Path: ...!eternal-september.org!feeder3.eternal-september.org!i2pn.org!i2pn2.org!.POSTED!not-for-mail From: joes Newsgroups: sci.math Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary) Date: Thu, 9 Jan 2025 20:17:39 -0000 (UTC) Organization: i2pn2 (i2pn.org) Message-ID: <8036bca97a855105d629273e992bdd66856a7ffc@i2pn2.org> References: <98519289-0542-40ce-886e-b50b401ef8cf@att.net> <8e95dfce-05e7-4d31-b8f0-43bede36dc9b@att.net> <53d93728-3442-4198-be92-5c9abe8a0a72@att.net> <9c18a839-9ab4-4778-84f2-481c77444254@att.net> <8ef20494f573dc131234363177017bf9d6b647ee@i2pn2.org> <66868399-5c4b-4816-9a0c-369aaa824553@att.net> <412770ca-7386-403f-b7c2-61f671d8a667@att.net> <56517e7b-7fa0-4ea8-88f4-bbd4c385e8d2@att.net> MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit Injection-Date: Thu, 9 Jan 2025 20:17:39 -0000 (UTC) Injection-Info: i2pn2.org; logging-data="2824885"; mail-complaints-to="usenet@i2pn2.org"; posting-account="nS1KMHaUuWOnF/ukOJzx6Ssd8y16q9UPs1GZ+I3D0CM"; User-Agent: Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2) X-Spam-Checker-Version: SpamAssassin 4.0.0 Bytes: 4840 Lines: 75 Am Thu, 09 Jan 2025 19:25:19 +0100 schrieb WM: > On 09.01.2025 18:52, Jim Burns wrote: >> On 1/8/2025 9:31 AM, WM wrote: >>> On 08.01.2025 14:31, Jim Burns wrote: >> >>>> ⦃k: k < ω ≤ k+1⦄ = ⦃⦄ >>>> ω-1 does not exist. >>> >>> Let us accept this result. >>> Then the sequence of endsegments loses every natnumber but not a last >>> one. >>> Then the empty intersection of infinite but inclusion monotonic >>> endsegments is violating basic logic. >>> (Losing all numbers but keeping infinitely many can only be possible >>> if new numbers are acquired.) >>> Then the only possible way to satisfy logic is the non-existence of ω >>> and of endsegments as complete sets. >> >>> (Losing all numbers but keeping infinitely many can only be possible >>> if new numbers are acquired.) >> No. > Losing all numbers but keeping infinitely many is impossible in > inclusion-monotonic sequences. This case doesn't occur. >> Sets do not change. > But the terms (E(n)) differ from their successors by one number. > >> Not all sets are finite. >> ⎛ By 'finite', I mean ⎝ 'smaller.than fuller.by.one sets' > Spare your gobbledegook. Finite means like a natural number. Especially not of the same cardinality as n+1. > Much waffle deleted. Honest thanks for the note. >>> It is useless to prove your claim as long as you cannot solve this >>> problem. > >>> (Losing all numbers but keeping infinitely many can only be possible >>> if new numbers are acquired.) >> No. > Don't be silly. It is possible with infinite sets, which can't be reduced by a finite number. >> Sets emptier.by.one than ℕ are not smaller. > They are. But that is irrelevant here. The sequence of endsegments loses > all numbers. If all endsegments remain infinite, we have a > contradiction. No, they are subsets of the same cardinality. There is no contradiction. >> In the sequence of end.segments of ℕ there is no number which empties >> an infinite set to a finite set. > Then there cannot exist a sequence of endsegments obeying > ∀k ∈ ℕ: E(k+1) = E(k) \ {k+1} for all k ∈ ℕ and getting empty. No term of the sequence is empty, if you mean that. >> and there is no number which is in common with all its end.segments. > Therefore all numbers get lost from the content and become indices. WDYM "become"? There is no point at which all naturals would be counted - N being infinite. >> ℕ has only infinite end.segments. > Then it has only finitely many, because not all numbers get lost from > the content. Huh? No. Then not all numbers would be "indices". >> The intersection of all (infinite) end.segments of ℕ is empty. > What is the content if all elements of ℕ have become indices? There is no such endsegment. >> Sets do not change. -- Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math: It is not guaranteed that n+1 exists for every n.