Timeline for Model-completeness of real exponential fields
Current License: CC BY-SA 4.0
11 events
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| Aug 27, 2022 at 8:18 | history | edited | Emil Jeřábek | CC BY-SA 4.0 |
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| Aug 26, 2022 at 14:01 | answer | added | user44143 | timeline score: 4 | |
| Aug 26, 2022 at 9:14 | history | edited | Emil Jeřábek | CC BY-SA 4.0 |
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| Aug 26, 2022 at 6:15 | comment | added | Emil Jeřábek | @JamesHanson E.g., the dense linear order with one end-point $M=([0,+\infty),<)$. Then $0$ is universally definable in $M$, and $\mathrm{Th}(M,0)$ has full quantifier elimination, but $T=\mathrm{Th}(M)$ is not model-complete, as $[1,+\infty)\models T$ (it is even isomorphic to $M$), but the inclusion $[1,+\infty)\subseteq[0,+\infty)$ is not elementary. Also, while all model-complete theories are $\forall\exists$-axiomatizable, $T$ is not, as $\bigcup_n[-n,+\infty)$ is not a model of $T$. | |
| Aug 26, 2022 at 4:24 | comment | added | James E Hanson | What is a good example of a structure $M$ such that $\mathrm{Th}(M)$ is not model-complete, but $\mathrm{Th}(M,a)$ is for some $a \in M$? | |
| Aug 25, 2022 at 15:15 | comment | added | Emil Jeřábek | @C7X As I wrote in the first sentence, I use $\mathbb R$ to denote the structure $(\mathbb R,0,1,+,\cdot,<)$, so the language of $(\mathbb R,b^x)$ includes $+$ by definition. | |
| Aug 25, 2022 at 14:36 | comment | added | C7X | Since we also need $+$ in the language in the construction of $1+x$, does this result apply to $(\mathbb R,b^x)$? | |
| Aug 25, 2022 at 14:11 | history | edited | Emil Jeřábek | CC BY-SA 4.0 |
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| Aug 25, 2022 at 14:10 | comment | added | Emil Jeřábek | @MattF. Almost. You need to include $1$ in the language, as otherwise $t(0,\dots,0)=0$ for all terms $t$ (under the above definition of $x^y$). On the other hand, you don’t need $-$ if you allow terms on both sides of the equation. That is, every existential formula $\phi(\vec x)$ in $(\mathbb R,x^y)$ is equivalent to a formula of the form $\exists\vec y\,t(\vec x,\vec y)=s(\vec x,\vec y)$ where $t$ and $s$ are terms in the language $(1,+,\cdot,x^y)$ (and you can limit the shapes of $t$ and $s$ further if you wish). | |
| Aug 25, 2022 at 13:22 | comment | added | user44143 | It looks like condition 3 is also equivalent to "there is a term $t(x_1,x_2,\dots,x_n)$ in the language $(+,-,\times,\,^\wedge)$ which has solutions to $t=0$ with $x_1=e$ and no solutions with $x_1 \neq e$ -- is that right? | |
| Aug 25, 2022 at 12:40 | history | asked | Emil Jeřábek | CC BY-SA 4.0 |