Timeline for Is this theory the complete theory of the real ordered field?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Mar 2, 2021 at 4:13 | comment | added | user44143 | @AlexKruckman, thanks! Steps 1, 2 and 4 are familiar...now it will take some staring at Step 3 to make sense of it. | |
| Mar 1, 2021 at 16:27 | comment | added | Alex Kruckman | Of course, it takes some work to show that this definition works! As I said in my previous comment, you can read a complete proof in the book by Jensen and Lenzing. | |
| Mar 1, 2021 at 16:25 | comment | added | Alex Kruckman | @MattF. Let $R$ be real closed and $t$ transcendental over $R$. Step 1: $x\in R$ is definable in $R(t)$ by $\exists y (1+x^4 = y^2)$. Step 2: Define $x\leq_2 y$ if $y-x$ is a sum of two squares in $R(t)$. Step 3: $n\in \mathbb{N}$ is definable in $R(t)$ using $t$ as a parameter by $\varphi(n,t)$: $$n\in R\land \exists x (x\neq 0 \land x^2\leq_2 t^2\land \forall y\in R ((y\neq n \land x^2\leq_2(t-y)^2)\to x^2\leq_2 (t-y-1)^2))).$$ Step 4: Remove the parameter for $t$ by $$n\in \mathbb{N}\iff \forall x(\varphi(0,x)\land \forall y\in R((\varphi(y,x)\to\varphi(y+1,x)) \to \varphi(n,x))).$$ | |
| Mar 1, 2021 at 14:23 | comment | added | user44143 | At the risk of being a broken record....it'd be interesting to see the definition of $\mathbb{N}$ in $\mathbb{Q}^{\text{alg}}(r)$ explicitly. | |
| Mar 1, 2021 at 3:39 | comment | added | Noah Schweber | @AlexKruckman Did not know that theorem - that definitely ruins my suspicion! | |
| Mar 1, 2021 at 3:35 | comment | added | Alex Kruckman | I think $\mathbb{Q}^{\text{alg}}(r)$ cannot be an example for any transcendental $r$. It's a theorem that $\mathbb{N}$ is definable in $\mathbb{Q}^{\text{alg}}(r)$ (see Theorem 3.29 in Model Theoretic Algebra by Jensen and Lenzing). Using the arithmetic on $\mathbb{N}$, we can carry out computations, so the cut corresponding to any computable real number should be definable. But there are always computable reals which are not in $\mathbb{Q}^{\text{alg}}(r)$. | |
| Feb 21, 2021 at 1:40 | comment | added | Will Sawin | @MattF. Is it obvious there is no example with higher $x$? | |
| Feb 12, 2021 at 1:50 | comment | added | user44143 | For that example, taking $x=y=z\in\{n\pi-\lfloor n\pi\rfloor:n\in\mathbb{Z}^+\}$ will be enough for $(x,y,z)$ to satisfy the condition, so the supremum will just be $1$. I think finding an example may be trickier. | |
| Feb 8, 2021 at 14:04 | comment | added | Will Sawin | @Matt F. I don't have a great example, but you can pick a rational curve that intersects this square, like $x=y=z$, and then choose a general polynomial of degree $>5$ or so that vanishes on the curve, like $x^6+y^6- 2z^6 + x+ z - 2y$, and if there's some trick that works for that one keep adding more terms at random until it stops working. | |
| Feb 8, 2021 at 13:21 | comment | added | user44143 | Do you have an example of $f$ where it is easy to prove that the set is inhabited but hard to prove that it has a supremum? | |
| Feb 6, 2021 at 15:46 | history | edited | Will Sawin | CC BY-SA 4.0 |
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| Feb 6, 2021 at 15:13 | history | answered | Will Sawin | CC BY-SA 4.0 |