10

We know that free ultrafilter exist on natural numbers. I would like to see an example of a free ultrafilter on natural numbers.

asv
  • 911

2 Answers2

19

I'm assuming you meant "free ultrafilter."

This is a very reasonable question! Unfortunately, in a very real sense we can't exhibit a concrete example of a free ultrafilter on $\mathbb{N}$ - it is consistent with ZF (= set theory without the axiom of choice) that there are no free ultrafilters on $\mathbb{N}$. Any free ultrafilter has to be "hard to define" in various precise ways (this gets a bit technical - the relevant field is "descriptive set theory").

The situation is similar with regard to a number of other kinds of object whose existence relies on the axiom of choice, like non-measurable sets (as in the Banach-Tarski paradox).

Noah Schweber
  • 260,658
  • Yes I meant "free ultrafilter", thank you. – asv Aug 17 '18 at 14:11
  • If I am not wrong, it is like to construct a choice function (if someone crates it, we would have that ZF implies AC but it is impossible) – asv Aug 17 '18 at 14:14
  • The hyperreal numbers are constructed by any free ultrafilter. It is possible to give a canonical hyperreal numbers also if it is not possibile to give explicitly a free ultrafilter? – asv Aug 17 '18 at 14:22
  • Do you have a reference for the consistency of ZF+ ¬UF(ℕ)? – Lukas Juhrich Jul 15 '19 at 21:05
  • 1
    @Luke See the references at this old MSE question. – Noah Schweber Jul 15 '19 at 21:11
  • From my understanding, the reason why "this gets a bit technical" is that this is a little more involved than the fact that it is consistent with that there are no free filters on $\mathbb{N}$ --- a priori it might be that one could define an object explicitly that is only provably a free ultrafilter in ZFC, not in ZF. I'm trying to edit Wikipedia, so I would appreciate it if you could give me a reference to how this issue is addressed in descriptive set theory. – Imperishable Night May 29 '23 at 04:32
  • 1
    @ImperishableNight What I had in mind is that a nonprincipal ultrafilter on the naturals does not have the Baire property (see here). This means that no nonprincipal ultrafilter can be defined as a countable Boolean combination of sets which are existential-second-order-definable with real parameters (this is a fancy way of saying that analytic sets, and countable Boolean combinations thereof, do have the Baire property). – Noah Schweber May 29 '23 at 04:52
  • Actually, now that I've thought about this more, could a free ultrafilter be "explicitly defined" in ZF + V=L? If I'm understanding correctly, V=L implies the axiom of choice since one could always choose the set "constructed the earliest", and by consistently following this principle one can point to a definite example of a free ultrafilter in ZF + V=L. Of course this just shows that the phrase "concrete example" is an especially subtle one. – Imperishable Night Jun 07 '23 at 06:56
  • The impossibility of exhibiting a free ultrafilter on $\Bbb N$ perplexes me slightly, particularly in relation to my following belief: Define the filter at $y$ as the set of natural numbers $x$ such that some $n^{th}$ iterate of compositions the function $f(x)=(3x+2^{\nu_2(x)})/2$ satisfies $f^n(x)=y$, then if the Collatz conjecture is true, this yields a free ultrafilter on the natural numbers. Am I doing something wrong here? Or am I simply proving that it would require the axiom of choice to prove the Collatz conjecture ? – Robert Frost Sep 19 '23 at 11:04
  • Actually FWIW I made it a question: https://math.stackexchange.com/questions/4771715/ – Robert Frost Sep 19 '23 at 11:29
7

Edit: The question originally asked for an example of a free filter on the natural numbers.

The filter $\{X\subseteq \mathbb{N}\mid \mathbb{N}\setminus X\text{ is finite}\}$ is free. This is known as the cofinite filter, or the Fréchet filter.

Maybe you meant to ask for an example of a free ultrafilter? No truly explicit example can be given - there is no constructive proof of the existence of a free ultrafilter on $\mathbb{N}$.

Alex Kruckman
  • 86,811
  • Yes I meant "free ultrafilter", thank you. – asv Aug 17 '18 at 14:10
  • The hyperreal numbers are constructed by any free ultrafilter. It is possible to give a canonical hyperreal numbers also if it is not possibile to give explicitly a free ultrafilter? – asv Aug 17 '18 at 14:20
  • 2
    I'm assuming that by "hyperreal numbers", you mean an ultraproduct of the ordered field $\mathbb{R}$ by a free ultrafilter on $\mathbb{N}$. If you assume the continuum hypothesis, then any two such ultraproducts are isomorphic (see here), so there is a canonical ordered field of hyperreals (up to isomorphism). But in the absence of CH, or if you want to do something like work in a language with all subsets of the reals named by predicates, the hyperreals won't be canonical. – Alex Kruckman Aug 17 '18 at 14:43
  • I assume that this means that this means that the existence of a free ultrafilter on Z follows in ZF from the ultrafilter lemma without the axiom of choice then?

    So ZF plus the boolean prime ideal theorem is a sufficient axiomatization to find a free ultrafilter on the natural numbers, to use the ultrapower construction, etc etc.

     – saolof Aug 14 '20 at 02:37
  • 1
    @saolof Yes, the boolean prime ideal theorem (BPIT) / ultrafilter lemma immediately implies (over ZF) that there is a free ultrafilter on the natural numbers. Using this ultrafilter, you can use the ultrapower construction. But note that when working with ultrapowers, the main tool is Łoś's theorem, which can't be proven in ZF + BPIT. For more on the relationship between Łoś's theorem, BPIT, and choice, see here. – Alex Kruckman Aug 14 '20 at 13:34
  • @AlexKruckman Ah right, Łoś's theorem is necessary to prove the transfer principle.

    Can Łoś's theorem be weakened to work up to a single cardinality though? For example, Zorn's lemma has a weaker form that can be proven using the axiom of dependent choice over the ordinals.

    Then again, BPIT already implies non-measurable sets, which is the main disadvantage of a foundation for analysis that uses AOC. But I'd love to see if there is a simple-enough foundation for NSA that works in a Solovay-like model.

     – saolof Aug 14 '20 at 22:28
  • 1
    @saolof Sure - Łoś's theorem for ultraproducts by ultrafilters on $\omega$ only requires countable choice. You might get some interesting (and more nuanced) answers if you asked about NSA in a Solovay-like model here or in Math Overflow. – Alex Kruckman Aug 14 '20 at 23:42