508 reputation
412
bio website peter.krautzberger.info
location Germany
age
visits member for 4 years, 1 month
seen Jul 10 at 14:01

Mathematician by training. Works for MathJax.


Feb
8
awarded  Good Answer
Aug
4
comment Idempotent ultrafilters and the Rudin-Keisler ordering
No problem at all, glad I could help.
Aug
4
accepted How “much” does (Grigorieff) forcing destroy an ultrafilter?
Aug
4
comment Idempotent ultrafilters and the Rudin-Keisler ordering
Todd, see my comment to Andreas's answer -- given any two ultrafilters $p,q$ you'll find an idempotent u with $f(u)=p, g(u)=q$.
Aug
4
comment Idempotent ultrafilters and the Rudin-Keisler ordering
(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
comment Idempotent ultrafilters and the Rudin-Keisler ordering
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
Jul
2
awarded  Yearling
Jan
26
awarded  Nice Answer
Sep
15
revised Are there q-filters which are not ultrafilters?
added 63 characters in body
Sep
15
answered Are there q-filters which are not ultrafilters?
Aug
14
awarded  Nice Answer
Jul
10
comment Awfully sophisticated proof for simple facts
Wouldn't you rather look at the power of an arbitrary element and cycles therein?
Jul
3
awarded  Yearling
Jun
14
awarded  Scholar
Jun
14
accepted Destroying the P-filter-property
Jun
14
comment Destroying the P-filter-property
Martin, thanks again. I was really hoping for a positive answer, but hank you for sharing these two partial ones.
Jun
6
comment Destroying the P-filter-property
Thank you, Martin.
Jun
4
revised Destroying the P-filter-property
small correction: filters on $\omega$
Jun
4
comment Destroying the P-filter-property
Martin, thanks! I hope you don't mind that I'll wait a bit to see if more (partial) answers turn up.
Jun
4
comment Destroying the P-filter-property
Yes, I mean filters on $\omega$. I'll add that to my question.