Cheat Engine

[KeNnZ]

The Banach-Tarski Paradox is the famous "doubling the ball" paradox, which claims that by using the axiom of choice it is possible to take a solid ball in 3-dimensional space, cut it up into finitely many pieces and, moving them using only rotation and translation, reassemble the pieces into two balls the same size as the original. Or short: the ball is equi-decomposable with two copies of itself. For the ball, five pieces are sufficient to do this; it cannot be done with fewer than five. There is an even stronger version of the paradox: Any two bounded subsets (of 3-dimensional Euclidean space R3) with non-empty interior are equi-decomposable. In other words, a marble can be cut up into finitely many pieces and reassembled into a planet. Or you could disassemble a telephone and reassemble it to make a water lily. OK, you can't do this with a real marble or a real water lily, which are made up of atoms, but you can do it with their shapes. It is interesting to note that this theorem depends on three dimensions; while intuitively the two-dimensional case seems to be easier, it is in fact not true that all bounded subsets of the plane with non-empty interior are equi-decomposable. Still, there are some paradoxical decompositions in the plane: a circle can be cut into finitely many pieces and reassembled to form a square of equal area; see Tarski's circle-squaring problem. Note that in the decomposition, the pieces won't be measurable, and so they will not have "reasonable" boundaries nor 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.

In 1924, Stefan Banach and Alfred Tarski described this paradox, building on earlier work by Felix Hausdorff who managed to "chop up" the unit interval into countably many pieces which (by translation only) can be reassembled into the interval of length 2. He did this in order to show that there can be no non-trivial translation invariant measure on the real line which assigns a size to all sets of real numbers. 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. A sketch of the proof follows. Essentially, the paradoxical decomposition of the ball is achieved in four steps: Find a paradoxical decomposition of the free group in two generators. Find a group of rotations in 3-d space isomorphic to the free group in two generators. Use the paradoxical decomposition of that group and the axiom of choice to produce a paradoxical decomposition of the hollow unit sphere. Extend this decomposition of the sphere to a decomposition of the solid unit ball. 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 gets reduced to abaab-1a.

One can check that the set of those strings with this operation forms a group with neutral element the empty string, here denoted e. We will call this group G. The group G can be "paradoxically decomposed" as follows: let S(a) be the set of all strings that start with a and define S(a-1), S(B) and S(b-1) similarly. Clearly, G = {e} ? S(a) ? S(a-1) ? S(B) ? S(b-1) but also G = a S(a-1) ? S(a), and G = b S(b-1) ? S(B) (The notation a S(a-1) means: take all the strings in S(a-1) 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 G into four pieces (forget about e for now, it doesn't pose a problem), then "rotated" some of them by multiplying with a or b, then "reassembled" two of them to make G and reassembled the other two to make another copy of G. That's exactly what we want to do to the ball. This finishes step 1. In order to find a group of rotations of 3-d space that behaves just like (or "isomorphic to") the group G, 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 G. 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 number 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'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 S2 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 S2. (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.) Finally, connect every point on S2 with a ray to the origin; the paradoxical decomposition of S2 then yields a paradoxical decomposition of the solid unit ball (minus the origin, but that can be dealt with easily). A paradox that defeats normal logic.

There is also the paradox of the man who is his own mother, though it requires that you assume time travel is possible. “Jane” is left at an orphanage as a foundling. When “Jane” is a teenager, she falls in love with a drifter, who abandons her but leaves her pregnant. Then disaster strikes. She almost dies giving birth to a baby girl, who is then mysteriously kidnapped. The doctors find that Jane is bleeding badly, but, oddly enough, has both *** organs. So, to save her life, the doctors convert “Jane” to “Jim.” “Jim” subsequently becomes a roaring drunk, until he meets a friendly bartender (actually a time traveler in disguise) who whisks “Jim” back way into the past. “Jim” meets a beautiful teenage girl, accidentally gets her pregnant with a baby girl. Out of guilt, he kidnaps the baby girl and drops her off at the orphanage. The game. You just lost it. Later, “Jim” joins the time travelers corps, leads a distinguished life, and has one last dream: to disguise himself as a bartender to meet a certain drunk named “Jim” in the past. Will this loop ever end? No, it can't. Something different must happen in between Jane being dropped off at the orphanage and Jim bringing his former self (before he went back in time) to the past for something to change.

However, if time travel brings Jim back to the EXACT past, then nothing will change, ever. An infinite paradox. Zeno of Elea also created a paradox involving an arrow in flight. An arrow in flight is really at rest. At every point in it's flight, the arrow must occupy a length of space exactly equal to it's own length. After all, it cannot occupy a greater length, nor a lesser one. But the arrow cannot move within this length it occupies. It would need extra space in which to move, and it, of course, has none. So, at every point in it's flight, the arrow is at rest. And if it is at rest at every moment in it's flight, then it follows that it is at rest during the entire flight. A paradox that defeats normal logic.


inb4tl;dr. ctrl f

[KeNnZ]

5th paragraph, 6th sentence.

[Fastlv]

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