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

Some of you already saw this new preprint on arXiv, and some of you even emailed me about it. I'm reading the paper, but I decided to do something drastic and join Twitter, temporarily, so I can more easily have discussions about this paper.

I will update this post on occasion while I read it, to reflect new understanding, kind of like a live journal, if you will.

## 9 October, 2020


The way to think about this, so far, which will probably be wrong later, is that the functions are class functions on the universe, only that they can act on one another. Kind of like how if $$j\colon V\to V$$ is an elementary embedding, then we have $$j\circ j$$ and $$j(j)$$ as two different things.

In this sense, $$\fz$$ is kind of like the empty set, whereas $$\fo$$ is the identity. There is the notion of action, $$f[t]$$ which says that $$f$$ has a nontrivial action on $$t$$, which in turn means that $$\fz$$ is really a "filler object" which stands for "undefined". Using this understanding, I think, a lot things are somehow clearer.

Hopefully this will start to make more sense soon.

## 10 October, 2020

I didn't read anything today, it's the weekend and I have other things to do. But I wanted to document what I felt are the big open questions from myself (and others online) about this at the moment:

1. Is Flow consistent relative to a known theory? It feels like it should be somewhere in the vicinity of "many inaccessible cardinals" to the point of "inaccessible limit of inaccessible cardinals". But of course we need to understand it in order to see if it can be interpreted there. Maybe it's weaker, or maybe it's just inconsistent?

2. Is the proof $$\axiom{Flow}\proves\ZF+\axiom{PP}\not{\proves}\AC"$$ sound? If it is, then we move the question of correctness back to the question of consistency of Flow itself, and the methods can illuminate a way to use forcing-based methods (ahemiteratedsymmetricextensionsahem maybe?) to obtain this proof directly in familiar set theory.

3. In the proof, the model obtained of $$\ZF$$ is in fact a model of $$\ZF$$ without Regularity. In that case, one has to ask whether or not the failure of $$\AC$$ happens in the well-founded sets of that model? If not, then this is not quite a proof from $$\ZF$$, but just almost. And this would be a very interesting reverse case of "Every vector basis has a basis implies Choice", where we know the proof in $$\ZF$$, but it is open in $$\ZF$$ without Regularity.

I hope to have time on Monday to get back to this project.

## 12 October, 2020

I want to file a complaint about F8, the Coherence axiom. I feel that says something to the effect of $$\Bbb E(f)\implies\Bbb E(\fp(f))$$ and $$\Bbb E(\bigcup f)$$, where $$\bigcup f$$ denotes somehow the union over $$f$$, whatever that means. So in other words, if $$f$$ was emergent, then going up the rank, and going down in the rank, will preserve emergence.

Going on, restriction and F9 seem to be meant as an analogue of Separation, and very much so when applied to ZF-sets. The restricted power, $$\wp(f)$$ is the real analogue of the power set, whereas $$\fp(f)$$ is more of a "all partial functions $$f\to f$$".

The authors remark that $$\wp(f)\subseteq\fp(f)$$ for all $$f$$, even though there is no claim that either of these is a function. Rather the claim is that these are shorthand for formulas. I realise now that nowhere it is claimed that $$\fp(f)$$ even exists, let alone $$\wp(f)$$, or that if $$\fp(f)$$ exists, then $$\wp(f)$$ exists. These are interesting points, and I'd be happy to see them cleared up. But hopefully they will be later on in the text.

I'm peeking ahead, and it seems to me that the understanding of $$f[g]$$ as "$$g\in\dom f$$" is quite important.

## 17 October, 2020

Sorry for the silence over the last few days. I've been busy, and every time I peaked in the paper I felt the need to backtrack a bit, and had nothing significant to say. But now I read all the way to the formulation of "$$\mathfrak{F}$$-Choice".

And there's something a bit odd here. Look at the following.

Proposition. $$\AC$$ is equivalent to "If $$f$$ is a function, there is $$g\subseteq f$$ such that $$g$$ is injective and $$\rng f=\rng g$$.

Proof. Assume $$\AC$$, then simply consider the relation $$f^{-1}$$; it contains a function with the same domain, say $$h$$. But by the fact that $$f$$ was a function, $$h$$ must be injective (since the fibres of $$f$$ are pairwise disjoint). Let $$g=h^{-1}$$, and we're done. In the other direction, let $$\{A_i\mid i\in I\}$$ be a family of pairwise disjoint non-empty sets, and let $$f(a)=A_i$$ if and only if $$a\in A_i$$, then if $$g\subesteq f$$ is injective, we have that $$g^{-1}$$ is a choice function. $$\square$$

But the formulation of $$\mathfrak{F}$$-Choice explicitly states that the choice function is not a restriction of $$f$$. Indeed, it can "mix scramble" the actions. In other words, we are merely stating that given a function with domain $$X$$ and range $$Y$$, there is an injective function from a subset of $$X$$ onto $$Y$$. This is just restating the Partition Principle itself!

I don't have time to keep reading, but it seems to me now that $$\axiom{Flow}$$ is somehow designed to include $$\axiom{PP}$$. The question, of course, is whether or not it really negates the correct formulation of the Axiom of Choice? Hopefully, we'll find out more next week.

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