Re: Zenkin's paper on Cantor

From: Ross A. Finlayson (raf_at_tiki-lounge.com)
Date: 09/29/04 

Date: 29 Sep 2004 11:47:14 -0700

Use your strong analytical capabilities to either a) refute, or b)
validate that EF is a bijection from N to R[0,1]. Show your work, and
mark any derivations.

Here's one way to consider EF, n/d, start with small values of d. For
d = 2, the range includes zero, one half, and one. For d = 3, the
range includes 0, 1/3, 2/3, and 1. For d = 4, the range includes 0,
1/4, 1/2, 3/4, and 1.

You might claim that for any finite d that each of those elements of
the range is rational. The variable d is unbounded. The range
includes zero and one for any positive value of d.

You might claim that you can generate an antidiagonal, ie, use the
"diagonal argument", that is an element of R[0,1] and different in its
representation than each element of the range. You may be able to
generate a single one, but not more than that, as the list is sorted.
The single one you generated, were it not already on the list in a
different representation, then used to generate another, leads to the
inability to generate another, because it goes in its place in the
sorted order. There is not an antidiagonal that doesn't exist in the
range.

You might claim that Cantor's first proof does not allow the mapping
of the integers to reals if the mapping would lead to a contradiction.
 You construct an interval based upon alternately raising the lower
bound and lowering the upper bound according to his method, and find
that the first two element of the range are zero and iota. Iota is
defined to be the least positive real, and the "next" element after
zero on the real number line. Thus you are left with no
contradiction.

You might claim that via Cantor's result that a set may not biject
with its powerset, and the Cantor-Bernstein result that the existence
of bijections A<->B and B<->C implies the bijection A<->C, that
because N<-/->P(N) and R<->P(N), that N<-/->R because otherwise then
N<->P(N). For that, I encourage you to consider a theory of
ubiquitous ordinals, and that P(N) is the successor, order type, and
powerset of N.

You can integrate EF, it has an antiderivative of sorts, you can use
it get positive, finite results.

In my analysis of EF, it appears that Cantor's diagonal argument and
first proof about nested intervals do not apply to it, and indeed, it
appears that any function to avoid the nested intervals contradiction
would be monotonically increasing or decreasing, where there are a
variety of ways to consider the antidiagonal argument with respect to
the necessity of the binary case or leading zeroes.

To validate it in light of the powerset result leads to quite a
different consideration, and one which appears to be usable in a
foundationless set theory with a universal set.

It does lead to some consideration of "scalar" infinities and
infinitesimals.

So, if you care about mappings of the natural numbers to the real
numbers, take a while and examine the mathematical basis of these
claims. A bijection between the naturals and reals is not accepted by
many. Use your rational thinking to form an open-minded opinion.

Warm regards,

Ross F.

Relevant Pages

  • Re: Zenkins paper on Cantor
    ... validate that EF is a bijection from N to R. ... different representation, then used to generate another, leads to the ... of the integers to reals if the mapping would lead to a contradiction. ... A bijection between the naturals and reals is not accepted by ...
    (sci.math)
  • Re: Cantor Confusion
    ... >> So you can not show a bijection, ... Why isn't it sufficient to collect the shares of two edegs for every ... Paths can be bijected with the reals. ... Bis the set of all paths to which edge e belongs. ...
    (sci.math)
  • Re: Uncountable sets in CZF?
    ... It doesn't follow that there's a bijection. ... > naturals to the reals. ... What that means is that one of the reasons that people call the reals ... That implies it is not a mathematical fact and to promote the other ...
    (sci.math)
  • Re: Skolems Paradox and why is math the way it is?
    ... ZF proves the LACK of a bijection IN the ZF system, ... > set of reals is uncountable is perfectly precisely rendered into ... If you look at the axiom of equality two sets are the same IFF they ... since all you need to do is add the existance of a bijection B from N ...
    (sci.math)
  • Re: infinity ...
    ... >>> But I have constructed a bijection between the two using an intermediate binary ... That means that imagining any completed bijection also leads to a ... contradiction. ... in the powerset of N, and if no natural maps to N, then ...
    (sci.math)
Loading