Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:

  1. 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.)
  2. For every $G_\delta$ that contains $u$, $u$ is in the interior of the $G_\delta$. (A bit hipster.)
  3. 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?

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
I am amazed by how simple the proof is, and yet how far I was from it. Thank you very much. – SantiagoC Apr 3 '14 at 13:29

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.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.