding $\Pi^1_3$ formula that works in every generic extension by a poset of size less than $\delta$, as desired.


If we want trees for $\Pi^1_3$ formulas that work in generic extensions by posets of arbitrary size, we can assume that there is a proper class of Woodin cardinals, but this is overkill in terms of consistency strength. It suffices to assume that $M_1^\sharp(X)$ exists for every set $X$. This $M_1^\sharp(X)$ is a mouse over $X$ with a Woodin cardinal and a partial measure on top.

If you only want to show that the existence of such trees for $\Pi^1_3$ formulas is consistent, rather than showing that it holds in $V$, you can weaken the hypothesis further: given a strong cardinal $\kappa$, it follows from a theorem of Woodin that forcing with $\text{Col}(\omega,2^{2^\kappa})$ creates trees for $\Pi^1_3$ formulas that work in all further generic extensions.

Some references for this material are Steel's paper "The derived model theorem" (as Yizheng mentioned,) Larson's book "The Stationary Tower", and Kanomori's book "The Higher Infinite" (Section 32 in particular.)

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