Am 25.03.2026 um 22:17 schrieb Alan Mackenzie:These sorts of accounts are mostly back-and-forth win-lose gamesWM <wolfgang.mueckenheim@tha.de> wrote:>Am 25.03.2026 um 18:55 schrieb Alan Mackenzie:>>Look at the short Binary Tree of only two levels L1 and L2:>N0
/ \
L1 N1 N2
/ \ / \
L2 N3 N4 N5 N6
...>The paths of the set P = {P(n) | n ∈ ℕ}, where node N(n) is mapped to
path P(n) ....>...., which you haven't defined, ....That is not necessary for my proof.>>.... fill the tree such that no further path can be distinguished.>If you define path P(n) as the finite path between N0 and Nn, that
subset of paths will indeed cover the tree.My paths are infinite. P(n) runs from the root node to node Nn and then>
continues infinitely.
You have failed to define your paths. Just what nodes does path P(n)
pass through below Nn?
Please choose the continuation of the paths as you like it best to
contradict my proof.
>Again, you haven't defined your paths. But a strategy for finding a path>
different from all in your countable set is as follows:
>
(i) Start from node N0.
(ii) At each node we pass through:
(a) identify the first path P(m) in the countable collection which
passes through the current node.
(b) Extend the new path by going to node in the next level different
from where P(m) went.
Through that node (like every node) also a path of my set goes. As I
told you, *every* node lies on one (or more) of my paths.
>(v) Repeat steps (ii)(a) and (ii)(b) indefinitely.>
Fail infinitely often.
>It will be seen that the new path is different from each path P(n) in the>
countable collection.
All nodes and their children and their grandchildren are already occupied.
>
Again, I recommend that you choose the path continuations according to
your taste. Then you can try to construct the diagonal path.
>
Regards, WM
contradicting each other, sort of like the account of incessant
inductive inference seeking gain and not estimating loss,
so these of the invincible ignorance of inductive inference
mayhaps don't even know they're at a loss.
"Talking past each other, ...", is the usual idea that they're
neither speaking to each other, nor to the matters both raised.
Anyways in an account like mine paradoxes get _resolved_ instead
of _forestalled_, where they stall forever, slowly filling up
the stall. This is a, "win-win", situation, instead.
Why lose?
One can definitely prove you're each proving each other losers.
Why lose?
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