Path: ...!weretis.net!feeder9.news.weretis.net!news.nk.ca!rocksolid2!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: Fri, 10 Jan 2025 09:03:03 -0000 (UTC) Organization: i2pn2 (i2pn.org) Message-ID: <1faeb6a4fe626d6d8795f11995f0391f8b4f8e58@i2pn2.org> References: <53d93728-3442-4198-be92-5c9abe8a0a72@att.net> <9c18a839-9ab4-4778-84f2-481c77444254@att.net> <8ef20494f573dc131234363177017bf9d6b647ee@i2pn2.org> MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit Injection-Date: Fri, 10 Jan 2025 09:03:03 -0000 (UTC) Injection-Info: i2pn2.org; logging-data="2904003"; 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: 4763 Lines: 71 Am Thu, 09 Jan 2025 23:27:21 +0100 schrieb WM: > On 09.01.2025 22:03, joes wrote: >> Am Thu, 09 Jan 2025 11:54:27 +0100 schrieb WM: > >>> If none is empty, then other numbers must be inside. Contradiction. >> No number is an element of all segments. > But what numbers are inside if all natural numbers are outside? Inside what? Inside each segment are infinitely many naturals. Mind your quantifiers: It is not required that the intersection be nonempty. >>> Infinitely many endsegments need infinitely many indices. Therefore no >>> natural number must remain as content in the sequence of endsegments. >> Only insofar as every number eventually "leaves" the endsegments. > But what remains? In what? The intersection is empty. >> This however does not imply and empty endsegment, > What remains? Nothing "remains". There is no end, only a limit. >> since there inf. many of both naturals and therefore endsegments. > Infinitely many numbers leave. All elements of ℕ leave. Yes, in the limit. >>>>> Unless you claim that the general law does not hold for ∀k ∈ ℕ. >>>> It does not hold for the infinite intersection. >>> Why not? >> Sorry, I lost track. Which law? > The law that the intersection gets empty but only by one element per > term. Well, in the limit (sigh) infinitely many numbers have been "lost". >>>>> It is trivially true that only one element can vanish with each >>>>> endsegment. >>>> Which noone contradicted. >>> Then the empty intersection is preceded by finite intersections. >> Does not follow. > It follows from the law that only one element per term can leave. Care to write out that deduction? >>>>> Jim "proved" that when exchanging two elements O and X, one of them >>>>> can disappear. His "proofs" violate logic which says that lossless >>>>> exchange will never suffer losses. >>>> Wrong. The limit of the harmonic series is zero, even though none of >>>> the terms are. >>> There is no exchange involved. >> Same reason: the limit may have different properties than the terms. > There is no limit involved when counting the fractions. It is an infinite "process". >>>>> That's a simple fact. The sequence of natural numbers 1, 2, 3, ..., >>>>> n, n+1, ... >>>>> cannot be cut into two actually infinite sequences, namely indices >>>>> and contents. >>>> Why should it? >>> Because an infinite sequence of indices followed by an infinite >>> sequence of contentent would require two infinite sequences. >> But this never happens. > An infinite set of infinite endsegments happens according to matheology. > It requires two infinite sequences. No, it doesn't. There is no "infinitieth" segment. >>> The claim of infinitely many infinite endsegments is false. >> There are infinitely many naturals though. > But all must be in the set of indices, if there are infinitely many > endsegments. But infinitely many must be in the set of contents, if all > endsegments are infinite. Those are the same set N. -- 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.