Quote:
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Originally Posted by maddog
I can help a little bit. You acknowledge that there are an infinite number of integers. In fact you can
enumerate them. The same for the rationals. There is a proof that I am unable to give there are as many
rationals as integers. However, the quantity of reals are more than the integers. Cantor came up with a
notion of transfinite numbers or "levels of infinity". The number of integers are assigned Aleph 0 (naught),
for the subscripted Hebrew character. Number of reals is Aleph 1. The number of continuous functions over
the reals Rn is Aleph 2. And so on. For more info, do a google search on Cantor and Transfinite numbers.
Maddog
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Well, i know that the set of natural , rational, intergers are denumerable so then they share the same cardinality, and i believe the book called it "aleph nunght"( called it N).
Then the book show that there is another set of number
c(stand for continum) that are of greater cardiinary than aleph nunght.
Definition:
c is the cardinaity of the set of
real numbers
in other word
N <
c
How to find a cardinary larger than
c?
In order to do so, the book proof a general statement that is ---a set is always a proper subset of it `s power set.
in other word
Let `s denote suppose "K" as a set. then it`s power set, called it P[k].
then this relationship always hold:
A theorm: "cordinarity K < cordinarity P[K] "
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briefing on Power set:
suppose there are n element in a set K ( or the cardinity of K)
then the cardinairty of set K, P[K] is 2^n
example:
K={ a, b, c}
P[K] ={ {}, {a}, {b}, {c}, {a,b}, {a, c} , {b, c}, {a, b, c}}
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In other word
By thinking of the cardinarity of
c and it power set P[
c]
then the relationship must also hold
c < P[
c]
using the same type of reasoning
c< P[c]< P[P[c]]< P[P[P[c]]]<.............................
thus estable the order of the infinite
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I Think
If say the set of P is {a, b }, then the number of element in the power set of p mush be 2^2
listing them: {{}, {a}, {b}, {a,b}}
If the cardinaity of the real is
c, then the power set of
c, P[
c] is 2^c
continue the same reasoning
N <
c < 2^
c < 2^
c^
c< 2^
c^
c^
c <....................................
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How do you list the element of 2^c or 2^c^c or........?
To tell you the truth, it is soo counter intuitive that i have a hard time accepting it. The whole thing invoke the image of god to me. what are other people s thoughts? philosophies?...
perhaps it should be in the philosophy section.
It is such a amusement that it should be worth wild to discuss the philosophy or even God.
If human can tame the infinite, then can god be tame? or by reason alone, can we understand god itself??????????????[/B]