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The point is that the ultrapower of any structure $\mathcal{M}$ by a nonprincipal ultrafilter $\mu$ on $\mathbb{N}$ is countably saturated, that is, it realizes any finitely satisfiable $n$-type with …

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

The answer is Yes, and this is the ultraproduct construction. Let U be any nonprincipal ultrafilter on the set of primes. This is simply the dual filter to a maximal ideal on the set of primes, contai …

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

I assume that ultrafilters for you cannot be the whole lattice (since otherwise the ultrafilter assertion would become trivialized). …

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 …

My preferred account of extenders is the following: an elementary embedding $j:V\to M$ is an extender embedding if every element of $M$ can be expressed in the form $j(f)(\alpha)$, where $f:\kappa\to …

It follows that every projective set in $V[G]$ is Lebesgue measurable, and consequently, there can be no nonprincipal projective ultrafilters in $V[G]$, since the existence of a nonprincipal ultrafilter … [Update:] And indeed, by a result of Shelah, it is equiconsistent with ZFC that every projective set has the property of Baire, and in that model, there can be no projective nonprincipal ultrafilters, …

Easy differences arise if one allows principal ultrafilters, since the ultrapower of $X$ by a principal filter is canonically isomorphic to $X$, but other ultrapowers are not. … Without the GCH, it is consistent with ZFC to have ultrafilters on the same set leading to nonisomorphic ultrapowers. …

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. …

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. …

The answer is: zero. The reason is that every ultrafilter has zero as the infimum of the upper density of its members. To see this, observe that if a set $U$ is in the ultrafilter $\mathcal{U}$, with …

The answer is that the two models are related in the most natural way: The ultraproduct and forcing extension constructions commute, in the sense that the ultraproduct of a sequence of forcing extensi …