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my definition of p-point in $\beta\mathbb{N}$ is: $\mathcal{U}$ is a p-point if and only if every $\{A_n:n\in\mathbb{N}\}\subset\mathcal{U}$ has a pseudo-intersection, i.e. a $B\in\mathcal{U}$ such that $B\backslash A_n$ is finite for every $n$.

In topology the definition is: the intersection of a countable family of neighbourhood of $\mathcal{U}$ is a neighbourhood of $\mathcal{U}$.

How can I prove that these two definitions are equivalent?

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I think that this is true only for non-principal ultrafilters and working in the topology on $\beta\omega^*=\beta\omega\setminus\omega$. – Martin Sleziak Apr 5 '11 at 17:04
up vote 1 down vote accepted

Probably someone will come up with a simpler solution, but I will give it a try.

I will work on $\beta\omega^*$ (the question was probably intented in this way). I.e., we only work with non-principal ultrafilters and the topology is given by the base consisting of sets $U_A=\{\mathcal F; A\in \mathcal F\}$ for $A\subseteq\omega$.

The key property, which I use very often is $$\mathcal F\in U_A \Leftrightarrow A\in\mathcal F.$$

$\boxed{\Rightarrow}$ Suppose that every system of sets from $\mathcal U$ has a pseudointersection. Given any system of basic neighborhoods $U_{A_n}$ of $\mathcal U$, there exists a pseudointersetction $B\in\mathcal U$. We only need to show $U_B\subseteq U_{A_n}$ for all $n$. But if $B\in \mathcal F$ for some non-principal ultrafilter $\mathcal F$ and $B\setminus A_n=F$ is finite, then $F'=\omega\setminus F\in\mathcal F$ and $B\cap F'=B\setminus F\in\mathcal F$ and $B\setminus F\subseteq A_n$. Hence, $A_n\in\mathcal F$.

$\boxed{\Leftarrow}$ Suppose we are given a sequence of sets $A_n\in\mathcal U$. If the intersection $\bigcap A_n$ is open, this means that there exists $B\in\mathcal U$ with $U_B \subseteq \bigcap U_{A_n}$. The last inclusion is equivalent to the condition that $$(\forall \mathcal F\in\beta\omega^*) [B\in\mathcal F \Rightarrow (\forall n) A_n\in\mathcal F].$$ We want to show that $B$ is pseudointersection. Suppose not, i.e., there exists some $n_0$ such that $B\setminus A_{n_0}$ is not finite. Then there exists a non-principal ultrafilter $\mathcal F_0$ such that $B\setminus A_{n_0}\in \mathcal F_0$. (Use the fact that the system $\{B\setminus A_{n_0}\}\cup\{\omega\setminus F; F \text{ is finite}\}$ has finite intersection property.) Now, this ultrafilter contains $B$, but it does not contain $A_{n_0}$, contradicting the above property.

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thanks, it works, but I can't find a counterexample if we work with also the principal ultrafilters in the topology of $\beta\omega$ – Jacob Fox Apr 5 '11 at 18:13

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