Timeline for End elementary extension in infinitary logic of some $L_\alpha$ producing a $L_\beta$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 21 at 22:46 | comment | added | C7X | @Johan Thank you! I will try to complete it. | |
| Mar 21 at 10:09 | vote | accept | Johan | ||
| Mar 21 at 10:09 | comment | added | Johan | Hi @C7X, sorry I forgot to answer your last comment. I think the error you pointed out plus the theoretical limitation that Gabe Goldbard suggested already builds a good case against what I was trying to do. Maybe you could add their observation in your answer for the sake of completeness? | |
| Mar 21 at 5:58 | answer | added | C7X | timeline score: 1 | |
| Mar 4 at 3:13 | comment | added | C7X | @Johan I was also going to write an answer mentioning that well-foundedness is not expressible in $L_{\omega_1\omega}$ (according to this comment it is not even expressible in $L_{\infty\omega}$) and attempting to produce an $M$ whose ordinals are non-well-founded, however if $\Sigma$-recursion does not apply, is it still worth writing? | |
| Feb 21 at 18:16 | comment | added | Johan | Yes you're right. Which makes the definition actually $\Sigma_2$ and so we can't use $\Sigma$-recursion. Thanks, that explains the contradiction with Gabe Goldberg comment. | |
| Feb 15 at 10:25 | comment | added | C7X | Should the definition of $L(x,\alpha+1)$ instead be something along the lines of $L(x,\alpha+1)\leftrightarrow\bigvee_{n\in\omega}\bigvee_{\varphi}\exists p_1,\ldots,p_n \, \forall y(y\in x\leftrightarrow (L(y,\alpha)\land\varphi(y,p_1,\ldots,p_n)))$, with a $L(p_1,\alpha)\land\ldots\land L(p_n,\alpha)$ restriction and comprehension for $y$? As is currently written, if $L(x,\alpha+1)$ there does not seem to be anything forbidding $L(x',\alpha+1)$ where $x'$ is an arbitrary subset of $x$. | |
| May 19, 2023 at 15:07 | history | edited | Johan | CC BY-SA 4.0 |
deleted 61 characters in body
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| May 18, 2023 at 16:10 | comment | added | Gabe Goldberg | If $\mathcal A$ is a countable fragment and $\alpha$ is a countable ordinal, then by $\mathbf{\Sigma}^1_1$-boundedness, either there is an illfounded $\mathcal A$-elementary end extension of $L_\alpha$ or the $\beta < \omega_1$ such that $L_\beta$ is an $\mathcal A$-elementary extensions of $L_\alpha$ are bounded below $\omega_1$. So it seems there must be something wrong with your argument (or mine, I guess). I haven't looked closely at your argument though. | |
| May 18, 2023 at 15:41 | history | asked | Johan | CC BY-SA 4.0 |