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

On 09.12.2024 21:07, Python wrote:

Le 09/12/2024 à 20:40, Crank Wolfgang Mückenheim from Hochschule Augsburg aka WM a écrit :

On 09.12.2024 20:18, Python wrote:

Le 08/12/2024 à 23:34, Crank Wolfgang Mückenheim from Hochschule Augsburg aka WM a écrit :

On 08.12.2024 19:01, Jim Burns wrote:

On 12/8/2024 5:50 AM, WM wrote:

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∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n)
What can't you understand here?

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{E(i):i} is the set.of.all non.empty end.segments.
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⋂{E(i):i} is the intersection.of.all

non.empty end.segments.

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∀n ∈ ℕ:
{E(i):i}∪{E(n+1)} = {E(i):i}
Each is "already" in.

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Not the empty endsegment.
∀n ∈ ℕ: E(n) is non-empty. But not every E(n+1).

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You could hardly write something worse and more wrong that that.
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The very core property of N is that if n ∈ ℕ then n+1 ∈ ℕ.

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That is correct for definable natural numbers and even for almost all dark natural numbers.
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The very core property of analysis is that equal sequences have equal limits if they have limits at all.

 E(1)∩E(2)∩...∩E(n) = E(n)
 Lim E(1)∩E(2)∩...∩E(n) = {}
Lim E(n) = {}
 The are equal.

Not in a set theory where every endsegment is infinite.
An empty limit endsegment requires finite predecessors because there is only one way to emptiness allowed, namely this one:
∀k ∈ ℕ : E(k+1) = E(k) \ {k}.
Regards, WM