Re: Incompleteness of Cantor's enumeration of the rational numbers (doubling-spaces)

On 12.11.2024 23:03, joes wrote:

Am Tue, 12 Nov 2024 22:38:51 +0100 schrieb WM:

On 12.11.2024 21:03, Jim Burns wrote:

On 11/12/2024 1:06 PM, WM wrote:

On 12.11.2024 17:47, Jim Burns wrote:

 

3) Then we could first cover all naturals and then all halves and then
all quarters and so on. But we know that already after covering all
naturals no further intervals are available.

Well, because this order has the type
omega + omega + … omega = omega*omega = omega^2.
This amounts to saying that the naturals are a subset of the rationals.
It goes back to the lines (or columns) of your tired matrix.
That is not a bijection between N and Q. That doesn’t prove there is none.

The geometry of covering intervals does not depend on Cantor's trick.
The impossibility to cover the real line by intervals
I(n) = [n - 1/10, n + 1/10]
proves that there is no bijection possible.
By the way, when in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
...
the X are accumulated around the upper left corner, we recognize the same: By re-ordering the X there is no covering of the matrix by X possible.
Believers in matheology must doubt geometry.
Regards, WM