Date: Dec 13, 2012 3:26 AM Author: Virgil Subject: Re: On the infinite binary Tree In article

<1251b926-8319-4feb-b0ef-8fd4d886f612@r14g2000vbd.googlegroups.com>,

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

> > Ah I see, so you are imposing another condition on the definition of a

> > path,

>

>

> No, that is *the* definition of a path in a Binary Tree.

>

> > that it must begin with 0 since it is representing a number in

> > the interval [0,1], that's OK, then according to that I can see that

> > you are correct since the number of paths do corresponds to the number

> > of ending node, and with my calculation it would be less than the

> > total number of nodes by one. I'll need to check this again, but let

> > me agree with you on that for the current moment.

> >

> > But still you have a problem. What you really managed to prove is that

> > the total number of FINITE undirectional paths beginning from 0 in the

> > infinite binary tree is countable, since we can simply draw an

> > injective map from those paths to their ending nodes.

>

> No. I proved that the number of infinite paths is countable by

> constructing all nodes of the Binbary Tree by a countable set of

> infinite paths.

WM is again, or should I say still, self-deluded in all sorts of ways.

The only way WM could actually have CONSTRUCTED all nodes of a INFINITE

binary tree is by completing infinitely many construction steps himself

which he has often claimed that no one can ever do.

Such trees can exist only in the imagination, as is the case with a

great many mathematical "constructions".

But the set of paths of such an imagined tree, to be consistent, must

have a different path for every different subset of the set of all

naturals numbers, being the set of levels at which that path branches

left, and there are uncountably many such subsets of N.

>

>

> > This is clear.

> > But the problem that you will face is with the INFINITE undirectional

> > paths beginning from 0 of the infinite binary tree, now those paths

> > cannot be identified with the nodes at their ends because they simply

> > have no ending nodes because they are infinite, so you need to

> > calculate another way to map those kinds of paths to nodes.

>

> I did it already.

Not in any way satisfactory to anyone else besides yourself.

>

> > The only

> > one I can think of really is to map each infinite path to the

> > COLLECTION of all nodes in it, which is to say we map each infinite

> > path to the subset of all nodes of the binary tree that simply have

> > all and only those nodes in that path as elements of. Now this will

> > bring us again to Cantor, how many subsets of nodes of the infinite

> > binary tree do we have?

>

> No again. The number of subsets of the set of naural numbers is of

> interest. Every subset of N is represented by one and only one single

> infinite path. It is not interesting to find out the set of subsets of

> nodes - at least not for our question.

Quite right about the bijection between paths and subsets of N, but it

is well know that outside of Wolkenmuekenheim there exist uncountalby

many subsets of N.

>

>

> > Cantor says it is uncountable, therefore the

> > total number of INFINITE paths in the binary tree can be uncountable!

>

> And it is countable too. Therefore countability is not well defined.

I have never seen any convincing proof of that alleged countability, but

have often seen Cantor's convincing proof of uncountability, which seems

to me valid, at least everywhere outside Wolkenmuekenheim .

>

>

> > However I think you are claiming that each infinite path is to be

> > mapped to the Godel number of the defining formula of it (which is a

> > finite parameter free formula), and of course by then we'll be having

> > only countably many such infinite paths of the binary tree. And also

> > this is well known since definable sets in this way are always

> > countable (this is pretty much standard). Now the real point is what

> > if there is a path for which we cannot find a finite parameter free

> > formula that define it, by then those kinds of paths cannot be proved

> > countable in this way. However you seem to dismiss that possibility as

> > nonsensical.

>

> It is nonsensical because the same could be assumed for Cantor's

> diagonal. It would be undefinable and it would be impossible to prove

> that it differs from all lines of the list - in particular if

> undefinable reals exist and are members of the list.

A set is defined as countable ONLY IF it is possible to define a

surjection from N to that set. So that if WM wishes to say that the set

of all paths in a complete infinite binary tree is countable, he must be

able define a mapping from N to that set of paths or some set

bijectable to it.

But it has already been shown that there is a set, the set of all

subsets of H, which bijects with the set of all paths, but which the set

of naturals cannot be surjected to, so that WM's goal is

self-contradictory, at least to everyone not a WMytheologist.

>

> However, this question is irrelevant. It stands that the complete

> Binary Tree can be constructed by countably many paths. That's enough

> to have a contradiction.

It certainly would contradict a good deal of standard mathematics if it

were a valid construction, but it is not, as has been demonstrated often

enough to convince anyone but WMytheologists.

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