bio | website | peter.krautzberger.info |
---|---|---|
location | Germany | |
age | ||
visits | member for | 5 years |
seen | Mar 3 at 9:03 | |
stats | profile views | 722 |
Mathematician by training. Works for MathJax.
Sep 24 |
awarded | Autobiographer |
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. |