Re: WM and end segments...

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

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

We've known your game for years; it is to obfuscate, confuse, and lie.

       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.

That is really uncalled for.

"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.

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.

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.....

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.

It is clear that the number of such paths is the same as the power set of
the natural numbers.  There are more elements in any power set than in
the original set.  So there are more infinite paths than can be indexed
by the natural numbers.

Regards, WM

--
Alan Mackenzie (Nuremberg, Germany).