I need some help on two exercise:
Here is the first:
Let $\mathcal{U}$ be a non-principal ultrafilter. Suppose that every $f:\mathbb{N}\rightarrow\mathbb{N}$ is $\mathcal{U}$-equivalent to a constant or to a finite-to-one function. I want to prove that if $\mathbb{N}=\bigsqcup_{n\in\mathbb{N}} A_n$ where $A_n\not\in\mathcal{U}$ then there exists $B\in\mathcal{U}$ such that $B\cap A_n$ is finite for every $n$.
Here what I did until now:
using the axiom of choice suppose to have a function $f:\mathbb{N}\rightarrow\mathbb{N}$ such that $f(n)\in A_n$. If $f$ is $\mathcal{U}$-eq. to a constant then $\{n:f(n)=c\}\in\mathcal{U}$ for some $c$, but this implies that $c\in A_n$ for all $n$, contradiction. So $f$ is $\mathcal{U}$-eq. to a finte-to-one function $g$. Set $B:=\{n:g(n)=f(n)\}\in\mathcal{U}$, I'd like to prove that $B\cap A_k$ is finite for all $k$ (I don't even know if it is true).
Here is the second exercise: we have a non-principal ultrafilter $\mathcal{U}$. Suppose that for every partition of $\mathbb{N}=\bigsqcup A_n$ with $A_n\not\in\mathcal{U}$ then there exists $B\in\mathcal{U}$ such that $|B\cap A_n|<\infty$ for all $n$. Then if we have $\{X_n:n\in\mathbb{N}\}\subset\mathcal{U}$ then there exists $B\in\mathcal{U}$ such that $|B\backslash X_n|<\infty$.
Here what I did until now:
let $\{X_n:n\in\mathbb{N}\}\subset\mathcal{U}$ then $\bigcap X_n=\emptyset$ because $\mathcal{U}$ is not principal. So $\bigcup X_n^c=\mathbb{N}$, if this union would have been disjoint then I could have applied the hypothesis and it would have been ok. I don't know how to go on.