Timeline for Is there a pair of non-isomorphic structures each of which is isomorphic to an ultrapower of the other?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Nov 23, 2018 at 16:24 | comment | added | YCor | I'd tend to guess that under the negation of CH (in ZFC), two suitable ultraproducts of the real field could work (one needs the negation of CH to have them non-isomorphic). | |
| Nov 22, 2018 at 21:29 | vote | accept | James E Hanson | ||
| Nov 21, 2018 at 15:57 | comment | added | Asaf Karagila♦ | Since we can assume that $I=J$, this gives us a reasonable notion of iterating the ultrapowers as an ultrapower itself, moreover the iteration is just taking the product of the ultrafilters in the Stone space (modulo some reasonable bijection of $I$ with $I^2$, that is). Which goes back to my remarks about solving some equations in the Stone space (although with multiplication rather than with addition), and with discerning properties of ultrapowers from properties of filters, etc. | |
| Nov 21, 2018 at 15:54 | answer | added | Gabe Goldberg | timeline score: 8 | |
| Nov 21, 2018 at 15:15 | comment | added | Miha Habič | A (trivial) restatement of the question: is there a structure $\mathfrak{A}$ and ultrafilters $\mathcal{U},\mathcal{F}$ such that $\mathfrak{A}$ is isomorphic to $\mathfrak{A}^{\mathcal{U}\cdot\mathcal{F}}$ but not $\mathfrak{A}^{\mathcal{U}}$? | |
| Nov 21, 2018 at 14:15 | comment | added | James E Hanson | I meant it in the sense of "as well as sets $I$, $J$ and...". I've edited the question. | |
| Nov 21, 2018 at 14:12 | history | edited | James E Hanson | CC BY-SA 4.0 |
Clarified quantification of $I$ and $J$.
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| Nov 21, 2018 at 14:06 | comment | added | Andrés E. Caicedo | Hmm... actually, my first reply to @YCor is wrong, sorry, of course there are upper bounds on the size of $I,J$. Apologies. So $\mathfrak A,\mathfrak B$ must be reasonably saturated and yet not isomorphic... | |
| Nov 21, 2018 at 13:52 | comment | added | Andrés E. Caicedo | @YCor (That comment about quizzes seems unnecessarily uncharitable on your part.) In any case, just as in the Keisler-Shelah result, I would expect precise bounds on the smallest possible size of $I,J$ to depend on set-theoretic considerations. | |
| Nov 21, 2018 at 13:46 | comment | added | YCor | @AndrésE.Caicedo yes of course, but this site is no quiz to guess what a sensible interpretation is (and the question is meaningful, say with $I,J$ infinite countable). | |
| Nov 21, 2018 at 13:15 | comment | added | Andrés E. Caicedo | @YCor The first interpretation is the sensible one at this stage. Just as in the proof of the Keisler-Shelah theorem, one is not going to get the required situation if $I,J$ are given to us beforehand. (But, once one has an affirmative answer, this changes, as it then suffices that $I,J$ are large enough.) | |
| Nov 21, 2018 at 8:20 | comment | added | YCor | Quantifiers on $I$ and $J$ are missing. I guess the question means "as well as sets $I,J$ and...". But it also makes sense fixing $I$ and $J$ beforehand. Please clarify. | |
| Nov 21, 2018 at 7:16 | comment | added | Asaf Karagila♦ | I guess I don't know. But i would imagine it is something about solving the equation $U+x=F+y$ in $\beta\omega$, or something. | |
| Nov 21, 2018 at 7:15 | comment | added | James E Hanson | What do you mean by 'complementing ultrafilters'? | |
| Nov 21, 2018 at 7:13 | comment | added | Asaf Karagila♦ | One goal post would be to look at the characteristics of ultrapowers which follow from properties of the ultrafliter, then try to take two ultrafilters which produce non-isomorphic ultrapowers (say of $\Bbb N$) and then iterate that with some complementing ultrafilters to get the wanted result. | |
| Nov 21, 2018 at 7:11 | comment | added | Asaf Karagila♦ | We can assume $I=J$ by taking their union if necessary. Also, since both are $\frak A$ and $\frak B$ satisfy this being isomorphic to an ultrapower of the other, they have the same cardinality. | |
| Nov 21, 2018 at 7:02 | history | asked | James E Hanson | CC BY-SA 4.0 |