The Banach-Tarski Paradox - Concordia Collegefaculty.cord.edu/ahendric/presentations/HendricksonBanachTarski.pdf · The Banach-Tarski Paradox Anders O.F. Hendrickson Department of

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



Conclusion

Outline

1 Introduction

2 Paradoxes

3 Free Group

4 Doubling the Ball

5 Conclusion




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.




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:

−→




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:

−→




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 .




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 .




Conclusion

Transitivity of ∼

TheoremIf A ∼ B and B ∼ C, then A ∼ C.

Proof sketch.

−→ −→




Conclusion

Transitivity of ∼


Proof sketch.

−→ −→




Conclusion

Transitivity of ∼


Proof sketch.

−→

−→




Conclusion

Transitivity of ∼


Proof sketch.

−→ −→




Conclusion

Transitivity of ∼


Proof sketch.

−→

−→




Conclusion

Transitivity of ∼


Proof sketch.

−→ −→




Conclusion

Transitivity of ∼


Proof sketch.

−→ −→




Conclusion

Transitivity of ∼


Proof sketch.

−→ −→




Conclusion

Transitivity of ∼


Proof sketch.

−→ −→




Conclusion

Transitivity of ∼


Proof sketch.

−→ −→�




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.)




Conclusion


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!




Conclusion


Alice Bob Carlos David Edgar Fred · · ·

Thomas





Conclusion



Thomas





Conclusion



Thomas





Conclusion


Alice Bob Carlos David Edgar · · ·

Thomas





Conclusion


Alice Bob Carlos David Edgar · · ·

Thomas





Conclusion


Thomas Alice Bob Carlos David Edgar · · ·

Thomas





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.




Conclusion



m−n ∈ Q.




Conclusion



m−n ∈ Q.




Conclusion



m−n ∈ Q.




Conclusion



m−n ∈ Q.




Conclusion



m−n ∈ Q.




Conclusion



m−n ∈ Q.




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.




Conclusion

Our first paradox





Conclusion

Our first paradox





Conclusion

Our first paradox





Conclusion

Our first paradox





Conclusion

Our first paradox

−→

LemmaC ∼ C − {point}




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




Conclusion

The free group F2






Conclusion

The free group F2






Conclusion

The free group F2






Conclusion

The free group F2






Conclusion

Decomposition of F2

Dissect F2 into R ∪̇ S ∪̇ T ∪̇ U.




Conclusion

Decomposition of F2

Dissect F2 into R ∪̇ S ∪̇ T ∪̇ U.




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?




Conclusion

Reassembly of F2

−→





Conclusion

Reassembly of F2

−→ &





Conclusion

Reassembly of F2

−→ &





Conclusion

Reassembly of F2

−→ &





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.




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.




Conclusion








Conclusion








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 .




Conclusion

F2 on the sphere





Conclusion

F2 on the sphere





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 .




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 .




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.




Conclusion

Bad Points


r1x = r2x =⇒ (r2)−1r1x = x ,






Conclusion

Bad Points


r1x = r2x =⇒ (r2)−1r1x = x ,






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!




Conclusion






Conclusion






Conclusion






Conclusion


LemmaS ∼ S − X.




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 ).




Conclusion


LemmaS − X ∼ 2(S − X ).




Conclusion


LemmaS − X ∼ 2(S − X ).




Conclusion


LemmaS − X ∼ 2(S − X ).




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?




Conclusion




Proof.S ∼ S − X ∼ 2(S − X ) ∼ 2S.





Conclusion




Proof.S ∼ S − X ∼ 2(S − X ) ∼ 2S.





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}.




Conclusion

Filling in the ball







Conclusion

Filling in the ball







Conclusion

Filling in 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}.




Conclusion

The central point





Conclusion

The central point





Conclusion

The central point





Conclusion

The Banach-Tarski Theorem, Weaker Form

Theorem (Weak Banach-Tarski)B ∼ 2B.

Proof.B ∼ B − {center} ∼ 2(B − {center}) ∼ 2B.




Conclusion

The Banach-Tarski Theorem, Weaker Form

Theorem (Weak Banach-Tarski)B ∼ 2B.

Proof.B ∼ B − {center} ∼ 2(B − {center}) ∼ 2B.




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.




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.




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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.




Conclusion






Conclusion






Conclusion






Conclusion

Questions

Questions?


Documents

The Banach-Tarski Paradox - Concordia Collegefaculty.cord.edu/ahendric/presentations/HendricksonBanachTarski.pdf · The Banach-Tarski Paradox Anders O.F. Hendrickson Department of