Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)

On 12/21/24 4:58 PM, WM wrote:

On 21.12.2024 20:32, Jim Burns wrote:

On 12/21/2024 6:34 AM, WM wrote:

On 20.12.2024 19:48, Jim Burns wrote:

On 12/19/2024 4:37 PM, WM wrote:

>

That means all numbers are lost by loss of
one number per term.
>
That implies finite endsegments.

>
Q. What does 'finite' mean?

 Finite endsegments have a natural number of elements.

SO, none of your E(n) are finite endsegments, since they all have an INFINITE number of elements, being the INFINITE set of
{ n+1, n+2, n+3, ... } by your definition.

>
Consider end.segments of the finite cardinals.
>
Q. What does 'finite' mean,
'finite', whether darkᵂᴹ or visibleᵂᴹ?

 Finite endsegments cannot be visible

Finite endsegments do not exist.

>

Here is a new and better definition of endsegments
>
E(n) = {n+1, n+2, n+3, ...} with E(0) = ℕ.
>
∀n ∈ ℕ : E(n+1) = E(n) \ {n+1}
means that the sequence of endsegments can decrease only by one natnumber per step.

>
E(n+1) is larger.than each of
the sets for which there are smaller.by.one sets.
E(n+1) isn't any of
the sets for which there are smaller.by.one sets.

 No. E(n+2) is smaller than E(n+1) by one element, namely n+2.

But Aleph_0 - 1 is not smaller than Aleph_0. You can show that E(n+2) is a proper subset of E(n+1), but that doesn't make it smaller, as it can be equalinemerous if the size is infinite.

>
E(n+1) isn't smaller.by.one than E(n).
E(n+1) is emptier.by.one than E(n)

 It is also smaller, but we cannot distinguish ℵ₀ and ℵ₀ - 1.
{1, 2, 3, ..., ω} is smaller by one element than {0, 1, 2, 3, ..., ω}.

ω is not a member of the natural numbers.

|{1, 2, 3, ..., ω}| < |{0, 1, 2, 3, ..., ω}|

Which isn't a true statement, just shows you don't understand the nature of infinte sets.

>

Therefore the sequence of endsegments
cannot become empty

>
Yes, because
the sequence of end.segments
can become emptier.one.by.one, but
it cannot become smaller.one.by.one.

 Both happens. Cantor's bijections are nonsense.

No, YOUR logic in nonsense.

>

(i.e., not all natnumbers can be applied as indices)

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Each finite.cardinal can be applied,
which makes the sequence emptier.by.one
but does not make the sequence smaller.by.one.

 Rest deleted because it is wrong. There are |ℕ|^2 + 1 fractions.

Which is Aleph_0, same as |ℕ|.
You are just to ignorant to understand that,

 Regards, WM