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echo: philos
to: BOB SEWELL
from: WILLIAM ELLIOT
date: 1998-03-29 02:50:00
subject: Infinity Proofs

 >>> Bob Sewell on Infinity 
 BS> This also shows that Z+ = w and Z = w and that w (the infinite
 BS> set of positive integers) equals what looks like 2 * w (all the
 BS> integers, positive and negative). 
The bijection that proves that Z+ = Z, is x  x+1 over non-negative x.
What's w, an arbitrary infinite number?  Which one?
 BS> I can also show that the set of all even integers 2Z is countably
 BS> infinite, which by the definition of cardinality, means that this
 BS> infinite set has the same cardinality as Z+, and by the last proof,
 BS> the same cardinality as Z.  This is one way of saying that, as far as
 BS> cardinality goes, Z = Z+ = 2Z.
Elementary.
 BS> If you want, I'll show that the cardinalities of the set of
 BS> positive rational numbers equals that of the set of *all* rational
 BS> numbers equals that of Z+.  I can show that the set of all real
 BS> numbers between 0 and 1 is uncountable, and yet this set has the same
 BS> cardinality as the set of *all* real numbers, which is a way of
 BS> stating that w ^ w = w.
Set D = the denumberable cardinality of the integers, C = the cardinality of 
the continuum.  So the cardinality of the interval (0,1) = C.  Now the real 
numbers can be broken into a denumberable number of intervals (n,n+1) for all 
integers positive and negative.  Hence the cardinality of the the real number 
line is D * C = C.  It is not C^C, that is something else.
A plane has one horizontal line thru each point of a vertical axis.  Hence 
the cardinality of the plane is C * C = C.  Continuing, the cardinality of 
space is C^3 = C and of n dimensional space C^n = C.
 BS> I'm not following where this is leading or how it relates to
 BS> whether or not w ^ w > w or not.
You agree that A^A > 2^A?  I maintain that 2^A > A.  Hence A^A > A and cannot 
equal A.  By decimal or binary notation for real numbers 10^D = 2^D = C and 
C > D as you say you can prove.  Indeed you can by a Cantor diagonal 
argument.  Hence as I maintain, 2^D > D.
What is a proof for A + B = max(A,B) when either A or B is infinite?
Actually A + A = A for infinite A would suffice.
Card(S) + Card(T) = max(Card(S), Card(T)) for arbitrary infinite sets S & T.
Proofs for integers and reals are easy, but for any sets, what's the proof?
Actually Card(S) + Card(S) = Card(S) for infinite set S would suffice.
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