Re: Cantor and the binary tree

In article <1116939502.814879.192170@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
> If we accept that, in binary digits, SUM{n = 1 ... oo} 2^-n = 0.111...
> = 1

You may note that that is *not* an infinite sum...

> .
> 0 1
> 0 1 0 1
> ..................
....
> But we find that, up to line number n, there are -1 + 2^(n+1) nodes
> whereas 2^n different paths arrive at and 2^(n+1) different paths
> spring off from line number n. Hence, in the enumerated domain, there
> is at most one more path than nodes. After leaving any finite line
> number n (if it is reasonable to make such a distinction) we can no
> longer apply these formulae. But we know that any new branching
> increases the number of paths by 1 and, by definition, the number of
> nodes by 1 too (because any branching is a node). Therefore, the number
> of paths always equals that of the nodes + 1. It is simply impossible
> to assume that one of these numbers becomes uncountably infinite while
> the other remains countably infinite.

Also we find that up to line n, summing up to that node along the path
gives a value for k/(2^n) for some integer k. Therefore a number along
a node is always equal to a rational with a denominator that is a power
of two. It is simply impossible to assume that one of these numbers
becomes 1/3.
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