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This is not intuitive at all. I have a masters in math. Concrete numbers are irrelevant to me, and their explicit representation is even less relevant. Any num
by tomp 4y ago
This is not intuitive at all.
I have a masters in math. Concrete numbers are irrelevant to me, and their explicit representation is even less relevant. Any number can easily be described by x.
Btw this is just the standard reason for the existence of the Axiom of Choice. If there exists a procedure to describe which number to choose (i.e. an algorithm/formula), you don't need AC at all.
IMO the only "intuitive" anti-AC argument is, that you can get most of math with Axiom of Countable Choice, while you avoid paradoxes like Banach-Tarski.
- davrosthedalek 4y agoQuestion: While the standard construction of the set for Banach-Tarski requires AOC, is it proven that without AOC, Banach-Tarksi isn't true?
- tomp 4y agoNot sure, but if you replace AC with a weaker axiom, Axiom of Dependent Choice, you can construct a model in which all subsets of reals are measurable (called the Solovay model), which precludes the Banach-Tarski paradox (which requires non-measurable sets). Could BT be constructed without AC but with other way of ensuring there exist non-measurable sets? No idea.
- throwaway81523 4y agoBy "without AC" I think you mean "with the negation of AC". It turns out that BT+ZF is equivalent to ZF + a weaker version of AC called the Ultrafilter Lemma, iirc. There is more info in the Wikipedia article about BT. Also, the Hahn-Banach theorem (plus ZF) proves BT.
- Kranar 4y agoDepends what you mean by without AOC. If you just mean whether Banach-Tarski can be proven in ZF the answer is no. But you also don't need the full power of AOC to prove Banach-Tarski.
- lisper 4y ago> If there exists a procedure to describe which number to choose (i.e. an algorithm/formula), you don't need AC at all. The problem here is not that there does not exist a procedure to describe which number to choose, the problem is that there does not exist any way to describe which number you chose. Those are not the same thing.
- mckeed 4y agoWhen you say "Any number can easily be described by x" aren't you already using Choice? And doesn't "let x be one of those numbers from the set of numbers that cannot be described with finite symbols" seem a bit like a paradox?
- NotYourLawyer 4y agoHow do you feel about this example? https://www.solipsys.co.uk/new/APointAgainstTheAxiomOfChoice.html https://www.solipsys.co.uk/new/APointAgainstTheAxiomOfChoice...
- pdpi 4y ago> you can get most of math with Axiom of Countable Choice Wait, I think you just made me realise that my intuition for AoC is wrong and actually matches AoCC instead, and that I'd never stopped to consider that the collection of sets you're choosing from could be uncountable. If that's correct, that explains so much.
- throwaway81523 4y agoI don't think you can get the Hahn-Banach theorem with just countable choice. So you have to re-do a whole lot of functional analysis that depends on it.