3 ms·

Here is a potentially daft question that I nonetheless would appreciate if someone could answer. Is it possible to deny the axiom of choice for the purposes of

Here is a potentially daft question that I nonetheless would appreciate if someone could answer. Is it possible to deny the axiom of choice for the purposes of measures while accepting it for vector spaces? I am wondering if you could say, "there are two kinds of sets, ones equipped with a choice function and ones without it, and measurable sets are of the latter kind."

joerichey 4y ago
If you accept that _all_ vector spaces have a (Hamel) basis, you can then prove the Axiom of Choice: http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf This means if you want to deny the Axiom in some cases, you will also have to allow for the existence of vector spaces without a basis.
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contravariant 4y ago
Not in a meaningful way I think. I mean you could weaken it to 'all finite vector spaces have a basis', but I think regular induction is enough to prove that, you don't need the axiom of choice.
QuesnayJr 4y ago
You probably could, though you would probably have to give up or weaken other axioms as well. You would need to prevent there being any bijections between the good sets and the bad sets, because you then you could use the bijection to define a choice function on the bad set. What set theorists usually do is just distinguish between when you use the axiom of choice and when you don't. There's one super-subtle point, though: even if you ban using the axiom of choice, you can still construct sets where it is independent of the axioms of set theory whether they are measurable. You have to add "these sets are measurable" as an axiom. The simplest way to do it is to consider a class of sets where you know you didn't use the axiom of choice, and then declare that these sets are measurable. One common choice is the class of "projective sets". Usually set theorists impose a stronger property called determinacy, so they add to set theory the Axiom of Projective Determinacy. Then you can say "If I stick to the good sets -- projective sets -- nothing bad happens. Arbitrary choice functions take me out of the projective sets, where bad things can happen."

joerichey 4y ago

If you accept that _all_ vector spaces have a (Hamel) basis, you can then prove the Axiom of Choice: http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf This means if you want to deny the Axiom in some cases, you will also have to allow for the existence of vector spaces without a basis.

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contravariant 4y ago

Not in a meaningful way I think. I mean you could weaken it to 'all finite vector spaces have a basis', but I think regular induction is enough to prove that, you don't need the axiom of choice.

QuesnayJr 4y ago

You probably could, though you would probably have to give up or weaken other axioms as well. You would need to prevent there being any bijections between the good sets and the bad sets, because you then you could use the bijection to define a choice function on the bad set. What set theorists usually do is just distinguish between when you use the axiom of choice and when you don't. There's one super-subtle point, though: even if you ban using the axiom of choice, you can still construct sets where it is independent of the axioms of set theory whether they are measurable. You have to add "these sets are measurable" as an axiom. The simplest way to do it is to consider a class of sets where you know you didn't use the axiom of choice, and then declare that these sets are measurable. One common choice is the class of "projective sets". Usually set theorists impose a stronger property called determinacy, so they add to set theory the Axiom of Projective Determinacy. Then you can say "If I stick to the good sets -- projective sets -- nothing bad happens. Arbitrary choice functions take me out of the projective sets, where bad things can happen."