On 19 Dez., 21:14, Virgil <Vir...@home.esc> wrote:
> > Yes. A very convincing and often required proof of completenes of a > > linear set is to know the last element. > > Knowing the allegedly last element in a set does not work unless one has > imposed a linear ordering on the set, which is in no wise necessary for > sethood. Which is the "last" point of the set of points on a circle?
I am interested in linear sets.
> > It is rather silly to argue about the uncountability of the set of > > paths. > > Then stop doing it. There is a perfectly adequate proof that one cannot > have a list of paths that contains all paths, or , equivalently, one > cannot have a list of all subset of N.
You are wrong. The "perfectly adequate proof" is wrong.
> > Only minds completely disformed by set theory could try to > > defend the obviously false position that there were uncountably many > > paths. > > Minds thus "deformed" by set theory are often capable of creating valid > proofs that WM not only cannot create but also cannot even follow.
There is one simple proof: Every node of the binary tree splits one path into two paths. the number of paths cannot surpass the number of nodes. Nobody would ever have doubted that without the unfortunate development of mathematics induced by Cantor. > > > > > But "10 Questions" will give you the answer why there are not > > uncountably many paths. There are no infinite decimal expansions of > > real numbers. There are not due paths in the tree. > > 1 = 1.000... is a real number with an infinite decimal expansion, and in > ZF there are infinitely many more of them.
From the sequence of digits you cannot obtain 1. First you must be sure that there never will appear a digit other than 0. But unless you know the last digit, you cannot be sure. And you cannot know the last digit.
Of course you can know a formula saying digit d_n = 0 for all n in N. But that is a finite formula. It is not an infinite sequence of digits that informes you about the due number.