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Solovay's model (which assumes $I$ = "existence of inaccessible cardinal") will be a well-known construction to produce a model of ZF where only the full Axiom of Choice ($AC$) fails, but the Axiom of Countable Choice ($AC_\omega$) remains true.

For obvious reasons, it assumes the existence of an inaccessible cardinal. My question is about "what are those obvious reasons?" part.

I am aware that both $AC$ and a weaker $AC_\omega$ are independent of $ZF$. Apparently this has even been nicely answered here on MO. I am also aware of other answers like:

A bit of background - after reading "Axiom of Choice" by H. Herrlich I got an initial impression that since $AC_\omega$ is weaker than $AC$, it can be more attractive or lead to less "catastrophes" (to quote the author). But as said, there are models where $AC_\omega$ fails equally dramatically.

For the same obvious reasons assuming the existence of some form of large cardinals might be the way forward, where "the obvious reasons" would be that this is also the only way.

So, I would try to break down and phrase my question more succinctly:

  1. Are there any known attempts to build "$ZF + X - I$" where $AC$ fails but $AC_\omega$ holds? What is that $X$?
  2. Is $X$ a weaker version of $AC_\omega$? Is it possible?
  3. Is $X$ a bit stronger like $DC$? Any evidence its a good fit to be less risk in causing "catastrophes"? What is the intuition here?
  4. Alternatively, is there more evidence to justify the existence of large cardinals? Is this always "ZF + Y", where Y is always "there exists a large cardinal"?

To quote D.Scott "if you want more you have to assume more". But I mean is there more to this? And why would one even ignore existence of large cardinals assuming they solve more problems than they manage to create?

I guess the idea behind "In search of Ultimate L" is all about constructing a program to provide more for the point (4) (and also the (*), Omega-consitency and all of the amazing work by Woodin). But then $AC_\omega$ is replaced with stronger $AD(\mathbb{R})$ (although still consistent with $AC$ on $L(\mathbb{R})$ - so it seems to be even possible to show this in $ZFC$).

Bottom line, I understand that large cardinals in their expressiveness are equivalent to assuming maximality principles like $MM$ which will be required for all kind of SSP forcings (apparently, there is a 2021 paper by Asper´o and Schindler showing that "$MM^{++} implies (*)$"). I clearly see how encouraging this is.

But is this known to be the only way forward? Maybe the real question for me here is:

  1. Is $MM^{++}$ in some way $X$? Assuming the BoundedMM implies $AC$ (this needs clarification).
  2. Is there a way to show that axiom $X$ (if exists) is always equivalent in its "expressiveness" to some $Y$? And also having stationary set preserving (SSP) forcings or Omega-Consistency will be the only rich enough way of showing this?

PS

I genuinely hope my question makes sense. If you feel that this question is more appropriate for any other forum, please let me know. Set theory is my hobby and I am very new here, so I am more than happy to move or repost this question if downvoted. I am also grateful for any guidance to rephrase the question and happy to update it if it is "too broad", "too naive" or "too far stretched". I understand this.

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    $\begingroup$ No large cardinals are needed to get the consistency of ZF + DC + not AC. Thomas Jech's book "The Axiom of Choice" has a chart near the end showing (non)implications between AC and various weaker axioms. For more details, there's the book "Consequences of the Axiom of Choice" by Paul Howard and Jean Rubin. $\endgroup$ Commented Aug 29, 2023 at 13:26
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    $\begingroup$ Since you mentioned Solovay's model, it may be relevant to mention that Shelah proved that the cardinal hypothesis cannot be removed (if you want all sets to be L-measurable). $\endgroup$ Commented Aug 29, 2023 at 13:51
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    $\begingroup$ @YauhenYakimovich My copy of "The Axiom of Choice" is definitely older than 2003, but I would expect the chart to survive from one edition to another. (Note that this book is different from Jech's book "Set Theory".) $\endgroup$ Commented Aug 29, 2023 at 14:55
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    $\begingroup$ I have to say, this reads like there's a lot of misconceptions that should be cleared out. I'm afraid I don't have a lot of time these days, but hopefully one of the excellent explainers that we have on the site can help. I will try to find some time this week to post an answer if not. $\endgroup$
    – Asaf Karagila
    Commented Aug 29, 2023 at 16:24
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    $\begingroup$ (Also, at least according to Wikipedia, the quote you've attributed to Solovay is actually due to Dana Scott. Although it is certainly a thought that many people probably had independently of each other. In fact, until now, I didn't realise it was someone's quote...) $\endgroup$
    – Asaf Karagila
    Commented Aug 29, 2023 at 16:26

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