Timeline for Is this theory the complete theory of the real ordered field?
Current License: CC BY-SA 4.0
11 events
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| Feb 11, 2021 at 17:49 | comment | added | Spencer Dembner | @EmilJeřábek yes, that's what I had in mind (and as you said, it extends to subfields of $\mathbb{R}$ but maybe not further). | |
| Feb 11, 2021 at 15:08 | comment | added | Emil Jeřábek | No, sorry, it’s not so simple. The argument does need $L$ to be archimedean (that is, $K\subseteq L\subseteq\mathbb R$), as otherwise it’s not clear that if $\phi(K)=(a,b)$, then $\sup\phi(L)=b$. I’m not sure if it actually holds for nonarchimedean $L$. | |
| Feb 11, 2021 at 9:39 | comment | added | Emil Jeřábek | Presumably, you mean $\exists_1$ when you write $\Sigma_1$ (ordered fields have no useful notion of bounded quantifiers). Note that your argument that the least upper bound property holds for parameter-free $\exists_1$ formulas uses nothing special about $K(\pi)$: this is true for all ordered fields $L\supseteq K$. (In place of $\mathbb R$, you can take the real closure of $L$.) | |
| Feb 10, 2021 at 22:28 | history | edited | Spencer Dembner | CC BY-SA 4.0 |
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| Feb 10, 2021 at 21:51 | comment | added | Spencer Dembner | @user107952 no, unfortunately the argument doesn't work because of the issue Emil brought up. | |
| Feb 10, 2021 at 21:50 | comment | added | user107952 | @SpencerDembner So, does this settle the issue? You answered the question with a correct argument? | |
| Feb 10, 2021 at 21:28 | history | edited | Spencer Dembner | CC BY-SA 4.0 |
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| Feb 10, 2021 at 21:27 | comment | added | Spencer Dembner | Nevermind, I see the issue. My apologies | |
| Feb 10, 2021 at 21:25 | comment | added | Spencer Dembner | Right, but I don't believe that my proof uses quantifier elimination / o-minimality for $K(\pi)$ (again, I'm probably missing something). It's true that the set you identify isn't a finite union of intervals in $K(\pi)$, but it is in $\mathbb{R}$ which I think is all I use. | |
| Feb 10, 2021 at 20:59 | comment | added | Emil Jeřábek | $K$ has quantifier elimination, but $K(\pi)$ has not. You are supposed to show the least upper bound property for sets of the form $\{x\in K(\pi):K(\pi)\models\phi(x)\}$ where $\phi$ is an arbitrary formula, and this cannot be reduced to the case of just quantifier-free $\phi$. For example, the set defined by the formula $\phi(x)=\exists y\,x=y^2$ is dense in $K(\pi)_{\ge0}$ with a dense complement, hence it certainly is not a finite union of intervals. | |
| Feb 10, 2021 at 20:40 | history | answered | Spencer Dembner | CC BY-SA 4.0 |