3 ms·

If one doesn’t accept the Axiom of Choice but uses instead Dependent Choice the paradox no longer holds. Is it the case in this situation that the 4 desirable

If one doesn’t accept the Axiom of Choice but uses instead Dependent Choice the paradox no longer holds. Is it the case in this situation that the 4 desirable properties hold?

CaptainNegative 2y ago
Assuming the usual consistency caveats, the paradox is no longer a theorem of ZF+DC, but its complement isn't either. So in that case the analogue to the fifth postulate is even stronger, as there are both models in which you get the counterintuitive results of unmeasurable sets and those in which you don't, and the axioms are not strong enough to distinguish the two.
- skhunted 2y ago
  In ZF+DC is it true that measures satisfy the desirable properties mentioned by Colin? I think the sticking point is isometry invariance. Are there measures in ZF+DC of R^3 that are finitely (countably?) additive and isometry invariant?
fmap 2y ago
You don't get new theorems if you remove assumptions. Rather, you get the ability to add different assumptions. The Banach-Tarski paradox shows that classical set theory makes the wrong assumptions to intrinsically model measure theory and probability. There are other systems which don't suffer from this paradox and hence don't need all the machinery of sigma algebras and measurable sets. I wish there was a good accessible book/article/blog post about this, but as is you'd have to Google point-free topology or topos of probability (there are several).
- deleted 2y ago
  [deleted]
- skhunted 2y ago
  I think the following is a valid question. At least I hope it isn’t completely stupid. Is there a known set theory of the form ZF+(something) which relatively consistent with ZFC in which additive, isometry invariant measures exist? I guess what you are saying is that the only known, reasonable way around this is the topos notion you mentioned.

3 ms·

If one doesn’t accept the Axiom of Choice but uses instead Dependent Choice the paradox no longer holds. Is it the case in this situation that the 4 desirable

by skhunted 2y ago

If one doesn’t accept the Axiom of Choice but uses instead Dependent Choice the paradox no longer holds. Is it the case in this situation that the 4 desirable properties hold?

CaptainNegative 2y ago
Assuming the usual consistency caveats, the paradox is no longer a theorem of ZF+DC, but its complement isn't either. So in that case the analogue to the fifth postulate is even stronger, as there are both models in which you get the counterintuitive results of unmeasurable sets and those in which you don't, and the axioms are not strong enough to distinguish the two.
- skhunted 2y ago
  In ZF+DC is it true that measures satisfy the desirable properties mentioned by Colin? I think the sticking point is isometry invariance. Are there measures in ZF+DC of R^3 that are finitely (countably?) additive and isometry invariant?
fmap 2y ago
You don't get new theorems if you remove assumptions. Rather, you get the ability to add different assumptions. The Banach-Tarski paradox shows that classical set theory makes the wrong assumptions to intrinsically model measure theory and probability. There are other systems which don't suffer from this paradox and hence don't need all the machinery of sigma algebras and measurable sets. I wish there was a good accessible book/article/blog post about this, but as is you'd have to Google point-free topology or topos of probability (there are several).
- deleted 2y ago
  [deleted]
- skhunted 2y ago
  I think the following is a valid question. At least I hope it isn’t completely stupid. Is there a known set theory of the form ZF+(something) which relatively consistent with ZFC in which additive, isometry invariant measures exist? I guess what you are saying is that the only known, reasonable way around this is the topos notion you mentioned.