Cantor Confusion
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From: William Hughes on 7 Feb 2007 08:35 On Feb 7, 7:36 am, mueck...(a)rz.fh-augsburg.de wrote: > On 6 Feb., 13:28, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > Let E be the sequence of the sets F(e_i). There are two statements. > > > i. For each of the sets F(e_i) in the sequence E, there is > > an element of F(e_i) which is greater than the cardinal > > number of F(e_i) > > > ii There an element of one of the F(e) that is greater > > than the cardinal number of E. > > > i is true, ii is false, E is a counterexample. > > E is not an example, because it is self contradictory. > > 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. What *is* true is that it is impossible to have for finite natural numbers n= |{2,4,6,...,2n}| > 2n Does this cause a problem in the limit? For all for finite natural numbers n = |{2,4,6,...,2n}| < 2n 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. Can you answer the following question yes or no? Is the potentially infinite set of finite even numbers a set of finite even numbers? - Wiliam Hughes
From: mueckenh on 7 Feb 2007 09:29 On 7 Feb., 14:35, "William Hughes" <wpihug...(a)hotmail.com> wrote: > On Feb 7, 7:36 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > > > On 6 Feb., 13:28, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > Let E be the sequence of the sets F(e_i). There are two statements. > > > > i. For each of the sets F(e_i) in the sequence E, there is > > > an element of F(e_i) which is greater than the cardinal > > > number of F(e_i) > > > > ii There an element of one of the F(e) that is greater > > > than the cardinal number of E. > > > > i is true, ii is false, E is a counterexample. > > > E is not an example, because it is self contradictory. > > > 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. > What *is* true is that it is impossible to have for finite natural > numbers > > n= |{2,4,6,...,2n}| > 2n > > Does this cause a problem in the limit? In fact there is no limit. The only statement we can make is that the inequality below is true and remains true in every conceivable case. > > For all for finite natural numbers > > n = |{2,4,6,...,2n}| < 2n > > Now take the limit as n-->oo (the < becomes <= in a limit, No. n < 2n is always true for natural numbers (which property excludes 0). There is no reason to assume that 1/2 becomes 1 in "the limit". > lim[n-->oo] n = oo, lim[n-->oo] 2n = oo) > > oo <= oo > > There is no contradiction. Wrong for all finite numbers. > > Can you answer the following question yes or no? > > Is the potentially infinite set of finite even numbers > a set of finite even numbers? Yes, I can. The answer is yes. Probably set theory gives another answer. Regards, WM
From: Franziska Neugebauer on 7 Feb 2007 09:43 mueckenh(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 K�che kochen will, mu� Hitze vertragen. > Wer Unsinn als Zahl verkaufen will, darf sich nicht wundern, wenn der > Zahlcharakter auch gepr�ft 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 f�r jede nat�rliche Zahl n? In English, please! F. N. -- xyz
From: William Hughes on 7 Feb 2007 10:02 On Feb 7, 9:29 am, mueck...(a)rz.fh-augsburg.de wrote: > On 7 Feb., 14:35, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > On Feb 7, 7:36 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > On 6 Feb., 13:28, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > Let E be the sequence of the sets F(e_i). There are two statements. > > > > > i. For each of the sets F(e_i) in the sequence E, there is > > > > an element of F(e_i) which is greater than the cardinal > > > > number of F(e_i) > > > > > ii There an element of one of the F(e) that is greater > > > > than the cardinal number of E. > > > > > i is true, ii is false, E is a counterexample. > > > > E is not an example, because it is self contradictory. > > > > 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. > > What *is* true is that it is impossible to have for finite natural > > numbers > > > n= |{2,4,6,...,2n}| > 2n > > > Does this cause a problem in the limit? > > In fact there is no limit. The only statement we can make is that the > inequality below is true and remains true in every conceivable case. > > > > > For all for finite natural numbers > > > n = |{2,4,6,...,2n}| < 2n > > > Now take the limit as n-->oo (the < becomes <= in a limit, > > No. n < 2n is always true for natural numbers (which property excludes > 0). There is no reason to assume that 1/2 becomes 1 in "the limit". > > > lim[n-->oo] n = oo, lim[n-->oo] 2n = oo) > > > oo <= oo > > > There is no contradiction. > > Wrong for all finite numbers. And since the limit is not a finite number the fact that there is a contradiction for all finite numbers does not mean that there is a contradiction for the limit. > > > > > Can you answer the following question yes or no? > > > Is the potentially infinite set of finite even numbers > > a set of finite even numbers? > > Yes, I can. The answer is yes. Probably set theory gives another > answer. Let E be the potentially infinite set of finite even numbers You have now made three claims -every set of finite even numbers contains numbers which are larger than the cardinal number of the set - E is a set of finite even numbers - the statment "E contains numbers which are larger than the cardinal number of the E" is false - William Hughes
From: Han de Bruijn on 7 Feb 2007 10:12
William Hughes wrote: > And since the limit is not a finite number the fact that there > is a contradiction for all finite numbers does not mean that > there is a contradiction for the limit. In my calculus class, the fact that the "limit is not a finite number" just always meant: the limit _does not exist_, i.e. there IS NO limit. Why does contemporary mathematics employ two measures for the infinite? Why doesn't the left hand know what the right hand (calculus) is doing? Han de Bruijn |