Timeline for Tarski-Seidenberg for strict inequalities and bounded quantification
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 3 at 21:52 | comment | added | Dima Pasechnik | Solutions to finitely many polynomial equations are the solutions to just one equation, the product of these finitely many ones. And it's a basic fact in characteristic 0 algebraic geometry that it's a hypersurface, it cannot contain an open ball. | |
| Jan 2 at 16:39 | comment | added | Paul Taylor | The Bochnak-Coste-Roy book is freely downloadable here. I think this site is legitimate, but am open to correction. | |
| Dec 10, 2022 at 15:24 | comment | added | Paul Taylor | @AndrejBauer has told me privately that Theorem 2.7.2 of "Real algebraic geometry" by Jacek Bochnak, Michel Coste and Marie-Françoise Roy (Springer, 1998, ISBN 978-3-642-08429-4) gives an explicit proof this result. It is more complicated than my attempt and I don't follow it. | |
| Dec 10, 2022 at 14:06 | comment | added | Paul Taylor | Am I correct that the solutions to finitely many non-zero polynomial equations cannot cover an open ball? | |
| Dec 10, 2022 at 13:59 | comment | added | Paul Taylor | @EmilJeřábek: I was hoping that you would take the challenge and not just dismiss compactness. So, we have eliminated the equation from some of the disjuncts, so let $a$ avoid (the inequalities in) those but satisfy some other equation. I think my argument eliminates another one of those. | |
| Dec 10, 2022 at 13:55 | comment | added | Emil Jeřábek | Compactness would help if you could also show the existence of such a ball around any point $a$ that does not satisfy $\phi$. | |
| Dec 10, 2022 at 13:42 | comment | added | Paul Taylor | @EmilJeřábek: Of course, I mean that the equations can be eliminated, and I showed that that is true for some open ball around each point. The word compactness comes to mind, but there should be algebraic ideas that get us from here to a proof. I reckon you would be a good person to do that! | |
| Dec 10, 2022 at 13:37 | comment | added | Emil Jeřábek | Expressing $\phi$ only on the ball around $a$ is pointless. You assumed that $\phi$ was true on that ball, that’s how you chose the ball. So $\phi$ is trivially equivalent to $\top$ on the ball. | |
| Dec 10, 2022 at 13:31 | comment | added | Emil Jeřábek | It does not contribute $\top$. It contributes $\top\land\bigwedge_jq_{i,j}>0$, that is, $\bigwedge_jq_{i,j}>0$. You need all the $p_i$ to be identically $0$ to express $\phi$ as a monotone combination of strict polynomial inequalities. | |
| Dec 10, 2022 at 13:28 | comment | added | Paul Taylor | @EmilJeřábek: no, just one, which contributes $\top$ to a disjunction, at least within the ball. Maybe it needs more thought for the rest of the space. | |
| Dec 10, 2022 at 13:16 | comment | added | Emil Jeřábek | All right, so one of the polynomials is identically $0$. And then what? You need all the polynomials to be identically $0$ to get the required characterization, right? | |
| Dec 10, 2022 at 12:06 | history | edited | Paul Taylor | CC BY-SA 4.0 |
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| Dec 10, 2022 at 11:10 | history | edited | Paul Taylor | CC BY-SA 4.0 |
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| Dec 10, 2022 at 11:02 | history | edited | Paul Taylor | CC BY-SA 4.0 |
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| Dec 10, 2022 at 10:56 | history | answered | Paul Taylor | CC BY-SA 4.0 |