Cantor Confusion
in [Math]
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From: Virgil on 23 May 2007 16:29 In article <1179933246.747971.247250(a)q66g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 22 Mai, 04:58, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1179750984.210277.71...(a)n15g2000prd.googlegroups.com> WM > > <mueck...(a)rz.fh-augsburg.de> writes: > > > On 21 Mai, 04:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > Pardon, I meant "countably many definitions". This is implied by > > > > > the > > > > > finity of every definition. If there were infinite definitions, > > > > > then > > > > > there were uncountably many definitions. > > > > > > > > Yup. So the question remains: "you disagree with common mathematical > > > > terminology?" > > > > > > No, it is obvious is that a finite definition means a definition which > > > is given by a finite number of words. Infinite definitions are > > > nonsense, according to Cantor. > > > > Yup, you are still living some hundred years ago. Mathematics has gone > > forward since that time. > > Unfortunately today "finite" is meaning "infinite" only when "forward" > is meaning "backwards". Finite today has several possible meanings, depending on context. In the context of set theory, the two most common meanings are: (1) Where N is the set of naturals, a set S is finite if and only if there is n in N such that S bijects with {m in N: m < n}, or (2) A set S is finite if and only if there does not exist any injection from S to a proper subset of S. The meaning of "infinite" is then defined to mean not finite. The meaning of "infinite" in most standard set theories will be equivalent, or very nearly so, to one of these. And in some, such as ZFC or NBG, the above two are themselves equivalent.
From: WM on 23 May 2007 16:40 On 22 Mai, 22:48, Virgil <vir...(a)comcast.net> wrote: > In article <1179863445.912400.167...(a)k79g2000hse.googlegroups.com>, > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 22 Mai, 03:53, Virgil <vir...(a)comcast.net> wrote: > > > In article <1179777652.039258.88...(a)y2g2000prf.googlegroups.com>, > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > On 20 Mai, 21:09, Virgil <vir...(a)comcast.net> wrote: > > > > > In article <1179668114.743217.104...(a)p47g2000hsd.googlegroups.com>, > > > > > > There cannot be in any set theory of repute any infinite set which is > > > > > not actually infinite. > > > > > > Whatever WM insists on as being potentially but not actually infinite, > > > > > it cannot be a set. Call it a pre-set or a quasi-set of a pseudo-set, > > > > > but not a set. > > > > > So the union U(T(n)) of all finite trees T(n), formed in an obvious > > > > way, is an actually infinite binary tree AIBT with actually countable > > > > sets of paths and nodes and edges. > > > > I will presume that "AIBT" should be "CIBT". and the set of paths is not > > > countable, but otherwise, you are correct. > > > AIBT = actually infinite binary tree AIBT . The set of paths in the > > AIBT is countable, because a countable union of countable sets is > > countable. > > WM has yet to show that the set of paths of an an AIBT is a countable > union of countable sets. All paths in a binary tree T(n) with n levels are finite. Now take the union of all these finite paths. By means of AC we can prove that this countable union of at most countable sets is finite. > > Others have shown directly that a CIBT does not have a merely countable > set of all paths. > > WM is asserting a difference between an AIBT and a CIBT. As CIBTs are > what occur in mathematics, one must presume that AIBTs only occur in > MathUnRealism. > > > > > > > >> (2) ==> p is nothing but a union of finite paths. > > > > > > Not neccessarily. That is the case only if one requires that a > > > > > path be no more than a set of nodes. > > > > > But every path has a soul in addition? > > > > It has a set of edges as well as a set of nodes. > > > Neither of the edges or the nodes is increased when switching from the > > AIBT to the CIBT. > > But, apparently, the set of paths is increased, since in a CIBT, as > previously defined, that set of paths is not countable. Apparently this assertion is nonsense. > > In a CIBT, there is, for every subset of N, a path unique to that > subset, which branches left from the level of each member of that subset > of N, and right from the level of each non-member. > Thus the set of paths bijects with the power set of N. > > And in ZF and NBG that is of cardinality greater than that of N. Perhaps the same is true in Disneyland. But why do you mention that here? > > > > > > > But, again, that is *not* the diagonal proof of Cantor. > > > > > > > The following wm-proof certainly even in your opinion belongs to the > > > > > > diagonal proofs considered by Cantor: > > > > > > > 0) mmm... > > > > > > 1) wmmm... > > > > > > 2) wwmmm... > > > > > > 3) wwwmmm... > > > > > > 4) wwwwmmm... > > > > > > ... .......... > > > > > > > And if the list can be considered as a completed entity, then there > > > > > > must be all natural numbers in the first column. > > > > > > As each "column" can contain only w's or m's, there are no natural > > > > > numbers in any column. > > > > > The first column consists of natural numbers only. > > (and of the unnatural 0) > > > > They are not a part of the strings, but only arguments to the function > > > which defines the list. > > > My statement is nevertheless correct. > > Not if constrained to string members. > > mmm... > wmmm... > wwmmm... > wwwmmm... > wwwwmmm... > ... ........ > > Is essentially the same list, but without the numbers. And what are you trying to say? That my statement now is incorrect concerning this list? Note however, if removing the first line, then we have a bijection between the w's on the diagonal and the w's in the first line and the w's in the last line for every initial sequence of lines. > wmmm... > wwmmm... > wwwmmm... > wwwwmmm... > ... ........ If the infinite actually and completely does exist as a finished entity, then the set of lines is finished in the same sense as the set of diagonal elements and in the same sense as the line elements, no? Regards, WM
From: Virgil on 23 May 2007 19:28 In article <1179952825.344411.38650(a)q66g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > > > AIBT = actually infinite binary tree AIBT . The set of paths in the > > > AIBT is countable, because a countable union of countable sets is > > > countable. > > > > WM has yet to show that the set of paths of an an AIBT is a countable > > union of countable sets. > > All paths in a binary tree T(n) with n levels are finite. Now take the > union of all these finite paths. The union of all those paths, each being itself merely a set of nodes, is merely a set of nodes, not a set of paths. SO WM is looking at the wrong thing again. > > > > Others have shown directly that a CIBT does not have a merely countable > > set of all paths. > > > > WM is asserting a difference between an AIBT and a CIBT. As CIBTs are > > what occur in mathematics, one must presume that AIBTs only occur in > > MathUnRealism. > > > Neither of the edges or the nodes is increased when switching from the > > > AIBT to the CIBT. > > > > But, apparently, the set of paths is increased, since in a CIBT, as > > previously defined, that set of paths is not countable. > > Apparently this assertion is nonsense. Not to those who can grok the proof that the set of paths is equinumerous with the power set of the set of levels in a CIBT. > > > > In a CIBT, there is, for every subset of N, a path unique to that > > subset, which branches left from the level of each member of that subset > > of N, and right from the level of each non-member. > > Thus the set of paths bijects with the power set of N. > > > > And in ZF and NBG that is of cardinality greater than that of N. > > Perhaps the same is true in Disneyland. But why do you mention that > here? Because it appears to be true everywhere outside of WM's world of MathUnRealism, and Disneyland is much too sane to be in WM's world. > > > > > My statement is nevertheless correct. > > > > Not if constrained to string members. > > > > mmm... > > wmmm... > > wwmmm... > > wwwmmm... > > wwwwmmm... > > ... ........ > > > > Is essentially the same list, but without the numbers. > > And what are you trying to say? That my statement now is incorrect > concerning this list? Note however, if removing the first line, then > we have a bijection between the w's on the diagonal and the w's in the > first line and the w's in the last line for every initial sequence of > lines. > > > wmmm... > > wwmmm... > > wwwmmm... > > wwwwmmm... > > ... ........ > > If the infinite actually and completely does exist as a finished > entity, then the set of lines is finished in the same sense as the set > of diagonal elements and in the same sense as the line elements, no? That they are all "finished" does not mean that they are all finished in exactly the same way.
From: Dik T. Winter on 25 May 2007 21:52 In article <1179864527.794567.66170(a)p77g2000hsh.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 22 Mai, 04:17, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1179746990.290370.268...(a)z24g2000prd.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > On 21 Mai, 04:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > for what is now called "simply normal". > > > > > > Please do not conclude from your lacking knowledge on that of others. > > > Weakly normal is a common definition. See fort instance. > > >http://eom.springer.de/N/n067560.htm > > > Or do you think that "weakly normal" is not common because it is > > > missing in Wikipedia? I would estimate Springer somewhat higher than > > > Wikipedia. > > > > I do not use Wikipedia in general for mathematical knowledge. My main > > online source is Mathworld. And of course the large library of my > > institute. > > Springer Online is edited by a man from CWI, Amsterdam. Yes. So what? Was he the particular editor of that article? But, whatever, "weakly normal" is *not* the common term. The common term is "simply normal". > > > > Yes, so what? The Champerowne constants and the Copeland-Erdos > > > > constants are *not* rational. Read just below your quote, where > > > > they give a Champerowne constant. > > > > > > So what? I never said that every however normal number must be > > > rational. But 0.01234567890123456789... (weakly normal to base 10) > > > and 0,012012012... (weakly normal to base 3) *are* rational. > > > > So your statement was irrelevant as a response to my statement. > > Your statement was irrelevant to my book. You think so. > > But from > > your book: > > "also eine sogenannte normale irrationale Zahl, die keine erkennbares > > Muster der Ziffernfolge aufweist" > > Note the last sentence! Yes, I note it. And I note also that the first part of that sentence is nonsense, and that is what I do remark on. Whether or not pi is normal has *no* relationship with the remainder. > > > In the first place, a normal number can have a well defined sequence of > > digits (as the Champerowne numbers show). > > Of course, but then we could compute the numerals. Yes, we could indeed. > > And, in the second place, there > > are numbers without a well defined sequence of digits that are *not* > > normal. > > And there are red cars, which are not normal cars. > > > (Normal here meaning normal to a particular base.) The two are not > > equivalent. So even if no rule can be given for the digits of pi, that > > does *not* mean that it is normal. > > But if pi is normal with no recognizable pattern, then no rule can be > given. (If I say A ==> B then I do not imply B ==> A.) Normality of pi has nothing to do with it. If pi is not normal and with no recognizable pattern, then also no rule can be given. To wit: normal => no recognizable pattern and no recognizable pattern => normal are *both* false. So even if it can be proven that pi is normal that does *not* mean that there is no revognizable pattern. That requires a different proof. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 25 May 2007 22:15
In article <1179931587.894431.146050(a)u30g2000hsc.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 22 Mai, 04:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1179747854.168331.52...(a)x35g2000prf.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > On 21 Mai, 04:37, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1179654356.461792.242...(a)n59g2000hsh.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > > > Its is more. You cannot answer the question whether the numbers P = > > > > > [pi*10^10^100] and P' = P with the last digit replaced by 3 > > > > > nsatisfy P < P'. > > > > > > > > Yes, so what? Your distinction is just terminology, and not more than > > > > that. > > > > > > It is by far more than a difference in terminology if one will never > > > or always be able to answer a question. > > > > Makes absolutely no sense. You have some implicit notion of number in mind > > that I do not have. > > Trichotomy. But there is trichotomy. We know that either P < P' or that P = P' or that P > P'. But we do not have the means to determine which of the three holds. > > You are obviously confused. For an injection from paths to nodes you need > > a catalogue of all paths. > > The tree *is* a catalogue of path, namely of *all* existing paths > which represent existing real numbers of the interval [0, 1]. And I thought that a catalogue in your sense numbered the paths from 1 onwards. Apparently you used a different meaning, again. > Each of > these paths p can only be distinguished from another path p' when both > have been separated from each other. > > You know that each node separates one more path, the number of > separated paths minus number of nodes is 2-1-1+2-1-1+2-1-1 +-... No. Each node splits a set of paths in two sets of paths. When you come in at a node with a set of K paths, you come out with two sets of K/2 paths (this is a bit informal). If K is infinite, so is K/2. And if K is uncountable, so is K/2. So all this splitting does show nothing. And it is not the case that each node separates one more path. If that were the case you should be able to indicate what path is separated by the root node. > Therefore, there must be as many separation points, or nodes, as > separated paths. Wrong. > > > > I would think that that is *not* an injection from paths to nodes. > > > > Moreover, at each node there are many paths that leave it on the > > > > left-hand side, so it is not even an injection. > > > > > > Here is another mapping: Map the node on one of the many paths, call > > > it A. The others will get their nodes when they separate from A. > > > > Again, you are mapping nodes to paths. Why can't you read? > > Nodes can be bijected to paths. Therefore the direction is > unimportant. The direction *is* important. If you want to prove bijection you have to show an injection in both ways. I ask you for an injection from paths to nodes, but you are either unable or unwilling to give one. An injection from nodes to paths is trivial. What you are trying to do here is to prove the set of paths is countable by giving an injection from the set of nodes (countable) to the set of paths. That is not a proof for a bijection. You either have to show a surjection from the nodes to the paths or an injection from the paths to the nodes. You fail. > > > You do not find any path. You find at most path bunches. > > > > As each path bunch starts at the root, I find that at each node the > > number of path bunches does not increase but diminish. > > Interesting. Not more than that? > > > > > The number of paths cannot be larger than the number of edges. > > > > > > > > Why not? > > > > > > Because every path contains more edges than it has in common with any > > > other path. > > > > That is not a proof. > > No? And was so much covinced ... Why do you think this is not a proof? > But don't come up with male and female dancers. You are thinking that any means all in your statement. It does not, at least in mathematical terminology. If you want it you have to prove it. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |