Re: Well Ordering the Reals

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.

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.

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.

.

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