3 ms·
No one not working on foundations has any problem with axiom of choice. It has weird implications but so what? Banach Tarski just means physical shapes aren't a
by lanstin 10mo ago
No one not working on foundations has any problem with axiom of choice. It has weird implications but so what? Banach Tarski just means physical shapes aren't arbitrarily subdividable.
- black_knight 10mo agoBanach Tarski is not about physical shapes. The thing is, the foundations negating axiom of choice are just as consistent as those with. So, how do mathematicians justify their faith in AC?
- fpoling 10mo agoAC makes things much easier as it allows to play God powers. Negating AC is not significantly different from constructing mathematics that avoids AC (no assumption about validity of AC). And that makes things way harder with longer proofs and only in sub-cases of classical theorems.
- black_knight 10mo agoSimply assuming the negation of AC is boring, as negations often are. But there are stronger statements, implying the negation of AC which might be as useful. I think for instance one could assume all subsets of the plane to be measurable. Seems convenient to me. Same with law of the excluded middle. Tossing it out we can assume all functions are computable and all total functions in the real are continuous. Seems nice and convenient too!
- jesuslop 10mo agoMy 2 cents is they do justify it by the interest of the consequences, as Tychonoff or Nullstellensatz. I wouldn't call that faith: Best practices is to state Tychonoff as "AC implies Tychonoff" and that last is logically valid. Sometimes the "AC implies..." is missing, buried in the proof or used unawaredly or predates ZFC, and is a bad thing. But very ofen one now see asterisks on theorems needing it.