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 a 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.

Diagonalization Definition: P(N) is the set of all subsets of N. 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).

Consider the following table: f: N => P(N) 0 => P0, 1 => P1, 2 => P2, … *f is “onto” so every set in P(N) is in this list. Consider the following table: 0 1 2 3 … P0 d00 d01 d02 d03 P1 d10 d11 d12 d13 P2 d20 d21 d22 d23 : dij = *The table for f provides a 2 dimensional bit vector. Example: Set {2, 5, 6} => P* =0010011000…. , 1’s only for the present number

Consider/define the set D such that for each j ≥ 0: if and only if (*) Note that D is represented by the complement (1 0) of the diagonal. Observations: D is a subset of P Since P0, P1, P2, … is a list of all the subsets of P, it follows that D = Pi (**), for some i ≥ 0. Question: Is ? By definition of D given in *, if and only if But D = Pi by **, and substitution gives if and only if A contradiction. Hence, P(N) is uncountable. 

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.

f: N => R 0 => r0, 1 => r1, 2 => 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 zero’s to the right) Consider real numbers over [0, 1), or xi is 0 *The table is a 2 dimensional vector of digits.

Consider/define the real number: y = 0.y0y1y2…(infinite) where: yi = (dii +1) mod 10 for all i ≥ 0 (*) 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 j ≥ 0. This means that y = rj = 0.dj0 dj1 dj2 … djj-1 djj djj+1 … (from the table) This also means that yi = dji, for all i ≥ 0, 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. So, real numbers with 0 as the integer part is uncountably infinite. Similarly, reals with 1 as the integer part in uncounatbly infinite, and so on. 

Rational number set is countable: represent with integers p and q (for p/q) over a matrix, and index them in a special cross-diagonal pattern Is the set of 2-tuples of integers (p, q) countable? 3-tuples? n-tuples, for constant n?