# Non-principal ultrafilters and Ultrafilter Lemma

$\textbf{Definition}$ A family $\mathscr{F}$ of subsets of $I$ has the finite intersection property if for each $S_1, \ldots,S_n\in\mathscr{F}$ it holds that $S_1\cap\ldots\cap S_n\neq\varnothing$.

$\textbf{Ultrafilter Lemma}$ If a family $\mathscr{F}$ has the finite intersection property, then there is an ultrafilter $\mathcal{U}\supset\mathscr{F}$.

Thus there exist non principal ultrafilters......

I can't understand how one can deduce the existence of non-principal ultrafilters from the ultrafilter lemma.

## 2 Answers

Let $I$ be an infinite set. Let $\mathscr{F}$ be the family of all subsets of $I$ whose complement is finite. Then $\mathscr{F}$ has the finite intersection property. Therefore it is contained in an ultrafilter $\mathscr{U}$.

For any $a \in I$, $\mathscr{U}$ cannot be the principal ultrafilter determined by $a$, since $I - \{a\} \in \mathscr{F} \subseteq \mathscr{U}$.

Edit: I've just re-read the definition of "finite intersection property." This answer can be simplified slightly by saying that $\mathscr{F}$ should be the family of all sets $I - \{a \}$ for $a \in I$.

• Or with the lemma as stated here, consider the set of all $I\setminus\{a\}$, which has the finite intersection property iff $I$ is infinite. – Carsten S Jan 8 '14 at 17:45

HINT: Consider the co-finite subsets of $I$.