Recently I have been exposed to the concept of P-Points. I have three definitions: A non-principal ultrafilter $u$ is a P-Point if:
- For every sequence $\left < A_n \right >_{n\in \omega}$ of elements of $u$ there exists some set $B \in u$ that is almost contained in every $A_n$. (Most common one.)
- For every $G_\delta$ that contains $u$, $u$ is in the interior of the $G_\delta$. (A bit hipster.)
- For every function $f:\mathbb{N}\to \mathbb{N}$ there is some $A\in u$ such that $f| A$ is finite-to-one or constant. (Paper specific.)
Does anyone know how to prove that these definitions are equivalent? Or maybe could point out where some of the proofs may be?
$\endgroup$2 Answers
$\begingroup$To see that (1) is equivalent to (2), all you really need to know is the following translation. A basic open set in $\beta\mathbb{N}$ has the form $N_A = \{v : A\in v\}$, where $A$ is some subset of $\mathbb{N}$; and when $A \subseteq B$, $N_A\subseteq N_B$. When $A$ is almost-contained in $B$, then we can only say $$N_A\cap (\beta\mathbb{N}\setminus\mathbb{N}) \subseteq N_B\cap (\beta\mathbb{N}\setminus\mathbb{N})$$ But since we're dealing primarily with non-principal ultrafilters, the proof still works out.
I like the hipster version. (But I only learned about it after it was already cool.)
$\endgroup$ $\begingroup$Observe that $1$ is equivalent to
For all partition of $\omega = \bigsqcup_{n \in \omega} A_n$ with $A_n \notin U$, there exists a $X \in U$ such that $|X \cap A_n| < \aleph_0$ for all $n \in \omega$.
To show $(1) \Rightarrow (3)$. Let $f : \omega \rightarrow \omega$. If there exists a $n$ such that $f^{-1}(n) \in U$, then $f \upharpoonright f^{-1}(n)$ is constant. Suppose no such $n$ exist. Define $A_n = f^{-1}(n)$. By the version of 1 above, there exists $A \in U$ such that $A \cap A_n = A \cap f^{-1}(n)$ is finite for all $n$. Hence $f \upharpoonright A$ is finite to one.
To show $(3) \Rightarrow (1)$. Suppose $\omega = \bigsqcup A_n$ with $A_n \notin U$ for all $n \in \omega$. Define $f(x) = n$ if and only if $x \in A_n$. By (3), either there is a $A \in U$ such that $f \upharpoonright A$ is constant or $f \upharpoonright A$ is finite to one. The former can not occur because this would imply $A \subseteq A_n$ for some $n$ which would imply $A_n \in U$. Contradiction. Thus $f \upharpoonright A$ is finite to one. This means $A \cap A_n$ is finite for all $n \in \omega$. (1) has been verified.
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