Cantor Confusion
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From: mueckenh on 13 Nov 2006 09:55 David Marcus schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > David Marcus schrieb: > > > > > > > What is wrong with my reasoning? > > > > > > > > Nothing. Wrong is only the assumption that "goes on forever" could be > > > > considered a finished infinity, i.e., could be denoted by a fixed > > > > cardinal number being larger than any natural number. > > > > > > I'm not sure what you mean. In particular, I'm not sure what > > > "considered", "finished" (or "finished infinity"), "denoted", and > > > "fixed" mean here. > > > > I see. But recently you used the word "completed infinity". > > I don't think I ever said that. Do you have a quote? Here it is: ================ 2306 Von: David Marcus - Profil anzeigen Datum: Sa 11 Nov. 2006 21:29 E-Mail: David Marcus <DavidMar...(a)alumdotmit.edu> Gruppen: sci.math Noch nicht bewertetBewertung: Optionen anzeigen Antworten | Antwort an Autor | Weiterleiten | Drucken | Einzelne Nachricht | Original anzeigen | Missbrauch melden | Nachrichten dieses Autors suchen - Zitierten Text ausblenden - - Zitierten Text anzeigen - Lester Zick wrote: > On 10 Nov 2006 15:18:07 -0800, "MoeBlee" <jazzm...(a)hotmail.com> wrote: > >Lester Zick wrote: > >> Your way or the highway huh, Moe(x). > >If he has an argument that he thinks can be put in set theory, then I'm > >interested in his argument; If he doesn't think his argument can be put > >in set theory, then I'm not interested. He can post about his argument > >all he wants, but I'm not obligated to study his argument. > No one suggests you are. The problem I see is that one might cast an > argument in such terms as are acceptable to you and still not satisfy > exactly the same criteria on the part of others. I mean unless you are > the generally acknowledged expert in the field. Otherwise it would > look to me like you're just trying to take control of the discussion > in terms you find acceptable whether or not others do. > Let me see if I can simplify how the issue is or ought to be argued. > You posit certain properties of an infinite however you define it. So > the question then becomes whether your claim is or can be true. What does "is or can be true" mean? In mathematics, we are normally only concerned with provability (unless discussing philosphy). > Now > one way to show it's actually true would be to produce some entity > with the properties you posit of an infinite. Otherwise you'd have to > find some other way to get at the truth of what you claim unless you > just intend to claim it's true because you or others say so. > Now as I understand WM's argument he suggests you can never actually > produce any physical infinite because the physical universe is finite. > However he then apparently concludes from this that there can be no > infinites at all because there can be no physical infinites if the > universe is finite. That doesn't seem to be what WM is saying. He seems to be saying that the notion of a completed infinity .... ======================== > > > Be sure that my finished infinity means the same. > > > > > > If you look at the diagonal, you always see that it is only there where > > > > lines are. Therefore you will always see that it is of finite length. > > > > "It goes on forever" does not mean actually infinite length. The > > > > lengths of the lines also increase from line to line forever. > > > > Nevertheless, an infinite length will never be reached. > > > > > > To me, "infinite length" just means "goes on forever". To you, the two > > > phrases have different meanings, it seems. So, what do you mean by the > > > phrase "infinite length"? > > > > "Goes on forever" is a property of the set of natural numbers, as well > > of the elements n as of initial the segments 1,2,3,...,n. > > Yes. > > > It does not > > make you believe that there are infinite natural numbers n. > > If "infinite natural number" means a natural number which is larger than > every natural number, then I don't believe there are infinite natural > numbers. But there is a complete initial segment N of N which is larger than any other segment of N? > > > Why does it > > make you believe that there is an infinite initial segment 1,2,3... ? > > Sorry. I don't understand. "1,2,3,..." is the set of natural numbers. > You just wrote a few lines above that the natural numbers "go on > forever", and I agreed to it. You seem to be asking me why I believe the > set of natural numbers goes on forever. But, we just agreed that was > true. So, what are you asking me? If it only goes on forever without being completet anywhere, then there is no chance to find out whether all natural numbers are sufficient to enumerate all real numbers or not. Regards, WM
From: mueckenh on 13 Nov 2006 10:07 William Hughes schrieb: > Well, since I already agree there is no infinite column this does > not change anything. If you add only one element to each column, you get the order type omega + 1 for the length of he matrix. If you add only one element to each line, the order type of the width of the matrix remains omega. > > > There > > > are an infinite number of lines. Each line is finite. > > > > There is an infinite number of initial segments of columns. Each one is > > finite > > Which is just what I said (changing names from "line" to "initial > segments > of columns" changes nothing). Wrong. By lines I understand the horizontal rows of a matrix. By columns I understand the vertical rows. > > There are no omega lines. > > The number of lines is greater than any finite natural number. > What would you call the number of lines? Unbounded. > > > > 3,4,5,...,omega,1,2 > > > > > > > > > > An infinite set consisting of the finite integers and > > > the non-natural number omega. This represents > > > omega+2 > > > > No, the sequence above represents omega + 3. But you see the problem. > > > > No. There are two representations of omega + 3. Why is this a > problem? There are many representations of omega + 3, but above there is not omega + 2 (as you wrote). Regards, WM
From: Dik T. Winter on 13 Nov 2006 10:08 In article <1163353613.745777.63980(a)f16g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > > The first falseness is his assumption that every set can be > > > > well-ordered. > > > > > > This assumption is as wrong as the official assumption made by Zermelo. > > > > Assumption? > > AC is taken to be an axiom. That is nothing but an arbitrary > assumption. Or it is taken not to be an axiom. So you eather reason with AC or without AC. > Well-ordering is so easy connected with AC that we can > state: Well-ordering has been assumed. Not in set theory. AC is an axiom that may or may not hold, depending on the branch you are following. > > Do you know what that term means in mathematics? It means from a > > chosen set of axioms using ordinary logic. > > The chosing of the set is an arbitrary act. It is not reduced to first > principles in modern math. Cantor had first principles! They were in fact axioms, although he did not state it as such. And the first principles he did chose where of course arbitrary. > > > Cantor considered well-ordering as a first principle, > > > Zermelo introduced it at a first principle = axiom. Cantor was wrong, > > > Zermelo was right? > > > > Cantor did state it without suggesting either that it was a first > > principle or something else. He just assumed it. And he was wrong > > with that assumption. > > He was as wrong as Zermelo, not a bit more. Oh, well, do set theory without AC. No problem. There is a number of people doing it. And I do not understand why taking AC is wrong. As far as is known it does not lead to an inconsistency. > > > At the most in the ridiculous axiom faith of "modern > > > mathematics". > > > > Oh. Whatever that means. > > It means you attitude: After a set of axioms has been chosen, we an be > sure to make true mathematics. We do not make "true mathematics". There is not a single "true mathematics". After a set of axioms has been chosen we can make a valid or an invalid theory. And the theory is invalid if and only if there is an inconsistency. > > > > His second falsehood is when he states that a set of first > > > > cardinality (meaning sets of cardinality aleph-0) can only be > > > > counted with use of numbers of the second class (meaning omega > > > > and larger). And I think that especially this quote has lead > > > > Wolfgang Mueckenheim astray. Also see my discussion about this > > > > quote with Dave Seaman. The falsity is apparent if you realise > > > > that quote means that every set with cardinality aleph-0 has an > > > > omega-th element. > > > > > > No explicitly stated but implied. > > > > Explicitly stated. See my discussion with Dave Seaman where I give the > > quote, and where there is a long discussion about the meaning. > > I hoped to settle this question with my quote. How can your quote settle whether it was explicitly stated or not in my quote? > > For the non-German speaking, Cantor explains that > > the ordered set has ordinal > > 1, 2, 3, ... omega > > n+1, n+2, ..., 1, 2, ... n omega+n > > 2, 4, 6, ..., 1, 3, 5, ... 2*omega > > (Note, in current mathematics the last is noted as omega*2, I will use > > the old notation in the sequel.) He is right and indeed, > > 1, 2, 4, 8, ..., 3, 6, 12, 24, ..., 5, 10, 20, 40, ... > > has ordinality omega^2. A set of cardinality aleph-0 can be ordered > > according to every countable ordinality. > > That did he mean with countable by numbers of class II. > You see, he did not use an omegath element. Yes, so what? > > No. The above statement is not implied by the letter of Cantor to > > Mittag-Leffler. It is explicitly stated elsewhere. And it is wrong. > > To "count" a set of ordinality omega you do not need omega. In most > > cases you need omega, but there is one exception. Can you find it? > > To "count" a set of ordinality a with ordinals you need only the > > ordinals smaller than a. (And note that in counting with ordinals > > you start at 0, because that is the first ordinal.) > > The latter statement is true in modern mathematics. But nonsense > nevertheless. What is the ordinal number of the empty set? > "To count a set of ordinality omega > you do not need omega" is just my position. To count the natural > numbers, you need not omega, because every set of natural numbers is > cunted by natural numbers. Cantor's position, however, was the > opposite. Yes, and Cantor was wrong in that. As I have argued already. > > > In the examples above, we have no omega. Introducing it in fact as > > > number which follows on all (even) natural numbers, we get > > > 2, 4, 6, ..., 2nu, ..., omega, 1, 3, 5, ..., (2nu + 1), ..., 2omega > > > That is false. > > > > That ordered set has ordinality 2*omega + 1. What is the problem? > > That each omega can be substituted by 1,2,3,... or say a,b,c,... > yielding 4 omega. You are again confusing a set with its contents. If you want to look at ordinals as sets you can substitute omega with {1,2,3,...} or {a,b,c,...}, not with what you write. And in the sequence above they still remain a single element. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Daniel Grubb on 13 Nov 2006 10:00 >> make it into this matrix: >> >> 1 0 0 0 0 0 .... >> 2 3 0 0 0 0 ... >> 4 5 6 0 0 0 ... >> .... >> .... >> >> In which case, it is a square matrix and every line >> has infinite length. So the diagonal, which has infinite >> length, has the same length as every row and every column. >There is no line without trailing zeros. The diagonal has no zeros. I don't see the relevance of this. The question wasn't about whether there are zeros, the question was about the length of the diagonal versus the number of rows or columns. >> An alternative is to look at the incomplete, triangular array >> that we started with and notice that in triagular arrays the >> length of the diagonal is always at least as large as the >> length of each row. In the situation where there are finitely >> many rows, the length of the diagonal is the same as the number >> of rows. >> >> The sticking point is whether the length of the diagonal is >> the same as the length of some row even when there are infinitely >> many rows. You seem to be claiming the answer is yes. >As we have not the least idea of what "actual infinity" could be, we >can do nothing but extrapolate from the finite domain. That's a pretty bold claim. Why do you assert this? While *you* might not have any ideas what an 'actual infinity' is, I think that I do. >Would you believe that the diagonal of our infinite triangle can be >longer than the first column? Until I saw that there was a bijection from the entries of the first column to the entries of the diagonal, I would entertain the idea. After I saw that bijection, I would know they are the same cardinality. >> While >> it is clear that the length of the diagonal is at least the length of >> any row, it seems clear that the length of the diagonal is actually >> longer than the length of each row in this case since it is at >> least as long as the next row down. Since there is always a >> 'next row down', the length of the diagonal is longer than >> the length of every row. >For every line there is always a next line longer than the first. >Nevertheless you do not believe that there are lines with infinite >length. Right. The n-th line has length n, which is finite. So all lines have finite length. 1. For every line L, the length of L is finite. Also, 2. For every line L, there is another line L' so that the length of L' is larger than the length of L. Your problem seems to be that you think that 1. and 2. are contradictory. This seems to be because you think that 2. implies that there is a line of infinite length. If this is what you think, would you be kind enough to give a proof? If it is not what you think, would you clarify your stance? The final part is that if D is the diagonal, 3.For every line L, the length of D is at least as large as the length of L. From 2., this implies that 4. For every line L, the length of D is larger than the length of L. We also have 5. For every finite number n, there exists a line L so that the length of L is n. (Just take the n-th line). Together, 4. and 5. show that the length of D is larger than any finite number n. Hence the length of D is infinite. --Dan Grubb
From: mueckenh on 13 Nov 2006 10:16
David Marcus schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > David Marcus schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > David Marcus schrieb: > > > > > > > Okay, so for you there is no point to draw about your quote other than > > > > > > > understanding Cantor's own writings. And, I'll add that in this > > > > > > > particular instance, Cantor, as you translated, is not in conflict with > > > > > > > the theorem of current set theory that omega is a limit ordinal and the > > > > > > > first oridinal that is greater than all natural numbers. > > > > > > > > > > > > That is what I said. omega is a limit. In modern set theory there are > > > > > > limits. > > > > > > > > > > Moe said that "omega is a limit ordinal". He did not say that "omega is > > > > > a limit". The two statements are not the same. > > > > > > > > Moe said "Cantor, as you translated, is not in conflict with the > > > > theorem of current set theory". Cantor said "omega can be understood as > > > > a limit". > > > > > > Cantor said "understood". That is not the same as saying "is". > > > > Cantor also said: "is". Therefore it is clear that here he meant "is". > > Are you saying that even though Cantor wrote "understood", he really > meant "is"? No. At other places he said "is". Cantor called them "Grenzzahlen" and denoted them by Lim_n etc. Because by his second creation principle these numbers are created by limit processes. "wir nennen sie Zahlen zweiter Art, sind so beschaffen, daÃ? es für sie eine nächstkleinereï? gar nicht gibt; diese gehen aber aus Fundamentalreihen als deren Grenzzahlen hervor" "a = Lim_n (a_n) a ist hier die auf sämtliche Zahlen a_n der GröÃ?e nach nächstfolgende Zahl." > > > No. The hidden errors can better be recognized at the roots. > > Perhaps, but irrelevant. If you can't find the "hidden errors" in the > modern formulations, a likely explanation is that the "hidden errors" > have been removed in the process of changing the formulations. > Regardless, if the "hidden errors" are still there, your only hope of > convincing people is to point to them in the formulations that they > know. The most fundamental hidden error is that sets are described a potentially infinite (going on for ever) but taken and treated as if they were actually infinite. Regards, WM |