To install click the Add extension button. That's it.

The source code for the WIKI 2 extension is being checked by specialists of the Mozilla Foundation, Google, and Apple. You could also do it yourself at any point in time.

Kelly Slayton
Congratulations on this excellent venture… what a great idea!
Alexander Grigorievskiy
I use WIKI 2 every day and almost forgot how the original Wikipedia looks like.
What we do. Every page goes through several hundred of perfecting techniques; in live mode. Quite the same Wikipedia. Just better.

Banach–Tarski paradox

From Wikipedia, the free encyclopedia

"Can a ball be decomposed into a finite number of point sets and reassembled into two balls identical to the original?"

The Banach–Tarski paradox is a theorem in set-theoretic geometry, which states the following: Given a solid ball in three-dimensional space, there exists a decomposition of the ball into a finite number of disjoint subsets, which can then be put back together in a different way to yield two identical copies of the original ball. Indeed, the reassembly process involves only moving the pieces around and rotating them without changing their shape. However, the pieces themselves are not "solids" in the usual sense, but infinite scatterings of points. The reconstruction can work with as few as five pieces.[1]

An alternative form of the theorem states that given any two "reasonable" solid objects (such as a small ball and a huge ball), the cut pieces of either one can be reassembled into the other. This is often stated informally as "a pea can be chopped up and reassembled into the Sun" and called the "pea and the Sun paradox".

The theorem is called a paradox because it contradicts basic geometric intuition. "Doubling the ball" by dividing it into parts and moving them around by rotations and translations, without any stretching, bending, or adding new points, seems to be impossible, since all these operations ought, intuitively speaking, to preserve the volume. The intuition that such operations preserve volumes is not mathematically absurd and it is even included in the formal definition of volumes. However, this is not applicable here because in this case it is impossible to define the volumes of the considered subsets. Reassembling them reproduces a set that has a volume, which happens to be different from the volume at the start.

Unlike most theorems in geometry, the mathematical proof of this result depends on the choice of axioms for set theory in a critical way. It can be proven using the axiom of choice, which allows for the construction of non-measurable sets, i.e., collections of points that do not have a volume in the ordinary sense, and whose construction requires an uncountable number of choices.[2]

It was shown in 2005 that the pieces in the decomposition can be chosen in such a way that they can be moved continuously into place without running into one another.[3]

As proved independently by Leroy[4] and Simpson,[5] the Banach–Tarski paradox does not violate volumes if one works with locales rather than topological spaces. In this abstract setting, it is possible to have subspaces without point but still nonempty. The parts of the paradoxical decomposition do intersect a lot in the sense of locales, so much that some of these intersections should be given a positive mass. Allowing for this hidden mass to be taken into account, the theory of locales permits all subsets (and even all sublocales) of the Euclidean space to be satisfactorily measured.

YouTube Encyclopedic

  • 1/5
    41 655 854
    7 953
    18 392
    241 681
    13 798
  • The Banach–Tarski Paradox
  • The Banach-Tarski paradox
  • The Banach-Tarski paradox
  • Infinity shapeshifter vs. Banach-Tarski paradox
  • The Banach Tarski paradox - is it nonsense? | Sociology and Pure Mathematics | N J Wildberger


Hey, Vsauce. Michael here. There's a famous way to seemingly create chocolate out of nothing. Maybe you've seen it before. This chocolate bar is 4 squares by 8 squares, but if you cut it like this and then like this and finally like this you can rearrange the pieces like so and wind up with the same 4 by 8 bar but with a leftover piece, apparently created out of thin air. There's a popular animation of this illusion as well. I call it an illusion because it's just that. Fake. In reality, the final bar is a bit smaller. It contains this much less chocolate. Each square along the cut is shorter than it was in the original, but the cut makes it difficult to notice right away. The animation is extra misleading, because it tries to cover up its deception. The lost height of each square is surreptitiously added in while the piece moves to make it hard to notice. I mean, come on, obviously you cannot cut up a chocolate bar and rearrange the pieces into more than you started with. Or can you? One of the strangest theorems in modern mathematics is the Banach-Tarski paradox. It proves that there is, in fact, a way to take an object and separate it into 5 different pieces. And then, with those five pieces, simply rearrange them. No stretching required into two exact copies of the original item. Same density, same size, same everything. Seriously. To dive into the mind blow that it is and the way it fundamentally questions math and ourselves, we have to start by asking a few questions. First, what is infinity? A number? I mean, it's nowhere on the number line, but we often say things like there's an infinite "number" of blah-blah-blah. And as far as we know, infinity could be real. The universe may be infinite in size and flat, extending out for ever and ever without end, beyond even the part we can observe or ever hope to observe. That's exactly what infinity is. Not a number per se, but rather a size. The size of something that doesn't end. Infinity is not the biggest number, instead, it is how many numbers there are. But there are different sizes of infinity. The smallest type of infinity is countable infinity. The number of hours in forever. It's also the number of whole numbers that there are, natural number, the numbers we use when counting things, like 1, 2, 3, 4, 5, 6 and so on. Sets like these are unending, but they are countable. Countable means that you can count them from one element to any other in a finite amount of time, even if that finite amount of time is longer than you will live or the universe will exist for, it's still finite. Uncountable infinity, on the other hand, is literally bigger. Too big to even count. The number of real numbers that there are, not just whole numbers, but all numbers is uncountably infinite. You literally cannot count even from 0 to 1 in a finite amount of time by naming every real number in between. I mean, where do you even start? Zero, okay. But what comes next? 0.000000... Eventually, we would imagine a 1 going somewhere at the end, but there is no end. We could always add another 0. Uncountability makes this set so much larger than the set of all whole numbers that even between 0 and 1, there are more numbers than there are whole numbers on the entire endless number line. Georg Cantor's famous diagonal argument helps illustrate this. Imagine listing every number between zero and one. Since they are uncountable and can't be listed in order, let's imagine randomly generating them forever with no repeats. Each number regenerate can be paired with a whole number. If there's a one to one correspondence between the two, that is if we can match one whole number to each real number on our list, that would mean that countable and uncountable sets are the same size. But we can't do that, even though this list goes on for ever. Forever isn't enough. Watch this. If we go diagonally down our endless list of real numbers and take the first decimal of the first number and the second of the second number, the third of the third and so on and add one to each, subtracting one if it happens to be a nine, we can generate a new real number that is obviously between 0 and 1, but since we've defined it to be different from every number on our endless list and at least one place it's clearly not contained in the list. In other words, we've used up every single whole number, the entire infinity of them and yet we can still come up with more real numbers. Here's something else that is true but counter-intuitive. There are the same number of even numbers as there are even and odd numbers. At first, that sounds ridiculous. Clearly, there are only half as many even numbers as all whole numbers, but that intuition is wrong. The set of all whole numbers is denser but every even number can be matched with a whole number. You will never run out of members either set, so this one to one correspondence shows that both sets are the same size. In other words, infinity divided by two is still infinity. Infinity plus one is also infinity. A good illustration of this is Hilbert's paradox up the Grand Hotel. Imagine a hotel with a countably infinite number of rooms. But now, imagine that there is a person booked into every single room. Seemingly, it's fully booked, right? No. Infinite sets go against common sense. You see, if a new guest shows up and wants a room, all the hotel has to do is move the guest in room number 1 to room number 2. And a guest in room 2 to room 3 and 3 to 4 and 4 to 5 and so on. Because the number of rooms is never ending we cannot run out of rooms. Infinity -1 is also infinity again. If one guest leaves the hotel, we can shift every guest the other way. Guest 2 goes to room 1, 3 to 2, 4 to 3 and so on, because we have an infinite amount of guests. That is a never ending supply of them. No room will be left empty. As it turns out, you can subtract any finite number from infinity and still be left with infinity. It doesn't care. It's unending. Banach-Tarski hasn't left our sights yet. All of this is related. We are now ready to move on to shapes. Hilbert's hotel can be applied to a circle. Points around the circumference can be thought of as guests. If we remove one point from the circle that point is gone, right? Infinity tells us it doesn't matter. The circumference of a circle is irrational. It's the radius times 2Pi. So, if we mark off points beginning from the whole, every radius length along the circumference going clockwise we will never land on the same point twice, ever. We can count off each point we mark with a whole number. So this set is never-ending, but countable, just like guests and rooms in Hilbert's hotel. And like those guests, even though one has checked out, we can just shift the rest. Move them counterclockwise and every room will be filled Point 1 moves to fill in the hole, point 2 fills in the place where point 1 used to be, 3 fills in 2 and so on. Since we have a unending supply of numbered points, no hole will be left unfilled. The missing point is forgotten. We apparently never needed it to be complete. There's one last needo consequence of infinity we should discuss before tackling Banach-Tarski. Ian Stewart famously proposed a brilliant dictionary. One that he called the Hyperwebster. The Hyperwebster lists every single possible word of any length formed from the 26 letters in the English alphabet. It begins with "a," followed by "aa," then "aaa," then "aaaa." And after an infinite number of those, "ab," then "aba," then "abaa", "abaaa," and so on until "z, "za," "zaa," et cetera, et cetera, until the final entry in infinite sequence of "z"s. Such a dictionary would contain every single word. Every single thought, definition, description, truth, lie, name, story. What happened to Amelia Earhart would be in that dictionary, as well as every single thing that didn't happened to Amelia Earhart. Everything that could be said using our alphabet. Obviously, it would be huge, but the company publishing it might realize that they could take a shortcut. If they put all the words that begin with a in a volume titled "A," they wouldn't have to print the initial "a." Readers would know to just add the "a," because it's the "a" volume. By removing the initial "a," the publisher is left with every "a" word sans the first "a," which has surprisingly become every possible word. Just one of the 26 volumes has been decomposed into the entire thing. It is now that we're ready to investigate this video's titular paradox. What if we turned an object, a 3D thing into a Hyperwebster? Could we decompose pieces of it into the whole thing? Yes. The first thing we need to do is give every single point on the surface of the sphere one name and one name only. A good way to do this is to name them after how they can be reached by a given starting point. If we move this starting point across the surface of the sphere in steps that are just the right length, no matter how many times or in what direction we rotate, so long as we never backtrack, it will never wind up in the same place twice. We only need to rotate in four directions to achieve this paradox. Up, down, left and right around two perpendicular axes. We are going to need every single possible sequence that can be made of any finite length out of just these four rotations. That means we will need lef, right, up and down as well as left left, left up, left down, but of course not left right, because, well, that's backtracking. Going left and then right means you're the same as you were before you did anything, so no left rights, no right lefts and no up downs and no down ups. Also notice that I'm writing the rotations in order right to left, so the final rotation is the leftmost letter. That will be important later on. Anyway. A list of all possible sequences of allowed rotations that are finite in lenght is, well, huge. Countably infinite, in fact. But if we apply each one of them to a starting point in green here and then name the point we land on after the sequence that brought us there, we can name a countably infinite set of points on the surface. Let's look at how, say, these four strings on our list would work. Right up left. Okay, rotating the starting point this way takes us here. Let's colour code the point based on the final rotation in its string, in this case it's left and for that we will use purple. Next up down down. That sequence takes us here. We name the point DD and color it blue, since we ended with a down rotation. RDR, that will be this point's name, takes us here. And for a final right rotation, let's use red. Finally, for a sequence that end with up, let's colour code the point orange. Now, if we imagine completing this process for every single sequence, we will have a countably infinite number of points named and color-coded. That's great, but not enough. There are an uncountably infinite number of points on a sphere's surface. But no worries, we can just pick a point we missed. Any point and color it green, making it a new starting point and then run every sequence from here. After doing this to an uncountably infinite number of starting point we will have indeed named and colored every single point on the surface just once. With the exception of poles. Every sequence has two poles of rotation. Locations on the sphere that come back to exactly where they started. For any sequence of right or left rotations, the polls are the north and south poles. The problem with poles like these is that more than one sequence can lead us to them. They can be named more than once and be colored in more than one color. For example, if you follow some other sequence to the north or south pole, any subsequent rights or lefts will be equally valid names. In order to deal with this we're going to just count them out of the normal scheme and color them all yellow. Every sequence has two, so there are a countably infinite amount of them. Now, with every point on the sphere given just one name and just one of six colors, we are ready to take the entire sphere apart. Every point on the surface corresponds to a unique line of points below it all the way to the center point. And we will be dragging every point's line along with it. The lone center point we will set aside. Okay, first we cut out and extract all the yellow poles, the green starting points, the orange up points, the blue down points and the red and purple left and right points. That's the entire sphere. With just these pieces you could build the whole thing. But take a look at the left piece. It is defined by being a piece composed of every point, accessed via a sequence ending with a left rotation. If we rotate this piece right, that's the same as adding an "R" to every point's name. But left and then right is a backtrack, they cancel each other out. And look what happens when you reduce them away. The set becomes the same as a set of all points with names that end with L, but also U, D and every point reached with no rotation. That's the full set of starting points. We have turned less than a quarter of the sphere into nearly three-quarters just by rotating it. We added nothing. It's like the Hyperwebster. If we had the right piece and the poles of rotation and the center point, well, we've got the entire sphere again, but with stuff left over. To make a second copy, let's rotate the up piece down. The down ups cancel because, well, it's the same as going nowhere and we're left with a set of all starting points, the entire up piece, the right piece and the left piece, but there's a problem here. We don't need this extra set of starting points. We still haven't used the original ones. No worries, let's just start over. We can just move everything from the up piece that turns into a starting point when rotated down. That means every point whose final rotation is up. Let's put them in the piece. Of course, after rotating points named UU will just turn into points named U, and that would give us a copy here and here. So, as it turns out, we need to move all points with any name that is just a string of Us. We will put them in the down piece and rotate the up piece down, which makes it congruent to the up right and left pieces, add in the down piece along with some up and the starting point piece and, well, we're almost done. The poles of rotation and center are missing from this copy, but no worries. There's a countably infinite number of holes, where the poles of rotations used to be, which means there is some pole around which we can rotate this sphere such that every pole hole orbits around without hitting another. Well, this is just a bunch of circles with one point missing. We fill them each like we did earlier. And we do the same for the centerpoint. Imagine a circle that contains it inside the sphere and just fill in from infinity and look what we've done. We have taken one sphere and turned it into two identical spheres without adding anything. One plus one equals 1. That took a while to go through, but the implications are huge. And mathematicians, scientists and philosophers are still debating them. Could such a process happen in the real world? I mean, it can happen mathematically and math allows us to abstractly predict and describe a lot of things in the real world with amazing accuracy, but does the Banach-Tarski paradox take it too far? Is it a place where math and physics separate? We still don't know. History is full of examples of mathematical concepts developed in the abstract that we did not think would ever apply to the real world for years, decades, centuries, until eventually science caught up and realized they were totally applicable and useful. The Banach-Tarski paradox could actually happen in our real-world, the only catch of course is that the five pieces you cut your object into aren't simple shapes. They must be infinitely complex and detailed. That's not possible to do in the real world, where measurements can only get so small and there's only a finite amount of time to do anything, but math says it's theoretically valid and some scientists think it may be physically valid too. There have been a number of papers published suggesting a link between by Banach-Tarski and the way tiny tiny sub-atomic particles can collide at high energies and turn into more particles than we began with. We are finite creatures. Our lives are small and can only scientifically consider a small part of reality. What's common for us is just a sliver of what's available. We can only see so much of the electromagnetic spectrum. We can only delve so deep into extensions of space. Common sense applies to that which we can access. But common sense is just that. Common. If total sense is what we want, we should be prepared to accept that we shouldn't call infinity weird or strange. The results we've arrived at by accepting it are valid, true within the system we use to understand, measure, predict and order the universe. Perhaps the system still needs perfecting, but at the end of day, history continues to show us that the universe isn't strange. We are. And as always, thanks for watching. Finally, as always, the description is full of links to learn more. There are also a number of books linked down there that really helped me wrap my mind kinda around Banach-Tarski. First of all, Leonard Wapner's "The Pea and the Sun." This book is fantastic and it's full of lot of the preliminaries needed to understand the proof that comes later. He also talks a lot about the ramifications of what Banach-Tarski and their theorem might mean for mathematics. Also, if you wanna talk about math and whether it's discovered or invented, whether it really truly will map onto the universe, Yanofsky's "The Outer Limits of Reason" is great. This is the favorite book of mine that I've read this entire year. Another good one is E. Brian Davies' "Why Beliefs Matter." This is actually Corn's favorite book, as you might be able to see there. It's delicious and full of lots of great information about the limits of what we can know and what science is and what mathematics is. If you love infinity and math, I cannot more highly recommend Matt Parker's "Things to Make and Do in the Fourth Dimension." He's hilarious and this book is very very great at explaining some pretty awesome things. So keep reading, and if you're looking for something to watch, I hope you've already watched Kevin Lieber's film on Field Day. I already did a documentary about Whittier, Alaska over there. Kevin's got a great short film about putting things out on the Internet and having people react to them. There's a rumor that Jake Roper might be doing something on Field Day soon. So check out mine, check out Kevin's and subscribe to Field Day for upcoming Jake Roper action, yeah? He's actually in this room right now, say hi, Jake. [Jake:] Hi. Thanks for filming this, by the way. Guys, I really appreciate who you all are. And as always, thanks for watching.

Banach and Tarski publication

In a paper published in 1924,[6] Stefan Banach and Alfred Tarski gave a construction of such a paradoxical decomposition, based on earlier work by Giuseppe Vitali concerning the unit interval and on the paradoxical decompositions of the sphere by Felix Hausdorff, and discussed a number of related questions concerning decompositions of subsets of Euclidean spaces in various dimensions. They proved the following more general statement, the strong form of the Banach–Tarski paradox:

Given any two bounded subsets A and B of a Euclidean space in at least three dimensions, both of which have a nonempty interior, there are partitions of A and B into a finite number of disjoint subsets, , (for some integer k), such that for each (integer) i between 1 and k, the sets Ai and Bi are congruent.

Now let A be the original ball and B be the union of two translated copies of the original ball. Then the proposition means that you can divide the original ball A into a certain number of pieces and then rotate and translate these pieces in such a way that the result is the whole set B, which contains two copies of A.

The strong form of the Banach–Tarski paradox is false in dimensions one and two, but Banach and Tarski showed that an analogous statement remains true if countably many subsets are allowed. The difference between dimensions 1 and 2 on the one hand, and 3 and higher on the other hand, is due to the richer structure of the group E(n) of Euclidean motions in 3 dimensions. For n = 1, 2 the group is solvable, but for n ≥ 3 it contains a free group with two generators. John von Neumann studied the properties of the group of equivalences that make a paradoxical decomposition possible, and introduced the notion of amenable groups. He also found a form of the paradox in the plane which uses area-preserving affine transformations in place of the usual congruences.

Tarski proved that amenable groups are precisely those for which no paradoxical decompositions exist. Since only free subgroups are needed in the Banach–Tarski paradox, this led to the long-standing von Neumann conjecture, which was disproved in 1980.

Formal treatment

The Banach–Tarski paradox states that a ball in the ordinary Euclidean space can be doubled using only the operations of partitioning into subsets, replacing a set with a congruent set, and reassembling. Its mathematical structure is greatly elucidated by emphasizing the role played by the group of Euclidean motions and introducing the notions of equidecomposable sets and a paradoxical set. Suppose that G is a group acting on a set X. In the most important special case, X is an n-dimensional Euclidean space (for integral n), and G consists of all isometries of X, i.e. the transformations of X into itself that preserve the distances, usually denoted E(n). Two geometric figures that can be transformed into each other are called congruent, and this terminology will be extended to the general G-action. Two subsets A and B of X are called G-equidecomposable, or equidecomposable with respect to G, if A and B can be partitioned into the same finite number of respectively G-congruent pieces. This defines an equivalence relation among all subsets of X. Formally, if there exist non-empty sets , such that

and there exist elements such that

then it can be said that A and B are G-equidecomposable using k pieces. If a set E has two disjoint subsets A and B such that A and E, as well as B and E, are G-equidecomposable, then E is called paradoxical.

Using this terminology, the Banach–Tarski paradox can be reformulated as follows:

A three-dimensional Euclidean ball is equidecomposable with two copies of itself.

In fact, there is a sharp result in this case, due to Raphael M. Robinson:[7] doubling the ball can be accomplished with five pieces, and fewer than five pieces will not suffice.

The strong version of the paradox claims:

Any two bounded subsets of 3-dimensional Euclidean space with non-empty interiors are equidecomposable.

While apparently more general, this statement is derived in a simple way from the doubling of a ball by using a generalization of the Bernstein–Schroeder theorem due to Banach that implies that if A is equidecomposable with a subset of B and B is equidecomposable with a subset of A, then A and B are equidecomposable.

The Banach–Tarski paradox can be put in context by pointing out that for two sets in the strong form of the paradox, there is always a bijective function that can map the points in one shape into the other in a one-to-one fashion. In the language of Georg Cantor's set theory, these two sets have equal cardinality. Thus, if one enlarges the group to allow arbitrary bijections of X, then all sets with non-empty interior become congruent. Likewise, one ball can be made into a larger or smaller ball by stretching, or in other words, by applying similarity transformations. Hence, if the group G is large enough, G-equidecomposable sets may be found whose "size"s vary. Moreover, since a countable set can be made into two copies of itself, one might expect that using countably many pieces could somehow do the trick.

On the other hand, in the Banach–Tarski paradox, the number of pieces is finite and the allowed equivalences are Euclidean congruences, which preserve the volumes. Yet, somehow, they end up doubling the volume of the ball! While this is certainly surprising, some of the pieces used in the paradoxical decomposition are non-measurable sets, so the notion of volume (more precisely, Lebesgue measure) is not defined for them, and the partitioning cannot be accomplished in a practical way. In fact, the Banach–Tarski paradox demonstrates that it is impossible to find a finitely-additive measure (or a Banach measure) defined on all subsets of a Euclidean space of three (and greater) dimensions that is invariant with respect to Euclidean motions and takes the value one on a unit cube. In his later work, Tarski showed that, conversely, non-existence of paradoxical decompositions of this type implies the existence of a finitely-additive invariant measure.

The heart of the proof of the "doubling the ball" form of the paradox presented below is the remarkable fact that by a Euclidean isometry (and renaming of elements), one can divide a certain set (essentially, the surface of a unit sphere) into four parts, then rotate one of them to become itself plus two of the other parts. This follows rather easily from a F2-paradoxical decomposition of F2, the free group with two generators. Banach and Tarski's proof relied on an analogous fact discovered by Hausdorff some years earlier: the surface of a unit sphere in space is a disjoint union of three sets B, C, D and a countable set E such that, on the one hand, B, C, D are pairwise congruent, and on the other hand, B is congruent with the union of C and D. This is often called the Hausdorff paradox.

Connection with earlier work and the role of the axiom of choice

Banach and Tarski explicitly acknowledge Giuseppe Vitali's 1905 construction of the set bearing his name, Hausdorff's paradox (1914), and an earlier (1923) paper of Banach as the precursors to their work. Vitali's and Hausdorff's constructions depend on Zermelo's axiom of choice ("AC"), which is also crucial to the Banach–Tarski paper, both for proving their paradox and for the proof of another result:

Two Euclidean polygons, one of which strictly contains the other, are not equidecomposable.

They remark:

Le rôle que joue cet axiome dans nos raisonnements nous semble mériter l'attention
(The role this axiom plays in our reasoning seems to us to deserve attention)

They point out that while the second result fully agrees with geometric intuition, its proof uses AC in an even more substantial way than the proof of the paradox. Thus Banach and Tarski imply that AC should not be rejected solely because it produces a paradoxical decomposition, for such an argument also undermines proofs of geometrically intuitive statements.

However, in 1949, A. P. Morse showed that the statement about Euclidean polygons can be proved in ZF set theory and thus does not require the axiom of choice. In 1964, Paul Cohen proved that the axiom of choice is independent from ZF – that is, it cannot be proved from ZF. A weaker version of an axiom of choice is the axiom of dependent choice, DC, and it has been shown that DC is not sufficient for proving the Banach–Tarski paradox, that is,

The Banach–Tarski paradox is not a theorem of ZF, nor of ZF+DC.[8]

Large amounts of mathematics use AC. As Stan Wagon points out at the end of his monograph, the Banach–Tarski paradox has been more significant for its role in pure mathematics than for foundational questions: it motivated a fruitful new direction for research, the amenability of groups, which has nothing to do with the foundational questions.

In 1991, using then-recent results by Matthew Foreman and Friedrich Wehrung,[9] Janusz Pawlikowski proved that the Banach–Tarski paradox follows from ZF plus the Hahn–Banach theorem.[10] The Hahn–Banach theorem does not rely on the full axiom of choice but can be proved using a weaker version of AC called the ultrafilter lemma. So Pawlikowski proved that the set theory needed to prove the Banach–Tarski paradox, while stronger than ZF, is weaker than full ZFC.

A sketch of the proof

Here a proof is sketched which is similar but not identical to that given by Banach and Tarski. Essentially, the paradoxical decomposition of the ball is achieved in four steps:

  1. Find a paradoxical decomposition of the free group in two generators.
  2. Find a group of rotations in 3-d space isomorphic to the free group in two generators.
  3. Use the paradoxical decomposition of that group and the axiom of choice to produce a paradoxical decomposition of the hollow unit sphere.
  4. Extend this decomposition of the sphere to a decomposition of the solid unit ball.

These steps are discussed in more detail below.

Step 1

Cayley graph of F2, showing decomposition into the sets S(a) and aS(a−1). Traversing a horizontal edge of the graph in the rightward direction represents left multiplication of an element of F2 by a; traversing a vertical edge of the graph in the upward direction represents left multiplication of an element of F2 by b. Elements of the set S(a) are green dots; elements of the set aS(a−1) are blue dots or red dots with blue border. Red dots with blue border are elements of S(a−1), which is a subset of aS(a−1).

The free group with two generators a and b consists of all finite strings that can be formed from the four symbols a, a−1, b and b−1 such that no a appears directly next to an a−1 and no b appears directly next to a b−1. Two such strings can be concatenated and converted into a string of this type by repeatedly replacing the "forbidden" substrings with the empty string. For instance: abab−1a−1 concatenated with abab−1a yields abab−1a−1abab−1a, which contains the substring a−1a, and so gets reduced to abab−1bab−1a, which contains the substring b−1b, which gets reduced to abaab−1a. One can check that the set of those strings with this operation forms a group with identity element the empty string e. This group may be called F2.

The group can be "paradoxically decomposed" as follows: Let S(a) be the subset of consisting of all strings that start with a, and define S(a−1), S(b) and S(b−1) similarly. Clearly,

but also


where the notation aS(a−1) means take all the strings in S(a−1) and concatenate them on the left with a.

This is at the core of the proof. For example, there may be a string in the set which, because of the rule that must not appear next to , reduces to the string . Similarly, contains all the strings that start with (for example, the string which reduces to ). In this way, contains all the strings that start with , and , as well as the empty string .

Group F2 has been cut into four pieces (plus the singleton {e}), then two of them "shifted" by multiplying with a or b, then "reassembled" as two pieces to make one copy of and the other two to make another copy of . That is exactly what is intended to do to the ball.

Step 2

In order to find a free group of rotations of 3D space, i.e. that behaves just like (or "is isomorphic to") the free group F2, two orthogonal axes are taken (e.g. the x and z axes). Then, A is taken to be a rotation of about the x axis, and B to be a rotation of about the z axis (there are many other suitable pairs of irrational multiples of π that could be used here as well).[11]

The group of rotations generated by A and B will be called H. Let be an element of H that starts with a positive rotation about the z axis, that is, an element of the form with . It can be shown by induction that maps the point to , for some . Analyzing and modulo 3, one can show that . The same argument repeated (by symmetry of the problem) is valid when starts with a negative rotation about the z axis, or a rotation about the x axis. This shows that if is given by a non-trivial word in A and B, then . Therefore, the group H is a free group, isomorphic to F2.

The two rotations behave just like the elements a and b in the group F2: there is now a paradoxical decomposition of H.

This step cannot be performed in two dimensions since it involves rotations in three dimensions. If two rotations are taken about the same axis, the resulting group is the abelian circle group and does not have the property required in step 1.

An alternate arithmetic proof of the existence of free groups in some special orthogonal groups using integral quaternions leads to paradoxical decompositions of the rotation group.[12]

Step 3

The unit sphere S2 is partitioned into orbits by the action of our group H: two points belong to the same orbit if and only if there is a rotation in H which moves the first point into the second. (Note that the orbit of a point is a dense set in S2.) The axiom of choice can be used to pick exactly one point from every orbit; collect these points into a set M. The action of H on a given orbit is free and transitive and so each orbit can be identified with H. In other words, every point in S2 can be reached in exactly one way by applying the proper rotation from H to the proper element from M. Because of this, the paradoxical decomposition of H yields a paradoxical decomposition of S2 into four pieces A1, A2, A3, A4 as follows:

where we define

and likewise for the other sets, and where we define

(The five "paradoxical" parts of F2 were not used directly, as they would leave M as an extra piece after doubling, owing to the presence of the singleton {e}.)

The (majority of the) sphere has now been divided into four sets (each one dense on the sphere), and when two of these are rotated, the result is double of what was had before:

Step 4

Finally, connect every point on S2 with a half-open segment to the origin; the paradoxical decomposition of S2 then yields a paradoxical decomposition of the solid unit ball minus the point at the ball's center. (This center point needs a bit more care; see below.)

N.B. This sketch glosses over some details. One has to be careful about the set of points on the sphere which happen to lie on the axis of some rotation in H. However, there are only countably many such points, and like the case of the point at the center of the ball, it is possible to patch the proof to account for them all. (See below.)

Some details, fleshed out

In Step 3, the sphere was partitioned into orbits of our group H. To streamline the proof, the discussion of points that are fixed by some rotation was omitted; since the paradoxical decomposition of F2 relies on shifting certain subsets, the fact that some points are fixed might cause some trouble. Since any rotation of S2 (other than the null rotation) has exactly two fixed points, and since H, which is isomorphic to F2, is countable, there are countably many points of S2 that are fixed by some rotation in H. Denote this set of fixed points as D. Step 3 proves that S2D admits a paradoxical decomposition.

What remains to be shown is the Claim: S2D is equidecomposable with S2.

Proof. Let λ be some line through the origin that does not intersect any point in D. This is possible since D is countable. Let J be the set of angles, α, such that for some natural number n, and some P in D, r(nα)P is also in D, where r(nα) is a rotation about λ of nα. Then J is countable. So there exists an angle θ not in J. Let ρ be the rotation about λ by θ. Then ρ acts on S2 with no fixed points in D, i.e., ρn(D) is disjoint from D, and for natural m<n, ρn(D) is disjoint from ρm(D). Let E be the disjoint union of ρn(D) over n = 0, 1, 2, ... . Then S2 = E ∪ (S2E) ~ ρ(E) ∪ (S2E) = (ED) ∪ (S2E) = S2D, where ~ denotes "is equidecomposable to".

For step 4, it has already been shown that the ball minus a point admits a paradoxical decomposition; it remains to be shown that the ball minus a point is equidecomposable with the ball. Consider a circle within the ball, containing the point at the center of the ball. Using an argument like that used to prove the Claim, one can see that the full circle is equidecomposable with the circle minus the point at the ball's center. (Basically, a countable set of points on the circle can be rotated to give itself plus one more point.) Note that this involves the rotation about a point other than the origin, so the Banach–Tarski paradox involves isometries of Euclidean 3-space rather than just SO(3).

Use is made of the fact that if A ~ B and B ~ C, then A ~ C. The decomposition of A into C can be done using number of pieces equal to the product of the numbers needed for taking A into B and for taking B into C.

The proof sketched above requires 2 × 4 × 2 + 8 = 24 pieces - a factor of 2 to remove fixed points, a factor 4 from step 1, a factor 2 to recreate fixed points, and 8 for the center point of the second ball. But in step 1 when moving {e} and all strings of the form an into S(a−1), do this to all orbits except one. Move {e} of this last orbit to the center point of the second ball. This brings the total down to 16 + 1 pieces. 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.

Obtaining infinitely many balls from one

Using the Banach–Tarski paradox, it is possible to obtain k copies of a ball in the Euclidean n-space from one, for any integers n ≥ 3 and k ≥ 1, i.e. a ball can be cut into k pieces so that each of them is equidecomposable to a ball of the same size as the original. Using the fact that the free group F2 of rank 2 admits a free subgroup of countably infinite rank, a similar proof yields that the unit sphere Sn−1 can be partitioned into countably infinitely many pieces, each of which is equidecomposable (with two pieces) to the Sn−1 using rotations. By using analytic properties of the rotation group SO(n), which is a connected analytic Lie group, one can further prove that the sphere Sn−1 can be partitioned into as many pieces as there are real numbers (that is, pieces), so that each piece is equidecomposable with two pieces to Sn−1 using rotations. These results then extend to the unit ball deprived of the origin. A 2010 article by Valeriy Churkin gives a new proof of the continuous version of the Banach–Tarski paradox.[13]

Von Neumann paradox in the Euclidean plane

In the Euclidean plane, two figures that are equidecomposable with respect to the group of Euclidean motions are necessarily of the same area, and therefore, a paradoxical decomposition of a square or disk of Banach–Tarski type that uses only Euclidean congruences is impossible. A conceptual explanation of the distinction between the planar and higher-dimensional cases was given by John von Neumann: unlike the group SO(3) of rotations in three dimensions, the group E(2) of Euclidean motions of the plane is solvable, which implies the existence of a finitely-additive measure on E(2) and R2 which is invariant under translations and rotations, and rules out paradoxical decompositions of non-negligible sets. Von Neumann then posed the following question: can such a paradoxical decomposition be constructed if one allows a larger group of equivalences?

It is clear that if one permits similarities, any two squares in the plane become equivalent even without further subdivision. This motivates restricting one's attention to the group SA2 of area-preserving affine transformations. Since the area is preserved, any paradoxical decomposition of a square with respect to this group would be counterintuitive for the same reasons as the Banach–Tarski decomposition of a ball. In fact, the group SA2 contains as a subgroup the special linear group SL(2,R), which in its turn contains the free group F2 with two generators as a subgroup. This makes it plausible that the proof of Banach–Tarski paradox can be imitated in the plane. The main difficulty here lies in the fact that the unit square is not invariant under the action of the linear group SL(2, R), hence one cannot simply transfer a paradoxical decomposition from the group to the square, as in the third step of the above proof of the Banach–Tarski paradox. Moreover, the fixed points of the group present difficulties (for example, the origin is fixed under all linear transformations). This is why von Neumann used the larger group SA2 including the translations, and he constructed a paradoxical decomposition of the unit square with respect to the enlarged group (in 1929). Applying the Banach–Tarski method, the paradox for the square can be strengthened as follows:

Any two bounded subsets of the Euclidean plane with non-empty interiors are equidecomposable with respect to the area-preserving affine maps.

As von Neumann notes:[14]

"Infolgedessen gibt es bereits in der Ebene kein nichtnegatives additives Maß (wo das Einheitsquadrat das Maß 1 hat), das gegenüber allen Abbildungen von A2 invariant wäre."
"In accordance with this, already in the plane there is no non-negative additive measure (for which the unit square has a measure of 1), which is invariant with respect to all transformations belonging to A2 [the group of area-preserving affine transformations]."

To explain further, the question of whether a finitely additive measure (that is preserved under certain transformations) exists or not depends on what transformations are allowed. The Banach measure of sets in the plane, which is preserved by translations and rotations, is not preserved by non-isometric transformations even when they do preserve the area of polygons. The points of the plane (other than the origin) can be divided into two dense sets which may be called A and B. If the A points of a given polygon are transformed by a certain area-preserving transformation and the B points by another, both sets can become subsets of the A points in two new polygons. The new polygons have the same area as the old polygon, but the two transformed sets cannot have the same measure as before (since they contain only part of the A points), and therefore there is no measure that "works".

The class of groups isolated by von Neumann in the course of study of Banach–Tarski phenomenon turned out to be very important for many areas of Mathematics: these are amenable groups, or groups with an invariant mean, and include all finite and all solvable groups. Generally speaking, paradoxical decompositions arise when the group used for equivalences in the definition of equidecomposability is not amenable.

Recent progress

  • 2000: Von Neumann's paper left open the possibility of a paradoxical decomposition of the interior of the unit square with respect to the linear group SL(2,R) (Wagon, Question 7.4). In 2000, Miklós Laczkovich proved that such a decomposition exists.[15] More precisely, let A be the family of all bounded subsets of the plane with non-empty interior and at a positive distance from the origin, and B the family of all planar sets with the property that a union of finitely many translates under some elements of SL(2, R) contains a punctured neighborhood of the origin. Then all sets in the family A are SL(2, R)-equidecomposable, and likewise for the sets in B. It follows that both families consist of paradoxical sets.
  • 2003: It had been known for a long time that the full plane was paradoxical with respect to SA2, and that the minimal number of pieces would equal four provided that there exists a locally commutative free subgroup of SA2. In 2003 Kenzi Satô constructed such a subgroup, confirming that four pieces suffice.[16]
  • 2011: Laczkovich's paper[17] left open the possibility if there exists a free group F of piecewise linear transformations acting on the punctured disk D \{0,0} without fixed points. Grzegorz Tomkowicz constructed such a group,[18] showing that the system of congruences ABCB U C can be realized by means of F and D \{0,0}.
  • 2017: It has been known for a long time that there exists in the hyperbolic plane H2 a set E that is a third, a fourth and ... and a -th part of H2. The requirement was satisfied by orientation-preserving isometries of H2. Analogous results were obtained by John Frank Adams[19] and Jan Mycielski[20] who showed that the unit sphere S2 contains a set E that is a half, a third, a fourth and ... and a -th part of S2. Grzegorz Tomkowicz[21] showed that Adams and Mycielski construction can be generalized to obtain a set E of H2 with the same properties as in S2.
  • 2017: Von Neumann's paradox concerns the Euclidean plane, but there are also other classical spaces where the paradoxes are possible. For example, one can ask if there is a Banach–Tarski paradox in the hyperbolic plane H2. This was shown by Jan Mycielski and Grzegorz Tomkowicz.[22][23] Tomkowicz[24] proved also that most of the classical paradoxes are an easy consequence of a graph theoretical result and the fact that the groups in question are rich enough.
  • 2018: In 1984, Jan Mycielski and Stan Wagon [25] constructed a paradoxical decomposition of the hyperbolic plane H2 that uses Borel sets. The paradox depends on the existence of a properly discontinuous subgroup of the group of isometries of H2. Similar paradox is obtained by Grzegorz Tomkowicz [26] who constructed a free properly discontinuous subgroup G of the affine group SA(3,Z). The existence of such a group implies the existence of a subset E of Z3 such that for any finite F of Z3 there exists an element g of G such that , where denotes the symmetric difference of E and F.
  • 2019: Banach–Tarski paradox uses finitely many pieces in the duplication. In the case of countably many pieces, any two sets with non-empty interiors are equidecomposable using translations. But allowing only Lebesgue measurable pieces one obtains: If A and B are subsets of Rn with non-empty interiors, then they have equal Lebesgue measures if and only if they are countably equidecomposable using Lebesgue measurable pieces. Jan Mycielski and Grzegorz Tomkowicz [27] extended this result to finite dimensional Lie groups and second countable locally compact topological groups that are totally disconnected or have countably many connected components.

See also


  1. ^ Tao, Terence (2011). An introduction to measure theory (PDF). p. 3. Archived from the original (PDF) on 6 May 2021.
  2. ^ Wagon, Corollary 13.3
  3. ^ Wilson, Trevor M. (September 2005). "A continuous movement version of the Banach–Tarski paradox: A solution to De Groot's problem". Journal of Symbolic Logic. 70 (3): 946–952. CiteSeerX doi:10.2178/jsl/1122038921. JSTOR 27588401. S2CID 15825008.
  4. ^ Olivier, Leroy (1995). Théorie de la mesure dans les lieux réguliers. ou : Les intersections cachées dans le paradoxe de Banach-Tarski (Report). arXiv:1303.5631.
  5. ^ Simpson, Alex (1 November 2012). "Measure, randomness and sublocales". Annals of Pure and Applied Logic. 163 (11): 1642–1659. doi:10.1016/j.apal.2011.12.014.
  6. ^ Banach, Stefan; Tarski, Alfred (1924). "Sur la décomposition des ensembles de points en parties respectivement congruentes" (PDF). Fundamenta Mathematicae (in French). 6: 244–277. doi:10.4064/fm-6-1-244-277.
  7. ^ Robinson, Raphael M. (1947). "On the Decomposition of Spheres". Fund. Math. 34: 246–260. doi:10.4064/fm-34-1-246-260. This article, based on an analysis of the Hausdorff paradox, settled a question put forth by von Neumann in 1929:
  8. ^ Wagon, Corollary 13.3
  9. ^ Foreman, M.; Wehrung, F. (1991). "The Hahn–Banach theorem implies the existence of a non-Lebesgue measurable set" (PDF). Fundamenta Mathematicae. 138: 13–19. doi:10.4064/fm-138-1-13-19.
  10. ^ Pawlikowski, Janusz (1991). "The Hahn–Banach theorem implies the Banach–Tarski paradox" (PDF). Fundamenta Mathematicae. 138: 21–22. doi:10.4064/fm-138-1-21-22.
  11. ^ Wagon, p. 16.
  13. ^ Churkin, V. A. (2010). "A continuous version of the Hausdorff–Banach–Tarski paradox". Algebra and Logic. 49 (1): 81–89. doi:10.1007/s10469-010-9080-y. S2CID 122711859. Full text in Russian is available from the page.
  14. ^ On p. 85. Neumann, J. v. (1929). "Zur allgemeinen Theorie des Masses" (PDF). Fundamenta Mathematicae. 13: 73–116. doi:10.4064/fm-13-1-73-116.
  15. ^ Laczkovich, Miklós (1999). "Paradoxical sets under SL2(R)". Ann. Univ. Sci. Budapest. Eötvös Sect. Math. 42: 141–145.
  16. ^ Satô, Kenzi (2003). "A locally commutative free group acting on the plane". Fundamenta Mathematicae. 180 (1): 25–34. doi:10.4064/fm180-1-3.
  17. ^ Laczkovich, Miklós (1999). "Paradoxical sets under SL2(R)". Ann. Univ. Sci. Budapest. Eötvös Sect. Math. 42: 141–145.
  18. ^ Tomkowicz, Grzegorz (2011). "A free group of piecewise linear transformations". Colloquium Mathematicum. 125 (2): 141–146. doi:10.4064/cm125-2-1.
  19. ^ Adams, John Frank (1954). "On decompositions of the sphere". J. London Math. Soc. 29: 96–99. doi:10.1112/jlms/s1-29.1.96.
  20. ^ Mycielski, Jan (1955). "On the paradox of the sphere". Fund. Math. 42 (2): 348–355. doi:10.4064/fm-42-2-348-355.
  21. ^ Tomkowicz, Grzegorz (2017). "On decompositions of the hyperbolic plane satisfying many congruences". Bulletin of the London Mathematical Society. 49: 133–140. doi:10.1112/blms.12024. S2CID 125603157.
  22. ^ Mycielski, Jan (1989). "The Banach-Tarski paradox for the hyperbolic plane". Fund. Math. 132 (2): 143–149. doi:10.4064/fm-132-2-143-149.
  23. ^ Mycielski, Jan; Tomkowicz, Grzegorz (2013). "The Banach-Tarski paradox for the hyperbolic plane (II)". Fund. Math. 222 (3): 289–290. doi:10.4064/fm222-3-5.
  24. ^ Tomkowicz, Grzegorz (2017). "Banach-Tarski paradox in some complete manifolds". Proc. Amer. Math. Soc. 145 (12): 5359–5362. doi:10.1090/proc/13657.
  25. ^ Mycielski, Jan; Wagon, Stan (1984). "Large free groups of isometries and their geometrical uses". Ens. Math. 30: 247–267.
  26. ^ Tomkowicz, Grzegorz (2018). "A properly discontinuous free group of affine transformations". Geom. Dedicata. 197: 91–95. doi:10.1007/s10711-018-0320-y. S2CID 126151042.
  27. ^ Mycielski, Jan; Tomkowicz, Grzegorz (2019). "On the equivalence of sets of equal measures by countable decomposition". Bulletin of the London Mathematical Society. 51: 961–966. doi:10.1112/blms.12289. S2CID 209936338.


External links

This page was last edited on 7 November 2023, at 17:36
Basis of this page is in Wikipedia. Text is available under the CC BY-SA 3.0 Unported License. Non-text media are available under their specified licenses. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc. WIKI 2 is an independent company and has no affiliation with Wikimedia Foundation.