Cantor Confusion
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From: Virgil on 13 Nov 2006 16:24 In article <1163433510.518520.69770(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > > They were in fact axioms, although he did not state it as such. And the > > first principles he did chose where of course arbitrary. > > Oh, he would rotate in his grave if he heard you. The exercise will do him good. >Of course he only > assumed those first principles which were true in nature or reality. Perhaps in WM's "reality" but WM's reality is quite different from everyone elses'.
From: Virgil on 13 Nov 2006 16:26 In article <1163433653.585487.297660(a)e3g2000cwe.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1163426814.926776.54020(a)i42g2000cwa.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > ... > > > May be so with some sets. A set of natural numbers includes its > > > cardinal number. > > > > I did not know that. Care to prove that the set {1, 3, 5, 7} includes > > the cardinal 4? > > Thank you for your attention. Correction: > An initial segment of natural numbers includes its cardinal number. An initial segment of ordinal numbers does not.
From: mueckenh on 13 Nov 2006 16:33 Dik T. Winter schrieb: > In article <1163353613.745777.63980(a)f16g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > > > 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? Meanwhile I checked your quotes and could not find any hint that Cantor explicitly or implicitly stated that a set of cardinality aleph_0 has an omega-th element. Could you repeat it please, perhaps I have overlooked it. But I cannot imagine that Cantor said so (and anticipated me). I studied all of his papers and most of his correspondence but I do not remember that he ever said so. It would have been in opposition to his general views. omega is the order type of a set which has *no* last element. > > > > 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? Why then do you assert he said that every set with cardinality aleph-0 had an omega-th element? > > > > 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.) No. It is outdated to start with the first. > > > > The latter statement is true in modern mathematics. But nonsense > > nevertheless. > > What is the ordinal number of the empty set? It is zero and zero is the zero-th ordinal, not the first (or 1-st). > > > "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. He used omega to count N, but omega was not considered by him as an element of N. > > > > > 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. A set is its contents - and nothing more! There are no ghosts in mathematics. > 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. There is a difference with Cantor' s notation. But that is outdated. Regards, WM
From: mueckenh on 13 Nov 2006 16:34 Daniel Grubb schrieb: > >> 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. The diagonal consists of numbers =/= 0. Therefore it must touch every line at such a position =/= 0. Please drop the zeros. Please increase every column by 1 digit. You will obtain omega +1 digits and the ordinal of the length of the matrix will then be omega + 1. Please increase every line by 1 digit. You will obtain less than omega digits and the ordinal of the width of the matrix will then be omega. But if you do not drop the zeros, then every line will contain zeros before the last (added)element, such that the diagonal will not be possible without zeros. This contradicts the fact that it consists completely of natural numbers (zero is not a natural number). > >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. What is your idea about the actual infinite? > > >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. The bijection is d_nn <--> d_n1. The diagonal d_11, d_22, ...,d_nn, ... with omega digits is bijected to the first column d_11, d_21, ...,d_n1, ... with omega digits. Please try the same now with each line and each column and the diagonal extended by 1 digit. The diagonal d_11, d_22, ...,d_nn, ..., d_ww is bijected to the first column which represents the length of the matrix d_11, d_21, ...,d_n1, ..., d_w1 The diagonal, however, cannot be bijected to the first or any other line because there is no line with omega + 1 elements. Neither is the width of the matrix omega + 1, but only d_11, d_12, ...,d_1n, ... Regards, WM
From: mueckenh on 13 Nov 2006 16:48
Lester Zick schrieb: > >> Now as I understand WM's argument he suggests you can never actually > >> produce any physical infinite because the physical universe is finite. Even if it is not finite, the domain accessible to us will remain finite forever. > >> 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. Everything of our thinking including mathematics relies on physical fundaments (writing, speaking, thinking). Therefore physical restrictions imply mathematical restrictions. > > > >That doesn't seem to be what WM is saying. I do not use the physical argument against set theory. It only makes me ultimately sure that I am right. > > I don't consider myself to be an expert on what WM is saying. The > above is just what I've gleaned from past comments. If it's incorrect > I apologize. I've just tried to make the best argument for whatever he > actually may be saying. Well done. > > > He seems to be saying that > >the notion of a completed infinity leads to either absurdities or > >contradictions. "Completed infinity" is an absurdity. > > Well as far as I can tell it does. Most notably the containment of > sets and subsets. I don't remember as the calculus ever requires > infinites. > > > Perhaps he thinks the way to avoid these absurdities is > >to only consider things that can be physically produced. > > Well that would certainly be one way. Another would be to stop using > finite infinites. Infinites only make sense in relation to one another > and not in relation to finites. > > >> Now personally I find most of the arguments disingenuous on both > >> sides. > > > >what is a standard mathematical argument that you find disingenuous? > > Insistence on the rectitude of arguments which can't be demonstrated > true. WM's arguments seem to rely on a finite universe. Opposition to > his arguments seem to rely on problematic assumptons of set theory. > Neither seems very convincing yet adherents of each pretend they are. Try to construct as many numbers as you can using only 100 bits. Then increase to 10^10 bits, then increase to 10^100 bits. More is not available. I find this very convincing. Regards, WM |