Re: Incompleteness of Cantor's enumeration of the rational numbers

On 11/15/2024 5:04 AM, WM wrote:

On 14.11.2024 19:31, Jim Burns wrote:

On 11/14/2024 5:20 AM, WM wrote:

Here is a single claim which is true:

>
You don't say what reason you (WM) have
for knowing that that single claim is true.

>
It can be proven

for every finite geometric figure
that covering it by small pieces or intervals
does not depend on the individuality and therefore on the order of the pieces.
That means
if there is a configuration where
the figure is not covered completely,
every possible shuffling will also fail.

For infinite figures
we use the analytical limit
as is normal in mathematics.

The reason you (WM) give is that
☠⎛ whatever is true of all finite
☠⎝ is also true of the infinite.
You (WM) are misinterpreting the infinite as
☠( just like the finite, but bigger.
The finite are the countable.to from.nothing.
Anything countable.to from.nothing is finite.
It's what.we.mean.
It isn't a boundary which distinguishes
the infinite from the finite,
not even a dark, inaccessible boundary.
It is a property which distinguishes
the infinite from the finite,
being countable.to from nothing.
Each countable.to from.nothing
is countable.past
  to a further countable.to from.nothing.
and thus
is not.last countable.to from.nothing.
There is no last.countable.to from.nothing,
not even
☠( a dark last.countable.to from.nothing.
Each finite is countable.to from.nothing.
All the finites are not countable.to from.nothing.
All the finites are infinitely.many.
⎛ A finite set A can be ordered so that,
⎜ for each subset B of A,
⎜ either B holds first.in.B and last.in.B
⎜ or B is empty.
⎜
⎜ An infinite set is not finite.
⎜ An infinite set C can _only_ be ordered so that
⎜ there is a non.empty subset D of C, such that
⎜ either first.in.D or last.in.D or both don't exist,
⎜ not visibly and not darkly.
⎜
⎝ And our sets do not change.
It's what.we.mean.