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Every z-filter $\mathscr{F}$ on a topological space $X$ is the intersection of all z-primefilters refining it. Is it also true that it is the intersection of all z-ultrafilters refining it? Can you give a proof or counterexample?

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No, this is not the case.

Let's first remark that, while each z-ultrafilter is prime, not every z-primefilter is ultra. To see this, we will give a counterexample on the space $X = \mathbb{R}$.

First take a free ultrafilter $\mathscr{U}$ on the set $$N:=\left\{\frac{1}{n}\mid n\in \mathbb{N}\right\}$$ And define the z-filter $$\mathscr{F}:=\{Z\in \mathcal{Z}(\mathbb{R})\mid Z\cap N \in \mathscr{U}\}$$ This is clearly a z-filter, and it is also prime: take $Z_1, Z_2\notin \mathscr{F}$, then $Z_1\cap N \not \in \mathscr{U}$ and $Z_2\cap N \notin \mathscr{U}$. Because $\mathscr{U}$ is ultra we then have $$(Z_1\cap N) \cup (Z_2\cap N) = (Z_1 \cup Z_2)\cap N \notin \mathscr{U}$$ so $Z_1\cup Z_2\notin \mathscr{F}$ and $\mathscr{F}$ is prime. But it is not ultra because it is stricly contained in the z-ultrafilter $\mathscr{G}$generated by $0$: $$\mathscr{G}=\{Z\in \mathcal{Z}(\mathbb{R})\mid 0 \in Z\}$$ This is because for each $Z\in \mathscr{F}$, $Z\cap N$ is a sequence converging to $0$, and $Z$ is closed so $0\in Z$ and $Z \in \mathscr{G}$. The inclusion is strict as witnessed by the singleton $\{0\}$.

In fact, $\mathscr{G}$ is the only z-ultrafilter that refines $\mathscr{F}$, so the intersection of all z-ultrafilters that refine $\mathscr{F}$ is equal to $\mathscr{G}$ and not $\mathscr{F}$. We will now check that this claim holds.

Let $\mathscr{H}$ be another z-ultrafilter that refines $\mathscr{F}$. It suffices to show that each element of $\mathscr{H}$ contains $0$. Suppose not, let $Z \in \mathscr{H}$ with $0\notin Z$. Because $Z$ is closed, it can also not contain any sequence converging to $0$, so there must be some $m\in \mathbb{N}$ such that $$N_m:=\left\{\frac{1}{n}\mid n\geq m\right\}\cup \{0\}$$ is disjoint form Z. $N_m$ has a finite complement in $N$ and it is a zeroset, so $N_m \in \mathscr{F}\subset\mathscr{G}$. This is a contradiction because then $\mathscr{G}$ contains the disjoint sets $Z$ and $N_m$.

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