Why can no one in sci.math understand my simple point?
in [Logic]
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From: WM on 18 Jun 2010 12:29 On 18 Jun., 13:09, "Peter Webb" > The computable numbers are countable. > > And similarly Cantor's proof does not show that there are an uncountable > number of Reals. What do you understand by "uncountable"? > It proves exactly what Cantor claimed it did, which is that > you cannot list all Reals. Cantor said that there are 2^aleph_0 reals and aleph_0 rationals. And he "proved" that 2^aleph_0 > aleph_0. And he said that there are an uncountable number of reals because countable means listable. Regards, WM
From: WM on 18 Jun 2010 12:33 On 18 Jun., 14:17, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Newberry <newberr...(a)gmail.com> writes: > > On Jun 15, 9:46 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) > > wrote: > >> WM says... > > >> >On 15 Jun., 16:32, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > >> >> The proof does not make use of any property of infinite lists. > >> >> The proof establishes: (If r_n is the list of reals, and > >> >> d is the antidiagonal) > > >> >> forall n, d is not equal to r_n > > >> >As every n is finite, it belongs to a finite initial segment of the > >> >infinite list. > > >> I'm not sure what you are saying. The fact is, we can prove > >> that for every real r_n on the list, d is not equal to r_n. > >> That means that d is not on the list. > > > How do you know that it does not prove that an anti-diagonal does > > exist i.e. that an antidiagonal is a contradiction in terms? > > Because every infinite sequence of digits represents a real number? And > the antidiagonal is one such sequence? No. An infinite sequence of digits does not represent a number. In general it does not even converge. In order to have convergence, you need the powers of 10 or 2 or so. But without a finite definition there are no infinite sequences at all, neither with nor without powers. Regards, WM
From: Jesse F. Hughes on 18 Jun 2010 12:52 WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 18 Jun., 14:17, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> Newberry <newberr...(a)gmail.com> writes: >> > On Jun 15, 9:46 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) >> > wrote: >> >> WM says... >> >> >> >On 15 Jun., 16:32, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: >> >> >> The proof does not make use of any property of infinite lists. >> >> >> The proof establishes: (If r_n is the list of reals, and >> >> >> d is the antidiagonal) >> >> >> >> forall n, d is not equal to r_n >> >> >> >As every n is finite, it belongs to a finite initial segment of the >> >> >infinite list. >> >> >> I'm not sure what you are saying. The fact is, we can prove >> >> that for every real r_n on the list, d is not equal to r_n. >> >> That means that d is not on the list. >> >> > How do you know that it does not prove that an anti-diagonal does >> > exist i.e. that an antidiagonal is a contradiction in terms? >> >> Because every infinite sequence of digits represents a real number? And >> the antidiagonal is one such sequence? > > No. An infinite sequence of digits does not represent a number. In > general it does not even converge. In order to have convergence, you > need the powers of 10 or 2 or so. But without a finite definition > there are no infinite sequences at all, neither with nor without > powers. Yes, very enlightening. What a unique and remarkable grasp of mathematics. And what a shame that anyone allows you to teach. -- Jesse F. Hughes "I think the problem for some of you is that you think you are very smart. I AM very smart. I am smarter on a scale you cannot really comprehend and there is the problem." -- James S. Harris
From: WM on 18 Jun 2010 13:03 On 17 Jun., 21:59, MoeBlee <jazzm...(a)hotmail.com> wrote: > I never said a real number can be defined by an infinite sequence. All finite words belong to a countable set. If you exclude infinite words (sequences) then there is no chance for uncountability. > > > A real number > > can be defined only by a finite word. But there is no diagonalization > > over finite words. > > Without even commenting on what you mean or whether it is true, it > does not refute that the formalized argument is first order logic > applied to axioms and incontrovertible Incontrovertible is religion. Because its adherents exclude refutations from their perception. > You've not said what "wrong" assumption I've "started with". The possibility of an infinite sequence of infinite sequences that can be completed in order to obtain a completed "anti-diagonal" sequence. > All of this business of yours does not refute what is simply > introvertible, that a formal proof exists in the manner I've > mentioned. There may be a proof. But as the result is wrong the proof is not worth much. > > > The translation of these notions into your "incontrovertible" theory > > is the weak point. > > NO, you did not listen to what I said. I did NOT say anything about an > incontrovertible THEORY. Rather, I said it is incontrovertible that a > certain finite sequence of finite sequences of symbols exists. But this finite sequence leads to the result that an uncountably infinite sequence of infinite sequences exists. And that is wrong. > > > > Does ZFC not prove that all constructible numbers are countable? > > I don't know. What is the definition IN THE LANGUAGE of ZFC of > 'constructible number'? > > Anyway, I have no idea how you think that bears upon what I just > wrote. So there seems to be a gap in ZFC. But it is easy to prove that in fact there are only countably many constructible numbers. > > > Is Cantor's diagonal not a constructible number? > > This reveals your misunderstanding of the very basics of Cantor's > argument (which I prefer to take in it's formalization in Z set > theory). > > There is no object in ZFC that is "Cantor's diagonal". > > Rather, for any given enumeration of a set of binary sequences, there > is a diagonal and anti-diagonal for that enumeration. > > So, your question should be, for any given enumeration of a set of > denumerable binary sequences, is the diagonal and/or the anti-diagonal > constructible? Just that is what I expressed in short words. > > Or you might mention some particular enumeration of some particular > set of denumerable binary sequences and ask whether its diagonal and/ > or anti-diagonal is constructible? > > But the answer would depend on a *ZFC* definition of 'constructible'. And that is not possible to obtain? > > Cantor > > uses some x of R and shows that R is an uncountable set by > > constructing an x_0 of R. > Also we show that there is a bijection between the real interval [0 1] > and the set of denumerable binary sequences. So [0 1] is uncountable. > > And if R is countable then, a fortiorti, [0 1] is countable. Hence you agree that uncefinable, unconstructable and so on reals muts exist. > > Are you or are you not familiar with various certain ordinary Z theory > formalizations of Cantor's argument? I am, but I do not want to discuss in Z here, but I am interested in its results. > > Please answer that question. I cannot address you very well on this > subject if I don't know whether you are or are not familiar with some > treatment in which Cantor's argument is formalized in Z set theory. > > > We can show, probably also in ZFC, that the constructible elements of > > R are not an uncountable set. > > That requires your definition IN THE LANGUAGE of ZFC of > 'constructible'. > > Do you know what is meant by 'the language of ZFC'? > > Please answer that question, as again, I need to know in order to > guage my discussion with you. Yes, I know. But I do not want to use it. I only want to see the question below answered: > > > ZFC proves a FORMULA that we READ or RENDER in ENGLISH as "there > > > exists an x such that x is uncountable". > > > This proof is probably done by constructing another element called the > > anti-diagonal? > > Actually, there is another proof in which the anti-diagonal method is > not used. If you mean Hessenberg's proof, then let it aside for a moment. It would deserve further but different discussion. > > And again, just to be clear, there is not THE anti-diagonal, but > rather there is the anti-diagonal for each given enumeration of a set > of denumerable binary sequences. And how many such given enumerations of a set of denumerable binary sequences do exist? > > > So, concerning the constructible elements alone, ZFC cannot be able to > > prove their countability? Or how else could you avoid a contradiction? > > Please, you're wearing my patience. > > I have no doubt that you may show whatever contradictions between ZFC > and your own notions. That is not at issue with me. > > On the other hand, you have not shown any ZFC derivation of a formula > P & ~P nor have you shown that one exists. I am not about to show that. I am only interested in the question: Can ZFC define "constructible reals"? Can it prove their countability? Regards, WM
From: WM on 18 Jun 2010 13:11
On 17 Jun., 22:14, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jun 17, 2:00 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > That means, even undefinable real numbers are in trichotomy with each > > other and in particular with definable real numbers? > > You said trichotomy is not important in this discussion. Oh, I cannot remember and I am surprised. Trichotomy is the most important issue when dealing with numbers. It is always important. > > Anyway, if 'definable real number' is given a definition in ZFC and > every 'definable real number' is a real number, then of course, > trichotomy holds. > > Listen, you don't need to waste our time. > > If I say trichotomy holds among all real numbers than I mean just what > I said. But if there are more than countably many real numbers, then there are most of them so called moonlightr numbers. We do not know anything about them. How can they obey trichotomy? So when you say what you said, then you should argue how you can do what you said. > > > How can that be? There is no way to name an undefinable real (because > > there are only countably many names). And it is impossible to define a > > real number by an infinite sequence, because onl finite sets and > > sequences can be defined by listing the elements or terms. > > We prove a theorem that the ordering satisfies trichotomy. > > All the rest of your perceived difficulties with this have no bearing > on that. Meanwhile, if you think we prove two contradictory theorems, > then just state the exact P such that you think P and ~P are proven in > ZFC. You would not agree that proving trichotomy for numbers that have no definition and cannot be computed and cannot be constructed, is a contradiction? > > > Have you ever tried to put an undefinable real number in order with > > other, definable real numbers? > > Whatever I have or have not tried has no bearing on what is or is not > a theorem of ZFC. But you could find out whether the theorem is sensible or is nonsense, if you only tried to apply it. > > > > Or to be pedantic: > > > > <x y> e <_r > > > <y x> e <_r > > > x=y > > > > And there is no ordering on the cardinals, > > > OK. Let us stay with the reals. Say you have two undefinable reals > > between 0 an 1. How can you manage to find out which one is less than > > the other? > > To say that trichotomy is a theorem is not to say also that we have a > way to find out anything at all. We have found out that there is a > proof of the trichotomy of the reals. That's all I claim in this > immediate regard. But don't you think that it would be better to prove trichotomy by putting two numbers in correct order by magnitude than to prove that this can be done whereas in fact it cannot be done? Regards, WM |