Path: ...!weretis.net!feeder8.news.weretis.net!news.szaf.org!news.karotte.org!news.space.net!news.muc.de!.POSTED.news.muc.de!not-for-mail From: Alan Mackenzie Newsgroups: sci.math Subject: Re: How many different unit fractions are lessorequal than all unit fractions? (infinitary) Date: Wed, 9 Oct 2024 15:11:31 -0000 (UTC) Organization: muc.de e.V. Message-ID: References: <4bc3b086-247a-4547-89cc-1d47f502659d@tha.de> MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: quoted-printable Injection-Date: Wed, 9 Oct 2024 15:11:31 -0000 (UTC) Injection-Info: news.muc.de; posting-host="news.muc.de:2001:608:1000::2"; logging-data="42405"; mail-complaints-to="news-admin@muc.de" User-Agent: tin/2.6.3-20231224 ("Banff") (FreeBSD/14.1-RELEASE-p3 (amd64)) Bytes: 5775 Lines: 129 WM wrote: > On 09.10.2024 16:13, Alan Mackenzie wrote: >> WM wrote: >>> On 08.10.2024 23:08, Alan Mackenzie wrote: >>>> WM wrote: >>>>> On 07.10.2024 18:11, Alan Mackenzie wrote: >>>>>> What I should have >>>>>> written (WM please take note) is: >>>>>> The idea of one countably infinite set being "bigger" than another >>>>>> countably infinite set is simply nonsense. >>>>> The idea is supported by the fact that set A as a superset of set B= is >>>>> bigger than B. >>>> What do you mean by "bigger" as applied to two infinite sets when on= e of >>>> them is not a subset of the other? >>> That is not in every case defined. But here are some rules: >>> Not all infinite sets can be compared by size, but we can establish s= ome >>> useful rules. >> Possibly. But these rules would require proof, which you haven't >> supplied. > These rules are self-evident. They're anything but self-evident for infinite sets. Presumably the muddle you're in is much the same as what mathematicians were in a couple of hundred years ago. Things have advanced since then. >>> =EF=83=A8 The rule of subset proves that every proper subset has fewe= r elements >>> than its superset. So there are more natural numbers than prime numbe= rs, >>> |N| > |P|, and more complex numbers than real numbers, |C| > |R|. Ev= en >>> finitely many exceptions from the subset-relation are admitted for >>> infinite subsets. Therefore there are more odd numbers than prime >>> numbers |O| > |P|. >> This breaks down in a contradiction, as shown by Richard D in another >> post: To repeat his idea: >> The set {0, 2, 4, 6, ...} is a subset of the natural numbers = N, >> {0, 1, 2, 3, 4, 5, 6, ...}, thus is smaller than it. >> We can replace the second set by one of the same "size" by multiplying >> each of its members by 4. We then get the set >> {0, 4, 8, ...}. >> Now this third set is a subset of the first hence is smaller than it. > No. When we *in actual infinity* multiply all |=E2=84=95|natural number= s by 2,=20 > then we keep |=E2=84=95| numbers but only half of them are smaller than= =CF=89, i.e.,=20 > are natural numbers. The other half is larger than =CF=89. Ha ha ha ha! This is garbage. If you think doubling some numbers gives results which are "larger than =CF=89" you'd better be prepared to give a= n example of such a number. But you're surely going to tell me that these are "dark numbers" (which I've proved don't exist). >>> Theorem: If every endsegment has infinitely many numbers, then >>> infinitely many numbers are in all endsegments. >> That is simply false. You cannot specify a single number which is in >> all endsegments. > True. This proves dark numbers. Dark numbers don't exist, or at least they're not natural numbers. There is no number in each and every end segment of N. [ .... ] >>> Note: The shrinking endsegments cannot acquire new numbers. >> An end segment is what it is. It doesn't change. > But the terms of the sequence do. Here is a simple finite example: > {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} > {2, 3, 4, 5, 6, 7, 8, 9, 10} > {3, 4, 5, 6, 7, 8, 9, 10} > {4, 5, 6, 7, 8, 9, 10} > {5, 6, 7, 8, 9, 10} > {6, 7, 8, 9, 10} > {7, 8, 9, 10} > {8, 9, 10} > {9, 10} > {10} > { } . Example of what? The reasoning you might do on finite sets mostly isn't applicable to infinite sets. > Theorem: Every set that contains at least 3 numbers (call it TN-set)=20 > holds these numbers in common with all TN-sets. Not even an ignorant schoolboy would maintain this. The two TN-sets {0, 1, 2} and {3, 4, 5} have no numbers in common. > Quantifier shift: There is a subset of three elements common to all > TN-sets. Understood? Yes, I understand completely. What you've written is garbage. > Now complete all sets by the natural numbers > 10 and complete the > sequence. You can't "complete" a set. A set is what it is, defined by its well defined members and is not subject to change. > Then we get: All sets which have lost at most n elements have the=20 > remainder in common. Note: All sets which are infinite have lost at mos= t=20 > a finite number of elements. More gobbledegook which isn't maths. > Regards, WM --=20 Alan Mackenzie (Nuremberg, Germany).