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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)
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)
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)
I've put in Felix Hausdorff's first name, but there's no article for him yet.
Doubling the unit interval would be impossible if the doubled interval didn't have the same number of points as the original. But there's more to it than that, because only countably many pieces are used, whereas breaking it up into individual points would involve uncountably many pieces.
Zundark, 2001-08-09
---
""Doubling the ball" by dividing it into parts and moving them around by rotations and translations, without any stretching, bending ..." - the word 'bending' is inapropriate here. Stretching and adding points 'change the volume', but bending does not.
Can you please clarify (in the article) the following basic points:
Steven G. Johnson 04:07, 22 Mar 2004 (UTC)
-- Zundark 10:13, 22 Mar 2004 (UTC)
I've tried to clarify the above points; please check. I'm still a bit confused by your saying that "size" is meant in the sense of Lebesgue measure, since later in the article it talks about there being no non-trivial "measure" for arbitrary sets. I guess the point is that the pieces are not measurable, but their combination is? Steven G. Johnson 21:53, 22 Mar 2004 (UTC)
The revised definition is much more clear, thanks! It is great to have a formal definition of what is actually being proved. Steven G. Johnson 03:12, 23 Mar 2004 (UTC)
Thanks, I think we should not push too hard on these clarifications, it is very nice article, I belive further clarifications might make too havy. Tosha 04:58, 23 Mar 2004 (UTC)
Towards the end of Step 2 "This step cannot be performed in two dimensions since it involves rotations in three dimensions. If we take two rotations about the same axis, the resulting group is commutative and doesn't have the property required in step 1." What exactly is the property required in step 1? Dgrinstein ( talk) 05:13, 19 February 2017 (UTC)
To do Step 1, you need a free group, which is a group whose only relations are the ones that allow you to cancel a with a−1 and b with b−1. In particular, in a free group there is no relation that allows you to reorder the elements in a product of as and bs. This implies that a group isn't free if a and b commute, that is, if ab=ba. Since rotations in a two-dimensional plane do commute, they cannot form a free group. Will Orrick ( talk) 12:50, 19 February 2017 (UTC)
As I understand the definitions of in the section about paradox decomposition, we mean . Now note that there exist strings in which start with "aa..." which means that by we would get the whole free group back and not only the part without as indicated by the picture. To point this out: -- 74.236.150.135
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)
By the way, there is a new page Hausdorff paradox. I don't feel qualified to work on the content; but it is clearly very close to this page. If this is becoming a featured article candidate, perhaps including tha material might make this page more complete.
Charles Matthews 07:57, 5 May 2004 (UTC)
I came across an interesting anagram of "BANACH TARSKI".
It's "BANACH TARSKI BANACH TARSKI". — Ashley Y 10:45, 2004 Jul 9 (UTC)
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:
I wonder if replacing "cut it up into" with "divide it into" would be clearer, since the division isn't a knife cut style topological division. I also think the last sentence of the introduction starting "actually, as explained below" is confusing and arguably not neutral, since the previous sentence provides both points of view. Any objections to replacing that phrase and deleting that sentence? Warren Dew 23:53, 12 March 2007 (UTC)
The proof here is closer to Hausdorff paradox I think to move it there and leave this page with no proof. Tosha
I can provide an illustration of Step 1 of the proof sketch, namely the paradoxical decomposition of . I envision something like the picture at free group, but with the sets , and marked and labelled. Is there interest? -- Dbenbenn 01:35, 6 Dec 2004 (UTC)
I think it wold be great (with colors?)... Tosha 02:50, 6 Dec 2004 (UTC)
Very nice I think Tosha 07:18, 13 Dec 2004 (UTC)
Anyone want this picture on the page?
The fact that the free group can be so decomposed follows from the fact that it is non-amenable. I think we should put this in - It makes the discussion a little more transparent.
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)
Banach–Tarski paradox is indeed counter intuitive but I'm not sure that what makes it counter intuitive is related to the axiom of choice. I would say that by using the axiom of choice one can get a counter intuitive result from another one which is already counter intuitive :
Think about a subset of the ball as something "covering" some part of the ball. The intuition tells us that when you move by a rotation a subset of the ball into the ball you will "uncover" some points and also "cover" some points which where "uncovered" before applying the rotation, but that these two phenomena will compensate each other. In fact this is not true in the non measurable world and this is the heart of the Banach–Tarski paradox.
More precisely the intuition tells us that it is impossible to find a subset X of the unit sphere (the sphere bounding the ball), apply a rotation to it and get a subset strictly included in X.
Similarly intuition tells us that it is impossible to find a subset X of the unit sphere which is strictly included in the sphere, apply a rotation to X and get something which stricly contains X.
This is nevertheless true :
Using the same notation as in the article take some x lying in the sphere (not on the rotation axis) and consider the subset of the sphere . If one applies the rotation to it we get something stricly included in X and if one apply the rotation to X one get something that strictly contain X. This is due to the fact that
and the way H is acting on the unit sphere.
You should realize that the Banach-Tarski paradox is only somewhat more strange than the fact that e.g. applying a shift to the left by 10 units of length to the set of all integers greater than 10 on the real line (and labeling corresponding points with their new values) produces a strictly bigger set - the set of all positive integers.
Someone recently moved this article from Banach–Tarski paradox to Banach-Tarski paradox (substituting the endash with a hyphen). Really that's the name I'd prefer too; I don't like all these Unicodes that look almost like ASCII characters and can easily be mistyped. But the current WP standard seems to be endashes when the names of two workers are combined, so I moved it back. -- Trovatore 23:57, 4 January 2006 (UTC)
This is a popular paradox, and I believe that there are (relatively) many laypersons who may have a passing interest in what this thing is all about. In my opinion, it would be nice if there was a section here that would require very little background, intuitively explaining what the thing is about and wouldn't require much time to read (no equations!).
Sketch of how such a section could be done:
I don't know how exactly to phrase this, and my knowledge of the matter is rather superficial (I'm pretty much a layperson myself!), so I didn't dare actually try to modify the article.
The goals I was thinking of are:
Too much to ask? Maybe. But in my opinion, an explanation like stated above would exist in an ideal encyclopedia entry, as not all interested readers know much of anything about modern math. (and an ideal encyclopedia is for everyone, right? :-)
It may be that some of the above would be better to put in a general article about geometrical paradoxes (and a link from here to that), but I think that merely referring to the articles about measure theory, related paradoxes, etc. would scare many away. Reading a layperson's section should require just about the same amount of effort as reading a popular science magazine article, and chasing hyperlinks in search of comprehensible and relevant information is a far cry from that.
130.233.22.111 16:35, 6 February 2006 (UTC)
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)
Done. I also removed a confused paragraph about it not being a real paradox, which in addition to being dubious was also in a completely inappropriate section. -- Trovatore 06:40, 15 October 2006 (UTC)
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
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)
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)
Wow. I thought exactly the same thing about Hilbert's Hotel. I think you're absolutely right. Both theorems work because we are allowed to "push" extra stuff away to infinity to make room for something new, or vice versa. Danielkwalsh ( talk) 09:31, 28 April 2011 (UTC)
The article currently says
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)
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) (reworded slightly 18:47, 21 December 2007 (UTC))
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)
I undid an edit by Likebox. There are three specific parts that were undone:
— Carl ( CBM · talk) 15:42, 11 March 2008 (UTC)
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)
I noticed Arcfrk removed this sentence:
While there are a very few mathematicians who dislike the axiom of choice, the consensus in the field is that the axiom of choice is a perfectly valid axiom and theorems proved with it are not "suspect" or "contingent" (no more than any other theorem). The lede does cover the minority point of view: "The existence of nonmeasurable sets, such as those in the Banach–Tarski paradox, has been used as an argument against the axiom of choice, although most mathematicians accept that nonmeasurable sets exist." That is about as much weight as the minority viewpoint should be given here, as it is exceedingly uncommon among contemporary mathematicians. — Carl ( CBM · talk) 11:39, 13 March 2008 (UTC)
Banach-Tarski Theorem should redirect to Banach-Tarski Paradox. The "paradox" is the counter-intuitive demonstration that an object can be divided into a small number of parts, just five, that lose their volume and be reassembled into a form that doubles the original volume. The fact that this theorem requires the Axiom of Choice is no more relevant than the fact that the Pythagorean Theorem requires Euclid's Fifth postulate. I understand that Tarski originally thought of this paradox/theorem as a counterexample to the Axiom of Choice, but later regarded it as evidence that "measure" was a more subtle concept than previously thought. Alan R. Fisher ( talk) 23:04, 27 August 2008 (UTC)
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)
Would it be appropriate to add the following sentence to the first paragraph?
A humorous name for this strong form is the General Suitcase Packing theorem, the joke being the non-constructive nature of the proof.
(I have no source to cite, but I recall reading something similar many years ago.)
Alan R. Fisher (
talk) 05:39, 1 September 2008 (UTC)
The pea and the Sun example is terribly misleading, however popular it is, since the pea and the Sun are roughly spherical, but the construction is about turning one ball into two balls of the same radius. I added "Alternatively, an anagram for Banach-Tarski is Banach-Tarski Banach-Tarski" but that was reverted, along with excisions of some of the nonsense in the first paragraph. As a mathematician, I have no idea what "infinite scattering of points" is supposed to mean, and I doubt it tells much of anything correct to anyone else. Note that the rationals are a measurable subset of the reals, so the paradox is not that we have decompositions into dense subsets. I added a link to non-measurable sets which was reverted, too. I guess I'll give up trying to improve this tripe. —Preceding unsigned comment added by 66.30.116.104 ( talk) 19:36, 15 February 2010 (UTC)
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 . 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)
I would also be interested if anyone could provide a counterexample where two rotations by irrational multiples of π around orthogonal axes do not generate a free group. Jaan Vajakas ( talk) 20:52, 12 December 2008 (UTC)
All right, I found an example myself: if α and β are such that cos(α/2)*cos(β/2)=0.5 and if a denotes rotation about x-axis by angle α and b is rotation about z-axis by β then ababab is the identity transform. Since there is a continuum of such pairs (α, β) but the set of the pairs, where either component is a rational multiple of π, is countable, it follows that in some of the pairs both components must be irrational multiples of π. Jaan Vajakas ( talk) 08:40, 14 December 2008 (UTC)
Since nobody replied, I made the change to the article. Jaan Vajakas ( talk) 12:58, 14 December 2008 (UTC)
Hi Guys,
I'm confused by the word 'almost' and 'majority of' in step 3 of this section. I've traced back the origin of these words to an anon and a helpful user, so I don't think it's vandalism. But I'm fairly fluent in mathematics and the wording doesn't make sense to me.... Anyone understand their purpose?
Isaac ( talk) 17:48, 29 December 2008 (UTC)
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)
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)
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)
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)
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)
The text that was changed:
An IP editor changed it to:
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)
Bringing in "define" just asks for trouble. The notion of "definability" has not been defined in the article yet, crops up nowhere else in the article, and really is off-topic. Too much detail is wrong, and too much biassed detail is even wronger. I changed this to something more like the original, and vaguer: not solids in the usual sense etc...which is true even if they were arcwise connected since they're not measurable.... 98.109.239.253 ( talk) 03:10, 5 February 2012 (UTC)
I see how the coastline of whichever island you live on cannot be given a single length - you have to specify how long your measuring-stick is ... Likewise, I see that it is pretty meaningless to ask whether your lungs have more surface area than a football pitch - although we have probably all seen such claims ! Volume, however seems intuitively much more measurable - any real-world examples, or discussion of why the 3rd dimension is different ?
It all seems like counting how many angels can dance on the head of a pin?, though !
-- 195.137.93.171 ( talk) 04:18, 13 October 2010 (UTC)
For those not familiar with Cayley graphs, adding labels for a inverse and b inverse on the branches opposite a and b would make the diagram more easily comprehensible. Ross Fraser ( talk) 22:06, 18 April 2011 (UTC)
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)
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)
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)
OK, I added the clarification in Step 3. JanBielawski ( talk) 23:14, 8 December 2011 (UTC)
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:
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:
Stevie-O ( talk) 21:17, 11 January 2012 (UTC)
"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)
Could the two identical spheres be decomposed and reassembled into the original first sphere? KaiQ ( talk) 00:54, 31 October 2013 (UTC)
I very much enjoyed the proof sketch (never thought I'd understand this proof!), but this statement in Section 4.5 confused me:
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)
The following is copied from an old thread (Layman's section). New stuff should always go to the bottom of the talk page.
The recent contribution by user Garfield Garfield sounds intriguing but it is more of an illustration of a non-conservation in physics than B--T. If we had a discussion of similarity to non-conservation (of mass, parity, etc.) we could mention this also, but reliable sources would be needed. Tkuvho ( talk) 08:11, 18 August 2014 (UTC)
I think that the set equations in step 3 of the proof sketch are wrong. It is correct that , but it does not mean that triples . The problem is that is a finished set, which is then operated on by . For to be tripled, each element would have to be operated on by first, and then by any other . — Preceding unsigned comment added by 37.24.252.113 ( talk) 07:51, 1 November 2014 (UTC)
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)
For the individual behind an IP address that reverted my edit:
/info/en/?search=Wikipedia:Manual_of_Style/Layout#See_also_section states, "The links in the "See also" section might be only indirectly related to the topic of the article because one purpose of "See also" links is to enable readers to explore tangentially related topics." The content at the Hindu page qualifies, as it tangentially related, essentially amounting to a religious usage of the paradox. Please note I am not Hindu and am by no means trying to weasel it in for whatever reason. 0nlyth3truth ( talk) 23:16, 23 July 2015 (UTC)
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)
What up
Whack! You've been whacked with a wet trout. Don't take this too seriously. Someone just wants to let you know that you did something silly. |
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)
came up from reference desk...if someone mathematically knowledgeable to properly phrase a brief explanation in the intro of this article that this kind of taking apart and reassembly couldn't actually be accomplished in the real-world, but is only valid within the mathematical realm etc....this topic gets some attention in the popular press and creates confusion etc... 68.48.241.158 ( talk) 16:49, 19 May 2016 (UTC)
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)
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)
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)
In the picture of F_2, the free group on two letters, you have the label S() on the northwest side; the label aS() on the southwest side; and the label S(a) on the northeast side.
This is hopelessly confusing and completely wrong. The diagram should be labelled as follows, by the four "quadrants" to the north, east, south, and west.
East: S(a)
North: S(b)
West: S()
South: S()
Now if you click on this diagram you are taken to the .svg page. Then below the picture there's a link that says "Qef - Paradoxical decomposition F2.svg" which brings you to this beautiful depiction of the effect of aS(a-inverse), which magically eats up two other quadrants. This is the very heart of the entire proof. https://commons.wikimedia.org/wiki/File:Paradoxical_decomposition_F2.svg
It would be far better for people trying to understand this proof to have both diagrams easily available (how many readers will find the second diagram?) and label the quadrants correctly in the first picture.
189.223.226.82 ( talk) 00:46, 19 April 2017 (UTC)
The F_2 graphic offers a more or less straightforward illustration of a paradoxical decomposition of an orbit of a point. Imagine the center point (where the label is) of the graph being a point in 3D space. Call this center point . Rotate it "to the left" by multiplication and you get the red-blueish point at the -label. Rotate it further by anything but . You will get to another point in the left quadrant. The whole left quadrant is the set of all points that originate from with as first rotation. But the left quadrant itself is meaningless to the paradox. If we rotate it "to the right" by , we only get as a new point. However, if we look at the set of all red dots, i.e. , we obtain the set of all blue dots. And if we count the blue and the red dots, there are roughly three times as many blues as reds. — Preceding unsigned comment added by Damluk ( talk • contribs) 16:28, 4 May 2017 (UTC)
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:
Luzian ( talk) 10:54, 20 April 2017 (UTC)
189.223.226.82 ( talk) 20:44, 22 April 2017 (UTC)
I think the formula
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:
or:
These show 3 simple conditions that the subsets of and have to meet, to be -equidecomposable:
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 ""?
Sven.st ( talk) 11:50, 11 July 2017 (UTC)
I have just removed the statement in the article that the sets in the Banach–Tarski decomposition have "large porosity" as I believe it may be misleading or incorrect. If I am mistaken, it can be added back, but it would be good to add some explanation somewhere, perhaps in the article Porous set and/or in the Banach–Tarski article. My understanding is that "porous", both in the mathematical and physical definitions, requires having finite-sized voids, which is something that I believe is not true of the Banach–Tarski sets. It may be that I have misunderstood something, either about the definition of "porous" or about the Banch–Tarski sets, in which case some enlightening explanation would be greatly appreciated. and I would be happy to have my change reverted. Will Orrick ( talk) 17:13, 26 June 2018 (UTC)
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)
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)
Article sez:
In 1991, using then-recent results by Matthew Foreman and Friedrich Wehrung, Janusz Pawlikowski proved that the Banach–Tarski paradox follows from ZF plus the Hahn–Banach theorem. The Hahn–Banach theorem does not rely on the full axiom of choice but can be proved 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. [emphasis mine]
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)
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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)
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)
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)
I've put in Felix Hausdorff's first name, but there's no article for him yet.
Doubling the unit interval would be impossible if the doubled interval didn't have the same number of points as the original. But there's more to it than that, because only countably many pieces are used, whereas breaking it up into individual points would involve uncountably many pieces.
Zundark, 2001-08-09
---
""Doubling the ball" by dividing it into parts and moving them around by rotations and translations, without any stretching, bending ..." - the word 'bending' is inapropriate here. Stretching and adding points 'change the volume', but bending does not.
Can you please clarify (in the article) the following basic points:
Steven G. Johnson 04:07, 22 Mar 2004 (UTC)
-- Zundark 10:13, 22 Mar 2004 (UTC)
I've tried to clarify the above points; please check. I'm still a bit confused by your saying that "size" is meant in the sense of Lebesgue measure, since later in the article it talks about there being no non-trivial "measure" for arbitrary sets. I guess the point is that the pieces are not measurable, but their combination is? Steven G. Johnson 21:53, 22 Mar 2004 (UTC)
The revised definition is much more clear, thanks! It is great to have a formal definition of what is actually being proved. Steven G. Johnson 03:12, 23 Mar 2004 (UTC)
Thanks, I think we should not push too hard on these clarifications, it is very nice article, I belive further clarifications might make too havy. Tosha 04:58, 23 Mar 2004 (UTC)
Towards the end of Step 2 "This step cannot be performed in two dimensions since it involves rotations in three dimensions. If we take two rotations about the same axis, the resulting group is commutative and doesn't have the property required in step 1." What exactly is the property required in step 1? Dgrinstein ( talk) 05:13, 19 February 2017 (UTC)
To do Step 1, you need a free group, which is a group whose only relations are the ones that allow you to cancel a with a−1 and b with b−1. In particular, in a free group there is no relation that allows you to reorder the elements in a product of as and bs. This implies that a group isn't free if a and b commute, that is, if ab=ba. Since rotations in a two-dimensional plane do commute, they cannot form a free group. Will Orrick ( talk) 12:50, 19 February 2017 (UTC)
As I understand the definitions of in the section about paradox decomposition, we mean . Now note that there exist strings in which start with "aa..." which means that by we would get the whole free group back and not only the part without as indicated by the picture. To point this out: -- 74.236.150.135
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)
By the way, there is a new page Hausdorff paradox. I don't feel qualified to work on the content; but it is clearly very close to this page. If this is becoming a featured article candidate, perhaps including tha material might make this page more complete.
Charles Matthews 07:57, 5 May 2004 (UTC)
I came across an interesting anagram of "BANACH TARSKI".
It's "BANACH TARSKI BANACH TARSKI". — Ashley Y 10:45, 2004 Jul 9 (UTC)
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:
I wonder if replacing "cut it up into" with "divide it into" would be clearer, since the division isn't a knife cut style topological division. I also think the last sentence of the introduction starting "actually, as explained below" is confusing and arguably not neutral, since the previous sentence provides both points of view. Any objections to replacing that phrase and deleting that sentence? Warren Dew 23:53, 12 March 2007 (UTC)
The proof here is closer to Hausdorff paradox I think to move it there and leave this page with no proof. Tosha
I can provide an illustration of Step 1 of the proof sketch, namely the paradoxical decomposition of . I envision something like the picture at free group, but with the sets , and marked and labelled. Is there interest? -- Dbenbenn 01:35, 6 Dec 2004 (UTC)
I think it wold be great (with colors?)... Tosha 02:50, 6 Dec 2004 (UTC)
Very nice I think Tosha 07:18, 13 Dec 2004 (UTC)
Anyone want this picture on the page?
The fact that the free group can be so decomposed follows from the fact that it is non-amenable. I think we should put this in - It makes the discussion a little more transparent.
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)
Banach–Tarski paradox is indeed counter intuitive but I'm not sure that what makes it counter intuitive is related to the axiom of choice. I would say that by using the axiom of choice one can get a counter intuitive result from another one which is already counter intuitive :
Think about a subset of the ball as something "covering" some part of the ball. The intuition tells us that when you move by a rotation a subset of the ball into the ball you will "uncover" some points and also "cover" some points which where "uncovered" before applying the rotation, but that these two phenomena will compensate each other. In fact this is not true in the non measurable world and this is the heart of the Banach–Tarski paradox.
More precisely the intuition tells us that it is impossible to find a subset X of the unit sphere (the sphere bounding the ball), apply a rotation to it and get a subset strictly included in X.
Similarly intuition tells us that it is impossible to find a subset X of the unit sphere which is strictly included in the sphere, apply a rotation to X and get something which stricly contains X.
This is nevertheless true :
Using the same notation as in the article take some x lying in the sphere (not on the rotation axis) and consider the subset of the sphere . If one applies the rotation to it we get something stricly included in X and if one apply the rotation to X one get something that strictly contain X. This is due to the fact that
and the way H is acting on the unit sphere.
You should realize that the Banach-Tarski paradox is only somewhat more strange than the fact that e.g. applying a shift to the left by 10 units of length to the set of all integers greater than 10 on the real line (and labeling corresponding points with their new values) produces a strictly bigger set - the set of all positive integers.
Someone recently moved this article from Banach–Tarski paradox to Banach-Tarski paradox (substituting the endash with a hyphen). Really that's the name I'd prefer too; I don't like all these Unicodes that look almost like ASCII characters and can easily be mistyped. But the current WP standard seems to be endashes when the names of two workers are combined, so I moved it back. -- Trovatore 23:57, 4 January 2006 (UTC)
This is a popular paradox, and I believe that there are (relatively) many laypersons who may have a passing interest in what this thing is all about. In my opinion, it would be nice if there was a section here that would require very little background, intuitively explaining what the thing is about and wouldn't require much time to read (no equations!).
Sketch of how such a section could be done:
I don't know how exactly to phrase this, and my knowledge of the matter is rather superficial (I'm pretty much a layperson myself!), so I didn't dare actually try to modify the article.
The goals I was thinking of are:
Too much to ask? Maybe. But in my opinion, an explanation like stated above would exist in an ideal encyclopedia entry, as not all interested readers know much of anything about modern math. (and an ideal encyclopedia is for everyone, right? :-)
It may be that some of the above would be better to put in a general article about geometrical paradoxes (and a link from here to that), but I think that merely referring to the articles about measure theory, related paradoxes, etc. would scare many away. Reading a layperson's section should require just about the same amount of effort as reading a popular science magazine article, and chasing hyperlinks in search of comprehensible and relevant information is a far cry from that.
130.233.22.111 16:35, 6 February 2006 (UTC)
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)
Done. I also removed a confused paragraph about it not being a real paradox, which in addition to being dubious was also in a completely inappropriate section. -- Trovatore 06:40, 15 October 2006 (UTC)
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
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)
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)
Wow. I thought exactly the same thing about Hilbert's Hotel. I think you're absolutely right. Both theorems work because we are allowed to "push" extra stuff away to infinity to make room for something new, or vice versa. Danielkwalsh ( talk) 09:31, 28 April 2011 (UTC)
The article currently says
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)
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) (reworded slightly 18:47, 21 December 2007 (UTC))
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)
I undid an edit by Likebox. There are three specific parts that were undone:
— Carl ( CBM · talk) 15:42, 11 March 2008 (UTC)
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)
I noticed Arcfrk removed this sentence:
While there are a very few mathematicians who dislike the axiom of choice, the consensus in the field is that the axiom of choice is a perfectly valid axiom and theorems proved with it are not "suspect" or "contingent" (no more than any other theorem). The lede does cover the minority point of view: "The existence of nonmeasurable sets, such as those in the Banach–Tarski paradox, has been used as an argument against the axiom of choice, although most mathematicians accept that nonmeasurable sets exist." That is about as much weight as the minority viewpoint should be given here, as it is exceedingly uncommon among contemporary mathematicians. — Carl ( CBM · talk) 11:39, 13 March 2008 (UTC)
Banach-Tarski Theorem should redirect to Banach-Tarski Paradox. The "paradox" is the counter-intuitive demonstration that an object can be divided into a small number of parts, just five, that lose their volume and be reassembled into a form that doubles the original volume. The fact that this theorem requires the Axiom of Choice is no more relevant than the fact that the Pythagorean Theorem requires Euclid's Fifth postulate. I understand that Tarski originally thought of this paradox/theorem as a counterexample to the Axiom of Choice, but later regarded it as evidence that "measure" was a more subtle concept than previously thought. Alan R. Fisher ( talk) 23:04, 27 August 2008 (UTC)
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)
Would it be appropriate to add the following sentence to the first paragraph?
A humorous name for this strong form is the General Suitcase Packing theorem, the joke being the non-constructive nature of the proof.
(I have no source to cite, but I recall reading something similar many years ago.)
Alan R. Fisher (
talk) 05:39, 1 September 2008 (UTC)
The pea and the Sun example is terribly misleading, however popular it is, since the pea and the Sun are roughly spherical, but the construction is about turning one ball into two balls of the same radius. I added "Alternatively, an anagram for Banach-Tarski is Banach-Tarski Banach-Tarski" but that was reverted, along with excisions of some of the nonsense in the first paragraph. As a mathematician, I have no idea what "infinite scattering of points" is supposed to mean, and I doubt it tells much of anything correct to anyone else. Note that the rationals are a measurable subset of the reals, so the paradox is not that we have decompositions into dense subsets. I added a link to non-measurable sets which was reverted, too. I guess I'll give up trying to improve this tripe. —Preceding unsigned comment added by 66.30.116.104 ( talk) 19:36, 15 February 2010 (UTC)
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 . 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)
I would also be interested if anyone could provide a counterexample where two rotations by irrational multiples of π around orthogonal axes do not generate a free group. Jaan Vajakas ( talk) 20:52, 12 December 2008 (UTC)
All right, I found an example myself: if α and β are such that cos(α/2)*cos(β/2)=0.5 and if a denotes rotation about x-axis by angle α and b is rotation about z-axis by β then ababab is the identity transform. Since there is a continuum of such pairs (α, β) but the set of the pairs, where either component is a rational multiple of π, is countable, it follows that in some of the pairs both components must be irrational multiples of π. Jaan Vajakas ( talk) 08:40, 14 December 2008 (UTC)
Since nobody replied, I made the change to the article. Jaan Vajakas ( talk) 12:58, 14 December 2008 (UTC)
Hi Guys,
I'm confused by the word 'almost' and 'majority of' in step 3 of this section. I've traced back the origin of these words to an anon and a helpful user, so I don't think it's vandalism. But I'm fairly fluent in mathematics and the wording doesn't make sense to me.... Anyone understand their purpose?
Isaac ( talk) 17:48, 29 December 2008 (UTC)
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)
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)
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)
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)
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)
The text that was changed:
An IP editor changed it to:
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)
Bringing in "define" just asks for trouble. The notion of "definability" has not been defined in the article yet, crops up nowhere else in the article, and really is off-topic. Too much detail is wrong, and too much biassed detail is even wronger. I changed this to something more like the original, and vaguer: not solids in the usual sense etc...which is true even if they were arcwise connected since they're not measurable.... 98.109.239.253 ( talk) 03:10, 5 February 2012 (UTC)
I see how the coastline of whichever island you live on cannot be given a single length - you have to specify how long your measuring-stick is ... Likewise, I see that it is pretty meaningless to ask whether your lungs have more surface area than a football pitch - although we have probably all seen such claims ! Volume, however seems intuitively much more measurable - any real-world examples, or discussion of why the 3rd dimension is different ?
It all seems like counting how many angels can dance on the head of a pin?, though !
-- 195.137.93.171 ( talk) 04:18, 13 October 2010 (UTC)
For those not familiar with Cayley graphs, adding labels for a inverse and b inverse on the branches opposite a and b would make the diagram more easily comprehensible. Ross Fraser ( talk) 22:06, 18 April 2011 (UTC)
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)
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)
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)
OK, I added the clarification in Step 3. JanBielawski ( talk) 23:14, 8 December 2011 (UTC)
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:
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:
Stevie-O ( talk) 21:17, 11 January 2012 (UTC)
"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)
Could the two identical spheres be decomposed and reassembled into the original first sphere? KaiQ ( talk) 00:54, 31 October 2013 (UTC)
I very much enjoyed the proof sketch (never thought I'd understand this proof!), but this statement in Section 4.5 confused me:
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)
The following is copied from an old thread (Layman's section). New stuff should always go to the bottom of the talk page.
The recent contribution by user Garfield Garfield sounds intriguing but it is more of an illustration of a non-conservation in physics than B--T. If we had a discussion of similarity to non-conservation (of mass, parity, etc.) we could mention this also, but reliable sources would be needed. Tkuvho ( talk) 08:11, 18 August 2014 (UTC)
I think that the set equations in step 3 of the proof sketch are wrong. It is correct that , but it does not mean that triples . The problem is that is a finished set, which is then operated on by . For to be tripled, each element would have to be operated on by first, and then by any other . — Preceding unsigned comment added by 37.24.252.113 ( talk) 07:51, 1 November 2014 (UTC)
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)
For the individual behind an IP address that reverted my edit:
/info/en/?search=Wikipedia:Manual_of_Style/Layout#See_also_section states, "The links in the "See also" section might be only indirectly related to the topic of the article because one purpose of "See also" links is to enable readers to explore tangentially related topics." The content at the Hindu page qualifies, as it tangentially related, essentially amounting to a religious usage of the paradox. Please note I am not Hindu and am by no means trying to weasel it in for whatever reason. 0nlyth3truth ( talk) 23:16, 23 July 2015 (UTC)
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)
What up
Whack! You've been whacked with a wet trout. Don't take this too seriously. Someone just wants to let you know that you did something silly. |
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)
came up from reference desk...if someone mathematically knowledgeable to properly phrase a brief explanation in the intro of this article that this kind of taking apart and reassembly couldn't actually be accomplished in the real-world, but is only valid within the mathematical realm etc....this topic gets some attention in the popular press and creates confusion etc... 68.48.241.158 ( talk) 16:49, 19 May 2016 (UTC)
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)
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)
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)
In the picture of F_2, the free group on two letters, you have the label S() on the northwest side; the label aS() on the southwest side; and the label S(a) on the northeast side.
This is hopelessly confusing and completely wrong. The diagram should be labelled as follows, by the four "quadrants" to the north, east, south, and west.
East: S(a)
North: S(b)
West: S()
South: S()
Now if you click on this diagram you are taken to the .svg page. Then below the picture there's a link that says "Qef - Paradoxical decomposition F2.svg" which brings you to this beautiful depiction of the effect of aS(a-inverse), which magically eats up two other quadrants. This is the very heart of the entire proof. https://commons.wikimedia.org/wiki/File:Paradoxical_decomposition_F2.svg
It would be far better for people trying to understand this proof to have both diagrams easily available (how many readers will find the second diagram?) and label the quadrants correctly in the first picture.
189.223.226.82 ( talk) 00:46, 19 April 2017 (UTC)
The F_2 graphic offers a more or less straightforward illustration of a paradoxical decomposition of an orbit of a point. Imagine the center point (where the label is) of the graph being a point in 3D space. Call this center point . Rotate it "to the left" by multiplication and you get the red-blueish point at the -label. Rotate it further by anything but . You will get to another point in the left quadrant. The whole left quadrant is the set of all points that originate from with as first rotation. But the left quadrant itself is meaningless to the paradox. If we rotate it "to the right" by , we only get as a new point. However, if we look at the set of all red dots, i.e. , we obtain the set of all blue dots. And if we count the blue and the red dots, there are roughly three times as many blues as reds. — Preceding unsigned comment added by Damluk ( talk • contribs) 16:28, 4 May 2017 (UTC)
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:
Luzian ( talk) 10:54, 20 April 2017 (UTC)
189.223.226.82 ( talk) 20:44, 22 April 2017 (UTC)
I think the formula
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:
or:
These show 3 simple conditions that the subsets of and have to meet, to be -equidecomposable:
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 ""?
Sven.st ( talk) 11:50, 11 July 2017 (UTC)
I have just removed the statement in the article that the sets in the Banach–Tarski decomposition have "large porosity" as I believe it may be misleading or incorrect. If I am mistaken, it can be added back, but it would be good to add some explanation somewhere, perhaps in the article Porous set and/or in the Banach–Tarski article. My understanding is that "porous", both in the mathematical and physical definitions, requires having finite-sized voids, which is something that I believe is not true of the Banach–Tarski sets. It may be that I have misunderstood something, either about the definition of "porous" or about the Banch–Tarski sets, in which case some enlightening explanation would be greatly appreciated. and I would be happy to have my change reverted. Will Orrick ( talk) 17:13, 26 June 2018 (UTC)
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)
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)
Article sez:
In 1991, using then-recent results by Matthew Foreman and Friedrich Wehrung, Janusz Pawlikowski proved that the Banach–Tarski paradox follows from ZF plus the Hahn–Banach theorem. The Hahn–Banach theorem does not rely on the full axiom of choice but can be proved 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. [emphasis mine]
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)