Cantor Confusion
in [Math]
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From: Virgil on 23 Feb 2007 16:54 In article <1172267009.642042.165340(a)k78g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 23 Feb., 20:21, Virgil <vir...(a)comcast.net> wrote: > > In article <1172253288.305584.116...(a)j27g2000cwj.googlegroups.com>, > > > > mueck...(a)rz.fh-augsburg.de wrote: > > > On 23 Feb., 14:35, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > Every tree T(n) contains only finite paths, > > > namely such with n nodes. There is no tree containing an infinite > > > path. > > > > Then there can be no infinite trees. > > Correct! > > > > > Every set of finite subsets of N is countable. The set of all finite > > > subsets of N is countable. > > > > > > > There are countably many unions like U(p(n)). > > > > > > Prove it. > > > > > If A is countable, then the set of all finite sequences of elements of > > > A is countable. > > > > But neither extends to the infinite subsets of A or infinite sequences > > of elements of A. > > The theorem extends to the finite subsets or finite sequences of > elements of of U(T(n)) . As in case of the tree all elements of P(oo) > like p(oo) are unions of finite pathes, p(oo) = Up(n), q(oo) = Uq(m), > r(oo) = Ur(i), and the sets of finite paths like p(n), are countable, > the unions of sets of finite paths like U(p(n)) = p(oo) belong to a > countable set. > > In short: Everything in the countable union of finite nodes and finite > paths of U(T(n)) is countable. Therefore the set of all possible > unions of finite paths is countable too. P(oo) is a subset of this set > of all possible unions of finite paths. Then the complete infinite binary tree has paths not in the union of those finite binary trees, as the set of paths of the complete infinite binary tree has a distinct path for every possible endless binary string, and there are uncountably many such strings. WM seems to ignore that obvious fact in his perpetual failure to prove his allegations.
From: William Hughes on 23 Feb 2007 17:49 On Feb 23, 12:26 pm, mueck...(a)rz.fh-augsburg.de wrote: > On 22 Feb., 13:26, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > On Feb 22, 2:42 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > On 20 Feb., 23:26, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > On Feb 20, 4:40 pm, mueck...(a)rz.fh-augsburg.de wrote: > > > > > 1,2,3,...,n <---> n > > > > > i: every initial segment of the natural numbers can be mapped > > > > to a natural number. > > > > > ii: the set of all natural numbers does not contain an element > > > > that is not in an initial segment of the natural numbers. > > > > > iii: The set of all natural numbers can be mapped to a natural > > > > number > > > > > No i: and ii: do not imply iii: > > > > Why not? > > > Because, as you have noted, there are times when i and ii are true > > and iii is false. > > There are times when it is false and there are times when it is true. Which means that knowing that i and ii are true does not tell you whether iii is true. You need something more. > The conclusion from the finity of finite segments on the finity of > infinite segments is as false as the conclusion from the evenness on > evenness, because all these are individual properties of segments. > > Another kind of property concerns the fact that every segment of even > positive numbers contains numbers which are greater than the cardinal > number.For sets which have a cadinal number which is n trichotomy with > even positive numbers this holds. Saying "For sets which have a cadinal number" is just making statement i. Statement iii does not follow, and in this case statement iii is a statement about E which does not have a cardinality. If we use the Wolkenmueckenheim definition of cardinality, then the statment is true but trivial. Noting that no strictly potentially infinite set has a cardinality in Wolkenmeukenheim, we conclude that this says nothing about sets without a fixed maximum. >Otherwise you should explain how the > set with n numbers > {2,4,6,...,2n} can be filled up by other (namely greater) even > numbers , without increasing the sizes of numbers. It is similar to > create uncountably many paths in the binary tree withoutthe presence > of uncountably many nodes or edges. > > Both tasks can only be managed by pure belief in these results and > refraining from any attempt to understand it. As nobody has made either claim, this is not a problem. You should start reacting to what people claim, not what you think follows from the claim. > > > > > > > > But look here for a direct proof: > > > > Every natural number n can be mapped on its finite initial segment > > > (i.e. that one where n is the largest number). > > > Therefore, there are as > > > many finite initial segments of natural numbers as are natural numbers > > > (one-to-one). > > > So set of initial segments of natural numbers is potentially > > infinite. > > > > Conclusion: Should there be an infinite initial segment, we had one > > > more initial segments than numbers. > > > There is no infinite initial segment (statement i). > > This does not tell you whether the set > > of initial seqments is potentially infinite or not. > > > > How could the infinite segment be > > > distinguished from the finite ones? > > > What infinite initial segment? The fact that there > > are only finite initial segments does not mean that the > > set of initial segments is not potentially infinite. > > I told you already. Potential infinity is not concerned, because there > is no cardinal number, at least no cardinal number which could be in > trichotomy with finite numbers. If there is a "number of initial segments" it must be the cardinality of the set of initial segments. If the set of initial segments is strictly potentially infinite, then, in Wolkenmuekenheim, this set has no cardinality, and hence there is no "number of initial segements". Saying that something that does not exist is a natural number is nonsense, even in Wolkenmuekenheim. > > What distinguishes finite sets and pot. infinite sets? As every set > can be extended by elements, every set is potentially infinite. There are two type of sets of natural numbers in Wolkenmuekenheim, those sets that have a cardinality (e.g. {1,2,3}) and those sets that do not have a cardinality (e.g. E, the set of all even numbers) outside of Wolkenmuekenheim they are called bounded and unbounded. Now, you do not like the idea of two types of sets, because people keep saying things like "showing something is true of sets of type I does not mean that the same thing is true for sets of type II". So you try to make two types of sets go away by changing the nomenclature. So there are no unbounded sets, but some sets have a bound that changes. Now we get sets that are potentially infinite and those that are not. To solve this we will call all sets potentially infinite. However, we still have two type of sets, those that must be potentially infinite, call them strictly potentially infinite, (e.g. the set E) and those that can be described as not potentially infinite (e.g. {1,2,3}). You cannot change properties by changing names. Whatever you do you still have two types of sets of natural numbers. > > If we say n, then we point to a finite set. If we say N, then we point > to all those finite sets in general. > "all those finite sets in general" is not the same as "a finite set" Saying that N is composed of finite sets is statement i. Statement iii does not follow for *ANY* property (whether or not N has this property, or whether or not there is a proof using statement i and some other fact). - William Hughes
From: William Hughes on 23 Feb 2007 18:06 On Feb 23, 5:49 pm, "William Hughes" <wpihug...(a)hotmail.com> wrote: > On Feb 23, 12:26 pm, mueck...(a)rz.fh-augsburg.de wrote: > > >Otherwise you should explain how the > > set with n numbers > > {2,4,6,...,2n} can be filled up by other (namely greater) even > > numbers , without increasing the sizes of numbers. It is similar to > > create uncountably many paths in the binary tree withoutthe presence > > of uncountably many nodes or edges. > > > Both tasks can only be managed by pure belief in these results and > > refraining from any attempt to understand it. > > As nobody has made either claim, this is not a problem. > You should start reacting to what people claim, not > what you think follows from the claim. Sorry, I misread the second claim. This claim has in fact been made. (no problem, paths are sets of nodes and edges. Saying that there are only countably many nodes and edges is statement i, saying that a path is composed of nodes and edges is statement ii. Saying that there are a countable number of paths is statement iii. Statement iii does not follow from statements i and ii. Statments i and ii are true. Statement iii is false.) - William Hughes
From: Dik T. Winter on 23 Feb 2007 21:06 In article <1172251940.593004.240900(a)h3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 23 Feb., 05:31, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > On 21 Feb., 16:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > like infinity. But it is impossible that both follows, infinity > > > > > and naturality. The definition a = z for an infinite chain of > > > > > inequalities a < b < c < ... < z is a *wrong definition*. > > > > > > > > What are you *talking* about? > > > > > > Simply about 2n - n < 2(n+1) - (n+1). > > > > I do still not understand. What are you *talking* about? > > The difference 2n - |{2,4,6,...,2n}| steadily inceases with incresing > n. Only after infinitely many increasings it drops to -aleph_0. Eh? I would think that 2n - |{2,4,6,...,2n}| equals n, so it increases without bounds. Why do you think it drops suddenly drops? > > The (only) infinite initial segment differs from each finite initial > > segment by having an element that is not in the finite initial segment. > > How does it differ from all finite segments? Or is here an occasion > where the simultaneous consideration of all (segments of) natural > numbers is inappropriate (quite contrary to Cantor's diagonal proof)? There is no single number where it differs from *all* finite segments. But you can do your comparison in parallel, but the comparisons do *not* give a single number. Why you think that should be the case escapes me. > > > > Wrong, there is no induction involved. > > > > > > Either induction or belief. > > > > I think you are on "induction and belief". What induction is there in > > f(n) = 2 * n? Moreso, what induction is there in (when considering the > > reals) f(x) = x^3? > > You need to find the position of such numbers as [pi*10^10^100] in the > sequence of natural numbers, but without always having such an easily > computable short hand as [pi*10^10^100]. There are at least > [pi*10^10^90] natural numbers which cannot be determined other than by > counting along the sequence of natural numbers. Therefore inductive > counting is required. For reals the same is valid, further you have to > count the digit positions behind the point. This makes not much sense mathematically. At least I can not determine any sensible meaning here. -- 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 23 Feb 2007 21:08
In article <1172252004.929693.56250(a)j27g2000cwj.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 23 Feb., 05:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1172134355.420444.86...(a)k78g2000cwa.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: .... > > > No? P(oo) = {p(oo), q(oo), r(oo), ...} > > > > > > p(oo) = Up(n), q(oo) = Uq(m), r(oo) = Ur(i) > > > > > > How many unions Up(n), Uq(m), Ur(i), ... can be in a union of finite > > > trees? > > > > Uncountably many, > > How can the set of finite subsets of a countable set be uncountable? In what way is p(oo) a finite subset? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |