Virgil
Posts:
8,833
Registered:
1/6/11


Re: Cantor's first proof in DETAILS
Posted:
Nov 30, 2012 12:51 AM


In article <fcaa9d750b4c4ccfabb6e4ee1f3e91b2@i7g2000pbf.googlegroups.com>, "Ross A. Finlayson" <ross.finlayson@gmail.com> wrote:
> On Nov 29, 9:13 am, Marshall <marshall.spi...@gmail.com> wrote: > > On Thursday, November 29, 2012 7:18:59 AM UTC8, Ross A. Finlayson wrote: > > > On Nov 28, 4:58 pm, Marshall <marshall.spi...@gmail.com> wrote: > > > > Um, so EF is a restriction of division? > > > > > > The domain of x depends on the value of d. I don't recall having seen > > > > that sort of thing before, but I guess I do know what that means. > > > > But I can't figure out what the domain of d is. It sorta looks like the > > > > domain of d depends on what d is, but what the heck would that mean? > > > > > > And it's just a name, but what about EF has anything to do with > > > > equivalency? > > > > > > Marshall > > > > > Mr. Spight, it's about the equivalency or equipollency or equipotency > > > of infinite sets. > > > > Ok. > > > > > EF(n) = n/d, d>oo, n>d. > > > > I'm not very good with limits, so I'm not sure exactly what this means. > > Are you saying that the value of EF is a limit? A double limit? Also > > I'm not sure how n can approach d when n is a parameter of EF. > > > > I guess also you're saying EF is *not* a restriction on division. > > > > > Properties include: > > > EF(0) = 0 > > > EF(d) = 1 > > > EF(n) < EF(n+1) > > > The domain of the function is of those natural integers 0 <= n <= d. > > > > Whoa, you lost me. Here, EF has only one parameter, and you show > > 1. EF(0) = 0 > > 2. EF(d) = 1 > > but also > > 3. EF(n) < EF(n+1) > > > > But it seems you have done some hidden binding of d that I'm not > > clear about. Given 2. above, EF applied to any nonzero number yields > > 1, which contradicts your 3. above. What gives? > > > > > It's very simple this. Then, not a real function, it's standardly > > > modeled by real functions: > > > EF(n,d) = n/d, d E N, n>d > > > with each having those same properties. > > > > > Then, the coimage is R[0,1] as is the range. > > > > I'm not following. > > > > Marshall > > > Basically this reads that d is unbounded and n ranges from zero > through d. > > EF(n,d) is a family of functions, with d unbounded it's a particular > EF(n) with properties modeled by those standard real functions, in a > similar way as to how, for example, Dirac's delta or Heaviside's step > are so modeled. > > Being dense in the reals, in its range, leads to a variety of > considerations of anywhere dense elements in their natural order.
A family of functions or its alleged limit function can hardly be "dense in the reals" according to any standard topology or linear ordering, at least in any proper mathematical sense.
But then nothing about Ross' "EF" makes any sort of mathematical sense. > > The reals: wellordered.
Not by Ross. While by some considered theoretically possible, no one has yet come up with any explicit wellordering of the reals, and for Ross to suggest that he has done it is the arrogance of abject ignorance. > 

