4 ms·

What do you mean? If ZF without choice shows that a statement P holds then certainly ZFC also proves the same statement. Do you mean that there is a consistent

by fmap 9y ago

What do you mean? If ZF without choice shows that a statement P holds then certainly ZFC also proves the same statement. Do you mean that there is a consistent extension of ZF with the statement "There exists a surjection from R to P(R)"?

yequalsx 9y ago
There is a theorem that states that either there is a nonmeasurable subset of the reals or that there is a surjection of the reals to a larger set. The theorem is by Sierpinski. This is way out of my area but I'm guessing ZF is not strong enough to show either all sets are measurable or the existence of a non measurable set. The common criticism of AC is Banach-Tarski. So if you don't agree with Banach-Tarski and want every subset of R to be measurable then you have to conclude an equally bizarre result. As an algebraist I accepted AC. Instead of constantly saying, "let V be a vector space over k with a basis" it's easier to just assume all vector spaces have a basis. I think analysts need AC more than other branches, The Intermediate Value Theorem isn't provable without AC. I think.
- fmap 9y ago
  What I found after Googling is that there is a theorem by Sierpinski which shows that if we make an additional assumption about the cardinality of [R]^\omega, which seems to be approximately the same as the boolean prime ideal axiom, then there is a non-measurable set. Thus the existence of non-measurable sets is implied by principles weaker than full choice, but that's not too surprising. The case of measure theory is particularly interesting, though, since you don't even need to change your underlying logic to get a better model. For instance, if you base your "measure theory" on valuations on locales instead of measures on sigma algebras then the theory itself becomes simpler and the Banach-Tarski "paradox" goes away. Briefly, in locale theory, your "measure" is defined on sublocales instead of subsets. While there are more sublocales than subsets, the condition for when two sublocales are disjoint is stronger. This is what breaks the Banach-Tarski construction. The orbit subsets used in Banach-Tarski still exist, but while they are disjoint as sets, they are not disjoint as sublocales and thus don't decompose the volume of the sphere.
  - yequalsx 9y ago
    I'm not familiar with locale theory. Thanks for the reference. I'm guessing there will be some non-intuitive results. My non-expert impression is that whatever one chooses in terms of logic and set theory there will be bizarre results when dealing with sets of cardinality of the continuum and not dealing with the continuum leaves out too much. Here is a reference to a mathiverflow comment. https://mathoverflow.net/a/22935 https://mathoverflow.net/a/22935

by fmap 9y ago

yequalsx 9y ago

There is a theorem that states that either there is a nonmeasurable subset of the reals or that there is a surjection of the reals to a larger set. The theorem is by Sierpinski. This is way out of my area but I'm guessing ZF is not strong enough to show either all sets are measurable or the existence of a non measurable set. The common criticism of AC is Banach-Tarski. So if you don't agree with Banach-Tarski and want every subset of R to be measurable then you have to conclude an equally bizarre result. As an algebraist I accepted AC. Instead of constantly saying, "let V be a vector space over k with a basis" it's easier to just assume all vector spaces have a basis. I think analysts need AC more than other branches, The Intermediate Value Theorem isn't provable without AC. I think.

fmap 9y ago
What I found after Googling is that there is a theorem by Sierpinski which shows that if we make an additional assumption about the cardinality of [R]^\omega, which seems to be approximately the same as the boolean prime ideal axiom, then there is a non-measurable set. Thus the existence of non-measurable sets is implied by principles weaker than full choice, but that's not too surprising. The case of measure theory is particularly interesting, though, since you don't even need to change your underlying logic to get a better model. For instance, if you base your "measure theory" on valuations on locales instead of measures on sigma algebras then the theory itself becomes simpler and the Banach-Tarski "paradox" goes away. Briefly, in locale theory, your "measure" is defined on sublocales instead of subsets. While there are more sublocales than subsets, the condition for when two sublocales are disjoint is stronger. This is what breaks the Banach-Tarski construction. The orbit subsets used in Banach-Tarski still exist, but while they are disjoint as sets, they are not disjoint as sublocales and thus don't decompose the volume of the sphere.
- yequalsx 9y ago
  I'm not familiar with locale theory. Thanks for the reference. I'm guessing there will be some non-intuitive results. My non-expert impression is that whatever one chooses in terms of logic and set theory there will be bizarre results when dealing with sets of cardinality of the continuum and not dealing with the continuum leaves out too much. Here is a reference to a mathiverflow comment. https://mathoverflow.net/a/22935 https://mathoverflow.net/a/22935