Re: abundance of irrationals!)
- From: Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx>
- Date: Tue, 10 May 2005 13:19:16 -0400
Randy Poe said:
>
> aeo6 Tony Orlow wrote:
> > I agree WM. We obviously have more naturals than squares, and more
> naturals
> > than even naturals. There are obviously more rationals than naturals.
>
> > Bijections based on counting paradigms fail miserably at
> distiguishing anything
> > about infinite sets.
>
> To criticize is to volunteer. Define set size please.
>
> In what sense does the bijective definition of cardinality
> "fail miserably"? It fails to satisfy Tony? When did that
> get written down as a requirement of mathematics?
It fails to distinguish between the sizes of the naturals and rationals, the
evens and naturals, the naturals and squares, and yet distinguishes between the
reals and the rationals for unjustified reasons, even though the rationals
constitute a dense subset of the reals, and essentially a type of enumeration
of them. It tried to have a set with a finite range, consisting of the sum of
an infinite number of finite ranges, which is patenetly impossible. Failure to
the nth degree.
>
> - Randy
>
>
The size of a set of real numbers defined using a function f(n) on the naturals
from x to y, is defined as the largest whole number less than the value of the
complementary function g(n) at y, minus the largest whole number less than the
value of the complementary function at x-1. The complementary function of f(n)
is defined as the function g(n) for which f(g(n))=g(f(n))=n.
Let's look at the set {2,3,4.....}. We map this to the naturals s.t. f(n)=n+1,
the complement of which is g(n)=n-1. We take the entire range, up to N (aleph_0
if you must), so if the entire set of naturals is N, this set is N-1. If we try
a finite range, say x=3,y=10, we get g(10)-g(2)=8 members of the set in that
range. Not convinced? Let's try another.
The set of all even naturals is defined by a mapping function on the naturals
of f(x)=2*x. The complementary function is g(x)=x/2, so over the entire range
of N, there are N/2 even naturals. For a finite range of [3,10], we have 10/2-
2/2=8/2, or 4 even numbers in the range of [3,10]: 4,6,8,10.
Okay, one more. The set of all squares is defined by a mapping function of f(n)
=n^2, so g(n)=sqrt(n). Therefore, there are sqrt(N) squares in N (this is only
comparable to other functions on N, not to finites, so calm down. It's no
specific number). Over a finite range [3,10] we have floor(sqrt(10))-floor(sqrt
(2))=3-1=2 squares in the range [3,10]: 4 and 9. In this case we needed to use
floor() because the complementary function results in non-whole numbers, and
must be mapped to whole numbers.
Okay, one more. Let's try the set of all powers of 2, so the mapping function
is f(n)=2^n. The complementary function is log2(n). Therefore, the set of all
powers of two in N is the log2(N). The set of all powers of 2 in [3,10] is
floor(log2(10))-floor(log2(2))=3-1=2 powers of two in the range [3,10]: 4 and
8.
Do you still think bijections are the most robust possible generalization from
finite to infinite sets? Do you still think I am a crank? Go "play" with it and
see if you can break it, and if you can't figure out why it works, put your
picturing cap on and ask.
--
Smiles,
Tony
.
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