Timeline for When do two ultrafilters yield isomorphic ultrapowers?
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| Aug 7, 2020 at 0:33 | history | edited | tomasz | CC BY-SA 4.0 |
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| Aug 7, 2020 at 0:29 | vote | accept | tomasz | ||
| Aug 6, 2020 at 23:57 | answer | added | Gabe Goldberg | timeline score: 10 | |
| Aug 6, 2020 at 23:39 | comment | added | tomasz | @GabeGoldberg: I see. My knowledge of the Rudin-Keisler order is rather superficial --- I suppose it is a standard fact that $j_U=j_W$ implies that $U$ and $W$ are equivalent. But if they are not $\sigma$-complete, then $j_U$ does not make sense, as $V^U$ is not well-founded, no? Anyway, a definitive answer even in the case of $\lambda=\aleph_0$ would be interesting, I guess. | |
| Aug 6, 2020 at 23:17 | comment | added | Gabe Goldberg | No you're right, I meant in the countable case. In large cardinal theory, if $U$ is countably complete, then $j_U : V\to M_U$ denotes the transitive collapse of the ultrapower of the universe of sets by $U$. | |
| Aug 6, 2020 at 23:13 | comment | added | tomasz | @GabeGoldberg: What do you mean by $j_U, j_W$? Regarding signatures: the question I was pondering was without any restriction on the signature, though I guess it would be interesting to ask about the restriction to smaller signatures (although I was more inclined to think of signatures of size $\lambda$, $2^{\lambda}$ or $2^{<\lambda}$, but if you say it already trivializes for continuum, then maybe that does not make sense.). | |
| Aug 6, 2020 at 22:48 | comment | added | Gabe Goldberg | Do you want to restrict to signatures that are finite or maybe countable or maybe of size less than the continuum? Otherwise there is an unsatisfying positive answer. | |
| Aug 6, 2020 at 22:37 | comment | added | Gabe Goldberg | For countably complete ultrafilters $U$ and $W$, this is true, because for any set $x$, it $y$ is its transitive closure, one has $j_U(y,x,\in)\cong j_W(y,x,\in)$, and by wellfoundedness, we may act as though these two structures are actually equal. Hence $j_U$ and $j_W$ agree on all sets, so they are Rudin-Keisler equivalent, which is what you're asking about. For the countably incomplete case, I will need some aspirin. | |
| Aug 6, 2020 at 22:19 | history | edited | YCor |
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| Aug 6, 2020 at 20:44 | history | asked | tomasz | CC BY-SA 4.0 |