Re: Well Ordering the Reals
- From: Tony Orlow <aeo6@xxxxxxxxxxx>
- Date: Mon, 31 Oct 2005 08:26:51 -0500
The Ghost In The Machine said:
> In sci.math, Tony Orlow
> <aeo6@xxxxxxxxxxx>
> wrote
> on Fri, 28 Oct 2005 15:50:27 -0400
> <MPG.1dcc4b0a16b579eb98a590@xxxxxxxxxxxxxxxxxxxxxxxxx>:
> > Hi All -
> >
> > I am told it is impossible to well order the real numbers such that there is a
> > first and a successor operation which defines each other. However, I believe I
> > have such an enumeration. I have posted a web page at
> > http://www.people.cornell.edu/pages/aeo6/WellOrder/ .
> >
> > Please feel free to comment, either here or by email. Is there a valid
> > objection to this well-ordering, or does it do something not proviously though
> > possible?
>
> Several problems.
>
> [1] +oo and -oo aren't real. This is easily fixable by enumerating the
> extended reals, of course, instead of the real numbers, or by
> tacking the reals onto the front of the mapping, as you
> apparently have done.
Yes, Stephen also dislikes the presence of those two. You can see my comment to
him. They can certainly be eliminated as elements, and used only as a starting
point for the enumeration of the finite reals, if that makes you feel better.
That simplifies the quantitative portion of the mapping to the naturals,
slightly.
>
> [2] What, precisely, is 'x'?
x is a real in (0,1). An alternative enumeration can use 1<x<oo. The precise
values of the elements beyond -1 and 1 depends on x. So, one can do the
enumeration with any number which is in those ranges. I have updated my page
with a more detailed description of the enumeration. The rightmost column of
the second table displays the values for x=1/2.
>
> [3] For any mapping m : N -> R, I can find a number d such that
> m(i) != d for all i in N. There are various methods by which
> one might accomplish this; the simplest is to assume
> m2 : N -> [0,1), (or construct m2(x) = m1(i(x)), where i() is
> a mapping from [0,+oo) to [0,1)), explicitly construct a
> d2's decimal expansion somewhere in [0,1) (the Carnot Diagonal
> Proof is or should be well known to most mathematicians), and
> then prove d2 is not in m2's image. Since there's a 1-1
> correspondence between [0,1) and [0,+oo) -- there's a few,
> in fact; the simplest one might be y=x/(1-x), and finding
> the inverse yields:
>
> y+1=1/(1-x); 1/(y+1)=(1-x); x=1-(1/(y+1))=y/(y+1)
>
> one can then compute d = d2/(1-d2) which is not in m's image.
Cantor's diagonal proof does not hold much water for me regarding this. There
are several reasons for this. His proof says nothing about real numbers, but
addresses the enumeration of a digital number system with base greater than 1.
The result essentially boils down to the fact that for any set of symbols of
size S, the set of all unique strings of length L constructed from S symbols
has a size S^L, which is greater than L for any S>1. The list of digital
strings is always longer than it is wide. For n binary digits, we have 2^n
strings. So the diagonal from which he constructs his antidiagonal does not
traverse the entire set of real that it purports to. If there are N digits in
each real, and we traverse 1 digits for each real through the diagonal, then
the diagonal traverse N of the 2^N strings. The antidiagonal is not missing
from the list. It is simply below the diagonal. This general proof of the
"uncountability" of the real numbers, and the impossibility of a well ordering,
is simple conflation of what is a propertiy of digital number systems, and more
generally, symbolic languages.
>
> [4] Near as I can figure, the first 5 numbers of your enumeration
> are -oo, +oo, 0, -1, and +1. Beyond that, I can't tell what's
> going on, especially since x^(-oo) = 1 for any real x > 1;
> this isn't exactly a midpoint in any logical sense of the term.
Excuse me, but x^0=1 for any finite real. x^-oo=0, for any finite real >1.
Similarly, for any real in (0,1), x^oo=0.
>
> It may depend on what one defines as "midpoint"; one simple
> one, for example, might be the harmonic mean m = 2/(1/x+1/y).
> The main problem with this midpoint is that one only gets
> certain members of Q. There are other midpoint defs one
> can try but in light of [3] there's not much point.
Limitations of the digital number systems are not to be conflated with
properties of real numbers. The digital enumeration of the reals is actually
valid, when you take into account N=S^L, as above. However, it does suffer from
base-dependency in its structure, requiring trees with 2 children for every
node in binary, ten in decimal. This enumeration does not suffer that drawback,
and enumerates the entire line at once, large and small elements together, in a
linear order.
You suggestion above, 2/(1/x+1/y), looks like it would generate rational
numbers, not all reals. If the majority of reals are irrational, this won't
work. Of course, what is rational probably ultimately depends on the method of
enumeration. In this system, perhaps sqrt(2) can be considered rational, at
least where x=2. It can even be considered something of an integer, the set
being in linear order. But that is not what is going on here. Here, we are
identifying the relationship that ties {-oo,0,oo} and {0,1,oo} and {1,x oo}. I
think I have identified the salient relationship. Yes, no?
>
>
--
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
.
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