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I still don't understand how is that true. Here are the definitions: Let P be a class of closed subsets of a topological space X which is closed under finite intersections and finite unions. A closed filter F is a collection of non-empty elements of P with the properties: 1. F is closed under finite intersections 2. If A is in F and B is a superset of A belonging to P, then B belongs to F.

A closed ultrafilter is a closed filter F satisfying: Whenever A belongs to P and A meets each element of F, then A is in F.

A prime closed filter is a closed filter F satisfying : If A and B belong to P and their union belongs to F, then either A is in F or B is in F.

I don't understand why a prime closed filter can't be a closed ultrafilter, yet a closed ultrafilter is always a prime closed filter.

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The proof that a closed ultrafilter is prime is just as in the case of normal filters: if $A\cup B$ belongs to the filter then one of $A$ and $B$ (possibly both) must intersect all members of the ultrafilter.

The converse is not always true as can be seen by a counterexample. Take an ultrafilter $u$ on the set $\mathbb{N}$ of natural numbers. Let $\mathcal{F}$ be the family of closed sets, $F$, in $\mathbb{R}$ with the property that $\{n:2^{-n}\in F\}$ belongs to $u$. Then $\mathcal{F}$ is prime (use that $u$ is maximal) but not a closed ultrafilter because it is contained in the larger closed filter $\{F:0\in F\}$, which is ultra.

The reason that `normal' prime filters are maximal is that in the power set of a set you have complements; that facilitates that proof. .

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