In article <eda6be39-323e-4b70-b40c-a43519b16e73@33g2000vbe.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 22 Jun., 15:07, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > In article > > <fd4c6395-51cf-4045-883e-f5c986b91...@z7g2000vbh.googlegroups.com> WM > > <mueck...@rz.fh-augsburg.de> writes: > > > On 19 Jun., 15:42, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > > > > > > > If every terminating path of the form 111...1000... is used, then > > > > > every node is covered by the last 1. > > > > > > > > But that still does not show that all nodes are covered by a countable > > > > set > > > > of paths. It only shows that every node is covered by a set of paths > > > > that > > > > contains a countable subset, not that the set itself is countable. > > > > > > It is easy to construct a bijection between a countable set of paths > > > and all nodes. These nodes can even be defined by the paths that lead > > > to them. That shows that all nodes of the tree can be covered by a > > > countable set of paths. > > > > Yes, they can be covered by a countable set of paths. Nothing more. It > > does *not* show a bijection, and not a construction of a bijection. > > I do not argue that there is a bijection with all real numbers. The > reason is that real numbers with infinite bit sequences do not exist > *as bit sequences". pi exists in form of many formulas, Vielta, > Wallis, Gregory-Leibniz, Euler, for instance. But it does not exist > asa a subject to Cantor's proof.
Cantor's proof only speaks of infinite binary sequences, not real numbers, and proves that there is no such thing as a complete listing of them.
> All existing sequences are in my tree.
Then your tree is incomplete, so not a maximal infinite binary tree at all.
> And obviously there are only countably many.
It is not obvious to anyone who can understand Cantor's proof to the contrary. > > > > > > Yes, they are in the tree, so you have to prove that they are also > > > > used in > > > > the covering, which you have not done. You do not have to show that > > > > each > > > > node is covered, you have to show that with your covering you use each > > > > path. You did show the former. > > > > > > If every node is covered, then there remains no path to be covered. > > > > What do you mean with "covering a path"? Until now you were talking about > > "covering nodes". And whatever it may mean, in what way does "every node > > is covered" show that there remains "no path to be covered"? > > Try to find a path p that contains a node not yet covered by a path q > used to construct the tree.
Try to discover a node of any maximal infinite binary tree not covered by uncountably many paths.
> In order to prove that there is such a > path p
We do not claim any such path, we only claim there will be paths not used in any construction which only uses only countably many paths, which is quite a different issue.
> you must show for every path q that there is a node not covered > by q.
WRONG! WE only need show a path not used in the construction. If the set of paths used in the construction is cunable, then any proof of its countability allows, by the Cantor construction, proof of existence of a path not used.
> But that is impossible.
No its just the Cantor proof.
> If you think, you have accomplished it > at some node, there will always be another q that is identical with p > to that node.
WM'S nonsense about nodes just shows how little he understands about maximal infinite binary trees.
> That means, you cannot diagonalize the tree.
Since the set of paths in a maximal infinite binary tree is already known to be uncountable, there is no point to even trying to "diagoanlize" it.
> It contains > already all paths that can be distinguished by nodes
I wonder what sort of paths WM has in mind which cannot be "distinguished by nodes"? Or at least by sets of nodes.
In a real world maximal infinite binary tree, any path can be distinguished from all others by the set of all its nodes. > > > > > > No. From that we can obtain that the number 1/3 is also in a list of > > > terminating rationals. There does not exist a sequence 0.010101... > > > that is longer than *every* finite sequence of that form. > > > > Ah, so now you contend that 1/3 is a terminating rational. > > No. It is a rational that has no decimal and no binary representation.
In the larger world, outside of WM's mathUnrealism, where non-terminating decimals and binaries reside, it has both.
