The following text is a quote from p.180 of Halbeisen's book Combinatorial Set Theory. This book is also available on website of a course taught by the author. (As mentioned in Asaf's comment, it is also available on the author's website.)

For two sets $x,y\subseteq\omega$ we say that $x$ is almost contained in $y$, denoted $x\subseteq^*y$, if $x\setminus y$ finite.

A pseudo-intersection of a family $\mathscr F \subseteq [\omega]^\omega$ of infinite subsets of $\omega$ is an infinite subset of $\omega$ that is almost contained in every member of $\mathscr F$.

Furthermore, a family $\mathscr F \subseteq [\omega]^\omega$ has the strong finite intersection property (sfip) if every finite subfamily has infinite intersection.

For example any filter $\mathscr F \subseteq [\omega]^\omega$ has the sfip, but no ultrafilter on $[\omega]^\omega$ has a pseudo-intersection.

  • How can we show that a free ultrafilter cannot have an infinite pseudointersection?

This fact is used to show that $\mathfrak p$ is well-defined and $\mathfrak p \le \mathfrak c$, where the pseudo-intersection number $\mathfrak p$ is the smallest cardinality of any family $\mathscr F \subseteq [\omega]^\omega$ which has the sfip but which does not have a pseudo-intersection.

I will post my proof below; but I wonder whether there are different ways to show this.

  • $\begingroup$ The book is also available on the author's website. $\endgroup$ – Asaf Karagila Jan 25 '14 at 17:02
  • 1
    $\begingroup$ You mean here, right? $\endgroup$ – Martin Sleziak Jan 25 '14 at 17:04
  • $\begingroup$ Yes, exactly there. $\endgroup$ – Asaf Karagila Jan 25 '14 at 17:04
  • 2
    $\begingroup$ Martin, in that last paragraph, $\frak p\leq c$ regardless to anything, because every family in $[\omega]^\omega$ has cardinality $\leq\frak c$. $\endgroup$ – Asaf Karagila Jan 25 '14 at 22:15
  • $\begingroup$ Of course, you are right. I have changed the wording to $\mathfrak p$ is well-defined and $\mathfrak p \le \mathfrak c$ which is the same wording as in the book. (Maybe I should have left there only the first part.) $\endgroup$ – Martin Sleziak Jan 26 '14 at 6:45

Let $\mathscr F$ be an arbitrary ultrafilter, which contains no finite sets.

Suppose that $A$ is an infinite pseudointersection of $\mathscr F$.

Then we have $A\subseteq^* G$ for each $G\in\mathscr F$.

Since $\mathscr F$ is an ultrafilter, we have either $A\in\mathscr F$ or $\omega\setminus A\in\mathscr F$. But $\omega\setminus A$ cannot be in $\mathscr F$, since $A\setminus (\omega\setminus A)=A$ is infinite and thus $A\not\subseteq^* \omega\setminus A$.

So we get that $A\in\mathscr F$.

Now denote $A=\{a_n; n\in\omega\}$ and let $B=\{a_{2n}; n\in\omega\}$ and $C=\{a_{2n+1}; n\in\omega\}$. Since $A=B\cup C$ and $\mathscr F$ is an ultrafiter; one of the sets $B$, $C$ belongs to $\mathscr F$. But neither $A\subseteq^* B$ nor $A\subseteq^* C$ holds, which yields a contradiction.

  • $\begingroup$ This looks right, but there are seemingly a bunch of unnecessary moves (see my answer below). $\endgroup$ – GME Jan 25 '14 at 18:12

Let $\mathcal U$ be an ultrafilter on $\omega$, and let $x$ be a pseudo-intersection of $\mathcal U$. Furthermore, let $y$ be such that both it and its compliment contain infinitely many elements of $x$ (for instance, $y$ could be your $B$). Then, either $y$ or $\omega\backslash y$ are in $\mathcal U$, so either $x\backslash y$ or $x\backslash (\omega\backslash y) = x\cap y$ is finite, which is impossible.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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