Re: Well Ordering the Reals

Sean West said:
>
A simple proof that your well ordering does not work is the fact that each
listing (when you list out the reals according to your ordering) is finite.
Therefore your representations of the reals are a subset of the set of all
strings, which is countable...and the reals are uncountable. I haven't read
every message here, but I'm assuming some have pointed this out in a similiar
way.
>

Well, if the strings are allowed to be infinitely long, then they can specify
every real precisely, but then constitute an uncountable set. If not, and they
are restricted to finite lengths, then one can only be guaranteed to come
arbitrarily close to any desired value with the binary string, and the set
would be countable.
-------------------------

>
But even so, I think there is a simpler way to accomplish what you are trying
to do. First off, there is no need to try and ever actually reach the largest
bound of the reals (infinity). The natural numbers can be well-ordered, simply
by counting up. The ordering of the natural numbers never attempts to reach
the ends of the number line, the fact that it increases forever will accomplish
that task. In the same vein, this can be done with the reals. The problem
with the reals is not how big they get (the reals aren't larger than the
natural numbers), but rather how "dense" they are. Thus you kind of have to
count infinitely up and infinitely inwards. Your way accomplishes this,
however it tries to deal with infinity in a "number" way, which is a big nono
(i.e. 2*infinity is insensible).
>
Oh, poppy***. We are defining the real numbers here, and the real numbers are
points on the real number line, so defining the real line itself by declaring
oo and -oo is a natural first step in defining that which the line contains.
For 0<x<1, x^oo=0 and x^-oo=oo, which is shorthand for lim(n->oo: x^n)=0 and
lim(n->-oo: x^n)=oo. These are perfectly resonable concepts that lead to a
progression:

x^-oo=oo
x^oo=0
x^0=1
x^1=x

Now, as I said below, you can avoid the ends of the line and mention of
infinity, if you wish, although I think that's a mistake. Still, for the faint
of infinite, I offer the Peano-esque definition:

1) 0 is a real number
2) If x is a real number 2^x and -2^x are real numbers.

There, now you don't have to touch that dirty oo.
--------

Heres a refined way to do what your doing (again this won't list all reals,
but does what your way does without dealing with infinity). I'll only deal
with the positives, just deal with the negatives in a similiar way in parallel
with this procedure. Start off with 1, then list its "first split", 0.5. Then
list 2 and list 1's "second splits" 0.25 and 0.75 and 2's first split, 1.5.
Then list 3 and 1's third splits, 2's second splits, and 3's first split. And
so on. This will seem to eventually cover the entire number line.
>
That's not the way this set works. Take a look at the simplified description
without infinity, below.
----------------------------

This, however, only lists the rationals (it will effectively list them all).
There are even simpler ways to list the rationals, such as listing n/m where
you start with n=1, keep incrementing n, and every time you increment n you
list out all m's from 1 to n (although this is repetetive).
>
! Um, yeah, those would be rationals, but my set produces irrationals and
transcendentals early on in its genesis, things like 2^2^1/2 and such.
-------------------------------------

My basic critique...theres no reason for your procedure to ever deal with
infinity, and besides...it doesn't work!
>
! Well it does work, and if you find oo distasteful then you can start with
root element 0, and produce successive elements using 2^x and -2^x where x is
in the set. Then you don't have to look at oo.
-- ---------
Smiles,

Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
.

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