Timeline for Does there exist a section of $\mathcal{P}(\kappa)\to\mathcal{P}(\kappa)/(\text{fin})$ that is "nearly Boolean"?
Current License: CC BY-SA 4.0
7 events
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| 2 days ago | comment | added | David Gao | I did also make this somewhat more confusing by calling $f$ “almost Boolean”, since condition 3 forces $f$ to not be a Boolean algebra homomorphism (as otherwise the pullback of the principal ultrafilter at any $\lambda < \kappa$ will be an ultrafilter on $\mathcal{P}(\kappa)/I$ that does not satisfy condition 3). | |
| 2 days ago | comment | added | Will Brian | @DavidGao: Oh right, good point. I was just thinking of condition 1. | |
| 2 days ago | comment | added | David Gao | Thanks! But if $\mathcal{P}(\omega)/I$ is trivial, no section can satisfy condition 3, no? The unique ultrafilter is $\{[\omega]\}$ and $f([\omega]) = \omega$. | |
| 2 days ago | comment | added | Will Brian | @DavidGao: I think it depends on the ideal. If the ideal is all sets of cardinality $<\!\lambda$ for some $\lambda \leq \kappa$, then the answer is still no, for essentially the same reason. But here's a silly example where the answer is yes: let $u$ be an ultrafilter on $\mathcal P(\omega)/(\mathrm{fin})$, and let $I$ be the corresponding ideal. Then $\mathcal P(\omega)/I$ is trivial, and you can get a section like what you want. Off the top of my head, I don't have any examples of ideals where you get a nice section for interesting reasons. | |
| 2 days ago | comment | added | David Gao | Thanks a lot! Just in case: are you aware if this is still impossible should $\mathcal{P}(\kappa)/(\text{fin})$ be replaced by a further quotient? | |
| 2 days ago | vote | accept | David Gao | ||
| 2 days ago | history | answered | Will Brian | CC BY-SA 4.0 |