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(1) In the second example in Section 3.1 of the Wikipedia article on filter, the last sentence says:

A nonprincipal filter on an infinite set is not necessarily free.

On the other hand, Martin Sleziak's answer to this question (and also the second sentence of this question) says:

Filter, which is not free is called principal. Hence every filter is either free or principal and the same is true for ultrafilters.

Clearly, these people contradict Wikipedia. I am guessing Wikipedia is correct. Let $N_x$ be the neighborhood filter at $x$ in a Euclidean space. Then $N_x$ is not free since $\bigcap N_x = \{x\} \neq \varnothing$, and $N_x$ is not principal since $\{x\} \notin N_x$. Am I correct?

(2) I suppose Martin Sleziak's statement is true for ultrafilters (that they are either free or principal). Suppose an ultrafilter $\mathcal{F}$ on $S$ is not free. Then $\mathcal{F}$ does not include the Fréchet filter, so it contains some finite subset $P$ of $S$. Since an ultrafilter is prime, it follows that $\{a\} \in \mathcal{F}$ for some $a \in P$. Then $\mathcal{F}$ must coincide with the principal filter generated by $\{a\}$, as it is an ultrafilter and included in $\mathcal{F}$. Is this correct?

(3) In the third example in the same section of the same Wikipedia article, the second sentence says:

A filter on $S$ is free if and only if it contains the Fréchet filter.

Shouldn't this sentence begin with "An ultrafilter on $S$ ..."? If a filter includes the Fréchet filter, then it is obviously free. But does the reverse implication hold? If not, what counterexamples are there?

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  • $\begingroup$ I parsed the title, "Free filter on Wikipedia", several different ways, all wrong. $\endgroup$ – Gerry Myerson Apr 3 '16 at 7:19
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Your first two points are correct. Another simple counterexample to Martin Sleziak’s assertion can be obtained as follows. Let $E$ be the set of even integers, let $\mathscr{F}$ be the Fréchet filter on the set of odd integers, and let $\mathscr{G}=\{E\cup F:F\in\mathscr{F}\}$; then $\mathscr{G}$ is a filter on $\Bbb Z$, and $\bigcap\mathscr{G}=E\notin\mathscr{G}$, so $\mathscr{G}$ is neither free nor fixed.

However, the statement from Wikipedia that you quote in your third point is correct as it stands. To see that the other implication is also true, suppose that $\mathscr{F}$ is a filter on $S$ that does not contain the Fréchet filter on $S$; then there is some cofinite $A\subseteq S$ such that $A\notin\mathscr{F}$. It follows that $F\nsubseteq A$ for each $F\in\mathscr{F}$, so if $D=S\setminus A$, then $F\cap D\ne\varnothing$ for each $F\in\mathscr{F}$. Suppose that for each $x\in D$ there is an $F_x\in\mathscr{F}$ such that $x\notin F_x$. Then $\bigcap_{x\in D}F_x$ is an element of $\mathscr{F}$ disjoint from $D$, which is impossible. Thus, $D\cap\bigcap\mathscr{F}\ne\varnothing$, and in particular $\mathscr{F}$ is not free.

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  • $\begingroup$ Thank you so much! So the maximality condition is unnecessary. This may not be a mathematical question, but do you have any idea of why I keep encountering the statement that an ultrafilter is free iff it includes the Fréchet filter, rather than the stronger Wikipedia statement? $\endgroup$ – Usagi Apr 3 '16 at 7:15
  • $\begingroup$ @Usagi: Probably because the free/principal distinction usually isn’t important unless one is dealing with ultrafilters. You’re welcome! $\endgroup$ – Brian M. Scott Apr 3 '16 at 7:17
  • $\begingroup$ You have solved all my questions brilliantly. You've made me very happy and want to study more. Thank you a lot! $\endgroup$ – Usagi Apr 3 '16 at 7:24
  • $\begingroup$ @Usagi: You’re very welcome, and thank you! $\endgroup$ – Brian M. Scott Apr 3 '16 at 7:25
  • $\begingroup$ @Usagi I agree with you and I am tempted to coin the word "Briantly" to qualify the outstanding answers of Brian M. Scott. $\endgroup$ – J.-E. Pin Apr 3 '16 at 13:50

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