Timeline for Is $\mathsf{ZFC+V=L}$ consistently $\omega$-complete?
Current License: CC BY-SA 4.0
12 events
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| Sep 30, 2022 at 18:04 | comment | added | James E Hanson | Is this answer enough to conclude that there is a specific sentence $\varphi$ such that $T$ proves 'if there are $\omega$-models of $T$, then there are $\omega$-models of $T + \varphi$ and $T+\neg \varphi$? | |
| Mar 8, 2021 at 21:38 | vote | accept | Noah Schweber | ||
| Mar 8, 2021 at 21:38 | history | edited | Farmer S | CC BY-SA 4.0 |
Gave a detailed argument for $x\leq_{\mathrm{T}} T^+$.
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| Mar 8, 2021 at 21:34 | comment | added | Farmer S | Okay, I added it in. | |
| Mar 8, 2021 at 21:33 | history | edited | Farmer S | CC BY-SA 4.0 |
Gave a detailed argument for $x\leq_{\mathrm{T}} T^+$.
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| Mar 8, 2021 at 21:24 | comment | added | Noah Schweber | Ah, I see. I find it a bit easier to think in terms of coding structures, but you're right, that works; if you modify the answer to include this, I'll accept it! | |
| Mar 8, 2021 at 21:23 | comment | added | Farmer S | (I'm not saying to computably build (the theory of) $L_\alpha$ from the wellorder, but that for each $x$, there are integers $e,n$ such that $m\in x$ iff $T^+$ contains the statement "Let $\alpha$ be the ordertype of the wellorder given by $e$, and $y$ be the $n$th real of the theory of $L_\alpha$; then $m\in y$".) | |
| Mar 8, 2021 at 21:15 | comment | added | Noah Schweber | I think the following works: grab a whole model instead of just the theory. Specifically, working within $V\models S$, the $L$-least real $r$ coding an $\omega$-model of $\mathsf{ZFC+V=L}$ is hyperarithmetic and hence its jump is an element of the structure it codes, which it consequently computes - an obvious absurdity. | |
| Mar 8, 2021 at 21:15 | comment | added | Farmer S | Hmm, I guess it also doesn't really make sense to say $T^+$ models "$W$ is a wellorder''; it should be that we let $e$ be some integer which is the index of a wellorder of $\omega$ of the right length in whatever coding we're using, and from that integer, $T^+$ recovers $x$. (That's more formally what I had in mind.) | |
| Mar 8, 2021 at 21:12 | comment | added | Noah Schweber | Very nice! However, I think your proof that $x\le_TT^+$ isn't complete. Specifically, we cannot computably build $L_\alpha$ from a copy of $\alpha$, so "[we] can recover $x$ from $W$" isn't really correct on the nose. (In fact, given a well-ordering $A$, the only reals computable from all copies of $A$ are the computable ones.) I think I see how to get around this though. | |
| Mar 8, 2021 at 21:08 | vote | accept | Noah Schweber | ||
| Mar 8, 2021 at 21:10 | |||||
| Mar 8, 2021 at 20:56 | history | answered | Farmer S | CC BY-SA 4.0 |