Skip to main content
14 events
when toggle format what by license comment
Aug 23, 2022 at 2:43 comment added lyrically wicked "I'm not sure what your objection is." — there was no objection; I just wanted to clarify a few things that are somewhat unrelated to the answer. "Which do you actually want to know about here?" — I wanted to know about the heights of the minimal rank of set models of uncomputably axiomatizable theories (such theories are mentioned in your comment on Math.SE). (Maybe this question needs a separate post.) Sorry for the possible confusion and thank you for the explanations!
Aug 22, 2022 at 16:46 comment added Noah Schweber @lyricallywicked I'm not sure what your objection is. In your comment you asked about ZFC+supercompact, so in my comment I focused on that. In your question you asked about general theories, so in my answer I focused on that. Which do you actually want to know about here?
Aug 22, 2022 at 7:18 comment added lyrically wicked "this has nothing to do with the details of T (beyond its computable axiomatizability)" — wait, but the question of this post implies that there is no such limitation. Pick any consistent theory T from the set of all infinite binary sequences (note that T may be uncomputably axiomatizable, and the complexity of axioms is unrestricted in the Lévy hierarchy). Then T must have a set model $S$. How large can the minimal rank of $S$ be? Will the answer depend on the definition of rank?
Aug 22, 2022 at 6:28 comment added Noah Schweber @lyricallywicked Depending on the model $S$ it could be quite small: every computably axiomatizable consistent theory has a "not-too-incomputable" model (e.g. $<_T0'$), so in particular if $T$ is consistent then it has a model in $L_{\omega+1}$. And again, this has nothing to do with the details of $T$ (beyond its computable axiomatizability). Of course, such a model won't be well-founded, or even an $\omega$-model.
Aug 22, 2022 at 6:27 comment added lyrically wicked Then if $S$ is a set that models such T, what is the minimal rank of $S$?
Aug 22, 2022 at 6:25 comment added Noah Schweber @lyricallywicked Yes, due to the completeness theorem, and this has nothing to do with the details of $T$ here. $T$ may not have a transitive set model, though, let alone one of the form $V_\alpha$ for some ordinal $\alpha$.
Aug 22, 2022 at 6:25 comment added lyrically wicked I have read the answers to this question on Mathoverflow, and now I have another question: if T = ZFC + "there exists a supercompact cardinal" is consistent, does T have a set model?
Aug 21, 2022 at 21:03 comment added Noah Schweber @lyricallywicked I think that $\alpha$ (assuming it exists) has no definition snappier than "The smallest ordinal $\alpha$ such that $V_\alpha\models T$." If you really want a different type of definition, you should specify what you're looking for (and note that it may not exist).
Aug 20, 2022 at 12:50 comment added lyrically wicked @ArvidSamuelsson: thank you for the explanation. Yes, my statement "$\alpha$ must be much smaller than $\beta_{\mathcal{L}}$" should be "$\alpha$ (if it exists) must be much smaller than $\beta_{\mathcal{L}}$"...
Aug 20, 2022 at 9:50 comment added Arvid Samuelsson @lyricallywicked If there is no cardinal that is supercompact up to the next worldly cardinal, there is no $\alpha$ such that $V_\alpha \vDash T$, even if there is a cardinal that is supercompact in $V$, for that supercompact cardinal may be the greatest worldly cardinal.
Aug 20, 2022 at 9:43 comment added Arvid Samuelsson More precisely, $\beta_{\mathcal{L}}$ is less than or equal to the least $\Sigma_2$-correct cardinal.
Aug 20, 2022 at 7:04 comment added lyrically wicked It seems that I need an explanation for the claim that $\beta_{\mathcal{L}}$ is less than the least supercompact. Consider the theory T = ZFC + "there exists a supercompact cardinal", which corresponds to some infinite binary sequence (existence of a supercompact cardinal is $\Sigma_3$ in the Lévy hierarchy). What is the smallest ordinal $\alpha$ such that $V_{\alpha} \models \text{T}?$ Note that $\alpha$ must be much smaller than $\beta_{\mathcal{L}}.$
May 12, 2022 at 8:39 vote accept lyrically wicked
May 6, 2022 at 4:59 history answered Noah Schweber CC BY-SA 4.0