```Date: Dec 5, 2012 11:06 PM
Author: ross.finlayson@gmail.com
Subject: Re: Cantor's first proof in DETAILS

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.  Thereis a constant monotonic difference between each EF(n) and EF(n+1), forexample 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.  Theconstant 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 issimply via symmetry, that the range of EF is [0,1] and the range ofthe 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 ofsand is a dune?  All of them.And that would be infinitely out through the integers, because it issomewhere, and not for any finite is it.  That's the way it is:  viareason.Work.Zero:  a real quantity.Regards,Ross Finlayson
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