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Without the axiom of choice you get similarly bizarre results. For instance wothout choice there exists a surjection from the reals to a set of greater cardina

Without the axiom of choice you get similarly bizarre results. For instance wothout choice there exists a surjection from the reals to a set of greater cardinality.

BeetleB 9y ago
That's because the reals don't exist, which is Doron's point. We need the Axiom of Choice because we invented the reals.
- yequalsx 9y ago
  The existence of the reals is independent of the axiom of choice. The collection of mathematicians that don't believe infinite sets exist has maybe a few members. Constructive mathematicians believe the reals exists. They can be constructed.
  - GregBuchholz 9y ago
    >Constructive mathematicians believe the reals exists. I think when people mention the "reals" they are usually referring to the uncountable reals, which you can't construct, compute, name, or know: https://arxiv.org/abs/math/0404335 https://arxiv.org/abs/math/0404335 ...(maybe skip to chapter 5 to get to the meat of it). So constructivists don't believe in those types of numbers, and so get "choice" as a theorem instead of an axiom.
    - yequalsx 9y ago
      Constructive mathematicians definitely do believe the reals exist and are uncountable. Not all reals are computable or definable. The set of computable reals is countable. There are different varieties of constructive mathematics. A very large majority of constructive mathematicians believe the reals are uncountable. In intuitionistic mathematics the reals are uncountable. In some constructive versions of math you get weird things like there being an injection from R to N but there not being a bijection due to diagonalization. I think you are confusing computable/nameable with constructive as that term is used by most mathematicians. Note that Chaitan's Constant is not computable.
    - yequalsx 9y ago
      This link may be of interest to you: https://mathoverflow.net/a/30694 https://mathoverflow.net/a/30694
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.

Without the axiom of choice you get similarly bizarre results. For instance wothout choice there exists a surjection from the reals to a set of greater cardinality.

BeetleB 9y ago

That's because the reals don't exist, which is Doron's point. We need the Axiom of Choice because we invented the reals.

yequalsx 9y ago
The existence of the reals is independent of the axiom of choice. The collection of mathematicians that don't believe infinite sets exist has maybe a few members. Constructive mathematicians believe the reals exists. They can be constructed.
- GregBuchholz 9y ago
  >Constructive mathematicians believe the reals exists. I think when people mention the "reals" they are usually referring to the uncountable reals, which you can't construct, compute, name, or know: https://arxiv.org/abs/math/0404335 https://arxiv.org/abs/math/0404335 ...(maybe skip to chapter 5 to get to the meat of it). So constructivists don't believe in those types of numbers, and so get "choice" as a theorem instead of an axiom.
  - yequalsx 9y ago
    Constructive mathematicians definitely do believe the reals exist and are uncountable. Not all reals are computable or definable. The set of computable reals is countable. There are different varieties of constructive mathematics. A very large majority of constructive mathematicians believe the reals are uncountable. In intuitionistic mathematics the reals are uncountable. In some constructive versions of math you get weird things like there being an injection from R to N but there not being a bijection due to diagonalization. I think you are confusing computable/nameable with constructive as that term is used by most mathematicians. Note that Chaitan's Constant is not computable.
  - yequalsx 9y ago
    This link may be of interest to you: https://mathoverflow.net/a/30694 https://mathoverflow.net/a/30694

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.