Extending complete filters - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T21:51:23Z http://mathoverflow.net/feeds/question/117440 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/117440/extending-complete-filters Extending complete filters Tomek Kania 2012-12-28T20:48:52Z 2012-12-28T22:37:38Z <p>Suppose $\kappa$ is a measurable cardinal and let $\mathcal{F}\subset\wp(\kappa)$ be a $\kappa$-complete non-principal filter. Can we extend $\mathcal{F}$ to a $\kappa$-complete ultrafilter?</p> <p>My motivation comes from a problem I have just encountered. I need a $\kappa$-complete ultrafilter whereas the best I can do is to construct only some filter. </p> <p>Of course, the standard Zorn argument does not work here by a simple meta-argument. Take the filter on $\omega_1$ consisting of sets with countable complement. It is $\omega_1$-complete but there is no $\omega_1$-complete ultrafilter on $\omega_1$, since this cardinal is not measurable (in ZFC). </p> http://mathoverflow.net/questions/117440/extending-complete-filters/117453#117453 Answer by Joel David Hamkins for Extending complete filters Joel David Hamkins 2012-12-28T22:32:35Z 2012-12-28T22:37:38Z <p>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.</p> <p><strong>Theorem</strong>. The following are equivalent, for any uncountable regular cardinal $\kappa$.</p> <ol> <li>$\kappa$ is a measurable cardinal.</li> <li>Every $\kappa$ complete filter $F$, generated by at most $\kappa$-many sets, extends to a $\kappa$-complete ultrafilter.</li> </ol> <p>Proof: It is easy to see that $2$ implies $1$, since the filter of co-bounded sets in $\kappa$ is $\kappa$-complete and generated by the tails, so there is a $\kappa$-complete non-principal ultrafilter on $\kappa$.</p> <p>For the main direction, assume $\kappa$ is measurable and $F$ is a $\kappa$-complete filter on a set $D$ with $F$ generated by at most $\kappa$ many sets $X_\alpha$, for $\alpha\lt\kappa$. Let $j:V\to M$ be an elementary embedding with critical point $\kappa$. By applying $j$ to $\vec X=\langle X_\alpha\lt\kappa\rangle$ and restricting to $\kappa$, we see that $\langle j(X_\alpha)\mid\alpha\lt\kappa\rangle$ is in $M$. And since this is fewer than $j(\kappa)$ many elements of $j(F)$, which is $j(\kappa)$-complete in $M$, it follows that $\bigcap_{\alpha\lt\kappa}j(X_\alpha)\in j(F)$, and in particular, there is some $a\in \bigcap_\alpha j(X_\alpha)$. Define $U=\{X\subset D\mid a\in j(X)\}$. It is easy to verify that $U$ is a $\kappa$-complete ultrafilter on $D$ and $F\subset U$, as desired. QED</p> <p>For $\theta$-generated filters, one generally needs $\theta$-strong compactness, as mentioned in the comments, and this is in fact equivalent to $\theta$-strong compactness. The essence of the argument above, then, is that a cardinal $\kappa$ is measurable if and only if it is $\kappa$-strongly compact. </p> <p>That said, if you want this filter extension property, then I encourage you to go ahead and make the strong compactness assumption. There are many beautiful theorems using strongly compact cardinals. </p>