For (2), I suggest writing $a$ as the Turing join of its even and odd parts, call them $a_0$ and $a_1$. Use $a_0$ to compute a perfect subtree $T$ of $2^{<\omega}$ such that every infinite path in $T$ is mutually Cohen generic with $a_1$. When considered as a Sacks condition in $L[a]$, $T$ forces that $a$ is not an element of $L[b]$.
2 of 2
Correction: $T$ should be a subtree of $2^{<\omega}$ not $2^omega$.
Theodore Slaman
- 1.8k
- 12
- 13
Theodore Slaman
- 1.8k
- 12
- 13