Re: Cantor's diagonal proof wrong?

From: Ross A. Finlayson (raf_at_tiki-lounge.com)
Date: 11/22/04 

Date: 22 Nov 2004 13:21:38 -0800

Hi Bob,

That's pretty convincing.

You know I argue the other way, with variously about the binary case
being sufficient, for the dual representation argument, and necessary,
for the case of the one true binary antidiagonal.

Your argument is too short, there is the slippery-slope notion about
dual representation of rationals and why your antidiagonal algorithm
must be done with the consideration of necessarily forming an
irrational or non-dually-representable antidiagonal (everywhere
non-diagonal), "multiple antidiagonals in non-binary."

http://groups.google.com/groups?selm=3c6b9c1e.0308030119.6cfb38b3%40posting.google.com

With the decimal expansions, and other cases with base necessarily
greater than three, as the antidiagonal argument can not be seen to
hold in the case of base less than four, I still struggle to provide a
convincing argument against it.

One reason I can do, believe, that infinite sets are equivalent is
because infinite sets are inexhaustible, unending, always containing
more elements as finitely or even often infinitely many elements are
removed, etcetera, and that upon uniquely (canonically, naturally)
ordering a pair of infinite sets that each is in a very simple way
enumerable.

With the binary antidiagonal, if the list is an enumeration of the
reals, arbitrarily on a given interval, then a bijection would yield
the antidiagonal being dually represented, or arguably in-, un-, or
non-constructible. I argue that the binary case is sufficient and
necessary.

The averral to leading zeroes and how the expansions would as well
represent the same set of each of some interval of real numbers with
the zeroes leading to an antidiagonal not in the set of reals under
any circumstances is useful to that extent. Unfortunately it doesn't
change that the expansion, with a rule defining its construction as
there is not an infinitely long tape upon which to write the
infinitely long expansion, is the same number without all the leading
zeroes.

It seems obvious that for a list of rationals that the antidiagonal,
regardless of construction, would share the dual representation with a
list element, or be irrational. The irrationals, interleaved among
the rationals, share a property with the rationals in being the dense
complement of a set dense in the reals in the reals, as do the
algebraics and transcendentals, and, that is about it.

Anyways, I feel for my argument that I need a stronger dispute against
the contrived decimal case. Basically the contrivance is to avoid the
construction of a dually represented rational number under any
circumstances because of the dual representation of rational numbers
in a base with a factor not coprime to the denominator. As one of my
sheaf of arguments about mathematical foundations, I want to determine
a concise explanation of why the decimal antidiagonal argument may
fail, to file behind why the binary antidiagonal argument may fail.

For the general case with the coded powerset there are other slightly
more fundamental statements of duality and the interpretation of sets
as ordinals, etcetera, and with regards to the naturals and reals and
pigeonhole/nested intervals style arguments of eg Cantor's or Megill's
type, which apply to rationals in the standard models of rational
numbers, there are the considerations of the one-sided reals on the
line and other notions of nonstandard definitions of the reals. I
digress.

Here's a notion, it's about the contrivance basically, with the
decimal case. The antidiagonal contains a five unless the list
element contains a five at the index in which case the antidiagonal
(non-diagonal) contains a four. The notion with that contrivance is
to avoid any rational form that ends with all fives in decimal, for
example 5/9. It does not because the rational might be of the form
.000..00555... early in the list, where to generate an everywhere
non-diagonal would require expression of the expansion in an infinite
base. That is not about leading zeroes so much as the leading part of
the finite expansion, and the same expansion upon discarding
significant leading zeroes. That notion does not well address the
contrivance, rather it is still about the leading zeroes, and then
about significant and conversely insignificant leading zero(e)s.
Being still about the leading zeroes, significant leading zeroes, any
antidiagonal for the range [0,1/b) for some antidiagonal in base b
would not be in the range. There's that, and through composition any
interval of reals can be mapped to the interval [0, 1/b] for any
positive integer base b.

That seems to make some sense, for open-minded math enthusiasts with
much exposure to the antidiagonal argument and the fundamentals of the
problems of the antidiagonal, for example myself.

If the expansion is represented in an infinite base, then the mapping
for the unit intervals is as so: f(x)= . x, because in an infinite
base the expansion for any real number is of the form b1 . b2, the
expansion is not infinite, instead the base is infinite. (x+1)

http://groups.google.com/groups?selm=3c6b9c1e.0403132237.4e57ea6e%40posting.google.com

There are many rational numbers that you can write in a finite
expansion with very simple declarative structural rules of the
expansion, any rational number in the unit interval can be written in
the form . b1 b2 b3 .. (two dot finite semi-ellipsis of arbitrary
elements) bn (c1 c2 c3 .. cn)..., the initial finite expansion that
has to do primarily with the coprimality of the base of the expansion
and the rational number in question and some repeating terminator that
is often (0)... when the denominator of the reduced fraction is not
coprime to some factor of the base.

Anyways the point there is being that there are no irrational numbers
that you can write in any finite expansion for a rational base for the
expansion with only the capability of representing one unique, finite,
repeating terminating sequence of the rational number, where every
rational number can be represented in that form, in any rational base,
and no irrationals can be represented in that form.

That's not saying you can't add iota to each rational, that would be
changing the form.

About EF and the decimal algorithm, it is similar to Megill's
construction and EF, with the non-monotonic behavior. That is where
the antidiagonal is dually represented, and there's only one possible
antidiagonal for any base to generate for any algorithm the list is
always sorted and maintains order. Between zero and one there are
real numbers, and only real numbers, and everywhere real numbers.

While that is so, it also leads to the conundrum of x/x ~= .555..., at
worst .111..., for the generic antidiagonal algorithm where 1 is
replaced with 0 and anything else replaced with 1. Thankfully
regardless of that x/x > 0, x > 0, and 9/5 and b-1 are real numbers,
so x = x. In the necessary and sufficient binary case x/x = 1. That
means exactly that.

Consider Hodges' exposition on typical Cantor cranks:

"Ax Ey f(x,y) |- Ey Ax f(x,y)" -
http://groups.google.com/groups?selm=y8zekr82iiz.fsf%40nestle.csail.mit.edu

Perhaps rediagonalization is not so far off the mark. Does anyone
here but me get that?

Can Mssrs. Borel and Combinatorics get along? That seems more likely
now. Is N not compact? Is the direct sum of infinitely many copies
of N not empty? Consider the convergent sequence .9, 1, .99, 1, .999,
1, ..., that's not strictly convergent.

"In examining the process itself, it's necessary to explain how
something can be different than everything, including itself, or
perhaps more agreeably, that it's not."

The set of all those sets is the empty set.

Anyways, if this post represents a specific facet, it's about the
antidiagonal argument as the expansion's base b diverges.

Also, consider the interval [0,1/b), the antidiagonal of any list is
always greater than or equal to 1/b, thus not on that interval.

Regards,

Ross F.