Cantor Confusion
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From: mueckenh on 26 Oct 2006 07:16 Dik T. Winter schrieb: > As you allow constructions by lists, and as there are uncountably many > lists, there are also uncountably many WM-constructible numbers. Where are those list? How are uncountably many different lists are constructed by countably many words? > Yes, that is precisely what I wrote, quite sometime ago already, about > the computable numbers. It is easy to show that the computable numbers > are countable. But the theory has been developed a bit since then. > The problem with what König wrote is that not every finite definition > also gives a number. If you mean that there are more finite definitions than numbers, you are right. Therefore we have a countable set of constructable numbers. > When you substitute "computable" for "constructable", both are indeed > right. But Cantor was wrong. It is only after Turing that this was > solved. The finite definitions are countable, but not every finite > definition gives a number. So there exists a list of finite definitions > (this has been formalised using Turing machines), but this is not a > list of computable numbers, because there will be non-halting Turing > machines in the list, and you do not know how to separate them from > the rest. Look up the halting problem. There is no computable list > of computable numbers. Why do you insist on this obvious fact? It does not support your position. There is a list of all finite constructions or definitions (encoded by numbers, Gödel). This is the definition of countablity. The diagonal number of this list shows the listed numbers are uncountable. Contradiction. Regards, WM
From: jpalecek on 26 Oct 2006 07:37 mueckenh(a)rz.fh-augsburg.de wrote: > David Marcus schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > David Marcus schrieb: > > > > > Binary Tree > > > > Unfortunately, it was described in a way that I can't understand it. A > > > > wild guess on my part is that you mean to set up a correspondence > > > > between edges and sets of paths. > > > > > > I am sorry, but if you need a wild guess to understand this text, then > > > we should better finish discussion. Observe just how the discussion > > > runs with all those who understood it, like Han, William, jpale. > > > > Han doesn't understand it (although he probably thinks he does). William > > and jpale simply pick the mathematically meaningful statement that is > > closest to what you write and go from there. > > The model is that simple that any student in the first semester could > understand it. Every paths which branches into two paths necessarily > needs two additional edges for this sake. It is only your formalistic > attitude that blocks your understanding. But you must not think that > anybody is blocked like you. This looks like a proof by induction. Indeed, you can prove your formula by induction for FINITE paths. But for infinite ones, you must do one more transfinite step. This is no unnecessary formalism, but something you need to understand to get your proof working. And it is not intuitive, and it is not simple. Regards, JP
From: mueckenh on 26 Oct 2006 07:42 David Marcus schrieb: > Lester Zick wrote: > > On Mon, 23 Oct 2006 22:01:25 -0400, David Marcus > > <DavidMarcus(a)alumdotmit.edu> wrote: > > > > >Han de Bruijn wrote: > > > > [. . .] > > > > >> > > >> There are many readers here who DO understand Mueckenheim's binary tree. > > >> And no, binary trees will not be found in Halmos' "Naive Set Theory". > > >> Because it's too naive, I suppose .. > > > > > >You mean all the cranks think they understand it. > > > > So this set of all cranks. Would that be those who disagree with you? > > That's not the definition of "crank". Wikipedia has a definition of > crank. > > You can readily detect cranks because they never clearly state what > their words mean, they keep repeating the same things rather than > understanding and replying to what others say to them, but always asking for an explanation in the only anguage they believe to understand. And if delivering an answer, then they don't know how to interpret it, neither in their nor in another language. > Since Mueckenheim claims his binary tree shows an inconsistency in > standard mathematics (at least he sometimes claims that--at other times > he claims to not be discussing standard mathematics), It shows that there is a countable set which has not less elements than the set of real numbers. > if anyone really > understood what Mueckenheim was saying, they could translate it into > standard mathematics and demonstrate the inconsistency. > > Some people claim they understand it, but fail to give their > translation. I guess I can see now the reason of your problems: You don't understand because you are looking for a tanslation always. There is no translation required. The tree is that mathematics which deserves this name. It is outside of your model, independent of ZFC, but generally valid and, therefore; covering ZFC too. Regards, WM
From: Tony Orlow on 26 Oct 2006 12:43 mueckenh(a)rz.fh-augsburg.de wrote: > Tony Orlow schrieb: > > >>> How would you construct an actually infinite set? Pair, power, union? >>> They all stay in the finite domain if you start with existence of the >>> empty (or any other finite) set. Comprehension or replacement cannot go >>> further. So, how would you like to achieve it? >>> >>> Regards, WM >>> >> Inductive subdivision of the unit continuum? We certainly seem to be >> able to specify, or approximate arbitrarily closely, some values with >> infinite strings of digits. It seems obvious that any finite interval in >> the continuum has more than any finite number of points within it. So, >> isn't that an actually infinite set, albeit with linear finite measure >> and bounds? > > > Sorry Tony, you are in error. We cannot approximate sqrt(2) arbitrarily > close. We can visualize it by the diagonal of a square and we can name > it. But we cannot approximate it better than to an epsilon of > 1/10^10^100. It woud be nice if we could, but assuming we can manage > it, only because otherwise mahematics becomes too difficult, is a bit > too simple. > > Regards, WM > Can you justify that limit of accuracy in mathematical terms, without resorting to discussions of the size of the universe? I see no theoretical, mathematical reason for it. And, whether we can specify every point or not, especially if we can't, there are more than any finite number of reals in any unit interval. Or, do you claim there are a finite number? TO
From: Randy Poe on 26 Oct 2006 13:07
mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > Within the *real* numbers the limit does exist. And a decimal number is > > nothing more nor less than a representative of an equivalence classes. > > So we are agian at this point: The real numbers do exist. Not in your worldview. According to you, the only numbers that exist are a subset of the rational numbers, those with finite decimal expansions < 10^100 digits. - Randy |