Re: The complete infinite binary tree has only countably many infinite paths.

On 29 Mrz., 04:46, "Peter Webb" <webbfam...@xxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
Dear Calvin,
there is no point of formalizing such an obvious truth as I have shown
by my three proofs in the initial posting. But I think that this
thread has yielded a lot of new insights for all parties. Perhaps you
will agree that when you first answered my FOM-contribution, you had
no clue what a multitude of facets the binary tree may hide, won't
you?

Regards, WM

*******************************
Personally speaking, I learned nothing from this thread, and it provided no
new insights.

That need not be blamed to the thread.

I have heard this argument about binary trees many times
before, and its probably been floating around for 100+ years. Anybody who
has spent any time in sci.math or sci.logic would realise instantly that the
counter-example of 1/3 will be raised, but instead of the crank admitting
they have not formed a bijection between N and R, they will try and confuse
their way out of it with some bull*** about 1/3 mapping to some number
which is not part of N.

Happens every couple of months here. The informational or educational value
of such threads is zero for anybody with the slightest knowledge of set
theory.

You could have learned the following: There is no path 0.111... in the
tree, namely a path that is distinct by a certain number (aleph_0) of
ones from all paths used to construct the tree. There is only, for
every path with n bits of value 1, another path with one more bits of
value 1. That is the whole infinite story.

But the same holds for Cantor's list. There is no line that shows you
that the diagonal number is not in the list. There is only a line that
shows you that the diagonal number in not in the first n lines of the
list but that does not exclude that it could be in the next line.

Regards, WM
.

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