Investigation of discontinuities with Fourier series.
Set Theory Begins.

  • Any infinite subset of the natural numbers or the integers is countable.

  • The rational numbers are a countable set.
  • 1/1 2/1 3/1 4/1 5/1 6/1 ...
    1/2 2/2 3/2 4/2 5/2 6/2 ...
    1/3 2/3 3/3 4/3 5/3 6/3 ...
    1/4 2/4 3/4 4/4 5/4 6/4 ...
    1/5 2/5 3/5 4/5 5/5 6/5 ...
    1/6 2/6 3/6 4/6 5/6 6/6 ...
    ... ... ... ... ... ... ...

    0, 1/1,   1/2, 2/1,     3/1, 2/2, 1/3,   1/4, 2/3, 3/2, 4/1,  ...

  • The algebraic numbers are countable. [ Another first type of diagonal argument.] (1874)



    Definition 1. An algebraic number is the solution to a polynomial equation with integer coefficients.

    Fact 2. A polynomial equation of degree N can have at most N distinct real number solutions.

    Fact 3. There are a finite number of integer polynomial equations of degree N where the sum of the absolute values of the coefficients is less than a given number K.

    Proof Procedure: (modified slightly from Cantor's)
    List algebraic number roots from integer polynomials of degree 1 with sum of the absolute values of the coefficients equal to 1. (This is just 1x=0 and -1x=0.)

    use degree 1, sum 2;
    (This is 2x=0, -2x=0, 1+1x=0, -1+1x=0, 1-1x=0, -1-1x=0.)
    then degree 2, sum 1;
    then degree 3, sum 1;
    then degree 2, sum 2;
    then degree 1, sum 3;. ...

  • Cantor's proof that the number of points on a line segment are uncountable. (1874)

  • A decimal based proof that there are an uncountable set of real numbers.

  • (A variation similar to 1891 proof)

    Consider a sequence of integers, with each integer between 0 and 9.
    If we let kth integer of the sequence determine the 2k+1 th decimal place of a number then we can determine a unique real number by specifying that the 2k th decimal be a 5.

    Let S be the set of of all real numbers that are specified in this fashion.
    S is a subset of [0,1].

    Suppose there is a function,  f : N ->  S.
    Then let b=.b1 5 b3 5 b5 5 b7 5 b9 5 ... where we choose  b2k+1 so that it is not equal to the 2k+1st digit of f(k).

    Thus b is a member of s but b not equal to f(k) for any k, so f is not "onto".
    Hence S is not a countable set, and consequently neither is [0,1].

    Note: The set of {0,1} - valued  sequences is also "uncountable."

  • There is no onto function from R, the set of real numbers, to P(R), the set of all subsets of the real number.



    Proof: Suppose f: R -> P(R).
    Let  B = {x such that x is not an element of f(x)}.
    Suppose B = f(b) for some b.
    If b is in B then b is not in f(b)=B, which is a contradiction.
    So b is not in B, but then b is not in f(b), so b is in B!

    Thus B is not f(b) for any b, and f is not onto.

    Thus: There are sets which are larger than the reals.

    Footnote  on measuring sets of real numbers:

    The rational numbers between 0 and 1 have  "measure" zero.
    Proof:  List the rationals a1,a2,a3,...  and for any number s,  use an interval of length s/2k about ak. Then the set of rational numbers has measure less than s. Since s can be any number, the set of rationals between 0 and 1 has measure 0.

    Generalization:  Any countable set of real numbers has "measure" zero.