Path: ...!weretis.net!feeder9.news.weretis.net!news.nk.ca!rocksolid2!i2pn2.org!.POSTED!not-for-mail From: joes Newsgroups: sci.math Subject: Re: The reality of sets, on a scale of 1 to 10 [Was: The non-existence of "dark numbers"] Date: Sun, 13 Apr 2025 08:20:10 -0000 (UTC) Organization: i2pn2 (i2pn.org) Message-ID: References: <9e0c7e728f7de44e13450d7401fe65d36c5638f3@i2pn2.org> MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit Injection-Date: Sun, 13 Apr 2025 08:20:10 -0000 (UTC) Injection-Info: i2pn2.org; logging-data="139062"; 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: 6736 Lines: 112 Am Thu, 03 Apr 2025 22:17:33 +0200 schrieb WM: > On 03.04.2025 16:40, Alan Mackenzie wrote: >> WM wrote: >>> On 28.03.2025 17:32, Alan Mackenzie wrote: >> >>>> So tell us, O wise one, how many elements are there in {1, 3, 5, 7, >>>> 9, ...}? And how many elements in {0, 4, 8, 12, 16, ...}? Which of >>>> these two numbers is bigger, and why? >>> |ℕ|/2 > |ℕ|/4. >> Start out with a set of natural numbers. Multiply each member by four, >> giving a new set. You'd have us believe that the new set contains >> fewer elements than the original set. > Fact. Ifff the natural numbers are an actually infinite set, then its > elements are invariable and fixed. How else could it be. > By multiplication no larger numbers > can be created. What you have in mind is a potentially infinite set. > > Let me explain in detail: > Cantor created the sequence of the ordinal numbers by means of his first > and second generation principle: > 0, 1, 2, 3, ..., ω, ω+1, ω+2, ω+3, ..., ω*2, ω*2+1, ω*2+2, > ω*2+3, ..., ω*3, ... . > This sequence, except its very first terms, has no relevance for > classical mathematics. But it is important for set theory that in actual > infinity nothing fits between ℕ and ω. Likewise before ω*2 and ω*3 > there is no empty space. What do „fits” and „space” mean? > According to Hilbert we can simply count beyond the > infinite by a quite natural and uniquely determined, consistent > continuation of the ordinary counting in the finite. But we would > proceed even faster, when instead of counting, we doubled the numbers. > This leads to the central issue: Multiply every element of the set ℕ by > 2 > {1, 2, 3, ...}*2 = {2, 4, 6, ...} . > > The density of the natural numbers on the real axis is greater than the > density of the even natural numbers. In which sense? There are infinitely many of either. > Therefore the doubled natural numbers cover twice as many space than > before. Again, how? How much? > What is the result of > doubling? Either all doubled numbers are natural numbers, then not all > natural numbers have been doubled. Why? Every even number has a natural half its value. > Natural numbers not available before > have been created. This is possible only based on potential infinity. Or > all natural numbers have been doubled, then the result stretches > farther, namely beyond all natural numbers. No, there is no natural number whose double is larger than ω. Your mistake is to think of infinity as a really big number of the same kind as naturals, but it is rather more like a type or a special point. > It is more suggestive to double the set ℕ U {ω} = {1, 2, 3, ..., ω} > with the result > {1, 2, 3, ..., ω}*2 = {2, 4, 6, ..., ω*2} . > What elements fall between ω and ω*2? Where is ω? > What size has the interval between 2ℕ and ω*2? What is the interval between two sets? > The natural answer is (0, ω]*2 = (0, ω*2] with ω or ω+1 > amidst. No, that would be two consecutive infinities. You can do that, but not with f(x)=2x and not without violating the order. > The number of doubled natural numbers is precisely |ℕ|. But half > of the doubled numbers are no longer natural numbers; they surpass ω. If > all natural numbers including all even numbers are doubled and if > doubling increases the value for all natural numbers because n < 2n, > then not all doubled even numbers fit below ω. Yes they do. The product of two naturals is also a natural. > Natural numbers greater > than all even natural numbers however are not possible. > > Every other result would violate symmetry and beauty of mathematics, for > instance the claim that the result would be ℕ U {ω, ω*2}. All numbers > between ω and ω*2, which are precisely as many as in ℕ between 0 and ω, > would not be in the result? Every structure must be doubled! The „structure” (order type) of N is ω and not ω*2. > Like the > interval [1, 5] of lengths 4 by doubling gets [1, 5]*2 = [2, 10] of > length 8, the interval (0, ω]*2 gets (0, ω*2] with ω*2 = ω + ω =/= ω > where the whole interval between 0 and ω*2 is evenly filled with even > numbers like the whole interval between 0 and ω is evenly filled with > natural numbers before multiplication. On the ordinal axis the numbers > 0, ω, ω*2, ω*3, ... have same distances because same number of ordinals > lie between them. How do you define subtraction for ordinals? > This means that contrary to the collection of visible > natural numbers ℕ_def which only are relevant in classical mathematics > the whole set ℕ is not closed under multiplication. Some natural numbers > can become transfinite by multiplication. No, the union of the set called N and those numbers you have added is not closed. That’s your problem, not that of mathematics. > This resembles the displacement of the interval (0, 1] by one point to > the left-hand side such that the interval [0, 1) is covered. What is „one point”? Can you give an explicit function? -- 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.