Re: WM and end segments...

WM <wolfgang.mueckenheim@tha.de> wrote:

Le 22/07/2024 à 21:45, Alan Mackenzie a écrit :

WM <wolfgang.mueckenheim@tha.de> wrote:

Only a matheologian fixed in his views can claim that after knowing my
game

We've known your game for years;

You have not understood it. Otherwise if not agreeing you could show an
error. But you can only curse:

 it is to obfuscate, confuse, and lie.

"You seem to be ignoring the fact that, after you have colored a
countable family of pathes, say P0, P1, ..., Pn, ..., there may be
other paths Q that are not on this countable list but have,
nevertheless, had all their nodes and edges colored. Perhaps the
first node and edge of Q were also in P1, the second node and edge of
Q were in P2, etc. [...] by choosing the sequence of Pn's
intelligently, you can, in fact, ensure that this sort of thing
happens for every path Q." [Andreas Blass, loc cit]

It can happen for every FINITE path Q.

Not for finite and not for infinite paths. If the second node is in P2,
then also the first node is in P2. That is the principle of the Binary
Tree.

An infinite path in an infinite binary tree can be coded as an infinite
sequence of Ls and Rs, corresponding to whether at the next node one goes
left or right.  So, for example, the very first path might be
LLLLLLLL.....

It is impossible to use infinite sequences of Ls or Rs. What can be used
is a finite abbreviation like "LLLLLLLL.....". But there are only
countably many finite

I have just used infinite sequences of Ls and Rs.  It's clear you do not
have a degree in mathematics.

But, supposing these infinite paths can be mapped to the integers, what
is the second path?  And the third one?  There is no systematic way of
numbering these paths.

There is no way to enumerate the rationals either. See
https://osf.io/preprints/osf/tyvnk, 4 pages English or 4 pages German,
according to your preference.

Strawman.  There is no way to enumerate infinite paths in that binary
tree.  As for the rationals, you're just being an idiot, as it is common
knowledge how to enumerate the rationals.

It is clear that the number of such paths is the same as the power set of
the natural numbers.

Yes.

There are more elements in any power set than in
the original set.

Yes, but that has not the least to do with countability.

Wrong.

So there are more infinite paths than can be indexed by the natural
numbers.

There are more fractions than can be indexed.

Wrong.

Nevertheless my game shows a contradiction. Can you understand that?
The "explanation" of Andreas Blass is absolute nonsense because of the
principle of the Binary Tree. Can you understand that?

I'm sure I could, if I could be bothered.  Most of your contradiction
games come from not understanding the differences between the finite and
the infinite.  This one is likely another following this pattern.

Regards, WM

--
Alan Mackenzie (Nuremberg, Germany).