Re: WM and end segments...

Le 22/07/2024 à 17:08, Alan Mackenzie a écrit :

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

Le 22/07/2024 à 01:10, Ben Bacarisse a écrit :

 

That's a can of worms in WMaths.  WM has written 734,342,120 nonsense
posts about binary trees over the years.

 

It appears so to poors who cannot think straight.

 

Paths in the Binary Tree can be distinguished by nodes only. There must be at least as much nodes as paths.

 I explained this too you a while back.  Infinity is complicated.

No reason to go without logic.

What
is true in the infinite case is that there are as many FINITE paths as
nodes, namely a countable infinity of them.  However there are an
uncountable infinity of INFINITE paths.

Only a matheologian fixed in his views can claim that after knowing my game
Conquer the Binary Tree
Here is a variant of the construction by infinite paths, a game that only can be lost if set theory is true: You start with one cent. For a cent you can buy an infinite path of your choice in the Binary Tree. For every node covered by this path you will get a cent. For every cent you can buy another path of your choice. For every node covered by this path (and not yet covered by previously chosen paths) you will get a cent. For every cent you can buy another path. And so on. Since there are only countably many nodes yielding as many cents but uncountably many paths requiring as many cents, the player will get bankrupt before all paths are conquered. If no player gets bankrupt, the number of paths cannot surpass the number of nodes. [Hippasos: "What can we learn from the new game CTBT that I devised for my students?", MathOverflow (2 Jul 2010). W. Mückenheim: "History of the infinite", HI12.PPT, current lecture]
But there are more nutcakes like you.
"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] My reply: If the second node is in P2 then also the first node is in P2, and so on for all n   – for every path Q of the Binary Tree. No way to get rid of already coloured paths by choosing "intelligently"! Here not even the antidiagonal is constructed.
Regards, WM