Asaf Karagila
I don't have much choice...

# Open Problems

Here is a list of open problems that I am interested in. Some problems are posted with a small prize to them. Just to prevent this from being eternally binding, the statute of limitations is exactly ten years from the date of posting (unless specified otherwise). Exact nature of the prize may change depending on unforeseen circumstances, your mileage may vary.

Please send me updates on these problems, in case you know anything. I will add below each problem any progress that was made towards solving it.

You may also send me problems to post here. However, external problems must come with a prize, and I am not responsible for any disputes as to the nature of the prize between the winners and whomever suggested the problem.

### 1. The Partition Principle

The Partition Principle states: suppose that there is a surjective map $$f\colon A\to B$$, then there an injective function $$g\colon B\to A$$.

Well. Assuming the axiom of choice, we can prove there is even $$g$$ which splits $$f$$. And if we require the injection to actually split $$g$$, then we can actually prove the axiom of choice. However, without assuming the axiom of choice, we cannot prove a whole lot about the partition principle. We know it implies $$\DC$$, and therefore not provable from $$\ZF$$ itself. But not much more.

Question: Is the Partition Principle equivalent to the Axiom of Choice over $$\ZF$$?

Prize: A bottle of whisky aged 15 or older for a negative answer; a fine gin/vodka for a positive answer.

(April 8th, 2018)

### 2. Iterations of Symmetric Extensions on non-finite supports

In my Ph.D. thesis I have developed a method to iterate symmetric extensions transfinitely. However, a significant part of the definition relies on the finiteness of the support of the iteration. This means that at limit steps countable choice principles are likely to fail.

Question: Are there sufficiently flexible limitations that let us define the iterations for countable, mixed, Easton, or otherwise-of-interest supports? Particularly, are there limitations which still allow adding reals, and thus obtain simpler constructions of choiceless models of $$\ZF+\DC$$ where certain choice-consequences fail for the reals?

(April 8th, 2018)

### 3. Can you force PFA with a closed forcing?

It is a theorem of König–Yoshinobu that you cannot violate the Proper Forcing Axiom with an $$\omega_2$$-closed forcing (improving upon Larson's earlier result with $$\omega_2$$-directed closed). So if PFA holds, after forcing with a sufficiently closed forcing, it continues to hold.

Question: What about the converse? If you force with a sufficiently closed forcing, and PFA holds in the extension, does it hold in the ground model?

Prize: A shirt.

This can be varied to ask the following natural question.

Question: If $$\PP$$ is a proper forcing, and $$\QQ$$ is sufficiently closed. Is it true that $$\forces_\QQ\check\PP$$ is proper?1

Prize: A different shirt.

Solution: Yasuo Yoshinobu has kindly provided a negative answer to the question. Given a regular $$\kappa$$, he provided two examples of proper forcings whose properness is not preserved by $$\kappa$$-closed forcings. (Solutions provided on June 5th and 11th, 2018.)

(April 26th, 2018)

1. June 3rd, 2018: Thanks to Yasuo Yoshinobu for pointing out that $$\QQ$$ and $$\PP$$ were reversed in the second question.

### 4. Lower consistency strength of consecutive triviality of the approachability ideal

Shelah's approachability ideal has been presented in many places (e.g. in this post by Assaf Rinot), $$I[\lambda]$$ denotes the approachability ideal on $$\lambda$$. It is a theorem of Mitchell that from a $$\kappa^+$$-Mahlo, we can force that $$S\cap E^{\aleph_2}_{\aleph_1}$$ is non-stationary for all $$S\in I[\aleph_2]$$. Boban Velickovic is currently trying to get that in addition to that, also for all $$S\in I[\aleph_3]$$ the intersection $$S\cap E^{\aleph_3}_{\aleph_2}$$ is non-stationary.

Question: We know that consecutive singulars or consecutive cardinals with the tree proprety give you an inner model with a Woodin cardinal. Is this the same here?

(April 26th, 2018)

### 5. Does any nontrivial forcing preserve properness?

Shelah gave a neat construction of a $$\sigma$$-closed forcing which is not proper after adding a Cohen real. Yoshinobu, in a similar manner, provided a solution to Problem 3(b), where adding a Cohen subset to $$\kappa$$ violates properness of a forcing (working from an observation of Shelah in "Proper and Improper Forcing", but also a different solution based on MRP-style forcings).

It seems, if so, that $$\Add(\kappa,1)$$ always violates properness of some forcing for any $$\kappa$$.

Question: If $$\PP$$ is a nontrivial forcing (read: it admits no atoms), is there provably a forcing $$\QQ$$ such that $$\QQ$$ is proper, but $$\forces_\PP\check\QQ$$ is not proper?

Prize: A nice bottle of port (or equivalent).

(November 23rd, 2018)

### 6. Is the Krein–Milman Theorem related to a Łoś's Theorem for continuous logic?

This is something that I came up with a couple of years ago, and I have brought this up now and then with people, but recently someone suggested I put it on my site. Which makes sense.

We know that for first-order logic Łoś's Theorem + Compactness Theorem imply the axiom of choice. This is due to Paul Howard, and the proof goes through extension of filters to ultrafilters, rather than compactness, but the two are equivalent.

I know practically nothing about continuous logic from a model theoretic point of view. But I know a bit about equivalents of the axiom of choice.

The Krein–Milman Theorem, which states that every compact convex set is the closure of the convex hull of its extremal points.

Assuming the Krein–Milman Theorem and the Compactness Theorem for FOL also imply the axiom of choice. And we can strengthen Krein–Milman slightly and weaken the Compactness Theorem to the Hahn–Banach Theorem.

Now. The reason that Łoś's Theorem requires choice is when dealing with the existential quantifier. Similarly, Krein–Milman's proof seem to mainly utilize choice when proving that extremal points even exist to begin with. Moreover, the Compactness Theorem is itself equivalent to the Banach–Alaoglu theorem, also dealing with compactness of the closed unit ball in the weak* topology.

Question: How is the basic model theory of continuous logic connected to the axiom of choice (and specifically to the classical logic related results)? More specifically, is the Krein–Milman Theorem in some way a geometrization of Łoś's Theorem?

Prize: Why does everything needs to have a prize? Having a nice paper is not enough for you?

(June 8th, 2019)

### 7. How many normal measures can a measurable cardinal have?

Let $$\kappa$$ be a measurable cardinal, by which I mean there is a $$\kappa$$-complete free ultrafilter on $$\kappa$$, and in this context we are going to add uncountable as well (and just to clarify, I am using $$\kappa$$ to hint that I only consider $$\aleph$$ cardinals for this definition). Assuming the axiom of choice, $$\kappa$$ carries anywhere between $$1$$ and $$2^{2^\kappa}$$ normal measures, with all the options possible. (I am cheating here a bit, since I'm implicitly assuming $$2^{2^\kappa}=\kappa^{++}$$, but let's leave that aside, since this will soon be rendered irrelevant.) We also know from the work of Spector, and more recently Bilinsky and Gitik, that it is also consistent that a measurable cardinal has no normal measures. And Arthur Apter had shown that it is consistent for $$\aleph_{\omega+1}$$ to be measurable with an arbitrarily large, but well-ordered, number of normal measures.

A normal measure is a subset of $$\mathcal P(\kappa)$$, there are at most $$2^{2^\kappa}$$ measures, but without the axiom of choice, $$2^\kappa$$, and certainly $$2^{2^\kappa}$$ need not be well-orderable. Even more alarming, $$2^{2^\kappa}$$ need not be linearly orderable, and it may contain an amorphous subset.

Question: What are the limitations on the cardinal of the set of normal measures on a measurable cardinal $$\kappa$$? Does it have to be well-orderable? Linearly orderable? Does it have to have certain properties to it? Can it be amorphous (to some degree, e.g. $$\aleph_1$$-amorphous, if we assume some $$\DC$$), or anything else?

Prize: A bottle of wine from me. Also an incredible paper.

(September 21st, 2020)

### 8. An explicit ultrafilter in Cohen's first model

We know from the work of Blass and Pincus that $$L(X)$$ satisfies the Boolean Prime Ideal theorem if and only if $$X^{<\omega}$$ has carries ultrafilter $$U$$ such that for all $$x\in X$$, $$\{p\in X^{<\omega}\mid x\in\operatorname{rng}(p)\}\in U$$, call this property "fine".

We know that Cohen's first model satisfies the Boolean Prime Ideal theorem, and that it has the form $$L(A)$$ where $$A$$ is the set of Cohen reals.

Question: Can we describe a fine ultrafilter on $$A^{<\omega}$$ in Cohen's first model?

(November 9th, 2020)