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Subject: Re: WM and end segments...
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Date: Mon, 22 Jul 24 20:14:33 +0000
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From: WM <wolfgang.mueckenheim@tha.de>
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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 
> 
> 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. 
> 
> 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.

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

There are more fractions than can be indexed. 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?

Regards, WM