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Partition Principle ($PP$) is the following statement:

For all sets $a$, $b$ there is an injection $f:a\rightarrow b$ iff there is a surjection $g:b\rightarrow a$

It is known that $ZF\vdash AC\rightarrow PP$ but as far as I know it is still open that $ZF\vdash PP\rightarrow AC$.

Question 1: What are references for partial results on proving or refuting $ZF\vdash PP\rightarrow AC$?

Let's assume this conjecture is false. Then the main question is:

How to prove $Con(ZF)\Longrightarrow Con(ZF+PP+\neg AC)$?

Miller has a construction for $Con(ZF)\Longrightarrow Con(ZF+\neg AC)$ using a special inner model of $ZF$ in a generic extension. The main idea of this construction is based on the following facts:

Definition: Let $M$ be a ground model and $K$ an inner model of $ZF$ (not necessarily a model of $ZFC$) definable in any model of $ZF$. The pair of forcing notions $\mathbb{P}, \mathbb{Q}$ in $M$ have Miller Property over inner model $K$ iff for all formula $\varphi(x)$ and for all $\alpha\in o(M)$ we have

$$1\Vdash_{\mathbb{P}}(\varphi(\alpha^{\vee}))^K~~~\text{iff}~~~1\Vdash_{\mathbb{Q}}(\varphi(\alpha^{\vee}))^K$$

Which $\alpha^{\vee}$ is the canonical name of $\alpha$ in $M$.

Fact 1: For $I,J$ uncountable in $M$, the forcing notions $Fn(I,2)$ and $Fn(J,2)$ have Miller property over inner model $L(P(\omega))$ of $ZF$.

Fact 2: Using forcing notions like what we introduced in step 1, one can build a generic extension $M[G]$ such that $(L(P(\omega)))^{M[G]}\models ZF+\neg AC$ and the fact that $(L(P(\omega)))^{M[G]}\models \neg AC$ comes from Miller Property of the special type of forcing notions which we used.

Question 2: Is it possible to find a reformulation of Miller's construction to prove consistency of Partition Principle and negation of Axiom of Choice? For example, may something similar to the following approach work? If not, what is wrong?

Step 1: Finding some inner model $K$ such that $K\models ZF+PP$

Question 3: What are examples of such inner models? Surely in this case I am searching for inner models of $ZF$ that doesn't satisfy $AC$ necessarily. (e.g. $L(P(\omega))$) But the point is that finding such an inner model doesn't complete the proof because the situation of $AC$ in this inner model depends on its background model which is a generic extension of a ground model ($M[G]$). Thus one should go to another step by building an appropriate background model using a nice forcing notion over a nice ground model ($M$) which possibly contains something more than merely $ZF$. Perhaps it should satisfy some large cardinal axioms too.

Step 2: Finding some type of forcing notions which have Miller Property on inner model $K$.

Step 3: Forcing with this type of forcing notions and considering $K^{M[G]}$ which is a model of $ZF+PP$

Step 4: Proving $K\models \neg AC$ using Miller Property of forcing notions.

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    This question seems to be pure speculation. Question 3, considered in isolation, has lots of answers, like Gödel's $L$, but these are useless for the purpose at hand because they satisfy AC. If we had an inner model that satisfies PP but is not known to satisfy AC, then we'd probably just work directly on that model and try to find a failure of AC in it. It looks very improbable to me that the approach in the question could succeed if direct approaches (once a reasonable $K$ is proposed) fail. – Andreas Blass Aug 22 '14 at 01:56
  • @AndreasBlass Are there any other suggested approaches to the problem of $Con(ZF+PP+\neg AC)$? As it seems a very basic question by its simple statement, it looks like a bit strange that it is still open! –  Aug 22 '14 at 02:03
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    Sam, I wrote an answer on MathOverflow some while ago about what we know about the Partition Principle. In a nutshell, there hasn't been much progress in the past 20 years, and in fact besides two papers, there hasn't been much progress at all for much longer. The problem is that the Partition Principle gives us "sufficient" structure that all the known tricks for proving $\sf AC$ fail, but its failure lies in places where we cannot yet reach with forcing. – Asaf Karagila Aug 22 '14 at 02:13
  • @AsafKaragila Would you please add a link to your MathOverflow post here? Who isolated this problem for the first time? I guess it should be a very old open question. –  Aug 22 '14 at 02:16
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    This is the link to the answer with the references; here is another about the history of the problem (indeed it is the oldest open problem in set theory). – Asaf Karagila Aug 22 '14 at 02:17
  • @AsafKaragila Thank you very much. I read here that you are working on this problem. Do you think there is any hope to reach some satisfactory partial result? –  Aug 22 '14 at 02:19
  • Sam, my long term goal is to disprove the equivalence between the two. But I don't expect to reach that goal within the next 30 years, or ever. It does give me a good lighthouse for coming up with new ideas. My plan for a Ph.D. is based on such ideas. If they work out as I hope they do, I will later sit down and think about "what else do I need?" and try to push more towards that direction. But I don't expect to have any partial results for a while, just an outline of a plan to attack the problem. And that too still require much work. – Asaf Karagila Aug 22 '14 at 02:22
  • @AsafKaragila I share the idea of non-equivalence between these two with you. I hope you succeed in the near future. –  Aug 22 '14 at 02:31
  • @Sam: I don't see this happening within the next five years to begin with. But in mathematics five years isn't that long, I guess. Then again, I don't see it happening within the next ten either. But thanks. I'll keep working. – Asaf Karagila Aug 22 '14 at 02:45

1 Answers1

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Yes, currently the problem is open whether or not the Partition Principle and the Axiom of Choice are equivalent. There are two major factors for this (in my opinion):

  1. Many people become less interested in choiceless results. So while they might be very happy to hear about them, they prefer to put their research efforts towards other directions.

  2. The Partition Principle gives us just enough structure on the cardinals so the obvious approaches fail pretty fast; but it also indicates that the failure of the axiom of choice might be far away from the ordinals. In other equivalences we can usually show that if a principle fails somewhere, then it will often fail on something which is "close to the ordinals" in some sense.

    This is a vague statement, and I don't intend to clarify on it further. To understand it, feel free to try and apply some of the usual approaches to this problem.


Forcing. Forcing is a great tool for proving consistency results. The "Miller property" which is better described as absoluteness, is not the reason why $L(\Bbb R)$ fails to satisfy the axiom of choice. This is actually not that difficult, once you add "enough" reals which "look the same" you can easily confuse the axiom of choice in a model like $L(x)$ where $x$ is any set including sufficiently many reals.

This was observed by Feferman/Solovay a long time ago. If you add $\omega_1$ Cohen reals to $L$, then $L(\Bbb R)$ fails to satisfy the axiom of choice (but it does satisfy the principle of dependent choice).

But models of the form $L(x)$ will almost usually fail to satisfy the partition principle if the axiom of choice fails. For example there is a definable surjection from $\Bbb R$ onto $\omega_1$; but if there are not $\omega_1$ sequences of reals in $L(x)$ then there is no injection back. Another example is that if there is an infinite Dedekind-finite set, then there is one which can be mapped onto $\omega$, but by definition there will be no injection in the other direction.

And this is true in many of these constructions of inner models of generic extensions. We will usually be able to single out that the set we are interested in has this sort of property which witnesses the failure of the Partition Principle.

So what about forcing? Forcing works fine without the axiom of choice. But we are so used to thinking about it in terms of the axiom of choice, that it becomes very difficult to use forcing as arbitrarily as we do when choice fails. We lose the ability to talk about chain conditions (because some forcings don't have maximal antichains at all), the mixing lemma can fail, and closure does not imply distributivity.

Then when we force over models where the axiom of choice fails we lose the ability to use these tools, and again we can often find examples which are either very hard to understand (so we don't know much about them yet), or witness the catastrophes which may befall on us (we may add unintentional real numbers in the process, and so on and so forth).

So we can try and appeal to absoluteness, but this will usually work for forcings which are associated with ordinals, forcings which are well-ordered inherently (like collapsing something to $\omega$) or has sufficiently "pretty" structure. Not to mention that if the Principle fails, it can fail in many places, which would require you to work much harder than you would have. There is a reason why so little people throughout history understood Sageev's proof for "For all infinite $a$, $a+a=a$ does not imply the axiom of choice". It's a monumentally difficult proof which is spanned over 180 pages (and that's an edited version of his Ph.D.)

So what can we do? Not much currently. As I wrote in the comment, I find this problem to be a lighthouse in the bogs of mathematics. I think to myself what sort of thing would have helped me to prove such a result. But there is a huge distance from my ideas to implementations, and even when the tools I currently craft will be ready for use, it might still take a long time before I will have any idea on how to apply them properly to this problem.

Asaf Karagila
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  • Thanks for your nice answer. What about large cardinals? Do they come to the story of your choiceless journey in 'bogs' of mathematics? (Your description looks like Odysseus' choiceless journey!) Have you ever thought about the positive side of this conjecture? Maybe it has a silly simple solution! ;-) –  Aug 22 '14 at 03:24
  • Not yet. I imagine that I might later on consider them. First axes and swords must be forged, then things like armor, and a helmet. Only then I'll consider going out to kill Trojans. That been said models of $\sf AD$ have the inherent problem that I wrote about, no $\omega_1$ sequences of real numbers, but that's all about large cardinals for now. As for a positive solution? Sure. Every couple of months I realize I can prove the equivalence, I start writing it, and a few minutes later I come to the same realization that there's a huge chasm which I can't quite cross yet. – Asaf Karagila Aug 22 '14 at 03:31