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Topic: Cantor's first proof in DETAILS
Replies: 85   Last Post: Dec 10, 2012 7:23 AM

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 ross.finlayson@gmail.com Posts: 2,720 Registered: 2/15/09
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 UTC-8, 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."

>
> > "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 co-image is R[0,1] as is the range.

Regards,

Ross Finlayson

Date Subject Author
11/25/12 ross.finlayson@gmail.com
11/25/12 Virgil
11/25/12 ross.finlayson@gmail.com
11/25/12 Graham Cooper
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11/25/12 Virgil
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