Re: Binary Tree and Pairs of Nodes

On 10 Okt., 06:55, Virgil <Vir...@xxxxxxxxx> wrote:
In article
<871188cc-f053-4ab5-870f-0b57ce13e...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,





 WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:
On 9 Okt., 16:49, LauLuna <laureanol...@xxxxxxxx> wrote:

     o
      |
      o
     /  \
    o    \
   /   \    \
  /     /     \
 /     \       \
/     /          \ ...

And so on.

Your game is just a countable sequence of choices, a countable
choice function. We have no guarantee that such a device will
'finally' give us the entire tree.

Of course there is no "finally". All we can say is that "iff there
was an infinite countable set of nodes" and "iff infinite sets could
be worked through completely", then every node would get its turn.

But every node occurs in infinitely many paths, as many paths as in the
entire infinite complete binary tree. So that every node gets infinitely
many
"turns".

Not according to the rule of my game. Consider the figure above. Node
x buys you one of the path leaving the path
0.111... at position x. You may say that there are many paths
deviating from 0.111..., and you are right. But if one of them
deviates from said path, it will get get another node. There is no
escape.

o
|
o
/ \
o \
/ \ \
/ / \
/ \ x
/ / / \
... /
\


With every won path you get infinitely many nodes. Each of them helps
you to get another path, one and only one path per node. In this game
you have the ratio of paths to nodes P/N = 0 at every step. The
infinite binary tree covers every step, but not more.

Similarly the usual decimal expansion of real numbers covers every
digit, but not more. There is no last digit of pi. There is every
digits of pi - unless one looks to close, but that is not in question
here. Set theorists insistuing of uncountably many reals must insist
that there are more than every digit of pi. I am not inclined to
discuss about such folly.



But if every node of the tree has been used to mark one path

Unless that node is a terminal (leaf) node, it marks more than one path,

Try to understand the rule of the game (there is only one). Try to
understand that there is no leaf node.

Regards, WM
.

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