Re: (Not quite) Cantor's diagonal proof

From: Ross A. Finlayson (raf_at_tiki-lounge.com)
Date: 10/29/04 

Date: 28 Oct 2004 21:40:47 -0700

Norman Megill <nm@nospam.see.signature.domain.invalid> wrote in message news:<XY-dnRTa4ISmyhzcRVn-vw@rcn.net>...
> Ross A. Finlayson wrote:
> > EF(1) is defined to be the least positive real
> > number.
>
> Here is a formal proof of the fact that "there does not exist a positive
> real such that there is no other number between it and 0."
>
> http://us.metamath.org/mpegif/nominpos.html

Hi,

Between zero and one there are only real numbers, and every value
between zero and one is a real number.

The hyperreals, from which are drawn Robinson's infinitesimals, are
each real numbers.

A strict ordering condition exists upon the natural numbers thus that
trichotomy exists in the naive natural numbers.

The natural numbers are placed in a correspondence with the numbers
between zero and one thus that each value in the unit interval is
assigned to a natural number, and m>n implies EF(m)>EF(n).

As the naturals diverge towards infinity, the free variable n going to
oo, EF(n) goes to one.

For each real number r on [0,1], there is a natural number n thus that
EF(n)=r. If r is definite and known, then n is indefinite, except for
zero, and for any finite n you select it is nearer to zero than some
"definite" real number r on the unit interval. It is by the
characteristic that over all of the naturals that the value of EF(n)
does go to one that the range of EF(n) is dense within the unit
interval.

You basically argue that do to the closure of the reals that when you
start with two you can iteratively redivide forever and never reach
some smallest real nor zero, where obviously in the limit the result
is zero. In step 8, you argue that any positive number is greater
than its half, where do you have that the half is greater than zero?

I on the other hand say EF(1) is greater than EF(0), and is a real
number, and is less than EF(2), and EF(n) < EF(n+1), etcetera.

Here's one way to consider this: N is mapped to values between zero
and one inclusive, in order. Whether they are rationals, reals, or
another set dense in the reals, any set dense in the reals will
contain infinitely many what I call "definite" values between any
other two "definite" values, where you can define the "definite" real
as a Cauchy sequence or Dedekind cut. The "indefinite" values are
those described by the acceptance of the geometric line as a
contiguous, and even continuous, sequence of points, each representing
a real number.

Zero is a real number, and a rational number, and algebraic, and a
root of various power series with integer coefficients, and an
integer, as is one. Between zero and one the only surety of any value
is that it is a real number, because there only, and everywhere, real
numbers between zero and one.

I believe that your assertion that there is no smallest real was first
taken as a position in regards to EF by Randy Poe or perhaps Dave
Seaman, about "what is EF(1)/2?" Basically the assertion is that if
you divide EF(1) by two then the result was actually EF(1) and you had
the wrong infinitesimal.

Obviously enough, the infinitesimals are totally dense, with allusion
to the fluents and fluxions and each fluxion being a fluent to further
fluxions. One is basically a fluent and its fluxions are multiples of
EF(1).

EF(1): iota, dx, the least positive real, irrational, transcendental,
and various other non-zero things.

Is not the integrand bar an S for Summation? Newton and Leibniz
thought so.

So, you have that no real number that you get by dividing two by two
over and over again is the least positive real. I agree. The same is
so for the rationals and the irrationals if you start with Pi instead
of two. The least positive real is synthesized or rationalized by its
very definition couched within terms of the unit interval consisting
only of real numbers, and basically naive trichotomy, that is to say
a<b, a>b, or a=b.

I guess you maintain that infinitesimals are not real numbers. Is not
every number between zero and one an infinitesimal?

Every number between zero and one is a real number.

I'm interested in why you don't encode the antidiagonal argument or
nested intervals result. I would expect similar informal arguments on
my part, already presented, to be applied. Do you not think the
assembled arguments would be readily crushed by the many
mathematicians that read sci.math?

If the argument does not collapse, whither then? The range of EF is
the unit interval, any subintervals of equal length contain equal
amounts of points via f(x)=x+C, the distance between two points EF(n)
and EF(n+1) is infinitesimal and non-zero.

It appears that EF is integrable and that it integrates to equal to
one. That leads to an analytical tool that the integration of f(x)=1
over the naturals is equal to two.

Select a positive real number, inductively divide by two forever. In
a sense (in the limit), the result is zero. The result is always
positive at any finite iteration.

Basically there's iota, i, and then there's I, the scalar infinity,
and Ii=1.

Also, lim 1/oo = 0.

Good luck, warm regards,

Ross Finlayson

Relevant Pages

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  • Re: Calculus XOR Probability
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