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I was reading this answer that no free ultrafilter can be exhibited on the natural numbers.

I have as a theorem that if the Collatz conjecture is true then the following is a free ultrafilter on the natural numbers:

Let $f(x)=(3x+2^{\nu_2(x)})/2$

where $2^{\nu_2(x)}$ indicates the highest power of $2$ that divides $x$ while still giving an integer quotient so for example $2^{\nu_2(24)}=8$.

Let $f^n(x)$ indicate the $n^{th}$ iterate of $f(x)$

Let $F_y$ be the filter at $y$ and be defined as the set of integers $x$ such that there is some $n\in\Bbb N$ such that $f^n(x)=y$, ordered by $n$. So for example $3\preceq5\preceq8\preceq16\ldots$

Then it is a theorem that if the Collatz conjecture is true, the set of all subsets of $N$ with the order inherited from $F_y:y\in\Bbb N$ is a free ultrafilter on $\Bbb N$.

I'm using this definition of free ultrafilter. Is it compatible with the one in use in Noah's answer?

Question

It appears to me that I have explicitly exhibited a free ultrafilter on $\Bbb N$ (modulo the small matter of a proof of the Collatz conjecture), yet such a thing can not be explicitly exhibited. What implications does this have for a proof of the Collatz conjecture? Does it mean nothing more than; such a proof will necessarily require the axiom of choice? Does it mean the proof cannot be explicitly exhibited?

I was curious a good long while ago about the possibility some apparently hard theorems could be unprovable or equivalent to the Godel sentence of a system. This reminds me of that hypothesis. Certainly the Collatz conjecture is "an arithmetical statement which claims that no number exists with a particular property", and therefore among the candidates for a Godel sentence.

Is it possible that this proves no arithmetical statement expressible in Peano arithmetic can disprove the Collatz conjecture, and therefore that the conjecture is true?

it's a hire car baby
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    It could also mean that the Collatz conjecture is false which would be compatible with your result. This would however be quite surprising (Personally, I would enjoy it however!) , so please check the proof again whether you really established this result. – Peter Sep 19 '23 at 11:37
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    Of course , it is possible that Collatz is unprovable in PA , it could even be unprovable in ZFC , assuming it is true. Even more , it could be not disprovable in ZFC if it is false. I heard from claims that this can be ruled out , but with no argument. It is only ASSUMED that ZFC is strong enough to prove all "natural" statements. – Peter Sep 19 '23 at 11:41
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    @Peter the proof is quite simple. Think of the filter as simply the set of natural numbers which get to $1$ in fewer than $n$ "odd" steps. There is no upper bound on Collatz sequence length, which is what makes it free. I take your point about Collatz possibly being false, but I want to understand what does it mean for the proof, assuming the Collatz conjecture is true. Simply because, I am working on it on the assumption it's true in which case the answer to this signposts methods for me. – it's a hire car baby Sep 19 '23 at 11:42
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    Noah says that what you call "exhibit" is rather technical. It may well be the case that your alleged free ultrafilter is "hard to define". I have absolutely no idea. @Noah Schweber – FShrike Sep 19 '23 at 11:46
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    how is your $F_y$ even a filter? isn't it a set of natural numbers? – univalence Sep 19 '23 at 12:28
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    1/ The $F_y$ you exhibit is not a free ultrafilter. You pick some natural $y$ and define $F_y$ as the set ${ S \subseteq \mathbb{N} \mid \forall x \in S. \exists n \in \mathbb{N}. f^n(x) = y }$. If this set was an ultrafilter at all, then we'd either have the even numbers contained inside it, $2\mathbb{N} \in F_y$, or else their complement, the odd numbers with $2\mathbb{N} + 1 \in F_y$. But it's immediate that neither of these sets belong to $F_y$, no matter how you choose $y$. – Z. A. K. Sep 19 '23 at 12:29
  • 2/ To show that $2\mathbb{N} \not\in F_y$, pick the smallest integer $2^k$ that is larger than $y$. You have $f(2^k) = \frac{3\cdot 2^k + 2^k}{2} = {4\cdot 2^k}{2} = 2^{k+1}$. By induction, $f^n(2^k) = 2^{k+n}$. Thus, it's clear that $f^n(2^k)$ will never hit $y$. If $2\mathbb{N}$ was in your purported ultrafilter, $f^n(e) = y$ would eventually happen for any even $e$. But since $2^k$ is even, and we established that it doesn't happen for $2^k$, this means that the set $2\mathbb{N}$ is not contained in your ultrafilter. – Z. A. K. Sep 19 '23 at 12:30
  • 3/ To show that $2\mathbb{N} + 1 \not\in F_y$, we can argue as follows. If $y$ itself is even, then $q = \frac{2^y - 1}{3}$ is odd, and if $3 < y$ then $y < q$. Moreover, $f(q)$ is a power of two, so again $f^n(q)$ only hits $q$ and powers of $2$. Hence, it never hits $y$, so the set of odd numbers does not belong in $F_y$. If $y$ is odd, the same argument works with $2^{y+1} -1$ instead. The special case $y=1$ is handled easily. All in all, we see that $F_y$ cannot be an ultrafilter at all, for any choice of $y$. – Z. A. K. Sep 19 '23 at 12:31
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    @it'sahirecarbaby: Well, your view was mistaken. There are plenty of non-free ultrafilters that fail to contain $2\mathbb{N}$, take e.g. ${S \subseteq \mathbb{N} \mid 67 \in S }$. – Z. A. K. Sep 19 '23 at 12:37
  • @Z.A.K. thanks for your help. $F_y$ are proper filters and therefore cannot be free filters at all let alone free ultrafilters. But the maximal filter over all $F_y$ is a free ultrafilter. – it's a hire car baby Sep 19 '23 at 14:02
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    What do you think the definition of "ultrafilter" is? I don't see any way to interpret the phrase "the set of all integers with the order inherited from $F_y:y\in\mathbb{N}$" so that it describes an ultrafilter. – Alex Kruckman Sep 19 '23 at 14:08
  • @AlexKruckman I'm using this definition - I'll add to the question: https://en.wikipedia.org/wiki/Ultrafilter#Ultrafilters_on_partial_orders and I have edited slightly to make build the ultrafilter out of members of the power set rather than members of $\Bbb N$ itself. – it's a hire car baby Sep 19 '23 at 15:39
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    Oh, I think I understand (part of) the confusion here. You've linked to the definition of "ultrafilter on a poset". When people talk about an ultrafilter on a set $X$, they don't mean "equip $X$ with some partial ordering and consider an ultrafilter on that poset". Instead, they mean "ultrafilter on the Boolean algebra $\mathcal{P}(X)$" (which is the set of all subsets of $X$ ordered by $\subseteq$). So the definition relevant for the notion of "free ultrafilter on $\mathbb{N}$" is a bit further down on the Wikipedia page you link to, in the section on ultrafilters on power sets. – Alex Kruckman Sep 19 '23 at 15:47
  • @AlexKruckman thanks I knew there were nuances in the types of Ultrafilter. Nevertheless I don't yet see that this is fatal to the premise of the question. – it's a hire car baby Sep 19 '23 at 15:57
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    Well, it remains the case that what you've written does not seem to define an ultrafilter. Can you try again to define it? Precisely which sets of natural numbers are in the ultrafilter? – Alex Kruckman Sep 19 '23 at 16:26
  • @AlexKruckman it's going to take me time to define it carefully and without error using that boolean algebra definition I am currently unable to parse but I continue to think the set of all orbits of $f$ ordered by their orbit direction conceptually fits the model of an ordered set of subsets of $\Bbb N$ and is a free ultrafilter. – it's a hire car baby Sep 19 '23 at 16:30
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    You don't need to understand the general definition of an ultrafilter on a Boolean algebra. An ultrafilter on $\mathbb{N}$ is a set $U$ of subsets of $\mathbb{N}$ such that: (1) if $A$ and $B$ are in $U$, then $A\cap B$ is in $U$, (2) if $A\in U$ and $A\subseteq B$, then $B\in U$, and (3) for every set $A$ or natural numbers, exactly one of $A$ or its compliment is in $U$. The set of all orbits of $f$ is obviously not an ultrafilter, since it is not the case that every set is either an orbit of $f$ or the complement of an orbit of $f$. – Alex Kruckman Sep 19 '23 at 23:14
  • @AlexKruckman I'm struggling to pin it down - maybe you can help me? https://math.stackexchange.com/q/4772341/334732 . Surely there is a standard model for an ultrafilter which describes the orbit of a function which acts transitively on a set? – it's a hire car baby Sep 20 '23 at 08:51
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    It's not enough to feel like some construction "conceptually fits the model" of a free ultrafilter or say things like "surely this can be done". You have you actually check the definition. As someone who has worked a lot with ultrafilters, I do not share your intuition - I see no reason at all why the concepts of "orbits of a function" and "free ultrafilter on $\mathbb{N}$" should have anything to do with one another. An ultrafilter $U$ on $\mathbb{N}$ has to decide for EVERY set $A$ whether $A$ or the complement of $A$ is in $U$. Making all these choices is where AC is useful... – Alex Kruckman Sep 20 '23 at 13:31

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The claim that the proof of the Collatz conjecture would imply the existence of a free ultrafilter would establish that the Collatz conjecture cannot be proved in ZF. This is because ZF does not prove the existence of a free ultrafilter. Similarly, ZF+ACC does not prove the existence of a free ultrafilter, and therefore (if your claim is correct) would not be able to prove the Collatz conjecture, either (here ACC is the axiom of countable choice). Even the stronger system ZF+ADC cannot prove the existence of a free ultrafilter and therefore would be unable to prove Collatz (here ADC is the axiom of dependent choice). One would need a stronger version of the axiom of choice, such as for example the axiom of choice for families indexed by the continuum, if one is to have any hope of establishing Collatz. Of course, as mentioned in the comments it is not known whether even the full ZFC would prove it. I notice that there is some doubt in the comments whether your $F_y$ can be used to get an ultrafilter (without using choice), which I haven't checked, so my remarks are contingent on the possibility of doing so, which needs to be established.

Mikhail Katz
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  • Thanks for this. The commenter simply thought I was claiming each proper $F_y$ was a free ultrafilter for some individual $y$ but it clearly cannot even be free as it is a principal filter. Rather than I meant the maximal filter over all $F_p$, which is a free ultrafilter. – it's a hire car baby Sep 19 '23 at 14:03
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    What is a "maximal filter over all $F_p$" exactly? Does its construction involve the axiom of c choice? Clearly if it does, this changes the situation considerably. @it's – Mikhail Katz Sep 19 '23 at 14:06
  • Not sure if this answers your query but if the Collatz conjecture is true then the filter over all $F_p$ has a partial order over all of $\Bbb N$. If Collatz fails, not all natural numbers are connected to a common successor by the order relation. – it's a hire car baby Sep 19 '23 at 14:14
  • Is $F_y$ supposed to be a member of your "ultrafilter"? @it's – Mikhail Katz Sep 19 '23 at 14:16
  • The order relations of every $F_y$ are inherited by my free ultrafilter. – it's a hire car baby Sep 19 '23 at 14:17
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    If I understand your definition correctly, $F_y$ is a subset of $\mathbb N$, in other words an element of $\mathcal P(\mathbb N)$. Since $y$ ranges over $\mathbb N$, there are only countably many such $F_y$. Therefore what you constructed is a countable collection of sets. However, an ultrafilter is necessarily an uncountable collection of members of $\mathcal P(\mathbb N)$. Therefore what you constructed cannot be an ultrafilter. If it is a filter, then one could find a maximal filter containing it, i.e., an ultrafilter. However, the existence of a maximal filter containing a given .. – Mikhail Katz Sep 19 '23 at 14:21
  • ... one depends on the axiom of choice. @it's – Mikhail Katz Sep 19 '23 at 14:23
  • Thank-you for that. Are you using this definition of ultrafilters: https://en.wikipedia.org/wiki/Ultrafilter#Ultrafilters_on_partial_orders – it's a hire car baby Sep 19 '23 at 14:23