Cantor Confusion
in [Math]
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From: Dik T. Winter on 21 Mar 2007 11:09 In article <1174481309.873623.92900(a)e1g2000hsg.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 20 Mrz., 17:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1174327803.614563.300...(a)o5g2000hsb.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > 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 cntributes > to he path. As there is no last term of the geometric series this makes no sense. > > > Can we agree upon the following facts? > > > (1) The infinite paths 0.000... and 0.111... are completely contained > > > in the complete tree T(oo), i.e., every node belonging to one of > > > these paths is a node belonging to the tree T(oo). > > > > Indeed. > > > > > (2) The set of nodes of the tree T(oo) is countable. In particular, > > > the number of nodes belonging to one level L(n) is countable for > > > every n. > > > > Indeed. > > > > > (3) There is no node in the tree (including every node of the paths > > > 0.000... and 0.111...) which belongs to a level where uncountably > > > many paths arrive at or start off from or cross through. > > > > No. I would state that at the root node of the tree there start > > uncountably many paths. Why do you think that number is not uncountable? > > (Remember, paths are non-terminating.) > > (3, clarified) There is no node in the tree (including every node of > the paths > 0.000... and 0.111...) which belongs to a level where uncountably > many paths have separated. At every node in the tree uncounably many paths go to the left and uncountably many paths go to the right. I do now know what you are meaning here. > > > But it is connected to the root node like every other path. The > > > branching offs are countable. > > > > Yes, I never contested that the number of branching offs is countable. So > > what? > > In the whole tree there can be no more separated paths than nodes. Pray give a *proof*, not just a statement. > > > > You state it again. at no finite distance from the root. But you are > > > > wrong. At each finite distance from the root there are uncountably > > > > many paths that go through that level. That number is the same for > > > > each level. > > > > > > That number is not visible. Let us take what can be proved: The number > > > of separated paths *is* countable because, within the tree, there > > > *cannot be more than countably many separated paths - at no finite > > > distance from the root! > > > > And so what, what does that prove about the complete tree? Indeed, at no > > finite distance from the root. What this *means* is that the number of > > terminating paths is countable. > > No. It means that the number of paths which consist only of nodes with > natural indexes is countable. Not at all. Pray provide a *proof*. > > But there is no finite distance from > > the root where all non-terminating paths are separated from each other, > > so the reasoning does not apply to them. > > The "distance" is measured by the index of the due node, i.e., by the > number n of the corresponding level. "At no finite distance from the > root" means at no node which can be enumerated by a natural number. Right. There is *no* node where all non-terminating paths are separated from each other. -- 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 21 Mar 2007 11:16 In article <1174481667.234070.21900(a)e65g2000hsc.googlegroups.com> mueckenh(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. Moreover, it is also not an element of a set of all finite paths, because all elements are finite. So an infinite path is not an element of it. And I would prefer if you refrain from the use of the ambiguous word "contains". That can mean different things. > > > The axiom of infinity grants the existence of all natural numbers, but > > > not the existence of an infinite natural number. > > > > Indeed. > > 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. > > > > Now an > > > > infinite paths is, for instance, the set {(i, 1) | i in N}, which is > > > > a subset of U(T(n)). > > > > > > You try to get an infinite element out of a union of finite elements. > > > That does not work. > > > > Where is the infinite element? > > You said: "The path-length of a non-terminating path is *not* a > natural number." And you find it in a set of serts of finite paths. Eh? N is the union of the sets of initial segments, the cardinal number of N is not a natural number. Where is the difference? The path-length is the cardinal number of the set that represents a path (actually one less if the path is finite). > > > (3) There is no node in the tree (including every node of the paths > > > 0.000... and 0.111...) which belongs to a level where uncountably > > > many paths arrive at or start off from or cross through. > > > > No. From the root node of the tree there start uncountably many paths. > > But they do never separate into uncountably many paths. At each node, uncountably many paths go to the left and uncountably many paths go to the right. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 21 Mar 2007 13:53 In article <1174481309.873623.92900(a)e1g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 20 Mrz., 17:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1174327803.614563.300...(a)o5g2000hsb.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > 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 cntributes > to he path. The "last term" of a geometric series does not exist, so that WM is saying that the contribution of the root to any path does not exist. > > > > > Can we agree upon the following facts? > > > (1) The infinite paths 0.000... and 0.111... are completely contained > > > in the complete tree T(oo), i.e., every node belonging to one of > > > these paths is a node belonging to the tree T(oo). > > > > Indeed. > > > > > (2) The set of nodes of the tree T(oo) is countable. In particular, > > > the number of nodes belonging to one level L(n) is countable for > > > every n. > > > > Indeed. > > > > > (3) There is no node in the tree (including every node of the paths > > > 0.000... and 0.111...) which belongs to a level where uncountably > > > many paths arrive at or start off from or cross through. > > > > No. I would state that at the root node of the tree there start > > uncountably many paths. Why do you think that number is not uncountable? > > (Remember, paths are non-terminating.) > > (3, clarified) There is no node in the tree (including every node of > the paths > 0.000... and 0.111...) which belongs to a level where uncountably > many paths have separated. This is less relevant to the number of paths in a CIBT than the price of peas in Prussia. > > > > > > But it is connected to the root node like every other path. The > > > branching offs are countable. > > > > Yes, I never contested that the number of branching offs is countable. So > > what? > > In the whole tree there can be no more separated paths than nodes. How do "separated paths" differ from other paths? There are certainly uncountably many paths altogether, so that must mean that there are uncountably many "unseparated paths", whatever they may be. > > And so what, what does that prove about the complete tree? Indeed, at no > > finite distance from the root. What this *means* is that the number of > > terminating paths is countable. > > No. It means that the number of paths which consist only of nodes with > natural indexes is countable. That would require that P(N) be countable, which is provably false. > > > But there is no finite distance from > > the root where all non-terminating paths are separated from each other, > > so the reasoning does not apply to them. > > The "distance" is measured by the index of the due node, i.e., by the > number n of the corresponding level. "At no finite distance from the > root" means at no node which can be enumerated by a natural number. For any two different infinite paths, there is a last node (futhest from the root) which they have in common, and this is treu in every tree, If there is a node through which one and only one path goes, then that path and that tree are finite, and that node is a leaf node. So that all WM's talk about "separated paths" can only be relevant in finite trees. In a CIBT, there is no one node after which a given path is separated from all others. That only happens in finite trees. But for any two different paths in a CIBT, there is a node after which the two different paths are separated. WM is conflating properties of finite trees with those of a CIBT again. CIBT = Complete Infinite Binary Tree, in case WM has forgotten.
From: Virgil on 21 Mar 2007 14:09 In article <1174481667.234070.21900(a)e65g2000hsc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 20 Mrz., 17:07, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > 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? If a nested set of finite paths, each representable by its leaf node, is taken to represent the infinite path through each of those leaf nodes, one has a fair representation of an infinite path. How else but as a set of nodes would WM represent an infinite path? > > > > > The axiom of infinity grants the existence of all natural numbers, but > > > not the existence of an infinite natural number. > > > > Indeed. > > why then do you believe that an infinite path (number) is contained in > the union of sets of finite paths (numbers). See above as to how an infinite nested sequence of finite paths can "be" an infinite path. > > > > > > Now an > > > > infinite paths is, for instance, the set {(i, 1) | i in N}, which is > > > > a subset of U(T(n)). > > > > > > You try to get an infinite element out of a union of finite elements. > > > That does not work. > > > > Where is the infinite element? > > You said: "The path-length of a non-terminating path is *not* a > natural number." And you find it in a set of serts of finite paths. Exactly! See above for how it works. > > > > (3) There is no node in the tree (including every node of the paths > > > 0.000... and 0.111...) which belongs to a level where uncountably > > > many paths arrive at or start off from or cross through. > > > > No. From the root node of the tree there start uncountably many paths. > > But they do never separate into uncountably many paths. At each point of "separation", one has uncountably many in each "bundle". For any finite set of nodes in an infinite tree, there are either no paths at all through all these nodes or there are uncountably many. > > (3, clarified) There is no node in the tree (including every node of > the paths > 0.000... and 0.111...) which belongs to a level where uncountably > many paths arrive at or start off from or cross through. Wrong! There are no nodes in any CIBT at which any paths end, but uncountably many start at the root and uncountably many "cross through" every other node. Note that for any CIBT, any node serves as the root of a subtree which is tree-isomorphic to the original tree, so as many paths pass through any node as start at the root.
From: Virgil on 21 Mar 2007 14:28
In article <1174482399.743448.268060(a)p15g2000hsd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 20 Mrz., 20:46, Virgil <vir...(a)comcast.net> wrote: > > In article <1174396356.841673.191...(a)n59g2000hsh.googlegroups.com>, > > > > > > > > Can we agree upon the following facts? > > > > > > > (1) The infinite paths 0.000... and 0.111... are completely contained > > > > > in the complete tree T(oo), i.e., every node belonging to one of these > > > > > paths is a node belonging to the tree T(oo). > > > > > > > (2) The set of nodes of the tree T(oo) is countable. In particular, > > > > > the number of nodes belonging to one level L(n) is countable for every > > > > > n. > > > > > > > (3) There is no node in the tree (including every node of the paths > > > > > 0.000... and 0.111...) which belongs to a level where uncountably many > > > > > paths have been separated or will separate in the next level. > > > > > > I fail to see that any of the points WM presents has any relevance > > > > whatsoever to the issue of whether the number of paths in a CIBT is > > > > countable or uncountable. > > > > > I am not so much interested in what you may feel relevant to prove > > > your opinion concerning the multitude of paths. I am only interested > > > in your boldness to refute the points 1, 2, or 3. > > > > WHY are you 'only interested' in my attitude with respect to such > > irrelevancies? > > Because ever objective observer may form his own opinion on your > further claims. Since many of my "claims" have been accompanied by valid proofs, and will continue to be, those proofs are much better evidence of the validity of those claims that my opinion on irrelevancies. > > >The truth of all three does [not] allow disproof of the > > uncountability of P(N) > > That is obviously impossible to debate with you. Since I am more concerned with proofs that debate, but WM seems more concerned with debate than proofs, that leaves WM in sad shape. > > > > > Does WM agree that EVERY endless sequence of 0's and 1's following that > > > > header of 0 is an infinite path belonging to the tree T(oo)? > > > > > Every existing path belongs to the complete tree. > > > > That carefully avoids answering my question, > > to which an affirmative answer would justify the uncountability of the > > set of paths > > Perhaps this is the proof of inconsistency of actual infinite and, > therefore, the proof of non-existence of those paths? And perhaps it is just WM blowing smoke again. And I judge it to be more smoke. > > > and a negative answer would confirms WM's misrepresentation > > of the issue. > > > > > And at no point of > > > the tree more than countably many infinite paths have separated. At every "point" of the tree, if that means node, uncountably many paths through the left branch separate from uncountably many through the right branch. Wm is again applying methods only suitable to finite trees to the CIBT with idiotic results. > > > > There is also no point at which any infinite path has yet become > > infinite, but that does not prevent the existence of paths having no > > last point. > > For you, it would not even pervent the existence of the non existing > last point, I assume. Wm assumes wrongly, as usual. > > Let us consider only the points (or better nodes) which in fact do > exist (because they can be enumerated or indexed by natural numbers): > > Do you think that the set of those paths, which consist only of > existing nodes, is countable? In what tree? In a finite tree in which every path has n edges, the set of all paths has cardinality 2^n In a tree in which every path has Aleph_0 edges , the set of all paths has cardinality 2^Aleph_0 >Aleph_0 > > > > And since there is no end of "points", that no more proves the > > countability of the set of all paths than the finiteness of each natural > > proves the finiteness of the set of all naturals. > > But you believe that the complete path 0.000... as well as the > complete path 0.010101... etc. including their missing ends are > contained in the complete tree? What "missing ends"? When a circle has no end, does it necessarily have a "missing end"? When an infinite sequence has no end, is there a "missing end". Do I believe that there exists a CIBT? Yes! It exists as surely as the set of all naturals in such systems as ZF and NBG. If a CIBT does not exist in any system that WM can deal with, that is purely his loss, not ours. |