Timeline for Explicit counter example to Vopěnka's principle in the constructible universe?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| May 8, 2023 at 15:44 | comment | added | Noah Schweber | @Ândsonjosé Since $\kappa=crit(j)$, that can't happen. | |
| May 8, 2023 at 13:29 | comment | added | Ândson josé | Supose that $\kappa= j(\gamma)$ for some $\gamma$. So $U$ is the trivial ultrafilter and $\kappa$ is not necessary measurable. | |
| Apr 22, 2015 at 19:42 | vote | accept | Chris Schommer-Pries | ||
| Apr 22, 2015 at 16:06 | comment | added | Noah Schweber | Fun aside: this argument - together with something that looks like the ultrafilter construction in my answer - shows that the existence of a measurable is equivalent to the existence of an elementary embedding of $V$ into some transitive class $M$. This is sort of a "phase transition" in large cardinals: whereas small large cardinals are defined combinatorially, it turns out that elementary embeddings are the right tool for defining/studying large large cardinals, and measurable cardinals sit right in between the two levels. | |
| Apr 22, 2015 at 16:04 | comment | added | Noah Schweber | Now, we have a problem: we can show that $W$ is a transitive model of ZFC containing all the ordinals, and that $W$ is a proper subclass of $L$. But this is impossible, since $L$ is (provably) the minimal inner model. (Why is $W\subsetneq L$? Well, the image of $\kappa$ under the natural elementary embedding from $V$ into $W$ is strictly larger than $\kappa$ - fun exercise - but if we assume $\kappa$ is the least measurable, this means that $W$ can't tell that $\kappa$ is measurable, so $W$ must not contain $\mathcal{U}$.) See Jech's book for the details of this argument. | |
| Apr 22, 2015 at 16:01 | comment | added | Noah Schweber | Ah, this is a beautiful argument! Roughly: suppose $\mathcal{U}$ is a countably complete ultrafilter on $\kappa$. We may take the ultrapower of $V$ (the whole universe!) mod $\mathcal{U}$; since $\mathcal{U}$ is countably closed, the result object $W$ - as a class-sized $\{\in\}$-structure - is well-founded. Now, this means we can apply the Mostowski collapse and view $W$ as a subclass of $V$. (cont'd) | |
| Apr 22, 2015 at 15:05 | comment | added | Chris Schommer-Pries | Right, of course. Great, so then it is clear to me that the category of structures for this language is a locally presentable category (the morphisms between the $L_\alpha$ are elementary embeddings since they satisfy extensionality, right?). You have constructed a proper class of objects and shown that if that class is not discrete, then there exists a measurable cardinal. Now the main thing that I still don't follow is why $V= L$ is inconsistent with the existence of measurable cardinals. | |
| Apr 22, 2015 at 14:30 | comment | added | Noah Schweber | In my answer the language was the language $\{\in\}$ of set theory, that is, by "$L_\alpha$" I meant "$(L_\alpha, \in)$". | |
| Apr 22, 2015 at 14:12 | comment | added | Chris Schommer-Pries | My guess would be that the locally presentable category would be something like the category of "first order structures" (not necessarily of the form $L_\alpha$). But now I am going to show the fact that I am not a set theorist because I don't quite know what this means. When I've seen "structure" used it has been in the following context: there is some (first order) language L, and then a structure is a set M with an interpretation of L in M. In your formulation what was the language L? or is it supposed to very? | |
| Apr 22, 2015 at 13:34 | comment | added | Noah Schweber | This might be silly, but: I think condensation lets us show that the category whose objects are the $L_\alpha$s, and whose morphisms are elementary embeddings, is locally presentable? | |
| Apr 22, 2015 at 13:15 | comment | added | Chris Schommer-Pries | So how can I turn this counter example into a locally presentable category with a large full discrete subcategory, like I originally wanted? | |
| Apr 22, 2015 at 12:57 | comment | added | Joel David Hamkins | Meanwhile, it is interesting to note that if Vopenka's principle holds in $V$, when it is larger than $L$, then the class of all $L_{\kappa^+}$ still exists in $V$, and in this case there is an elementary embedding from one $L_{\kappa^+}$ to another. It is just that the embedding does not exist in $L$. So the "counterexample" to VP still exists in $V$, but it is no longer a counterexample there. | |
| Apr 22, 2015 at 12:09 | history | edited | Noah Schweber | CC BY-SA 3.0 |
added 174 characters in body
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| Apr 22, 2015 at 12:08 | comment | added | Noah Schweber | Re: your second comment, yes, the punchline is that measurables can't exist in $L$; I've edited to make that explicit. | |
| Apr 22, 2015 at 11:48 | comment | added | Noah Schweber | $S$ is in $L_{\kappa^+}$ via condensation: take some elementary submodel $M$ of $L_\beta$ (where $S\in L_\beta$ - such a $\beta$ exists since $V=L$) with $L_{\kappa+1}\subseteq M$, $S\in M$, and $\vert M\vert=\kappa$. Now by condensation, the transitive collapse $N$ of $M$ is isomorphic to some $L_\gamma$; by cardinality considerations, we have $\gamma<\kappa^+$. But since $L_\kappa\subset M$, we know that the image of $S$ in $N$ is just $S$ itself - so $S\in L_\gamma$, and hence in $L_{\kappa^+}$. | |
| Apr 22, 2015 at 10:13 | comment | added | Neil Strickland | Also, I guess that the proof is supposed to finish via the argument explained at math.stackexchange.com/a/226616/146477. Is that what you had in mind? | |
| Apr 22, 2015 at 10:11 | comment | added | Neil Strickland | Can you explain why $S$ exists in $L_{\kappa^+}$? | |
| Apr 22, 2015 at 8:50 | history | answered | Noah Schweber | CC BY-SA 3.0 |