Infinity and Its Paradoxes

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Infinity and Its Paradoxes
Jonathan Choate Groton School jchoate_at_groton.org w
ww.zebragraph.com
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The Tortoise and the Hare

• A Tortoise and a Hare have a race. Being sure he
  will win since he can run at a rate of 10 m/sec
  and the Hare can only run at a rate of 1 m/sec,
  the Hare gives the Tortoise a 10 meter head
  start.

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• This was a big mistake since
• The Hare must first get to where the Tortoise
  started but during that time the Tortoise has
  moved forward to some position P1.

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• Now the Hare has to get to P1, but the Hare has
  now moved to P2.
• Now the Hare has to get to P2, but the Hare has
  now moved to P3.
• Now the Hare has to get to P3, but the Hare has
  now moved to P4.
• on and on and on and . .

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• Therefore, the Hare never catches the
  Tortoise!!!!!!

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• But lets look at how long the race lasted.

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A Monster Curve
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• The Koch Curve is the limit of this process
• and has
• -an infinite number of sides with length 0
• -an infinite perimeter.
• A bounded curve with an infinite perimeter!
• Wow!

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A Monster Shape

• Start with a tetrahedron with edges of length
  1, total surface area S, volume V and perimeter 6.

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• Make 4 copies of the tetrahedron scaled by a
  factor of ½ and you get a shape which has
• - a surface area of 4 (1/4)S S
• - a volume of 4(1/8)V (1/2)V
• - a perimeter of 4(1/2)6 12

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• Make 4 copies of the previous shape scaled by
  a factor of ½ and you get a shape which has
• - a surface area of 4 (1/4)S S
• - a volume of (1/2) (1/2)V (1/4)V
• -a perimeter of 4(1/2)12 24

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• Make 4 copies of the previous shape scaled by a
  factor of ½ and you get a shape which has
• - a surface area of 4 (1/4)S S
• - a volume of (1/2) (1/4)V (1/8)V
• - a perimeter of 4(1/2)24 48

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• Repeat this process several more times and
  you get

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• The limiting shape of this process will have
• - a surface area of S
• - a perimeter of length
• - volume of 0

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• Welcome to Hilberts Hotel
• The hotel which is full but never full because it
  always has rooms.
• How does Heir Hilbert do it?

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• If you arrived and needed a room, he would get
  on the PA system and announce

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• Attention Everybody add 1 to your room number
  and move.
• And then he would give you the key to room 1.
• 1

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• If 100 new people arrived and needed rooms,
  Heir Hilbert would get on the PA system and
  announce

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• Attention Everybody add 100 to your room number
  and move.
• And then he would hand out the keys to the first
  100 rooms.
• 100

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• If a bus with new people arrived and
  needed rooms, Heir Hilbert would get on the PA
  system and announce

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• Attention Everybody double your room number
  and move to that room.
• He would then turn to all the newcomers who
  had just arrived in a bus with numbered seats
  and announce.

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• Please take your seat number and subtract 1
  from it, multiply the result by 2 and add 1. This
  is your room number.
• He would then hand out all the odd numbered
  keys.

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• If buses arrived each with
  people all who needed rooms, Heir Hilbert would
  get on the PA system and announce

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• Attention Please subtract 1 from your room
  number, multiply the result by 2 and add 1.

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• He then would say to the new arrivals take your
  reservation number and multiply it by 2. This is
  your room number.
• He would then hand keys for all the even
  numbered rooms

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• Unfortunately, no one had given the new arrivals
  their reservation number so Heir Hilbert had to
  figure out a way of assigning each new arrival a
  unique reservation number.

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• Can you come up with a formula that given a
  persons bus number M and seat number N will
  provide a unique number R?

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Seat Number
Bus
31
Seat Number
Bus
32
x
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The Continuum Chateau

• Could there exist a hotel that has more rooms
  than Hilberts Hotel. Consider the following
• Assume that there is a hotel whose rooms are
  labeled with all possible infinite decimal
  expansions of the form
• 0 .d1d2d3d4. . .

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• Form the list where dij is the digit in the
• i-th number in the list in the j-th place.Here
  is the list with where they are in the list
• Room Number
• 1 0.d11d12d13d14d15. . . .. .
• 2 0.d21d22d23d24d25.. . .
• 3 0.d31d32d33d34d35.. . .
• 
  Etc.

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• There is a problem here..
• The room whose assigned number differs from
• -the first number in the first decimal place
  i.e it is not equal to .d11 .
• -the second number in the second decimal place
  i.e it is not equal to .d22 .

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• -the third number in the third decimal place i.e
  it is not equal to .d33 .
• -the n-th number in the n-th place i.e it is not
  equal to .dnn .
• The number 0.x1x2x3x4. . . where xi is some
  integer between 0 and 9 such that xi ? dii is not
  in the list.

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• Contradiction! Therefore the set of rooms with
  room numbers of the form 0. 0.d1d2d3d4. . . is
  not countable and its cardinality must be greater
  than .
• The preceeding is an argument for the fact that
  there are more positive reals on the interval ( 0
  , 1 ) than there are rationals.
• But is the source of a theorem that still has
  no proof.

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The Continuum Hypothesis

• There is no set whose size is strictly between
  that of the integers and that of the real
  numbers.
• This is one of David Hilberts 10 Great
  Problems which he stated in 1900. Prove it and
  youll be very, very famous!

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The Cantor Set
The Cantor Set is the limit of this process. How
many points in the Cantor Set?
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• Calculate how much of the Unit Interval you
  have removed as you create the Cantor Set.

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Binary Numbers

• In our base 10 system the number 357.45 is a
  shorthand for the polynomial
• 310 2 510 1 7 10 0 410-1 510-2
• In base 2, system the number 101.11 is a
  shorthand for the polynomial
• 12 2 02 1 12 012 -112-2

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• In base 10
• 122 021 12012-1 12-2
• equals
• 4 1 ½ ¼ 5 .75
• Every point in the interval 0, 1 can be
  expressed as a binary decimal.

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Ternary Numbers

• These are numbers whose digits can be 0, 1, or
  2.
• -If a ternary decimal starts with a 0, it must
  lie in the first third of the unit interval.
• -If a ternary decimal starts with a 1, it must
  lie in the middle third of the unit interval.
• -If a ternary decimal starts with a 2, it must
  lie in the second third of the unit interval.

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• Going back to the Cantor Set
• When you knock out the middle third, you have
  knocked out all decimals which have a leading
  decimal of 1. This means no point in the Cantor
  Set can have a leading digit of 1. Put another
  way the leading digit of any point in the Cantor
  Set has to be either a 0 or a 2

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• By a similar argument, you can show that when you
  knock out the middle thirds of the
• intervals 0,1/3 and 2/3, 1 , you are left
  with the intervals
• When the middle thirds of these two intervals are
  removed , you have eliminated all decimals which
  have a 1 in the second place.

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• In a similar manner, you can argue that the
  Cantor Set is the set of all decimals of the form
  0.x1x2x3x4 where ti is either 0 or 2.
• Take the entire set of points in the Cantor Set
  and replace every 2 with a 1 and you have all
  possible decimals of the form
• 0.b1b2b3b4. . .
• Where each bi is either 0 or 1.

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• But this means that there is 1-1
  correspondence between the points in the Cantor
  Set and the points in the Unit interval. Which
  means they have the same cardinality!
  YIKES!!!!!!!