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Topic: Cantor's first proof in DETAILS
Replies: 3   Last Post: Nov 30, 2012 1:00 AM

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Alan Smaill

Posts: 1,103
Registered: 1/29/05
Re: Cantor's first proof in DETAILS
Posted: Nov 29, 2012 10:35 AM
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"Ross A. Finlayson" <> writes:

> On Nov 28, 4:58 pm, Marshall <> 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.

Is this a version of the natural density of a subset of the natural

> Regards,
> Ross Finlayson

Alan Smaill

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