Let $\mathcal{P}_0(X)$ the Power set of $X$ without the empty set and let $\dot{x}:=\{A\subseteq X: x \in A\}$ the one point filter generated by $x$. Furthermore let $$ \mathcal{A} := \{ f \in X^{\mathcal{P}_0(X)} : \ \forall A \in \mathcal{P}_0(X): f(A) \in A\} $$ be the set of the functions mapping subsets of the power set to elements of them. Let $\varphi$ be a filter on $\mathcal{P}_0(X)$ with $$\forall f \in \mathcal{A} \exists x_f \in X : f[\varphi]=\dot{x}_f.$$ Here $f[\varphi]$ stands for the filter generated from the filter basis: $$ \{ \operatorname{image}f|_M \ : \ M \in \varphi\}$$

On the bottom of the post I included some notation, I hope this avoids notational issues.

At first we should show that $\varphi$ is already an ultrafilter.

Now we shall show that if $X=\mathbb{Z}$, then it follows that $\varphi$ is a one point filter.

For the first part my attempt was to use that a filter is a ultra filter when every set or its complement is in the filter. At first we look at the case where $x$ is independent of $f$ and then (somehow) conclude that $\varphi$ must already be an ultrafilter. If this is done we could look at the equivalence relation $\sim$ given by \[ f \sim g \iff x_f=x_g \] Now we use the equivalence classes given by this equivalence relation. If I am right, this reduce the problem again to the case where $x$ is independent of $f$.

For the second part my only idea was looking the the properties of $\mathbb{Z}$ which could help us (that the ultrafilter is already a one point filter). I think one of the properties which help is that $\mathbb{Z}$ is countable, but I don't see how it helps.

Let $S$ be a set (an arbitrary one), and $A,B\subseteq S$. We call $\varphi\subset \mathcal{P}(S)$ a filter when the following holds

- $\varnothing\notin \varphi$
- $A,B\in \varphi \implies A\cap B \in \varphi$
- $A\subset B $ and $A\in \varphi \implies B\in \varphi$

An ultrafilter is a filter, such that there is no bigger filter, e.g. when $G$ is a filter and $F$ is an ultra filter and $G\supseteq F\implies G=F$. More convenient is the equivalence that for every subset $A$ of $S$ it holds that $A\in F \wedge S\setminus A \in F$.

Maybe this question is not appropiate to MathOverflow cause it is actually not a research question. I faced this problem on the group exercises of my topology class I am in (so nothing which should be handed in). If it is not welcome I will delete it just leave a comment. I already posted that question on Math Stack Exchange