Timeline for Three old questions on the Sacks forcing
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Sep 1, 2014 at 21:47 | comment | added | Theodore Slaman | 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 $x$ | |
| Sep 1, 2014 at 20:08 | comment | added | Vladimir Kanovei | I wonder can the construction be strengthened so that $L[a]\cap L[b]=\emptyset$? | |
| Aug 31, 2014 at 22:37 | history | edited | Theodore Slaman | CC BY-SA 3.0 |
Correction: $T$ should be a subtree of $2^{<\omega}$ not $2^omega$.
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| Aug 31, 2014 at 19:50 | history | answered | Theodore Slaman | CC BY-SA 3.0 |