closed set and z-ultrafilter on normal space - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T06:37:56Z http://mathoverflow.net/feeds/question/91869 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91869/closed-set-and-z-ultrafilter-on-normal-space closed set and z-ultrafilter on normal space Douglas Somerset 2012-03-21T22:51:03Z 2012-03-22T01:57:56Z <p>Let $X$ be a completely regular, Hausdorff topological space and let $\cal F$ be a $z$-ultrafilter on $X$. Then for each zero set $W$ in $X$, either $W\in \cal F$ or there exists $Z\in \cal F$ such that $Z$ does not meet $W$ (this is the $z$-ultrafilter property). Now suppose that $X$ is additionally normal. Then is it true that for every closed set $W$ in $X$ either $W$ contains an element $Z$ of $\cal F$ or there exists $Z\in \cal F$ such that $Z$ does not meet $W$?</p> http://mathoverflow.net/questions/91869/closed-set-and-z-ultrafilter-on-normal-space/91873#91873 Answer by Todd Eisworth for closed set and z-ultrafilter on normal space Todd Eisworth 2012-03-22T00:39:52Z 2012-03-22T00:39:52Z <p>No: think of what happens with $\omega_1$ in the usual topology. This is certainly normal (even hereditarily normal), and since every real-valued continuous function on $\omega_1$ is eventually constant, the co-bounded sets form a $z$-ultrafilter. Now let $W$ be the set of countable limit ordinals.</p>