Timeline for Extending complete filters
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 29, 2012 at 0:13 | vote | accept | Tomasz Kania | ||
| Dec 28, 2012 at 22:49 | comment | added | Asaf Karagila♦ | Andres, thank you for the correction! I was under the impression that supercompact were known to be stronger (in consistency) than strongly compact. I guess this is just one of these common cases where I remembered something slightly wrong. | |
| Dec 28, 2012 at 22:43 | comment | added | Andrés E. Caicedo | Asaf, what you said is not quite correct. Any supercompact is strongly compact, but the general expectation is that the theories ZFC+"There is a supercompact cardinal" and ZFC+"There is a strongly compact cardinal" should be equiconsistent, so neither should be able to prove the existence of (set) models of the other. | |
| Dec 28, 2012 at 22:41 | comment | added | Joel David Hamkins | François, just $2^\kappa$-compactness suffices. If $j:V\to M$ is a strongly $\theta$-compact embedding, and $F$ is a $\kappa$-complete filter generated by a base $F_0$ of size at most $\theta$, then by the strong compactness cover property, $j"F_0$ is covered by a set $F_1\in M$ of size less than $j(\kappa)$, and without loss $F_1\subset j(F)$. So as in my answer, take $a\in \bigcap F_1\in j(F)$ and generate the ultrafilter $U$. | |
| Dec 28, 2012 at 22:32 | answer | added | Joel David Hamkins | timeline score: 9 | |
| Dec 28, 2012 at 22:29 | comment | added | Joseph Van Name | Those cardinals that you are talking about are called strongly measurable cardinals in the book "The Theory of Ultrafilters" by Comfort and Negrepontis if I remember correctly(I do not have the book with me at the moment). | |
| Dec 28, 2012 at 22:25 | comment | added | François G. Dorais | Wouldn't you just need $(2^\kappa)^+$-compactness for this? | |
| Dec 28, 2012 at 22:10 | comment | added | Asaf Karagila♦ | (Unless, of course, it would turn out to have an inherit inconsistency, a-la Reinhardt cardinals. So far, however, we haven't found such inconsistency.) | |
| Dec 28, 2012 at 22:09 | comment | added | Asaf Karagila♦ | If the theory "ZFC+There is a strongly compact cardinal" is consistent then it is impossible that it would be a theorem of ZFC. But depending on your background assumptions (which are generally thought of as plain ZFC, I suppose) we cannot prove nor disprove the existence of large cardinals. If your theory is, for example, ZFC+"There is a supercompact cardinal" then we can prove that there is a model in which there is a strongly compact cardinal. So we cannot disprove it from ZFC. | |
| Dec 28, 2012 at 22:04 | comment | added | Tomasz Kania | So, might it happen that "there are no strongly comapct cardinals" is a theorem of ZFC? | |
| Dec 28, 2012 at 21:52 | comment | added | Asaf Karagila♦ | We know it doesn't. This is why large cardinals are often called "Strong infinity axioms". | |
| Dec 28, 2012 at 21:41 | comment | added | Tomasz Kania | Do we know whether Con(ZFC) implies Con(ZFC + there is a strongly compact cardinal)? | |
| Dec 28, 2012 at 21:07 | comment | added | Tomasz Kania | Actually, this counts as an answer since the filter I am dealing with is rather mysterious and, consequently, I have no chance for a "painless" extension, by the above. | |
| Dec 28, 2012 at 21:02 | comment | added | Emil Jeřábek | en.wikipedia.org/wiki/Strongly_compact_cardinal | |
| Dec 28, 2012 at 20:55 | history | edited | Tomasz Kania | CC BY-SA 3.0 |
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| Dec 28, 2012 at 20:48 | history | asked | Tomasz Kania | CC BY-SA 3.0 |