Show that a prime closed filter is not always a closed ultrafilter

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.

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.