# Ultrafilters and principal filters [closed]

Can someone give me an example of an ultrafilter which is not principal?

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## closed as off topic by HJRW, Mark Sapir, Simon Thomas, Andreas Blass, Gerald EdgarSep 27 '11 at 14:50

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## 3 Answers

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Do you mean on a set or on a Boolean algebra? You need some fragment of the axiom of choice to construct a non-principal ultrafilter on the natural numbers, for example. (See the link in Igor Rivin's answer.)

However, the Boolean algebra consisting of finite and cofinite subsets of $\mathbb N$ has a non-principal ultrafilter, namely the collection of all cofinite sets.

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As far as I know, the existence of non principal (= free) ultrafilters is independent on ZF. It follows from AC, but it's weaker: Ultrafilter's lemma, which says exactly that non principal ultrafilter exist, by using AC in the form of Zorn's lemma, does not imply AC.

So, an explicit construction cannot be given, since otherwise you would have proved the Axiom of Choice. If you want a non-constructive construction, you can also follow the proof of Ultrafilter's lemma: take the family of all filters and apply Zorn's lemma (see this discussion http://everything2.com/title/Proof+that+any+filter+can+be+extended+to+an+ultrafilter for some apparently right details).

In some cases one can find quite explicit examples, as suggested above for Boolean algebras. But this more difficult for ultrafilters on a set: the family of cofinite (the completementary is finite) subsets of a countable set is NOT an ultrafilter!

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In the second paragraph, "otherwise you would have proved the Axiom of Choice" should be "otherwise you would have proved a weak form (not provable in ZF) of the axiom of choice". As you noted in the first paragraph, the existence of a non-principal ultrafilter is a strictly weaker statement than AC. –  Andreas Blass Sep 27 '11 at 12:49
I should perhaps be mentioned that it's consistent with ZFC that there are definable non-principal ultrafilters on the natural numbers. For example, Gödel's axiom of constructibility (V=L) implies that there is a definable well-ordering of the reals. That suffices to produce a definable non-principal ultrafilter. In fact, V=L implies the existence of a $\Delta^1_2$ non-principal ultrafilter on the set of natural numbers. –  Andreas Blass Sep 27 '11 at 12:53
The OP did not say explicit –  Gerald Edgar Sep 27 '11 at 14:49