# A nonprincipal ultrafilter that is not a $p$-point

On pg 76 of Jech's Set Theory, he proves the existence of a nonprincipal ultrafilter on $$\omega$$ that is not a $$p$$-point.

Given a partition $$\{A_n\}$$ of $$\omega$$ into $$\aleph_0$$ infinite pieces, let $$F$$ be the following filter on $$\omega$$.
$$X \in F$$ if and only if except for finitely many $$n$$, $$X \cap A_n$$ contains all but finitely many elements of $$A_n$$.
If $$D$$ is any ultrafilter extending $$F$$, then $$D$$ is not a $$p$$-point.

I've been thinking why the ultrafilter $$D$$ isn't a $$p$$-point but I couldn't figure it out

• About the bare existence: it's a remark in Rudin's 1956 paper: a compact Hausdorff space with only P-points is finite [Proof: otherwise, the set of open subsets is closed under countable intersections, and hence all countable subsets are closed. In turn this implies that all countable subsets are discrete (since removing an accumulation point would yield a non-closed countable subset), and since a discrete compact set is finite, this implies that all countable subsets are finite, which obviously implies finite.]
– YCor
Commented Mar 18, 2020 at 16:08

## 1 Answer

First of all, it is clear that the given Partition $$\{A_n\;|\;n\in\omega\}$$ satisfies $$A_n\notin D$$ for any n, since $$\omega\smallsetminus A_n\in F\subseteq D$$ for any n (since $$(\omega\smallsetminus A_n)\cap A_k=A_k$$ for all $$k\neq n$$).

Furthermore, assume $$X$$ has the property that $$|X\cap A_n|<\omega$$ for any $$n$$. Then $$|A_n\cap(\omega\smallsetminus X)|=|A_n\smallsetminus X|=|A_n\smallsetminus(A_n\cap X)|=\omega$$ And therefore $$\omega\smallsetminus X\in F$$ and $$X\notin D$$.

• Thanks! that is exactly what i needed Commented Mar 19, 2020 at 7:55