show/hide this revision's text 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$.

show/hide this revision's text 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
show/hide this revision's text 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.

show/hide this revision's text 3 Fixed error with V; added 2 characters in body
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