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

On 12/12/24 4:12 AM, WM wrote:

On 12.12.2024 01:32, Richard Damon wrote:

On 12/11/24 9:32 AM, WM wrote:

On 11.12.2024 03:04, Richard Damon wrote:

On 12/10/24 12:30 PM, WM wrote:

On 10.12.2024 13:17, Richard Damon wrote:

On 12/10/24 3:50 AM, WM wrote:

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Two sequences that are identical term by term cannot have different limits. 0^x and x^0 are different term by term.

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Which isn't the part I am talking of, it is that just because each step of a sequence has a value, doesn't mean the thing that is at that limit, has the same value.

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Of course not. But if each step of two sequences has the same value, then the limits are the same too. This is the case for
  (E(1)∩E(2)∩...∩E(n)) and (E(n)).

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But the limit of the sequence isn't necessary what is at the "end" of the sequence.

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The end of the sequence is defined by ∀k ∈ ℕ : E(k+1) = E(k) \ {k}.

 

None of which are an infinite sets, so trying to take a "limit" of combining them is just improper.

 Most endsegments are infinite. But if Cantor can apply all natural numbers as indices for his sequences, then all must leave the sequence of endsegments. Then the sequence (E(k)) must end up empty. And there must be a continuous staircase from E(k) to the empty set.
 Regards, WM
 

Note, "inifinite" isn't a Natural Number, or a Real Number, so NO segement, specified by values, can have an "infinte endsegment".
You seem to have a definitional problem.
Of course, since you logic says that infinite sets act just like finite seqments, and thus no set is actually infinite, as well as that 0 is equal to 1 since they are both limits going to the same target, such a confusion is only natural.