5
$\begingroup$

There are several well-studied orderings on the set $\omega^*$ of ultrafilters on the natural numbers. Three popular ones are $\le_i$ for $i = 1,2,3$. We define $\mathcal U \le_i \mathcal V$ to mean there exists a function $f \colon \omega \to \omega$ of type $(i)$ such that $f(\mathcal V) = \mathcal U$, where the types are explained below.

$(1)$ Arbitrary functions

$(2)$ Finite-to-one functions

$(3)$ Finite-to-one and weakly increasing functions

I am interested in whether there can there be immediate successors in any of the orderings above. We say $\mathcal V$ is an $(i)$-immediate successor to $\mathcal U$ to mean $\mathcal U <_i \mathcal V$ and there does not exist $\mathcal W$ such that $\mathcal U <_i \mathcal W <_i\mathcal V$.

Not being too fluent in this area, it's not easy for me to go into the literature, and see what large results can be made relate to these successors. Could anyone suggest relevant results or point me in the right direction please?

One application is that an element with no $(3)$-immediate successor would make the layers of the corresponding standard subcontinua behave a little like they were hereditarily indecomposable, and also a little like a composant of $\mathbb H^*$ that corresponds to a near coherence class without a $Q$-point.

More generally I am interested in descending chains. Some results I am aware of are as follows.

(1) If a near-coherence class $\mathscr E$ has no $Q$-point then it has a cofinal descending $(3)$-chain.

(2) There exist $(1)$-chains isomorphic to $\mathbb R$ under Martin's axiom. Indeed these chains consist of $P$-points, so they are also $(2)$-chains. (Andreas Blass, The Rudin-Keisler ordering of $P$-points, Trans. Amer. Math. Soc. 179 (1973), 145--166.).

$\endgroup$
2
  • $\begingroup$ In the case of the corresponding question for measurable cardinals $\kappa$, the answer is yes. In the canonical inner model $L[\mu]$ of one measurable cardinal $\kappa$, every measure on $\kappa$ is isomorphic to a finite product of $\mu$, and in this case the R-K classes form exactly an increasing $\omega$-sequence. So there are successors. $\endgroup$ – Joel David Hamkins Apr 13 '16 at 12:21
  • $\begingroup$ And actually, in that context, the R-K order for measures on a measurable cardinal is well-founded, and so one always has immediate successors. $\endgroup$ – Joel David Hamkins Apr 13 '16 at 12:31
1
$\begingroup$

This isn't a definitive answer (Andreas Blass should be able to provide that!) but the immediate successor phenomenon shows up in the study of weakly Ramsey ultrafilters and their relatives. I know a couple of references for this, but the real experts should be able to say more:

Andreas Blass, Ultrafilter Mappings and Their Dedekind Cuts, Transactions of the American Mathematical Society, Volume 188, Issue 2, 1974.

Ned Rosen, Weakly Ramsey P-points, Transactions of the American Mathematical Society, Volume 269, Number 2 February 1982.

$\endgroup$
2
  • 1
    $\begingroup$ I'm a little rusty. Does Andreas's original union ultrafilter paper help? IIRC, everything below a stable ordered union ultrafilter is either $\min(u)$, $\max(u)$, or $(\min(u),\max(u))$ so isn't the last one immediately before the first two? $\endgroup$ – Peter Krautzberger Apr 15 '16 at 9:00
  • 2
    $\begingroup$ @PeterKrautzberger In the case of a stable ordered-union ultrafilter $u$, we have that $u$ is an immediate successor of $(\min(u),\max(u))$, in the strong sense that all ultrafilters $<u$ are $\leq(\min(u),\max(u))$. In turn, $(\min(u),\max(u))$ is an immediate successor of two incomparable selective ultrafilters $\min(u)$ and $\max(u))$ (and there are no other nonprincipal ultrafilters below $(\min(u),\max(u))$. $\endgroup$ – Andreas Blass May 5 '16 at 20:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.