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Re: Cantor's first proof in DETAILS
Posted:
Nov 29, 2012 10:35 AM


"Ross A. Finlayson" <ross.finlayson@gmail.com> writes:
> On Nov 28, 4:58 pm, Marshall <marshall.spi...@gmail.com> wrote: >> On Monday, November 26, 2012 11:33:02 PM UTC8, Virgil wrote: >> >> > I find a citation from r 9/22/99 In which Ross states, what may well be >> > Ross' original "definition" of his alleged "Equivalency Function" which >> > as any mathematician can plainly see is not a function at all, and is >> > only equivalent to nonsense:: >> >> > " Consider the function >> > f(x, d)= x/d >> > for x and d in N. The domain of x is N from zero to d and the domain of >> > d is N as d goes to >> > infinity, d being greater than or equal to one. >> > I term this the Equivalency Function, and note it EF(x,d), also EF(x), >> > assuming d goes to >> > infinity." >> >> >http://groups.google.com/group/sci.math/msg/af29323d694cf89e1999  >> > "Equivalency Function" >> >> 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. > EF(n) = n/d, d>oo, n>d. > > 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. > > 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.
Is this a version of the natural density of a subset of the natural numbers?
http://en.wikipedia.org/wiki/Natural_density
> Regards, > > Ross Finlayson
 Alan Smaill



