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I've been doing a lot of work with ultrapowers and saturation recently. In particular, I am reading chapter 6 of Chang and Keisler as well as Keisler's paper on "Ultraproducts which are not saturated". In chapter 6 of C&K, they give define what it means to be a $\alpha-good$ ultrafilter and spend a long time on the existence of such types of filters. But then it hits me, I know about all these "different types" of ultrafilters, but I don't $know$ any $examples$ of such ultrafilters.

Is it provable that one cannot actually give an explicit example of a non-principal ultrafilter of $\mathbb{N}$? Or can someone give an explicit example of one (defined recursively)?

Edit: As André Nicolas pointed out, $ZF\not\vdash $ "There exists a non-principal ultrafilter". How do you construct a model of ZF which does not have any non-principal ultrafilters?

Kyle Gannon
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    The existence of a non-principal ultrafilter on $\mathbb{N}$ is not provable in ZF. – André Nicolas Jul 10 '14 at 18:39
  • @AndréNicolas So, I guess what's the proof of that? – Kyle Gannon Jul 10 '14 at 18:42
  • Construction of a model. – André Nicolas Jul 10 '14 at 18:44
  • @KyleGannon as far as I know existence of such ultrafilter is equivalent to AC which is independent of ZF. – Norbert Jul 10 '14 at 18:45
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    @Norbert: it's equivalent to a weak form of AC, namely the Boolean Prime Ideal Theorem. – Paul McKenney Jul 10 '14 at 18:45
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    @KyleGannon: In answer to the new question, you could look at the Solovay model, which is a well-known model of ZF + DC + "all sets of reals are Lebesgue measurable and have the Baire property". A nonprincipal ultrafilter cannot be measurable or have the Baire property. – Paul McKenney Jul 10 '14 at 18:47
  • @PaulMcKenney, thank you! I knew I remembered sommething wrong – Norbert Jul 10 '14 at 18:48
  • The last bit, namely that an ultrafilter can't be measurable or have the Baire property, doesn't take too much work; the construction of the Solovay model, on the other hand, requires a decent background in forcing, as far as I know. – Paul McKenney Jul 10 '14 at 18:49
  • I also feel obligated to point out that a model of "all sets of reals are Lebesgue measurable" requires the consistency of an inaccessible cardinal, whereas "all sets of reals have the BP" doesn't have any large cardinal strength. – Paul McKenney Jul 10 '14 at 18:56
  • If I may tag along: Does the Ultrafilter Lemma imply the Axiom of Choice? – Forever Mozart Jul 10 '14 at 19:08
  • @TomCruise: I believe this post has the answer to that question: http://math.stackexchange.com/questions/211617/ultra-filter-and-axiom-of-choice – Kyle Gannon Jul 10 '14 at 19:12
  • @Norbert: Apparently I am also guilty of perpetuating false information; the existence of a nonprincipal ultrafilter is weaker than the Boolean Prime Ideal Theorem, which implies that every filter can be extended to an ultrafilter. BPIT is still weaker than full AC, though. – Paul McKenney Jul 10 '14 at 19:20
  • @PaulMcKenney, no problem. Anyway existence of ultrafilters is not equivalent to AC. This is what I must remember. – Norbert Jul 10 '14 at 19:23
  • @Paul: It perhaps should be pointed out that Shelah's model where $\sf ZF+DC+BP$ holds is considerably more complicated than Solovay's construction. – Asaf Karagila Jul 10 '14 at 21:36

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Constructing any model of $\sf ZF$ where the axiom of choice fails requires some preliminary amount of knowledge. Here are some of the usual perquisites, some constructions may require only part of them.

  1. Permutation models of set theory with atoms. These are not quite models of $\sf ZF$, but they are relatively simpler to understand. Some statements can be transfered from such models to models of $\sf ZF$, and that makes a lot of examples relatively easy to grasp.

  2. Forcing. Forcing is an important tool for constructing any models of set theory in modern set theory, and one should be comfortable with the basics of forcing. Class forcing is worth mentioning, because sometimes it is needed in these contexts.

  3. Symmetric extensions. This is an extension of the method of forcing, but it deserves a separate mention. This is one of the two major tools of constructing models of $\sf ZF$ without the axiom of choice. The transfer theorems mentioned in the first point make use of this technique.

  4. Relative constructibility. This is another method of constructing models where the axiom of choice fails. In some cases it allows us to avoid using forcing, and in other cases it just allows us to avoid using symmetric extensions. But in either case this method is very useful and very common in papers where the axiom of choice fails.

For some purposes you might need more tools. Things like large cardinals can come into play, or infinitary combinatorics, and so on. There is also the things about understanding mathematical ideas that you wish to interact with. If you wish to construct a model without free ultrafilters on $\Bbb N$, then it is good to understand what are the properties of such sets, and what sort of things follow from their existence.

Now we can approach the examples for models where there are no free ultrafilters. The case where we are only interested in "No free ultrafilters over $\Bbb N$" is somewhat simpler, and this is a problem given in Jech The Axiom of Choice (Ch. 5, Problem 24, p.82).

This, as Paul McKenney pointed out a consequence of statements like "all sets of reals are measurable" and "all sets of reals have the Baire property". Since both properties become cumbersome in models where $\Bbb R$ is a countable union of countable sets, let me talk about them in conjunction of $\sf DC$ (or at least "$\omega_1$ is regular").

Both the statements hold in Solovay's model, which as far as models of $\sf ZF$ go, is relatively simple to construct. The proofs that both these properties hold require deeper and better understanding of set theory, though. Shelah proved that to obtain a model where all sets of reals are measurable requires the existence of an inaccessible cardinal, while the construction of a model of $\sf ZF+DC+BP$ does not require an inaccessible. Naturally, his construction is considerably more difficult than that of Solovay.

The general case of "There are no free ultrafilters" is more difficult, as expected. The proof is due to Andreas Blass, and can be found in his paper:

Blass, Andreas "A model without ultrafilters." Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 25 (1977), no. 4, 329–331.

Asaf Karagila
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  • As always, Thank you. – Kyle Gannon Jul 11 '14 at 08:35
  • I do not have Jech's "Axiom of Choice", but apparently there is no HOD(A∪{A}) free ultrafilter in the Cohen extension of a model M obtained by adding ω-many Cohen reals (equivalently one Cohen real), where A is the adjoined set of Cohen reals. Thus HOD(A∪{A})^M[G] models "there is no free ultrafilter on ω". Can you please briefly hint why? I am familiar with symmetric models but don't see how to use them here. Thanks. – Kyle Russ Oct 12 '14 at 05:36
  • @KyleR: I actually wrote a short post on the topic, http://karagila.org/2014/no-uniform-ultrafilters/ -- here I do it on $\omega_1$, but you can replace each instance of $\omega_1$ by $\omega$ and get the result (which is the hint for the problem from Jech's book). – Asaf Karagila Oct 12 '14 at 05:40
  • Thank you Asaf. I actually realized a direct proof stemming from the observation that any ultrafilter U over ω in HOD(A∪{A})^M[G] must both include a real r_1 and exclude a real r_2 both generic over the parameters from which U is defined. By the symmetry of the model, this gives a contradiction. – Kyle Russ Oct 13 '14 at 02:47
  • @KyleR: Admittedly, I never really saw the symmetric argument from the HOD construction, and I always preferred the former anyway. In any case, you're welcome. – Asaf Karagila Oct 13 '14 at 02:51