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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.

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1 Answer 1

up vote 3 down vote accepted

For problem 1, let $f$ be the unique function satisfying $f^{-1}(n) = A_n$. Since each $A_n \not \in \mathcal{U}$, we know $f$ is not equivalent to a constant function. Hence it's equivalent to a finite-to-one function $g$. Let $B$ be the set where $f$ and $g$ agree. Then $B \cap A_n \subset g^{-1}(n)$ which is finite.

What you said, about $c \in A_n$ for all $n$ isn't quite true. What you'll get is $c \in A_n$ for almost all $n$, i.e. the set of $n$ such that $c \in A_n$ will belong to $\mathcal{U}$. This is good enough though, since the $A_n$ are supposed to be disjoint, so you can't even have $c$ belonging to 2 of the $A_n$, let alone almost all of them. But the problem is as you pointed out, how can one tell if $B\cap A_k$ is finite for all $k$?

For your second question, you can't claim that $\bigcap X_n = \emptyset$, nor do you need to. Indeed, if each $X_n$ belongs to $\mathcal{U}$, then so does each $X_n \cup \{0\}$, but the intersection of these guys is not empty. Your observation is correct, if the union $\bigcup X_n^c$ were a disjoint union, you'd be done. What happens if you disjointify it?

$Y_0 = X_0^c,\ Y_{n+1} = X_{n+1}^c \setminus Y_n$

Let $B$ meet all the $Y_n$ at only finitely many points. Then $B$ meets $Y_0 = X_0^c$ at only finitely many points, and for each $n$ it meets $Y_{n+1} \cup Y_n \supset X_{n+1}^c$ at only finitely many points.

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