The Banach-Tarski Paradox - Concordia...
Transcript of The Banach-Tarski Paradox - Concordia...
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
The Banach-Tarski Paradox
Anders O.F. Hendrickson
Department of Mathematics and Computer ScienceConcordia College, Moorhead, MN
Math/CS Colloquium Series
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Outline
1 Introduction
2 Paradoxes
3 Free Group
4 Doubling the Ball
5 Conclusion
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
The Banach-Tarski Paradox
Our goal: to dissect a solid ball into finitely many pieces andreassemble the pieces into two balls of the original size.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Dissection
We want to know if two shapes are “equidecomposable,”meaning that one can be cut up and rearranged to become theother. A common example:
−→
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Dissection
We want to know if two shapes are “equidecomposable,”meaning that one can be cut up and rearranged to become theother. A common example:
−→
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Definitions
DefinitionWe say that sets X and Y in Euclidean space are congruent,and we write X ∼= Y , if X can be moved to fill the same spaceas Y using only rigid motions (sliding and rotating).
DefinitionWe say that two sets A and B are equidecomposable, andwrite A ∼ B, if there are disjoint sets A1, . . . ,An and disjoint sets
B1, . . . ,Bn such that A =n⋃̇
i=1Ai and B =
n⋃̇i=1
Bi with each
Ai∼= Bi .
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Definitions
DefinitionWe say that sets X and Y in Euclidean space are congruent,and we write X ∼= Y , if X can be moved to fill the same spaceas Y using only rigid motions (sliding and rotating).
DefinitionWe say that two sets A and B are equidecomposable, andwrite A ∼ B, if there are disjoint sets A1, . . . ,An and disjoint sets
B1, . . . ,Bn such that A =n⋃̇
i=1Ai and B =
n⋃̇i=1
Bi with each
Ai∼= Bi .
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Transitivity of ∼
TheoremIf A ∼ B and B ∼ C, then A ∼ C.
Proof sketch.
−→ −→
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Transitivity of ∼
TheoremIf A ∼ B and B ∼ C, then A ∼ C.
Proof sketch.
−→ −→
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Transitivity of ∼
TheoremIf A ∼ B and B ∼ C, then A ∼ C.
Proof sketch.
−→
−→
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Transitivity of ∼
TheoremIf A ∼ B and B ∼ C, then A ∼ C.
Proof sketch.
−→ −→
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Transitivity of ∼
TheoremIf A ∼ B and B ∼ C, then A ∼ C.
Proof sketch.
−→
−→
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Transitivity of ∼
TheoremIf A ∼ B and B ∼ C, then A ∼ C.
Proof sketch.
−→ −→
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Transitivity of ∼
TheoremIf A ∼ B and B ∼ C, then A ∼ C.
Proof sketch.
−→ −→
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Transitivity of ∼
TheoremIf A ∼ B and B ∼ C, then A ∼ C.
Proof sketch.
−→ −→
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Transitivity of ∼
TheoremIf A ∼ B and B ∼ C, then A ∼ C.
Proof sketch.
−→ −→
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Transitivity of ∼
TheoremIf A ∼ B and B ∼ C, then A ∼ C.
Proof sketch.
−→ −→�
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Infinity is Weird, Part 1
Some infinities are more infinite than others.If you can list all the elements of an infinite set one by one,it is called countably infinite. Examples:
N = {1,2,3,4,5,6, . . .}Z = {0,1,−1,2,−2,3,−3, . . .}
But a continuum like R or [2,3] or R2 has more elementsthan N does. It is uncountably infinite.(Cantor’s diagonal argument.)
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Infinity is Weird, Part 2
Alice Bob Carlos David Edgar
· · ·
Thomas
Suppose you own a (countably) infinite hotel.All the rooms are occupied.Now Thomas shows up and wants a room.Move every existing guest to the right.Now there’s room for Thomas!
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Infinity is Weird, Part 2
Alice Bob Carlos David Edgar Fred · · ·
Thomas
Suppose you own a (countably) infinite hotel.All the rooms are occupied.Now Thomas shows up and wants a room.Move every existing guest to the right.Now there’s room for Thomas!
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Infinity is Weird, Part 2
Alice Bob Carlos David Edgar Fred · · ·
Thomas
Suppose you own a (countably) infinite hotel.All the rooms are occupied.Now Thomas shows up and wants a room.Move every existing guest to the right.Now there’s room for Thomas!
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Infinity is Weird, Part 2
Alice Bob Carlos David Edgar Fred · · ·
Thomas
Suppose you own a (countably) infinite hotel.All the rooms are occupied.Now Thomas shows up and wants a room.Move every existing guest to the right.Now there’s room for Thomas!
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Infinity is Weird, Part 2
Alice Bob Carlos David Edgar · · ·
Thomas
Suppose you own a (countably) infinite hotel.All the rooms are occupied.Now Thomas shows up and wants a room.Move every existing guest to the right.Now there’s room for Thomas!
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Infinity is Weird, Part 2
Alice Bob Carlos David Edgar · · ·
Thomas
Suppose you own a (countably) infinite hotel.All the rooms are occupied.Now Thomas shows up and wants a room.Move every existing guest to the right.Now there’s room for Thomas!
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Infinity is Weird, Part 2
Thomas Alice Bob Carlos David Edgar · · ·
Thomas
Suppose you own a (countably) infinite hotel.All the rooms are occupied.Now Thomas shows up and wants a room.Move every existing guest to the right.Now there’s room for Thomas!
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Infinite hotel in finite space
Pick a starting point and an angle θ = απ, where α 6∈ Q.Rotate our starting point by θ repeatedly.We never hit the same point twice, since otherwisemθ = nθ + 2kπ =⇒ (m − n)απ = 2kπ =⇒ α = 2k
m−n ∈ Q.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Infinite hotel in finite space
Pick a starting point and an angle θ = απ, where α 6∈ Q.Rotate our starting point by θ repeatedly.We never hit the same point twice, since otherwisemθ = nθ + 2kπ =⇒ (m − n)απ = 2kπ =⇒ α = 2k
m−n ∈ Q.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Infinite hotel in finite space
Pick a starting point and an angle θ = απ, where α 6∈ Q.Rotate our starting point by θ repeatedly.We never hit the same point twice, since otherwisemθ = nθ + 2kπ =⇒ (m − n)απ = 2kπ =⇒ α = 2k
m−n ∈ Q.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Infinite hotel in finite space
Pick a starting point and an angle θ = απ, where α 6∈ Q.Rotate our starting point by θ repeatedly.We never hit the same point twice, since otherwisemθ = nθ + 2kπ =⇒ (m − n)απ = 2kπ =⇒ α = 2k
m−n ∈ Q.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Infinite hotel in finite space
Pick a starting point and an angle θ = απ, where α 6∈ Q.Rotate our starting point by θ repeatedly.We never hit the same point twice, since otherwisemθ = nθ + 2kπ =⇒ (m − n)απ = 2kπ =⇒ α = 2k
m−n ∈ Q.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Infinite hotel in finite space
Pick a starting point and an angle θ = απ, where α 6∈ Q.Rotate our starting point by θ repeatedly.We never hit the same point twice, since otherwisemθ = nθ + 2kπ =⇒ (m − n)απ = 2kπ =⇒ α = 2k
m−n ∈ Q.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Infinite hotel in finite space
Pick a starting point and an angle θ = απ, where α 6∈ Q.Rotate our starting point by θ repeatedly.We never hit the same point twice, since otherwisemθ = nθ + 2kπ =⇒ (m − n)απ = 2kπ =⇒ α = 2k
m−n ∈ Q.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Our first paradox
Now divide circle into two subsets: all those points, andeverything else.Rotate the points by θ. Each will fit into the next one’s slot.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Our first paradox
Now divide circle into two subsets: all those points, andeverything else.Rotate the points by θ. Each will fit into the next one’s slot.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Our first paradox
Now divide circle into two subsets: all those points, andeverything else.Rotate the points by θ. Each will fit into the next one’s slot.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Our first paradox
Now divide circle into two subsets: all those points, andeverything else.Rotate the points by θ. Each will fit into the next one’s slot.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Our first paradox
Now divide circle into two subsets: all those points, andeverything else.Rotate the points by θ. Each will fit into the next one’s slot.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Our first paradox
−→
LemmaC ∼ C − {point}
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
The free group F2
F2 = the set of all “words” using the “letters” a, b, a−1, and b−1,provided we agree to cancel aa−1, a−1a, bb−1, b−1b.
e.g. abab−1,a3b−2,baa−1b = b2 ∈ F2Note that F2 is countably infinite
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
The free group F2
F2 = the set of all “words” using the “letters” a, b, a−1, and b−1,provided we agree to cancel aa−1, a−1a, bb−1, b−1b.
e.g. abab−1,a3b−2,baa−1b = b2 ∈ F2Note that F2 is countably infinite
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
The free group F2
F2 = the set of all “words” using the “letters” a, b, a−1, and b−1,provided we agree to cancel aa−1, a−1a, bb−1, b−1b.
e.g. abab−1,a3b−2,baa−1b = b2 ∈ F2Note that F2 is countably infinite
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
The free group F2
F2 = the set of all “words” using the “letters” a, b, a−1, and b−1,provided we agree to cancel aa−1, a−1a, bb−1, b−1b.
e.g. abab−1,a3b−2,baa−1b = b2 ∈ F2Note that F2 is countably infinite
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
The free group F2
F2 = the set of all “words” using the “letters” a, b, a−1, and b−1,provided we agree to cancel aa−1, a−1a, bb−1, b−1b.
e.g. abab−1,a3b−2,baa−1b = b2 ∈ F2Note that F2 is countably infinite
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Decomposition of F2
Dissect F2 into R ∪̇ S ∪̇ T ∪̇ U.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Decomposition of F2
Dissect F2 into R ∪̇ S ∪̇ T ∪̇ U.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Reassembly of F2
−→
Dissect F2 into R ∪̇ S ∪̇ T ∪̇ U.Then b−1R ∪ T = F2 and aS ∪ U = F2, so F2 ∼ 2F2!Of course, this isn’t geometric. Can we fit F2 into finiteEuclidean space?
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Reassembly of F2
−→
Dissect F2 into R ∪̇ S ∪̇ T ∪̇ U.Then b−1R ∪ T = F2 and aS ∪ U = F2, so F2 ∼ 2F2!Of course, this isn’t geometric. Can we fit F2 into finiteEuclidean space?
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Reassembly of F2
−→ &
Dissect F2 into R ∪̇ S ∪̇ T ∪̇ U.Then b−1R ∪ T = F2 and aS ∪ U = F2, so F2 ∼ 2F2!Of course, this isn’t geometric. Can we fit F2 into finiteEuclidean space?
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Reassembly of F2
−→ &
Dissect F2 into R ∪̇ S ∪̇ T ∪̇ U.Then b−1R ∪ T = F2 and aS ∪ U = F2, so F2 ∼ 2F2!Of course, this isn’t geometric. Can we fit F2 into finiteEuclidean space?
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Reassembly of F2
−→ &
Dissect F2 into R ∪̇ S ∪̇ T ∪̇ U.Then b−1R ∪ T = F2 and aS ∪ U = F2, so F2 ∼ 2F2!Of course, this isn’t geometric. Can we fit F2 into finiteEuclidean space?
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
F2 on the sphere
Take θ = cos−1 35 .
Consider a sphere, and let a = rotation around x-axis by θand b =rotation around z-axis by θ.Like 2 moves of a Rubik’s cube,Let G be all moves you can get by repeatedly doing anysequence of a, b, a−1, or b−1.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
G is isomorphic to F2
LemmaNo two sequences of rotations in G result in the same rotationof the sphere.
Proof.A boring computation with matrices.
Thus G acts just like F2, and from now on we’ll identify the two.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
G is isomorphic to F2
LemmaNo two sequences of rotations in G result in the same rotationof the sphere.
Proof.A boring computation with matrices.
Thus G acts just like F2, and from now on we’ll identify the two.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
G is isomorphic to F2
LemmaNo two sequences of rotations in G result in the same rotationof the sphere.
Proof.A boring computation with matrices.
Thus G acts just like F2, and from now on we’ll identify the two.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
F2 on the sphere
Pick a point x ∈ S, and subject it to all the rotations.If we get a different point for each rotation, then we have a“copy” F2x of F2 on the sphere!Then we can divide F2x into four pieces,F2x = Rx ∪̇ Sx ∪̇ Tx ∪̇ Ux , move them by rigid motions(rotations), and reassemble them into 2 copies of F2x .
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
F2 on the sphere
Pick a point x ∈ S, and subject it to all the rotations.If we get a different point for each rotation, then we have a“copy” F2x of F2 on the sphere!Then we can divide F2x into four pieces,F2x = Rx ∪̇ Sx ∪̇ Tx ∪̇ Ux , move them by rigid motions(rotations), and reassemble them into 2 copies of F2x .
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
F2 on the sphere
Pick a point x ∈ S, and subject it to all the rotations.If we get a different point for each rotation, then we have a“copy” F2x of F2 on the sphere!Then we can divide F2x into four pieces,F2x = Rx ∪̇ Sx ∪̇ Tx ∪̇ Ux , move them by rigid motions(rotations), and reassemble them into 2 copies of F2x .
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Lots of F2’s on the sphere
If we take a set W of such “good” points, then we get lotsof copies of F2.So we can divide F2W into 4 pieces,F2W = RW ∪̇ SW ∪̇ TW ∪̇ UW ,and reassemble them into 2 copies of F2W .
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Lots of F2’s on the sphere
If we take a set W of such “good” points, then we get lotsof copies of F2.So we can divide F2W into 4 pieces,F2W = RW ∪̇ SW ∪̇ TW ∪̇ UW ,and reassemble them into 2 copies of F2W .
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Bad Points
What if when we rotate a point x by all the rotations in F2,we don’t get all distinct points?If so, then
r1x = r2x =⇒ (r2)−1r1x = x ,
so x is unmoved by the rotation (r2)−1r1—so it’s a pole!
Let X = {all bad points}. Every rotation has two poles, so|X | = 2|F2| is countably infinite.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Bad Points
What if when we rotate a point x by all the rotations in F2,we don’t get all distinct points?If so, then
r1x = r2x =⇒ (r2)−1r1x = x ,
so x is unmoved by the rotation (r2)−1r1—so it’s a pole!
Let X = {all bad points}. Every rotation has two poles, so|X | = 2|F2| is countably infinite.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Bad Points
What if when we rotate a point x by all the rotations in F2,we don’t get all distinct points?If so, then
r1x = r2x =⇒ (r2)−1r1x = x ,
so x is unmoved by the rotation (r2)−1r1—so it’s a pole!
Let X = {all bad points}. Every rotation has two poles, so|X | = 2|F2| is countably infinite.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Eliminating Bad Points
Since |X | < |S|, pick an axis through no point of X .Rotate through an angle απ so that all points in X vanish.If our rotation sends one point of X to another point in X ,then that’s a bad angle. But there are only countably manybad angles, so we can choose a good angle!
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Eliminating Bad Points
Since |X | < |S|, pick an axis through no point of X .Rotate through an angle απ so that all points in X vanish.If our rotation sends one point of X to another point in X ,then that’s a bad angle. But there are only countably manybad angles, so we can choose a good angle!
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Eliminating Bad Points
Since |X | < |S|, pick an axis through no point of X .Rotate through an angle απ so that all points in X vanish.If our rotation sends one point of X to another point in X ,then that’s a bad angle. But there are only countably manybad angles, so we can choose a good angle!
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Eliminating Bad Points
Since |X | < |S|, pick an axis through no point of X .Rotate through an angle απ so that all points in X vanish.If our rotation sends one point of X to another point in X ,then that’s a bad angle. But there are only countably manybad angles, so we can choose a good angle!
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Eliminating Bad Points
LemmaS ∼ S − X.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Now S − X is the union of infinitely many sets like F2.Pick just one point from each of these sets; call thatcollection Z . [Axiom of Choice!]Now S − X = F2Z = RZ ∪̇ SZ ∪̇ TZ ∪̇ UZ , and we canreassemble: S−X = a−1RZ ∪̇SZ and S−X = bTZ ∪̇UZ .
LemmaS − X ∼ 2(S − X ).
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Now S − X is the union of infinitely many sets like F2.Pick just one point from each of these sets; call thatcollection Z . [Axiom of Choice!]Now S − X = F2Z = RZ ∪̇ SZ ∪̇ TZ ∪̇ UZ , and we canreassemble: S−X = a−1RZ ∪̇SZ and S−X = bTZ ∪̇UZ .
LemmaS − X ∼ 2(S − X ).
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Now S − X is the union of infinitely many sets like F2.Pick just one point from each of these sets; call thatcollection Z . [Axiom of Choice!]Now S − X = F2Z = RZ ∪̇ SZ ∪̇ TZ ∪̇ UZ , and we canreassemble: S−X = a−1RZ ∪̇SZ and S−X = bTZ ∪̇UZ .
LemmaS − X ∼ 2(S − X ).
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Now S − X is the union of infinitely many sets like F2.Pick just one point from each of these sets; call thatcollection Z . [Axiom of Choice!]Now S − X = F2Z = RZ ∪̇ SZ ∪̇ TZ ∪̇ UZ , and we canreassemble: S−X = a−1RZ ∪̇SZ and S−X = bTZ ∪̇UZ .
LemmaS − X ∼ 2(S − X ).
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Doubling the surface
Putting these pieces together, we now can double the surfaceof a sphere!
PropositionS ∼ 2S.
Proof.S ∼ S − X ∼ 2(S − X ) ∼ 2S.
Can we double a solid ball?
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Doubling the surface
Putting these pieces together, we now can double the surfaceof a sphere!
PropositionS ∼ 2S.
Proof.S ∼ S − X ∼ 2(S − X ) ∼ 2S.
Can we double a solid ball?
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Doubling the surface
Putting these pieces together, we now can double the surfaceof a sphere!
PropositionS ∼ 2S.
Proof.S ∼ S − X ∼ 2(S − X ) ∼ 2S.
Can we double a solid ball?
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Filling in the ball
Now, to each piece of the sphere, glue all the points between itand the center!
LemmaB − {center} ∼ 2(B − {center}).
The last step: we need to show that B ∼ B − {center}.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Filling in the ball
Now, to each piece of the sphere, glue all the points between itand the center!
LemmaB − {center} ∼ 2(B − {center}).
The last step: we need to show that B ∼ B − {center}.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Filling in the ball
Now, to each piece of the sphere, glue all the points between itand the center!
LemmaB − {center} ∼ 2(B − {center}).
The last step: we need to show that B ∼ B − {center}.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Filling in the ball
Now, to each piece of the sphere, glue all the points between itand the center!
LemmaB − {center} ∼ 2(B − {center}).
The last step: we need to show that B ∼ B − {center}.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
The central point
Pick any circle within the ball through the center.Use our first trick: C ∼ C − {point}.Thus B ∼ B − {center}.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
The central point
Pick any circle within the ball through the center.Use our first trick: C ∼ C − {point}.Thus B ∼ B − {center}.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
The central point
Pick any circle within the ball through the center.Use our first trick: C ∼ C − {point}.Thus B ∼ B − {center}.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
The central point
Pick any circle within the ball through the center.Use our first trick: C ∼ C − {point}.Thus B ∼ B − {center}.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
The Banach-Tarski Theorem, Weaker Form
Theorem (Weak Banach-Tarski)B ∼ 2B.
Proof.B ∼ B − {center} ∼ 2(B − {center}) ∼ 2B.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
The Banach-Tarski Theorem, Weaker Form
Theorem (Weak Banach-Tarski)B ∼ 2B.
Proof.B ∼ B − {center} ∼ 2(B − {center}) ∼ 2B.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
The Banach-Tarski Theorem, Full Strength
Theorem (Banach-Tarski)
Let A and B be two bounded subsets of Rn with nonemptyinteriors. Then A ∼ B.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Morals of the Story: The Axiom of Choice
Some people don’t like the “fishy” Axiom of Choice, andthis is one reason why.However,
We can still have some paradoxes without it!The Axiom of Choice is needed to prove
that every vector space has a basisthat every field has an algebraic closurelots of other important results
Logicians have proved that if the Axiom of Choice couldcause a contradiction, then so would all mathematics evenwithout it!
So I have no qualms about using it.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Morals of the Story: The Axiom of Choice
Some people don’t like the “fishy” Axiom of Choice, andthis is one reason why.However,
We can still have some paradoxes without it!The Axiom of Choice is needed to prove
that every vector space has a basisthat every field has an algebraic closurelots of other important results
Logicians have proved that if the Axiom of Choice couldcause a contradiction, then so would all mathematics evenwithout it!
So I have no qualms about using it.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Morals of the Story: The Axiom of Choice
Some people don’t like the “fishy” Axiom of Choice, andthis is one reason why.However,
We can still have some paradoxes without it!The Axiom of Choice is needed to prove
that every vector space has a basisthat every field has an algebraic closurelots of other important results
Logicians have proved that if the Axiom of Choice couldcause a contradiction, then so would all mathematics evenwithout it!
So I have no qualms about using it.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Morals of the Story: The Axiom of Choice
Some people don’t like the “fishy” Axiom of Choice, andthis is one reason why.However,
We can still have some paradoxes without it!The Axiom of Choice is needed to prove
that every vector space has a basisthat every field has an algebraic closurelots of other important results
Logicians have proved that if the Axiom of Choice couldcause a contradiction, then so would all mathematics evenwithout it!
So I have no qualms about using it.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Morals of the Story: The Axiom of Choice
Some people don’t like the “fishy” Axiom of Choice, andthis is one reason why.However,
We can still have some paradoxes without it!The Axiom of Choice is needed to prove
that every vector space has a basisthat every field has an algebraic closurelots of other important results
Logicians have proved that if the Axiom of Choice couldcause a contradiction, then so would all mathematics evenwithout it!
So I have no qualms about using it.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Morals of the Story: The Axiom of Choice
Some people don’t like the “fishy” Axiom of Choice, andthis is one reason why.However,
We can still have some paradoxes without it!The Axiom of Choice is needed to prove
that every vector space has a basisthat every field has an algebraic closurelots of other important results
Logicians have proved that if the Axiom of Choice couldcause a contradiction, then so would all mathematics evenwithout it!
So I have no qualms about using it.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Morals of the Story: The Axiom of Choice
Some people don’t like the “fishy” Axiom of Choice, andthis is one reason why.However,
We can still have some paradoxes without it!The Axiom of Choice is needed to prove
that every vector space has a basisthat every field has an algebraic closurelots of other important results
Logicians have proved that if the Axiom of Choice couldcause a contradiction, then so would all mathematics evenwithout it!
So I have no qualms about using it.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Morals of the Story: The Axiom of Choice
Some people don’t like the “fishy” Axiom of Choice, andthis is one reason why.However,
We can still have some paradoxes without it!The Axiom of Choice is needed to prove
that every vector space has a basisthat every field has an algebraic closurelots of other important results
Logicians have proved that if the Axiom of Choice couldcause a contradiction, then so would all mathematics evenwithout it!
So I have no qualms about using it.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Morals of the Story: Measure Theory
Why is this called a “paradox”? Because rigid motionsshould preserve volume, but we’ve doubled volume!The real conclusion is there is no way to measure thevolume of all subsets of R3.So if you want a reasonable notion of volume you need torestrict your attention to “nice” sets.This also leads to the study of amenable groups—groupsthat avoid such paradoxes.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Morals of the Story: Measure Theory
Why is this called a “paradox”? Because rigid motionsshould preserve volume, but we’ve doubled volume!The real conclusion is there is no way to measure thevolume of all subsets of R3.So if you want a reasonable notion of volume you need torestrict your attention to “nice” sets.This also leads to the study of amenable groups—groupsthat avoid such paradoxes.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Morals of the Story: Measure Theory
Why is this called a “paradox”? Because rigid motionsshould preserve volume, but we’ve doubled volume!The real conclusion is there is no way to measure thevolume of all subsets of R3.So if you want a reasonable notion of volume you need torestrict your attention to “nice” sets.This also leads to the study of amenable groups—groupsthat avoid such paradoxes.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Morals of the Story: Measure Theory
Why is this called a “paradox”? Because rigid motionsshould preserve volume, but we’ve doubled volume!The real conclusion is there is no way to measure thevolume of all subsets of R3.So if you want a reasonable notion of volume you need torestrict your attention to “nice” sets.This also leads to the study of amenable groups—groupsthat avoid such paradoxes.
Hendrickson The Banach-Tarski Paradox
IntroductionParadoxes
Free GroupDoubling the Ball
Conclusion
Questions
Questions?
Hendrickson The Banach-Tarski Paradox