Lattice of differences between ultrafilters - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T05:41:24Z http://mathoverflow.net/feeds/question/105684 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105684/lattice-of-differences-between-ultrafilters Lattice of differences between ultrafilters Noah S 2012-08-28T05:51:07Z 2013-02-18T15:40:09Z <p>Consider two ultrafilters, $U$ and $V$, on the same cardinal $\kappa$. Let $D(U, V)=\lbrace X\subseteq \kappa: X\in U-V\rbrace$; clearly $D(U, V)$ is a lattice under $\subseteq, \cap, \cup$ since the intersection of two $U$- or $V$-large sets is $U$- or $V$-large, and the union of two $U$- or $V$-small sets is $U$- or $V$-small; by the same reasoning, $D(U, V)$ is a $\lambda$-complete lattice, where $\lambda$ is the minimum of the completeness of $U$ and the completeness of $V$. </p> <p>My general question is, does this lattice have any interesting properties?</p> <p>In particular, I'm interested in the following: let $M\models ZFC^-$, let $U\in M$ be a countably complete ultrafilter on some $M$-measurable cardinal $\kappa$, and let $j: M\rightarrow \prod M/U$ be the elementary embedding of $M$ into the ultrapower via $U$. Let $V=\lbrace X\in\wp^M(\kappa): \kappa\in j(X)\rbrace$; then $V\in M$ and $V$ is a normal ultrafilter on $\kappa$. In particular, if $U$ is not normal, then $U\not=V$. Intuitively, however, the difference between $U$ and $V$ is "small" (to be fair, this "intuition" may just be a figment of my not understanding inner model theory); is this somehow reflected by the lattice $D(U, V)$? In general, can anything about the relationship between $U$ and $V$ be read off of the lattice $D(U, V)$?</p> <p>(Also, is this notion studied somewhere? I've googled around, unsuccessfully.)</p> <p>EDIT: to clarify, I'm most interested in properties which can be determined from the isomorphism type of the lattice $D(U, V)$ alone.</p> <p>Thanks in advance; hopefully this isn't too open-ended.</p> http://mathoverflow.net/questions/105684/lattice-of-differences-between-ultrafilters/105700#105700 Answer by Joel David Hamkins for Lattice of differences between ultrafilters Joel David Hamkins 2012-08-28T10:59:44Z 2012-08-29T01:39:06Z <p>I've got it! </p> <p><b>Theorem.</b> 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 $U\neq V$. </p> <p>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 $U$.</p> <p>Proof. Notice first that when $U\neq V$ we can recognize whether $U$ is principal from the lattice $D(U,V)$, which will have a least element exactly in this case; similarly, we can recognize whether $V$ is principal from $D(U,V)$, which will have a greatest element exactly in this case.</p> <p>Next, let me recall from my earlier post how the filters $U$ and $V$ can be reconstructed from $D(U,V)$, using $D(U,V)$ not just as a lattice but specifically as a collection of subsets of $\kappa$. Namely, 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 $X\in D(U,V)$.</p> <p>But let me now explain how one can get access to essentially the same information up to isomorphism, just from knowing $D(U,V)$ as a lattice, and not knowing how these elements sit as subsets of $\kappa$. Assume $U$ is non-principal, and let $x$ be an arbitrary element of $D(U,V)$, viewed now only as a lattice. Let $A$ be the collection of immediate predecessors of $x$ in $D(U,V)$, namely, the set of $a\in D(U,V)$ such that $a\lt x$ and there is nothing between $a$ and $x$. (We know that these $a$ represent the removal of one element of the set representing $x$.) Now, I can define an ultrafilter $U'$ on $A$, by saying that $B\subset A$ is in $U'$ just in case there is a greatest lower bound to $A-B$ in $D(U,V)$. If $x$ is representing the set $X$ in $D(U,V)$, then I claim that $U\upharpoonright X\cong U'\upharpoonright A$, by the map that maps each point in $X$ to the set in $A$ obtained by omitting that point from $X$. Thus, $U'$ is Rudin-Keisler equivalant to $U$, and was constructed purely from viewing $D(U,V)$ as a lattice. Similarly, assume $V$ is non-principal and let $C$ be the set of immediate successors of $x$ in the lattice $D(U,V)$. (These lattice elements correspond exactly to the sets obtained by adding one additional point to the set that $x$ is representing.) Define the ultrafilter $V'$ on $C$ by $D\subset C$ is in $V'$ just in case there is no least upper bound of $D$ in $D(U,V)$. The map that sends the elements of $C$ to the corresponding points of $\kappa$ actually used in $D(U,V)$ is a Rudin-Keisler isomorphism of $V\upharpoonright(\kappa-X)$ with $V'\upharpoonright C$. So again, from $D(U,V)$ viewed purely as a lattice, we are able to extract $V$ up to isomorphism.</p> <p>Conversely, if $U\cong U'$ and $V\cong V'$, where $U\neq V$ and $U'\neq V'$, then we may find a single function $f$ witnessing the isomorphisms simultaneously (working partly on a $U$-big set and partly on its complement, a $V$-big set), thereby showing that $D(U,V)\cong D(U',V')$ as lattices. So this is a complete classification of $D(U,V)$ up to isomorphism as a lattice. QED</p> <p>Thus, the properties about $U$ and $V$ that we can determine from the lattice isomorphism class of the lattice $D(U,V)$ are precisely the properties that are determined by the isomorphism classes of $U$ and $V$ themselves, plus the knowledge of whether $U\neq V$. </p>