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Re: Cantor's first proof in DETAILS
Posted:
Dec 5, 2012 11:06 PM
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On Dec 4, 1:22 pm, Virgil <vir...@ligriv.com> wrote:
> In article
> <42cabcca-089d-456f-837a-c1d789bda...@jj5g2000pbc.googlegroups.com>,
> "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
>
> > And, m > n implies EF(m) > EF(n), because here m > n. It is true,
> > that. The functions there modeling it are all constant monotone
> > increasing, and they all go to one.
>
> IN your origin definitions, you had EF_n(m) = 1/n, for 1 <= m <= n,
> so that SUM_(x=1..n) EF_n(x) = 1, for all n and then defined you EF()
> also the limit function of the EF_n() as n -> oo.
>
> That did not work then and does not work now.
> --
EF_d(n) = n/d, and EF = EF_oo.
It does work: n+1 > n => (n+1)/m > n/m, for non-negative m, n. There
is a constant monotonic difference between each EF(n) and EF(n+1), for
example where zero is less than one, in trichotomy of the integers.
Standardly in the limit EF(0) = 0 and lim_n->d EF(n) = 1. The
constant monotonic differences sum to one. (The limit is the sum.)
And yes, we're all aware that lim_d->oo n/d = 0, and also that n+1 > n
=> (n+1)/d > n/d, for all positive n, d (d strictly). Then, It is
simply via symmetry, that the range of EF is [0,1] and the range of
the complementary or reverse EF is [1,0], that they cross at 1/2,
their average, constant over all values.
Simply it accumulates, the value: empty: full. How many grains of
sand is a dune? All of them.
And that would be infinitely out through the integers, because it is
somewhere, and not for any finite is it. That's the way it is: via
reason.
Work.
Zero: a real quantity.
Regards,
Ross Finlayson
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