Timeline for Is this theory the complete theory of the real ordered field?
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| May 5, 2021 at 15:41 | comment | added | Dmytro Taranovsky | @Ris It is not used, but I wanted the example to be a subfield of $\mathbb{R}$. | |
| May 5, 2021 at 5:12 | comment | added | Ris | One more question: where is the $\omega$-model requirement used? For some possible mismatch between "formula"s? | |
| May 5, 2021 at 4:54 | comment | added | Dmytro Taranovsky | @Ris Right (using automorphisms of the poset or its boolean algebra). | |
| May 5, 2021 at 4:35 | comment | added | Ris | So because it is a Cohen real, it is shifted correctly as $+ q$ and $\times q$ by automorphisms induced by $+ q$ and $\times q$ on conditions, and since the field doesn't change, every conditions forces the same supremum. Right? | |
| May 4, 2021 at 22:44 | comment | added | Dmytro Taranovsky | @Ris Using it, every extension of the reals here is compatible with every forcing condition, and so we cannot force different $\sup(X)$. | |
| May 4, 2021 at 22:00 | comment | added | Ris | Where is the fact that $qr$ is also a Cohen real used in the proof? | |
| Mar 24, 2021 at 16:38 | vote | accept | user107952 | ||
| Mar 1, 2021 at 22:28 | comment | added | Emil Jeřábek | I suppose you already know this, but let me mention explicitly in connection to the last question that no model elementarily equivalent to $\mathbb R_M(r)$ is computable, and the first-order theory of $\mathbb R_M(r)$ is highly undecidable. Indeed, the results on definability in simple transcendental extensions quoted in comments by @AlexKruckman show that $\mathbb R_M$ and $\mathbb N$ are definable in $\mathbb R_M(r)$, from which it follows easily that the full second-order arithmetic structure of the ground model $(\mathbb N,\mathcal P_M(\mathbb N))$ is interpretable in $\mathbb R_M(r)$. | |
| Mar 1, 2021 at 12:11 | comment | added | Emil Jeřábek | Thank you. A nice argument, by the way. | |
| Mar 1, 2021 at 12:05 | comment | added | Dmytro Taranovsky | @EmilJeřábek I used $ℚ$ by analogy with $ℚ^\text{alg}(r)$, but I just changed it $ℝ_M$. | |
| Mar 1, 2021 at 12:04 | history | edited | Dmytro Taranovsky | CC BY-SA 4.0 |
changed notation from Q_M to R_M
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| Mar 1, 2021 at 10:13 | comment | added | Emil Jeřábek | Since $\mathbb Q_M(r)$ is generated by the reals of $M$, rather than the rationals of $M$, shouldn't it be denoted $\mathbb R_M(r)$? (This kept tripping me when trying to understand the argument.) | |
| Mar 1, 2021 at 2:56 | comment | added | Dmytro Taranovsky | @AlexKruckman Correct. (Also, forcing is useful in model theory.) | |
| Mar 1, 2021 at 2:41 | comment | added | Alex Kruckman | Apologies for the naive questions - as a model theorist, I typically try to avoid thinking about forcing! So the other interesting remaining question (to me) is whether there is a purely algebraic / model-theoretic proof. | |
| Mar 1, 2021 at 2:37 | comment | added | Alex Kruckman | Ok, so what you're saying is: $M$ defines the set of reals $a\in \mathbb{R}^M$ such that that there exists a condition which forces "there exists $b\in \mathbb{Q}_M(r)$ such that $\mathbb{Q}_M(r) \models \varphi(b)$ and $a < b$", and the supremum of this set in $\mathbb{R}^M$ is also the supremum of set defined by $\varphi(x)$ in $\mathbb{Q}_M(r)$. Right? I suppose one needs to check that this set is actually bounded in $\mathbb{R}^M$, but it's probably a standard theorem that the field generated over $\mathbb{R}$ by a Cohen real is Archimedean, so it follows from boundedness in the extension. | |
| Mar 1, 2021 at 2:24 | comment | added | Dmytro Taranovsky | @AlexKruckman The reason is that (for each formula) the forcing relation is definable in $M$, and by ZFC\P, in $M$ we have least upper bounds for definable sets of reals (which can be proper classes in $M$). | |
| Mar 1, 2021 at 2:12 | comment | added | Alex Kruckman | I don't follow the argument that $\text{sup}(X)$ is in $M$. Fix a first-order formula $\varphi(x)$, let $X\subseteq \mathbb{Q}_M(r)$ be the set defined by $\varphi(x)$ in this field, and assume $X$ is bounded in $\mathbb{Q}_M(r)$. I agree that for rational $p < q$, the property $p<\sup(X)<q$ is determined by a forcing condition, and all conditions agree on all properties of this form. Why does this imply that $\sup(X)\in M$? | |
| Mar 1, 2021 at 0:42 | history | answered | Dmytro Taranovsky | CC BY-SA 4.0 |