Diagonalization - PowerPoint PPT Presentation

1 / 7
About This Presentation
Title:

Diagonalization

Description:

The Special Book is defined to be that book that contains the titles of all ... A similar contradiction is known as The Barber of Seville Paradox. 2. Diagonalization ... – PowerPoint PPT presentation

Number of Views:340
Avg rating:3.0/5.0
Slides: 8
Provided by: fit
Category:

less

Transcript and Presenter's Notes

Title: Diagonalization


1
Diagonalization
  • Fact Many books exist.
  • Fact Some books contain the titles of other
    books within them.
  • Fact Some books contain their own titles within
    them.
  • Consider the following book with title The
    Special Book.
  • The Special Book is defined to be that book that
    contains the titles of all books that do not
    contain their own titles.
  • Question Does the special book exist? Could one
    write the special book?
  • A similar contradiction is known as The Barber of
    Seville Paradox.

2
Diagonalization
  • Definition
  • P(N) is the set of all subsets of N.
  • P(N) , 0, 0, 1, 1, 2, 3, 2, 5, 9,
    13,
  • Theorem P(N) is uncountable.
  • Proof (by contradiction) Suppose that P(N) is
    countable. Then by definition it is either
    finite or countably infinite. Clearly, it is not
    finite, therefore it must be countably infinite.
    By definition, since it is countably infinite it
    has the same cardinality as N (the natural
    numbers) and, by definition, there is a bijection
    from N to P(N).

3
  • f N gt P(N)
  • 0 gt N0, 1 gt N1, 2 gt N2, f is onto so
    every set in P(N) is in this
  • list.
  • Consider the following table
  • 0 1 2 3
  • N0 d00 d01 d02 d03
  • N1 d10 d11 d12 d13
  • N2 d20 d21 d22 d23
  • dij The table is a 2 dimensional bit vector.

4
  • Consider/define the set D such that for each j gt
    0
  • if and only if ()
  • Note that D is represented by the complement of
    the diagonal.
  • Observations
  • D is a subset of N
  • Since N0, N1, N2, is a list of all the subsets
    of N, it follows that D Ni (), for some i gt
    0.
  • Question Is ?
  • By definition of D given in , if and
    only if
  • But D Ni by , and substitution gives
    if and only if

5
Diagonalization
  • Theorem The real numbers are uncountable.
  • Proof (by contradiction) Let R denote the set of
    all real numbers, and suppose that R is
    countable. Then by definition it is either
    finite or countably infinite. Clearly, it is not
    finite, therefore it must be countably infinite.
    By definition, since it is countably infinite it
    has the same cardinality as N (the natural
    numbers) and, by definition, there is a bijection
    from N to R.

6
  • f N gt R
  • 0 gt r0, 1 gt r1, 2 gt r2, f is onto so
    every real number is in this list.
  • Consider the following table
  • 0 1 2 3
  • r0 d00 d01 d02 d03
  • r1 d10 d11 d12 d13
  • r2 d20 d21 d22 d23
  • where ri xi.di0 di1 di2 dim (padded with
    zeros to the right)
  • The table is a 2 dimensional vector of digits.

7
  • Consider/define the real number
  • y 0.y0y1y2(infinite)
  • where
  • yi (dii 1) mod 10 for all igt0 ()
  • Observations
  • y is a real number
  • Since r0, r1, r2, is a list of all real
    numbers, it follows that y must be in this list,
    i.e., y rj, for some jgt0.
  • This means that y rj 0.dj0 dj1 dj2 djj-1
    djj djj1 (from the table)
  • This also means that yi dji, for all igt0, and,
    in particular, that
  • yj djj
  • But by
  • yj (djj 1) mod 10
  • a contradiction. Therefore, no such one-to-one
    and onto function exists, and therefore the real
    numbers are uncountable.
Write a Comment
User Comments (0)
About PowerShow.com