In a deposition submitted under oath, William Elliot said:
 BS> I've just been learning it the last year or so in college.
 WE> What course?  Set theory?
    I believe this was covered in the two Discrete Math courses, as well
 as a bit in Linear Algebra.  My Computer Science major requires so much
 bloody math, I'll end up with a math minor by default.
 WE> What college level?
    I don't know.  I believe Discrete is sophomore, while Linear is
 Junior level.	I'm a senior going part-time, so I'm not taking much
 in sequence.  I'll be a senior until about 2001.  Hopefully, I'll
 graduate before I retire.	;^)
 WE> Now 2^A > A for all
 WE> cardinal numbers as proven originally by Cantor.  You yourself claim
 WE> that demunmerable is greater than the cardinality of the continuum,
 WE> which is an example of 2^A > A.  State your position on this
 WE> contentious point, that you can prove A^A = A, for infinite A as
 WE> claimed earlier.
    I don't know any more.  It all gets fuzzy in my brain when we define
 infinity as an endless set, then define a set larger than that.  How
 can you have something bigger than something that never ends?	I
 don't actually believe it; I got most of this from a book called
 Infinity and the Mind by Rudy Rucker.
    But, Mr Rucker says that, unlike finite ordinals, infinite ordinals
 are not commutative, therefore 1 + A = A, but A + 1 = A + 1, and 2 * A
 = A, but A * 2 = A + A.  Furthermore, starting with 0 and continuously
 adding 1, you count through the ordinals to get:
 0, 1, 2, ..., A, A + 1, A + 2, ..., A * 2, A * 2 + 1, A * 2 + 2, ...,
 A * 2 + A (aka A * 3), and continue through A * n for each finite n,
 and on to A * A, which is also A ^ 2, then to A ^ 3, A ^ 3, to A ^ A
 and on even further.
    So, I guess you are right and I was mistaken.  But, like I said, I
 really don't believe there is any number past A because the very
 definition of A seems to me to include any and all numbers greater than
 A.
 WE> Don't remember Cantor's proof
    I'm not sure I can get it into ASCII format on the screen here, but
 I'll try.  First you put all the rational numbers on a grid as shown
 below:
    1    1    1    1    1    1
    - -> -    - -> -    - -> - . . .
    1    2    3    4    5    6
      /      ^    /    ^    /
     /      /    /    /    /
    v      /    /    /    /
          /    v    /    v
    2    x    2    x    2    2
    -    -    -    -    -    - . . .
    1    x    3    x    5    6
        ^    /    ^    /
    |  /    /    /    /
    v /    /    /    /
          v    /    v
    3    3    x    3    3    3
    -    -    -    -    -    - . . .
    1    2    x    4    5    6
        /    ^    /
       /    /    /
      /    /    /
     v    /    v
    4    x    4    4    4    4
    -    -    -    -    -    - . . .
    1    x    3    4    5    6
        ^    /
    |  /    /
    v /    /
     /    v
    5    5    5    5    5    5
    -    -    -    -    -    - . . .
    1    2    3    4    5    6
        /
       /
      /
     v
    6    6    6    6    6    6
    -    -    -    -    -    - . . .
    1    2    3    4    5    6
    |
    v
    .    .    .    .    .    .
    .    .    .    .    .    .
    .    .    .    .    .    .
    Equivalent fractions are x'ed out as we come to them because we
 don't want to count duplicates (1/2 = 2/4, 6/2 = 3, etc.).
    Anyway, define a function F from Z+ to Q+ (positive rational
 numbers) by starting a count at 1/1 and following the arrows (if you
 can; it ain't easy in ASCII!) as indicated, skipping over any numbers
 already counted.  To be specific, set F(1) = 1/1, F(2) = 1/2, F(3) = 2/1,
 skip 2/2 since 2/2 = 1/1, then F(4) = 3/1, F(5) = 1/3, F(6) = 1/4, etc.
 Continue in this way, defining F(n) for each positive integer n.
    Note that every positive rational number appears somewhere in the
 grid, and the counting procedure is set up so that every point in the
 grid is reached eventually.  Thus the function is ONTO.  Also, by
 skipping numbers that have already been counted, no number is counted
 twice, hence F is one-to-one.	Because F is a one-to-one correspondence
 from Z+ to Q+, then Q+ (the positive rational numbers) is countably
 infinite and therefore is countable.
... A conclusion is simply the place where you got tired of thinking.
--- PPoint 2.05
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