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July 30

Axiom of choice, axiom of countable choice, any others?

We have the Axiom of countable choice, which is weaker than Axiom of choice, so is it possible, meaningful or even already done to have an Axiom of aleph 1 choice or Axiom of cardinality of continuum choice, weaker than the Axiom of choice? Also, could the Axiom of dependent choice actually be such an axiom, corresponding to some particular aleph?Rich (talk) 20:20, 30 July 2024 (UTC)[reply]

The notion of cardinality is a fragile one without AC, but you can certainly define the axiom "If

(A_{\alpha })_{\alpha <\omega _{1}}

is a sequence of nonempty sets, the product is nonempty", and that could be reasonably thought of as

\aleph _{1}

choice. That's stronger than dependent choice (DC), since it's enough to construct Aronszajn trees, and you can arrange that the Solovay model satisfies DC + no Aronszajn trees. I'm confident it's weaker than full choice, but I don't know enough about symmetric extensions to show it.--Antendren (talk) 06:40, 31 July 2024 (UTC)[reply]

Interesting. it makes me wonder if a hierarchy of AC axioms as adjuncts to ZF, akin to what i've read about a hierarchy of Large cardinal axioms added to ZFC could be discovered.Rich (talk) 02:29, 1 August 2024 (UTC)[reply]

Even for finite cardinals this is non-trivial; for example, AC₂ constructively implies AC₄. (Andrzej Mostowski. "Axiom of choice for finite sets". Fundamenta mathematicae, vol. 33 (1945), pp. 137–168.) --Lambiam 08:29, 1 August 2024 (UTC)[reply]

Going in the other direction, the Axiom of global choice is stronger than the Axiom of choice. This axiom is used Bourbaki's Theory of Sets. I think these weaker forms of the AoC are efforts to find a "Goldilocks zone" where you can do standard real analysis but don't have to be bothered by pesky apparent paradoxes like non-measurable sets. But I don't know if the resulting plethora of different axioms is more helpful than confusing. --RDBury (talk) 15:46, 1 August 2024 (UTC)[reply]