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 $W$?
1 Answer
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No: think of what happens with $\omega_1$ in the usual topology. This is certainly normal (even hereditarily normal), and since every realvalued continuous function on $\omega_1$ is eventually constant, the cobounded sets form a $z$ultrafilter. Now let $W$ be the set of countable limit ordinals.

$\begingroup$ Thanks. I have been wondering about that for years. Credit to MathOverflow for getting the answer in three hours! $\endgroup$ Commented Mar 22, 2012 at 8:37