8 ms·

The Point of the Banach-Tarski Theorem

145 points by ColinWright 4y ago

CJefferson 4y ago
I always feel part of the confusion with Banach-Tarski is that lots of words don't use their "natural definitions", which makes the proof more surprising. People (not this article) often talking about "cutting" a sphere, which is really misleading. This result is, in many ways, quite similar to the idea I can "cut" the integers into the odd integers and even integers (but with many more fine details). This is still a nice article, which explains the actual result well.
- hgsgm 4y ago
  > (but with many more fine details). Uncountably many more fine details.
azeemba 4y ago
As the author points out that Banach-Tarski theorem is an example of hard-to-accept result that comes out of the easy-to-accept axiom of choice. There is a popular quote that related to this: > The axiom of choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma? From https://en.wikipedia.org/wiki/Axiom_of_choice https://en.wikipedia.org/wiki/Axiom_of_choice Axiom of choice, the well-ordering principle and Zorn's lemma are equivalent statements (any one proves the other two). But each has a very different "believability" feel to it.
- jiggawatts 4y ago
  The Axiom of choice has never felt completely self-evident to me. E.g.: what if you have sets where the elements are non-computable? How do you "choose" objects that cannot even be named? Something like: "the set of all programs that cannot be proven to halt" and the like can be used to create pathological sets where the set itself obviously exists, but you cannot name any of the members. Actually, an ever better example is: "The set of reals that are not the solution to any equation that can be written with a finite number of symbols." -- an infinite set that has no nameable members!
  - H8crilA 4y ago
    I think your second set doesn't exist (exactly like the set of all sets doesn't exist) under the axiom of choice, for I can use the axiom of choice to select an element, hence giving it a name :). I took some liberty with the allowable names, of course. You still have a point, though.
  - soVeryTired 4y ago
    Those are just subsets of R though (though I think your last example would take more work to make rigorous, if it's even possible). The weirdness of the axiom of choice only really comes through when you consider bigger and bigger collections of sets. For example: - sets indexed by the natural numbers: S_1, S_2, ... It seems totally reasonable that you should be able to make a new set by picking something from the first set, something from the second set, etc - sets indexed by continuous time (i.e. real numbers). Here it's a bit less 'obvious'. If I have sets S_t for _every_ time t > 0, can I really make choices 'fast' enough? What if the sets are so unstructured that I'm forced to stop and look at each set in turn to make my choice? - sets indexed by the power set of the real numbers. If you weren't convinced that I'd struggle to pick elements of S_t for all t > 0, what if I had to make a choice for every _possible combination_ of real numbers, infinite or otherwise? I feel like the last example demonstrates how powerful the full axiom of choice actually is. NB - I'm a dilettante rather than an actual logician, so there may be mathematical inaccuracies here.
    - lmm 4y ago
      > Those are just subsets of R though (though I think your last example would take more work to make rigorous, if it's even possible). You could form e.g. the non-algebraic reals, which is almost all of the reals. > sets indexed by the power set of the real numbers. If you weren't convinced that I'd struggle to pick elements of S_t for all t > 0, what if I had to make a choice for every _possible combination_ of real numbers, infinite or otherwise? To the extent to which you can form that indexed collection of sets in the first place, surely you can form a similarly indexed collection of elements of them the same way. How can you say you've formed a non-empty set if you can't select an element of it? If we permit ourselves to form this indexed collection "lazily", surely we can do the choice "lazily" as well. (Just my intuition about these things)
whatshisface 4y ago
Here is a potentially daft question that I nonetheless would appreciate if someone could answer. Is it possible to deny the axiom of choice for the purposes of measures while accepting it for vector spaces? I am wondering if you could say, "there are two kinds of sets, ones equipped with a choice function and ones without it, and measurable sets are of the latter kind."
- joerichey 4y ago
  If you accept that _all_ vector spaces have a (Hamel) basis, you can then prove the Axiom of Choice: http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf This means if you want to deny the Axiom in some cases, you will also have to allow for the existence of vector spaces without a basis.
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- contravariant 4y ago
  Not in a meaningful way I think. I mean you could weaken it to 'all finite vector spaces have a basis', but I think regular induction is enough to prove that, you don't need the axiom of choice.
- QuesnayJr 4y ago
  You probably could, though you would probably have to give up or weaken other axioms as well. You would need to prevent there being any bijections between the good sets and the bad sets, because you then you could use the bijection to define a choice function on the bad set. What set theorists usually do is just distinguish between when you use the axiom of choice and when you don't. There's one super-subtle point, though: even if you ban using the axiom of choice, you can still construct sets where it is independent of the axioms of set theory whether they are measurable. You have to add "these sets are measurable" as an axiom. The simplest way to do it is to consider a class of sets where you know you didn't use the axiom of choice, and then declare that these sets are measurable. One common choice is the class of "projective sets". Usually set theorists impose a stronger property called determinacy, so they add to set theory the Axiom of Projective Determinacy. Then you can say "If I stick to the good sets -- projective sets -- nothing bad happens. Arbitrary choice functions take me out of the projective sets, where bad things can happen."
moloch-hai 4y ago
I asked a mathematician about what a wacky conclusion it is. He said that whenever you allow infinity, you get results like that. It relies on uncountably-infinite division of an object, which corresponds to no real-world experience anywhere in the universe. Real objects have, you know, atoms. We use real numbers a lot, but we are careful never to rely on their more extreme properties anywhere it would matter. In practice, in fact, we use floating-point numbers, not reals, when doing actual calculations, and use numerical analysis to stay well clear of nonsensical results. If you tried to rely on BT in a real calculation, you would find a lot of NaNs and Infs.
- Aicy 4y ago
  In the real world we have atoms, but atoms are disturbances in the wave function of the universe, which may not be discrete. We know atoms pop in and out of existence in matter anti-matter pairs. Maybe the universe is performing Banach Tarski under the hood, to make something from nothing? > we use floating-point numbers, not reals, when doing actual calculations I don't believe only calculations made by computers to be "actual calculations". Humans were calculating thousands of years before computers were invented, using integer or real numbers (e.g pi).
  - moloch-hai 4y ago
    No human or computer has ever used an actual real, transcendental number in any computation. It would take infinitely long.
    - ttctciyf 4y ago
      I know what you mean, but Euler's identity is a counterexample to what you literally said, isn't it?
      - Someone 4y ago
        I don’t think so. Programs that use numbers approximate e and π. It’s only programs that manipulate symbols that can use their exact values.
karmakaze 4y ago
I'm glad to have found this post. I discovered the Banach-Tarski theorem via Vsauce[0]. It was interesting but I couldn't get the significance of it. It either didn't seem like an unexpected result or too esoteric to appreciate. There's phrasing in the post that could be misunderstood (later clarified) but can leave unclarity from assumed understanding of the earlier description. > In R3, given a solid ball B of radius R, it is possible to partition B into finitely many pieces such that those pieces can be reassembled to form two solid balls B1 and B2 each of radius R. "finitely many pieces" could be confusing because though we may be talking about 6 'pieces' those pieces have an uncountable infinity of radial line slices. It's also missing the rigid motion part which is key. The part that didn't seem surprising (and is considered trivial) was in handling uncountable infinities of things. The 'number' of points on a line [0, 1) is the same as the number on a line [0, 2) so I wouldn't be surprised to map points from [0, 1) to [0, 2) filling the latter without 'gaps'. Similarly for areas. But what the theorem is saying is that this kind of mapping doesn't work in R1 nor R2 but does work in R3 (with rigid motions). The part that makes B-T surprising is that the extra volume/ball can be constructed with rigid motions of those uncountably infinite sets. This is where it seems beyond me to appreciate: that one or two balls have the same uncountably number of radial line slices is considered trivial, and mapping using rigid motions is surprising. For example, if you do the same rearrangement but instead of sets of radial lines, consider the set of points at the surface of those radial lines, we're basically working in R2. The reason why it can't be done is because if we instead of being on the surface of a sphere we're on a plane, then those same motions aren't distance preserving. Thinking geometrically doesn't seem weird: an extra dimension let's you do something you can't in lower ones. I think the algebraic description that the post makes may be clearer to appreciate the difference, and I'll be giving it another read and more thought. [0] https://www.youtube.com/watch?v=s86-Z-CbaHA https://www.youtube.com/watch?v=s86-Z-CbaHA
- gerdesj 4y ago
  It is a remarkable theorem but you can't take a snooker ball and turn it into two snooker balls without being Paul Daniels (UK magician). This is an excellent example of language going astray and confusing mathematic rigour with some sort of "reality". You can loosely model a "sphere in R3" with a snooker ball in errr the universe thingie which is probably R3ish or perhaps R3T1 or whatevs. Besides someone has spilt a whole pint of Guinness on the table and the balls are chipped. Even if we get our match balls out that are not chipped, and are jolly shiney, they are still subject to things like Mr Planck's constant and the fine structure of matter. The point is quite literally in the title - don't balls it up!
soVeryTired 4y ago
Mathoverflow has some good discussion of Banach-Tarski [0]. The top-rated posts argue that - The 'problem' with B-T may be related to our notion of "space" (i.e. point-set topology) rather than any issues with the axiom of choice - Most practical uses of the axiom of choice could get by with the axiom of countable choice, under which B-T doesn't hold - Mathematics need not directly model the 'real' world or match our physical intuition [0] https://mathoverflow.net/questions/260057/axiom-of-choice-banach-tarski-and-reality https://mathoverflow.net/questions/260057/axiom-of-choice-ba...
- dkural 4y ago
  Thank you for the summary and the link. I'd also add that our physical intuition also need not match the "real world", it's a model "good enough" for evolutionary purposes that fails in myriad ways.
- msla 4y ago
  > Mathematics need not directly model the 'real' world or match our physical intuition There's so many other ways to demonstrate this. For example, it's possible to map the integers to the rational numbers one-to-one, which is a good definition for "as many as" when dealing with infinite sets, even though there are infinitely many rational numbers between any two rational numbers. Moreover, there are as many integer multiples of a trillion as there are integers total. This is to do with set cardinality: https://en.wikipedia.org/wiki/Cardinality https://en.wikipedia.org/wiki/Cardinality ... which again comes back to the axiom of choice: > If the axiom of choice holds, the law of trichotomy holds for cardinality. Thus we can make the following definitions: > * Any set X with cardinality less than that of the natural numbers [...] is said to be a finite set. > * Any set X that has the same cardinality as the set of the natural numbers [...] is said to be a countably infinite set.[10] > * Any set X with cardinality greater than that of the natural numbers [...] is said to be uncountable. (... with some mathematical notation that doesn't work here elided)
yellowapple 4y ago
"So for those of you who don't know the result, here it is in simple, non-technical terms:" proceeds to immediately use a character that can't even be copy/pasted due to needing MathJax to render it, refuses to elaborate further Thankfully, it ain't hard to find the Banach-Tarski theorem in actual simple, non-technical terms, so for those wondering what in tarnation that character is: it just means three-dimensional space (and is usually rendered as ℝ³ instead of 𝕽³, at least so Wikipedia claims).
- quchen 4y ago
  Had a laugh about that too. It’s sometimes hard to tell what’s »non-technical« when you’ve spent a lot of time on something, and even then what it means to be »technical« depends a lot on the audience. Non-technical here means knows basic math but is not familiar with measure theory; it does not mean _anyone_. I like that they added my favorite math joke in their (linked at the very top) limited audience jokes section [1] (bottom left, starts with »Two mathematicians are in a bar«), which is also a about what basic math knowledge is. [1]: https://www.solipsys.co.uk/new/LimitedAudienceJokes.html https://www.solipsys.co.uk/new/LimitedAudienceJokes.html
  - yellowapple 4y ago
    > Non-technical here means knows basic math but is not familiar with measure theory; it does not mean _anyone_. I guess "knows basic math" is relative; I tend to draw the "basic" line at "it's required for a high school diploma or GED", but I suppose I can blame my country's education system for that bar being much lower than what mathematicians seem to assume :)
- smcl 4y ago
  I did get a chuckle at that, like the running joke about how a monad is “just a monoid in the category of endofunctors”. To be fair, you could delete the “R^3” part and the explanation can still be understood.
  - yellowapple 4y ago
podiki 4y ago
After a semester of a senior math seminar with one student always bringing up "but you assumed the axiom of choice!" (mostly non-seriously), the gradstudent teaching it decided for his, and last, presentation to show this theorem. It was pretty wild as I had never heard of it before (being a physics-math person). Very cool. (The seminar was on knot theory but I had more fun with Banach-Tarski).
ur-whale 4y ago
So, Banach-Tarski says you can split a sphere of volume S into a finite number of pieces and reassemble the pieces into two spheres of volume S. Is the proof constructive? As in: does the proof actually show how to build the pieces? If yes, is the boundary of the pieces of measurable surface? Can the pieces be rendered in 3D? Or is it just another one of those "proofs" where if the set of pieces doesn't exist we land on a contraction?
- josephcsible 4y ago
  https://youtu.be/s86-Z-CbaHA?t=673 https://youtu.be/s86-Z-CbaHA?t=673 nicely visualizes it. It's an infinite number of infinitely thin filaments from the center of the sphere going outward in every direction.
  - hgsgm 4y ago
    Not just infinite, uncountably infinite!
- ColinWright 4y ago
  * The proof uses the Axiom of Choice, so no, it's not constructive; * The boundary of the pieces has to be unmeasurable, in that sense it's similar to Vitali's set[0]; * No, the pieces are more-or-less a "fog" of points. * I don't know why you have "proofs" in quotation marks, and I don't know what you mean by the second half of the sentence (even if I replace "contraction" with "contradiction"). I do feel like a careful reading, or perhaps re-reading, would let you ask more specific questions that we can help you with. [0] https://en.wikipedia.org/wiki/Vitali_set https://en.wikipedia.org/wiki/Vitali_set
  - ur-whale 4y ago
    > I don't know why you have "proofs" in quotation marks Because proof by contradiction is viewed by an entire category of mathematicians as "wrong" (please make sure to note the use of quotes in case you missed them). Read all about it here: https://en.wikipedia.org/wiki/Constructivism_(philosophy_of_mathematics) https://en.wikipedia.org/wiki/Constructivism_(philosophy_of_... In a very tangible sense, they're correct. "Proof" by contradiction, tertium non datur, proof by the absurd, whatever you want to call it, more often than not entirely fails to produce working examples, which makes the "proof" fare less useful than the one that explains how to build an actual exemplar.
sshine 4y ago
Th student revue at University of Copenhagen made a music video ten years ago: https://www.youtube.com/watch?v=uFvokQUHh08 https://www.youtube.com/watch?v=uFvokQUHh08 Instead of saying "Barbara Streisand", they say "Banach-Tarski" and a number of oranges double:
mikewarot 4y ago
Give every point in the Sphere B a room in the infinity hotel. If the room number is odd, it belongs to sub-sphere B1, otherwise sub-sphere B2. Odd things happen when you divide by infinity.
- harry8 4y ago
  This has always been my struggle with all of this "infinity" maths. Infinity/Infinity is anything you like and many things you don't and these things always seem to boil down to dividing by infinity. Visit each room in the hotel in turn, each one for half the total time spent going to and visiting the last. Without explanation of why we can divide infinity by infinity when we want to and not get total garbage. Banach tarski, infinity/infinity = 2 because the numerator is exactly equaly to (infinity 2) then you cancel the infinity from top and bottom. Also (infinity 3) so clearly 2=3 and this is useful. infinity also equals infinity * infinity so 1 = infinity. As you say odd things happen when you divide by infinity and mostly they don't seem to be helpful things. They seem to be wholly invalid things. But not here? I'm sure that's not it and I'm totally missing the point, but that point is being skipped over and handwaved away with a sneer and a muttering of "mathematical maturity" rather often.
  - schoen 4y ago
    What I got from this post is that the weirdness here comes in, not so much from the Hilbert's Hotel phenomenon about cardinalities and sets that can be put into correspondence with their own proper subsets, but from looking more deeply at something relatively familiar and something that we ordinarily use to tame infinities: volume. Even though there are infinitely many real numbers in [0,1], we have the idea that the unit cube [0,1]³ should have a finite volume of exactly 1, or the unit sphere { (x,y,z) | √(x²+y²+z²)≤1 } should have a finite volume of exactly 4π/3. Or indeed the unit line segment [0,1] should have a finite length of exactly 1, even though it contains infinitely many points. This stuff has felt totally normal and appropriate in mathematics ever since Euclid: Euclid would probably agree that you can't count how many points are in a line segment, but still endorses talking about lengths of line segments (or at least ratios of lengths of different line segments). While it feels like we know how to work with volumes, and that they're comfortably finite and well-behaved, things like Banach-Tarski suggest that if we want to have every set of points have a well-defined volume in any given number of dimensions, we're actually going to run into bad trouble. But the article suggests that there are several ways to avoid this trouble, including just saying that some weird geometric objects don't have a defined n-dimensional volume. Instead, maybe only some "nice" sets should have one?
rocqua 4y ago
The article states that all known versions of these 'paradoxes' arise from the axiom of choice. But I believe it isn't known yet whether denying the axiom of choice categorically prevents these paradoxes and allows 'perfect' measures to exist. Anyone care to clarify?
- pfortuny 4y ago
  Yes, see the math overflow question quoted above. Not choice gives you measures where all sets are measurable.
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hackandthink 4y ago
Banach-Tarski from a constructivist/intuitionistic point of view: "So it really seems that when we switch to constructive theory, it actually does something even better, it eliminates this pathology altogether, making balls behave more sensibly?" https://math.stackexchange.com/questions/175675/intuitionistic-banach-tarski-paradox https://math.stackexchange.com/questions/175675/intuitionist... I like theories without crazy theorems. And I like theories with crazy theorems. Edit: the link below from soVeryTired to Andrej Bauers post is better
SaintRomuald 4y ago
I never understood this example used to explain it, vsauce made a video on the banach tarski theorem. You make an infinite list of numbers between 0 and 1 chosen at random. Apparently you can make a new number that was never seen in the list before if you pick a digit from each number in the list and add one to it. Say the list has numbers 0.36285728.. 0.95825597.. 0.47264112.. .. I can make a new number by taking the 3 from the first number the 5 from the second and the 2 from the third num and so on. 0.463.. I never understood
- sukilot 4y ago
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- ColinWright 4y ago
  You have misunderstood, or misremembered. You don't take "3" from the first, you consider three, but choose something that's not "3", so the number you are constructing differs from that first number. Then in the second place you don't take "5", you choose something that's not "5", so the number you are constructing differs from that second number. And so on. So every time you have a list of real numbers, it cannot contain all real numbers ... you can always construct many, many, many numbers that are not in your list. But I'm not sure how this is related to what we've been talking about. If you understand what I've said here and are still confused, maybe you can be a bit more specific. If you have not understood what I have said here, perhaps you can ask more specific questions.
sukilot 4y ago
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145 points by ColinWright 4y ago

I always feel part of the confusion with Banach-Tarski is that lots of words don't use their "natural definitions", which makes the proof more surprising. People (not this article) often talking about "cutting" a sphere, which is really misleading. This result is, in many ways, quite similar to the idea I can "cut" the integers into the odd integers and even integers (but with many more fine details). This is still a nice article, which explains the actual result well.

hgsgm 4y ago
> (but with many more fine details). Uncountably many more fine details.

azeemba 4y ago

As the author points out that Banach-Tarski theorem is an example of hard-to-accept result that comes out of the easy-to-accept axiom of choice. There is a popular quote that related to this: > The axiom of choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma? From https://en.wikipedia.org/wiki/Axiom_of_choice https://en.wikipedia.org/wiki/Axiom_of_choice Axiom of choice, the well-ordering principle and Zorn's lemma are equivalent statements (any one proves the other two). But each has a very different "believability" feel to it.

jiggawatts 4y ago
The Axiom of choice has never felt completely self-evident to me. E.g.: what if you have sets where the elements are non-computable? How do you "choose" objects that cannot even be named? Something like: "the set of all programs that cannot be proven to halt" and the like can be used to create pathological sets where the set itself obviously exists, but you cannot name any of the members. Actually, an ever better example is: "The set of reals that are not the solution to any equation that can be written with a finite number of symbols." -- an infinite set that has no nameable members!
- H8crilA 4y ago
  I think your second set doesn't exist (exactly like the set of all sets doesn't exist) under the axiom of choice, for I can use the axiom of choice to select an element, hence giving it a name :). I took some liberty with the allowable names, of course. You still have a point, though.
- soVeryTired 4y ago
  Those are just subsets of R though (though I think your last example would take more work to make rigorous, if it's even possible). The weirdness of the axiom of choice only really comes through when you consider bigger and bigger collections of sets. For example: - sets indexed by the natural numbers: S_1, S_2, ... It seems totally reasonable that you should be able to make a new set by picking something from the first set, something from the second set, etc - sets indexed by continuous time (i.e. real numbers). Here it's a bit less 'obvious'. If I have sets S_t for _every_ time t > 0, can I really make choices 'fast' enough? What if the sets are so unstructured that I'm forced to stop and look at each set in turn to make my choice? - sets indexed by the power set of the real numbers. If you weren't convinced that I'd struggle to pick elements of S_t for all t > 0, what if I had to make a choice for every _possible combination_ of real numbers, infinite or otherwise? I feel like the last example demonstrates how powerful the full axiom of choice actually is. NB - I'm a dilettante rather than an actual logician, so there may be mathematical inaccuracies here.
  - lmm 4y ago
    > Those are just subsets of R though (though I think your last example would take more work to make rigorous, if it's even possible). You could form e.g. the non-algebraic reals, which is almost all of the reals. > sets indexed by the power set of the real numbers. If you weren't convinced that I'd struggle to pick elements of S_t for all t > 0, what if I had to make a choice for every _possible combination_ of real numbers, infinite or otherwise? To the extent to which you can form that indexed collection of sets in the first place, surely you can form a similarly indexed collection of elements of them the same way. How can you say you've formed a non-empty set if you can't select an element of it? If we permit ourselves to form this indexed collection "lazily", surely we can do the choice "lazily" as well. (Just my intuition about these things)

whatshisface 4y ago

Here is a potentially daft question that I nonetheless would appreciate if someone could answer. Is it possible to deny the axiom of choice for the purposes of measures while accepting it for vector spaces? I am wondering if you could say, "there are two kinds of sets, ones equipped with a choice function and ones without it, and measurable sets are of the latter kind."

joerichey 4y ago
If you accept that _all_ vector spaces have a (Hamel) basis, you can then prove the Axiom of Choice: http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf This means if you want to deny the Axiom in some cases, you will also have to allow for the existence of vector spaces without a basis.
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contravariant 4y ago
Not in a meaningful way I think. I mean you could weaken it to 'all finite vector spaces have a basis', but I think regular induction is enough to prove that, you don't need the axiom of choice.
QuesnayJr 4y ago
You probably could, though you would probably have to give up or weaken other axioms as well. You would need to prevent there being any bijections between the good sets and the bad sets, because you then you could use the bijection to define a choice function on the bad set. What set theorists usually do is just distinguish between when you use the axiom of choice and when you don't. There's one super-subtle point, though: even if you ban using the axiom of choice, you can still construct sets where it is independent of the axioms of set theory whether they are measurable. You have to add "these sets are measurable" as an axiom. The simplest way to do it is to consider a class of sets where you know you didn't use the axiom of choice, and then declare that these sets are measurable. One common choice is the class of "projective sets". Usually set theorists impose a stronger property called determinacy, so they add to set theory the Axiom of Projective Determinacy. Then you can say "If I stick to the good sets -- projective sets -- nothing bad happens. Arbitrary choice functions take me out of the projective sets, where bad things can happen."

moloch-hai 4y ago

I asked a mathematician about what a wacky conclusion it is. He said that whenever you allow infinity, you get results like that. It relies on uncountably-infinite division of an object, which corresponds to no real-world experience anywhere in the universe. Real objects have, you know, atoms. We use real numbers a lot, but we are careful never to rely on their more extreme properties anywhere it would matter. In practice, in fact, we use floating-point numbers, not reals, when doing actual calculations, and use numerical analysis to stay well clear of nonsensical results. If you tried to rely on BT in a real calculation, you would find a lot of NaNs and Infs.

Aicy 4y ago
In the real world we have atoms, but atoms are disturbances in the wave function of the universe, which may not be discrete. We know atoms pop in and out of existence in matter anti-matter pairs. Maybe the universe is performing Banach Tarski under the hood, to make something from nothing? > we use floating-point numbers, not reals, when doing actual calculations I don't believe only calculations made by computers to be "actual calculations". Humans were calculating thousands of years before computers were invented, using integer or real numbers (e.g pi).
- moloch-hai 4y ago
  No human or computer has ever used an actual real, transcendental number in any computation. It would take infinitely long.
  - ttctciyf 4y ago
    I know what you mean, but Euler's identity is a counterexample to what you literally said, isn't it?
    - Someone 4y ago
      I don’t think so. Programs that use numbers approximate e and π. It’s only programs that manipulate symbols that can use their exact values.

karmakaze 4y ago

I'm glad to have found this post. I discovered the Banach-Tarski theorem via Vsauce[0]. It was interesting but I couldn't get the significance of it. It either didn't seem like an unexpected result or too esoteric to appreciate. There's phrasing in the post that could be misunderstood (later clarified) but can leave unclarity from assumed understanding of the earlier description. > In R3, given a solid ball B of radius R, it is possible to partition B into finitely many pieces such that those pieces can be reassembled to form two solid balls B1 and B2 each of radius R. "finitely many pieces" could be confusing because though we may be talking about 6 'pieces' those pieces have an uncountable infinity of radial line slices. It's also missing the rigid motion part which is key. The part that didn't seem surprising (and is considered trivial) was in handling uncountable infinities of things. The 'number' of points on a line [0, 1) is the same as the number on a line [0, 2) so I wouldn't be surprised to map points from [0, 1) to [0, 2) filling the latter without 'gaps'. Similarly for areas. But what the theorem is saying is that this kind of mapping doesn't work in R1 nor R2 but does work in R3 (with rigid motions). The part that makes B-T surprising is that the extra volume/ball can be constructed with rigid motions of those uncountably infinite sets. This is where it seems beyond me to appreciate: that one or two balls have the same uncountably number of radial line slices is considered trivial, and mapping using rigid motions is surprising. For example, if you do the same rearrangement but instead of sets of radial lines, consider the set of points at the surface of those radial lines, we're basically working in R2. The reason why it can't be done is because if we instead of being on the surface of a sphere we're on a plane, then those same motions aren't distance preserving. Thinking geometrically doesn't seem weird: an extra dimension let's you do something you can't in lower ones. I think the algebraic description that the post makes may be clearer to appreciate the difference, and I'll be giving it another read and more thought. [0] https://www.youtube.com/watch?v=s86-Z-CbaHA https://www.youtube.com/watch?v=s86-Z-CbaHA

gerdesj 4y ago
It is a remarkable theorem but you can't take a snooker ball and turn it into two snooker balls without being Paul Daniels (UK magician). This is an excellent example of language going astray and confusing mathematic rigour with some sort of "reality". You can loosely model a "sphere in R3" with a snooker ball in errr the universe thingie which is probably R3ish or perhaps R3T1 or whatevs. Besides someone has spilt a whole pint of Guinness on the table and the balls are chipped. Even if we get our match balls out that are not chipped, and are jolly shiney, they are still subject to things like Mr Planck's constant and the fine structure of matter. The point is quite literally in the title - don't balls it up!

soVeryTired 4y ago

Mathoverflow has some good discussion of Banach-Tarski [0]. The top-rated posts argue that - The 'problem' with B-T may be related to our notion of "space" (i.e. point-set topology) rather than any issues with the axiom of choice - Most practical uses of the axiom of choice could get by with the axiom of countable choice, under which B-T doesn't hold - Mathematics need not directly model the 'real' world or match our physical intuition [0] https://mathoverflow.net/questions/260057/axiom-of-choice-banach-tarski-and-reality https://mathoverflow.net/questions/260057/axiom-of-choice-ba...

dkural 4y ago
Thank you for the summary and the link. I'd also add that our physical intuition also need not match the "real world", it's a model "good enough" for evolutionary purposes that fails in myriad ways.
msla 4y ago
> Mathematics need not directly model the 'real' world or match our physical intuition There's so many other ways to demonstrate this. For example, it's possible to map the integers to the rational numbers one-to-one, which is a good definition for "as many as" when dealing with infinite sets, even though there are infinitely many rational numbers between any two rational numbers. Moreover, there are as many integer multiples of a trillion as there are integers total. This is to do with set cardinality: https://en.wikipedia.org/wiki/Cardinality https://en.wikipedia.org/wiki/Cardinality ... which again comes back to the axiom of choice: > If the axiom of choice holds, the law of trichotomy holds for cardinality. Thus we can make the following definitions: > * Any set X with cardinality less than that of the natural numbers [...] is said to be a finite set. > * Any set X that has the same cardinality as the set of the natural numbers [...] is said to be a countably infinite set.[10] > * Any set X with cardinality greater than that of the natural numbers [...] is said to be uncountable. (... with some mathematical notation that doesn't work here elided)

yellowapple 4y ago

"So for those of you who don't know the result, here it is in simple, non-technical terms:" proceeds to immediately use a character that can't even be copy/pasted due to needing MathJax to render it, refuses to elaborate further Thankfully, it ain't hard to find the Banach-Tarski theorem in actual simple, non-technical terms, so for those wondering what in tarnation that character is: it just means three-dimensional space (and is usually rendered as ℝ³ instead of 𝕽³, at least so Wikipedia claims).

quchen 4y ago
Had a laugh about that too. It’s sometimes hard to tell what’s »non-technical« when you’ve spent a lot of time on something, and even then what it means to be »technical« depends a lot on the audience. Non-technical here means knows basic math but is not familiar with measure theory; it does not mean _anyone_. I like that they added my favorite math joke in their (linked at the very top) limited audience jokes section [1] (bottom left, starts with »Two mathematicians are in a bar«), which is also a about what basic math knowledge is. [1]: https://www.solipsys.co.uk/new/LimitedAudienceJokes.html https://www.solipsys.co.uk/new/LimitedAudienceJokes.html
- yellowapple 4y ago
  > Non-technical here means knows basic math but is not familiar with measure theory; it does not mean _anyone_. I guess "knows basic math" is relative; I tend to draw the "basic" line at "it's required for a high school diploma or GED", but I suppose I can blame my country's education system for that bar being much lower than what mathematicians seem to assume :)
smcl 4y ago
I did get a chuckle at that, like the running joke about how a monad is “just a monoid in the category of endofunctors”. To be fair, you could delete the “R^3” part and the explanation can still be understood.
- yellowapple 4y ago

podiki 4y ago

After a semester of a senior math seminar with one student always bringing up "but you assumed the axiom of choice!" (mostly non-seriously), the gradstudent teaching it decided for his, and last, presentation to show this theorem. It was pretty wild as I had never heard of it before (being a physics-math person). Very cool. (The seminar was on knot theory but I had more fun with Banach-Tarski).

ur-whale 4y ago

So, Banach-Tarski says you can split a sphere of volume S into a finite number of pieces and reassemble the pieces into two spheres of volume S. Is the proof constructive? As in: does the proof actually show how to build the pieces? If yes, is the boundary of the pieces of measurable surface? Can the pieces be rendered in 3D? Or is it just another one of those "proofs" where if the set of pieces doesn't exist we land on a contraction?

josephcsible 4y ago
https://youtu.be/s86-Z-CbaHA?t=673 https://youtu.be/s86-Z-CbaHA?t=673 nicely visualizes it. It's an infinite number of infinitely thin filaments from the center of the sphere going outward in every direction.
- hgsgm 4y ago
  Not just infinite, uncountably infinite!
ColinWright 4y ago
* The proof uses the Axiom of Choice, so no, it's not constructive; * The boundary of the pieces has to be unmeasurable, in that sense it's similar to Vitali's set[0]; * No, the pieces are more-or-less a "fog" of points. * I don't know why you have "proofs" in quotation marks, and I don't know what you mean by the second half of the sentence (even if I replace "contraction" with "contradiction"). I do feel like a careful reading, or perhaps re-reading, would let you ask more specific questions that we can help you with. [0] https://en.wikipedia.org/wiki/Vitali_set https://en.wikipedia.org/wiki/Vitali_set
- ur-whale 4y ago
  > I don't know why you have "proofs" in quotation marks Because proof by contradiction is viewed by an entire category of mathematicians as "wrong" (please make sure to note the use of quotes in case you missed them). Read all about it here: https://en.wikipedia.org/wiki/Constructivism_(philosophy_of_mathematics) https://en.wikipedia.org/wiki/Constructivism_(philosophy_of_... In a very tangible sense, they're correct. "Proof" by contradiction, tertium non datur, proof by the absurd, whatever you want to call it, more often than not entirely fails to produce working examples, which makes the "proof" fare less useful than the one that explains how to build an actual exemplar.

sshine 4y ago

Th student revue at University of Copenhagen made a music video ten years ago: https://www.youtube.com/watch?v=uFvokQUHh08 https://www.youtube.com/watch?v=uFvokQUHh08 Instead of saying "Barbara Streisand", they say "Banach-Tarski" and a number of oranges double:

mikewarot 4y ago

Give every point in the Sphere B a room in the infinity hotel. If the room number is odd, it belongs to sub-sphere B1, otherwise sub-sphere B2. Odd things happen when you divide by infinity.

harry8 4y ago
This has always been my struggle with all of this "infinity" maths. Infinity/Infinity is anything you like and many things you don't and these things always seem to boil down to dividing by infinity. Visit each room in the hotel in turn, each one for half the total time spent going to and visiting the last. Without explanation of why we can divide infinity by infinity when we want to and not get total garbage. Banach tarski, infinity/infinity = 2 because the numerator is exactly equaly to (infinity 2) then you cancel the infinity from top and bottom. Also (infinity 3) so clearly 2=3 and this is useful. infinity also equals infinity * infinity so 1 = infinity. As you say odd things happen when you divide by infinity and mostly they don't seem to be helpful things. They seem to be wholly invalid things. But not here? I'm sure that's not it and I'm totally missing the point, but that point is being skipped over and handwaved away with a sneer and a muttering of "mathematical maturity" rather often.
- schoen 4y ago
  What I got from this post is that the weirdness here comes in, not so much from the Hilbert's Hotel phenomenon about cardinalities and sets that can be put into correspondence with their own proper subsets, but from looking more deeply at something relatively familiar and something that we ordinarily use to tame infinities: volume. Even though there are infinitely many real numbers in [0,1], we have the idea that the unit cube [0,1]³ should have a finite volume of exactly 1, or the unit sphere { (x,y,z) | √(x²+y²+z²)≤1 } should have a finite volume of exactly 4π/3. Or indeed the unit line segment [0,1] should have a finite length of exactly 1, even though it contains infinitely many points. This stuff has felt totally normal and appropriate in mathematics ever since Euclid: Euclid would probably agree that you can't count how many points are in a line segment, but still endorses talking about lengths of line segments (or at least ratios of lengths of different line segments). While it feels like we know how to work with volumes, and that they're comfortably finite and well-behaved, things like Banach-Tarski suggest that if we want to have every set of points have a well-defined volume in any given number of dimensions, we're actually going to run into bad trouble. But the article suggests that there are several ways to avoid this trouble, including just saying that some weird geometric objects don't have a defined n-dimensional volume. Instead, maybe only some "nice" sets should have one?

rocqua 4y ago

The article states that all known versions of these 'paradoxes' arise from the axiom of choice. But I believe it isn't known yet whether denying the axiom of choice categorically prevents these paradoxes and allows 'perfect' measures to exist. Anyone care to clarify?

pfortuny 4y ago
Yes, see the math overflow question quoted above. Not choice gives you measures where all sets are measurable.

deleted 4y ago

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hackandthink 4y ago

Banach-Tarski from a constructivist/intuitionistic point of view: "So it really seems that when we switch to constructive theory, it actually does something even better, it eliminates this pathology altogether, making balls behave more sensibly?" https://math.stackexchange.com/questions/175675/intuitionistic-banach-tarski-paradox https://math.stackexchange.com/questions/175675/intuitionist... I like theories without crazy theorems. And I like theories with crazy theorems. Edit: the link below from soVeryTired to Andrej Bauers post is better

SaintRomuald 4y ago

I never understood this example used to explain it, vsauce made a video on the banach tarski theorem. You make an infinite list of numbers between 0 and 1 chosen at random. Apparently you can make a new number that was never seen in the list before if you pick a digit from each number in the list and add one to it. Say the list has numbers 0.36285728.. 0.95825597.. 0.47264112.. .. I can make a new number by taking the 3 from the first number the 5 from the second and the 2 from the third num and so on. 0.463.. I never understood

sukilot 4y ago
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ColinWright 4y ago
You have misunderstood, or misremembered. You don't take "3" from the first, you consider three, but choose something that's not "3", so the number you are constructing differs from that first number. Then in the second place you don't take "5", you choose something that's not "5", so the number you are constructing differs from that second number. And so on. So every time you have a list of real numbers, it cannot contain all real numbers ... you can always construct many, many, many numbers that are not in your list. But I'm not sure how this is related to what we've been talking about. If you understand what I've said here and are still confused, maybe you can be a bit more specific. If you have not understood what I have said here, perhaps you can ask more specific questions.

sukilot 4y ago

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