Timeline for Orderings of ultrafilters
Current License: CC BY-SA 2.5
24 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 10, 2011 at 11:20 | comment | added | porton | @Joel David Hamkins: Don't bother about the proof that $F\leq_1 \mu$ implies that $F$ is an ultrafilter. I proved it in an other way anyway. | |
| Apr 10, 2011 at 9:44 | comment | added | porton | @Joel David Hamkins: Your proof that $F\leq_1 \mu$ implies that $F$ is an ultrafilter seems erroneous. In the proof you say that $f^{-1}Y\notin\mu$ implies $f^{-1}(I-Y)\in\mu$. It seems to not imply. So it seems to me that this is with an error. Can it be corrected? | |
| Mar 2, 2011 at 22:09 | vote | accept | porton | ||
| Mar 1, 2011 at 20:33 | comment | added | Joel David Hamkins | I have edited the answer; I had reversed some inclusions, but it seems that the counterexample still works fine. | |
| Mar 1, 2011 at 20:32 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
Fixed some errors; added 2 characters in body; deleted 2 characters in body; Post Made Community Wiki
|
| Mar 1, 2011 at 18:18 | comment | added | porton | No, my attempt to show that $\leq_1 = \leq_2$ failed. Indeed we now have no proof that these are inequal. The problem is open. | |
| Mar 1, 2011 at 14:10 | comment | added | porton | You say "we see that $G\leq_1\mu$ as witnessed by the identity function on $I$. It seems you err here: You've missed $\supseteq$ and $\subseteq$, it is not witnessed. I have the idea how to indeed prove true that $\leq_1 = \leq_2$ for arbitrary filters. I have not yet written down the proof, so I may also mistake. | |
| Jan 24, 2011 at 21:17 | comment | added | Joel David Hamkins | Andres, it is in my upcoming book! I think it is also in my paper on Prikry trees and canonical seeds (I think I give the basic seed theory there). But let me sketch it. Suppose $j_\nu=h\circ j_\mu$ and observe $X\in\mu\iff [id]_\mu\in j_\mu(X)\iff h([id]_\mu)\in j_\nu(X)$. In other words, $\mu$ is generated by the seed $h([id]_\mu)=[f]_\mu$ via $\nu$ for some representing function $f$, and from this it follows that $\mu=f*\nu$. | |
| Jan 24, 2011 at 18:07 | comment | added | Andrés E. Caicedo | @Joel: Do you have a reference for this characterization? I.e., for the direction: If $j_\nu$ factors through $j_\mu$, then there us a reduction $\mu\le_{RK}\nu$. | |
| Jan 24, 2011 at 15:43 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
Corrected spelling
|
| Jan 24, 2011 at 15:38 | comment | added | Joel David Hamkins | I edited. The Rudin-Keisler order is defined on Wikipedia en.wikipedia.org/wiki/Ultrafilter the same way I defined it, but only for ultrafilters, which is how I believe it is usually considered. There are numerous equivalent characterizations. For example, one interesting characterization is that $\mu\leq_{RK}\nu$ iff the ultrapower $j_\nu$ of the universe factors through $j_\mu$. | |
| Jan 24, 2011 at 15:33 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
Fixed typos; edited body
|
| Jan 24, 2011 at 15:11 | comment | added | porton | What is $\mu\of f[\nu]$? | |
| Jan 24, 2011 at 15:09 | comment | added | porton | First correct "G\leq_{RK]" F into "G\leq_{RK}". Second, I don't understand what $f*F$ is because it is not mentioned in the formula $X\in G\leftrightarrow f^{-1}X\in F$ which should define it. Third, where you've got that definition of Rudin-Kiesler order? I am now reading Comfort and Negrepontis ``The Theory of Ultrafilters''. There that order is defined in a different way and sadly it is not shown that it is equivalent to your definition. What you'd recommend me to read about this order? | |
| Jan 24, 2011 at 14:43 | comment | added | Joel David Hamkins | Yes, I misunderstood you, and how have now edited my answer. | |
| Jan 24, 2011 at 14:40 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
deleted 930 characters in body
|
| Jan 24, 2011 at 12:35 | comment | added | porton | You misunderstood me. By $b=f[a]$ where $a$ is a filter I mean that $b$ is the filter generated by the base $\lbrace f[A] | A \in a \rbrace$. Sorry, my notation is ambiguous. $f[a]$ for $a$ a filter on $U$ is not the same as $f[a]$ where $a$ is a subset of $U$. You should infer the meaning from context. | |
| Jan 24, 2011 at 12:29 | comment | added | Joel David Hamkins | If $b=f[a]$, then $f$ must be surjective, in order that $J\in f[a]$, since $J\in b$ for sure. So if there is no surjection from $I$ to $J$, then $b=f[a]$ is impossible. (My remarks about measures are just about countably complete ultrafilters, which are the same as 2-valued measures measuring all sets.) | |
| Jan 24, 2011 at 11:41 | comment | added | porton | I don't understand you. What you say about measures? In my original problem measures were not mentioned. What is "the filter generated by $a$ on $J$"? My notion of isomorphism is not too strong, it is carefully balanced. You say "there is no surjection from $I$ to $J$" what is irrelevant because I require existence of a function not a surjection. It seems for me that your entire answer is pointless (or do I misunderstand?) | |
| Jan 23, 2011 at 17:27 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
deleted 42 characters in body
|
| Jan 23, 2011 at 17:03 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
added 20 characters in body
|
| Jan 23, 2011 at 16:57 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
added 1022 characters in body; added 15 characters in body; added 2 characters in body; added 25 characters in body
|
| Jan 23, 2011 at 16:34 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
added 389 characters in body
|
| Jan 23, 2011 at 16:28 | history | answered | Joel David Hamkins | CC BY-SA 2.5 |