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A Top Down Look at the Banach-Tarski Paradox Leonard M. Wapner El Camino College Infinity is where things happen that don’t. - an anonymous schoolboy Loosely Speaking ...

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## A Top Down Look at the Banach-Tarski Paradox

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**A Top Down Lookat the Banach-Tarski Paradox**Leonard M. Wapner El Camino College**Infinity is where things happen that don’t.**- an anonymous schoolboy**Loosely Speaking ...**The theorem states that it is possible to partition a solid ball into finitely many pieces and reassemble them to form two balls, each of the same shape and volume as the original. B1 B B2**Limerick**There once were two mathematicians Who split a lead ball with partitions Though with only five parts Our two masters of arts Reassembled it into munitions!**Georg Cantor**I see it, but I don’t believe it! I was forced by logic ...**Henri Poincare**… a disease from which mathematics would have to recover !**Stefan Banach Alfred Tarski1892 - 1945**1902 - 1983**Tarski Obit**An irate citizen once demanded of the Illinois legislature that they outlaw the teaching of this result in Illinois schools! From an obituary of Alfred Tarski in California Monthly, the UC Berkeley alumni magazine.**Axiom of Choice**A B C . . . . . . . a, b, c, ...**Rejoice!**Isn’t AC great? How else could we get such counterintuitive results?**Great-circle Distance**d M L O Find Find d Find OL & OM**Definitions and Notation**• Let S = {(x,y,z) : x2+y2+z2 = 1}. We refer to S as the unit sphere (centered at the origin). • Let B = {(x,y,z) : x2+y2+z2 < 1}. We refer to B as the unit ball (centered at the origin). • Given two sets C and D, we say C is congruent to D, and write C D, if C can be made to coincide with D using only**Definitions and Notation (cont)**• … translations and rotations of C. • Given two sets C and D, we say C is equivalent by finite decomposition to D, and write C D, if we can divide C into a finite number of disjoint parts, respectively translate and rotate these parts in such a way that they can be rearranged to form D.**Definitions and Notation (cont)**• Given any two rotations 1 and 2 of S, define the product rotation 12 as the rotation obtained by first rotating by 2 then by 1. Similarly, for any rotation and natural number n, define the rotation n to be the rotation equivalent to n sequential rotations.**Statement of Theorem**A closed unit ball B can be decomposed into two disjoint sets, B = B1 B2, such that B B1 and B B2 . B1 B B2**Proof of Theorem**• Form and partition a group of rotations G acting on S. • Use G to create two spheres, S1 and S2, from the original sphere S. This is the Hausdorff Paradox. • By “thickening” S1 and S2 , create B1 and B2 such that B ~ B1 and B ~ B2.**Rotations**• Define three basic rotations , 2 and . • Form the group G of all products of these rotations consisting of a finite number of factors. • Sort these products into three subsets G1, G2 and G3. • Note significant properties of the subsets.**Sphere**• Define P, the set of poles. • Partition S\P into three sets K1, K2 and K3 • Show K1 K2 K3 K2 K3. • Use “cookie cutter” methods to create two spheres, S1 and S2, from the original sphere S. • Absorb the poles.**Ball**• “Thicken” S1 and S2 to form B1 and B2. • Absorb the origin.**Anagram**Banach-Tarski Banach-TarskiBanach-Tarski**Equivalence By Finite Decomposition**Square Isosceles Triangle C D C ~ D**Basic Rotations** axis (z axis) axis (z = x) : 120° 2:240 45° : 180°**Matrix Definitions of and **-1/2 -3/2 0 3/2 -1/2 0 0 0 1 0 0 1 0 -1 0 1 0 0 = =**Partitioning G**1 G1, G2, G2, 2 G3 G1 G2 G3 or 2 G2 G1 G1 G2 G3 G1 begins with: 2 G3 2 G1 2 G2**Properties of G1 , G2 and G3**G1 G2 G3 , , 2, ... 1, , 2, ... 2, , 2, ... G1 = G2 2G1 = G3 G1 = G2G3**Partitioning S\P**• We call two points in S\P equivalent if there exists a rotation in G which sends one point to the other. This is an equivalence relation and allows us to partition S\P into equivalence classes. There are uncountably many such classes. • Using AC, choose one point from each class and define the choice set M as the set of these chosen points. • Let K1 = G1M , K2 = G2M and K3 = G3M.**P - The Set of Poles**Every rotation G has two poles. These are the two points of S which remain fixed under . Let P be the set of all such poles for all G . Since G is countable, the set P is also countable.**Hyperbolic Plane** K1 K2 K3 K1 K2 K3

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