Timeline for What strengthenings of measurability do the Mostowski collapses of ultrapowers possess?
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 7, 2017 at 23:52 | comment | added | Keith Millar | @MihaHabič I know, but to not confuse the people who may read my question, I edited the question to be more readable. | |
| Nov 7, 2017 at 23:51 | vote | accept | Keith Millar | ||
| Nov 7, 2017 at 8:40 | comment | added | Miha Habič | So, with your edits, you really seem to be asking what kind of closure properties the ultrapower by an ultrafilter on $\lambda$ might posses. Joel already answered that in his post below; such ultrapowers never contain $V_{\lambda+2}$ (and are thus not closed under $2^\lambda$-sequences), but if GCH fails at $\lambda$, they might be closed under $\lambda^+$-sequences, and more. | |
| Nov 6, 2017 at 23:51 | comment | added | Keith Millar | Ok, I edited the question and replaced it with an equivalent statement. | |
| Nov 6, 2017 at 23:51 | history | edited | Keith Millar | CC BY-SA 3.0 |
Redid question with equivalency
|
| Nov 6, 2017 at 23:06 | history | edited | Andrés E. Caicedo | CC BY-SA 3.0 |
edited title
|
| S Nov 6, 2017 at 22:47 | history | suggested | CommunityBot | CC BY-SA 3.0 |
Corrected typo in title.
|
| Nov 6, 2017 at 19:30 | review | Suggested edits | |||
| S Nov 6, 2017 at 22:47 | |||||
| Nov 6, 2017 at 18:32 | comment | added | Asaf Karagila♦ | By the way, I know that $MO$ denotes the Mostowski collapse. I am just sufficiently familiar with the fact that is stated in most introductory places: where we consider ultrapowers, we identify a well-founded ultrapower with its transitive collapse. And otherwise, many of these questions are oddly placed (as I explain in my first comment). | |
| Nov 6, 2017 at 17:44 | comment | added | Asaf Karagila♦ | You can't delete a question with an upvoted answer... | |
| Nov 6, 2017 at 16:38 | comment | added | Keith Millar | I'll just delete the question, it doesn't seem like my point was broadcasked properly. | |
| Nov 6, 2017 at 16:23 | answer | added | Joel David Hamkins | timeline score: 4 | |
| Nov 6, 2017 at 16:22 | comment | added | Andreas Blass | At first, I read the question as asking whether the existence of one $j$ with critical point $\kappa$ and with some property would imply that all ultrapowers by $\kappa$-complete non-principal ultrafilters would have the same property. But that reading doesn't match your claim about measurability being an example. Every measurable cardinal $\kappa$ has a $\kappa$-complete non-principal ultrafilter in whose ultrapower $\kappa$ is no longer measurable (equivalently: minimal in Mitchell order). | |
| Nov 6, 2017 at 16:14 | comment | added | Joel David Hamkins | Ah, good idea Miha. It isn't about Mostowski collapses at all. I would phrase the question as: which large cardinal embeddings can be realized by ultrapowers? And then one should say what kinds of ultrapowers one is talking about. | |
| Nov 6, 2017 at 16:09 | comment | added | Miha Habič | I think what you are asking is whether many (or all?) of the reflection properties captured by large cardinal embeddings can be realized by ultrapowers by ultrafilters. Is this correct? And if so, do you want to restrict the carrier set of the ultrafilter ($\kappa$-complete ultrafilters on $\kappa$? Or on $\mathcal{P}_\kappa(\lambda)$? Or something else?) | |
| Nov 6, 2017 at 16:02 | review | Close votes | |||
| Nov 6, 2017 at 22:50 | |||||
| Nov 6, 2017 at 16:00 | comment | added | Asaf Karagila♦ | Maybe it would be advisable to read again what people mean when they talk about ultrapower construction in set theory. Perhaps introductory books or papers. | |
| Nov 6, 2017 at 15:58 | comment | added | Keith Millar | @JoelDavidHamkins $M$ is a transitive inner model such that $j:V\rightarrow M$ is an elementary embedding with critical point $\kappa$. $M$ has no relation to $MO$. Am I misinterpreting your comment? | |
| Nov 6, 2017 at 15:55 | comment | added | Keith Millar | @AsafKaragila $MO$ is not the ultrapower, rather the Mostowski collapse of it. | |
| Nov 6, 2017 at 15:39 | comment | added | Joel David Hamkins | I'm confused by the question. When we say $M^\theta\subset M$ for an ultrapower $M$, we mean that $M$ is closed under $\theta$ sequences in a way that makes sense with thinking about $\langle M,\in^M\rangle$ as a structure in the language of set theory, which is of course not literally the same as saying it is closed under $\theta$-sequences. When $M$ is well-founded, as it is with complete ultrafilters, then this amounts to saying the corresponding thing about the Mostowski collapse of $M$, which trivializes the question. Set theorists usually identify the ultrapower with its collapse. | |
| Nov 6, 2017 at 15:33 | comment | added | Asaf Karagila♦ | This question is oddly phrased. How could $V_\alpha$ be a subset of the ultrapower anyway? Since the ultrapower is a collection of equivalence classes. Most sets are not equivalence classes. | |
| Nov 6, 2017 at 15:30 | history | asked | Keith Millar | CC BY-SA 3.0 |