Cantor Confusion
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From: David R Tribble on 8 Feb 2007 00:53 >> Then I take it you think the above is bijection from the real numbers >> to the naturals? Which natural number does the real number 1/3 map to? > Federico Ferreres Solana wrote: > It is 33333...(periodic) > > If you do not allow infinetly large numbers in naturals, And we don't... > you must come to the conclusion that fractionals are more, ... No, the correct conclusion is that your particular mapping between the rationals and the naturals does not count all of the rationals. But just because your particular mapping is not a bijection does not mean there is not such a bijection at all. Consider the rather elegant mapping: S(0) = 0 S(1) = 1 S(2n) = S(n)+1 S(2n+1) = 1/S(2n) This maps every natural to a unique rational. The inverse mapping (easily derived by noticing the mappings of 2^n) provides a unique natural for every reduced-form rational. Cantor also provided a mapping of all the rationals to all the naturals, though not quite as simple as the one above. So there do exist complete mappings between the rationals and the naturals, proving that they are equinumerous sets (i.e., they have the same cardinality).
From: David Marcus on 8 Feb 2007 01:05 Federico Ferreres Solana wrote: > > Then I take it you think the above is bijection from the real numbers > > to the naturals? Which natural number does the real number 1/3 map > > to? > > It is 33333...(periodic) > > If you do not allow infinetly large numbers in naturals, you must come to the conclusion that fractionals are more, because you have an expanded definition, allowing you to write infinetly large numbers such as 0.33...(periodic) or PI, but that you can NEVER see in practice, just imagine them. > > 1/3 is an abreviation for 1(all, base stuff, world, object to be parted into potentially infinite slides)/3(parts), which means in naturals, infinity/3, or 3333...(periodic). > > Natural number 1 is the representation of the smallest slides of pie called infinity. But 1 is also the largest fractional number imaginable, something that when we use naturals start to call infinity. > > A fractional can be seen as defining an integer. 1/2 defines, or names 2 slides of a pie, so it may be seen as a fraction only, but then it defines 2, as you have 1 slide, and another one. When you define 0.11...(periodic) you are basically thinking there is an infinetly large number of slides in the pie, but you can never know what it is, yet, you think you know what 0,50000...(periodic) is, because you relate it to the whole pie. 500000...(periodic) is exactly the same in relation to something we named "infinity", that is exactly similar to what we denote as "1" in fractionnals. Infinity is exatly the same in the naturals as the "1" is to fractionals. When we don't allow that, we think of the problems as separate sets. > > In one (fractionals) we part from infinity towards smaller and smaller parts, and allow infinetly large number of slices through notation (1/3), allowing us to use infinity as a number ("1"), and then in the naturals, we claim to not know how to divide infinity. > > The problem arises when only inadvertly allows infinetly large numbers of "fractions" in the fractionals (which is OK for me) but forbids from thinking about infinetly large natural number. > > If you defined fractionals, as parts of something (1), you had to reach by induction, you would see you could only aproximate PI, but that PI will not be in the set. And if you do prove it is in the set, then you have proven that the natural 1415...(rest of numbres of PI), also exists, as the fractional part of PI implicitly defines in the fractionals and infinitely large number of parts or slices of a pie. Why don't you try starting at the beginning? Please define "real number", "natural number". -- David Marcus
From: mueckenh on 8 Feb 2007 02:34 On 6 Feb., 22:30, cbr...(a)cbrownsystems.com wrote: > Do you now return to your claim that there exists a surjection from > 1/13 onto 1/64, but not from 1/64 onto 1/13? > > Cheers - Chas- Zitierten Text ausblenden - Didn't you receive my last answer? > What are the /members/ of the set 2 that you mean to act as the domain > of the surjection? Or is the set 2 the empty set? Let m, n be natural numbers. If n =/= m, then there is no one-to-one mapping of n onto m. If If n > m, then there is no injective mapping of n onto m. If If n < m, then there is no surjective mapping of n onto m. > I don't mean some finite collection of /subsets/ of the set 2, but the > /members/ of the set 2. You agree that there is in general a difference > between "x is a subset of Y" and "x is a member of Y", yes? n = {0,1,2,...,n-1} The members of the set 2 = {0, 1} are 0 and 1. Perhaps there is a gap in your understanding of numbers and their representation in set theory? Regards, WM
From: mueckenh on 8 Feb 2007 02:42 On 7 Feb., 14:35, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > The potentially infinite set of even numbers is *constructed* by its > > segments > > > {2,4,6,...,2n} > > > Every time we increase n by 1 we increase 2n by 2. This cannot be > > avoided. Therefore it is impossible to have for finite natural numbers > > > lim[n-->oo] |{2,4,6,...,2n}| > 2n > > The limit of finite natural numbers is not a finite natural number. There is no limit. The notation lim[n-->oo] says nothing but that there is *no* limit. The size of n is not bounded. > What *is* true is that it is impossible to have for finite natural > numbers > > n= |{2,4,6,...,2n}| > 2n The set of natural numbers has only finite natural numbers. Threore nothing of this set can change the inequality. Only your firm belief in matheology can accomplish this - in your eyes. > > Now take the limit as n-->oo (the < becomes <= in a limit, > lim[n-->oo] n = oo, lim[n-->oo] 2n = oo) > > oo <= oo > > There is no contradiction. Of course not. How could there be a contradiction? In the infinite "<" becomes even ">" if required to keep ZFC free of contradictions. How could it be otherwise? Regards, WM
From: mueckenh on 8 Feb 2007 02:46
On 7 Feb., 15:43, Franziska Neugebauer <Franziska- Neugeba...(a)neugeb.dnsalias.net> wrote: > mueck...(a)rz.fh-augsburg.de wrote: > > On 6 Feb., 11:42, Franziska Neugebauer <Franziska- > > Neugeba...(a)neugeb.dnsalias.net> wrote: > >> mueck...(a)rz.fh-augsburg.de wrote: > >> > On 6 Feb., 11:22, Franziska Neugebauer <Franziska- > >> > Neugeba...(a)neugeb.dnsalias.net> wrote: > >> >> mueck...(a)rz.fh-augsburg.de wrote: > >> >> > Franziska Neugebauer wrote: > >> >> [...] > >> >> >> > The simplest reason is that omega - n = omega for all n in N. > > >> >> >> Where did you get that from? Reference? EB? > > >> >> > You could even figure it out by yourself. > > >> >> I cannot find any reference. Perhaps, there is none. > > >> > Already Cantor knew it, but I am too lazy to look for the > >> > reference. Read his papers, then you will encounter it. > > >> You are certainly able to brief a proof of "omega - n = omega for all > >> n in N.", are you? And of course to define subtraction involving > >> non-natural numbers like omega. > > > Was soll die Aufregung, meine Dame? > > Wer in der Küche kochen will, muß Hitze vertragen. > > Wer Unsinn als Zahl verkaufen will, darf sich nicht wundern, wenn der > > Zahlcharakter auch geprüft wird > > > Cantor, Collected Works p. 323: Es kommt hier noch die Operation der > > Subtraktion hinzu. Sind a und > > b zwei Ordnungszahlen und a < b, so existiert immer eine bestimmte > > Ordnungszahl, die wir b - a nennen ... > > > Ist omega eine Ordnungszahl > n für jede natürliche Zahl n? > > In English, please! Cantor wrote in German. If someone really wants to understand the origins of set theory, she should learn German. (If you have forgotten it, look at my post to Virgil. There I translated the relevant paragraph.) Regards, WM |