6 complete classification of lattices D(U,V) ; added 266 characters in body

I've got it!

Theorem. The lattices of the form$D(U,V)$ admit a complete classification by the isomorphism classes of $U$ and $V$ and the question of whether $U\neq V$.

The point is that the lattice isomorphism class of $D(U,V)$, when$U\neq V$, determines and is determined by isomorphism classes of $U$ and $V$ (that is, by their Rudin-Keisler equivalence classes). Meanwhile, when the ultrafilters are the same, $D(U,U)$ is the empty lattice, independently of $U$.

Proof. Notice first that when $U\neq V$ we can recognize whether $U$ is principalfrom the lattice $D(U,V)$, which will have a least element exactly in thiscase; similarly, we can recognize whether $V$ is principal from$D(U,V)$, which will have a greatest element exactly in this case.

Next, let me recall from my earlier post how the filters $U$ and$V$ can be reconstructed directly from the $D(U,V)$, using $D(U,V)$ not just asa lattice but specifically as a collection of subsets of $D(U,V)$. To see this\kappa$. In this sense But let me now explain how one can get access to essentially thesame information up to isomorphism, just from knowing$D(U,V)$is just asa mashing-together lattice, and not knowing how these elements sit as subsets of$\kappa$. Assume$U$is non-principal, and let$V$, including essentially x$ be an arbitraryelement of $U$ on its lower cones and D(U,V)$, viewed now only as a lattice. Let$A$be thecomplements collection of elements immediate predecessors of$V$on its upper cones. Finallyx$ in $D(U,V)$, namely,you also inquired about the relationship set of $a\in D(U,V)$ such that $a\lt x$ and there is nothingbetween a measure $U$ a$and$x$. (We know that these$a$represent the induced measure removalof one element of the set representing$V$defined x$.) Now, I can define anultrafilter $U'$ on $A$, by saying that $X\in V\iff \kappa\in j_U(X)$. The relevant concept here B\subset A$is in$U'$just in case there is a greatest lower bound to$A-B$in$D(U,V)$.If$x$is representing the Rudin–Keisler order, since one has set$\kappa=[f]_U$for some function X$ in $f:\kappa\to \kappa$, and it follows D(U,V)$, then I claim that$V=f*U$and consequently U\upharpoonright X\cong U'\upharpoonright A$, by the map thatmaps each point in $V\leq_{RK} U$ under X$to the Rudin–Keisler order. You are using set in$\kappa$A$ obtained by omitting thatpoint from $X$. Thus, $U'$ is Rudin-Keisler equivalant to $U$, andwas constructed purely from viewing $D(U,V)$ as a seed for a measurelattice.

IncidentallySimilarly, it could be that assume $U$ V$is non-principal and let$V$C$ be the set ofimmediate successors of $x$ in your example are actually isomorphic as measures the lattice $D(U,V)$. (and give rise Theselattice elements correspond exactly to the same ultrapower embedding), sets obtained by applying a permutation of $\kappa$, since this does not preserve normality.

Regarding your edit addingone additional point to the question, here set that $x$ is something easy to say:

Theorem.representing.) Definethe ultrafilter $V'$ on $C$ by $D\subset C$ is in $V'$ just incase there is no least upper bound of $D$ in $D(U,V)$. The following are equivalent. mapthat sends the elements of $C$ to the corresponding points of$\kappa$ actually used in $D(U,V)$ is isomorphic to a Rudin-Keisler isomorphismof $D(U',V')$ by V\upharpoonright(\kappa-X)$with$V'\upharpoonright C$. Soagain, from$D(U,V)$viewed purely as a permutation lattice, we are able toextract$f:\kappa\to\kappa$. V$ up to isomorphism.

Conversely, if $U\cong U'$ and $V\cong V'$are isomorphic simultaneously via the , where $U\neq V$ and$U'\neq V'$, then we may find a single function $f$.

Prooff$witnessing theisomorphisms simultaneously (working partly on a$U$-big set andpartly on its complement, a$V$-big set), thereby showing that$D(U,V)\cong D(U',V')$as lattices. The backward implication So this is cleara completeclassification of$D(U,V)$up to isomorphism as a lattice. The forward implication follows from my observation aboveQED Thus, that the properties about$U$and$V$are constructible from$D(U,V)$. So that when we perform that same process on$D(U',V')$, we get$U'$and can determine from the lattice isomorphism class of the lattice$V'$, which therefore simply be D(U,V)$ are precisely the properties that are determined by the isomorphism classes of $U$ and $V$ transformed by themselves, plus the knowledge of whether $f$. QEDU\neq V$. 5 added 582 characters in body The filters$U$and$V$can be reconstructed directly from the lattice$D(U,V)$. To see this, let$X$be any element of the lattice, so that$X\in U$and$X\notin V$. It follows that the complement of$X$is in$V$, and also any larger set than the complement of$X$is in$V$. From this, it follows that for$Y\subset X$we have$Y\in U$if and only if$Y\in D(U,V)$. So the ultrafilter$U$and the lattice$D(U,V)$agree completely on the subsets of$X$. This is enough to reconstruct$U$, since a set is in$U$if and only if it has$U$-large intersection with$X$. Similarly, we can reconstruct$V$, namely, a set$Y$is in$V$if and only if$Y-X\in V$, since$X$is not in$V$; the complement of$Y-X$is$X\cup(\kappa-Y)$, and this is not in$V$, but containing$X$it is in$U$and hence in$D(U,V)$. In summary, $$Y\in U\ \ \ \iff\ \ \ Y\cap X\in D(U,V)$$ $$Y\in V\ \ \ \iff\ \ \ X\cup(\kappa-Y)\in D(U,V)$$ and this does not depend on the choice of$X\in D(U,V)$. In this sense,$D(U,V)$is just a mashing-together of$U$and$V$, including essentially$U$on its lower cones and the complements of elements of$V$on its upper cones. Finally, you also inquired about the relationship between a measure$U$and the induced measure$V$defined by$X\in V\iff \kappa\in j_U(X)$. The relevant concept here is the Rudin–Keisler order, since one has$\kappa=[f]_U$for some function$f:\kappa\to \kappa$, and it follows that$V=f*U$and consequently$V\leq_{RK} U$under the Rudin–Keisler order. You are using$\kappa$as a seed for a measure. Incidentally, it could be that$U$and$V$in your example are actually isomorphic as measures (and give rise to the same ultrapower embedding), by applying a permutation of$\kappa$, since this does not preserve normality. Regarding your edit to the question, here is something easy to say: Theorem. The following are equivalent. •$D(U,V)$is isomorphic to$D(U',V')$by a permutation$f:\kappa\to\kappa$. •$U\cong U'$and$V\cong V'$are isomorphic simultaneously via the function$f$. Proof. The backward implication is clear. The forward implication follows from my observation above, that$U$and$V$are constructible from$D(U,V)$. So that when we perform that same process on$D(U',V')$, we get$U'$and$V'$, which therefore simply be$U$and$V$transformed by$f$. QED 4 swapped U and V in seed example The filters$U$and$V$can be reconstructed directly from the lattice$D(U,V)$. To see this, let$X$be any element of the lattice, so that$X\in U$and$X\notin V$. It follows that the complement of$X$is in$V$, and also any larger set than the complement of$X$is in$V$. From this, it follows that for$Y\subset X$we have$Y\in U$if and only if$Y\in D(U,V)$. So the ultrafilter$U$and the lattice$D(U,V)$agree completely on the subsets of$X$. This is enough to reconstruct$U$, since a set is in$U$if and only if it has$U$-large intersection with$X$. Similarly, we can reconstruct$V$, namely, a set$Y$is in$V$if and only if$Y-X\in V$, since$X$is not in$V$; the complement of$Y-X$is$X\cup(\kappa-Y)$, and this is not in$V$, but containing$X$it is in$U$and hence in$D(U,V)$. In summary, $$Y\in U\ \ \ \iff\ \ \ Y\cap X\in D(U,V)$$ $$Y\in V\ \ \ \iff\ \ \ X\cup(\kappa-Y)\in D(U,V)$$ and this does not depend on the choice of$X\in D(U,V)$. In this sense,$D(U,V)$is just a mashing-together of$U$and$V$, including essentially$U$on its lower cones and the complements of elements of$V$on its upper cones. Finally, you also inquired about the relationship between a measure$V$U$ and the induced measure $U$ V$defined by$X\in U\iff V\iff \kappa\in j(X)$j_U(X)$. The relevant concept here is the Rudin–Keisler order, since one has $\kappa=[f]_V$ \kappa=[f]_U$for some function$f:\kappa\to \kappa$, and it follows that$U=f*V$V=f*U$ and consequently $U\leq_{RK} V$ V\leq_{RK} U$under the Rudin–Keisler order. You are using$\kappa$as a seed for a measure. Incidentally, it could be that$U$and$V$in your example are actually isomorphic as measures (and give rise to the same ultrapower embedding), by applying a permutation of$\kappa\$, since this does not preserve normality.

3 Fixed error with V; added 2 characters in body
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