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Namely, it is a theorem of Ketonen that if a cardinal $\kappa$ admits $\kappa$-complete uniform ultrafilters on every regular $\lambda\geq\kappa$, then $\kappa$ is strongly compact. …

For this, the situation would be that one runs into a problem strictly before $\kappa^+$ since otherwise the union of the ultrafilters on $\kappa$ in $V[G_\alpha]$ would be an ultrafilter in $V[G]$. …

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 …

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

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 …

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

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 …

Regarding your final request, here is an example of a forcing notion $\mathbb{P}$ that preserves all ground-model ultrafilters on $\omega$, but $\mathbb{P}\times\mathbb{P}$ destroys all ground model ultrafilters … In particular, forcing with $T$ once adds no reals, since it is a Suslin tree, and therefore preserves all ultrafilters on $\omega$. …

Here is a necessary and sufficient condition: Theorem. If $U$ is an ultrafilter in $V$, then the following are equivalent: $U\cup\{G\}$ is an ultrafilter base. $G$ is not disjoint from any element …

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 …

for ultrafilters on $\omega$. … In the large cardinal context of countably complete ultrafilters, we have normal measures, the Mitchell order and many other kinds of structure on the collection of ultrafilters. …

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 …

For question (1), the answer is the truth approximation property at $\delta$ implies the existence of a measurable cardinal. This is simply because the filter $\mathcal{F}$ witnessing your isomorphism …

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 …

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 …