# Three old questions on the Sacks forcing

I came across the two following Qs in 1970.

1. Find reals $a,b$ such that $a$ is Sacks over $L[b]$ and vice versa $b$ is Sacks over $L[a]$. Note that a Sacks $\times$ Sacks generic pair definitely does not work.

2. Suppose that $a$ is Cohen over $L$ and $b$ is Sacks over $L[a]$. Does $a$ belong to $L[b]$? Comment: this Q answers in the positive if $a$ is Sacks or Solovay-random over $L$.

The next Q is perhaps not that dead lock, but still I have no clue.

1. Let $\mathbf P$ consist of all perfect sets $P$ on the real plane $\mathbb R\times\mathbb R$ such that all vertical and all horisontal cross-sections of $P$ are perfect (in particular, non-countable). What does it force?

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 $L[b]$.
• I wonder can the construction be strengthened so that $L[a]\cap L[b]=\emptyset$? – Vladimir Kanovei Sep 1 '14 at 20:08
• Yes, I would say so ($L[a]\cap L[b]=L$). Choose $T$ so that any two infinite paths in $T$ are mutually Cohen generic. Note, for any perfect subtree $T_1$ of $T$ and any non-constructible set $x$ in $L[a]$, there can be at most one path $g$ in $T_1$ such that $x$ is in $L[g]$. Argue that for any $T_1$ and any term $\tau$, there is a Cohen condition $p$ in $T_1$ which Cohen forces an incompatibility between $\tau$ and $x$. The perfect subtree of $T_1$ of nodes compatible with $p$ Sacks forces an incompatibility between $b$ and $x$. So it's definable in $L[a]$ and dense that $\tau$ is not $x$ – Theodore Slaman Sep 1 '14 at 21:47