Date: Dec 5, 2012 11:53 PM
Subject: Re: Cantor's first proof in DETAILS
"Ross A. Finlayson" <email@example.com> wrote:
> 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.
From which EF_d(n) < EF_d(n+1) whenever n < d < oo.
So for which n in |N, if any, is Ef(n) < EF(n+1)?
EF_n(n) = 1 for all n in |N, so is EF_oo(oo) = 1 ?
Ross has left too many unanswered questions about the nature of his
alleged EF for it to be anything useful.
> 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.)
Which limit is which sum? A if anyone cares.
> 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.
If you could only make your nonsense funny, it might be worth something,
but as it is, dull as dishwater and corrupted by internal
contradictions, it is merely boring.