>>> Bob Sewell on Infinity 
 BS> How can you have something bigger than something that never ends? I
 BS> don't actually believe it; I got most of this from a book called
 BS> Infinity and the Mind by Rudy Rucker.
The integers and the real numbers.  The integers never end 1,2,3,.... but the 
reals never end either 1,2,3... but between every two are infinity many more. 
 between 1 and 2, 2 and 3, etc.  But between every two numbers in the 
infinity of numbers between 1 and 2 is yet another infinity and so on without 
limit infinitely more and more numbers cramming in between infinities of 
numbers.  These infinities within infinities are so dense that you can't list 
them one after another like you can integers.  What's the 'next' number after 
1/2?  .51, .501, .5001, ...?  See, you can't tell me the next.
 BS> But, Mr Rucker says that, unlike finite ordinals, infinite
 BS> ordinals are not commutative, therefore 1 + A = A, but A + 1 = A + 1,
 BS> and 2 * A = A, but A * 2 = A + A.  Furthermore, starting with 0 and
 BS> continuously adding 1, you count through the ordinals to get:
Be careful about notation as w is the ordinal number of the integers, A0, the 
cardinal.  w + 1 is 1,2,3,...; 1 while 1 + w is 1,1,2,3,4,...  So w + w is 
actually 1,2,3,....; 1,2,3.... and there is no 1-1 order preserving 
correspondence between w and w + 1 or between w and w + w.  But there is one 
between 1 + w and w.  W, capital omega, is the ordinal number of the next 
largest cardinal A1, conjectured to be cardinal number of the reals.
 BS> 0, 1, 2, ..., A, A + 1, A + 2, ..., A * 2, A * 2 + 1, A * 2 + 2, ...,
 BS> A * 2 + A (aka A * 3), and continue through A * n for each finite n,
 BS> and on to A * A, which is also A ^ 2, then to A ^ 3, A ^ 3, to A ^ A
 BS> and on even further.
w^w, w^w^w, etc until w^w^w^w^.....  Drives you crazy, polynomials of 
denumerable ordinals.  Finite ordinals and cardinals are the same, so that's 
why this distinction is so novel.
 BS> So, I guess you are right and I was mistaken.  But, like I said, I
 BS> really don't believe there is any number past A because the very
 BS> definition of A seems to me to include any and all numbers greater
 BS> than A.
Let's assume that all of the real numbers in the unit interval are 
denumerable.  So we can list all of them.  Let a1 a2 etc be digits as also 
b1, c1 and so on for as many digits as we need.  Note each real number in the 
unit interval is a denumerable series of digits.  So I list the numbers:
#1 = 0.a1 a2 a3 a4 ...
#2 = 0.b1 b2 b3 b4 ...
#3 = 0.c1 c2 c3 c4 ...
#4 = 0.d1 d2 d3 d4 ...
etc for 5,6,7...
Now I show there is a real number #x not in this list.
Chose a x1 other than a1.  Chose a x2 other than b2.  Chose a x3 other than 
c3.  Chose a x4 other than d4.  Continuing choosing x5, x6, etc such that x5 
is a different digit that the fifth digit of #5 and x6 is a different digit 
that the sixth digit of #6 and so on for a denumerable series of digits.  
Thus #x = 0.x1 x2 x3 x4 ... is a real number that is not included in the 
original list.  It can't be #1 because x1  a1.  It can't be #2 because x2 
 b2, and so on for all the numbers in the list.  Hence the list cannot be a 
1-1 correspondence of integers to real numbers.  So the cardinality of the 
reals cannot equal the cardinality of the integers.  Since, for every integer 
n there is a real number 1/n in the unit interval we now see that there are 
fewer integers than real numbers.
This is the diagonal argument.  Are you familiar with set theory?  Do you 
know about P(S) the power set of S which the set of all sets included in the 
set S, {x | x included in S}?  A simple generalized diagonal argument can be 
used to show the the cardinality of P(S) is greater than the cardinality of S 
itself.  So we can continue S, P(S), P(P(S)), etc with a never ending series 
of ever larger infinities.  But that's not the end, the whole series can be 
leap frogged to an even greater number than all of the previous numbers.  
There's no end to this series of ever greater infinities.
 BS> I'm not sure I can get it into ASCII format on the screen here,
 BS> but I'll try.  First you put all the rational numbers on a grid as
 BS> shown below:
 BS> 1    1    1    1    1    1
 BS> - -> -    - -> -    - -> - . . .
 BS> 1    2    3    4    5    6
This is not the Cantor diagonal argument.  It however is an enumeration of D 
* D showing that D * D = D where D is the denumerable cardinality of the 
integers.
 BS> Equivalent fractions are x'ed out as we come to them because we
 BS> don't want to count duplicates (1/2 = 2/4, 6/2 = 3, etc.).
Let R be the number of rationals.  Now D <= R <= D * D.  But D = D * D, so R 
= D.
 BS> Anyway, define a function F from Z+ to Q+ (positive rational
 BS> numbers) by starting a count at 1/1 and following the arrows (if you
 BS> can; it ain't easy in ASCII!) as indicated, skipping over any numbers
 BS> already counted.  To be specific, set F(1) = 1/1, F(2) = 1/2, F(3) =
 BS> 2/1, skip 2/2 since 2/2 = 1/1, then F(4) = 3/1, F(5) = 1/3, F(6) =
 BS> 1/4, etc. Continue in this way, defining F(n) for each positive
 BS> integer n. 
Correct, or you can use the short cut as above to avoid the skipping.
 BS> then Q+ (the positive rational numbers)
 BS> is countably infinite and therefore is countable.
I agree, and the reals are uncountable, hence more numerous than the 
ntegers.
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