Timeline for Idempotent ultrafilters and the Rudin-Keisler ordering
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 4, 2013 at 16:02 | comment | added | Peter Krautzberger | No problem at all, glad I could help. | |
| Aug 4, 2013 at 2:56 | comment | added | Andreas Blass | @PeterKrautzberger Thanks for correcting my erroneous guess, and apologies for not correctly remembering the result. | |
| Aug 4, 2013 at 0:04 | comment | added | Peter Krautzberger | (just to be clear: this is old news, see the book by Hindman&Strauss; it's just that my thesis is freely available.) | |
| Aug 3, 2013 at 23:53 | comment | added | Peter Krautzberger | There are idempotents RK-above any ultrafilter, e.g., you can extend the inverse filter under the map that maps each $n$ to the minimum (or maximum) of its binary expension. See my dissertation | |
| Aug 3, 2013 at 22:57 | vote | accept | Noah Schweber | ||
| Aug 3, 2013 at 21:37 | comment | added | Andreas Blass | @NoahS for your second comment: I don't know but I would expect that there are ultrafilters on $\omega$ with no idempotent ultrafilters RK-above them. I would be surprised if there exist (say under CH if it helps) any nonprincipal ultrafilter U and any function f such that the set of ultrafilters mapped to U by f is nonempty and closed under addition. | |
| Aug 3, 2013 at 21:32 | comment | added | Andreas Blass | @NoahS for your first comment: Under CH, the Ramsey ultrafilters below a given ultrafilter U don't characterize U (up to isomorphism). For one thing, there are lots of ultrafilters (including some P-points) with no Ramsey ultrafilters below them. Also, if you fix a Ramsey ultrafilter V, then there are several (undoubtedly $2^{\aleph_1}$, but I haven't checked carefully) isomorphism classes of ultrafilters that are RK-above U and no other Ramsey ultrafilters - in fact above no other ultrafilters at all (except of course themselves and principal ultrafilters). | |
| Aug 3, 2013 at 20:39 | comment | added | Noah Schweber | A more relevant question: do idempotent ultrafilters exist RK-above an arbitrary ultrafilter? I suspect the answer is yes, but I'm having trouble proving it: given a function $f: \omega\rightarrow\omega$ and an ultrafilter $U$, the set of $V$ RK-above $U$ via $f$ is compact as a subset of $\beta\mathbb{N}$, but it's not clear to me that it is closed under $\oplus$ (if it were, I could apply Ellis' theorem and be done). Is this known? | |
| Aug 3, 2013 at 18:25 | comment | added | Noah Schweber | I'm curious: is it known to what extent (assuming CH, say) the Ramsey ultrafilters below a given ultrafilter $U$ characterize $U$? That is, are there non-RK-equivalent $U, V$ which RK-bound the same Ramsey ultrafilters? Alternatively, is it consistent that there are no such pairs of ultrafilters? (It is certainly consistent that such a pair exists, since all pairs of ultrafilters satisfy this property if there are no Ramsey ultrafilters.) | |
| Aug 3, 2013 at 17:42 | history | answered | Andreas Blass | CC BY-SA 3.0 |