Timeline for Is the forcing relation defined for mathematical formulas?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 16, 2014 at 22:39 | vote | accept | Carlos | ||
| Feb 16, 2014 at 22:39 | comment | added | Carlos | I see! For each mathematical formula $\phi$, we can define another mathematical formula $p\Vdash\phi$ in such a way that $M\vDash p\Vdash \phi$ ($M$ a set) satisfies the fundamental theorem. Thanks! | |
| Feb 16, 2014 at 22:24 | comment | added | Andreas Blass | To amplify Monroe's point: The forcing predicate (for mathematical formulas) behaves just like the satisfaction predicate. Over proper classes, it can't be defined fully, but over set models (like $H_\lambda$) it can. So what Shelah is doing is fine, uniformly for all mathematical formulas. | |
| Feb 16, 2014 at 22:04 | comment | added | Monroe Eskew | Yes, but we instead use $\Vdash^{H_\lambda}_\mathbb{P}$ and avoid meta-mathematical worries. | |
| Feb 16, 2014 at 21:41 | comment | added | Carlos | Thank you. The problem is that in order to use the hypothesis that $N\prec H_\lambda$ to prove $N[G]\prec H_\lambda[G]$, using $\Vdash$ seems unavoidable, and, in view of the answer to Question 1, it seems we should restrict ourselves to consider meta-mathematical formulas. | |
| Feb 16, 2014 at 20:36 | history | answered | Monroe Eskew | CC BY-SA 3.0 |