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William
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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[T])$$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.

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

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.

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

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.

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.

Thanks for any information that can be provided.

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

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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William
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Universally Baire Tree Representation of Projective Sets

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.

Thanks for any information that can be provided.