Re: Well Ordering the Reals
- From: Tony Orlow <aeo6@xxxxxxxxxxx>
- Date: Wed, 16 Nov 2005 12:41:13 -0500
David R Tribble said:
> David R Tribble wrote:
> >> Yes, in any finite binary tree, there are 2^k paths and 2^(k+1)-1
> >> nodes.
> >>
> >> In an infinite binary tree, however, there are countably infinite
> >> (Aleph_0) nodes but uncountably infinite (2^Aleph_0) paths.
> >> The nodes can be denumerated with the finite naturals, but
> >> the paths correspond to infinite binary fractions; there are
> >> more fractions than naturals.
> >>
> >> Each path is an infinitely long bitstring, which can be considered
> >> as a binary fraction for one of the reals in [0,1]. And there are an
> >> uncountably infinite number (c) of those.
> >>
> >> We can also consider each path as a infinitely long bitstring
> >> corresponding to an (your) infinite naturals. There are an uncountably
> >> infinite number (c) of them, too.
> >>
> >> But there are only a countably infinite number of nodes. The infinite
> >> sum s = 1+2+4+8+... corresponds to a countable infinity (Aleph_0).
> >
>
> Tony Orlow wrote:
> > Actually, if each natural corresponds with a 1, we get the sum(n->N: 1)=N. If
> > each natural corresponds with a power of 2 we get sum(n->N: 2^n) =2^N-1, which
> > is the size of an uncountably infinite set. You are beign sloppy with your bit
> > counts here. That is my point regarding bijections and such.
>
> You are being sloppy with basic arithmetic.
>
> If your N and 2^N-1 were really different (infinite) numbers, then your
> ...111(2) and your "unit infinity" also must be different numbers.
They are. N=1:000...000. ....1111 might be considered a poorly formed version
of N-1, but one really can't tell how that relates to 1:000...000. I am not
sure why you say this anyway.
>
> However, it is provably true that your N, 2^N-1, and ...111(2) are all
> exactly the same (infinite) number. I've pointed this out to you
> several times, and you have never proven otherwise.
I certainly can't prove that within your system. Our systems are incompatible.
>
> You confuse ordinal 2^N and cardinal 2^card(N) all the time, oblivious
> to the fact that you are intermixing two completely different ideas.
Actually, I see tranfinite cardinals and limit ordinals both as unnecessary
when you allow infinite naturals, so we are simply on a different "wavelength".
>
>
--
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
.
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