Let $\mathbb{V}$ denote Prikry-Silver forcing. That is, $\mathbb{V}$ is forcing with partial functions $\omega \rightarrow 2$ with coinfinite domain or forcing with uniform trees.

Let $\dot x$ denote the canonical name for the unique real given by the generic filter. A real $r \in V[G]$ is a $\mathbb{V}$ generic real if and only if there exists some $H \subseteq \mathbb{V}$ and $H \in V[G]$ such that $V[H] \models r = \dot x_G$.

The question is: Is every new real in $V[G]$ a $\mathbb{V}$-generic real.

This meaning, if $p \in \mathbb{V}$ and $\tau \in V^{\mathbb{V}}$ and $p \Vdash_{\mathbb{V}} \tau \in {}^\omega 2 \wedge \tau \notin \check V$, then there exists a $q \leq_{\mathbb{V}} p$ such that $q \Vdash_{\mathbb{V}} \tau = \dot x$.

It is known that every new real in the Sacks extension, is a Sack generic real. Both Sacks forcing and Prikry-Silver forcing add reals of minimal degree (although Sacks actually is a minimal extension and Silver forcing is not).

Does having all new reals being generic follow from adding reals of minimal degree? If not, what is a counterexample of a forcing adding minimal degree such that a new real is not a generic real for that forcing?



1 Answer 1


No, not every new real in the Silver extension is generic for Silver forcing. To see this, take any new real $x$ in the extension, and form a new real $y$ by doubling every digit of $x$. So if $x=011010\ldots$, then $y=001111001100\ldots$. The real $y$ cannot be Silver generic, since it is dense in Silver forcing to violate the digits-doubling property. Given any condition, look at the first bit not specified, and then fill that bit (and possibly the next) in such a way that the double-digit property is violated.


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