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For example, if $\kappa$ is $\kappa+2$-strong, then there must be ultrafilters $U$ on $\kappa$ with Rudin-Kiesler rank $\omega$, giving rise to the situation of your question. … In the general case, or even in the case of ultrafilters on $\omega$, it remains easy to build increasing chains in the Rudin-Kiesler order. …

It is relatively consistent with ZFC plus the existence of a measurable cardinal $\kappa$ that all your cancellation laws hold for the $\kappa$-complete ultrafilters on $\kappa$. … The same idea would work for ultrafilters on other sets, such as $\omega$, provided that $\cal{U}$, $\cal V$ and $\cal W$ were all expressible as finite products of a common factor ultrafilter, and I seem …

Thus, the number of ultrafilters on an amorphous set $x$ is precisely $x+1$, which is strictly smaller than $2^x$. …

If $a$ and $b$ are principal ultrafilters, then so is the product filter as you have defined it. …

It follows that the Rudin-Keisler minimal ultrafilters are precisely the normal measures. … Meanwhile, the Rudin-Keisler order on this collection of ultrafilters is well-founded, which fulfills part of what you had requested. …

Under the Continuum Hypothesis, your solution space is all nonprincipal ultrafilters. …

No. Every nonprincipal ultrafilter $U$, considered as a partial under $\subseteq$, is a nontrivial product order. To see this, suppose that $U$ is a nonprincipal ultrafilter on $\kappa$. Partition $\k …

The hugeness of $\kappa$ is witnessed by an embedding $j:V\to M$ for which $M^\lambda\subset M$, where $\lambda=j(\kappa)$. In particular, for such an embedding we have $j''\lambda\in M$, and one may …

To get the ball rolling... One can show easily that $\chi(U)$ must be at least $\kappa^+$, since otherwise one can take the diagonal intersection of a $\kappa$-sized family and find a single set that …

This includes all nonprincipal ultrafilters on any set, unless there is a measurable cardinal, and in any case includes all nonprincipal ultrafilters on any set of size less than the least measurable cardinal … QED So the result is true for all $\sigma$-incomplete ultrafilters. …

If your filter is generated by $\kappa$ many sets, then indeed the conclusion you seek can be made, by a direct argument that does not go through strong compactness. Theorem. The following are equiva …

The existence of ultrafilters on every Boolean algebra (which implies non-principal ultrafilters on ω, since these come from ultrafilters on the Boolean algebra P(ω)/Fin) is a set-theoretic principle that … In this case, neither DC nor ACω would imply the existence of such ultrafilters. I'm less sure about finding models that have ultrafilters on ω, but not on all Boolean algebras. …

Let me offer a counterpoint to François's informative answer and explain how one also can look upon the nonstandard ∈ relation on pseudo finite sets as very rich indeed. I claim that every countable …

The ultrafilter $U$ is said to be Rudin-Kiesler below $G$, and this ordering on ultrafilters is intensely studied in large cardinal set theory. …

This kind of product ultrafilter occurs routinely in the set-theoretic large cardinal literature, since many large cardinals involve this kind of measure, particularly in the case where these ultrafilters …