Re: Two results of set geometry
- From: David R Tribble <david@xxxxxxxxxxx>
- Date: Thu, 04 Oct 2007 22:08:50 -0000
Tony Orlow wrote:
But, it doesn't make sense to have a
smallest infinity, especially when you have infinite sets that are
clearly smaller than N, like E.
David R Tribble wrote:
Smaller in what sense? We're still waiting for a convincing
argument from you that somehow quantifies these "relative
infinite set sizes".
Tony Orlow wrote:
Smaller in the sense that it's a proper subset, and especially such that
a countably infinite number of elements have been removed from N to get
E. E is obviously half of N.
David R Tribble wrote:
Yet it contains the same number (or "count") of elements.
Tony Orlow wrote:
No, it counts every second one, according to the inverse mapping N(e) = e/2.
No, it counts every single even, according to the mapping
f(k in N) = 2k. There is an even for every natural. So N
and E have the same "count" of elements.
The correct inverse is f(e) = e/2 for all e in E.
That's "e in E", not "e in N".
David R Tribble wrote:
So E is "half the size" of N, yet it's also the "same size"
as N. Which is why we keep asking you to be more specific
and define exactly what you mean by "size" and "count".
Tony Orlow wrote:
It's the same cardinality, half the Bigulosity (always capitalized).
So far, that's only because you say so, not because
you've proved it so.
David R Tribble wrote:
Natural density produces every result you've ever claimed
for your IFR, and then some. You claim that your IFR is
superior, so show us some cases where it yields more
meaningful results.
Tony Orlow wrote:
The set of square roots of naturals is omega^2, by IFR. By infinite-case
induction that's more than omega and less than 2^omega, since omega>1.
CH is solved. End of discussion. :)
Okay, so all you have to do now is prove that
omega < omega^2 < 2^omega, and you'll get that
Fields Medal you've always wanted. :-)
David R Tribble wrote:
I particularly like the result that d(P), the natural density
of the primes, is 0.
Tony Orlow wrote:
That's true. The natural density of N is 0, and P is a proper subset of N.
No, the natural density of N is d(N) = 1, by definition.
There is one natural for every natural, in a ratio of 1:1,
in other words.
The relative density of primes pi(n) tapers off as n gets larger,
so pi(n)/n approaches zero. Which makes it clear why
d(P) = 0.
David R Tribble wrote:
Which makes sense because
pi(n)/n ~ 1/log(n), which tends towards 0. Does IFR do any better?
Tony Orlow wrote:
Yes, IFR distinguishes other 0-measure sets from each other, with
infinite ratios described formulaically on an infinite variable.
Examples?
And what about non-zero-measure sets "described formulaically
on an infinite variable"?
.
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