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.


2 Answers 2


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 $V$.

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    $\begingroup$ 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. $\endgroup$ Jul 20, 2014 at 14:20
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    $\begingroup$ Actually, the assertion is $\Sigma^1_1$, if you already know that $c$ is a Borel code. $\endgroup$ Jul 20, 2014 at 15:47

Here is a property that differentiates between the product of two copies of Sacks forcing and the iteration. In the iterated Sack model (countable support iteration, the length any ordinal) the following is true:

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 $f(y)=x$.

After forcing with a product of two copies of Sacks forcing there is no Borel map coded in the ground model mapping one of the two generics to the other, because then the other generic would already be contained in the extension generated by the first generic.


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