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Nevertheless it feels like the point of Banach-Tarski is that it proves math went wrong somewhere. Evidently the sets it's talking about are not objects which a

by ajkjk 2y ago

Nevertheless it feels like the point of Banach-Tarski is that it proves math went wrong somewhere. Evidently the sets it's talking about are not objects which are interesting in reality.

moffkalast 2y ago
Math is not reality. Physics isn't even reality, they're just theoretical frameworks that are easy to work with and somewhat align with what we've observed. "All models are wrong, but some are useful" as they say. If you can prove something can or can't be done in math it doesn't mean shit, but it might end up being a useful guideline. Banach-Tarski assumes an infinite pointcloud (i.e. a mathematical sphere), which as you've realized, doesn't actually exist.
- pvillano 2y ago
  The best/worst part is that there are useful branches of mathematics which assume the Axiom of Choice is true AND there are useful branches of mathematics which assume the Axiom of Choice is false. That's one reason I take the view that mathematics does not exist: if math did exist, there would only be one set of axioms consistent with nature.
  - 9991 2y ago
    You conclude the problem is with math. I conclude the problem is with reality.
    - alkyon 2y ago
      In physical reality there is no infinity of anything* so Axiom of Choice is irrelevant for it. * Universe could be itself inifinite but locally everything is finite.
      - 9991 2y ago
        > locally everything is finite. You don't know that.
        ColinWright 2y ago
        >> Universe could be itself infinite but locally everything is finite. > You don't know that. I think they were saying that it could be that the Universe is infinite with everything still being locally finite. I do not think they were asserting that everything is locally finite, I think it was included in the "it could be" part. But certainly you're right that we don't know that.
  - alkyon 2y ago
    AC is an axiom (something assumed without proof) and not the result that would render mathematics itself inconsistent. Banach Tarski is not about AC, it is about existence of some pathological sets that have no measure. Edit: AC is generally assumed to be true because rejecting it leads to even stranger paradoxes than that of Banach-Tarski.
- ajkjk 2y ago
  Yes, all models are wrong, some are useful, and the model math is using is evidently not very good. Probably there is a better one out there. Physics isn't reality but it does use "closeness to reality" as a metric for quality of a theory, and that's by the metric by which Banach-Tarski is irritating.
  - lupire 2y ago
    The better (more realistic) model is constructive/intuitionist mathematics - ZF without C, or at least without Uncountable Choice. Naturally, you should be suspicious of a theory of mathematics that assumes the existence of "uncountable" things.
    - gjm11 2y ago
      You're conflating three separate things. 1. Accepting versus rejecting the axiom of choice. 2. Classical versus intuitionistic mathematics. Intuitionism goes much further than just rejecting the Axiom of Choice, and e.g. says that you haven't proved "p or q" until you have either proved p or proved q. It denies not just the axiom of choice but the "law of the excluded middle" which says that for any proposition p, either p is true or not p is true. 3. Accepting versus rejecting "large" sets -- in your case, you say we should be suspicious of anything that assumes the existence of uncountable things. (I think this is a bit unusual; there are finitists who deny that there are any infinite sets, and ultrafinitists who go further and say there aren't even arbitrarily big finite sets, but it's not so common to accept countable but not uncountable infinities.) I think intuitionism implies rejecting AC. I don't think any of the other possible implications between these three things holds; e.g., so far as I know most intuitionists have no particular problem with the existence of large infinite sets. (Actually, probably some theorems of the form "if there are no large sets, then the axiom of choice is true for boring reasons" are provable.)
  - isotypic 2y ago
    > and the model math is using is evidently not very good. There isn't really one "model of math". Plenty of people study alternative foundations of math or various axioms you can attach to ZF, etc. and its great that they do - choice, and lots of other set theoretic/foundational stuff, is very weird. But the reason mathematicians have largely settled on ZFC as a default is that it (and maybe nowadays + an inaccessible cardinal) also reduces some pathology and makes math more convenient. Let's list some things which are related to the axiom of choice: * The cartesian product of nonempty sets is nonempty (equivalent) * The reals can be partitioned into more parts than there are real numbers (consistent with the negation of choice) * All vector spaces have a basis (equivalent) * All commutative rings have a maximal ideal (equivalent) * All fields are contained in an algebraic closure (implied by choice) The first two, at least to me, are similar to Banach-Tarski in that they are things I would like to be true and false, respectively, and are not if we do not accept choice. The point here is that the weirdness of Banach-Tarski is as much related to the axiom of choice as it is to generally the fact that (uncountable) infinities are just very weird, and while introducing choice does introduce pathologies like Banach-Tarski, it also reduces some. The last three illustrate a more practical perspective. Suppose a universal decree, that math is no longer allowed to be done in ZFC, you could only use ZF, was imposed on mathematicians. This doesn't really change anything - all that mathematicians will do is, if they were perfectly fine with choice before, simply replace instances of "vector space" with "vector space with a basis", or "commutative ring" with "commutative ring with a maximal ideal", because the types of mathematical objects they care about are the ones with these desirable properties - they only use the axiom of choice because it ensures the general objects they work with also have those desirable properties. And mathematicians (often even the same ones as before, just wearing a different hat) who _do_ care about choice will continue studying those weird instances of the general object just as before. As a final point for the convenience of axiom of choice, there are lots of instances where a proof will use choice purely as a matter of convenience. Maybe with some technical set theory or a smarter argument its possible to completely eliminate the requirement of choice, or use a weaker, less-objectionable version of choice. Other times, while a general vector space having a basis requires choice, the vector spaces in your particular application have a basis regardless of your thoughts on the axiom of choice. But of course its much easier to simply assert "vector spaces have a basis" and introduce a (faux) dependence on the axiom of choice. For a concrete example of this last point, see https://mathoverflow.net/a/35772 https://mathoverflow.net/a/35772. While about inaccessible cardinals/"universes" rather than the axiom of choice, the principle I want to illustrate is the same. The proof of Fermat's last theorem, if you trace citations back, eventually depends on "universes". But the actual invocations of those theorems are applied to objects where that generality is not required. (Why not simply only have done the non-general scenario? Because "holds in general with additional axiom, holds in specificity without" is more knowledge than just "holds in specificity")
- lupire 2y ago
  Banach-Tarski assumes far more than the existence of a sphere. It assumes the existence of an uncountably infinitely complicated cut.
ajuc 2y ago
Would you say that about imaginary numbers too?
- andoando 2y ago
  Imaginary numbers are just 2D vectors. Its such a horrible name. The imaginary part isnt any more mysterious than the real part. Theyre just orthogonal properties.
  - ajuc 2y ago
    > Imaginary numbers are just 2D vectors. You meant complex numbers. I wrote "imaginary numbers" on purpose. > The imaginary part isnt any more mysterious than the real part. What's surprising is that imaginary numbers are required for QM. Not the 2d vector in one variable trick, actual imaginary numbers. See for example: https://www.nature.com/articles/s41586-021-04160-4 https://www.nature.com/articles/s41586-021-04160-4
    - lupire 2y ago
      I think that paper is saying that you have to use N independent 2D numbers, not 2N independent 1D numbers. Complex numbers are literally the same as 2D real numbers with the rotational symmetry constraint. https://www.reddit.com/r/Physics/comments/11ujjjd/quantum_theory_based_on_real_numbers_can_be/ https://www.reddit.com/r/Physics/comments/11ujjjd/quantum_th...
    - andoando 2y ago
      What are imaginary numbers if you remove its relation to the real numbers? Do they not become regular real numbers? How I am seeing it, real and imaginary numbers are both equivalent constructions with the same properties/construction (you can add/substract/multiply/divide them the same way), and its only in the context of using both (in the context of the complex numbers), that they can be differentiated.
  - fellowmartian 2y ago
    They’re not “just” 2D vectors, they have a bit more structure that makes them special, but sure there’s obviously nothing “imaginary” about them.

by ajkjk 2y ago

Nevertheless it feels like the point of Banach-Tarski is that it proves math went wrong somewhere. Evidently the sets it's talking about are not objects which are interesting in reality.

moffkalast 2y ago

Math is not reality. Physics isn't even reality, they're just theoretical frameworks that are easy to work with and somewhat align with what we've observed. "All models are wrong, but some are useful" as they say. If you can prove something can or can't be done in math it doesn't mean shit, but it might end up being a useful guideline. Banach-Tarski assumes an infinite pointcloud (i.e. a mathematical sphere), which as you've realized, doesn't actually exist.

pvillano 2y ago
The best/worst part is that there are useful branches of mathematics which assume the Axiom of Choice is true AND there are useful branches of mathematics which assume the Axiom of Choice is false. That's one reason I take the view that mathematics does not exist: if math did exist, there would only be one set of axioms consistent with nature.
- 9991 2y ago
  You conclude the problem is with math. I conclude the problem is with reality.
  - alkyon 2y ago
    In physical reality there is no infinity of anything* so Axiom of Choice is irrelevant for it. * Universe could be itself inifinite but locally everything is finite.
    - 9991 2y ago
      > locally everything is finite. You don't know that.
      - ColinWright 2y ago
        >> Universe could be itself infinite but locally everything is finite. > You don't know that. I think they were saying that it could be that the Universe is infinite with everything still being locally finite. I do not think they were asserting that everything is locally finite, I think it was included in the "it could be" part. But certainly you're right that we don't know that.
- alkyon 2y ago
  AC is an axiom (something assumed without proof) and not the result that would render mathematics itself inconsistent. Banach Tarski is not about AC, it is about existence of some pathological sets that have no measure. Edit: AC is generally assumed to be true because rejecting it leads to even stranger paradoxes than that of Banach-Tarski.
ajkjk 2y ago
Yes, all models are wrong, some are useful, and the model math is using is evidently not very good. Probably there is a better one out there. Physics isn't reality but it does use "closeness to reality" as a metric for quality of a theory, and that's by the metric by which Banach-Tarski is irritating.
- lupire 2y ago
  The better (more realistic) model is constructive/intuitionist mathematics - ZF without C, or at least without Uncountable Choice. Naturally, you should be suspicious of a theory of mathematics that assumes the existence of "uncountable" things.
  - gjm11 2y ago
    You're conflating three separate things. 1. Accepting versus rejecting the axiom of choice. 2. Classical versus intuitionistic mathematics. Intuitionism goes much further than just rejecting the Axiom of Choice, and e.g. says that you haven't proved "p or q" until you have either proved p or proved q. It denies not just the axiom of choice but the "law of the excluded middle" which says that for any proposition p, either p is true or not p is true. 3. Accepting versus rejecting "large" sets -- in your case, you say we should be suspicious of anything that assumes the existence of uncountable things. (I think this is a bit unusual; there are finitists who deny that there are any infinite sets, and ultrafinitists who go further and say there aren't even arbitrarily big finite sets, but it's not so common to accept countable but not uncountable infinities.) I think intuitionism implies rejecting AC. I don't think any of the other possible implications between these three things holds; e.g., so far as I know most intuitionists have no particular problem with the existence of large infinite sets. (Actually, probably some theorems of the form "if there are no large sets, then the axiom of choice is true for boring reasons" are provable.)
- isotypic 2y ago
  > and the model math is using is evidently not very good. There isn't really one "model of math". Plenty of people study alternative foundations of math or various axioms you can attach to ZF, etc. and its great that they do - choice, and lots of other set theoretic/foundational stuff, is very weird. But the reason mathematicians have largely settled on ZFC as a default is that it (and maybe nowadays + an inaccessible cardinal) also reduces some pathology and makes math more convenient. Let's list some things which are related to the axiom of choice: * The cartesian product of nonempty sets is nonempty (equivalent) * The reals can be partitioned into more parts than there are real numbers (consistent with the negation of choice) * All vector spaces have a basis (equivalent) * All commutative rings have a maximal ideal (equivalent) * All fields are contained in an algebraic closure (implied by choice) The first two, at least to me, are similar to Banach-Tarski in that they are things I would like to be true and false, respectively, and are not if we do not accept choice. The point here is that the weirdness of Banach-Tarski is as much related to the axiom of choice as it is to generally the fact that (uncountable) infinities are just very weird, and while introducing choice does introduce pathologies like Banach-Tarski, it also reduces some. The last three illustrate a more practical perspective. Suppose a universal decree, that math is no longer allowed to be done in ZFC, you could only use ZF, was imposed on mathematicians. This doesn't really change anything - all that mathematicians will do is, if they were perfectly fine with choice before, simply replace instances of "vector space" with "vector space with a basis", or "commutative ring" with "commutative ring with a maximal ideal", because the types of mathematical objects they care about are the ones with these desirable properties - they only use the axiom of choice because it ensures the general objects they work with also have those desirable properties. And mathematicians (often even the same ones as before, just wearing a different hat) who _do_ care about choice will continue studying those weird instances of the general object just as before. As a final point for the convenience of axiom of choice, there are lots of instances where a proof will use choice purely as a matter of convenience. Maybe with some technical set theory or a smarter argument its possible to completely eliminate the requirement of choice, or use a weaker, less-objectionable version of choice. Other times, while a general vector space having a basis requires choice, the vector spaces in your particular application have a basis regardless of your thoughts on the axiom of choice. But of course its much easier to simply assert "vector spaces have a basis" and introduce a (faux) dependence on the axiom of choice. For a concrete example of this last point, see https://mathoverflow.net/a/35772 https://mathoverflow.net/a/35772. While about inaccessible cardinals/"universes" rather than the axiom of choice, the principle I want to illustrate is the same. The proof of Fermat's last theorem, if you trace citations back, eventually depends on "universes". But the actual invocations of those theorems are applied to objects where that generality is not required. (Why not simply only have done the non-general scenario? Because "holds in general with additional axiom, holds in specificity without" is more knowledge than just "holds in specificity")
lupire 2y ago
Banach-Tarski assumes far more than the existence of a sphere. It assumes the existence of an uncountably infinitely complicated cut.

ajuc 2y ago

Would you say that about imaginary numbers too?

andoando 2y ago
Imaginary numbers are just 2D vectors. Its such a horrible name. The imaginary part isnt any more mysterious than the real part. Theyre just orthogonal properties.
- ajuc 2y ago
  > Imaginary numbers are just 2D vectors. You meant complex numbers. I wrote "imaginary numbers" on purpose. > The imaginary part isnt any more mysterious than the real part. What's surprising is that imaginary numbers are required for QM. Not the 2d vector in one variable trick, actual imaginary numbers. See for example: https://www.nature.com/articles/s41586-021-04160-4 https://www.nature.com/articles/s41586-021-04160-4
  - lupire 2y ago
    I think that paper is saying that you have to use N independent 2D numbers, not 2N independent 1D numbers. Complex numbers are literally the same as 2D real numbers with the rotational symmetry constraint. https://www.reddit.com/r/Physics/comments/11ujjjd/quantum_theory_based_on_real_numbers_can_be/ https://www.reddit.com/r/Physics/comments/11ujjjd/quantum_th...
  - andoando 2y ago
    What are imaginary numbers if you remove its relation to the real numbers? Do they not become regular real numbers? How I am seeing it, real and imaginary numbers are both equivalent constructions with the same properties/construction (you can add/substract/multiply/divide them the same way), and its only in the context of using both (in the context of the complex numbers), that they can be differentiated.
- fellowmartian 2y ago
  They’re not “just” 2D vectors, they have a bit more structure that makes them special, but sure there’s obviously nothing “imaginary” about them.