Paradox of the バナッハ = Tal ski

The paradox of the バナッハ = Tal ski: I divide a ball properly and can make two balls same as the cause by rearranging it.

A theorem (but each piece cannot define the volume in a normal meaning) to be able to make an original ball and the ball of the same radius two by the paradox (Banach-Tarski paradox) of the バナッハ = Tal ski dividing three dimensions of balls into the part of the limited unit in space, and rearranging them only using a turn, translation operation well. It is necessary to divide a ball into at least five to perform this operation.

By the proof of the バナッハ = Tal ski, paradox of ハウスドルフ was quoted, and optimization of the proof, expansion to various space were carried out by many people afterwards.

Because a result is against intuition, it is a theorem, but is called paradox. I use the axiom of choice in one place of the proof. When Stephane バナフ (バナッハ) and アルフレト タルスキ described this theorem for the first time in 1924, I caught axiom of choice affirmatively or arrested you negatively, or it is difficult to judge it. (they speak it only with (Le rôle que joue cet axiome dans nos raisonnements nous semble mériter l'attention.) where "the role that the axiom of choice for this study serves as deserves it if I pay attention.".)

I can give this theorem as follows.

• The ball is congruent with two balls same as itself separately.

But I am defined as follows to be congruent separately: I assume A and B a subset of the Euclidean space. As a merger of the subsets not to cross each other of the units limited A and B

${\displaystyle A=\bigcup _{i=1}^{n}A_{i},\quad B=\bigcup _{i=1}^{n}B_{i}}$

In other words,

A = A₁ ∪ ... ∪ An, B = B₁ ∪ ... ∪ B_n

Can express と; about all i,${\displaystyle A_{i}}$ と ${\displaystyle B_{i}}$ But, A and B are called division combination when it is congruent.

Furthermore, I can lead a system of the next stronger form from this theorem.

• If the inside is not empty and chose (the thing which, in other words, is not a curve and a curved surface with limited extent) two in the subset that is existence world of the three-dimensional Euclidean space optionally, they are congruent separately.

In other words, I am that I make the moon by dividing a glass marble into a limited unit, and rearranging it, and I rearrange a telephone and make a water lily, and it is possible (the materials are not changed for granted). The point meeting is comprised of a choice meeting made using axiom of choice, and, by the proof of this theorem, each piece is not ルベーグ measurability. In other words, each piece does not have the volume in a clear border and the normal meaning. Because I can make only the meeting that is measurability with the physical division, such division is impossible practically. However, such a conversion is possible for those geometric shapes.

These three dimensions of theorems hold good in all above-mentioned dimensions. Although it is not managed in the two-dimensional Euclid plane, the paradox about the division exists in the two dimensions: I can make the square of the same area by dividing Japanese yen into the part of the limited unit, and rearranging it. This is known as the quadrature of the circle (en:Tarski's circle-squaring problem) of the Tal ski.

John von Neumann proved that the theorem that was similar to the paradox of the バナッハ = Tal ski was established in 1929 when I relaxed a condition for conversion to keep an area not joint conversion in the two-dimensional Euclid plane. I can give this theorem as follows.

I do A and B two dimensions with the subset that is existence world with a point in Euclidean space. As a merger of the subsets not to cross each other of the units limited A and B

${\displaystyle A=\bigcup _{i=1}^{n}A_{i},\quad B=\bigcup _{i=1}^{n}B_{i}}$

I can express と. Conversion to keep an area about all i here ${\displaystyle f_{i}}$ But, exist

${\displaystyle B_{i}=f_{i}(A_{i})}$

とする 事 is made.

Table of contents

Summary of the proof

I give the proof of the theorem. The method here is similar to バナッハ and the thing with the Tal ski, but is not identical. The proof is divided into four steps essentially.

1. Freedom group having two generators${\displaystyle F_{2}}$ の finds "the division that is パラドキシカル".
2. Freedom group${\displaystyle F_{2}}$ I find the three-dimensional rotation group which is the と same model.
3. I make the division of the two dimensions spherical surface using the division and the axiom of choice which are パラドキシカル of the rotation group which I made with 2.
4. I expand three dimensions of division of 3 two dimensions spherical surfaces in the division of the ball.

I speak the details of each step.

Step 1

The free group produced by two generator a and b is comprised of character string to have four letter a, a^-1, b, limited length consisting of b^-1. The character string showing a just after a^-1 here is not permitted. It is similar about b. When there was two such character string, I define those products that I connected one of those character string. But I deal when "the character string that is not permitted" thereby occurred by replacing the part with "character string of sky". For example, the product of abab^-1a^-1 and abab^-1a becomes abab^-1a^-1abab^-1a, but I rearrange this part by "empty character string", and, as for this, it is in abaab^-1a to include "the character string that is not permitted" called a^-1a. As for the set of such a character string, it is checked by the operation that I defined here that it becomes the group having "character string of sky" in identity element e. I write this group as F₂.

 

Set of S(a^-1) and aS(a^-1) in the Cayley graph of F₂

Group${\displaystyle F_{2}}$ "The division that is パラドキシカル" is possible as if being less than は: It begins in a in S(a)${\displaystyle F_{2}}$ I assume it a set of the whole の character string. It is similar about S(a^-1), S(b), S(b^-1). Obviously,

${\displaystyle F_{2}= e \cup S(a)\cup S(a^{-1})\cup S(b)\cup S(b^{-1})}$ 

One

${\displaystyle F_{2}=aS(a^{-1})\cup S(a),\,}$ 

And

${\displaystyle F_{2}=bS(b^{-1})\cup S(b).\,}$ 

である. The notation called aS(a^-1) is the whole of the character string that risked a for the original left of S(a^-1).

The last line is the core of this proof. I gather, for example${\displaystyle aS(a^{-1})}$ は${\displaystyle aa^{-1}b}$ I include という character string.${\displaystyle a}$ は${\displaystyle a^{-1}}$ The character string this by a rule not to have to appear just after の${\displaystyle b}$ となる. In this way${\displaystyle aS(a^{-1})}$ は${\displaystyle b}$ I include all character string that begin to be given. Similarly${\displaystyle bS(b^{-1})}$ は${\displaystyle a}$ , ${\displaystyle a^{-1}}$ , ${\displaystyle b^{-1}}$ I include all character string that begin to be given.

Step 2

In a rotation group of the three-dimensional space just${\displaystyle F_{2}}$ I serve と in the same way${\displaystyle F_{2}}$ I take two axes, x at right angles and z to find) group which is the と same model. And that I express a turn of 1 radian that assumed z-axis an axis in turn, b of 1 radian that assumed x-axis an axis in a (even if an angle of the turns is not 1 radian, anything is all right if it is impossible several times of pi π). Two rotary a, b composition of the operation as the product${\displaystyle F_{2}}$ The proof of becoming the と same model is slightly complicated, but omits this part because it is not difficult. I assume a rotation group produced by a and b H. Then I can apply the division that is パラドキシカル which I got by step 1 for H.

Step 3

I can divide unit spherical surface S² into an orbital meeting by thinking about action of group H. In other words, two points of S² establish it when the turn that puts one point into the other exists in H when they belong to the same orbit only for again time (they warn that the orbit of a certain point becomes the dense set of S²). Using axiom of choice, I can make a new meeting from all orbit with a choice of just one point. I assume this meeting M. I can get all points of S² now by triggering the cause of a certain H at a point of a certain M. Therefore, the division that is パラドキシカル of H gives the division to four subset A₁, A₂, A₃, A₄ of S² as follows.

${\displaystyle A_{1}=S(a)M\cup M\cup B}$ 

${\displaystyle A_{2}=S(a^{-1})M\setminus B}$ 

${\displaystyle \displaystyle A_{3}=S(b)M}$ 

${\displaystyle \displaystyle A_{4}=S(b^{-1})M}$ 

Here

${\displaystyle B=\bigcup _{i=1}^{\infty }a^{-i}M}$ 

である.

The spherical surface is divided into four subsets now. I can obtain these spherical surfaces 2 times as large as a beginning in beating, and turning two meetings as follows.

${\displaystyle aA_{2}=A_{2}\cup A_{3}\cup A_{4}}$ 

${\displaystyle bA_{4}=A_{1}\cup A_{2}\cup A_{4}}$ 

Therefore

${\displaystyle A_{1}\cup aA_{2}=S^{2}}$ 

And

${\displaystyle A_{3}\cup bA_{4}=S^{2}}$ 

Step 4

Finally the division of S² which I thought about by step 3 is expanded naturally from a ball to the division of the meeting except the center point when I think about a segment of a line which links the origin to all points on S². (it is necessary to treat this center point a little more with caution, and, similarly, I omitted it by this summary, but it is necessary to give the thing which is on the axis of some kind of turns included in H in one of a point of S² the special treatment.)

References

• Koji Shiga "mathematicians Japan criticism company of the shaft of light Poland school from infinity", April, 1988. ISBN 4-535-78161-3。
• Toshikazu Sunada "paradox Iwanami Shoten 〈 Iwanami science library 〉, April, 1997 of the バナッハ Tal ski." ISBN 4-00-006549-1。
  • In Toshikazu Sunada "paradox Iwanami Shoten 〈 Iwanami science library 〉, December, 2009 of バナッハ-タルスキー", it is a new publication. ISBN 978-4-00-029565-9。
• Leonard M ワプナー "a paradox bean and the sun same size of バナッハ = タルスキ" Kaori Sato, Hiroki Sato reason, Aoto Corporation, December, 2009. ISBN 978-4-7917-6515-7。

Outside link

• Banach-Tarski Paradox - - From MathWorld (paradox of the バナッハ = Tal ski) (English)
• Paradox (Japanese sentence) of the バナッハ Tal ski

It is acquired by "paradox &oldid=58110332 of the https://ja.wikipedia.org/w/index.php?title= バナッハ = Tal ski"

This article is taken from the Japanese Wikipedia Paradox of the バナッハ = Tal ski

This article is distributed by cc-by-sa or GFDL license in accordance with the provisions of Wikipedia.

Wikipedia and Tranpedia does not guarantee the accuracy of this document. See our disclaimer for more information.

In addition, Tranpedia is simply not responsible for any show is only by translating the writings of foreign licenses that are compatible with CC-BY-SA license information.