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## Pictures

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)
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)

## Hausdorff, and doubling the interval

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.

## Clarifications requested

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)

• It means size as in Lebesgue measure. Which part of the article do you think implies something different?
• It says something about the pieces not being measurable, so combined with the fact that it never defines "size" I found this confusing. The main point is that it should clearly define "size". Steven G. Johnson
• It says at the beginning that the decomposition is into finitely many pieces. This clearly distinguishes it from a decomposition into uncountably many pieces. If you don't think this is clear enough, perhaps you could add a clarification yourself.
• I don't think the article as it stands is clear enough, simply because people might be confused into thinking that the mere identification of two spheres with one is the paradoxical part (most people aren't familiar with the fact that infinite sets can be identified with subsets of themselves). Steven G. Johnson
• The proof talks about S2, but at the very end it shows how to extend this to the ball (with a slight fudge). This is also explained at the beginning of the proof sketch, so I don't think it requires further clarification.

--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)

## S(a)

As I understand the definitions of ${\displaystyle S(a)}$ in the section about paradox decomposition, we mean ${\displaystyle S(a)=\{af|f\in F_{2}\}}$. Now note that there exist strings in ${\displaystyle F_{2}}$ which start with "aa..." which means that by ${\displaystyle aS(a^{-1})}$ we would get the whole free group back and not only the part without ${\displaystyle S(a^{-1})}$ as indicated by the picture. To point this out: ${\displaystyle aS(a^{-1})=\{ax|x\in S(a^{-1})\}=\{aa^{-1}x|x\in F_{2}\}=F_{2}}$ --74.236.150.135

If the first letter is ${\displaystyle a^{-1}}$, then the second letter can't be ${\displaystyle a}$ (because we're talking about words in reduced form). So ${\displaystyle aS(a^{-1})}$ doesn't contain any words starting with ${\displaystyle a}$. Note that the picture shows ${\displaystyle aS(a^{-1})}$ as the complement of ${\displaystyle S(a)}$, not the complement of ${\displaystyle S(a^{-1})}$. --Zundark 10:38, 27 November 2006 (UTC)

## end of proof

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)

## Curiosity

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)

## BTP & arguments against axiom of choice

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)
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)
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)

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

## Picture?

I can provide an illustration of Step 1 of the proof sketch, namely the paradoxical decomposition of ${\displaystyle F_{2}}$. I envision something like the picture at free group, but with the sets ${\displaystyle S(a^{-1})}$, ${\displaystyle aS(a^{-1})}$ and ${\displaystyle S(a)}$ 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)

Done! Is it clear, or can it be improved? --Dbenbenn 04:56, 13 Dec 2004 (UTC)

Very nice I think Tosha 07:18, 13 Dec 2004 (UTC)

## A simple picture

The Banach-Tarski "paradox": A sphere can be decomposed and reassembled into two spheres the same size as the original.

Anyone want this picture on the page?

• Looks good. I'll put it in. Eric119 23:34, 5 Feb 2005 (UTC)

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.

## Similar Result

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)

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)

## Is the Paradox related to the axiom of choice?

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 ${\displaystyle X=S(a^{-1}).x}$. If one applies the rotation ${\displaystyle a^{-1}}$ to it we get something stricly included in X and if one apply the rotation ${\displaystyle a}$ to X one get something that strictly contain X. This is due to the fact that

${\displaystyle a.S(a^{-1})=S(a^{-1})\cup S(b)\cup S(b^{-1})}$ 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.

Hi, new to wikipedia so not totally sure how to talk in here, apologies. Just to pick up on this; the set of all positive integers is not a bigger set than the set of all integers greater than 10. Intuitively it seems to be, but there is a bijection between the two sets given by the function, i.e. f(x) = x - 10. So you can pair off one number from each set together: (1,11), (2,12), (3,13),... etc. In this way, the two sets have the same size. So the Banach-Tarski result is actually much stronger. Bobmonkey 19:27, 13 November 2006 (UTC)
To expand on that, it's only with the Axiom of Choice that you can create those nonmeasurable sets that seem to become 'three times larger' when rigidly rotated. Without the Axiom of Choice, those sets don't exist, and the problem doesn't arise. Warren Dew 19:53, 12 March 2007 (UTC)
Um, that's a little confused. Axioms aren't something you use to create sets, nor do they have any bearing on whether sets exist. Without the axiom of choice you can't prove that the sets exist. This is an epistemological rather than an ontological limitation. --Trovatore 20:08, 12 March 2007 (UTC)
It is more than just a little confused, it is dead wrong on something much more basic than the AC. No one of the pieces seems to become larger when rotated or rigidly moved. The "seems to become larger" happens when these sets are re-assembled, not when they move. The resolution of this paradox is that they don't become larger since *they have no size* since they are non-measurable, so there is nothing to change or stay the same when rigidly moved or rotated. All the paradox is in the disassembly and the re-assembly, not in the rigid motions. 98.109.239.253 (talk) 03:05, 5 February 2012 (UTC)
Without the Axiom of Choice, you can't prove those sets exist. If you can't prove they exist, why would you assume that they do? As someone who is not a fan of the Axiom of Choice, I would assume the opposite, that they don't exist. Problem solved. Warren Dew 23:03, 12 March 2007 (UTC)
Whether they exist or not is independent of whether I can prove it, or of whether you assume it. It has nothing to do with you me at all. --Trovatore 23:19, 12 March 2007 (UTC)
Hey. Without the Axiom of Choice, you cannot show that nonmeasurable sets exist. This whole Banach-Tarski argument was originally intended (I believe) to show that the Axiom of Choice (AC) can be rejected as it does come up with such seemingly nonsensical consequences. Although nearly all mathematicians nowadays use AC, some do not accept it.
This is why the existence of these sets is in contention; if you do not accept AC, such sets cannot be shown to exist, so you do not accept the existence of such sets. I, on the other hand, assume AC is true, and so accept the existence of these sets. So it does make a big difference as to which mathematical perspective you look at this from!
I see your point that in that such sets existing ought to not be a matter of opinion, but whether or not you allow AC gives two different mathematical "worlds" - one where such sets can be shown to exist, and one where there is no reason at all to assume the existence of such sets. In this second case, you would assume that they do not exist, unless you could find a way to prove that they do without using AC. The article on Zermelo-Fraenkel set theory gives more information about all this. I hope this makes sense... :)Bobmonkeyofdoom 22:03, 4 April 2007 (UTC)
No, you're conflating epistemology and ontology here. Axioms don't give you worlds; they make assertions about those worlds. Now formalists will not accept the existence of those worlds at all, and that's fine; but even from the formalist point of view it's not accurate to speak about the axiom "creating" the sets. What the axiom does is enable a formal proof of a formalization of the claim that they exist; that formalization, for the formalist, is strictly speaking an uninterpreted string.
Of course this sort of syntax-only formalism is not the only alternative to realism; there are more nuanced intermediate views, like some sort of fictionalism, where the objects are considered not to exist in reality, but to be used by humans as a convenient guide to thought. Even in that view you shouldn't speak of assertions creating objects, though; it's a bad metaphor and therefore a bad guide to thought. --Trovatore 22:38, 4 April 2007 (UTC)
I agree with everything you say here! I don't ever talk about axioms "creating" sets (or at least I didn't mean to give that impression), just that whether or not such sets exist is predicated on whether or not you accept AC as true. Is this a fair comment? I bow to your superior knowledge as having a PhD in Set Theory :) But as AC is not dependent on ZF, can you not either accept or not accept the existence of nonmeasurable sets based on whether you accept AC?Bobmonkeyofdoom 00:07, 5 April 2007 (UTC)
Yes, that's fine. Except of course that it's not an if-and-only-if -- AC guarantees that there are nonmeasurable sets, but ~AC certainly doesn't guarantee that there aren't. --Trovatore 00:46, 5 April 2007 (UTC)

## endashes

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)

I moved it back to a hyphen. I don't know what "current WP standard" you're referring to, but I've never seen an en dash used in this type of case anywhere else. In particular, in TeX it's terribly easy to make an en dash (just type --), but even in TeX people use just a hyphen. dbenbenn | talk 12:01, 15 January 2006 (UTC)
Well, the case has been made that endashes are the correct way, so that we can know that Banach and Tarski are two people. I have come round to agreeing, despite the potential inconveniences. Birch–Swinnerton-Dyer then uniquely parses to two people. So, please don't move back like that. No one should get fanatical about format issues, but making a stand on them is just going to become a time-sink and distraction. Charles Matthews 19:30, 15 January 2006 (UTC)
The question is, which use is more common? It isn't up to us to determine whether an en dash is "correct". And in my experience, Banach-Tarski paradox is always written with a hyphen. I have never seen it written with an en dash. (Whereas mathematical papers written in TeX always use an en dash in something like Birch–Swinnerton-Dyer conjecture, precisely because using a hyphen would be ambiguous.) dbenbenn | talk 09:37, 16 January 2006 (UTC)
I have moved the page back to a hyphen for the aforestated reasons. What is more intuitive to type? wiki/Banach-Tarski paradox or wiki/Banach%E2%80%93Tarski_paradox?Kyle McInnes (talk) 15:40, 20 October 2006 (UTC)
It's irrelevant what's easier to type, as long as there's a redirect. Why do people keep bringing up the typing thing? That's what redirects are for. --Trovatore 17:28, 20 October 2006 (UTC)
Apologies. Thanks for creating the redirect. Kyle McInnes (talk) 17:41, 20 October 2006 (UTC)
No prob. For future reference, redirects are automatically created by page moves; it was there before this latest move-and-move-back. --Trovatore 18:59, 20 October 2006 (UTC)
The claim is that the endash is standard, in text prepared going through a professional copy-editing process. But, as I say, I don't think time should be spent on warring over such issues. Charles Matthews 10:14, 16 January 2006 (UTC)
As far as I understand, it should be with endash. (One can do anything in his TeX-file but for encilopedia should follow standards, also responsible editor would change it for publication). Tosha 16:17, 19 January 2006 (UTC)
It must be an en-dash because two people are involved. Hyphens are used for two-word names of a single person. So you have Mittag-Leffler and Levi-Civita but Banach–Tarski and Atiyah–Singer. JanBielawski (talk) 21:44, 8 December 2011 (UTC)
For what it is worth, the 1985 version of my book was The Banach-Tarski Paradox. But for the 2016 second edition (with G. Tomkowicz) just out I changed it to an n-dash, having come around to the view that the n-dash is always best in such cases. Stan Wagon --2601:284:8202:c0f5:f0cb:530b:721c:caf (talkcontribs) 03:59, 3 December 2016‎ (UTC)

## Layman's section

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:

• Size comparison. Explain that if there is a one-to-one correspondence (a bijective function) between two sets, they aren't necessarily of the same size in the ordinary sense. Example about [0, 1] <--y=2x--> [0,2]. This is why measure theory is necessary. Measure theory gives a notion of size (measure) that works much like Riemann integrals do (I assume the layperson knows some basics about integrals, if he doesn't, all explanation is futile), but is more general. Even so, the "problem" with measure theory is that not all sets are measurable, just like not all functions can be integrated. These non-measurable sets are weird beasts for which our intuition does not work. (Are all/most fractals measurable? Would be good example either way, as they're also famous "weird geometry" beasts) See also Cantor set for a set with infinite number of points but measure of 0 (infinite amount of "Cantor dust" is still just dust; similarly, a curve has measure of 0 in two dimensions, because it does not have area, except for weird beasts).
• Show that a set of points can easily be duplicated, if there are no restrictions as to what you can do with it. E.g. take a unit cube, cut it in two at x=0.5, then (translate and) multiply x of each point in the set with two, giving two exactly same sets of points as the original cube. This works the same way as one-to-one correspondence way of comparing set size: you might think that scaling x this way would leave "gaps" in the end result, but this is not so. Infinite sets are weird enough as they are. But the Banach-Tarski paradox is more paradoxical than just this.
• The paradox in Banach-Tarski paradox is that if you take a (solid) sphere, cut it in 5 pieces (although non-measurable, weird beast pieces), then move and rotate these to their places, you get two exact copies of the original sphere. Being exact copies, they of course have the same measure, too. Note that there is no scaling here, and no carving of the curve to infinite number of pieces, or anything. All the "funny stuff" is in what the 5 pieces look like. Being non-measurable, such pieces of course can't exist in the real world, as all real world objects are measurable (we can't even carve things up to infinite precision, much less somehow invoke the axiom of choice).
• Of the five pieces, one is for "fixing up" a small (countable) amount of gaps in the actual construction done with the other four pieces. The idea is to split the sphere in 4 parts in such a way that you can reassemble the original sphere from two of them, thus getting two original spheres. This reassembly is possible, because two of the pieces, which are supposed to be one-quarter of the size of the original sphere, actually become 3/4ths of the sphere when rotated by about 70.5 degrees. By performing this rotation on both of them, we get two 3/4ths of the sphere, and still have two 1/4 pieces to match. Combining these, we get two full spheres. However, this is merely an intuitive description, as the pieces are in fact non-measurable, and thus a notion that they are "quarter" or "3/4th" of the original sphere has no meaning. The construct thus escapes from the common fact that rotation preserves size by rotating pieces that do not have size. It just so happens, that when these pieces are combined to again produce measurable objects, the total size has magically doubled in the process.
• For details, see the actual proof.

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:

• Do not require much knowledge. (high school math is enough)
• Provide (non-obscure!) references for further reading, to connect all this to a larger background.
• Be clear in what is paradoxical and counterintuitive. This means explicitly pointing such things out.
• Be clear in what is not paradoxical or counterintuitive. This means explicitly stating what part the paradox is restricted to. "At least I don't have to think about these parts" makes for easier comprehension.
• Be clear in what the Banach-Tarski paradox is about, as opposed to what other (semi-)counterintuitive results there are.
• Try to keep reader's interest instead of merely stating the facts.

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)

I think David Morgan-Mar has done a pretty good explanation of the proof in lay man terms in the annotation of http://www.irregularwebcomic.net/2339.html . Maybe someone could work this into the article? PoiZaN (talk) 12:58, 21 June 2009 (UTC)
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)
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)
Well I'm not sure why we "can't" have a link to a comic, especially since it's possible to link directly to the annotation so that the comic isn't visible unless the user scrolls up the page. But I'll just assume it's a rule somewhere. Overall I think it is important not just to have a "layman's section", but to better organize the article in general. As it currently is, it goes from "this is a theorem that says you can duplicate spheres with math" to a thorough discussion of the history of it. But I'm not interested in the history of a theorem which hasn't even been described to me yet.
FYI here is how I would briefly paraphrase Banach-Tarski based solely on the comic annotation: first divide the sphere an infinite set of points; then create four new sets by rotating the points in four different directions; then because of how those four sets were constructed (using irrational numbers for the rotations) the only way to ever overlap points in the four sets is to completely reverse the rotation; with some relatively easy to explain logic the four sets can then be further rotated into sets that each contain three sets each; now that four sets each contain three quarters of the sphere, only two of those sets are required to reassemble the sphere, resulting in an extra two sets that can assemble another sphere; finally, there are parts of the sphere not accounted for by those four sets which will be accounted for by a fifth set using the same basic principles but a lot more complicated math. Is that accurate? I know that is not complete by any stretch, it is at least meaningful to me. With a summary like that in mind, I feel like I could start to delve into the actual proof and have some idea where it's going (or I could if I knew more than I do). Even without seeing the proof, I can trust that smarter people than myself agree that the proof is sufficient and start to draw conclusions like, "maybe this is why we can't actually divide a finite object into infinite pieces, unlike numbers", and, "infinity is weird". meustrus (talk) 17:02, 18 August 2014 (UTC)

## On hyphenated hyper-2-tuples

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)

## Doing the decomposition physically

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)

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)

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)

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)
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)

## Improving the text

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)

I've made substantial revisions, in particular, adding references to amenability in a few places. Arcfrk 09:50, 3 September 2007 (UTC)

## Banach-Tarski and Hilbert's Hotel

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)

Not much there. Hilbert's Hotel requires no skill, this requires ... skill. Charles Matthews (talk) 15:53, 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)

## exposition of Pawlikowski result could be misleading

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)

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) (reworded slightly 18:47, 21 December 2007 (UTC))

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)

## no general meaning of explicit construction

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)

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)
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)
("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)
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)
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)
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)
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)
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)
Ok, one last try, then I give up.Likebox (talk) 21:20, 15 January 2008 (UTC)

## Edit on 2008-3-11

I undid an edit by Likebox. There are three specific parts that were undone:

• The claim that the pieces are impossible to describe. This is simply false - the proof of the theorem gives a clear description of what the pieces are.
• The claim that an infinite number of choices is required. Any proof, being finite, will only invoke the axiom of choice finitely many times.
• That Alain Connes dislikes the theorem. This may be worth discussing somewhere, but not at the end of the very first paragraph. The vast majority of working mathematicians accept the axiom of choice, the existence of nonmeasurable sets, etc. A specific quote from Connes would establish context much better, in any case, than a generic reference to a book which is fundamentally about noncommutative geometry, not about the Banach-Tarski paradox. I would look up the quote by Connes if a more specific reference is given.

— Carl (CBM · talk) 15:42, 11 March 2008 (UTC)

"Infinitely many choices" doesn't mean "infinitely many uses of AC", it just means "at least one use of AC". (If there were only finitely many choices, AC wouldn't be needed.) --Zundark (talk) 16:06, 11 March 2008 (UTC)
Indeed, but the article already says that AC is required, lower in the lede: "What makes the paradox possible in set theory is the axiom of choice, which allows the construction of nonmeasurable sets, collections of points that do not have a volume in the ordinary sense and require an uncountably infinite number of arbitrary choices to specify." So unless something more than "AC is required" is intended, this doesn't need to also be in the first paragraph, which is already too long and detailed. — Carl (CBM · talk) 16:24, 11 March 2008 (UTC)
P.S. I do think a case could be made that each invocation of AC is a single choice - the choice of a set of representatives. — Carl (CBM · talk) 16:26, 11 March 2008 (UTC)
The article should clearly and quickly distinguish between a "theorem" like this and "all triangles have 180 degrees" or "there are infinitely many primes". Unlike the others, this "theorem" is a matter of opinion, and it's well understood that there is no right opinion.
I don't think anybody could sanely think that "uncountably many choices" means a proof with uncountably many invocations of the axiom of choice. But, kudos man, just thinking about a proof of uncountable length makes my brain get get all woozy!
I put in the caveats for the reader whose stomach becomes upset by counterintuitive results which contradict a correct intuition. This stuff is a bunch of set theoretic foolishness which has nothing to do with three dimensional geometry. It's really about the growth-rate at infinity for groups, which is a geometric notion on groups. The space-geometry part is just "the reals are an ordinal!" again, and there's a hundred years of this out there and its really getting tired. As for Alain Connes, bless his soul, his book explicitly advocates thinking of all subsets of R as measurable and you don't find many people fighting this fight. But it should be hashed and rehashed for the benefit of students, who need to be alerted as to when they are being sold a bill of goods.Likebox (talk) 09:07, 12 March 2008 (UTC)
Actually, I can see Likebox's point, but he's wrong in every detail. If he could propose an edit which deals with his concerns, but is not mathematically incorrect, it might be included. — Arthur Rubin (talk) 13:02, 12 March 2008 (UTC)
The problem with presenting things in a first-principles way is that the details look unfamiliar and the arguments need to be checked carefully because it is easy to make a mistake. I do it anyway, because the benefit is that it is much better pedagogically. In general, I don't think there is a good way of classifying people into "correct" and "incorrect". It is more useful to classify arguments into correct and incorrect. But everybody makes mistakes and I've made more than my share. In a forum like this, I don't worry so much about not getting corrected.Likebox (talk) 17:04, 12 March 2008 (UTC)
The article does already discuss the issue of choice and measurability - see the third and fourth paragraphs of the lede.
I still don't buy the "uncountably many choices", though. For one thing, the issue of countability/uncountability is a red herring - AC is needed to choose from countable collections of sets as well. But more importantly, the key property of the axiom of choice is that all the choice are made at the same time rather than one after another. The issue isn't whether it's possible to choose representatives separately, but whether it's possible to simultaneously select a representative from each of the sets in the collection. Saying "uncountably many choices" has a connotation that the choices are made one after another. — Carl (CBM · talk) 13:46, 12 March 2008 (UTC)
Yes, I know that the idea here is that the choices are made "at the same time", but that is the whole philosophical point. You shouldn't carelessly think of sets as already selected from a predefined universe of sets, that's leads to Cantor/Russell paradoxes. To actually make a theory, you need to think of them as part of a universe that's constructed step by step. In the step by step construction, choice functions are only "born" after uncountably many steps. The whole question revolves around whether you are going to view sets as preexisting, or constructed symbolically by a computational step-by-step procedure. The first point of view is, in my opinion, outdated.
Oh thank goodness, Likebox is here once again to save us from our "outdated" ways of thinking about mathematics. Can a 'computational' "proof" of the B/T Paradox be far behind? —Preceding unsigned comment added by 65.46.253.42 (talk) 20:13, 12 March 2008 (UTC)
The article discusses choice, but it doesn't do it with enough feeling in my opinion. The distinction between countable and uncountable choice is the essential one. Countable choice is not a big deal unless you're studying set theory, because you end up using it without thinking. It's really only choice for the continuum which is a problem. Uncountable choice on sets which are as big as the continuum is controversial because it's only purpose seems to be to prevent you from defining Lebesgue measure properly. This is central to how you view the real numbers, and it gums up any natural discourse about random objects.
I fortuitously stumbled on a respected source that says this (as an inconsequential aside)--- Alain Connes, and I think that justifies inclusion.Likebox (talk) 17:04, 12 March 2008 (UTC)
Could you give the page number for the reference to Connes? If he mentions it as an aside, why would that mean it should be included here? (In any case, as I pointed out, it is included here.) The axiom of choice in not controversial in the present day; it's accepted and used by the vast majority of mathematicians including noncommutative geometers. I'm interested to see what Connes has to say.
Also as an aside, in the canonical "constructed step by step" model of set theory, L, choice functions are all definable, and thus each can be constructed via a single application of comprehension. They aren't "born after an uncountable number of steps" any more than the entire set of real numbers is. — Carl (CBM · talk) 17:38, 12 March 2008 (UTC)
The pages is 51-52 and around there, and the discussion is very nice. It's an aside in the sense that it is outside the main line of development of the book.
You are right--- the step by step model has a choice function, but the construction of the step-by-step model makes it clear that it isn't constructing all of the universe. The reason is that L constructs "ordered" objects, the stuff that you can define iteratively in terms of text descriptions that are not "random". The intuition I have in this case is that it does not construct all of R, it "misses points", the random points.Likebox (talk) 20:41, 12 March 2008 (UTC)
Yep. And even some not-all-that-random ones. Like 0# for example. --Trovatore (talk) 20:49, 12 March 2008 (UTC)

## Refimprove tag

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)

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)
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)
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)

## The Banach-Tarski paradox is, indeed, a theorem

I noticed Arcfrk removed this sentence:

Because there are points of view in mathematics which reject constructions of this type, this theorem is not considered to be true in an absolute sense, but only true relative to a particular system of axioms for set theory

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)

The fact that the axiom of choice is accepted and used without question does not mean the results which absolutely require choice are uncontingent. They are not "suspect" in the sense that they are going to lead to logical problems, they are "suspect" in the sense that they are useless and silly. Nobody worries about BT when constructing measures, because mathematics evolved a machinery of measurable sets and random forcing which allows you to sidestep it, so that in every way it is more productive to think of it as false. But it took fifty years to acquire BT immunity, and the machinery evolved in a hostile environment, and that makes it cumbersome and alien to a student.
This has nothing to do with disliking choice on sets. It has to do with whether the real numbers are a small enough collection to be productively described as a well-orderable set at all. Nearly all theorems are absolutely true, meaning that once you understand the proof, it can be easily translated from any one reasonable axiomatic system to any other. But a very few theorems, like BT, are true only relative to a particular axiomatization. When you claim to prove a counterintuitive geometrical theorem and that theorem is not true in an absolute sense, then this theorem is just an opinion. While I personally think it is an outdated opinion, I respect that it is a majority opinion, and it should be presented in the way that the majority believes is best. But it should not be presented as an unqualified absolute mathematical truth.Likebox (talk) 14:58, 13 March 2008 (UTC)
There's no benefit in discussing "useless" and "silly" here. I don't follow your second paragraph above; all formalized theorems are only true relative to their axioms being true. In any case, the Banach-Tarski paradox is considered by nearly the entire mathematics community to be an (absolute) theorem, and Wikipedia is not the place to argue for a change in that attitude. If the opinion of the mathematics community begins to change, then this article should reflect that.
The fact that the theorem uses the axiom of choice is explicitly covered in the lede; I don't see the need to give it more weight than the reasonable weight it is already given. — Carl (CBM · talk) 15:51, 13 March 2008 (UTC)
There are different axiomatizations of the same objects, and nearly all of the time, you can translate theorems back and forth between different formalizations. For example, you can translate any theorem in finite set theory to the original Peano axioms and vice versa. There are axiomatizations of the reals as a complete ordered field, and you can translate most theorems into this formalization. Usually the objects that you have in mind when discussing geometry can be approached from the point of view of either second order logic or the theory of countable sets, and the different axiomatizations and points of view give results which are consistent with each other and with set theory. Those theorems which do not change from axiomatization to axiomatization are absolute, and they are not wedded to any particular formal system.
The Banach Tarski theorem is in no way accepted by nearly all the mathematical community as an absolute theorem. What is true is that is a theorem of ZFC, and by social convention a ZFC theorem is accepted. This is to avoid squabbling over foundations when trying to evaluate correctness of a proof. My guess of the breakdown in opinion in the professional community would be: 45% really true because ZFC is reality, 45% true as a matter of social convenience, 5% false, 5% no truth value. But I didn't take a formal poll, and I might be wrong.
You _of course_ have some kind of citation or reference to back up this astounding claim, I trust? —Preceding unsigned comment added by 65.46.253.42 (talk) 20:27, 13 March 2008 (UTC)
Of course not--- I didn't think it was an astounding claim.Likebox (talk) 23:15, 13 March 2008 (UTC)
Just to elaborate--- the Banach Tarski paradox is well known to not be an absolute theorem, in the sense that alternate axiomatizations have every subset of R measurable. The question is to what extent working mathematicians have a bias towards one axiomatization or another. From my experience, most of them don't care about questions of axiomatization that are not directly relevant to their work, so that means that they don't have a strong opinion one way or another. But having said that, the breakdown should follow the usual platonism/formalism/constructivism proportions, which I think are in the ratio 45/45/10.
Here's an unrepresentative informal poll I found [1].
And I don't see the Banach-Tarski paradox mentioned there once. As has been pointed out, you are of course welcome to your (wrong) opinions about the nature of mathematics, but they don't belong in Wikipedia unless they can be substantiated (by something other than a PHPnuke forum thread) and are deemed, by editorial consensus, to be relevant.
The point is that mathematicians, like any other human community, are aware that sometimes living in a community takes some compromise, and choice is not the most horrible compromise. These compromises, however, are sometimes mistinterpreted as a consensus about mathematical truth, when they are not.Likebox (talk) 18:20, 13 March 2008 (UTC)
I defer to Carl, who is an expert on set theory and logic, concerning the perceptions of AC within those fields. As a general impression, I cannot recall ever personally encountering a mathematician who was afraid to use Zorn's lemma, well-orderness principle, and a few other workhorses of mathematical proofs dealing with infinite sets (and my sample size is in the hundreds, if not thousands). Contrary to what has been stated above, their use is nearly universal: much of basic analysis, algebra, and geometry uses them without explicitly acknowledging this fact, which is a clear sign how fringe is the view that Likebox is trying to peddle. Concerning axiomatizations: as far as I know, the consistency of neither ZF nor ZFC has been proved. In this sense, all of mathematics is on shaky foundation. I do not, however, consider either prudent or practical to insert a disclaimer of this nature into every single mathematical book, paper, and encyclopaedic article.
On the subject of compromises: some time ago, Likebox and I worked out a compromise that we will keep the beginning of the article friendly to people just trying to learn a bit of interesting mathematics, without having to endure a mile long list of qualifiers (which I deeply cared about), and he can insert his onerous disclaimers elsewhere. However, he mistook this compromise for universal agreement with his misinformed interpretations, which it was not. Arcfrk (talk) 00:13, 14 March 2008 (UTC)
It is unfortunate that you are taking this as an attempt on my part to insert my own interpretation into an article. It isn't. First of all, I don't have any problem with choice or Zorn's lemma, they are fine. I am only saying that when you prove a theorem like this one, every sensible mathematician knows that it does not have the same truth value as "integer multiplication is commutative". It is a matter of axioms and opinions, and it is only a social compact that makes you think of it as true. This is the last I am saying on the subject.Likebox (talk) 01:46, 14 March 2008 (UTC)
"This is the last I am saying on the subject" -- Oh now President Likebox, I think we both know from experience that's just not true Dolores Landingham (talk) 18:26, 14 March 2008 (UTC)
I didn't mean to sound imperious--- sorry. I just don't have anything more to say--- I'm repeating myself.Likebox (talk) 14:29, 15 March 2008 (UTC)
Re Arcfrk: The status of AC in mathematics and its status in logic are both interesting philosophical questions and (in my opinion) have somewhat different answers. In mathematics outside of axiomatic set theory, as far as I can tell, the perception that there is any issue with AC is very small, and apparently diminished during the 20th century, possibly because contemporary mathematicians learned AC earlier in their education. Within axiomatic set theory, the study of determinacy is the main area in which axioms incompatible with choice are of interest. But this study is far removed from ordinary mathematics at the present time.
It is true that the consistency of ZF has not been proved in the same way the consistency of Peano arithmetic has been (and such a proof is very unlikely). However, the consistency of ZFC is equivalent to the consistency of ZF by work of Goedel. — Carl (CBM · talk) 02:59, 14 March 2008 (UTC)
Just a note on determinacy: I don't think that many set theorists who work in this area do so because they think the axiom of determinacy is actually true. Among those who are realists, and therefore think the question has a true or false answer, I'd bet the large majority think the axiom of choice is true. But the axiom of determinacy has some very interesting inner models, like L(R) for a fairly minimal example, and some interesting stuff happens in those models.
From this point of view, inner models satisfying AD are models that simply omit the "problematic" sets of reals like the ones that witness Banach–Tarski. Inside L(R) you can't do a paradoxical decomposition, because the sets making up the decomposition just aren't elements of L(R). L(R) is allowed to think that all sets of reals are Lebesgue measurable, not because it disagrees with V about what it means to be Lebesgue measurable, but just because it can't see the counterexamples.
There are lots of interesting questions about how closely an inner model of ZF+AD can resemble V. For example, which cardinals have to collapse? Steve Jackson has done a bunch of stuff on this. --Trovatore (talk) 04:58, 15 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)

{{sofixit}}. Redirects are cheap, and this one is unambiguous; I don't think anyone will challenge you. But you might want to observe our conventions on capitalization and endashes, and make it point to the official article title, which is Banach–Tarski paradox. --Trovatore (talk) 08:20, 28 August 2008 (UTC)
What Trovatore is saying is anyone can create redirects. I created these four just now: Banach–Tarski Theorem, Banach–Tarski theorem, Banach-Tarski Theorem Banach-Tarski theorem. — Carl (CBM · talk) 15:16, 28 August 2008 (UTC)

## the pea and the Sun

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)

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)
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)
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)

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)

## Generators of the free subgroup of SO(3)

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 ${\displaystyle \cos \theta \in \mathbb {Q} \setminus \left\{-1,\ -{\frac {1}{2}},\ 0,\ {\frac {1}{2}},\ 1\right\}}$. 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)

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)

## 'almost' in Step 3 of proof?

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)

I don't know what that's about, but I think it is some editorializing which could better be done with a discussion instead of with some curt comments. The point is probably that the process of construction of the sets is transfinite inductive, if you want to turn it into a construction, and it doesn't necessarily terminate by exhausting all the real numbers, or all the points in the ball. In a philosophical view that denies the well-ordering of the real numbers, the paradoxical decomposition can't happen because there aren't ordinals big enough to match one-to-one with the points in a ball. Then you can think of every set as measurable, and BT is false.Likebox (talk) 17:49, 16 March 2009 (UTC)
It's saying that the set of points that can be reached in more than one way is of measure 0 (see almost every). It has nothing to do with transfinite induction. This is then explained in more detail in the section "Some details, fleshed out." slightly below. — Carl (CBM · talk) 19:06, 16 March 2009 (UTC)
Oops! My biases are showing. I'm blushing.Likebox (talk) 19:13, 17 March 2009 (UTC)
Still, isn't this a contradiction to the statement that the set ${\displaystyle D}$ is countable? The set ${\displaystyle M}$ is uncountable, so with ${\displaystyle S(.)M}$ there are uncountable many points which were not rotated.

## quntitative bund

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)

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)

## This is impossible.

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)

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)
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)
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)
Conservation of mater only means conservation of mass. You can perfectly split a sphere of volume V in two V sphere if both sphere have half the volumic mass of the original sphere. — Preceding unsigned comment added by 78.232.116.198 (talk) 07:19, 25 July 2011 (UTC)

## Conservation of Matter?

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)

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)
If space itself is quantized, then this paradox is non-physical at an even deeper level. AManWithNoPlan (talk) 00:39, 5 March 2017 (UTC)
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)

## Does this describe it?

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)

No, the pieces are all disjoint. There are no common points between them.—Tetracube (talk) 15:39, 5 May 2010 (UTC)

## links to a pirat library

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)

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)

## Characterization of the "pieces"

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)

• First subtlety: I think the purpose of the infinite scattering of points formulation is to emphasize that the pieces are not contiguous regions that you could cut out with a knife, but more a sort of fog, likely to be both dense and codense inside (say) its convex hull. I think that's a good picture to get across; I'm not sure the bit about definability really covers it.
• On the other hand, if I recall correctly, it is possible to make the pieces connected and path-connected. I speculate that this is because the natural way of getting the pieces of the ball involves first getting the pieces of the sphere and translating them radially, and then you have to do some magic at the center. So you wind up with a bunch of radial line segments, and you somehow get them to hold together at the center. "Scattering" might not really convey that. --Trovatore (talk) 20:23, 24 July 2010 (UTC)
• Second subtlety: It is not in fact clear that the pieces are not definable. This by the way is a problem throughout WP on axiom-of-choice issues. For example, if V=HOD (that is, every set is hereditarily ordinal definable) then it is possible to get definable pieces, although not definable in any very nice or particularly useful way. V=HOD is consistent with ZFC; as to plausibility, it's probably not thought to be particularly plausible, but it's much more plausible than V=L. For example it's consistent with some fairly strong large cardinal axioms. --Trovatore (talk) 20:27, 24 July 2010 (UTC)
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)
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)
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)

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)

## Explaining non-measurable ?

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)

Actually, I'm not sure if you are talking about a hollow surface (ping-pong) or a solid sphere (snooker) ? Maybe you're so far from the real world that my question is silly ! I'l shut up now !
--195.137.93.171 (talk) 04:24, 13 October 2010 (UTC)
The paradox and proof refers to a solid sphere, not a hollow surface. I've clarified references to "centre of the ball" to ensure that the reader understands that this refers to the point at the ball's centre, not the interior of the sphere. Ross Fraser (talk) 04:00, 20 April 2011 (UTC)
Actually the main idea of the proof works on the sphere (the surface) rather than the ball (the solid thing). To run the proof for the ball-minus-center-point, you just take the paradoxical decomposition of the surface, and project it through the ball. Then there's some sort of fiddling to account for the center point; I never looked into it in enough detail to know just what you do. --Trovatore (talk) 04:39, 20 April 2011 (UTC)
Yes, I can see this now on closer reading of step 3. I've made further edits to clarify when the unit sphere as a surface is being referenced and when a solid ball of unit radius is being referenced. Ross Fraser (talk) 02:55, 21 April 2011 (UTC)

## Labels on Cayley graph

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)

Also, it might be clearer if the set labels were changed. The set aS(a−1) consists of the blue dots and the red dots. The red dots comprise the set S(a−1), which is a subset of aS(a−1). The blue dots alone comprise the set aS(a−1)∖S(a−1). Will Orrick (talk) 22:42, 3 October 2015 (UTC)
If you look closely, you can see that every red dot is surrounded by a blue border - every red dot is also a blue dot.

## Multiplication by a or b?

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)

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)

## splitting into parts?

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)

## Clarification would help

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)

## Implications in sketch step 1?

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.

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:

1. Let Z be the set of all integers.
2. Split Z into two sets, E (consisting of all integers evenly divisible by 2) and O (the odd integers, which are every other number)
3. 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 }
4. 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 }
5. E' = O' = Z. Hey, I just doubled the set of integers!

Stevie-O (talk) 21:17, 11 January 2012 (UTC)

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)

## 5 Pieces, the best ?

"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)

## Reversability

Could the two identical spheres be decomposed and reassembled into the original first sphere? KaiQ (talk) 00:54, 31 October 2013 (UTC)

Yes, of course. --Trovatore (talk) 01:10, 31 October 2013 (UTC)

## “Might cause some trouble”

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)

## Recent edits

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)
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)

## kaon decay

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)

Completely irrelevant in my opinion. I would change my mind if some physicist who knows what he's talking about said something like "the mathematics behind the Banach–Tarski paradox directly explains this anomaly in kaon decay", and this idea got significant attention in the physics community.
But if it's just someone making a vague analogy, then I don't care if that someone is Stephen Hawking; it's not relevant to this article. --Trovatore (talk) 08:36, 18 August 2014 (UTC)
This sort of speculation seems to originate with the article by Bruno Augenstein: Augenstein, Bruno W. Links between physics and set theory. Chaos Solitons Fractals 7 (1996), no. 11, 1761–1798. This got a serious review by Kreinovich there, but there are no citations at mathscinet. At google scholar there is a modest 14 citations. Tkuvho (talk) 08:35, 24 September 2014 (UTC)

## Implications in step 3

I think that the set equations in step 3 of the proof sketch are wrong. It is correct that ${\displaystyle bS(b^{-1})=S(b^{-1})\cup S(a^{-1})\cup S(a)\cup \{e\}}$, but it does not mean that ${\displaystyle bA_{4}}$ triples ${\displaystyle A_{4}}$. The problem is that ${\displaystyle S(b^{-1})M}$ is a finished set, which is then operated on by ${\displaystyle b}$. For ${\displaystyle M}$ to be tripled, each element ${\displaystyle m\in M}$ would have to be operated on by ${\displaystyle b}$ first, and then by any other ${\displaystyle s\in S(b^{-1})}$. — Preceding unsigned comment added by 37.24.252.113 (talk) 07:51, 1 November 2014 (UTC)

Hmmm. No, any point on the sphere (except those which are on an axis of rotation of an element ofH) can be uniquely represented as ${\displaystyle gp}$, where ${\displaystyle g\in \mathbf {H} }$ and ${\displaystyle p\in M}$. — Arthur Rubin (talk) 15:18, 1 November 2014 (UTC)
Im fine with this, but how does that contradict what I said previously? Im just saying that the proof sketch uses ${\displaystyle bA_{4}}$ as if it were ${\displaystyle (bS(b^{-1}))M}$ where in reality it is ${\displaystyle b(S(b^{-1})M)}$, which is a different set, namely ${\displaystyle S(b^{-1})M\cup M}$. Addendum: I think I got it (to some point). The annotation of the cayley graph is wrong. The red area is not ${\displaystyle S(a^{-1})}$, it is the set of all strings *ending* with${\displaystyle a^{-1}.}$The annotation of ${\displaystyle aS(a^{-1})}$ is wrong as well. I got confused and thought that the group acted from the right, ${\displaystyle (ab)(m)}$ being ${\displaystyle b(a(m))}$.
I guess we'll have to work on the annotation. But ${\displaystyle S(a^{-1})}$ is the set of all strings beginning with ${\displaystyle a^{-1}}$, which means ${\displaystyle A^{-1}}$ is the last operation performed on the element of M. — Arthur Rubin (talk) 00:08, 2 November 2014 (UTC)
If someone is interested: https://commons.wikimedia.org/wiki/File:Paradoxical_decomposition_F_2.svg. I'm not sure how to update the current image properly. — Preceding unsigned comment added by Damluk (talkcontribs) 18:23, 2 November 2014 (UTC)

## Is it responsible to say the Paradox needs the Axiom of Choice?

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)

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 (talkcontribs) 15:15, 4 April 2015 (UTC)
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)

## Vaisheshika - atomic theory

For the individual behind an IP address that reverted my edit:

https://en.wikipedia.org/wiki/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)

Also, please note the link to the Banach-Tarski paradox within that article section. 0nlyth3truth (talk) 23:20, 23 July 2015 (UTC)
To me this seems less like Banach–Tarski and more like the fact that all line segments, regardless of length, have the same number of points (cardinality of the continuum). The latter also strikes some people as paradoxical, but what's special about the Banach–Tarski paradox is that you only need a finite number of pieces.
So really the link to B–T from the Vaisheshika article is probably itself out of place.
Just the same, I personally have no hardened objection to the "See also" link. It is borderline plausible as something that readers might also want to see, when they're thinking about this sort of thing. --Trovatore (talk) 00:24, 24 July 2015 (UTC)
I agree that the analogy is inexact. However, I think the vital point of connection between the topics concerns not infinity vs finity, but rather that something can be carved up and then reassembled into something strictly larger. That the Banach-Tarski paradox obtains even with a finite number of pieces, however, would be a successful rebuttal to the Vaisheshika argument. Unfortunately, including this in either article would probably violate WP:OR, but I think the links unambiguously improve Wikipedia. Certainly, including the links allows individual, exploratory readers to make up their own mind about these tangentially related topics. 0nlyth3truth (talk) 01:52, 24 July 2015 (UTC)

## Banach–Tarski paradox and degenerate canceling of black hole quarks

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)

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)

What up

 Whack! You've been whacked with a wet trout.Don't take this too seriously. Someone just wants to let you know you did something silly.

## Is this a joke theorem?

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)

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)
There will be people more than happy to address your concerns at the refdesk. --Trovatore (talk) 23:54, 18 May 2016 (UTC)
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)

## brief explanation for pop audience

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)

Something along these lines could be useful if carefully sourced and worded. I thought there might already be something in the article, but I couldn't find it.
I say "carefully worded" because it's possible to overstate how sure we are that something "can't" be done physically; though in this case it might seem an abstract worry, it's something I'd want to be careful about. The direction I'd go is to point out that we don't have any physical way of cutting objects into pieces with arbitrary fineness. No one has a knife sharp enough to cut an electron in half. But I don't have a wording I'm comfortable with, and sourcing it to a good source (not a popularization) might be a challenge. --Trovatore (talk) 19:45, 19 May 2016 (UTC)
yes, I don't personally have any idea whether this kind of reassembly is even theoretically possible with unrealistically super advanced technology...JBL suggested it's not in that reference desk thread...but whatever is true as far as this goes would be worthwhile adding to the lead in this article...68.48.241.158 (talk) 16:01, 20 May 2016 (UTC)
Here are a couple of books asserting the physical impossibility or absurdity of these decompositions. They look reliable enough. --Mark viking (talk) 17:49, 20 May 2016 (UTC)
yes, perhaps..again, I'm not particularly comfortable crafting any addition to the article along these lines myself as do not have any technical expertise in this particular area..68.48.241.158 (talk) 15:29, 21 May 2016 (UTC)
The physical impossibility does not begin with the paradoxical decompositions. The mathematical model of 3D objects as a set of uncountably infinite dimensionless points is already not real-world accurate. Hence, no physical banach tarski, no fractals, cantor dust, etc. — Preceding unsigned comment added by Damluk (talkcontribs) 20:49, 26 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)

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)
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)

## Cosmology

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)

## Even more balls?

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)

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)

## Picture of F_2 is labelled incorrectly and confusingly

In the picture of F_2, the free group on two letters, you have the label S(${\displaystyle a^{-1}}$) on the northwest side; the label aS(${\displaystyle a^{-1}}$) 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(${\displaystyle a^{-1}}$)

South: S(${\displaystyle b^{-1}}$)

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)

I agree that the figure in the article is hard to interpret, but it isn't actually incorrect. The three labels aren't meant to label quadrants at all, but rather colors. Since the locations of the labels aren't intended to carry meaning, it is confusing to have them in different places. It would be clearer, I think, to have them all together in one location. Another confusing aspect is that the dots are small, which makes the colors hard to see unless one enlarges the figure. A third confusing aspect is that what look like red dots are actually red and blue—each red dot has a blue ring around it, but the blue is hard to see, even when the figure is enlarged.
The file you link to, "Paradoxical decomposition F2.svg", and the file in the article, "Paradoxical decomposition F 2.svg" (note the extra space between "F" and "2"), follow different conventions about the meaning of edges in the Cayley diagram. In the file you link to, edges represent right multiplication: edges directed to the right represent right multiplication by ${\displaystyle a}$, edges directed to the left represent right multiplication by ${\displaystyle a^{-1}}$, edges directed upwards represent right multiplication by ${\displaystyle b}$, and edges directed downwards represent right multiplication by ${\displaystyle b^{-1}}$. Hence membership of a group element in one of the sets ${\displaystyle S(a)}$, ${\displaystyle S(a^{-1})}$, ${\displaystyle S(b)}$, ${\displaystyle S(b^{-1})}$ is determined by the first edge in the path leading from the vertex ${\displaystyle e}$ to the vertex representing the group element. In the paradoxical decomposition, the set ${\displaystyle S(a)}$ consists of elements whose first edge is directed to the right; the set ${\displaystyle aS(a^{-1})}$ consists of elements whose first edge is directed up, down, or to the left; ${\displaystyle S(a^{-1})}$ is the subset of ${\displaystyle aS(a^{-1})}$ consisting of elements whose first edge is directed to the left.
In the file in the article, steps represent left multiplication: edges directed to the right represent left multiplication by ${\displaystyle a}$, edges directed to the left represent left multiplication by ${\displaystyle a^{-1}}$, edges directed upwards represent left multiplication by ${\displaystyle b}$, and edges directed downwards represent left multiplication by ${\displaystyle b^{-1}}$. Hence membership of a group element in one of the sets ${\displaystyle S(a)}$, ${\displaystyle S(a^{-1})}$, ${\displaystyle S(b)}$, ${\displaystyle S(b^{-1})}$ is determined by the last edge in the path leading from the vertex ${\displaystyle e}$ to the vertex representing the group element. In the paradoxical decomposition, the set ${\displaystyle S(a)}$ consists of elements whose last edge is directed to the right (these are the green dots); the set ${\displaystyle aS(a^{-1})}$ consists of elements whose last edge is directed up, down, or to the left (these are the red and blue dots); ${\displaystyle S(a^{-1})}$ is the subset of ${\displaystyle aS(a^{-1})}$ consisting of elements whose last edge is directed to the left (these are the red dots). The blue ring around the red dots is meant to indicate that elements of ${\displaystyle S(a^{-1})}$ are also elements of ${\displaystyle aS(a^{-1})}$.
I certainly find the file you link to to be much more easily interpreted, and am not sure why the choice was made to use the other file. Will Orrick (talk) 07:17, 23 April 2017 (UTC)
The file you link to and a related animation were removed from the article and replaced with the current figure in November 2014 in response to an earlier discussion. I wonder if there might be way of addressing the concerns raised in that discussion while still using the more intuitive figures. I don't have time right now to give it serious thought, unfortunately. Will Orrick (talk) 16:11, 23 April 2017 (UTC)
The old F2 graphic is
• easy interpretable with respect to a paradoxical decomposition of G-operations.
• hard interpretable with respect to a paradoxical decomposition of an orbit of some point.

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 ${\displaystyle e}$ label is) of the graph being a point in 3D space. Call this center point ${\displaystyle p}$. Rotate it "to the left" by ${\displaystyle a^{-1}}$ multiplication and you get the red-blueish point at the ${\displaystyle a^{-1}}$-label. Rotate it further by anything but ${\displaystyle a}$. You will get to another point in the left quadrant. The whole left quadrant is the set of all points that originate from ${\displaystyle p}$ with ${\displaystyle a^{-1}}$ as first rotation. But the left quadrant itself is meaningless to the paradox. If we rotate it "to the right" by ${\displaystyle a}$, we only get ${\displaystyle p}$ as a new point. However, if we look at the set of all red dots, i.e. ${\displaystyle S(a^{-1})\cdot p}$, 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 (talkcontribs) 16:28, 4 May 2017 (UTC)

I think the presentation would be easier to follow if we made it so that the quadrants were relevant to the paradoxical decomposition. My view is that the old F2 graphic was fine, and that the little animation that accompanied it should be restored as well. Perhaps we need to label several more points in the old graphic so that it is clear what the quadrants actually represent. At present, only a couple of points one step away from e are labeled, which leaves it ambiguous whether points two steps from e are obtained by left multiplication or right multiplication. If we go one step right to a and then one step up, and put the label ab on that point, it will be clear that right multiplication is meant. Then everything is fine, since b will be the first rotation applied to points of M, followed by a. That way it will be clear that the right quadrant contains group elements in which the last rotation to be applied to M is a. I can't see any downside. Will Orrick (talk) 14:28, 6 May 2017 (UTC)
The downside is, that rotation of 3D points does not follow the graph's edges. Consider the center point being rotated by abb. Where would we put that new point in the graphic? Apparently in the right quadrant, right-up-up. Now, if we rotate this point by ${\displaystyle a^{-1}}$, the new point's position is represented by ${\displaystyle bb}$, i.e. up-up. It is confusing that we have apparently used a single generator to rotate a point, yet the new point is not connected by an edge to it, it isn't even in the same quadrant. — Preceding unsigned comment added by Damluk (talkcontribs) 16:49, 6 May 2017 (UTC)
I see where you're coming from now, and hadn't really considered things from that point of view (wanting rotations to correspond to edges in the Cayley graph). I still do not see that as essential, however.
Any Cayley graph will privilege certain actions of generators over others. In this instance, one can have either left multiplication by generators or right multiplication by generators, and one has to decide.
Cayley
Although left multiplication does not correspond to moving along an edge in the old graphic, it does correspond to the readily visualizable move shown in the graphic at right, which makes use of the self-similarity of the graph. The analogous move in your example of multiplying abb on the left by a−1 would move the point that lies two steps up in the upward-pointing subtree of the right quadrant to the point two steps up in the upward-pointing subtree of the main figure. There would be an similar move for any point in the right quadrant, taking a point connected to a by a certain sequence of steps to the point connected to e by the same sequence of steps. I personally find this sort of global picture of the effect of left multiplication to be very helpful in developing an intuitive understanding of the paradoxical decomposition. Will Orrick (talk) 11:32, 7 May 2017 (UTC)
Ok, then we disagree here. I think that it is essential to retrace rotational movements by the means the caley graph provides. I also think that it is more important for the understanding to provide a visualisation of a paradoxical decomposition of points rather than of a paradoxical decomposition of G-actions. The gif animation is worth as much as it shows that the caley graph is self-similar. But there is no self-explanatory connection between G-actions and the gif's movements. Furthermore, these movements might create the impression of being of another quality or type than those that are encoded by the original graph's edges and vertices. — Preceding unsigned comment added by Damluk (talkcontribs) 20:03, 7 May 2017 (UTC)
Fundamentally any proof is a symbolic object, and no figure is ever essential. Different people understand things in different ways, so it's probably impossible to make general statements about whether one figure or another is more helpful to the understanding. It would, however, be nice to hear some additional opinions.
If the consensus is to stick with the current figure, I think some modifications for the sake of future readers would be helpful. Labeling some vertices beyond the nearest neighbors of the central point would clarify how the edges should be interpreted. Moving the labels to a corner of the figure would ensure that no one is tempted to interpret them as labels of quadrants. A caption explaining the color scheme would also help. It might be better to make the left half of the red/blue dots blue and the right half red rather than having a red center with a blue border. That would, at the very least, make the bi-coloring more visible and emphasize the equal importance of the red and blue colors. Will Orrick (talk) 20:31, 10 May 2017 (UTC)
Since there hasn't been any movement on this, I have, as a temporary measure, added some information to the picture caption. Will Orrick (talk) 09:36, 19 June 2017 (UTC)

## Simplified explanation

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:

1. take the set of real numbers between 0 and 1 which is uncountable (cf. Cantor)
2. 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
3. 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
4. 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.
5. do the same in reverse, we can make two identical sets out of one.

Luzian (talk) 10:54, 20 April 2017 (UTC)

Point is that the transformations are rigid, ie distance preserving.

189.223.226.82 (talk) 20:44, 22 April 2017 (UTC)

## Split the first formula in the section "Formal treatment" into three?

I think the formula

${\displaystyle A=\bigcup _{i=1}^{k}A_{i},\quad B=\bigcup _{i=1}^{k}B_{i},\quad A_{i}\cap A_{j}=\emptyset ,}$
${\displaystyle \quad B_{i}\cap B_{j}=\emptyset ,\quad 1\leq i for all,
${\displaystyle \quad 1\leq i\leq k,\quad g_{i}\in G,}$

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:

1. ${\displaystyle A=\bigcup _{i=1}^{k}A_{i},\ B=\bigcup _{i=1}^{k}B_{i}}$
2. ${\displaystyle A_{i}\cap A_{j}=B_{i}\cap B_{j}=\emptyset }$, for all ${\displaystyle 1\leq i
3. ${\displaystyle g_{i}(A_{i})=B_{i},\ g_{i}\in G}$, for all ${\displaystyle 1\leq i\leq k}$

or:

1. ${\displaystyle A=\bigcup _{i=1}^{k}A_{i}\land B=\bigcup _{i=1}^{k}B_{i}}$
2. ${\displaystyle \forall i,j}$ with ${\displaystyle 1\leq i
3. ${\displaystyle \forall i}$ with ${\displaystyle 1\leq i\leq k:\exists g_{i}\in G:g_{i}(A_{i})=B_{i}}$

These show 3 simple conditions that the subsets of ${\displaystyle A}$ and ${\displaystyle B}$ have to meet, to be ${\displaystyle G}$-equidecomposable:

1. They exactly cover their original set (${\displaystyle A}$ resp. ${\displaystyle B}$).
2. They are pairwise disjoint.
3. ${\displaystyle A_{i}}$ and ${\displaystyle B_{i}}$ 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 "${\displaystyle 1\leq i"?

Sven.st (talk) 11:50, 11 July 2017 (UTC)

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)
Looks good to me. Sven.st (talk) —Preceding undated comment added 10:04, 14 July 2017 (UTC)

## Porosity

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)

Thanks for catching that. I agree, "such a large porosity" doesn't tell you anything useful about why the sets are non-measurable, with or without the link. --Trovatore (talk) 17:19, 26 June 2018 (UTC)
According to the article a Porous set in ${\displaystyle \mathbb {R} ^{3}}$ has measure zero, so you were right to remove the statement.