Please observe the following diagram:
Let the (0,0) coordinate be the redix point of a complete infinite binary tree, where each element is a unique real number represented by 0;1 path of depth |
N|.
As I understand it, by such construction one can't define a given real number that is not already in that tree, so the diagonal argument can't be expressed and one can't conclude that there are more real numbers than natural numbers.
Moreover, each level of the complete infinite binary tree is of the form 2^
x, where in case of an infinite binary tree
x=|
N| so we have 2^|
N| unique 0;1 paths, that are completely included in the |
N|x|
N| matrix.
At level 2^
x, where
x=|
N| (
x is not restricted to finite-olny values) there are 2^|
N| nodes that are still ordered as 0 1 0 1 0 ... sequence. So the first 1 from the left side belongs to the real number .000...1 and its complement real number on the right side is .111...0
I am aware that by the agreed notion there can't be a real nember that has inftinitly many zeros and then 1, but since 2^
x holds for both finite or infinite values (in case that
x=|
N|) I don't see any problem to represent a real number by inftinitly many zeros and then 1.
Can you please help me to understand better what's going on?
