Re: Well Ordering the Reals

David R Tribble said:
> David R Tribble said:
> >> Your "well-ordering" of the reals appears to produce an uncountably
> >> infinite number of, um, "values", but what those values are is a
> >> complete mystery. What exactly is x^-oo supposed to be equal to?
> >
>
> Tony Orlow wrote:
> > Given 0<x<1, x^-oo is 1/x^oo is oo. Given 1<x<oo, x^-oo is 1/x^oo is 0. Any x
> > in those ranges will do for the enumeration.
>
> David R Tribble said:
> >> Also, you fail to provide even one example of how a given real is
> >> ordered by your scheme. Please show, for example, where 1/2 falls
> >> into your well-ordering. And what are the well-ordered successors to
> >> 1/2, 1, or 0?
> >
>
> Tony Orlow wrote:
> > Please look at the table at the bottom of the web page. The first 20 reals
> > using x=1/2 are listed with their binary representation and corresponding
> > binary naturals, their expoentitations and their values. The order goes 0, -1,
> > 1, -2, 2, -1/2, 1/2, -4, 4, -1/4, 1/4, -sqrt(2), sqrt(2), -sqrt(1/2), sqrt
> > (1/2), -16, 16, -1/16, etc.
>
> At each bifurcation level, you're doubling the number of previous
> "values",
> so that you end up with exactly
> s = 1 + 2 + 4 + 8 + 16 + ...
> values. This sum should be familiar; s is good ol' oo, which can be
> considered to be the number of of finite naturals, or Aleph_0.
Welllll, that all depends, now, doesn't it? In Cantors diagonal argument
concerning the reals, we have a grid of digits, which is aleph_0 wide, and what
we discover is that the list is NECESSARILY longer than it is wide, by the
antidiagonal not lying in the list traversed by the diagonal. It thereby proves
the set of reals is larger than the set of naturals. When you say the above
sequence equals "good ol' oo", how can you be so sure? Does it have log2(oo)
bits, or oo? If it has oo bits, then there are 2^oo numbers in the list, an
uncountable infinity, just like in Cantor's list.
>
> But the set of reals contains c (2^Aleph_0, or Beth_1) members.
> So your mapping omits a lot of reals, and does not biject N and R.
Hogwash. With a countably infinite number of bits, such that none is in an
infinite position, we have an uncountably infinite number of bit strings. Do
you actually disagree with that statement? Isn't it a tenet of standard
mathematics?
>
> In fact, it appears that your mapping only produces the powers
> of two (similar to the first w generations of the surreals).
You better look again. Eventually one gets into rather complex irrational
numbers. I have to work on what the best way is top present a deeper list of
numbers, so you can see better, but you are free to play on your own if you
like. I guarantee you find interesting numbers.
>
> What, for example, is the mapping for 1/3, and what is its successor?
Deja vu. I think the mapping for 1/3 is something like ....001001001. Dave
Rusin had confirmed that particular one months ago. The successor in this case
would be ....001001010, and would probably coincide with -1/3. I am only just
starting to really play with this system, since it seems necessary to figure
some things out to prove the well ordering to be well. :)
>
>

--
Smiles,

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

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