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

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Sujet : Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : sci.math
Date : 10. Dec 2024, 13:17:20
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <5805ad50ebff3400d1370d8c99790cbc727a340a@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
User-Agent : Mozilla Thunderbird
On 12/10/24 3:50 AM, WM wrote:
On 10.12.2024 01:45, Richard Damon wrote:
On 12/9/24 8:04 AM, WM wrote:
On 09.12.2024 13:03, Richard Damon wrote:
On 12/9/24 4:04 AM, WM wrote:
On 08.12.2024 19:01, Jim Burns wrote:
>
You (WM) are considering
infinite dark.finite.cardinals,
which do not exist.
>
Then analysis is contradicted in set theory.
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
The limit of the left-hand side is empty, the limit of the right- hand side is full, i.e. not empty.
I do not tolerate that.
>
By your logic, 1 equals 0,
>
No, that are two different sequences.
 
But since both 0^x and x^0 as x approaches 0 approach 0^0, your logic says that 0^0 is both 0 and 1.
 You should check your "logic". When two different sequences have different limits, this does not mean that the limits are identical. By the way 0^0 = 1 is simply a definition.
But that is just showing the error in YOUR logic.
The set of 0^x for all non-zero x is clearly 0, so since 0^0 is clearly the result of taking 0^x to the limit of x -> 0, that says, by your logic, that 0^0 must be 0.
The ser of x^0, for all non-zero x is clearly 1, so since 0^0 is clearly the result of taking 1^x to the limit of x -> 1, that says, by your logic, that 0^0 must be 1.
Thus 0^0 is both 0 and 1, but it must have just a value, so 0 must be 1.
(We can make it a limit to infinity, by replacing x with 1/n)
That is the error of assuming that just because a sequence apporaches a value as a limit (like your 1/10, 1/10, 1/10 ...) that this must be the "value" of the item at that limit.

>
Just because you have a sequence, doesn't mean you can talk about the end infinite state at the "end" of the sequence.
 The end infinite state is a set.
The end state is an infinite set, that doesn't have the same properties of the finite sets the form the sequence.

>
 you have two sequences that seem to go to the same infinte set at the end,
 Two sequences that are identical term by term cannot have different limits. 0^x and x^0 are different term by term.
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.
This ERROR seems to be a fondation of tour logic, a rule that tends to work in the finite realm, but fails at many infinities.

 REgards, WM
 

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