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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.

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  • $\begingroup$ The book is also available on the author's website. $\endgroup$ – Asaf Karagila Jan 25 '14 at 17:02
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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

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

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