Talk:Banach–Tarski paradox

message: Can this construction (perhaps the one using five pieces) be shown in a picture? Can it be animated on a computer screen (in simulated 3D)?
I find pictures much more intuitive than symbolic manipulation. David 16:10 Sep 17, 2002 (UTC)
:All Banach-Tarski constructions involve non-measurable pieces, so I don't think a useful picture (or animation) would be possible. --Zundark 16:46 Sep 17, 2002 (UTC)
What about the series of pictures in Scientific American magazine some years ago that showed how a ball can be sliced up and rearranged to become something else? (I forget what it was.) I believe that example involved non-measurable pieces as well. I think non-measurable pieces can be visualized, just not realized in nature because they may be infinitely thin or whatever. In any case, the pictures were interesting. David 17:05 Sep 17, 2002 (UTC)
A piece which is infinitely thin has Lebesgue measure 0. Non-measurable pieces are much worse than this, and cannot even be explicitly described.
I still maintain that a useful picture of a Banach-Tarski dissection is not possible, especially as it's impossible to even specify such a dissection (rather than merely prove one exists).
I can't comment on the Scientific American pictures, as I haven't seen them.
--Zundark 18:27 Sep 17, 2002 (UTC)
The paradoxical decomposition of the free group in two generators, which underlies the proof, could maybe be visualized by depicting its (infinite) Cayley graph and showing how it consists of four pieces that look just like the whole graph. AxelBoldt 18:41 Sep 17, 2002 (UTC)
:There's YouTube vidio that purports to explain this (The Banach–Tarski Paradox), that looks like it uses a "ball of lines" from the thumbnail pic.  I haven't watched yet (I don't binge-*watch* - I bing-*download* and then never watch the things I've downloaded :)  Is that what it does? Jimw338 (talk) 15:03, 25 August 2016 (UTC)
replies: text:  The animation in the video is not useful for the understanding of the rotations that lead to the doubling. It moves and just scales parts of the image of the caley graph. This could have been done with different geometrical shapes, explaining nothing.  — Preceding unsigned comment added by 178.203.109.161 (talk) 20:49, 26 August 2016 (UTC)
thread_title: Pictures

message: The following text has been copied from Talk:Banach Tarski Paradoxical Decomposition:
How about a full name for Hausdorff so it can be linked?
Is this "doubling the interval" thingie related to the fact that on the Real line there are the same number of points in any interval of any length?
Or am I simply showing my ignorance? Seems we need an article on infinity.
--Buz Cory
---Hey Buz bro I thought the same. It seems to me that this depends on fractal similarity for this to work.psic88 00:13, 28 May 2017 (UTC)
thread_title:  Hausdorff, and doubling the interval 

message: Can you please clarify (in the article) the following basic points:
* First, the article should explain what it means by saying that the two balls are of the "same size" as the original, since it implies later in the article that "size" is not meant in the "usual" sense of measure theory.
* Second, it should clearly distinguish the theorem from the "ordinary" fact that any infinite set (e.g. two spheres) can always be identified bijectively with a similarly infinite subset (e.g. one sphere) of itself.
*Third, there seems to be in a contradiction: in the introduction, it talks about a "solid ball" (which sounds like a sphere-bounded volume) whereas the proof talks about S2 (which usually denotes a spherical surface).
Steven G. Johnson 04:07, 22 Mar 2004 (UTC)
thread_title: Clarifications requested

message: As I understand the definitions of S(a) in the section about paradox decomposition, we mean
S(a) = \{af | f \in F_2\}.
Now note that there exist strings in F_2 which start with "aa..." which means that by aS(a^{-1}) we would get the whole free group back and not only the part without S(a^{-1}) as indicated by the picture. To point this out:
aS(a^{-1}) = \{ax | x \in S(a^{-1})\} = \{aa^{-1}x | x \in F_2\} = F_2
--74.236.150.135
:If the first letter is a^{-1}, then the second letter can't be a (because we're talking about words in reduced form). So aS(a^{-1}) doesn't contain any words starting with a. Note that the picture shows aS(a^{-1}) as the complement of S(a), not the complement of S(a^{-1}). --Zundark 10:38, 27 November 2006 (UTC)
thread_title:  S(a) 

message: I removed the last part from the proof, it is a sketch, not a prrof and I think such details should not be covered. Tosha 01:49, 5 May 2004 (UTC)
thread_title: end of proof

message: I came across an interesting anagram of "BANACH TARSKI".
It's "BANACH TARSKI BANACH TARSKI". —Ashley Y 10:45, 2004 Jul 9 (UTC)
:LOL! Loisel 00:29, 1 May 2007 (UTC)
thread_title:  Curiosity 

message: The usual argument against the idea that the BTP genuinely undermines the plausibility of the axiom of choice, is that the axiom of choice allows one to construct non-measurable subsets, which it is wrong to regard as "pieces" of the original ball: instead they interleave with each other to an infinitesimally fine degree, allowing a trick rather similar to Russell's Hotel to be carried out.  What's puzzling is the intrusion of a paradox of the infinite into what at first glance appeared to be a statement of geometry.
I'm not applying an edit directly, because this issue has ramifications elsewhere, and I haven';t time to properly formulate the text right now.  Changes are needed:
* cut it up into finitely many pieces -- very misleading description
* it is a paradox only in the sense of being counter-intuitive. Because its proof prominently uses the axiom of choice, this counter-intuitive conclusion has been presented as an argument against adoption of that axiom. -- needs counterargument here
* Changes needed in AofC page ---- Charles Stewart 20:07, 1 Sep 2004 (UTC)
replies: text: I agree, but it was better, it is all result of the second edit of ArnoldReinhold, so I have reverted it and added all later changes. Tosha 04:06, 2 Sep 2004 (UTC)

text: I think I've maybe not made my complaint clear: the article begins with an description of the TBP that is couched in intuitive but contentious terms: people who say the resolution of the paradox is that our intuitions about cutting up solids and applying spatial transformations only applies to measurable connected subsets (as indeed I do) will object to the way the topic is framed; the reversion hasn't changed much.  I'm not going to have much time for wikiing in the next ten days, but I plan on applying some changes then. ---- Charles Stewart 09:07, 2 Sep 2004 (UTC)
thread_title: BTP & arguments against axiom of choice

message: I can provide an illustration of Step 1 of the proof sketch, namely the paradoxical decomposition of F_2.  I envision something like the picture at free group, but with the sets S(a^{-1}), aS(a^{-1}) and S(a) marked and labelled.  Is there interest?  --Dbenbenn 01:35, 6 Dec 2004 (UTC)
replies: text: Done!  Is it clear, or can it be improved?  --Dbenbenn 04:56, 13 Dec 2004 (UTC)
thread_title:  Picture? 

message: Anyone want this picture on the page?
*Looks good. I'll put it in. Eric119 23:34, 5 Feb 2005 (UTC)
thread_title:  A simple picture 

message: I remember reading somewhere about a similar result, the name of which I can't remember. I don't know if the two are related theoretically, but they remind me of each other. The other result says that a set of points exists in euclidean space such that the projections of the set onto different planes can produce any set of 2 dimensional images you want. Does anyone know the name of this, or has anyone even heard of it at all? It seems like a link from this article to one on this other result might be useful. --Monguin61 10:00, 10 December 2005 (UTC)
replies: text: If anyone else is interested, the idea I was remembering was the digital sundial. Wikipedia's entry on sundials has an external link to the patent of an actual digital sundial, and as of now they are available for purchase from www.digitalsundial.com. The design of the physical device was the result of the work of one K. Falconer, and the idea is described, presumably in more detail, in his book on fractal geometry. --Monguin61 01:23, 15 December 2005 (UTC)
thread_title:  Similar Result 

message: * JA: You know I really hate being so WikiPiki, but there's supposedly a good reason to use ndashes instead of hyphens in names that refer to multiple persons rather than multiple ancestors of one person, so Banach-Tarski paradox should be Banach–Tarski paradox, and Hausdorff-Banach-Tarski paradox should be Hausdorff–Banach–Tarski paradox, and so on.  I would move the article ¼-with myself, but there's already a redirect from Banach–Tarski paradox to Banach-Tarski paradox, so it takes an admin-assisted deletion of Banach-Tarski paradox to do that.  Are you beginning to understand how the same content can get itself redistributed across two identical contents now?  Jon Awbrey 16:02, 9 March 2006 (UTC)
replies: text: I went ahead and moved it (no admin assistance required). You can't move an article onto another real article, but you can move article A onto article B when B is a redirect to A, unless B has some other history. Not sure what the rule is if B has at some point been a redirect to C. --Trovatore 17:13, 9 March 2006 (UTC)
thread_title:  On hyphenated hyper-2-tuples 

message: In a recient article in the Journal of Symbolic Logic, Trevor M. Wilson titled "A continuous movement version of the Banach—Tarski paradox: A solution to de Groot's Problem", it was proved that the dissaembly and reassembly of the balls can be performed by continuously and rigidly moving the pieces without the pieces ever intersecting at any point in time. Should this fact be mentioned and referenced? --Ramsey2006 22:46, 13 October 2006 (UTC)
replies: text: The pieces themselves are still infinitely convoluted and cannot actually be constructed physically (atoms are of a finite size, after all). So this still doesn't get anywhere near doing the decomposition "physically". But the fact that they can be translated, rotated, and reassembled without ever intersecting may be worth the mention.—Tetracube 06:01, 15 October 2006 (UTC)

text: I've changed the intro sentence to say "finitely many infinitely convoluted non-overlapping point sets", rather than "finitely many pieces" -- I think the core of the paradox is that the term "pieces" is misleading, since the normal common-sense intuitions regarding volume and solidity no longer apply to point sets of the type required. Although this wording is slightly awkward, I think it's appropriate in this case, because it avoids the far greater confusion that the use of the simple but misleading word "pieces" would generate, and the meaning is fully explained a couple of paragraphs below. -- Gigacephalus 08:25, 13 November 2007 (UTC)
replies: text: I like the use of "point sets", at least in conjunction with the rest of the edits you did.  I may break the first sentence into two in order to allow people to get a bit futher into the article before taking on much math.Warren Dew 21:37, 13 November 2007 (UTC)
thread_title:  Doing the decomposition physically 

message: This is a fantastic article, with quite accessible proof of the Banach–Tarski theorem about doubling the ball! Unfortunately, there are some inaccuracies and OR statements scattered around. For example, after reading Banach and Tarski's paper, I didn't get the impression that they intended it to somehow undercut the axiom of choice, as the text had claimed (hence I've removed that sentence). What they say is that
: Le rôle que joue cet axiome dans nos raisonnements nous semble mériter l'attention
and go on remarking how for two key results of their paper, the proof of the first uses the axiom of choice in a much weaker way then the proof of the second. 
I also regret that a clear explanation of the difference between one and two dimensional Euclidean spaces (where a paradoxical decomposition of this type is impossible) and the higher dimensional cases, related to amenability of the Euclidean group in low dimensions and non-amenability in high dimensions, is missing. While it is more relevant for the general theory of paradoxical decompositions, it seems prudent to have at least one section on this in the present article. Arcfrk 02:54, 3 September 2007 (UTC)
replies: text: I've made substantial revisions, in particular, adding references to amenability in a few places. Arcfrk 09:50, 3 September 2007 (UTC)
thread_title:  Improving the text 

message: I have a feeling like the Banach-Tarski paradox is something like the continuum equivalent to Hilbert's Hotel, where it's possible to put additional guests into a hotel that already has all rooms occupied. The hotel creates something (space for new guests) from seemingly nothing by exploiting its infinite nature, which seems counter-intuitive since such exploits don't work in the physical (finite) world - just like the Banach-Tarski thing. Can someone with a deeper understanding of how the proof works elaborate on this? wr 87.139.81.19 (talk) 12:45, 23 November 2007 (UTC)
replies: text: Not much there. Hilbert's Hotel requires no skill, this requires ... skill. Charles Matthews (talk) 15:53, 23 November 2007 (UTC)
thread_title:  Banach-Tarski and Hilbert's Hotel 

message: The article currently says
:In 1991, Janusz Pawlikowski proved that the Banach-Tarski paradox follows from ZF plus the Hahn-Banach theorem.  The Hahn-Banach theorem doesn't rely on the full axiom of choice but can be proven using a weaker version of AC called the ultrafilter lemma.  So Pawlikowski proved that the set theory needed to prove the Banach-Tarski paradox, while stronger than ZF, is weaker than full ZFC.
This phrasing, while it doesn't actually say so, kind of makes it sound as though it wasn't known before 1991 that full AC is not needed to prove Banach–Tarski. That is surely not the case. It's obvious from even a high-level description of the proof that the key use of AC is to choose elements from equivalence classes of points in R3 -- that is, essentially reals -- which means it follows from the existence of a wellordering of the reals. And it must have been known since the sixties that it's consistent with ZF that the reals can be wellordered but some larger set (say, the powerset of the reals) cannot.
Any suggestions on how to rephrase, while still getting the valid part of the point across?  One possibility would be to find a reference from earlier that specifically mentions that BT is weaker than full AC -- does anyone have one? --Trovatore (talk) 20:44, 20 December 2007 (UTC)
replies: text: I see what you mean, I went overboard slightly, sorry.  A good place to look for cites of earlier work is probably Pawlikowski's paper itself--does anyone have access?  As mentioned in the edit summary I got the cite from the Hahn-Banach talk page (plus Polish wikipedia), though I googled a little more info about the result before making that edit.  Btw what inspired this is I've been wondering for a while whether BT follows from ACA0, which is sort of plausible because Brown and Simpson proved that H-B for separable spaces (which would include R3, I'd expect, but I haven't examined the Brown&Simpson result) follows from WKL0 which I think is even weaker.  ACA0 is basically the system of Weyl's "Das Kontinuum" according to Feferman.  Warning: what follows after this is total OR and possibly nonsense.  But basically these systems are about the minimum needed for doing functional analysis and therefore for doing quantum mechanics.  What I'm getting at is that maybe it's hard to axiomatize physics in a way that doesn't lead to BT.  So instead of being a weird artifact of set theory, BT becomes in some sense a theorem of physics, a somewhat disturbing notion. 75.62.4.229 (talk) 12:48, 21 December 2007 (UTC)
replies: text: I found and added a pdf link to Pawlikowski's paper (which is just one page long) and a related one.  Still not sure how to describe the result's significance.  75.62.4.229 (talk) 21:55, 22 December 2007 (UTC)
thread_title:  exposition of Pawlikowski result could be misleading 

message: The article had claims about the lack of existence of an explicit construction. The proof of the Banach-Tarski decomposition begins by establishing a free action of F2 on the unit sphere, and then asks for a set containing exactly one point from each orbit of the group action. Although last step requires some form of choice, it would still be considered "explicit" by many mathematicians I have met; they would consider the entire proof an explicit construction, although not a canonical one.  While it would be possible to make those claims precise by referring to the definability of the decomposition, it isn't accurate to simply say that the construction is nonexplicit, because there is no generally accepted meaning of an explicit construction in mathematics. I rephrased a few sentences to remove the issue. Also, it isn't particularly accurate to talk about 'algorithms' when discussing constructions in set theory; it confuses set-theoretic constructions with computable constructions. — Carl (CBM · talk) 15:54, 14 January 2008 (UTC)
replies: text: These fairly charged statements had been introduced into the article just a few hours earlier, and frankly, I don't quite like most of the changes. Maybe we should just revert to the earlier version? Arcfrk (talk) 20:44, 14 January 2008 (UTC)
replies: text:  I hope the words are not charged, but neutral. The article before was biased towards the point of view that R is well-orderable. The BT construction is not at all explicit, because it requires a choice of an element from uncountably many sets. There is no algorithm to decide which points are in the sets, nor is there any procedure which allows the sets to be generated by any process other than transfinite induction to the continuum. The construction fails in a model where all subsets of R are measurable, and such models are exactly the same as far as computations or explicitly definable sets are concerned. I wanted to make the article neutral towards different set theory models, especially with regard to models which are probabilistically intuitive, meaning ones where you can consistently talk about choosing real numbers at random. Every time somebody talks about choosing a real numbers at random, they are implicitly imagining a universe where every subset of R has Lebesgue measure.Likebox (talk) 05:56, 15 January 2008 (UTC)
replies: text:  ("Charged" may not be the best term to use, but it appears that if you eliminated any bias, you've reintroduced a different form of bias and probably in larger quantities.) I am not sure that I get your point concerning algorithmic definability of the BT decomposition (which is not at all the same as "explicitness"), but then I don't have a PhD in Set Theory. The distinctions will likely be lost on the vast majority of the readers. Unfortunately, your edits removed a more plain word discussion of the meaning of the paradox and replaced them with quite technical and fairly polemical statements. The lead is not a good place for that! Arcfrk (talk) 06:15, 15 January 2008 (UTC)
replies: text:  Sorry, I might not have done a very good job. Perhaps "algorithmic" is better than explict? Just add the language that you feel is best, I was being bold, and perhaps I tilted the article the other way.Likebox (talk) 07:40, 15 January 2008 (UTC)
replies: text: I think generally that it's not too controversial that there's no explicit example, but the problem is that the statement itself has no precise agreed meaning. The problem with "algorithmic" is a different one. In most formulations all computable functions from the reals to the reals are continuous, so that there isn't a computable partition of the ball into five nonempty pieces at all, never mind whether you can rearrange them paradoxically. --Trovatore (talk) 07:45, 15 January 2008 (UTC)
replies: text:  You are right, algorithmic is ridiculous. Maybe the right phrase is "not constructible by joining and intersecting regions"? All I wanted the reader to get is that the sections are scatterings of points, which are constructed by a transfinite induction, not visualizable regions defined by any normal geometric operations.Likebox (talk) 08:01, 15 January 2008 (UTC)
replies: text: Well, it's certainly true that you can't do it with any countable collection of anything that could intuitively be described as "knife cuts" (that's a weak consequence of the fact that the pieces can't be Borel sets), and it would be worthwhile to get that point across. But it's tricky to think of any wording that conveys that intuitive idea, and still actually means something mathematically. --Trovatore (talk) 19:29, 15 January 2008 (UTC)
replies: text:  You're right. I give up. The only ways I can think of to phrase it are convoluted.Likebox (talk) 20:51, 15 January 2008 (UTC)
replies: text:  Ok, one last try, then I give up.Likebox (talk) 21:20, 15 January 2008 (UTC)
thread_title:  no general meaning of explicit construction  

message: I tagged this article with "Refimprove" to call for improved referencing.  More use of in-line citations (footnotes) would be appropriate.  For example, the first theorem stated is implied to be quote from the 1924 paper that appears in the bibliography below, but it would be appropriate to include a footnote at the location of the quote and to cite the article and its page number. doncram (talk) 22:11, 11 March 2008 (UTC)
replies: text: No, that would not be appropriate. A careful exposition of all material appearing in this article is contained in the book of Wagon in the bibliography. The reference to Banach–Tarski paper (a primary source) is included here both for its historical value and because the full paper is freely available. It is not helpful to cite random pages, and I find it most ironic that an article with an abundance of well chosen sources has been tagged as "insufficiently referenced". Arcfrk (talk) 23:09, 11 March 2008 (UTC)

text: That isn't a direct quote - note the lack of quotation marks. It's simply a statement of the theorem, set off from the main text. — Carl (CBM · talk) 23:28, 11 March 2008 (UTC)
replies: text: I'm sorry if the tag was inappropriate for this article.  I only visited because CBM has been engaging in a discussion elsewhere with me, with a degree of bullying that I was beginning to experience as a personal attack, over the fact that I had chosen to put copied text into quotation marks.  He was essentially taking the position there that full referencing of quotations is wrong.  I wondered what articles CBM works on, and looked at this one that CBM has contributed to.  Reading the article it seemed ambiguous to me, but perhaps I projected too much about CBM in assessing that this article was poorly referenced.  sincerely, doncram (talk) 17:01, 12 March 2008 (UTC)
thread_title:  Refimprove tag 

message: I have to agree with the anon here. The theorem is an assertion about R3, not about legumes and stars. Whether the physical universe, or its spacetime, is faithfully modeled/described by the real numbers, is an open question (and it's quite plausible that it always will be an open question).  I appreciate the desire to bring in some intuition (or in this case counter-intuition) but it should be done more carefully.  --Trovatore (talk) 19:04, 26 July 2008 (UTC)
replies: text: The pea and the Sun is a common analogy used in describing the paradox (just as frequent as talking about "hairy balls" when discussing indices of vector fields on a sphere), and we would be remiss in not including it (e.g. Google Scholar search).  I do it think it is misplaced however.  I would be much happier with it being in the very beginning.  After the more formal explanation has started, I don't think it adds much to all of a sudden bring up peas and Suns.  --C S (talk) 23:15, 26 July 2008 (UTC)

text: Ok, I cleaned it up along the lines of what I said.  I think the first paragraph is pretty good.  --C S (talk) 23:27, 26 July 2008 (UTC)
replies: text: Yeah, the pea/Sun thing works OK with that wording. I think what the anon objected to was claiming that the behavior of peas was part of the content of the theorem itself, and I agreed that that was problematic. --Trovatore (talk) 01:46, 27 July 2008 (UTC)
thread_title:  the pea and the Sun 

message: In the proof in the article it is said "let A be a rotation of some irrational multiple of π, take arccos(1/3), about the first, x axis, and B be a rotation of some irrational multiple of π, take arccos(1/3), about the second, z axis". 
This leaves the impression that every pair of rotations by irrational multiple of π around orthogonal axes generates a free group of rank 2. However, from Stan Wagon's "The Banach-Tarski Paradox" I could only find that it suffices to take a pair of rotations of the same angle θ such that \cos  \theta \in \mathbb{Q}\setminus\left\{-1,\ -\frac{1}{2},\ 0,\ \frac{1}{2},\ 1\right\}. So, unless someone can provide a reference to a proof that arbitrary irrational rotation angles suffice, I think the beforementioned sentence in the article needs to be made more precise.Jaan Vajakas (talk) 19:33, 12 December 2008 (UTC)
thread_title:  Generators of the free subgroup of SO(3) 

message: can sombody writ the number of pises one shold decompose the boll into in order to bols (if posibale - the best known, if posibale - a lower bund too)
thenks
132.77.4.43 (talk) 12:22, 4 July 2009 (UTC)
replies: text: The minimum number of pieces that works is 5. The article already says this (and has done so for years). --Zundark (talk) 14:28, 15 August 2009 (UTC)
thread_title:  quntitative bund 

message: If the first sphere has a volume of V, the two spheres that result have a volume of 2V. This is only true in the trivial case V=0.  —Preceding unsigned comment added by 75.28.53.84 (talk) 14:14, 15 August 2009 (UTC)
replies: text: The point is that the ball is broken into pieces that are so messy that they don't have a volume. So when the pieces are reassembled, there's no reason to expect the original volume to be restored. Doesn't the article already make this clear? --Zundark (talk) 14:21, 15 August 2009 (UTC)
replies: text: At first glance, it is certainly normal to find this paradoxical decomposition hard to believe. This was one of the early results which created controversy around the axiom of choice. --PST 07:37, 5 September 2009 (UTC)

text: I guess the trouble for someone who hasn't come across measure theory is to understand that there are subsets of R^3 that don't have a volume (or for which no volume can be assigned in a meaningful way). Certainly all subsets that can be physically constructed have a volume. But it is quite easy to construct subsets that have no equivalent in the real world. Take for example the all points within the unit sphere for which all coordinates are rational. There is no way that that set has any equivalent in the real world or could be constructed. Yet mathematically it is a perfectly admissable subset of R^3. Turns out this one does have a meaningful volume (zero), but it is not much harder to think of sets that have not (are not measurable), and with a bit more thinking come up with really weird sets that allow stuff like Banach-Tarsky213.160.108.26 (talk) 23:42, 11 March 2010 (UTC)
thread_title:  This is impossible. 

message: Wouldn't being able to duplicate one item into two items of identicle size of the original be tampering with the law of conservation of matter?
Also, if it does, we could make a huge ball of food, shatter it and reassemble it to make two huge balls of food. End of world hunger!
Maybe disregard that last part... --76.127.155.176 (talk) 13:45, 4 September 2009 (UTC)
replies: text: This paradox is talking about the mathematical sphere, not a physical sphere composed of atoms and molecules. Mathematical spheres are infinitely divisible, which is what is required to make this paradox work; physical spheres cannot be divided past a certain level (you get to fundamental particles and can no longer divide them), so they cannot actually be duplicated in this manner.—Tetracube (talk) 16:40, 4 September 2009 (UTC)
replies: text:  If space itself is quantized, then this paradox is non-physical at an even deeper level.   AManWithNoPlan (talk) 00:39, 5 March 2017 (UTC)

text:   To expand a little more on this topic and how it is not physical.  All the spheres are made up an an infinite number of infinitely small points, thus one sphere and two spheres have the same number of points (just happens to be an infinite number, but the same infinite number).   AManWithNoPlan (talk) 00:39, 5 March 2017 (UTC)
thread_title:  Conservation of Matter? 

message: There are no points in a piece that are not shared with another piece, so it is possible to have two sets of pieces that contain the same points. Though there are infinite number of points in a sphere, there can still be a finite number of pieces because you can divide infinity by a finite number, and get that many pieces of infinity. 
--66.66.187.132 (talk) 03:33, 5 May 2010 (UTC)
replies: text: No, the pieces are all disjoint. There are no common points between them.—Tetracube (talk) 15:39, 5 May 2010 (UTC)
thread_title: Does this describe it?

message: Is that ok to have (ine the bibliography) links to a Russian resource of illegal djvu and pdf ? Herve1729 (talk) 10:34, 14 June 2010 (UTC)
replies: text: In my opinion, and those of american judges, it is not legal and not even nice.  I removed it and contacted the author. 98.109.239.253 (talk) 03:08, 5 February 2012 (UTC)
thread_title:  links to a pirat library 

message: The text that was changed:
:However, the pieces themselves are complicated: they are not usual solids but infinite scatterings of points.
An IP editor changed it to:
:However, the pieces themselves are complicated: they are not usual solids but are actually pieces so complicated they are impossible to define.
I reverted pending discussion, because while I kind of see the point, I think there are some subtleties that need to be worked out.  I'll come back to this and explain. --Trovatore (talk) 20:17, 24 July 2010 (UTC)
replies: text: If you search history, you can find a few formulations that were better, in my opinion. There will always be a tension between people who want to put in a precise mathematical statement and others who insist on an intuitive description. While I am in the second camp, there is no question that everyone has his own intuition. But trying to make it too precise kind of defeats the purpose of being intuitive. Arcfrk (talk) 00:30, 25 July 2010 (UTC)
replies: text: I don't think I'm interested in searching the history.  Can you give us an idea of how to say it better? --Trovatore (talk) 01:23, 25 July 2010 (UTC)
replies: text:  Neither am I, so I just browsed through a few of my own edits. Here is a reasonable version of the lead. By the way, I am not happy with "SET THEORETIC geometry", please, remove it if you adjust the lead. Arcfrk (talk) 04:15, 25 July 2010 (UTC)
thread_title:  Characterization of the "pieces" 

message: At step 1 in the proof there is said that we need to divide group into four pieces and multiply them by a or b.
I have two questions: what are these four groups? How to multiply them with a or b?  Wojowu (talk) 10:29, 31 October 2011 (UTC)
replies: text: They aren't groups, they're just subsets: S(a), S(b), S(a-1) and S(b-1). The article explains how they are defined and how they are multiplied. I've reworded that paragraph to make it clearer that it's just giving an overview of what has already been done, rather than doing anything new. --Zundark (talk) 11:51, 31 October 2011 (UTC)
thread_title:  Multiplication by a or b? 

message: Can a ball be "split" into "parts" and re-assembled into two balls?  A few months ago I replaced a bold claim of "splitting" by a more moderate assertion of the existence of a decomposition, with a link to existence theorem to indicate the level of ontology we are dealing with here.  I suggest we replace the "part" by the more neutral "subset" with a link to subset, similarly to avoid controversial ontological commitments.   Tkuvho (talk) 14:40, 20 November 2011 (UTC)
thread_title:  splitting into parts? 

message: In Step 3 of the sketch of the proof it says: "the paradoxical decomposition of H then yields a paradoxical decomposition of S2". It would be nice to say what the parts of this decomposition are. I'm assuming they are M, S(a)M, S(a-1)M, S(b)M, S(b-1)M? But if that's the case then we more than double the sphere: S(a)M and S(a-1)M yield a sphere after S(a-1)M is rotated by a, and so do S(b)M and S(b-1)M. This means M is left over as an extra piece? This needs clarification, I think. JanBielawski (talk) 22:09, 8 December 2011 (UTC)
thread_title:  Clarification would help 

message: OK, I am by no means an expert on this subject (far from it), but something doesn't sit right with me.
Step 1 of the sketch seems to be saying:
:"Let F2 be a group consisting of the set of all strings satisfying <condition> with an operator *.
:
:Let's split the set portion of F2 into five disjoint subsets A, B, C, D, and E.
:We can reconstruct all of F2 using the * operator on the subsets A and B.
:We can also reconstruct all of F2 using the * operator on the subsets C and D.
:
:Tada! F2 has been doubled!"
To me, this isn't "doubling" F2. This is just arriving at F2 again two different ways, primarily by exploiting the infinite nature of F2, in conjunction with the fact that the * operator is not closed under the subsets (S(a) ∪ S(a')) and (S(b) ∪ S(b')).
Consider this:
# Let Z be the set of all integers.
# Split Z into two sets, E (consisting of all integers evenly divisible by 2) and O (the odd integers, which are every other number)
# Create a set E' by dividing all elements in E by 2.
#:E' = { x/2 : x in E }
#:Alternative: E' = { x : x in Z, 2x in E }
# Create a set O' by applying the expression (x-1)/2 to all elements in O.
#:O' = { (x-1)/2 : x in O } 
#:Alternative: O' = { x | x in Z, 2x+1 in O }
# E' = O' = Z. Hey, I just doubled the set of integers!
Stevie-O (talk) 21:17, 11 January 2012 (UTC)
replies: text: The point isn't really that we've "doubled" F2, but that we've "doubled" it and are now going to use this "doubling" to double a sphere. You couldn't use your "doubling" of Z in this way, because it's the wrong sort of doubling (you've multiplied by ½, instead of just shifting bits about and taking unions). The right sort of doubling is actually impossible with Z, because Z is amenable. --Zundark (talk) 13:14, 12 January 2012 (UTC)
thread_title:  Implications in sketch step 1? 

message: "With more algebra one can also decompose fixed orbits into 4 sets as in step 1. This gives 5 pieces and is the best possible."
Is there any sources ? link to a paper, name of the mathematician ?  — Preceding unsigned comment added by 109.26.131.236 (talk) 15:28, 27 May 2013 (UTC)
thread_title:  5 Pieces, the best ?  

message: Could the two identical spheres be decomposed and reassembled into the original first sphere? KaiQ (talk) 00:54, 31 October 2013 (UTC)
replies: text: Yes, of course. --Trovatore (talk) 01:10, 31 October 2013 (UTC)
thread_title:  Reversability 

message: I very much enjoyed the proof sketch (never thought I'd understand this proof!), but this statement in Section 4.5 confused me:
:[…] since the paradoxical decomposition of F_2 relies on shifting certain subsets, the fact that some points are fixed might cause some trouble.
This confused me for a while; it sounded like somehow F_2 isn't paradoxically decomposable anymore because we're using it wrong? At any rate, I don't think “shifting” is really the root of the problem. IIUC, it's more the fact that, since S(b) contains fixed points in M, it is not the case that S(b)M is disjoint from M, and so A_1 and A_3 are no longer disjoint. (CTTOI, shouldn't it suffice to remove all the fixed points from A_2, A_3, and A_4 rather than do the equidecomposability proof?)
At any rate, I think it would be good to be more specific about what “trouble” is lurking in the proof sketch.
Luke Maurer (talk) 07:28, 9 January 2014 (UTC)
thread_title:  “Might cause some trouble” 

message: The following is copied from an old thread (Layman's section). New stuff should always go to the bottom of the talk page.
:I've added a link to that comic in the summary (which is now anchored to the annotation so the web page now opens on the actual explanation instead of a funny comic). I'm not a mathematician but it explained the theorem well enough for me to at least understand that this is a thing that makes sense once all the assumptions are explained. I understand that this reference may not be adequate. To those wanting to remove it, however, PLEASE consider summarizing the explanation. Laying out the assumptions and theorems underlying the topic in simple terms is NOT a pointless exercise, ESPECIALLY with a topic like this with at least some recognition outside of mathematics. meustrus (talk) 17:11, 11 August 2014 (UTC)
replies: text: We can't have a link to a comic in the lead. We can't either write a summary of the whole (huge) linked page and put it into the article because it wouldn't count as a reliable source. Those are formal reasons. I believe too that there are other reasons. Some readers might think they understand the linked page better than the article. It is possible, but I doubt that they really understand the Banach–Tarski theorem any better. But yes, the article can be improved. YohanN7 (talk) 18:30, 11 August 2014 (UTC)
replies: text: I took some time to read the material in the linked page. It looks like the author have tried to make something out of it, but I can't see what's is explained better there than here. If you can think of anything that helps explaining either the theorem (the paradox if you want) or its proof, the go ahead and be bold, or discuss here. But by just avoiding what might be perceived as buzzwords, nothing is automatically improved. These buzzwords are almost always blue links to an article explaining that particular buzzword. Then, also, if you find the blue links completely inadequate, then the reason for this may be that articles normally link "only one step down the food chain". This article, for instance, should definitely link the axiom of choice, but probably don't go very deep into basics about cardinality. This article should probably not goo deeply into the difference between the mathematical solid ball and a physical one either, but it could be mentioned (finite "set" (collection of particles in space) vs uncountable set). YohanN7 (talk) 23:19, 11 August 2014 (UTC)
thread_title:  Recent edits 

message: The introduction says "It can be proven only by using the axiom of choice". But first of all, this is false. All that's required is the Ultrafilter Lemma (or Hahn-Banach, or the Order Extension Principle). Second, this claim is simply false if it turns out the ZF axioms are inconsistent. For the axioms are inconsistent, then we can prove Banach-Tarski (and its negation, and any other proposition we wish) from the ZF axioms. Since there is no proof of the consistency of the ZF axioms (and we could only have a proof of the consistency of ZF within ZF if ZF were inconsistent :-) ), neutrality suggests that Wikipedia should not take a stance on the question. A more nuanced statement is that *if* ZF is consistent, then the Paradox cannot be proved without *some version* of the Axiom of Choice. I realize that it's not as catchy as the formulation in the text, but accuracy seems more important than catchiness. But to be honest I really don't know what the standards for precision for the kind of expository writing that Wikipedia represents are, so maybe a case can be made for greater precision. Pruss (talk) 03:12, 31 March 2015 (UTC)
replies: text: Good point. In my opinion, "Banach-Tarski is not provable without AC" is as reasonable as "Banach-Tarski is not provable without the axiom of infinity". The real counterintuitive part in BT is not AC but the fact that Hilberts hotel can be implemented by rotations within a bounded 3D space. AC takes this only to a higher, measurable, level - but the magic has already happened.  — Preceding unsigned comment added by Damluk (talk • contribs) 15:15, 4 April 2015 (UTC)

text: The lead should not go into such painful detail as buying insurance against the inconsistency of ZF, nor should it delve into details about the precise strength of choice needed. The lead should be accessible and summarize the article, not confuse the reader with a tonne of blue links. (It is rather remarkable that axiomatic set theory has not rationalized its own usage of terminology to the point that premises are assumed consistent unless explicitly stated. This practice leads to awkward linguistic monotonicity and should not be allowed to spread into other areas of mathematics, much less into Wikipedia articles.) The flavor of choice actually needed could go into the body of the article (with citation to a reliable source.) All this, of course, provided that both the lead and the body of the article actually existYohanN7 (talk) 12:14, 5 April 2015 (UTC)
thread_title:  Is it responsible to say the Paradox needs the Axiom of Choice? 

message: A Steven Weinberg proposal over a Stephen Hawking hypothesis:
All quarks are sibling flavoured Banach–Tarski spheres. The degenerate pressure inside a black hole, compacts in the lowest possible volume quark groupings; thus for example two quarks, can be rotated in a way they become one and the same. So if for example start with two quarks regardless of their flavour, the degenerate pressure inside any black hole, manages to constitute them one and the same, because the best way to pack them is to force them fit one to the other. The result is one quark. The complete reaction requires 2 mesons, that merge and form a single one, but there are many other combinations. There is no data loss, because these quarks really become one and the same. There is no energy loss, because when the degenerate quark gluon plasma shrinks, it contributes a. to the black hole jets, b. to the degenerate particle at the core of the quark-gluon plasma of the black hole (inside the event horizon of a black hole, there is a quark-gluon plasma, that has as a core an indivisible quasi-fundamental degenerate particle, because after an energy threshold of degeneracy we simply have an indivisible particle, according to this proposal, all fundamental-indivisible particles are singularity flavours, always maintaining some probabilistic range of uncertainty, because actual points don't exist in the natural world). Of course this process lasts for zillion years, because quarks don't enjoy much that statistically rare (but not actually rare if we have zillion quarks) process. || Big Bangs due to spatiotemporal decohesion (quantum information cannot be transmitted inside a superluminally expanding universe, thus born out of the energy of that expansion virtual particles are forced to become actual because total decohesion is never allowed inside any universe) also known as "superluminal divergence" of all universal points, cause the exact opposite effect.— Preceding unsigned comment added by Special:www.utexas.edu (quark) 15:21, 12 February 2016 (UTC)
replies: text: Can't say I really follow that.  Could you please clarify whether Weinberg has actually proposed this, as you suggest in the first sentence?  If so, can you provide a ref? --Trovatore (talk) 20:17, 12 February 2016 (UTC)
thread_title:  Banach–Tarski paradox and degenerate canceling of black hole quarks 

message: Text removed; not relevant to improving the article.
The answer is, no, it's not a joke.  The explanations are not on-topic for this talk page (see WP:TALK).  But feel free to ask them at the mathematics reference desk, WP:RD/Math.  (You can go back into the history of this talk page and copy/paste your text to the reference desk -- ask on my talk page if you need help doing that.) --Trovatore (talk) 23:40, 18 May 2016 (UTC)
replies: text:  It must be the same thing as with also mentioned Von Neuman "paradox" that uses area-preserving transformations to change the area. If things like that aren't a joke in literal sense, they must be a joke in figurative sense, and a rather massive ones at that. Raidho36 (talk) 23:52, 18 May 2016 (UTC)
replies: text: There will be people more than happy to address your concerns at the refdesk. --Trovatore (talk) 23:54, 18 May 2016 (UTC)

text: While the theorem itself isn't a joke, jokes have been made about it; see for instance the xkcd comic. And there is a Banach-Tarski t-shirt. And a Futurama reference. Perhaps material for an In popular culture section. --Mark viking (talk) 00:24, 19 May 2016 (UTC)
thread_title:  Is this a joke theorem? 

message: Question prompted by this diff and this one that undid the first one.
I think it's clear there are no strong national ties, so per WP:RETAIN, the established variety should be kept, if we can identify one.  This is a methodology I support, in the absence of anything better, but here we see its weakness — this is a large article, edited by a lot of people not thinking about details of spelling, and there is not a consistent variety on the page.  There are occurrences of "centre" and "center" in the same section, which should be cleaned up in any case.
The only other word I could find that seemed specific to a variety was "neighbourhood".
So I went back through the history, fifty versions at a time.  After a while, I found that the original spelling appears to be "center"; some occurrences were changed to "centre" in 2008 or so (I didn't bother searching for the exact diff).  However, "neighbourhood" was already there in Feb 2008.  But if you go back to March 2007, there is no "neighbourhood" (or "neighborhood" either), and all occurrences are spelled "center".
It's a fairly meager amount of evidence for an article this size, but I think unless someone finds an earlier substantial version with a clear distinction (or a word I didn't notice), the rules in ENGVAR point to American English for this article. --Trovatore (talk) 03:47, 15 June 2016 (UTC)
replies: text: OK, it's been a couple weeks, and no one has responded.  I'll go ahead and standardize the page on American English per the analysis above. --Trovatore (talk) 07:33, 29 June 2016 (UTC)
replies: text: The above analysis and result seems appropriate. (In such edge cases, simply flipping a coin would probably be fine). Paul August ☎ 09:51, 29 June 2016 (UTC)
thread_title:  Does this page have an ENGVAR? 

message: Could the paradox illuminate the problem of the big-bang/expansion: perhaps at an 'infinite'-ly fine-grained level, the early universe, the primordial egg, the substance of sub-planck volumes involves volumes that not only conform to Banach-Tarski's spheres, not only can be duplicate/expanded, but also comprise the attribute that they must. We could go further back than that and conjecture that multiverses, the quantum 'foam' also have this fecundity, and that our present universe's pseudo-Riemannian 4-manifold expands under just such, or similar, circumstances. I've been trying to find references to these ideas, and would like to add such to the article, but can't find any at all. Anyone see anything like this? JohndanR (talk) 22:19, 26 December 2016 (UTC)
thread_title: Cosmology

message: Well if one can make two, then each of those can make two more,and so on, ad infinitum. That's more than a paradox, that's absurd.  — Preceding unsigned comment added by 121.216.107.2 (talk) 05:54, 31 March 2017 (UTC)
replies: text: Feel free to ask a question at Wikipedia:Reference Desk/Mathematics.  Discussion of the article subject matter is off-topic on this page, except in the context of discussing what should appear in the article. --Trovatore (talk) 06:29, 31 March 2017 (UTC)
thread_title:  Even more balls? 

message: Isn't this paradox just an elaborate and complicated way of demonstrating that "half" of an uncountable set is still the same uncountable set, similar to Cantor's uncountability proof of subsets of real numbers?
In other words, can't we demonstrate the same thing if we:
# take the set of real numbers between 0 and 1 which is uncountable (cf. Cantor)
# map it to the set of real numbers between 0 and 0.5 (divide every number by 2), so set 2 is uncountable and identical to set 1
# take another set identical to set 1 and map it to the set of real numbers between 0.5 and 1 (add 1 to every number, then divide by 2), so set 3 is uncountable and identical to set 1
# now combine sets 2 and 3, so we have the set of real numbers between 0 and 1, which is identical to set 1. so we have made one identical set out of two identical sets.
# do the same in reverse, we can make two identical sets out of one.
Luzian (talk) 10:54, 20 April 2017 (UTC)
replies: text: Point is that the transformations are rigid, ie distance preserving. 
189.223.226.82 (talk) 20:44, 22 April 2017 (UTC)
thread_title:  Simplified explanation 

message: I think the formula
:A=\bigcup_{i=1}^k A_i, \quad B=\bigcup_{i=1}^k B_i, \quad A_i\cap A_j=\emptyset,
: \quad B_i\cap B_j=\emptyset, \quad 1 \leq i < j \leq k,
\quad g_i(A_i) = B_i, for all,
:\quad 1 \le i \le k, \quad g_i \in G,
would be easier to understand if we split it into three parts and write it similarly to the way it is written in the original paper by Banach and Tarski (http://matwbn.icm.edu.pl/ksiazki/fm/fm6/fm6127.pdf, p. 3).
I propose using one of the following two formulations:
#  A=\bigcup_{i=1}^k A_i,\ B=\bigcup_{i=1}^k B_i
#  A_i\cap A_j=B_i\cap B_j=\emptyset , for all  1 \leq i < j \leq k 
#  g_i(A_i) = B_i,\ g_i  \in G , for all  1 \leq i \leq k 
or:
#  A=\bigcup_{i=1}^k A_i \land B=\bigcup_{i=1}^k B_i 
#  \forall i,j  with  1 \leq i < j \leq k: A_i\cap A_j=B_i\cap B_j=\emptyset
#  \forall i  with  1 \leq i \leq k: \exists g_i \in G: g_i(A_i) = B_i 
These show 3 simple conditions that the subsets of A and B have to meet, to be G-equidecomposable:
# They exactly cover their original set (A resp. B).
# They are pairwise disjoint.
# A_i and B_i are congruent (for each i).
The first proposal uses a notation similar to the one in the article, while the second one uses the notation,
I would have choosen, if I "translated" the formula used by Banach and Tarski. I am not very familiar with the
notations and formatting usually used in the Wikipedia, so feel free to change any part of the proposed 
formulation to make them consistent with other formulas.
PS: Does the formula in the article miss a "for all" before "1 \leq i < j \leq k"?
Sven.st (talk) 11:50, 11 July 2017 (UTC)
replies: text: I've made some edits following your suggestions.  Let me know if you think anything ought to be changed. Will Orrick (talk) 16:43, 11 July 2017 (UTC)
replies: text: Looks good to me. Sven.st (talk)  —Preceding undated comment added 10:04, 14 July 2017 (UTC)
thread_title:  Split the first formula in the section "Formal treatment" into three? 

message: The beginning of the article says: "Given a solid ball in 3‑dimensional space, there exists a decomposition of the ball into a finite number of disjoint subsets. . ."
From my conversational knowledge of the subject, and the sentence further along which says: ". . .the pieces themselves are not "solids" in the usual sense, but infinite scatterings of points," I would think that the first sentence should say "into a infinite number of disjoint subsets"
If I'm just misunderstanding something, then please excuse my ignorance. Peter J. Yost (talk) 01:09, 23 March 2020 (UTC)
replies: text: It's a finite number of sets, but each set is composed of infinite numbers of points. Volunteer Marek  03:04, 23 March 2020 (UTC)

text: It might help to point out that the "subsets" of the first passage are the "pieces" of the second passage.  A ball itself, being a continuous object, contains infinitely many points, so it is not surprising that the pieces of the ball also contain infinitely many points.  But the pieces in this case have to have some unusual properties. Will Orrick (talk) 03:17, 23 March 2020 (UTC)
replies: text: Right, I guess that’s supposed to be covered by the word “scattering”. Volunteer Marek  19:13, 23 March 2020 (UTC)
thread_title:  Possible typo in first paragraph? 

message: The text says
> there are countably many points of S2 that are fixed by some rotation in H. Denote this set of fixed points as D. Step 3 proves that S2 − D admits a paradoxical decomposition. 
I think it should instead define D to be the set of points that are fixed by some rotation, *and the orbits of all of those points*. We can easily apply "step 3" to all the orbits that contain no fixed points, but step 3 doesn't give us any mechanism for handling an orbit minus one point, so we should remove the entire orbit.
Is that correct?
I don't think this affects any of the subsequent proof, because all we require below is that D is countable, so it should be an easy patch. 135.180.43.140 (talk) 08:46, 19 January 2023 (UTC)
thread_title:  Nitpick/question about the "Details" section 

message: Article sez:
The bolded claim appears to be technically true, but it suggests that it wasn't known before 1991 that full ZFC was not needed to prove Banach–Tarski, which is certainly false even leaving aside the quibble that ZFC is not finitely axiomatizable so its full strength is never needed to prove any particular ZFC theorem.  It would have been obvious from the beginning that a wellordering of R is enough choice to prove Banach–Tarski, and it must have been known from the early days of forcing that a wellordering of R does not imply full AC. --Trovatore (talk) 22:58, 28 January 2024 (UTC)
replies: text: Update:  I happened to glance up the page and notice that I had made essentially the identical complaint sixteen years ago.  I hope I've had at least one original thought in that time.  Anyway I guess it's time to stop soliciting opinions and do something about it.   I just removed the last sentence of the quoted paragraph. --Trovatore (talk) 04:49, 30 January 2024 (UTC)
thread_title:  "Pawlikowski proved..."