Cantor Confusion
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From: mueckenh on 28 Jan 2007 07:27 On 28 Jan., 03:31, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1169905223.354212.199...(a)a75g2000cwd.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > On 26 Jan., 02:21, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > A path-length which has no upper bound may be sufficent to be called > > > > infinite. > > > > > > Sorry, a path-length is something fixed, so the wording is inadequate. > > > But if you call a path-length infinite you do not satisty (1), because > > > there you associate path-lengths with natural numbers. > > > > We can boil down the discussion about trees to the following simple > > question, considering only one path, for instance the path p on the > > outmost left hand side of the tree. This path p (in terms of nodes) is > > the union of all paths of finite trees with length n, n in N. > > If you consider the union of paths as the union of sets of nodes, you are > right. If I consider the path length as the union of path length: = {0,1,2,...,n-1}, I am right too. > > > Therefore all the path-*lengths* in the union are natural numbers. > > Yes, the cardinality of all those paths (as sets of nodes) is a natural > number. > > > Notwithstanding the question whether there are infinitely many paths > > in the union or not: If the union path p is infinite, > > Note that the union path is the union of sets of nodes. > > > then at least > > one of the paths in the union must be infinite. > > Wrong. The union of a collection of finite sets can be infinite, and in > this case is. Here we are not concerned with the cardinal number of the union of infinitely many elements but with the length of the union of all finite path lengths, where each path length is measured from the common origin, namely 0 at the root of the tree. > > > Is this so? > > No. You fail to see a crucial difference. Going back at natural numbers: > union(n in N) {1, 2, 3, ..., n} is N > but > union(n in N) {{1, 2, 3, ..., n}} does not contain N. > As your paths are equivalent to initial segments {1, 2, 3, ..., n} we find: > union(n in N) p_n is p_oo > but > union(n in N) {p_n} does not contain p_oo. Fine. Apply this knowledge to the paths of the infinite tree T(oo). > > > > And, again, that is fundamentally wrong. It does not show anything of > > > that kind. The union of all sets of paths from finite trees contains > > > only finite paths. That is pretty basic set theory. If none of the > > > sets used in the union contains an infinite path, there is also not an > > > infinite path in their union. > > > > That is set theory. But it is wrong. The union of all finite paths of > > length n is infinite, but not actually. It has no upper bound. > > A fundamentally misunderstanding. The union of a set of things is not > the set of the union of things. In case of path lengths measured as I defined, the union of all lengths is a length. > > And T(oo) does not contain any path? But how do you *define* "finite > without an upper bound"? Can you come up with mathematical definitions > of "actual infinitiy" and "potential infinity"? Read my chapter 8. There (nearly) all is said what can be said about that topic. > > > > A subset of P_C is something like { P(1), P(2) } with as elements > > > *sets* of paths. However, P is a set with as elements paths. So P > > > can *not* be a subset of P_C. > > It is a subset of the union set P(1) U P(2) U ... > > Wrong, and that is not what you claimed. You claimed that P was a subset > of P_C. You claimed: > > 4) The set of paths in T(oo) is a subset of the countable set > > of finite sets of all paths in the finite trees. > P was the set of paths in T(oo). P(k) was the set of paths in the finite > tree T(k). P_C was the set of finite sets P(k). If T(oo) is constructed as the union of all finite trees T(k), then every path in P is a path which is in the union P(1) U P(2) U ... of the elements of P_C. Therefore P is a subset of this union P(1) U P(2) U ... of elements of P_C. Is this correct? > But even this statement is indeed wrong: > > It is a subset of the union set P(1) U P(2) U ... > 0.010101... is a path in the complete tree, but is not a path in any of > the P(k). 0.010101... is in the complete tree T = T(oo).(Correct me, if I remeber that wrong, but here are as much opinions as are set theorists.). 0.010101... is not in the union of all T(n), n in N, as you say. Therefore T(oo) = {T(n) | n in N} =/= T(oo). Isn't that a contradiction? Regards, WM
From: mueckenh on 28 Jan 2007 07:32 Dik T. Winter schrieb: > Your notation is > ambiguous, to say the least. You wrote: > > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } > without explicit notational clarification it is not even clear what this > does mean. I already proposed a much clearer notation: > S = {[a_1k, a_2k, a_3k, ...]| a in {0, 1}, k in N } > where S consistes of the sequences: [a_1k, a_2k, a_3k, ...]. If you > want to present a sequence as a single unit, present it as a single > unit. In the notation: > S = { x_k, y_k | k in N } > I do *not* see a set of pairs, bit a set of elements x_k and y_k. When > it is a set of pairs, the proper notation is: > S = {(x_k, y_k)| k in N } > > > >You can also write: > > > S = { x_i, y_i | i in N } > > > meaning S contains as elements all x_i and y_i for i in N. Your notation > > > is a novelty invented by you. > > > > New ideas often require new notations. > > In that case you have to *explain* your notation. I did. I defined: L is a set of limits L_k L = { L_k | k in N } and S is the set of corresponding sequences S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } This well defined. In general one does not use further symbols to denote outwritten sequences. I could have used <a_1k, a_2k, a_3k, ...> but need not. > But your idea could > easily be formatted within standard notation, not new notation needed. I could have used <a_1k, a_2k, a_3k, ...> > And new notations should *never* be ambiguous. > > > > > > > > > S is a set of sequences. > > > > > > You say so, inventing completely new notation. > > > > Sets of sequences are not so new. > > Indeed, but your notation is new. > > > > > > Consider the > > > > > following: > > > > > L = { L_k | k in {0, 1, 2}} > > > > > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in {0, 1, 2} } > > > > > If S is a set the cardinality of S is at most 2. > > > > > > > > Cardinality is |S| = 3. Its elements are the sequences number k = 0, 1 > > > > , and 2. > > > > > > Wrong. > > > > Oh, I thought you had understood? A set of three sequences has > > cardinality 3. > > Yes, but in your notation of the set you do not use the proper notation for > a sequence as element of a set. replace by <a_1k, a_2k, a_3k, ...>, if that is all you have to object.> > So the sequences S_0, S_1, S_2 are unordered? > > Pray consider the notation > S = { [ a_1k, a_2k, ... ] | a in {0, 1}, k in N } > or somesuch. <a_1k, a_2k, a_3k, ...> Regards, WM
From: mueckenh on 28 Jan 2007 07:41 On 28 Jan., 02:51, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1169822674.776815.294...(a)q2g2000cwa.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > On 26 Jan., 13:06, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1169806783.611087.268...(a)a75g2000cwd.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > On 26 Jan., 02:41, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > It has nothing to do with sophisticated models at all. They only have > > > > to simulate the basic fact. A natural number in its most basic > > > > representation, namely unary or unadic, simply is the superset of all > > > > smaller numbers and a subset of all larger numbers. > > > > > > In that case you have first to *define* what you mean with "union of > > > representations". There is no definition for it in mathematics. > > > > For natural numbers c is the same as <. 2 c 3 is the same as 2 < 3. > > This is proven by II + I = III. > > Perhaps. You are focused at notation. Mathematics is *not* concerned with > notation, it is concerned with the basics behind it. An abuse which yields infinite finity and finite infinity. > And in what way is > III c IV c V > ? In no basic way. To see the basics, you must continue IIII, IIIII, etc. > > > > > That is the origin > > > > which has to be simulated by every correct theory of natural numbers. > > > > II is a subset of III. > > > > > > By what definition? In what way are II and III sets? > > > > In that way which need no further declaratiojn. Put three nuts in your > > hat. Then you can see the set. > > Yes, I have a set of three nuts. The set is a set of nuts. That does not > mean that three is a set. Three is a set of what exactly? 3 is the set of all existnig sets with 3 elements. (Yes, this set exists.) > > > > > This kowledge as to be taken into account when > > > > denoting these numbers by 2 and 3. It has been done by the notation 2 < > > > > 3. > > > > > > In the von Neumann model, indeed. a < b if a subset b. That is the > > > definition of <. In the Peano axioms < can also be defined, using the > > > successor function, but there it has not necessarily anything to do with > > > subsetting. > > > > In unary representation it has. In other representations it has to be > > defined. > > Oh. So you think that subsetting comes before ordering in unary, but it > requires definition in other representations? Remarkable. Subsetting is the most basic operation. You can do it with sets of match sticks before you know any ordinal number. III is more than II because II c III. But the meaning of c is, on this level, the same as the meaning of <. Both are closely connected. Therefore I informed Fanziska that here first impression about "what comes first" was wrong. > Set theory > and the Peano axioms are not interested in representations at all. They > are only concerned with the abstract concept behind it. Moreover, when > you want to apply set theory to unary representations you will get > problems. Consider: > 111 > how do you propose to define it as a set? There are different answers. Applying curly brackets, {111}, this can be understood as a set witht one element {1} only. In case without brackets, you have three elements, left, mddle, right, or first, second, third. Regards, WM
From: mueckenh on 28 Jan 2007 07:48 On 28 Jan., 11:05, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > G. Frege <nomail(a)invalid.?.invalid> writes>On 27 Jan 2007 21:10:06 -0800, davidmar...(a)alum.mit.edu wrote: > > >>>> I think that Cantor's diagonalisation argument has a terminal flaw. > > >>> You have earned crank status now. Congratulations! > > >> It would seem so. Andy keeps posting the same argument without taking > >> into account the comments made on his previous posts. > > >Right. I've also noticed that. Btw, a rather interesting experience: > >to meet a crank in statu nascendi! > Fair enough. But most cranks are probably convinced that they are right In particular this one. A typical attribute of this crank is his frequent change of (anonymous (--> coward)) names like Amicus, Gregor H. or G. Frege. It is pointless to discuss with him. Regards, WM
From: William Hughes on 28 Jan 2007 08:35
On Jan 28, 7:21 am, mueck...(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > You claim that if something is true for every set L_n, > > then it is true for N > I do not talk about N. This symbol has become the aim of heaviest > abuse. You talk about the union of all natural numbers. N is the union of all natural numbers. > > > > > We know something about the maximum that is true > > for every set L_n. > > > So you do want to prove something about the maximum > > of the set N. > I want to see whether the union of all finite numbers can be an > infinite number. A union of numbers is a set of numbers. It is also a number. The union of all finite numbers is an infinite number. > This question was raised in the framework of the > infinite tree. Set theorists asserted that a union of finite paths > cannot be / contain any infinite path. A union of finite paths is a set of finite paths. A set of finite paths is not a path. A set of finite paths does not contain an infinite path. Do not confuse paths and nodes, they are not the same thing. Each path p has a set of nodes N(p). The union of finite paths p1 and p2 contains the set of nodes (union of N(p1) and N(p2)). (note that both N(p1) and N(p2) are finite). The union of all finite paths contains a set of nodes that is the union of finite sets. We know that the union of finite sets can be a (potentially) infinite set. The union of all finite paths contains a (potentially) infinite set of nodes. - William Hughes |