Timeline for Can Assumptions about forcing produce Mice? [closed]
Current License: CC BY-SA 3.0
48 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 19, 2012 at 16:43 | history | locked | Kim Morrison | ||
| Nov 19, 2012 at 14:54 | comment | added | tweetie-bird | @Michael: it's not terribly clear from what you've written here what you are so upset about, but I'm sure everything will be fine if you just treat these mice fairly. I imagine these are grown-up mice you are talking about. | |
| Nov 19, 2012 at 10:26 | comment | added | Joel David Hamkins | I think it is a very natural and interesting question. | |
| Nov 19, 2012 at 8:27 | comment | added | S. Carnahan♦ | I don't know what exactly has happened here, but if Joel or Andreas answered one or more of your questions satisfactorily, it might be good to mark one of the answers as accepted, and leave the question in its most usable form. If you have other things to ask, it might be best to open a new question. | |
| Nov 19, 2012 at 7:42 | history | rollback | Andrés E. Caicedo |
Rollback to Revision 20
|
|
| Nov 19, 2012 at 7:19 | history | edited | Not Mike | CC BY-SA 3.0 |
deleted 87 characters in body
|
| Nov 19, 2012 at 7:08 | history | closed |
Not Mike Andrés E. Caicedo David Corwin Igor Rivin Todd Trimble |
no longer relevant | |
| Nov 19, 2012 at 7:08 | comment | added | Todd Trimble | Recent editing activity and comments make me feel that the question should probably be closed now as "no longer relevant". | |
| Nov 19, 2012 at 6:19 | comment | added | Not Mike | Mariano, sorry: Just learned life is a two player game, and the only sane ref is logic. | |
| Nov 19, 2012 at 5:35 | comment | added | Not Mike | Its very complicated sorry. | |
| Nov 19, 2012 at 4:54 | history | rollback | Mariano Suárez-Álvarez |
Rollback to Revision 18
|
|
| Nov 19, 2012 at 4:54 | comment | added | Mariano Suárez-Álvarez | PLEASE do not deface your own questions. People have put work in answering it, and removing the text makes their effort go to waste. | |
| Nov 19, 2012 at 4:33 | history | edited | Not Mike | CC BY-SA 3.0 |
deleted 2024 characters in body; edited title
|
| Nov 19, 2012 at 3:46 | history | edited | Not Mike | CC BY-SA 3.0 |
added 166 characters in body; deleted 166 characters in body
|
| Nov 19, 2012 at 0:55 | comment | added | Todd Trimble | Not my area of research, but: what's the question now? (You said the question had been changed based on the answers? or something has been changed based on the answers?) Also I don't follow the point of the edit. | |
| Nov 18, 2012 at 22:27 | history | edited | Not Mike | CC BY-SA 3.0 |
deleted 43 characters in body
|
| Nov 18, 2012 at 21:07 | history | edited | Not Mike | CC BY-SA 3.0 |
added 135 characters in body; added 9 characters in body; deleted 59 characters in body
|
| Nov 18, 2012 at 20:58 | history | edited | Not Mike | CC BY-SA 3.0 |
added 163 characters in body
|
| Nov 17, 2012 at 4:28 | history | edited | Not Mike | CC BY-SA 3.0 |
added 60 characters in body
|
| Nov 15, 2012 at 18:51 | answer | added | Andreas Blass | timeline score: 9 | |
| Nov 15, 2012 at 15:39 | answer | added | Joel David Hamkins | timeline score: 9 | |
| Nov 15, 2012 at 12:38 | history | edited | Not Mike | CC BY-SA 3.0 |
deleted 142 characters in body
|
| Nov 15, 2012 at 12:10 | history | edited | Not Mike | CC BY-SA 3.0 |
deleted 23 characters in body
|
| Nov 15, 2012 at 11:57 | comment | added | Andreas Blass | @Michael: Amit's comments about forcing that $\check\lambda$ is measurable refer to the last part of your question, in boldface, not to the specific things in the earlier part of your question. The underlying issue here is that "formal assertions about what can be forced" is probably not what you intended, since "you can force the existence of a measurable cardinal" is a formal assertion about what can be forced. | |
| Nov 15, 2012 at 11:54 | comment | added | Andreas Blass | I conjecture that Cof$(\mathbb P,\lambda)$ doesn't say what you intended, because the presence of $\beta$ and the requirement that $\delta\notin\beta$ seem irrelevant. If the range of $f$ is bounded by some $\delta\in\lambda$ then it's bounded by some $\delta\in\lambda\setminus\beta$, as we can just increase $\delta$ if necessary. | |
| Nov 15, 2012 at 11:50 | comment | added | Not Mike | Well, I didn't say $1 \forces_\mathbb{P} \lambda $ is measurable. I took a single consequence of a measurable cardinals existence: namely changing the cofinality of a regular cardinal without collapsing cardinals. and asked if that was the same as asserting a measurable exists (For what its worth singular cardinals, can't be measurable (club filter not being sufficiently closed and all.) so in the extension that cardinal is actually no longer measurable. Which is a subtle point you don't seem to be getting. | |
| Nov 15, 2012 at 11:41 | comment | added | Amit Kumar Gupta | ... because it says Con(ZFC + $\exists$ measurable)? | |
| Nov 15, 2012 at 11:40 | comment | added | Amit Kumar Gupta | I chose $\aleph_{\omega_1}$ because it's not regular. If $\lambda$ is singular then $\neg Cof(\lambda,\mathbb{P})$ holds trivially, since if $\lambda$ is singular, it can never be forced to be regular, by upwards absoluteness of singularity. Just so we're on the same page, do we agree that $Cof(\mathbb{P}, \lambda) \Leftrightarrow 1 \Vdash _{\mathbb{P}} \check{\lambda}$ regular, and that $\forall \alpha \in \lambda (cf(\alpha) < cf(\lambda)) \Leftrightarrow \lambda$ regular? Also does it make sense that "$1 \Vdash_{\mathbb{P}} \check{\lambda}$ measurable" has large cardinal strength... | |
| Nov 15, 2012 at 11:37 | history | edited | Not Mike | CC BY-SA 3.0 |
added 31 characters in body
|
| Nov 15, 2012 at 11:26 | comment | added | Not Mike | Key point here is the word Witness. In particular, a model does not satisfy a particular existential statement without first being able to produce a witness certifying that statement holds. | |
| Nov 15, 2012 at 11:21 | comment | added | Not Mike | I just want to know if a statement asserting the existence of a regular cardinal and particular partial order, has the same strength as the statement there exists a regular cardinal with a normal measure. | |
| Nov 15, 2012 at 11:19 | comment | added | Not Mike | $\aleph_{\omega_1}$ is not regular. | |
| Nov 15, 2012 at 11:18 | comment | added | Not Mike | Priky forcing using a normal measure for the least measurable cardinal \kappa satisfies the three statements I've outlined. The negation of a forcing statement is that below every p you densely witness the negation. The key point here is that the existential quantifier does not cross the forcing symbol without there being a $\mathbb{P}$-name in the ground-model which witnessed it. | |
| Nov 15, 2012 at 11:14 | comment | added | Amit Kumar Gupta | Also I'm not sure what the spirit of your bold question is. Trivially, "$\mathbb{P}$ forces $\lambda$ is measurable" has large cardinal strength. | |
| Nov 15, 2012 at 11:10 | comment | added | Amit Kumar Gupta | Let $\lambda = \aleph_{\omega_1}$ and let $\mathbb{P}$ be any forcing of size $\lambda^+$ which does nothing. Then the conjunction holds but $\lambda$ is clearly not measurable. | |
| Nov 15, 2012 at 11:07 | comment | added | Amit Kumar Gupta | Yes, the negation of $Cof(\mathbb{P}, \lambda)$ doesn't imply that $\lambda$ is not a cardinal, but I'm not sure why you're mentioning this. $Cof(\mathbb{P},\lambda)$ says that $\mathbb{P}$ forces (and hence preserves in this case) $\lambda$ to be a regular ordinal (and hence a cardinal in this case). The statement $\forall \alpha \in \lambda (cf(\alpha) < cf(\lambda))$ is a convoluted way of saying that $\lambda$ is a regular cardinal [since if $\lambda$ were not regular, then letting $\alpha = cf(\lambda)$ we'd get $cf(\lambda) = cf(cf(\lambda)) < cf(\lambda)$, contradiction]. | |
| Nov 15, 2012 at 11:04 | history | edited | Not Mike | CC BY-SA 3.0 |
deleted 92 characters in body
|
| Nov 15, 2012 at 10:51 | history | edited | Not Mike | CC BY-SA 3.0 |
added 46 characters in body; Post Made Community Wiki
|
| Nov 15, 2012 at 10:46 | history | edited | Not Mike | CC BY-SA 3.0 |
deleted 84 characters in body
|
| Nov 15, 2012 at 10:43 | comment | added | Not Mike | Also the negation of Cof(\mathbb{\po}, \lambda) only asserts that there is a sequence cofinal in \lambda indexed by an ordinal below \lambda. And should not imply that \lambda fails to be a cardinal. | |
| Nov 15, 2012 at 10:37 | history | edited | Not Mike | CC BY-SA 3.0 |
added 4 characters in body; deleted 175 characters in body
|
| Nov 15, 2012 at 10:25 | comment | added | Amit Kumar Gupta | Being forced by a dense set and being forced by $1$ are equivalent, as long as your poset has a $1$. Your formalization of $Cof(\mathbb{P},\lambda)$ says that $\mathbb{P}$ forces (and hence preserves) that $\lambda$ is regular. That's not the same as $\forall \alpha \in \lambda (cf(\alpha) < \lambda)$, which is just always true: $cf(\alpha) \leq \alpha < \lambda$. | |
| Nov 15, 2012 at 10:21 | history | edited | Not Mike | CC BY-SA 3.0 |
added 18 characters in body
|
| Nov 15, 2012 at 10:14 | history | edited | Not Mike | CC BY-SA 3.0 |
added 127 characters in body; deleted 147 characters in body
|
| Nov 15, 2012 at 9:39 | history | edited | Not Mike | CC BY-SA 3.0 |
added 31 characters in body
|
| Nov 15, 2012 at 9:22 | history | edited | Not Mike | CC BY-SA 3.0 |
deleted 52 characters in body
|
| Nov 15, 2012 at 9:03 | history | edited | Not Mike | CC BY-SA 3.0 |
added 86 characters in body; added 1 characters in body; edited title
|
| Nov 15, 2012 at 8:56 | history | asked | Not Mike | CC BY-SA 3.0 |