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 $x$. Furtermore 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: $$ \{ \operatorname{image} f|_M \ : \ M \in \varphi\}$$
It seems like our definition of filter is not the standard definition, to avoid confusion $\varphi \subset \mathcal{P}(\mathcal{P}(X))$
Now we shall show that if $X=\mathbb{Z}$, then it follows that $\varphi$ is a one point filter. From the given property we already can conclude that $\varphi$ is an ultrafilter but I don't know what I shall do next, could someone give me a hint?
As it was asked for in the comments, our definition of a filter is:
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 $A\in F \vee S\setminus A \in F$. The one point filter is already defined in the first sentence.
I posted the question on MathOverflow were someone wrote the answer.