
Re: Cantor's first proof in DETAILS
Posted:
Nov 29, 2012 10:18 AM


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.
Regards,
Ross Finlayson

