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[[FájlKép:Doubling of a sphere, as per the Banach-Tarski Theorem.png|bélyegkép|jobbra|350px| A Banach–Tarski-paradoxon „szemléltetése”. Egy gömböt fel lehet darabolni olyan darabokra, hogy abból két, ugyanakkora gömb rakható össze]]
[[Kép:Banach-tarski-síkban.jpg|bélyegkép|jobbra|350px|A paradoxon egyik legegyszerűbb síkbeli szemléltetése: a felső négyzet 64 (8x8) egység területű, az alsó téglalap 65 (13x5), a felületet viszont négy darab egybevágó alakzat hozza létre. A síkbeli paradoxon megoldása az, hogy az átrakott négyszögben a csúcsokra eső szögek összege nem egyenlő 90°-kal]]
A ('''Hausdorff–''')'''Banach–Tarski-paradoxon''' egy bizonyított matematikai [[tétel]], mely szerint egy 3 dimenziós, tömör gömböt a [[kiválasztási axióma]] felhasználásával fel lehet vágni véges sok olyan (nem [[Lebesgue-mérték|mérhető]]) darabra, amelyekből két, az eredeti gömbbel megegyező méretű tömör gömböt lehet összeállítani.
:A 3 dimenziós [[euklideszi tér]] bármely két belső ponttal rendelkező, korlátos részhalmaza egymásba átdarabolható.
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In other words, a marble could be cut up into finitely many pieces and reassembled into a planet, or a telephone could be transformed into a water lily. These transformations are not possible with real objects made of a finite number of [[atoms]] and bounded volumes, but it is possible with their geometric shapes. The Banach–Tarski paradox is made somewhat less bizarre by pointing out that there is always a function that can map one-to-one the points in one shape to another. For example, two balls can be transformed [[bijection|bijectively]] to a similarly infinite subset of itself (such as one ball). Likewise, we can make one ball into a larger or smaller ball by simply multiplying the radius of each point in the ball, using [[spherical coordinates]], by a constant. However, such transformations in general are non-isometric or involve an [[countable set|uncountably]] infinite number of pieces. The surprising consequence of the Banach-Tarski paradox is that it can be done with only rotation and translation (isometric mapping) of a finite number of pieces.
What makes the paradox possible is that the pieces are infinitely convoluted. Technically, they are not [[Lebesgue measure|measurable]], and so they do not have "reasonable" [[Boundary (topology)|boundaries]] or a "volume" in the ordinary sense. It is impossible to carry out such a disassembly physically because disassembly "with a knife" can create only measurable sets. This pure existence statement in mathematics points out that there are many more sets than just the measurable sets familiar to most people.
The paradox also holds in all dimensions higher than three. It does not hold for subsets of the [[Euclidean plane]]. (The statement above does not apply to a two-dimensional subset of three-dimensional space, since such a subset would have empty interior.) Still, there are some paradoxical decompositions in the plane: a [[disc (mathematics)|disc]] can be cut into finitely many pieces and reassembled to form a solid square of equal area; see [[Tarski's circle-squaring problem]].
The paradox shows that it is impossible to define "volume" on all bounded subsets of Euclidean space such that equi-decomposable sets will have equal volume.
The proof is based on the earlier work of [[Felix Hausdorff]], who found a [[Hausdorff paradox|closely related paradox]] 10 years earlier; in fact, the Banach–Tarski paradox is a simple corollary of the technique developed by Hausdorff.
Logicians most often use the term "paradox" for a statement in logic which creates problems because it causes contradictions, such as the [[Liar paradox]] or [[Russell's paradox]]. The Banach–Tarski paradox is not a paradox in this sense but rather a proven theorem; 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.
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== A bizonyítás vázlata ==
Figyelem! Ez a vázlat átugrik néhány részlet fölött.
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We now discuss each of these steps in more detail.
'''Step 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''<sup>−1</sup>, ''b'' and ''b''<sup>−1</sup> such that no ''a'' appears directly next to an ''a''<sup>−1</sup> and no ''b'' appears directly next to a ''b''<sup>−1</sup>. 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''<sup>−1</sup>''a''<sup>−1</sup> concatenated with ''abab''<sup>−1</sup>''a'' yields ''abab''<sup>−1</sup>''a''<sup>−1</sup>''abab''<sup>−1</sup>''a'', which gets reduced to ''abaab''<sup>−1</sup>''a''. One can check that the set of those strings with this operation forms a group with neutral element the empty string <math>e</math>. We will call this group <math>F_2</math>.
[[Fájl:Paradox felbontás F2.png|bélyegkép|jobbra|250px|A ''S''(''a''<sup>−1</sup>) halmaz és a ''aS''(''a''<sup>−1</sup>) halmaz ''F''<sub>2</sub>-nek a [[Cayley ábra]]ján]]
The group <math>F_2</math> can be "paradoxically decomposed" as follows: let ''S''(''a'') be the set of all strings that start with ''a'' and define ''S''(''a''<sup>−1</sup>), ''S''(''b'') and ''S''(''b''<sup>−1</sup>) similarly. Clearly,
:<math>F_2={e}\cup S(a)\cup S(a^{-1})\cup S(b)\cup S(b^{-1})</math>
but also
:<math>F_2=aS(a^{-1})\cup S(a)</math>, and
:<math>F_2=bS(b^{-1})\cup S(b)</math>.
(The notation ''a'' ''S''(''a''<sup>−1</sup>) means take all the strings in ''S''(''a''<sup>−1</sup>) and concatenate them on the left with ''a''.) Make sure that you understand this last line, because it is at the core of the proof. Now look at this: we cut our group <math>F_2</math> into four pieces (Forget about <math>e</math> for now, it doesn't pose a problem), then "shift" some of them by multiplying with ''a'' or ''b'', then "reassemble" two of them to make <math>F_2</math> and reassemble the other two to make another copy of <math>F_2</math>. That's exactly what we want to do to the ball.
'''Step 2.''' In order to find a group of rotations of 3D space that behaves just like (or "isomorphic to") the group <math>F_2</math>, we take two orthogonal axes and let ''A'' be a rotation of arccos(1/3) about the first and ''B'' be a rotation of arccos(1/3) about the second. (This step cannot be performed in two dimensions.) It is somewhat messy but not too difficult to show that these two rotations behave just like the elements ''a'' and ''b'' in our group <math>F_2</math>. We'll skip it. The new group of rotations generated by ''A'' and ''B'' will be called '''H'''. Of course, we now also have a paradoxical decomposition of '''H'''.
'''Step 3.''' The unit sphere ''S''<sup>2</sup> is partitioned into [[orbit (group theory)|orbit]]s by the [[group action|action]] of our group '''H''': two points belong to the same orbit if and only if there's a rotation in '''H''' which moves the first point into the second. We can use the [[axiom of choice]] to pick exactly one point from every orbit; collect these points into a set ''M''. Now (almost) every point in ''S''<sup>2</sup> can be reached in exactly one way by applying the proper rotation from '''H''' to the proper element from ''M'', and because of this, the paradoxical decomposition of '''H''' then yields a paradoxical decomposition of ''S''<sup>2</sup>.
'''Step 4.''' Finally, connect every point on ''S''<sup>2</sup> with a ray to the origin; the paradoxical decomposition of ''S''<sup>2</sup> then yields a paradoxical decomposition of the solid unit ball. (The center of the ball needs a bit more care, but we omit this part in the sketch.)
'''NB.''' This sketch glosses over some details. One has to be careful about the set of points on the sphere which happen to lie on an axis of rotation of some matrix in '''H'''. On the one hand, there are countably many such points so they "do not matter", and on the other hand it is possible to patch up even those points. The same applies to the center of the ball.
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==További információk==
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