Sujet : Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)
De : wolfgang.mueckenheim (at) *nospam* tha.de (WM)
Groupes : sci.mathDate : 04. Dec 2024, 14:31:12
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <viplj0$t1f8$1@dont-email.me>
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 26 27
User-Agent : Mozilla Thunderbird
On 04.12.2024 11:33, FromTheRafters wrote:
WM formulated the question :
On 03.12.2024 21:34, Jim Burns wrote:
On 12/3/2024 8:02 AM, WM wrote:
>
E(1)∩E(2)∩...∩E(n) = E(n).
Sequences which are identical in every term
have identical limits.
>
An empty intersection does not require
an empty end.segment.
>
A set of non-empty endsegments has a non-empty intersection. The reason is inclusion-monotony.
Conclusion not supported by facts.
In two sets A and B which are non-empty both but have an empty intersection, there must be at least two elements a and b which are in one endsegment but not in the other:
a ∈ A but a ∉ B and b ∉ A but b ∈ B.
Same with a set of endsegments. It can be divided into two sets for both of which the same is required.
Regards, WM
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