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

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}.

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

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Most endsegments are infinite.