Cantor Confusion
in [Math]
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From: Virgil on 23 Mar 2007 15:11 In article <1174654681.038185.299380(a)e65g2000hsc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 22 Mrz., 16:29, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1174562837.696384.225...(a)b75g2000hsg.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 21 Mrz., 16:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1174481309.873623.92...(a)e1g2000hsg.googlegroups.com> > > > > mueck...(a)rz.fh-augsburg.de > > > > > > What portion of the first node > > > > > > is assigned to your path? > > > > > > > > > > How much does the last term of the geometric series contribute to > > > > > the > > > > > value o the series? This is the value which the first node > > > > > contributes > > > > > to he path. > > > > > > > > As there is no last term of the geometric series this makes no sense. > > > > > > Nevertheless, it makes sense to calculate the sum of the series. > > > > You can calculate the sum of the series. But what the relation is to the > > contribution of the nodes to the paths is extremely unclear. From your > > statement I derive that the first node contributes nothing to the path. > > the first node contributes exactly as much as the last term of the > geometric series contributes to the sum 2 of the series. As this term > does not exist, it does not contribute anything. Ergo, the first term also contributes nothing! > The next one before > the last one also does not exist and not contribute. What is your immediate predecessor of the non-existent? > But somehow some > nodes manage to contribute enough to obtain the result 2 after all. Since in a tree, each node may be a geometric point in some plane, such a node, like a point, is that which has no part. So that WM's attempts to subdivide the indivisible come to naught. > > If we exchange infinitely many terms of the geometric series, its sum > remains 2, because it is absolutely converging. But as each term is a whole indivisible point or nothing at all, WM's sequence is all but two terms nothing at all. > > As the geometric series contains only the smallest possible infinity > of terms, it should not cause problems to read it from behind. A geometric series in which all terms are either 0 or 1, either has all terms 0 or only the first term 1 with common ratio 0, and thus has sum either 0 or 1 respectively. > > > > > > At no node in the tree there exists a single path. This means: There > > > are no single paths. > > > > Still unclear. Through each node go uncountably many paths. But what you > > mean with "there are no single paths" is unclear. Every two individual > > paths diverge at some node from each other. > > Consider all of them simultaneously. WM has often insisted that one cannot even consider all members of N sumultaneously, but now insists on considering all members of P(N) simultaneoulsly. WM now wants to swallow whole camels where he already must strain at gnats. > You like to do so when Cantor's > diagonal proof is concerned. On the contrary, one only considers one list at a time. > Every exchanged digit is followed by > infinitely many digits which have to be exchanged. Nevertheless you > say, it is possible to exchange all of them at one time. Consider the > paths o he tree - all at the same time. > > This is the big mistake of set theory. The unreasonable allowance or > prohibition of instantaneously possible actions. It is only in the physical world where actions take time. In the world of imagination, where mathematics lives, one escapes such physical limitations. > > > The number of separated paths up to level n is given by the number of > > > nodes of level n. No level of the tree has an uncountable number of > > > nodes. > > > > Makes no sense at all and is no proof. At each level n there are 2^(n-1) > > separated groups of paths, where each group contains uncountably many > > paths. > > That is wrong, because none of these paths ever gets isolated. Is it > similar to the quarks? Path-confinement? Every infinite path eventually gets separated from any other path, but it is only in a finite tree that any path ever gets simultaneously separated from all other paths, and that is only because finite paths have a leaf node which is separate from all other paths, but infinite paths have no leaf node, and every node in such a path is shared with other paths. For infinite ptahs, it is only the set of /all/ its nodes which is distinct from the set of all nodes of all other paths. > > We can safely state that at no level there are uncountably many > separated paths. We can confidently say that aat every level, WM is separated from common sense. > As every node is in the tree ad no node is outside, there are at no > place in the world uncountably many numbers. They exist, like all mathematics, in the world of imagination. > > Yes, so what? That is no proof of your statement. When we talk about the > > set of paths, we are *not* talking about any finite level. What happens at > > finite levels is irrelevant to the total result. > > Wrong. Every number has digits only at finite distance from the > decimal point. But 1/3, for instance, has non-zero digits at a greater than any given finite distance from the decimal point. > In the tree there is no level "infinite", but all the infinitely many > levels are there - each one in a finite distance from the root node. Which means that every such path, like N, is endless. > > > > As at each point in the tree there are uncountably many paths going through > > it, I wonder what you mean with "countably many paths are separated". > > At no point of the tree (and of the universe) more than countably many > path can be distinguished. There is a different path in any CIBT for every member of P(N).
From: Virgil on 23 Mar 2007 15:38 In article <1174655745.265882.25720(a)e65g2000hsc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 22 Mrz., 16:45, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1174563273.137227.294...(a)y66g2000hsf.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 21 Mrz., 16:16, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1174481667.234070.21...(a)e65g2000hsc.googlegroups.com> > > > > mueck...(a)rz.fh-augsburg.de writes: > > > > > > > > > On 20 Mrz., 17:07, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > ... > > > > > > > Yes. But the path-lengths are natural numbers, and there is no > > > > > > > infinite natural number. > > > > > > > > > > > > The path-length of a non-terminating path is *not* a natural > > > > > > number. > > > > > > > > > > How then can it be contained in a set which contains only sets of > > > > > finite paths? > > > > > > > > It is *not* an element of a set which contains only sets of finite > > > > paths. > > > > Pray get your statements correct. A set that contains only sets of > > > > finite > > > > paths has as elements sets of finite paths. Not paths. > > > > > > We can form the union of the sets. Then we get a set of finite paths. > > > > Yes, we can do that. And then we get a set of finite paths that does *not* > > contain a non-terminating path as an element, and also not the path-length > > of such a path as an element. So what now? > > This set of paths forms a tree which does not differ by any node from > the complete tree T(oo). A finite chain of nodes can, by definition, only be a path in a tree in which it is of maximal length, so no finite chain of nodes can be a path in a CIBT. Thus a CIBT either has all paths endless or no paths at all. > > > Again "contain". But, so what? I fail to see the relevance to the > > path-length of non-terminating paths. > > To spell it out clearly: If you take the infinite union U(T(n)) of all > finite trees T(n), then you do not have any infinite path. The you also do not have any CIBT, and in particular, you do not have T(oo). > > > > > > > why then do you believe that an infinite path (number) is contained > > > > > in > > > > > the union of sets of finite paths (numbers). > > > > > > > > It is not an element of the union of sets of finite paths. Why do you > > > > think that I do think so? The union of sets of finite paths is a set > > > > of finite paths, so there is no infinite path in it as element. > > > > > > Therefore, after having the union of all paths which are elements of > > > finite trees, there is no infinite path. > > > > Wrong. In the first place, paths are *not* elements of trees according to > > your definitions. Paths are *subsets* of trees according to them. Pray > > use the correct wording. > > OK. Subsets. > > > Second, the union of all paths etc., is *not* a > > path, it is just an infinite set of nodes, the same as the union of all > > trees. The paths are in it as subsets, as are the infinite paths. > > The union of all *finite* paths gives the same infinite set of nodes > as the union of all finite trees. But there is no infinite path > involved / required. Then there is no infinite tree required. If a set of nodes forms a tree at all, then every maximal chain of parent-to-child-edges nodes in /that/ set of nodes is a path in /that/ tree. > The union of all finite paths is an infinite union of finite paths, > but not an infinite path. The union of all finite paths is the node set of the entire tree. To get only its paths, one needs to consider special sets of paths. If one has a family of finite paths in which each path, as a set of nodes, is a subset of every longer path in the family, then the union of that family, as a set of nodes, is an infinite path in the tree formed by the "union" of those finite trees, as sets of nodes. > > > > Pray define "isolated". > > Two path are isolated at level k if they have digits m and n with m =/ > = n at a level i <= k. In other words, the paths have the same nodes up to some level i prior to level k, and at that level i they branch in different directions. For any finite tree, any path is isolated from /all/ others only at its leaf (last) node level. For a CIBT, there are no leaf nodes, so being isolated from all other paths at any node is no longer possible. But that is no more marvelous that the density of the rationals or reals, neither of which allows one member to be "isolated" from all others of its type.
From: Virgil on 23 Mar 2007 15:47 In article <1174664147.264581.197260(a)n76g2000hsh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 23 Mrz., 15:58, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1174654681.038185.299...(a)e65g2000hsc.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > How do you obtain that result? You have not shown a proof at all for it. > > > > > If we exchange infinitely many terms of the geometric series, its sum > > > remains 2, because it is absolutely converging. > > > > > > As the geometric series contains only the smallest possible infinity > > > of terms, it should not cause problems to read it from behind. > > > > If there were a last one. As there is not a last one I have some > > difficulty with it, because I do not know where to start. > > You need not to start. Reverse all terms of the series simultaneously. As nodes, like points, are that which has no part, talking of parts of nodes is nonsense. > > Yes, I do. What now? Every two paths separate from each other at some > > specific node, through all nodes go uncountably any paths, there is no > > level where all paths do separate. > > Because there is no "all paths". There are more things in the imagination, WM, than are drempt of in your philosophy. Including a set of all endless paths. > > > > > > > At no point of the tree (and of the universe) more than countably many > > > path can be distinguished. > > > > I would state that as "at no node in the tree more than countably many > > groups of paths can be distinguished", > > That is where we agree. > > or, equivalently "at no node in > > the tree terminate more than countably many terminating paths". > > No path terminates. They do in finite trees. > > This > > still does not say anything about the number of non-terminating paths. > > That is where we disagree. The difference being that we have proofs and WM only has assertions. > > I think we have cleared our positions sufficiently: We agree in: At no > level in the tree more than countably many groups of paths (including > single paths) can be distinguished. And outside of he tree there are > no paths. > > You nevertheless believe in the uncountable while I do not. Since there is an easy bijection between the power set of the set of non-leaf levels of a complete tree and the set of paths of that tree, If all paths of a binary tree have n edges there are 2^n paths. If all paths are endless, so that the set of levels has cardinality aleph_0, then the set of paths has cardinality 2^aleph_0 > aleph_0.
From: Virgil on 23 Mar 2007 15:50 In article <1174664333.103257.4920(a)y80g2000hsf.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 23 Mrz., 16:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > Pray define "isolated". > > > > > > Two path are isolated at level k if they have digits m and n with m =/ > > > = n at a level i <= k. > > > > So each two non-terminating paths are isolated from each other. And now > > what? > > Now we see that their number is countable, because there are never > more than countably many paths-bundles including such with only one > path isolated from one another. Wm sees what is not there and fails to see what is there in a CIBT or T(oo). The set of terminating decimals, like the set of path bundles as WM defines them, is countable, but the set of nonterminating decimals is not.
From: Virgil on 23 Mar 2007 16:06
In article <1174670803.869854.53820(a)o5g2000hsb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 23 Mrz., 17:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1174664147.264581.197...(a)n76g2000hsh.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 23 Mrz., 15:58, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > As the geometric series contains only the smallest possible > > > > > infinity > > > > > of terms, it should not cause problems to read it from behind. > > > > > > > > If there were a last one. As there is not a last one I have some > > > > difficulty with it, because I do not know where to start. > > > > > > You need not to start. Reverse all terms of the series simultaneously. > > > > Oh. How do I do that? What do I interchange the first one (1 with)? > > You are not so squeamish when exchanging every digit of Cantor's > diagonal. When one has a set of rules which says start here, one can start. What WM wants is to have us start without starting. > Cantor, unless working every digit simultaneously, will never finish. Cantor's rule is not time dependent. > In fact we need not know how to proceed, but it is sufficient to know > hat the sum remains 2 what ever we do. But to get there, WM insists on patitioning that which has no parts. Nodes are like points in that respect, they have no parts. > However, you may proceed as follows (but only after having read my > book) That lets us off that hook. NO one need read WM's propagandizing. > > > Because there is no "all paths". > > > > Pray, provide a proof (within set theory, where we are arguing). > > I am arguing within the tree. Which tree? In any CIBT or T(oo) there are sets of all paths, and those sets are uncountable. That WM has imagined some other tree is irrelevant. > There is never an uncountable number of > separated paths. As each path separates from every other at some node, there are uncountably many pairwise separable paths, which is all that counts. > You say all paths were uncountable and separable. > Conclusion. there is no "all paths" within the tree. By "separable" all we mean is that any two can be separated, wish allows uncountably many paths. That WM seems to want it to mean something else is irrelevant to the existence and uncountability of the set of all paths in a CIBT or T(oo). > > > > > I think we have cleared our positions sufficiently: We agree in: At no > > > level in the tree more than countably many groups of paths (including > > > single paths) can be distinguished. And outside of he tree there are > > > no paths. > > > > > > You nevertheless believe in the uncountable while I do not. > > > > Well, within set theory it can be proven. > > Within the tree it can be disproven. AS WM has, as yet, failed to prove anything, while his opposition has proved much, WM's claims of provability unaccompanied by proofs, must be rejected out of hand. > > > But you do not believe in > > set theory. On the other hand, you have *not* proven that set theory > > is inconsistent. That you do not believe in the uncountable is *not* > > an argument against set theory. > > > That you do believe in set theory is not a crime. But that you do > believe in uncountaly many separations without uncountably many > separations, that is hard to believe. What is unbelievable is that WM will reject sound proofs in favor of his unprovable beliefs. |