# Borel Sets in Sacks Generic Extension

Let $\mathbb{S}$ denote Sacks forcing. This is forcing with perfect trees or equivalently forcing with uncountable Borel subsets of ${}^\omega 2$ with the relation $\subseteq$.

Let $G \subseteq \mathbb{S}$ be a generic filter over $V$. Suppose $B$ is a Borel set coded in $V[G]$. This means $B$ is a subset of $({}^\omega 2)^{V[G]}$ definable by a $\Delta_1^1$ formula using a parameter in $({}^\omega 2)^{V[G]}$. Furthermore, suppose $B$ is an uncountable Borel set.

The question is: Does $B$ have a ground model coded uncountable Borel subset?

I believe that if the above is answered yes then in $V[G]$, the identity map is a dense embedding of $(\mathbb{S})^V$ and $\mathbb{S}$. Does this lead to any problems?

A one further question: Is there some forcing property that can be used to show that the product $\mathbb{S} \times \mathbb{S}$ is different (not equivalent in the forcing sense) from the iteration $\mathbb{S} * \dot{\mathbb{S}}$? Or to differentiate between countable support products and countable support iterations of Sacks forcing?

Thanks for any information or clarification.

No. For example, let $B$ be the set of reals Turing above $G$. Then $B$ has no nonempty Borel subset with a Borel code in $V$.
• The reason being, that the assertion that "the Borel set with code $c$ is nonempty" is a $\Sigma^1_2$ assertion, which is absolute between the extension and the ground model, but no ground model real can compute $G$. So the argument has nothing to do with Sacks forcing, and it works for any extension with a new real. – Joel David Hamkins Jul 20 '14 at 14:20
• Actually, the assertion is $\Sigma^1_1$, if you already know that $c$ is a Borel code. – Joel David Hamkins Jul 20 '14 at 15:47
For any two reals $x,y\in 2^\omega$ there is a continuous function $f:2^\omega\to 2^\omega$ coded in the ground model such that $f(x)=y$ or $f(y)=x$.