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Topic: UNCOUNTABILITY
Replies: 59   Last Post: Dec 24, 2012 2:06 PM

 Messages: [ Previous | Next ]
 Virgil Posts: 8,833 Registered: 1/6/11
Re: COUNTABILITY of a set requires N to surject to it
Posted: Dec 22, 2012 4:32 PM

In article
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 22 Dez., 06:05, Zuhair <zaljo...@gmail.com> wrote:
>

> > You are just not getting Cantor's argument.
>
> You have not got my argument. The list is complete with respect to its
> *enumeration*! There is no more place to insert another real number!

But there still provably exist at least as many unplaced reals as have
been placed in any given list.
>
>

> > Cantor's
> > argument is not about diagonalizing the set of ALL reals since that is
> > clearly not possible, Cantor's argument is about diagonalizing any
> > COUNTABLE set of reals. So again for any *countable* set S of reals
> > there is a diagonal r that differs from each real in S by a finite
> > initial segment n(r).

>
> And that shows that the Binary Tree should contain for every real a
> *finite* path that differs from a finite initial segment of the real.

Actually, one does not need Cantor's results to deduce that.

Given any real in, say, decimal format, EVERY real that differes from it
does so at some finite digit postion.

So similarly, any binary tree with more than one path does what WM says.

> Alas there are only countably many finite paths.

Irrelevant, as usual!

> Or do you object this
> simple truth?

Only to is being in any way relevant.

> On the other hand we know that the Binary Tree contains
> all possible finite initial segments. So there cannot be any finite
> initial segment of any real that differs from all finite paths of the
> Binary Tree. Hence there is a contradiction.

Only in WMytheology. The Cantor argument re binary sequences is totally
independent of WM's delusions about what it implies.
>
> > Thus the infinite binary tree itself cannot be
> > countable!

>
> Then you should be able to find out *by means of nodes* what is
> missing in the Binary Tree that I construct from countably many paths.

Once you tell me WHICH countably many paths, with proof of the set being
countable, I can easily provide you with just as many paths which you
missed, but until you tell me which ones you have, I cannot determine
which ones you do not have.
>
> >
> > Of course If you manage to prove
> > that R is as big as N by any argument, then of course you'll be wright
> > in saying that no completed infinite sets of naturals and reals could
> > ever exist. But so far, you couldn't.

>
> You are in error. Otherwise you could point out *by means of nodes at
> finite positions* (others do not exist, which paths are missing in the
> Binary Tree constructed from countably many paths). But you cannot.

Since you remain so careful not to tell anyone which paths you have, we
have no way of telling which ones you do not have, as the ones you
missed are determined entirely by the ones you did not miss,

> Therefore all your attempts to "prove" anything else are condemned to
> fail.

Only in WMytheology. In standard mathematics, standard proofs still
prove things.
>
> > The main problem with your binary tree proof is that you keep
> > insisting on the obvious error of identifying the complete infinite
> > binary tree by its nodes, while a tree is generally identified by the
> > ordered pair of the set of all its nodes and the set of all its EDGES.

>
> Every edge is in diect correspondence with the node at its end. You
> should be able to recognize that.
>

> > You think that once you've discerned the nodes of your tree then
> > that's it any tree that have those nodes would be it. Which is wrong,
> > I can simply construct another tree that completely differ from the
> > infinite complete binary tree and yet using the nodes of the complete
> > infinite binary tree themselves.

>
> But how would you prove that it is completely different? A Binary Tree
> is defined by its nodes only (or, what is tantamount) by its edges.
> Every node except the root node is the end of an edge).

It is neither the nodes in isolation nor thee edges in isolation, but
whether a node, other than the root node is a laft cild or a right child
of its parent node or whether an edge is a left branch or a right branch
from its parent node to its child node.
>
> > So what identifies the complete
> > infinite binary tree, and any tree actually, and even any graph is the
> > set of all nodes and the set of all Edges of it, you keep forgetting
> > the edges.

>
> Are you really insisting that the edges are not isomorphic to the
> nodes where they end?

Depends very much on what form of "isomorphism" you are claiming.

Certainly there is no ismorphism of any sort which bijects edges with
the nodes where they end.
> >
> > Now the reals correspond to PATHS of the complete infinite binary
> > tree, that's correct! And there is no proof whatsoever up till today
> > that proves the number of those paths to be countable.

>
> Again: The proof is that you and nobody else can discern which paths
> are missing when I construct or cover the complete Binary Tree by
> means of countably many paths.

Only because you refuse to tell us which paths you have.
The only way to tell which paths are missing is by knowing which paths
are not missing, but you are careful to keep that secret.

If you flip an honest coin and look at the result but keep it secret,
while I can know it to be either a head or a tail I cannot be sure which.

If WM creates a binary tree with only countably many infinite paths but
does not tell anyone else which paths he has included, then no one else
can tell WHICH paths are missing even though most of them will be.
>
>

> > I keep saying that: Mathematics is Discourse about POSSIBLE form.
>
> Not quite correct. Mathematics is *possible* discourse about possible
> froms.
>

> > Does uncountability of reals raise within this context of mathematical
> > interest. The answer is YES.

>
> As you are unable to discern, from the (kept secret) bunch of paths
> that I use to construct all possible finite paths of the Binary Tree,
> which further paths are missing, your belief in uncountably many paths
> is purely unmathematical nonsense.

It is your argument above which is, like so many of your other
arguments, unmathematical nonsense.

It is the fact of your keeping your set of paths SECRET that prevents
us from telling whether any particular path is or is not in it.

Any listing of them, and such a listing must be possible if there are
PROVABLY only countably many of them, itself provides proof that your
listing is incomplete.

That is why you must hide your list of paths!

Revealing it would prove you wrong!
--

Date Subject Author
12/19/12 george
12/20/12 Zaljohar@gmail.com
12/20/12 Barb Knox
12/20/12 ross.finlayson@gmail.com
12/20/12 Zaljohar@gmail.com
12/20/12 ross.finlayson@gmail.com
12/20/12 Graham Cooper
12/21/12 mueckenh@rz.fh-augsburg.de
12/21/12 Virgil
12/20/12 Graham Cooper
12/20/12 mueckenh@rz.fh-augsburg.de
12/20/12 Zaljohar@gmail.com
12/21/12 mueckenh@rz.fh-augsburg.de
12/21/12 Zaljohar@gmail.com
12/21/12 mueckenh@rz.fh-augsburg.de
12/21/12 Virgil
12/21/12 William Hughes
12/22/12 mueckenh@rz.fh-augsburg.de
12/22/12 William Hughes
12/22/12 Virgil
12/22/12 William Hughes
12/22/12 Graham Cooper
12/23/12 Virgil
12/23/12 William Hughes
12/24/12 Graham Cooper
12/24/12 mueckenh@rz.fh-augsburg.de
12/24/12 Virgil
12/23/12 mueckenh@rz.fh-augsburg.de
12/23/12 Virgil
12/23/12 mueckenh@rz.fh-augsburg.de
12/23/12 William Hughes
12/24/12 mueckenh@rz.fh-augsburg.de
12/24/12 Virgil
12/23/12 Virgil
12/22/12 Virgil
12/21/12 Zaljohar@gmail.com
12/22/12 mueckenh@rz.fh-augsburg.de
12/22/12 William Hughes
12/22/12 Virgil
12/21/12 Zaljohar@gmail.com
12/21/12 mueckenh@rz.fh-augsburg.de
12/21/12 Virgil
12/22/12 Zaljohar@gmail.com
12/22/12 Virgil
12/22/12 Zaljohar@gmail.com
12/22/12 mueckenh@rz.fh-augsburg.de
12/22/12 Virgil
12/22/12 mueckenh@rz.fh-augsburg.de
12/22/12 Virgil
12/23/12 mueckenh@rz.fh-augsburg.de
12/23/12 Virgil
12/21/12 Virgil
12/20/12 Virgil
12/20/12 Graham Cooper
12/20/12 Graham Cooper
12/21/12 mueckenh@rz.fh-augsburg.de
12/21/12 Virgil
12/21/12 Graham Cooper
12/21/12 Graham Cooper