Barna-Taskey Paradox Barna-Taskey Paradox (or Hausdorf-Barna-Taskey Paradox, also known as "the paradox of dividing the ball") is a mathematical theorem. Stefan Banach and Alfred Talsky first proposed this theorem in 1924. This theorem points out that a three-dimensional solid ball can be divided into finite (unmeasurable) parts under the condition of axiom of choice's establishment, and then it can be composed of two parts by simply rotating and translating to other places to recombine? 3? A complete ball with the same radius as the original ball. The original intention of Taskey's theorem is to refuse to choose the axiom of signs, but the proof is natural, so mathematicians think that this can only show that axiom of choice can lead to a minority? 3? Eight people were surprised by counterintuitive results. Let a and b be two subsets of Euclidean space. If they can be divided into the union of finite disjoint subsets in the form of sum, and the subset Ai is equal to Bi for any I, then the two subsets are said to be factorized. Thus, this paradox can be described as: a ball and its two copies are equally decomposed. For the ball, five dollars is enough to do this, but less than five dollars is not enough. There is even a stronger version of this paradox: any subset of two internally non-empty three-dimensional Euclidean spaces is equidistantly decomposed. In other words, a marble can be divided into a limited number of pieces and then reassembled into a planet, or a telephone can be deformed and hidden in a water lily. In real life, this deformation is not feasible, because the volume of atoms is not infinitely small, and the number is not infinitely large, but their geometric shapes can indeed be deformed in this way. If we know that there is always a way to map from the interior point of one geometry to the interior point of another geometry, maybe this paradox doesn't look so weird. For example, two balls can be bijected to an infinite subset of the same level (such as a ball). Similarly, we can map a ball to a ball with a big point or a small point, and we can map a point to another point according to the radius amplification factor. However, generally speaking, these transformations cannot preserve product, or need to divide geometry into uncountable infinite blocks. The unexpected point of Barnah-Taskey paradox is that the transformation can only be completed by rotating and translating a limited number of blocks. What makes this paradox possible is infinite entanglement. Technically, this is unpredictable, so they don't have a "reasonable" range or the usual "quantity". Physical methods such as knives can't complete this segmentation, because only measurable set can be segmented. This purely mathematical theorem points out that besides measurable set, which most people are familiar with, there are more and more measurable set. This paradox still holds true for situations above three dimensions. But it doesn't apply to Euclidean plane. The above statement does not apply to a two-dimensional subset of a three-dimensional space, because this subset may have an empty interior. At the same time, there are some specious decomposition combinations on the plane: a disk can be divided into finite blocks and reassembled into a solid square with the same area. See Taskey's cyclotomic problem. This paradox shows that if the subsets of equidistant decomposition are considered to have the same volume, it is impossible to define what is called "volume" for the bounded subsets of Euclidean space. This proof is based on Felix Hausdorff's early work. He discovered a similar paradox 10 years ago. In fact, Barna-Taskey Paradox is the extension and application of the technology used by Hausdorf. Logicians often use the word "paradox" to express logically inconsistent propositions, such as liar paradox or Russell paradox. Barnard-Taskey paradox is not a paradox in this sense, it is a proved theorem, and it is called a paradox only because it is counterintuitive. Because its proof clearly applies to axiom of choice, this abnormal conclusion is taken as a reason to oppose the use of this axiom. The summary of the proof is basically divided into four steps: find a special method to divide a free group with two generators, find a rotation group whose group is homomorphic to the two generators in three-dimensional space, and use the special division method of this group and axiom of choice to decompose the unit sphere. The details of each step are as follows: First, the free group with generators A and B consists of all finite strings containing symbols A, A and B, and there is no phenomenon that comb A is next to a or comb B is next to b ... For example, ababa connects ababa to get ababaababa, which can be simplified as ababa. We can verify that these strings form a group under this operation, and its unit element is the empty string E. We call this group F2. F_2 group can be divided into the following special segments: Let S(a) be all character strings starting with A, and define S(a), S(b) and S(b) in the same way. Obviously: f _ 2 = \ cups (a) \ cups (a) \ cups (b) \ cups (b) and: f _ 2 = as (a) \ cups (a), at the same time: f _ 2 = bs (b (a) means. This is the key to proof. Please read it carefully. Now let's divide the F_2 group into four pieces (e can be ignored) and multiply it by 1? 3? 4a or b to "rotate" them, then "recombine" three of them into F_2, and "recombine" the other two into another F_2. Is that what we want to do with the sphere? 0? 0 In the second step, in order to find the behavior of rotation group cage in three-dimensional space like F2, we take two coordinate axes, let A rotate arccos( 1/3) radian on the first axis and B rotate arccos( 1/3) radian on the other axis. (this? 3? Step 2 can be completed in two dimensions. ) it's a bit obscene, but? 3? Too difficult, these two kinds of rotation behaviors are just like the behaviors of two elements A and B in F2, so I won't go into details here. The rotation group generated by a and b is named. Of course, we can divide H according to the formula described in the first step. In the third step, the unit sphere S can be divided into several orbits through the operation in the H group: two points belong to the same orbit, and the first point will move to the second point if and only if H rotates. We can use axiom of choice to select a point on each track. Combine these cymbals into a set M. Now (almost) all points in S can be moved to M by the corresponding rotation of the appropriate elements in H. Therefore, the divisibility of H can also be applied to S. Fourth, finally, connect each point of S to the origin, and the divisibility of S can be applied to the solid unit ball? 0? There will be something special in the center of the ball, but it is ignored in this short proof. ) summary, this brief proof is deleted. Some matrix rotations in H just correspond to some positions and should be given special treatment. On the one hand, these points are countable, so they have no influence. On the other hand, what about them? 3? Conversely, even these points can be corrected. 0? The same applies to the center of the ball.